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VII-Mathematics
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Variable: A letter symbol which can take various numerical values is called a
variable or literal.
Example: x, y, z etc.
Constant: Quantities which have only one fixed value are called constants.
Examples:
*
*
In 2y + 7, 7 is a constant and ‘y’ is a variable.
If we say ‘a’ is a constant in bx + a, then it takes a fixed value.
Term:
Numericals or literals or their combinations by operation of multiplication
are called terms.
Example: 5n,
n
, − 5x2 ,7l 3m2 etc.
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Constant Term: A term of an expression having no literal is called a constant term.
7
−16
etc .
Example: 4, , 8,
3
5
Algebraic expression: A combination of terms obtained by the operations like
addition and subtraction is called an algebraic expression.
Example: 2a – 3b, 7x + 11y2, 7p − 8q +
7
r etc.
4
TYPES OF ALGEBRAIC EXPRESSIONS :
*
An expression containing only one term is called a monomial.
2
Example: 7x, −11a b,
*
−5
etc.
6
An expression containing two terms is called a binomial.
Example: 6y-x, 12x 2 + 4y,
*
An expression containing three terms is called a trinomial.
Example: 3p − 4q +
*
5 a b
, − + 4c, 2x + 3y − 4z etc.
6 2 3
An expression containing two or more terms is called a multinomial.
Example: 4x + 2y − 3z,7 +
*
a
+ 4 etc.
2
6
4
,3 x +
etc.
x
x
An expression containing one or more terms with non-negative integral indices
(powers) is called a polynomial.
2
Example: 8x − 2y − 3,2p − 3q +
5
r etc.
6
All polynomials are multinomials but every multinomial need not be a
polynomial.
Factors: In a product, each of the literal or numerical value is called a factor of
the product.
Example: 12 = 3 × 4, where 3, 4 are called the factors of 12, 6xy = 2 × 3 × x × y
where 2, 3, x, y are called the factors of 6xy.
Coefficient: In a product containing two or more than two factors, each factor is
called coefficient of the product of other factors.
Example: In 7xy, 7 is called numerical coefficient of ‘xy’ and ‘x’ is the literal
coefficient of ‘7y’ and ‘y’ is the literal coefficient of ‘7x’.
When the numerical coefficient of a term is +1 or -1, there is no need
to mention 1.
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Example : The coefficient of x 2 in x + 3x − 4 is 1, the coefficient of xy in 9xy − 10y 2 is 9.
Degree of a monomial: The degree of a monomial is the sum of the indices (powers)
in each of its variables.
Example : The degree of 8xy 2 z 4 is 1 + 2 + 4 = 7.
Every non-zero number is considered a monomial with degree zero.
Example : Degree of ‘34’ is ‘0’.
Degree of polynomial :
The highest power of terms in a polynomial is called the degree of a polynomial.
2
3
Example : Degree of 7x − 6x + 4x − 2 is ‘3’. The degree of 7x6 + 4x5 y6 + 9 is 11
Zero polynomial : If all the coefficients in a polynomial are zeroes, then it is
called a zero polynomial.
Zero of the polynomial : The number for which the value of a polynomial is zero,
is called zero of the polynomial.
Example : x =
−5
is called zero of 6x + 5.
6
Degree of zero polynomial is not defined.
Substitutions:
The method of replacing numerical values in the place of literal numbers is called
substitution.
Example: The value of 7z at z = 4 is 7 × 4 = 28.
LEVEL - 1
SINGLE CORRECT CHOICE TYPE:
1.
In given algebraic expression 12xy2z3, literal factors are
1)12
2) 2, 3
3) x, y, z
4) 12 x y z
2.
2x 2 − 7 , in the given algebraic expression, identify the terms
2) + 2x2
3) – 7
4) 2x2, –7
1) 2x2, 7
3.
The coefficient of x2 in
1) 2
4.
4 2
x is
3
4
3
2) 1
3)
2) m + n
3) mn
4)
3
4
m n
Degree of x y is
1) m ÷ n
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2
The value of a − bx + cx , when x =
a 
1) c  
b
2
a
is
b
ca 2
2)
b
3)
ca
b
4) 0
MULTI CORRECT CHOICE TYPE:
6.
Find the value of 4x 3 − 3x 2 + 5x − 6 when x = 3
1) 90
2) – 90
3) 32 × 10
4) 101
REASONING TYPE:
Statement I: Degree of the polynomial −8x 4 + 9x 2 + xe is 4.
Statement II: The highest power of variable terms in a polynomial is called degree
of a polynomial.
1)
Both Statements are true, Statement II is the correct explanation of
Statement I.
2)
Both Statements are true, Statement II is not correct explanation of
Statement I.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
7.
l m
−
+ 4n is an algebraic expression
2 3
8.
In the above expression, the variables are
1) m,n, 4
9.
10.
2) 2, 3
3)
[
1 1
,
2 3
Which type of expression is the above one?
1) monomial
2) binomial
3) trinomial
In the above expression, the constants are
1) 2, 3, 4
2) l, m, n
MATRIX MATCH TYPE:
11. COLUMN - I
a) Factors of 6xy is
b) Degree of 7x 2 − 6x + 4x 3 − 2 is
c) Degree of 8x2y is
d) Zero of 3x – 6 is
3)
1 1
,− , 4
2 3
]
4) l, m, n
[
]
4) Both 1 and 2
[
]
4) Both 2 and 3
COLUMN - II
1) 2
2) 3
3) x
4) y
5) 7
INTEGER ANSWER TYPE:
12.
Number of terms in the polynomial x 5 + x 2 + y 3 + y 2 + 1 __________
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LEVEL - 2
SINGLE CORRECT CHOICE TYPE:
13.
Which of the following is a multinomial?
1) 3 x +
14.
2)
1
x
3) 7xyz
2) 6
3) 5
]
[
]
[
]
4) −11a 2 b
Degree of a polynomial 7x 2 y 6 + x 5 is
1) 2
15.
4
x
[
4) 8
What will be the value of ‘a’ if 3x 2 + x + a equals to 8 when x = 1?
1) 12
2) 22
3) 32
4) 42
COMPREHENSION TYPE
In an algebraic expression the method of putting numerical numbers in place of
literal numbers is called substitution. Find the value of the given expression
16.
17.
18.
a 3 + b3 + c 3 − 3abc
When a = 2, b = 3, c = –4 is
1) 23
2) –27
When a = 1, b = 2, c = –1 is
1) – 22
2) 22
When a = –1, b = 0, c = 2 is
1) – 7
2) 7
3) 37
3) 14
3) 0
[
]
[
]
[
]
4) 43
4) 73
4) – 14
Addition of algebraic expressions :
Addition of algebraic expressions means adding the like terms of the expressions.
Unlike terms cannot be combined or added.
1.
2.
The sum of like terms is defined as a like term similar to each one of them whose
coefficient is equal to the sum of coefficients of the given terms.
Example: 2x + 5x - 3x = x (2 + 5 - 3) = 4x.
Combining the coefficients of like terms of an expression through addition or
subtraction is called simplification of an algebraic expression.
There are two methods of adding algebraic expressions. They are
i) Horizontal method
ii) Vertical method.
Horizontal method :
In this method, like terms should be added and unlike terms should be written
separately by using associative law of addition.
Example: Add 8x + 7y and 3x - 2y.
Solution: 8x + 7y + 3x – 2y = 8x + 3x + 7y – 2y = 11x + 5y.
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Vertical method :
Step 1: In this method the expressions to be added, are written one below the
other.
Step 2: The like terms of each type are placed in separate columns.
Step 3: The sum will be written below that column.
Step 4: If a particular like term is absent in an expression, the place is left
vacant.
Example:
5m2 − 3n2 + 4
7m2 − 8n2 − 10
_____________
12m2 − 11n2 − 6
______________
Subtraction of Algebraic expressions:
The additive inverse of a number:
1.
The additive inverse of any number is obtained by a simple change of its sign, so
additive inverse of a number is also called the negative of that number.
Example: Additive inverse of 7 is –7.
Additive Inverse of expression :
2.
The additive inverse or the negative of an expression is obtained by replacing each
term of the expression by its additive inverse.
Example: Additive inverse of -9x is 9x
3.
To subtract 1 st expression from the 2 nd expression, additive inverse of the 1 st
expression should be added to the 2nd expression.
If P and Q are two algebraic expressions then P – Q = P + (–Q).
Example: Subtract 11a – 6b from 7a + 4b
Solution: (7a + 4b) – (11a – 6b)
= 7a + 4b – 11a + 6b = –4a +10b
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Subtraction can also be done in two ways.
1.
Horizontal method :
Example: 2x + 3y – (y – 3x) = 2x + 3y – y + 3x = 2x + 3x + 3y – y = 5x + 2y
2.
Vertical method :
Example:
3x 2 − 5xy + y 2
− x 2 + xy + 2y 2
+ −
−
4x 2 − 6xy − y 2
LELVEL - 1
SINGLE CORRECT CHOICE TYPE:
1.
2
2
The sum of 6xy , − 2xy , −
5 2 2 2
xy , xy is
6
3
23 2
6
xy
xy 2
2)
6
23
Sum of the multinomials
3) −
1)
2.
23 2
xy
6
4) −
6
xy 2
23
–
1 − x − x − 3x ; 2x + x + 3; x + 5x − 2; x − x 2 − 3x
2
1) x + x 2 − x 3
3
2
3
2
2) − x + x 2 + x 3
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3) x − x 2 − x 3
4) − x − x 2 − x 3
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3.
4.
5.
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Add the 4x 2 + 7xy + 3y 2 + 1 and 2x 2 − 5xy − 2y 2 + 8 to 9x 2 − 8xy + 11y 2
1) 15x 2 − 6xy + 12y 2 + 9
2) 15x 2 + 6xy − 12y 2 + 9
3) 15x 2 − 6xy − 12y 2 − 9
4) 15x 2 + 6xy + 12y 2 + 9
2
2
2
2
If (3x y − 2xy + 7x − 2y ) + A = 7xy − 5x y + 4x − y then A =
1) −2x 2 y + 5xy 2 + 4x + y
2) 9xy 2 − 8x 2 y − 3x + y
3) −4x 2 y + 5xy 2 − 2x 2 y
4) 2x 2 y + 5xy 2 + 4x + y
Subtract the second from the first: a 2 + b2 − c2 , −a 2 + b2 − c 2
1) 2a2
2) 2b2
3) 2c2
4) 2a2 + 2b2 + 2c2
MULTI CORRECT CHOICE TYPE:
6.
How much 3x-4y-10z is greater than 13x-15y-19z ?
1) –10x + 11y + 9z
2) 9z – 10x + 11y
3) 10x – 11y – 9z
4) 10x – 11y + 9z
REASONING TYPE:
7.
Statement I: If you subtract 11a – 6b from 7a + 4b, then the resulting
expression is – 4a + 10b.
Statement II: To subtract 1st expression from the 2nd expression, additive inverse
of the 1st expression should be added to the 2nd expression.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
Adding algebraic expressions in vertical method the expressions to be added are
written one below the other in such a way that the like terms of each type are
placed in separate columns. Like terms in each column are then added and
sum written below that column. If a particular like term is absent in an expression,
the place is left vacant.
8.
The sum of the expressions 7x − 5y + 4a, 2b − x, − 3a + 2y − b in vertical method is
1) 6x − 3y + a + b
9.
3) −6x + 3y + a + b
4) 6x − 3y + a − b
The sum of the expressions 6xy + 10yz – 3xz, –2xy + 3yz + zx is
1) 4xy − 13yz − 2zx
10.
2) 6x + 3y − a − b
2) 4xy + 13yz − 2zx
The sum of the expressions
1)
1 2
l + lm − m 2
6
2)
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3) 4xy + 13yz + 2zx
4) −4xy − 13yz − 2zx
1 2 1
1
1
2
1
l + lm + m 2 , − l 2 + lm + m 2 in vertical method is
2
3
2
3
3
2
1 2
l − lm + m 2
6
3)
1 2
l + lm + m 2
6
4)
1 2
l − lm − m 2
6
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MATRIX MATCH TYPE:
11. Column - I
Column - I
a) If A = − x + y, B = x − y , then
1) A + B = 2x
b) If A = x + y, B = x − y ,then
2) C + D = 2x + 2y
c) If A = 2p + 3q − 4r, B = −2p − 3q + 4r , then
3) A + B = 0
d) If C = x − y, D = x + y , then
4) C + D = 2x
5) A + B = 4x
INTEGER ANSWER TYPE
12. Add 5x + 12, 32 – 5x ____________
LEVEL - 2
SINGLE CORRECT CHOICE TYPE:
13.
Perimeter of the following figure is
1) 2x + 9
14.
15.
2) 4x + 9
Add the following polynomials
1)
13 2 25 3 c3
a −
b +
6
12
20
3)
13 2 25 3 1 3
a +
b −
c
6
12
20
3) 4x – 9
4) 2x – 9
a 2 b3 c3 2a 2 3b3 4c3
+
− ,
+
−
, a 2 + b3 + c 3 .
2
3
4 3
4
5
2)
13 2 25 3 c3
a −
b −
6
12
20
4) None of these
2
2
2
2
2
2
Simplify (8x − 5xy + 3y ) + (3x + 2xy − 6y ) + ( −6x + xy + y ) , if x = 1, y = 1.
1) 0
2) 1
MATRIX MATCH TYPE:
16. COLUMN - I
3) 2
4) – 1
COLUMN - II
a) A = 5p2 + 2q 2 − 3c2 , B = 5q 2 + 2p2 + 3c2 , A + B =
1)
b) P = 5x 3 + 2x 3 − 6x 3 , Q = 4x 3 + 5x 3 − 10x 3 , P + Q =
2) 7q 2 + 7p2
c) x = p2 + 2q 2 , y = 5q 2 + 6p2 then x + y =
3) −9x 2 + 5xy − 3y 2
d) Add −6x 2 + 12xy − 9y 2 , −3x 2 − 7xy + 6y 2 =
2
2
4) 7 ( p + q )
0
5) 7q 2 + 14p2
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COMPREHENSION TYPE:
The additive inverse of any number is obtained by simply changing its sign that is
why additive inverse of a number is also called the negative of the number.
17.
18.
19.
What should be added to a 2 + 2 ab + b 2 to obtain 6a 2 + 7 ab ?
1) 5a 2 + 5ab + b 2
2) 5a 2 + 5ab − b 2
3) −5a 2 − 5ab + b 2
4) 5a 2 + 5ab + b 2
Subtract 2a 3 + 3a 2 − 5a + 6 from the sum of 4 a 3 + a 2 − 5 and −2a 2 + 5a − 6 is
1) 2 a 3 − 4 a 2 + 10 a − 17
2) 2 a 3 + 4 a 2 + 10 a + 17
3) 2a 3 − 4a 2 − 10a + 17
4) 2a 3 − 4a 2 − 10a − 17
Subtract 3x2 + 5x – 4 from 6x2 – 5x – 4 is
1) 3x2 – 10x – 8
2) 3x2 – 10x
4) –3x2 + 10x
3) 3x2 – 10x + 8
[
]
[
]
[
]
Variable: A symbol which can take various numerical values is called a variable.
Example : x, y, z, a, b, c etc.
Variables are also known as literals.
Constant: A symbol having a fixed value is called a constant.
Example: 3, 2010, –5,
3
etc.
7
Term: Numericals or literals or their combination formed by operation of
multiplication are called terms.
Example: 5n,
x
, −10x 2 ,7l 5 m , y, −5, x 2 , etc.
6
When we write a product that contains a variable, usually we omit the
multiplication sign. For example 6 × n can be written as 6n.
Coefficient: The numerical part of a term is called its coefficient.
Example: 1) In 2008y, 2008 is the coefficient of ‘y’.
2) In (–x), –1 is the coefficient of ‘x’
Exponential form: The product of a number ‘x’ with itself n times (n is natural
number) is given by x × x × x × x......... × x (n times) and is written as xn which is
called the exponential form. Here x is called base and n is called the exponent (or)
index of ‘x’. xn can be read as nth power of x (or) x raised to the power ‘n’.
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Exponential form is also called as power notation.
Example: 2 × 2 × 2 × 2 × 2 = 25
Here 2 × 2 × 2 × 2 × 2 is called the product form (or) expanded form and 25 is called
the exponential form.
*
*
The first power of a number is the number itself. i.e., a1 = a.
The second power is called square.
Example: Square of ‘3’ is 32
*
The third power is called cube.
Example: Cube of x is x3
*
1 raised to any integral power gives ‘1’
Example: 1100 = 1
*
(– 1)odd natural number = –1
Example: (– 1)375 = –1
*
(– 1)even natural number = 1
Example: (– 1)2010 = 1
DESCRIPTIVE TYPE QUESTIONS:
1.
Express the following numbers as a product of prime factors in the exponential
notation.
(i)
96
(ii) 2250
2.
Express in power notation
(i)
−64
125
(ii)
125
343
LEVEL-1
SINGLE CORRECT CHOICE TYPE:
3.
Which of the following is a variable?
1) 2010
2) 5
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[
3) x
]
4) 10
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In 5x2, the index of the base ‘x’ is
1) 5
5.
2) x
2) – 5
2) 4(x +y )
4
7.
3) x
3) 4 ( x + y )
4
7
2) 1
[
]
[
]
[
]
[
]
4) x4 + y4
The coefficient of (4x)7 is
1) 4
]
4) – 1
The exponential form of 4 ( x + y )( x + y )( x + y )( x + y ) is
1) 2 ( x + y )
[
4) 2
The coefficient of x in –5x is
1) 5
6.
3) x
2
3) 4
4) 7
MULTIPLE CORRECT CHOICE TYPE:
8.
Which of the following is a constant?
1) 5
3) π
2) 1008
4)
22
7
REASONING TYPE:
9.
Statement I:
( −1)
Statement II:
( −1)
1830447
= −1 .
odd number
= −1 .
1)
Both Statements are true, Statement II is the correct explanation of
Statement I.
2)
Both Statements are true, Statement II is not correct explanation of
Statement I.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
If
10.
l m
− + 5n is an algebraic expression, then
2 3
The variables are
1) m, n, 4
11.
3)
1 1
,
2 3
2) l, m, n
3)
1 −1
,
,5
2 3
2) 3
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3) 1
[
]
[
]
4) None
The number of coefficient is
1) 2
]
4) l, m, n
The coefficients are
1) 2, 3, 4
12.
2) 2, 3
[
4) 4
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MATRIX MATCH TYPE:
13.
COLUMN - I
COLUMN - II
a) Base of (– 2)8
1) a
b) The value of ( −1)
2010
+ ( −1)
2011
2) 84
c) Fourth power of 8 is
3) 4096
d) Base of a3 is
4) 0
5) –2
INTEGER ANSWER TYPE:
14.
The coefficient of ‘x’ in the expression 2009x – 2010 is___________.
LEVEL-2
SINGLE CORRECT CHOICE TYPE:
15.
The exponential form of 7 × 7 × 7....... × 7 (10 times)
1) 107
16.
2) 710
17.
2) 20 + (x – y)4
3) 20(x + y)4
2) – 1
[
]
4) 4(x – y)20
(– 1)2020 = ___________.
1) 2020
]
4) 73
3) 70
The product of 20 and fourth power of (x – y)
1) 20(x – y)4
[
3) 1
[
]
[
]
4) – 2020
MULTIPLE CORRECT CHOICE TYPE:
18.
The value of
( −1)
1) 1
1
+ ( −1) + ( −1) + ........... + ( −1)
2
2) 0
3
100
3) 5
is less than
4) 100
COMPREHENSION TYPE:
a × a × a .... × a (n times) = an. Here a is called the base and ‘n’ is called the exponent
19.
Square of 5 is
1) 5
20.
3) 10
2) r
3
3) r
2
2) 8
3) 10
[
]
[
]
4) r
Fourth power of ‘2’ is
1) 16
]
4) 125
Cube of r is
1) 3r
21.
2) 25
[
4) 2
INTEGER ANSWER TYPE:
22.
The value of 12x3 + 1, when x = 3 is___________.
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Laws of Exponents (or) Indices:
1.
The product of the two powers of the same base is a power of the same base with
the index equal to the sum of the indices.
i.e., If a ≠ 0 be any rational number and m, n be positive integers, then
a m × a n = a m+n
2.
3.
Power of Product
(ab )
n
= a n × bn where a ≠ 0, b ≠ 0 , and n is a positive integer
Example: (5 × 6 )
4
= (5 × 6 ) × (5 × 6 ) × (5 × 6 ) × (5 × 6 ) = (5 × 5 × 5 × 5 )(6 × 6 × 6 × 6 )
= 54 × 64.
4.
Quotient of powers of the same base.
a m − n
am 
= 1
a n  n−m
a
Example:
if
m>n
if
n>m
74 7 × 7 × 7 × 7
=
= 72 = 7 4 − 2 .
7×7
72
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Power of a quotient
m
am
a 
=
where a ≠ 0, b ≠ 0 , and m is a positive integer.
i.e.,  
bm
b
3
6.
5 5 5 53
5
=
× × =
Example:  
.
4 4 4 43
4
Powers with exponent zero:
If we apply the above laws of indices of evaluate
am
where m = n, and a ≠ 0
an
then
am
= a m−n = a 0 = 1
an
0
x
Example: 4 = 1,   = 1 etc ...
y
0
7.
8.
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DESCRIPTIVE TYPE QUESTIONS:
1.
By what number should (-2)-3 be multiplied so that the product may be equal to
10 -1?
−4
4
2.
3.
3a
2
2
2
If   ×   =   , then a ?
3
3
 
 
3
Find m so that 4 × 8m = 25.
−5
4.
5.
−10
5
5
5
Determine x so that   ×   =  
4
4
4
x(y - z)
y(z - x)
z(x - y)
Show that a
= 1
×a
×a
c
a
3x
.
b
7.
 xa   xb   xc 
Simplify  b  ×  c  ×  a  .
x  x  x 
Given that 5x = 1000, find 5x + 2 and 5x - 2?
8.
Simplify: (2x 2 y) × (−3x 2 y 2 )3 × (xy)−1 .
9.
If
6.
−3
p
p 2 3
=   ÷   , what is  
q 3  2
q
3
−10
?
LEVEL-1
SINGLE CORRECT CHOICE TYPE:
4
10.
5
2
3 2 3
The value of   ×   ×   is
4 3 8
1)
3
512
2)
512
3
3)
514
3
If (−4)3 × (−4)7 = (−4)2a , then the value of a is
1) 4
2) 5
3) 6
12. If 3a = 276, then the value of a is
1) 16
2) 17
3) 18
4)
3
514
11.
4) 7
4) 19
5
 −1 
 −1 
 
 
 2  + 8 
13. The value of
is
4
 −1 
 −1 
 
 
 4 
 2 
1) 0
2) 1
0
0
14. The value of (4 – 3 ) × 60 is
1) 0
2) 1
MULTIPLE CORRECT CHOICE TYPE:
15.
2
 −5 
The value of (2−1 ÷ 5 −1 ) × 

 8 
1) 0
2) 1
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3) 2
4) 3
3) 2
4) 3
−1
is less than
3) – 20
4) – 1
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REASONING TYPE:
16.
Statement I:
a −2010 = 3−2010 ⇒ a = 3 .
x
9
5
5
Statement II:   =   ⇒ x = 3 .
6
6
1)
Both Statements are true, Statement II is the correct explanation of
Statement I.
2)
Both Statements are true, Statement II is not correct explanation of
Statement I.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
For all non-zero rational numbers x, y and all positive integers m and n
xm
i) x n

 1

=  x m− n
 1
 n− m
x
if
if
m=n
m>n
if
n>m
m
x
xm
ii)   = m
y
y
7
17.
 − 2   −2 
The value of 
 ÷

 3   3 
1)
18.
2)
−n
y
= 
x
n
4
is
−27
8
3)
8
27
4)
19
64
2)
64
19
3)
27
16
4)
3
5
2) –
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3
5
3)
2
5
]
[
]
[
]
5
16
−1
−1
 3 
 −2  
÷


  is
The value of  
 5  
 2 
1)
[
27
8
 1 −3  1 −3   1  −3
The value of   −    ÷   is
 2    4 
 3 
1)
19.
−8
27
x
iii)  
y
4) –
4
15
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MATRIX MATCH TYPE:
20. COLUMN - I
a)
COLUMN - II
1
xa
xb
1)
b) ( x y )
−a
5
c)  
7
a
2) x a −b
−8
3)
 2009 
d) 

 2010 
(x y )
2020
 2009 
÷

 2010 
2021
2009
2010
7
4)  
5
5)
8
2010
2009
INTEGER ANSWER TYPE:
21.
 2020 
The value of 

 2000 
−2008
 2020 
×

 2000 
2008
× 2020 is___________.
LEVEL-2
SINGLE CORRECT CHOICE TYPE:
22.
  −1 2 −2 
  
The value of   3   
  
 

1)
23.
−1
81
2)
−3
is
1
813
3)
4) 2
x 2n − 3 × x 2n + 2
= x 27 , then n =
If
x −12
1) 1
2) 2
MULTI CORRECT CHOICE TYPE:
24.
3
27
The value of
4
56
21
2
3) 3
4) 4
3) 162
4) 256
2) ( xy )
3) x + y
4) ( x − y )
2) ( x )
3) ( y )
4) ( z )
3) z x + y
4) x y + z
is
1) 28
2) 44
COMPREHENSION TYPE:
If 5 x =125 , 6y = 36 and 7z = 7 , then
25.
(x + y + z )
2
=
2
1) xy
26.
( xyz )
xyz
=
1) ( xy )
x+ y + z
27.
z
z
x
x 2 y + y 2 x + y 2 z + yz 2 + zx 2 + z 2 x is
1) xy ( x + z )
2) 3 y
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x+z
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Introduction: Logarithm, in mathematics, is the ‘exponent’ or ‘power’ to which a
stated number called the base, is raised to yield a specific number. For
example, in the expression 102 = 100, the logarithm of 100 to the base 10 is 2.
This is written as log10100 = 2. Logarithms were originally invented to help
simplify the arithmetical process of multiplication, division, expansion to a power
and extraction of a ‘root’, but they are nowadays used for a variety of purposes in
pure and applied mathematics.
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Logarithm: If for a positive real number (a ≠ 1) , am = b, then the index m is called the
logarithm of b to the base a. We write this as
‘log’ being the abbreviation of the word ‘logarithm’. Thus,
where, am = b is called the exponential form and logab = m is called the
logarithmic form.
Illusration: Refer to the following Table
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Types of Logarithm:
* Natural Logarithm: Logarithm of numbers to base e are known as nature
logarithm.
* Common Logarithm: Logarithm of numbers to base 10 are known as
common logarithm.
* Logarithm of 1 to any base is equal to zero.
i.e., loga1 = 0, where a > 0, a ≠ 1 .
*
Logarithm of any number to the same base is 1.
i.e., loga a = 1, where a > 0, a ≠ 1 .
Laws of Logarithms:
* Product formula: The logarithm of the product of two numbers is equal to
the sum of their logarithms.
Generalization: In general, we have
log a ( mnpq...) = log a m + log a n + log a p + log a q + ...
*
*
*
Quotient formula: The logarithm of the quotient of two numbers is equal to
the difference of their logarithms.
where, a, m, n are positive and a ≠ 1 .
Power formula: The logarithm of a number raised to a power is equal to
the power multiplied by logarithm of the number.
loga(mn) = n logam,
where, a, m are positive and a ≠ 1 .
Base changing formula:
where, m, n, a are positive and n ≠ 1, a ≠ 1 .
*
Reciprocal relation: logba × logab = 1,
where, a, b are positive and not equal to 1.
*
log b a =
*
*
*
*
*
a
1
.
log a b
log a x = x , where, a and x are positive, a ≠ 1 .
If a > 1 and x > 1, then loga x > 0.
If 0 < a < 1 and 0 < x < 1, then loga x > 0.
If 0 < a < 1 and x > 1, then loga x > 0.
If a > 1 and 0 < x < 1, then loga x < 0.
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DESCRIPTIVE TYPE QUESTIONS:
1.
If log10 x = a, log10 y = b and log10 z = 2a – 3b, then express z in terms of x and y.
2.
Prove the log10 25 + log10 4 = log4 16.
3.
If log2 = 0.3010 and log3 = 0.4771, find the value of log5 + log8 – log 6 .
LEVEL-1
SINGLE CORRECT CHOICE TYPE:
4.
The value of log 9 81 =
1) 3
5.
2) 2
2) 125 = 55
7.
2) 8
The value of log
1) 2
2) 4
2
3) 2
3) –10
3)
]
[
]
[
]
[
]
[
]
4) 16
4) 10
4 is
2) 4
[
4) 5 = 1253
The value of log10 0.0001 is
1) –4
8.
3) 125 = 53
If log 2 x = 2 , then x =
1) 4
]
4) 81
Express log 5 125 = 3 in exponential form
1) 125 = 35
6.
3) 9
[
2
4) 3
MULTI CORRECT CHOICE TYPE:
9.
Which of the following is correct?
1) 26 = 64 ⇔ 6 = log 2 64
2) 82 = 64 ⇔ 2 = log 8 64
3) 43 = 64 ⇔ 3 = log 4 64
4) 641 = 64 ⇔ log 64 64 =1
REASONING TYPE:
10.
Statement I: The value of 57log5 2 =128 .
Statement II: If (a ≠1) be a positive real number, then a log a x = x .
1)
Both Statements are true, Statement II is the correct explanation of Statement I.
2)
Both Statements are true, Statement II is not correct explanation of Statement I.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
If ax = N (a ≠1 and a > 0 ) , then x = log a N
11.
If log 2 x = 3 , then x =
1) 8
2) 9
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[
3) –8
]
4) 16
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VII-Mathematics
12.
If log 81 x =
1) 243
13.
Techno Text book
3
, then x =
2
2) 729
If log10 x = a , then 10
a–1
3) 81
x
10
2)
10
x
MATRIX MATCH TYPE:
14. Column-I
a) If
]
[
]
[
]
[
]
[
]
[
]
4) 719
in terms of x is
3) 10 x
1)
[
4) x
Column-II
1) 0.1 =10−1
16 = 4 , then
1
2
1
3) log 2 32 =
5
4) 36 = 62
1
4
2) log16 =
b) If 325 = 2 , then
c) If log100.1= –1, then
d) If log636 = 2, then
5) log 32 2 =
1
5
INTEGER ANSWER TYPE:
15.
If log x
1 −1
=
, then x = _____________
2 5
5
LEVEL-2
SINGLE CORRECT CHOICE TYPE:
16.
The value of
log 8
is
log 2
1) 2
17.
2) 4
(
The value of log 3 27 3
)
3) 5
is
1
3
2)
2
2
COMPREHENSION TYPE:
1)
3)
5
2
If a ≠ 1, b ≠ 1 are positive real numbers, then log b a =
18.
4)
7
2
log b m
m
1
and log a = log a .
b
log a b
1
1
If log xy + log xy is
x
y
1)
19.
4) 6
x
y
2)
y
x
3) 1
4) –1
If log a x × log b y =
1) log b x .log a y
2) log b x × log y a
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3) log x b × log a y
4) log x b × log y a
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VII-Mathematics
20.
If
Techno Text book
log10 8
=
log10 2
1) 1
2) 2
MULTI CORRECT CHOICE TYPE:
21. The value of log 448 is
1) log (22 × 23 × 2) + log7
3) 6 log2 + log7
3) 3
[
]
[
]
[
]
[
]
[
]
[
]
[
]
[
]
4) 4
6
2) log2 + log7
4) log64 + log7
Revise Lows of Logarithms From Concept - 5
LEVEL-1
SINGLE CORRECT CHOICE TYPE:
1.
If logm + logn = log(m + n), then m =
n
n +1
3) n −1
n
If log (x + 1) + log (x – 1) = log 48, then x =
1) –7
2) 7
3) 1
If log2 = 0.3010, then log 1024 =
1) 0.6020
2) 1.5050
3) 3.010
n
1) n + 1
2.
3.
4.
3
4
2)
4
3
3) 1
n −1
n
4) 5
4) 30.10
4) log 3
If a 2 × b2 = 2ab , then loga =
b
2) logb + log2
3) logb – log2
2
MULTI CORRECT CHOICE TYPE:
6.
If log (10x + 5) – log (x – 4) = 2, then x =
1) log
9
2
REASONING TYPE:
1)
7.
4)
The simplified form of log 27 9 + log 8 4 is
1)
5.
2)
Statement I:
2)
5
3
3) 1.6
 2
4) log  
b
4) 4.5
 a 2 × b3

× b2  = 5 logb – loga.
log 
3
 a

Statement II: Logarithm of unity to any nonzero base greater than 0 and ≠1 is
zero.
1)
Both Statements are true, Statement II is the correct explanation of Statement I.
2)
Both Statements are true, Statement II is not correct explanation of Statement I.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
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COMPREHENSION TYPE:
If a = log20, b = log
8.
9.
10.
8
3
and c = 2 log
, then
3
2
Simplified form of a + b is
1) log3 + log5
2) log2
3) 5 log2 – log3 + log5
4) log7
Simplified form of b + c is
1) log3 + log 5
2) log 2
3) 5 log2 – log3 + log5
4) log7
Simplified form of c + a is
1) log3 + log5
2) log2
3) 5 log2 – log3 + log5
4) log7
MATRIX MATCH TYPE:
11.
Column-I
Column-II
a) log 32 =
1) 8 log2
b) log 100 =
2) 2
c) log 50 =
3) 1
d) log 256 =
4) 5log2
5) log2 + 2log5
INTEGER ANSWER TYPE:
12.
The value of
(0.05 )log
1
20
2010
= _____________
LEVEL-2
SINGLE CORRECT CHOICE TYPE:
13.
If log 6 n − log 6 n − 1 = log 6 3 , then the value of n is
1)
14.
3
4
2
3
If log10 x = a + b and y =10a − b , then log10 x
1) 5a + b
15.
2)
2) b – 5a
3)
2 3
y
3
2
4)
3
8
=
3) 5a + 2b
4) 5a – b
3) x2 = 1000y
4) y2 = 1000x
If 3 + log10 x = 2 log10 y then
1) x2 = 100y
2) y2 = 100x
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REASONING TYPE:
16.
Statement I:
 72 
log 
 = 3 log 2 + 2log 3 .
 10 
Statement II: log
a
= log a − log b;log a m = m log a log (ab ) = log a + log b .
b
1)
Both Statements are true, Statement II is the correct explanation of Statement I.
2)
Both Statements are true, Statement II is not correct explanation of Statement I.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
If x =1 + log abc ; y =1 + log b ca; z =1 + log c ab , then
17.
a=
1
1) (abc ) x
18.
1
3) (abc )z
4) 1
2) xyz
3) x + y + z
4) 0
2) xy + yz + zx
3) x + y + z
4) 0
1 1 1
+ + =
x y z
1) 1
19.
1
2) (abc ) y
xyz =
1) 1
MATRIX MATCH TYPE:
20.
Column - I
Column - II
a) log
ab
=
c
1) loga + logb + logc + logd
b) log
ab
=
cd
2) ploga – qlogb
c) log a p .bq =
3) ploga + qlogb
 ap 
d) log  q  =
b 
4) loga + logb – logc
5) loga + logb – logc – logd
INTEGER ANSWER TYPE:
21.
If log
5
128
5
+ log
+ log = log ___________
8
25
2
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Multiplication of Algebraic expressions :
Multiplication of Monomials :
We have ‘×’ sign for multiplication, it need not be written between the product of a
numeral and literal number.
Example: 3y = 3 × y
Steps to be followed to multiply monomials :
*
Numerical coefficient in the product = product of numerical coefficients in
the monomials.
*
Literal coefficient in the product = product of literal coefficients in the
monomials.
*These rules may be extended for the product of three or more monomials.
Example: – 7ab × 3b2 = (–7 × 3) (ab × b2) = –21ab3.
Multiplication of a Binomial and a Monomial:
We know the distributive property a (b + c) = ab + ac or (b + c) a = ba + ca
This property gives us the method of multiplication of a monomial and a binomial
by the monomial.
This can be done in two ways.
(i) Horizontal method
Ex: 4a (3a + 2b)
(ii) Column method
3a + 2b
= 4a × 3a + 4a × 2b
× 4a
__________
= 12a2 + 8ab
12a 2 + 8ab
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Multiplication of a Polynomial and a Monomial :
Multiply each term of the polynomial by the monomial.
Example:
4a (2a + 3b + 4c ) = 4a.2a + 4a.3b + 4a.4c = 8a 2 + 12ab + 16ac
Multiplication of two Binomials:
Multiply each term of the first binomial with each term of the second and add the
like terms in the product.
Suppose (a+b) and (c+d) are two binomials. By using the distributive law of
multiplication over addition, we can find their product as given below.
(a + b )× (c + d ) = a × (c + d )+ b × (c + d )
= (a × c ) + (a × d ) + (b × c ) + (b × d ) = ac + ad + bc + bd
The product of the two factors with the same sign is positive and the
product of the factors with the opposite sign is negative.
Examples: (i) 5x × 6y = 30xy
(ii) −3x × −2x = 6x 2
(iii) 6a × −4b = −24ab
Multiplication by using formulae:
*
*
( x + a )( x + b ) = x 2 + (a + b ) x + ab
Example: ( x + 2)( x + 3 ) = x 2 + x (2 + 3 ) + 2 × 3 = x 2 + 5x + 6
( x + a )( x − b ) = x 2 + (a − b ) x − ab
2
Example: ( x + 4 )( x − 5 ) = x 2 + (4 − 5 ) x − 20 = x − x − 20
*
*
( x − a )( x + b ) = x 2 − (a − b ) x − ab
Example: ( x − 2)( x + 1) = x 2 − (2 − 1) x − 2 × 1
( x − a )( x − b ) = x 2 − (a + b ) x + ab
Example: ( x − 3 )( x − 1) = x 2 − (3 + 1) x + 3 × 1
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= x2 − x − 2
= x 2 − 4x + 3
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DESCRIPTIVE TYPE QUESTIONS:
1.
Supply the missing terms in the following products:
(i)
(x2 – x + 1)(2x) = 2x3 – 2x2 _________
(ii) (2y + 1)(y – 4) = _________ –7y – 4.
(iii) (2a2 + 5) ( 6a2 - 7 ) = 12a4 + 16a2 _________
(iv) (4a + 15) (2a ......) = 8a2+ _________ –90.
2.
Find the following products:
(i)
(–24x + 15x3 – 16x5 + 6) and (–2x3).
(ii) (6a3 – 14a2 + 24a – 7) and 2ab.
(iii) (12m2n – 3mn2 – 1) and 6m2n2.
3.
Find the following products:
(i)
(x + 12) (x + 5)
(ii) (3p – 5) (5p – 6)
(iii) (4f – 3g)(5f + 6g)
(v)
4.
(iv)
k
 k

 − 5  − 3
3
 5

 4p
  5p

− 3
− 6

 5
 8

Find the following products and verify the results for the values given in each
problem.
(i)
11a2 + 3a + 2b – 4 and 3a, when a = 1, b = 2.
(ii) (7r – 2s)(12r + 5s) when r = –2, s = 2.
(iii)
 6p
  10p

− 2 
− 5  when p = 3.

 5
 3

LEVEL-1
SINGLE CORRECT CHOICE TYPE:
5.
The missing terms of the products: (3x – 7y) (5x – 9y)= 15x2 ....... + 63y2 is
1) (–62xy)
2) (62xy)
3) (45xy)
4) (–63xy)
4
3
2
2
6.
The products of (8p + 5p – 3p – 6p +7) and (–5p ) is
1) 40p6 – 25p5 + 15p4 + 30p3 – 35p2
2) – 40p6 – 25p5 + 15p4 + 30p3 – 35p2
6
5
4
3
2
3) 40p + 25p + 15p + 30p – 35p
4) – 40p6 – 25p5 – 15p4 + 30p3 + 35p2
7.
The products of (9a2 + b) and (4a2 – 3b) is
1) 36a4 + 23a2b – 3b2
2) 36a4 – 23a2b + 3b2
3) 36a4 – 23a2b – 3b2
4) 36a4 + 23a2b + 3b2
8.
The value of (8x3 – 4x + 3x2 + 7) (2x2) when x = 1 is
1) 28
2) 27
3) 26
4) 25
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MULTI CORRECT CHOICE TYPE:
9.
 l m  l m 
 +   +  = _______
 2 3  2 3 
 l m
1)  + 
2 3 
2
2)
l 2 lm m2
+
+
4 3
9
3)
m2
l2
+ 6lm +
4
9
4)
m2
l2
− 6lm +
4
9
REASONING TYPE:
10.
Statement I:
(2x + 3y )(4x − 5y ) = 8x 2 − 2xy − 15y 2 .
Statement II:
(a + b )(c + d ) = ac + ad + bc + bd .
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
The product of ( x + a )( x + b ) = x 2 + x (a + b ) + ab
( x − a )( x + b ) = x 2 + x (b − a ) − ab
( x − a )( x − b ) = x 2 − (a + b ) x + ab
( x + y )( x − y ) = x 2 − y 2
11.
The product of ( x + 3 ) and ( x + 7 ) is
1) x 2 + 7x + 2
12.
3) x 2 + 10x + 21
4) x 2 + 10x + 16
3) x 2 − 10x − 16
4) x 2 + 10x + 16
3) x 4 + 3x 2 − 4
4) x 4 − 3x 2 + 4
The product of ( x − 2) and ( x − 8 ) is
1) x 2 − (10 ) x + 16
13.
2) x 2 + 4x − 21
2) x 2 − 6x − 16
2
2
The product of ( x + 4 ) , ( x − 1) is
1) x 4 − 3x 2 − 4
2) x 4 + 3x − 4
MATRIX MATCH TYPE:
14. Column-I
Column-II
a) (x + y)(x – y) =
1) ( x − 5y )
b) (2a + b )(3b + a ) =
2) − y 2 + x 2
c) ( x − 5y )( x − 5y ) =
3) x 2 + x (a + b ) + ab
d) (x + a)(x + b) =
4) x 2 + 25y 2 − 10xy
2
5) ab + 6ab + 2a 2 + 3b2
INTEGER ANSWER TYPE.
15.
2
2
The value of (3n + 4n )(2n − 3n ) at n = 4 is _____________
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LEVEL-2
SINGLE CORRECT CHOICE TYPE:
16.
The value of a product (3x + 5y )(5x − 7y ) , if x = 2 and y = 3 is
1) 321
17.
18.
2) – 321
3) – 231
1 
1
2
2
The product of  x − y  and 5x − 4y
4 
5
(
)
4) 231
is
3
1) x +
5 2
4
4
x y − xy 2 − xy 2 + y 3
4
5
5
3
2) x −
5 2
4
x y − xy 2 + y 3
4
5
3
3) x −
5 2
4
x y + xy 2 + y 3
4
5
3
4) x −
5 2
4
x y − xy 2 − y 3
4
5
If you multiply (0.8x − 0.5y ) by (1.5x − 3y ) , then the product is
1) 1.2x 2 − 3.15xy + 1.5y 2
2) 1.2x 2 + 3.15xy + 1.5y 2
3) 1.2x 2 − 0.15xy + 1.5y 2
4) 1.2x 2 + 0.15xy + 1.5y 2
COMPREHENSION TYPE:
In the multiplication of a polynomial and a monomial, we use the distributive
laws given by (i) a × (b + c) = ab + ac
(ii) a × (b – c) = ab – ac
19.
20.
2 2
2
Multiplication of 3ab and (5a b − 3ab + 4ab ) is
1) 15a 3 b3 − 9a 3 b3 + 12a 2 b2
2) 15a 3 b3 − 9a 2b3 + 12a 2b2
3) 15a 3 b3 − 9a 2b3 − 12a 2 b2
4) 15a 3 b3 − 9a 2b3 + 12ab2
2
2
Multiplication of 3a b and ( −5ab + 6a b ) is
1) −15a 3 b2 − 18a 4 b2 2) −15a 3 b2 + 18a 3 b3 3) −15a 3 b2 + 18a 4 b2 4) −15a 3 b2 + 18a 3 b4
21.
Multiplication of 4ab and (6a − 2b ) is
2) 24ab – 8ab
1) 24a2b + 8ab2
MATRIX MATCH TYPE:
22. Column-I
a) 6p2q 2 (2pq )
(
4) –24ab + 8ab
Column-II
1) 12p3q3
)
3
2
b) 2x + 3x + x ×
c)
3) 24ab + 8ab
1
2
4p4 + 6p3 − 3p2 − 3p + 8
4p2
2 2
2
d) 3p q (2 pq )
3
2) x +
3 2 x
x +
2
2
2
3) p +
3
3 3
2
+ 2
p− −
2
4 4p p
4) 3 × 4 × p3q 3
5) 12p2q3
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Special Products Geometrical Proof:
1.
(a + b )
2
= a 2 + 2ab + b2
On multiplying the binomial (a + b) with itself, we get
(a + b )
2
= (a + b ) × ( a + b )
= a 2 + ab + ba + b2
= a 2 + 2ab + a 2
The square (see Figure 19.1) has been divided into two smaller squares (1)
and (3) and two rectangles (2) and (4) with their dimensions as shown. The
length of each side of the big square is a + b. The area of the big square is
the square of its sides or (a + b)2. The area of the big square
= Sum of the areas of square (1)
and (3) and the areas of
rectangles (2) and (4)
or
(a + b )
2
= a 2 + b2 + ab + ab = a 2 + b2 + 2ab
Thus, by applying geometry too, we arrive at the same conclusion viz.
(a + b )
2
2.
(a + b )
2
New,
= a 2 + 2ab + b2
= a 2 − 2ab + b2
(a − b )
2
= (a − b )(a − b )
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2
2
2
2
= a − ab − ba + b = a − 2ab + b
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In Figure 2(a – b)2 represents the area of the square (1) . Here, the area
of the big square is a × b = a2. (Length of each of its side = a)
Now, area of big square = area of square (1) + area of rectangle (2) + area of
rectangle (4) + area of square (3).
a 2 = (a − b ) + b (a − b ) + b (a − b ) + b2 = (a − b ) + ba − b2 + ba − b2 + b2
2
or
2
= (a − b ) + 2ba − b2
2
&&
⇒ (a − b ) = a 2 − 2ab + b2 x
2
Thus, by applying geometry as well, we arrive at the conclusion
(a − b )
2
= a 2 − 2ab + b2
DESCRIPTIVE TYPE QUESTIONS:
1.
Expand each of the following:
(i)
(x + 10)2
(iii)
2.
1 

 2x +

2x 

(ii)
(5m – 7)(5m – 7)
(iv)
1 
1

a − a − 
a 
a

2
Simplify the following expressions:
(i)
(a – 5)2 + (a – 3)2
(ii) (2a + 3b)2 + (4a + 5b)2
(iii) (l2 + m2)2 + (l2 – m2)2 (1 is replace as l)
2
(iv)
3.

1  
1 
 2x +
 −  2x −

3y  
3y 

Find the values of the following using either the formula (a + b)2 or (a – b)2
(i) (999)
4.
2
1

(ii)  299 
2

2
Find the value of x +
(i) x +
5.
2
1
=3
x
Find the value of 9y +
(i) 3y −
1
=6
2y
1
x
=4
(iii)
x+
1
x
=5
1
1
= 6 , 7 or 8.
2 if 3y 4y
2y
(ii) 3y −
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(iii) (0.96)2
1
1
if x + = 3,4 or 5 .
x2
x
(ii) x +
2
2
1
=7
2y
(iii) 3y -
1
= 8
2y
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VII-Mathematics
6.
7.
8.
9.
Techno Text book
3x + 4y = 16 and x y = 2 find 9x2 + 16y2
Fill in the following blanks:
(i)
(x + ____)2 = ____ + 2xy + ____
(ii) (____ + 5z)2 = ____ –70yz + ____
Fill in the following blanks so that each expression is the square of a binomial.
Find the binomial whose square is equal to the resulting expression.
(i)
4l2 - ......+ 9m2.
(ii) 64u2 + 80uv + .......
(iii) ......+ 2x2y2 + y4.
Find the value of ‘ k ’ in each expression so that it may be a perfect square.
Write the resulting expressions as the square of a binomial.
(i)
y2 – 14y + (k + 9).
(ii) 16p2 – (–k + 6)p + 49.
LEVEL-1
SINGLE CORRECT CHOICE TYPE:
10. The value of 0.768 × 0.768 – 2 × 0.768 × 0.568 + 0.568 × 0.568 is
1) 0.04
2) 0.4
3) 0.004
4) 0.0004
11.
12.
If 2x +
1)
104
3
If
y−
1
1
2
= 6, then the value of 4x + 2 is
3x
9x
2)
106
3
If v +
107
3
4)
109
3
1
= 2, then the value of y 2 + 12 is
y
y
1) 4
2) 5
MULTI CORRECT CHOICE TYPE:
13.
3)
3) 6
4) 7
1
= 5 , then
v
2
1) The value of v +
1
= 23
v2
3) The value of v 4 + 14 = 527
v
2
2) The value of v +
1
= 27
v2
4) The value of v 4 + 14 = 527
v
REASONING TYPE:
14.
Statement I: If A = 16x2 + 144y2, then the expression to be added to make it a
perfect square is ± 96xy .
Statement II: If ( A + B ) = A 2 + 2AB + B2 , ( A − B ) = A 2 − 2AB + B2 .
2
2
1)
2)
3)
Both Statement-I and Statement-II are true.
Both Statement-I and Statement-II are false.
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
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COMPREHENSION TYPE:
If ( A + B ) = A 2 + 2AB + B2 , ( A − B ) = A 2 − 2AB + B2 and ( A + B )( A − B ) = A 2 − B2 , then
2
15.
(l
2
2
+ m2 ) + ( l2 − m2 ) =
2
2
2
2
1) 2 l + m 
4
4
2) 2 l + m 
2
16.
4) 4l2m2

1 
3) 2 2x +
3y 

4)
2
3) 2  x +16
4) 16x
2

1  
1 
2x + 3y  − 2x − 3y  =

 

 2
1 
1) 2 4x + 2 
9y 

17.
3) l 4 + m4
( x + 4 ) + ( x − 2)
2
2
2
2) 4x +
1
9y 2
8x
3y
=
1) 2x 2 + 4x + 20
2) 2x 2 + 8x + 12
MATRIX MATCH TYPE:
18.
Missing terms in the perfect square.
Column-I
Column-II
a) a2 + 4a + ______
1) 4
b) y2 – 8yz + ______
2) 16z2
c) p2 + _____ + 4q2
3) 4pq
d) 4l2______ + 9m2
4) 12lm
5) –12lm
INTEGER ANSWER TYPE.
19.
If 9l2 – klm + 4m2 is a perfect square then k = ______________
LEVEL-2
SINGLE CORRECT CHOICE TYPE:
20.
The value of 687 × 687 – 313 × 313 is
1) 3, 50, 004
21.
2) 3, 74, 000
3) 5, 74, 000
2
When x = 2, the value of the expression 4x +
1)
4
25
2)
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−4
25
3)
4) 2, 74, 000
9
−12 is
x2
25
4
[
4)
]
−25
4
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VII-Mathematics
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MULTI CORRECT CHOICE TYPE:
2
22.
1
1 
1 
1


If  x +  − 2  x +   x −  +  x − 
x
x 
x 
x


2
is simplified then the result is
8
16
4
1) 2
2) 2
3) 2
x
x
x
MATRIX MATCH TYPE:
23. Column-I
a) If a + b = 6 and ab = 3, then a2 + b2 is
b) If 3x + 4y = 16 and xy = 2, then 9x2 + 16y2 is
c) If p – q = 5 and pq = 50, then p2 + q2
d) If 2l – 3m = –1 and lm = 20, then 4l2 + 9m2 =
2
4)  
x
[
]
2
Column-II
1) 251
2) 241
3) 125
4) 208
5) 30
To Find the value of (a + b)(a – b)
Now, (a + b )(a − b ) = a 2 − ab + ab − b2 = a2 – b2
in Figure, the product (a + b)(a – b) gives the area of the shaded rectangle consisting of square
(1) and rectangles (2) and (3) .
The area of the shaded portion can also be arrived at by subtracting the
areas of rectangle (4) { b(a – b)} and square (5) (b2) from the area of the big
square (a2) and then adding the area of rectangle (3) { b (a – b) } to it.
Thus,
(a + b )(a − b ) = a 2 − b (a − b ) − b2 + b (a − b )
or (a + b )(a − b ) = a 2 − b2
Thus, by applying geometry as well, we arrive at
(a + b )(a − b ) = a 2 − b2 .
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DESCRIPTIVE TYPE QUESTIONS:
1.
2.
Find the following products using the formula:
(i)
(l + m) (l – m)
(ii)
(2x2 + 5y3)(2x2 – 5y3)
(iii)
 c d  c d 
 −  + 
 5 4  5 4 
(iv)
(0.3r – 0.5s)(0.3r + 0.5s).
Find the following products and verify the results for the values given against
each:
(i)
(p + 5) (p – 5)(p2 + 25); when p = 1.
1 
1  2 1 

 a +   a −   a +  ; when a = 2.
2
2
4

(ii)
3.
(iii) (a – b) (a + b) ( a2 + b2 ) ( a4 + b4 ); when a = 0 , b = 1.
Find the values of the following products using the formula:
(i) 501 × 499
4.
(ii) (10.01) × (9.99)
(iii) 24
2
5
× 25
7
7
Fill in the following blanks:
(i)
(2x +5) (2x -5) = (........)2-(...........)2
(ii)
(....+ 4b) (......- 4b) = 25a2 - (.........)2
(iii)
999
9
10
×1000
1
10
= (.....)2 - (......)2
LEVEL-1
SINGLE CORRECT CHOICE TYPE:
5.
The product of (2p3 + 5q3)(2p3 – 5q3) is
1) – 4p6 + 25q6
6.
3) 4p6 + 25q6
4) 4p6 – 25q6
2 m2 
 l m  l m   l
+
−

The product of  2 3  2 3   4 + 9  ; when l = –2, m = + 2 is




1)
7.
2) – 4p6 – 25q6
65
81
2)
67
81
3)
62
81
4)
61
81
The product of 49.95 × 50.05 is
1) 2399.9975
2) 2499.9975
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4) 2199.9975
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VII-Mathematics
8.
Techno Text book
1
1
7
7
The product of  3 + 6  and  3 − 6  is




1
−1
2)
12
12
MULTI CORRECT CHOICE TYPE:
1)
9.
x+
1
18
4)
−1
18
1
1 1
= 3 and x − = , then
x
x 3
1
=1
x2
REASONING TYPE:
2
1) x −
10.
3)
Statement I :
2
2) x +
1
=7
x2
4
3) x −
1
=7
x4
4
4) x +
1
= 47
x4
If [a + b][a − b] = a 2 − b2 .
Statement II :
Area of rectangle whose length is ‘l’ units and breadth is ‘b’ units is
“lb” square units.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
1
1
1

2
If  x +   x −  = x − 2 , then
x
x
x

11.
[2x + 5][2x − 5]
=
1) 4x 2 − 25
12.
2
3) 4x −
2) 2x2
3)
1
25
2
4) 4x +
1
25
1 
1 1

 x + x   x − x  + x 2 =
1) x2
13.
2) 4x 2 + 25
[10.01]× [9.99]
1
x2
4)
2
x2
=
1) 99.9999
2) 90.9999
MATRIX MATCH TYPE:
14. Column-I
3) 99.9909
4) 99.0999
Column-II
a) [3x − 4y + z ][3x − 4y − z ] =
1) (2x + 3y ) − (5z )
b) [3x − 4y + z ][3x + 4y − z ] =
2) (3x ) − ( 4y − z )
c) [5a + 3b + 2c ][5a − 3b − 2c ] =
3) (5a ) − (3b + 2c )
d) [2x + 3y + 5z ][2x + 3y − 5z ] =
4) (3x − 4y ) − z 2
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2
2
2
2
2
2
2
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INTEGER TYPE:
15.
The value of
5718 × 5718 − 4135 × 4135
= _________
5718 + 4135
LEVEL-2
SINGLE CORRECT CHOICE TYPE:
16.
If [k + 4b][5a − 4b] = 25a 2 −16b2 , then k = _______
1) 4a
2) 5a
3) 4b
52 × 48 = (a)2 – (b)2, then the values of a and b are _____
1) 50, 2
2) 52, 48
3) 2, 50
REASONING TYPE:
4) 5b
17.
18.
Statement I:
4) 48, 52
[a + b] a 2 + b2  a 4 + b4  a 8 + b8  a16 + b16  a 32 + b32  =
a 34 − b64
a−b .
Statement II: [x + y][x – y] = x2 – y2.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
Algebraic Fractions:
*
Fractions involving polynomials with numerator or denominator or both, are
called algebraic fractions.
*
Algebraic fraction is said to be in its simplest form or in lowest terms if the
numerator and denominator have no common factors except 1.
*
An equation which is true for all real values of its variables, is called an
identity.
Literal factor (Divisor):
If two or more algebraic expressions are multiplied, their products are obtained.
The algebraic expressions which multiplied to form the product, are called the
factors of the product.
Example: 12xy = 3x × 4y
3x and 4y are factors of 12xy
Greatest/Highest common factor (G.C.F./H.C.F):
G.C.F./H.C.F of two or more monomials is the highest monomial which divides
each of the given monomials completely.
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Method of finding the G.C.F of given monomials:
*
*
*
Break each monomial into the simplest numerical and literal factors.
Find all numerical and literal factors that divide each of the given monomials.
Multiply such numerical and literal factors to get the G.C.D.
Example: Find the G.C.F of 9xy, 6x2, 3x
9xy = 3 × 3 × x × y
6x2 = 3 × 2 × x × x
3x = 3 × x
G.C.D = 3x
L.C.M of Monomials:
The L.C.M of two or more monomials is a monomial having the least powers of
constants and variables such that each of the given monomials is a factor of it.
The sign of the coefficient of the L.C.M of the monomials is the same as the sign of
the coefficient of the product of the monomials.
Example: Find the L.C.M of 12p2q2, 6p2q and 3pq
12p2q2 = 2 × 2 × 3 × p × p × q × q
6p2q = 2 × 3 × p × p × q
3pq = 3 × p × q
L.C.M = 2 × 2 × 3 × p × p × q × q = 12p2q2
Relation between H.C.F of polynomials and L.C.M of polynomials
By finding factors we find H.C.F & L.C.M of polynomials.
If p (x) and q(x) are two polynomials then their H.C.F & L.C.M are related as given
below.
H.C.F × L.C.M = p(x) × q(x)
1.
Find the L.C.M and H.C.F of 6xy and 3x.
And prove that product of two numbers = product of L.C.M and H.C.F
Sol: L.C.M
H.C.F
6xy = 2 × 3 × x × y
6xy = 2 × 3 × x × y
3x = 3 × x
3x = 3 × x
L.C.M = 3 × 2 × x × y = 6xy
H.C.F = 3 × x = 3x
Product of two numbers = 6xy × 3x = 18 x2y
Product of L.C.M and H.C.F = 6xy × 3x = 18x2y
Product of two numbers = Product of L.C.M and H.C.F.
DESCRIPTIVE TYPE QUESTIONS:
1.
Find the H.C.F of the following monomials:
(i)
16a2, 24b2.
(iii) 4a3b2c2, 16a4b2c, 20a3bc2
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(ii)
6l2m, 10lm2, 8lmn
(iv)
9a3b2c, 16a2b3d, 15ab3cd
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LEVEL-1
SINGLE CORRECT CHOICE TYPE:
2.
The H.C.F of 12x2y, 16xy2 is
1) 4xy
2) 16x2y2
3) 12x2y
4) 16xy2
3.
The H.C.F of 15pq, 20qr, 25rp is
1) 5pqr
2) 25pqr
3) 5
4) 25
2
2
4.
The H.C.F of 4x , 6xy, 8y x is
1) 8x2y2
2) 8y2x
3) 6xy
4) 2x
2 2 2
2 3 4
3
5.
The L.C.M of 30a b c , –18a b c , 6abc
1) 90a2b3c4
2) – 90a2b3c2
3) 30a2b2c
4) – 30a2b2c2
MULTI CORRECT CHOICE TYPE:
6.
If 14x2y3, 21x2y2, 35x4y5z are monomials, then
1) The H.C.F of numerical coefficients is 7.
2) The H.C.F of numerical coefficients is 35.
3) The H.C.F of literals is x2y2.
4) The H.C.F of literals is x4y5z.
REASONING TYPE:
7.
Statement I: The H.C.F of 4a2b3, –12a3b, 18a4b3 is 2a2b.
Statement II: The H.C.F of 9x2, 15x2y3, 6xy2 and 21x2y2 is 3xy3.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
For finding the H.C.F. and L.C.M. of given polynomials
H.C.F. = (H.C.F. of numerical coefficient) × (Each common factor raised to lowest power)
L.C.M. = (L.C.M. of numerical coefficients) × (Each factor raised to highest power)
8.
The G.C.F of the terms in the expression 5a4 + 10a3 – 15a2 is
1) 15a3
2) 5a2
3) 15a4
4) 5a4
9.
The G.C.F of the terms in the expression 2xyz + 3x2y + 4y2 is
1) xyz
2) z
3) y
4) x
2 2
2 2
10. The G.C.F of the terms in the expression 3a b + 4b c + 12a2b2c2 is
2) a2
3) c2
4) 12a2b2c2
1) b2
11. MATRIX MATCH TYPE:
Column-I
Column-II
3 2
2
a) The H.C.F of x y and –5y is
1) 5x
3
2
b) The H.C.F of 4x ,6y and 8z is
2) 2xy
2 3
3 2
c) The H.C.F of 2x y ,10x y and 14xy is
3) 2
3
2
d) The H.C.F of 5x ,–15x and 45x is
4) y2
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INTEGER ANSWER TYPE:
12. If 12ax2, 6a2x3, 2a3x5are monomials, then H.C.F of numerical coefficients is
______________
LEVEL-2
13.
14.
If A = 34x2a3b7 is the G.C.F of monomials B and C, then B, C are
1) 324x2a5b9; 405x4a3b7
2) 324x2a5b9; 405x4ab4
3) 324x2a2b3; 405x2ab4
4) 324x2ab2; 405x2a2b2
If A = 15x7y8z9 × 25x4y3z3 and B = 20x6y5z3 × 10x2y6z9, then their G.C.D. is
1) 25x8y11z10
2) 25x8y11z11
3) 25x8y11z12
4) 25x8y10z10
REASONING TYPE:
15. Statement I: H.C.F of 8a2b2, 12a3b2; 24a4b3c2 is 4a2b2.
Statement II: H.C.F of two or more monomials is the H.C.F of the numerals x
each common factor raised to the lowest power.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
H.C.F of two or more monomials is the H.C.F of the numerals x each common
factor raised to the lowest power.
16. H.C.F of 9a3b2c, 16a2b3d, 15ab3cd is
1) a3b2c
2) ab2
3) 3ab2
4) 3abcd
2 2
3 3 2
17. H.C.F of a b c, a b c is
1) a3b3c
2) a3b2c2
3) a2b2c
4) a2b2c2
18. H.C.F of 10x2y, 25x2y2 is
1) 10x2y
2) 5x2y2
3) 5x2y
4) 10x2y2
Factorization: The process of resolving the given expression into factors is called
factorization.
Different types of Factorization:
1.
Taking out common factor:
Steps :
*
Find the H.C.F. of all the terms of the given expression.
*
Divide each term by this H.C.F. and enclose the quotient within the brackets,
keeping the common factor outside the brackets.
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Example: Factorize 9ab + 6a
In this the H.C.F of 9ab and 6a is 3a.
∴ 9ab + 6a = 3a(3b + 2)
Here 3a and 3b + 2 are the factors of 9ab + 6a.
Finding factors of multinomials :
To factorize a multinomial, in general we have to express the multinomial as a
product of two or more expressions. These two or more expressions whose product
is equal to the given multinomial, are called the factors of the multinomial. This
is the reverse process of multiplication.
Example: 25a2 + 10ab + b2 = (5a + b)(5a + b)
Factors are (5a + b) (5a + b).
Prime multinomial: A multinomial is said to be prime if it is divisible by one and
itself only.
Example: (a + 7b), (m +
1
) etc.,
m
Common factor: A number or a number letter combination which divides all the
terms of a multinomial is called a common factor of the terms of the multinomial.
Factorizing when a monomial is a common factor: To write sum or difference of
terms as product, the G.C.F. of all the terms of the multinomial has to be found
and taken out as a common factor.
2
Example: 4x + 2x = 2x(2x + 1) .
Common Binomial factor: The greatest common factor of the terms of a multinomial
need not be a monomial always. It can also be a binomial.
Example: 4a(a + b) + 3(a + b) here (a + b) is a common binomial factor.
DESCRIPTIVE TYPE QUESTIONS:
1.
Factorise the following:
2ab + 6ad
(ii)
5l2 + 5lm – 5ln
(iii) –a3b4c5 + a2b5c4
(iv)
9l4m3 – 6l3m4+ 21l2m2
(vi)
5n – 5n–2
(i)
(v)
2.
3.
4.
5.
6.
7.
63a6 – 35a5 + 21a4 – 42a3
(vii) π r2 + π rl.
3x(l + m) – 5y (l + m).
5x (a + 2b) + (x – 2y)(a + 2b).
4x2(3a – 2b) + 3y2(2b – 3a).
(4a – b)(3x + y) – (2a – 3b)(3x + y).
3x(a + b) – x(a + b) + 2x(a + b).
(x + y)2 – (x – y)(x + y).
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LEVEL-1
SINGLE CORRECT CHOICE TYPE:
8.
The factors of 20x + 5x2 is
1) 5x(4 + x)
2) 5x(4 – x)
3)
9.
The factors of 19x – 57y is
1) – 19 (x + 3y)
2) 19 (x + 3y)
3)
3
3
2 3
10. The factors of – 10ab + 30ba – 50a b is
1) – 10ab (b2 – 3a2 + 5ab2)
2)
2
2
2
3) – 10ab (b + 3a + 5ab )
4)
3
2
2
11. The factors of 7x + 7xy + 7xz is
1) 7x(x2 + y2 + z2)
– 5x(4 + x)
4)– 5x(4 – x)
– 19 (x – 3y)
4) 19 (x – 3y)
10ab (b2 – 3a2 + 5ab2)
10ab (b2 + 3a2 + 5ab2)
2) – 7x(x2 + y2 + z2) 3) 7x(x2 – y2 + z2)
4) 7x(x2 + y2 – z2)
MULTI CORRECT CHOICE TYPE:
12.
The factors of 3x (x – 4)– 5 (x – 4) is/are
1) (3x + 5)
2)(3x – 5)
3) (x + 4)
4) (x – 4)
REASONING TYPE:
13.
Statement I: The factors of (x + 5)2 – 4 (x + 5) are (x + 5)(x + 1)
Statement II: The factors of 9a (6a – 5b) – 12a2 (6a – 5b) are 3a (6a – 5b)(3 – 4a).
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
If A = 6x2 – 12xy + 18y2, B =0.8ab – 0.2a2b2, C =
14.
15.
16.
2 2 4
8
x + xy − xyz ,then
3
9
9
The factors of A are
1) 6 (x2 – 2xy + 3y2)
2) 6 (x2 + 2xy + 3y2)
3) 6 (x2 – 2xy – 3y2)
4) – 6 (x2 – 2xy + 3y2)
The factors of B are
1) ab (0.8 – 0.2 a2b2)
2) ab (0.8 + 0.2 ab)
3) 0.2ab (4 – ab)
4) 0.2ab (4 + ab)
The factors of C are
1)
2 
2
4 
x  x + y + yz 
3 
3
3 
2)
3)
2 
2
4 
x  x − y − yz 
3 
3
3 
4) −
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2 
2
4 
x  x + y − yz 
3 
3
3 
2 
2
4 
x  x − y − yz 
3 
3
3 
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MATRIX MATCH TYPE:
17. Column-I
a) The factors of a(b – c)2 – d(c – b) is/are
b) The factors of x(x + y)3 – 3x2y(x + y) is/are
c) The factors of a(3x – 2y) + b(2y –3x) is/are
d) The factors of 8(3a – 2b)2 – 10(3a – 2b) is/are
INTEGER ANSWER TYPE:
18. The factor of 9a + 6b is ____________
Column-II
1)12a – 8b – 5
2) ab – ac + d
3) x2 + y2 – xy
4) a – b
LEVEL-2
SINGLE CORRECT CHOICE TYPE:
19.
The factors of 2a + 6b − 3 (a + 3b ) are
2
1) (a + 3b )(2 − 3a − 9b )
2) (a − 3b )(2 − 3a − 9b )
3) (a + 3b )(2 + 3a − 9b )
4) (a − 3b )(2 − 3a + 9b )
MULTI CORRECT CHOICE TYPE:
20.
The factors of a 2 − b2 − c 2 − 2bc is
1) a 2 − ( b + c )
2) (a − b + c )(a − b − c )
3) (a + b + c )(a − b − c )
4) (a − b + c )(a − b + c )
2
Factorisation and rearrangement of terms:
If we observe the terms of an algebraic expression in the order they are given,
they may not have a common factor. But by rearranging the terms in such a way
that each group of terms has a common factor, some algebraic expressions can be
factorized.
While rearranging the sign and value of each term should not be altered.
Example: ax − by − ay + bx
= x(a + b) − y(a + b) = (a + b)(x − y)
∴ The factors are (a + b) and (x – y)
DESCRIPTIVE TYPE QUESTIONS:
1.
Factorise the following expressions:
(i)
ax – by – ay + bx
(iii) cp + 6dq – 2cq – 3dp
(v) 5px – 10qy + 2rpx – 4qry
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(ii)
(iv)
3a3 – 4a2 +6a – 8
axy – bcz + bcxy – az
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LEVEL-1
SINGLE CORRECT CHOICE TYPE:
2.
The factors of ax + bx + ay + by are
1) (a + b)(x + y)
2) (a – b)(x + y)
3) (a + b)(x – y)
4) (a – b)(x – y)
3
2
3.
The factors of x + x + x +1 are
1) (x2 –1)(x + 1)
2) (x2 –1)(x – 1)
3) (x2 + 1)(x + 1)
4) (x2 + 1)(x – 1)
4.
The factors of xy – ab + bx – ay are
1) (y + b)(x + a)
2) (y – b)(x + a))
3) (y – b)(x – a)
4) (y + b)(x – a)
2
5.
The factors of 6ab – b +12ac – 2bc are
1) (2c + b)(–b + 6a)
2) (2c + b)(b + 6a)
3) (2c + b)(–b – 6a)
4) (2c – b)(–b + 6a)
MULTI CORRECT CHOICE TYPE:
6.
The factors of ax2 + by2 + bx2 + ay2 is/are
1) (a + b)
2) (a – b)
3) (x2 + y2)
4) (x2 – y2)
REASONING TYPE:
Statement I: The factors of a2 + 2a + ab + 2b are (a + 2)(a + b).
Statement I: The factors of x2 – xz + xy – yz are (x + y)(x + z).
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
If in a given expression it is not possible to take out a common factor directly .In a
such case ,we have to make a suitable arrangement of the terms and group them
in a such a manner as to have a common polynomial,then factorised.
7.
The factors of a (a + b – c) – bc are
1) (a + b)(a – c)
2) (a – b)(a + c)
3) (a + b)(a + c)
4) (a + b)(a + c)
8.
The factors of a2x2 + (ax2 + 1) x + a are
1) (x + a)(ax2 + 1)
2) (x + a)(ax2 – 1)
3) (x – a)(ax2 + 1)
4) (x – a)(ax2 – 1)
9.
The factors of 3ax – 6ay – 8by + 4bx are
1) (3a + 4b)(x + 2y) 2) (3a – 4b)(x – 2y) 3) (3a – 4b)(x + 2y) 4) (3a + 4b)(x – 2y)
MATRIX MATCH TYPE:
10. Column-I
Column-II
(expression)
(factors)
2
2
2
a) The factors of lm – mn – lm + n is/are
1) bx + y
b) The factors of x3 – 2x2y + 3xy2 – 6y3 is/are
2) a – c
c) The factors of a(a – 2b – c) + 2bc is/are
3) x – 2y
2
d) The factors of abx + (ay – b)x – y is/are
4) m – 1
5) m + 1
INTEGER TYPE:
11. The number of factors of the expression with degree 3 is________
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LEVEL-2
SINGLE CORRECT CHOICE TYPE:
12.
13.
The factors of a3x + a2 (x – y)–a(y + z) – z are
1) (a + 1)(a2x – ay – z)
2) (a + 1)(a2x + ay – z)
3) (a + 1)(a2x – ay + z)
4) (a + 1)(a2x + ay + z)
The factors of (x2 + 3x)2 – 5(x2 +3x) – y(x2 + 3x) + 5(x2 + 3x) are
1) (x2 + 3x + 5)(x2 + 3x – y)
2) (x2 + 3x – 5)(x2 + 3x + y)
3) (x2 + 3x + 5)(x2 + 3x + y)
4) (x2 + 3x – 5)(x2 + 3x – y)
REASONING TYPE:
14.
Statement I: The factors of 3ax – 6ay – 8by + 4bx are (x + 2y) (3a + 4b).
Statement II: The factors of x2 + ax + bx + ab are (x – a)(x – b).
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
To find the factors of difference of two squares:
The difference of two squares is always equal to the product of the sum and
difference of the square roots of the square terms in the expression.
Example: a 2 − b2 = (a + b)(a − b)
Example: factorise a4 b6 – x8.
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DESCRIPTIVE TYPE QUESTIONS:
1.
Factorise the following expressions:
2.
x2 y2
−
4
9
2
(v)
π R – π r2
(viii) 9f2 – (3f + 4g)2
(xi) (a + b)2 – (a – 2b
(i)
4a2 – 9b2
(iv)
(vii)
(x)
(xiii)
Find
d8 – 100
1.44x2 – 0.01y2
(a + b)2 – (a – b)2
x4 – x2
the values of each of the following :
(i)
 1  1
5  − 4 
 2  2
2
(ii)
2
(ii)
b2
25
2n
(vi) a – b2n
(ix) 4x2 – (2x – y )2
(xii) 2a9 – 32a
(iii) 4a2 –
402 − 172
442 − 252
LEVEL - 1
SINGLE CORRECT CHOICE TYPE:
3.
The factor of 5x2 – 80y2 are
1) 5(x + 4y)(x – 4y)
3) 5(x + 4y)(x + 4y)
4.
The factor of 0.09m2 – 0.25n2 are
1) (0.3m + 0.5n)(0.3m + 0.5n)
3) (0.3m – 0.5n)(0.3m – 0.5n)
5.
The factor of 4(a + b)2 – 9(a – b)2 are
1) (5a – b)(5b + a)
2) (5a + b)(5b – a)
6.
The value of (10.2)2 – (9.8)2 is
1) 6
2) 7
MULTI CORRECT CHOICE TYPE:
7.
The factors of
6 
1
1)  a + b 
5 
3
2) 5(x – 4y)(x – 4y)
4) 5(x – 4y)(x + 4y)
2) (0.3m + 0.5n)(0.3m – 0.5n)
4) (0.3m – 0.5n)(0.3m + 0.5n)
3) (5a – b)(5b – a)
4) (5a + b)(5b + a)
3) 8
4) 9
6 
1
3)  a − b 
5 
3
1 
6
4)  a − b 
3 
5
1 2 36 2
a −
b 1s/are
9
25
1 
6
2)  a + b 
3 
5
REASONING TYPE:
8.
Statement I: The factors of 2a5 – 32a are 2a(a2 + 4)(a + 2)2.
1 
 4
1 
1
 2 1 
Statement II: The factors of  a −
 are  a +
a + a −  .
625 

25
5
5



1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
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COMPREHENSION TYPE:
a2 – b2 = (a + b)(a – b).
9.
The factors of 1 – 2ab – (a2 + b2) are
1) (1 + a + b)(1 – a – b)
3) (1 + a + b)(1 + a – b)
10. The factors of a2(b + c) – (b + c)3 are
1) (b – c)(a + b + c)(a – b – c)
3) (b + c)(a – b + c)(a – b – c)
11. The fators of a(a – 1) – b(b – 1) are
1) (a + b)(a + b + 1) 2) (a – b)(a – b – 1)
MATRIX MATCH TYPE:
12. Column-I
a) The factors of a12x4 – a4x12 is/are
b) The factors of x8 – y8 is/are
c) The factors of 2 – 50x2 is/are
d) The factors of 3x4 – 243 is/are
2) (1 + a + b)(1 – a + b)
4) (1 + a + b)(1 + a + b)
2) (b + c)(a + b + c)(a – b – c)
(b + c)(a + b – c)(a – b – c)
3) (a – b)(a + b – 1)
4) (a + b)(a + b – 1)
Column-II
1) x2 + 9
2) 1 + 5x
3) x + y
4) (a − x )
INTEGER TYPE:
13. The value of 1572 – 432 is ____________
LEVEL - 2
SINGLE CORRECT CHOICE TYPE:
14. The factors of x2 – 4x + 4y – y2 are
1) (x – y)(x + y – 4)
2) (x – y)(x + y + 4)
3) (x – y)(x – y – 4)
4) (x + y)(x + y + 4)
15. The factors of x(x + z) – y(y + z) are
1) (x + y)(x + y + z)
2) (x – y)(x + y + z)
3) (x – y)(x – y – z)
4) (x + y)(x – y – z)
MULTI CORRECT CHOICE TYPE:
16. The factors of xy9 – yx9 is/are
1) xy
2) x2 + y2
3) x2 – y2
4) x4 + y4
REASONING TYPE:
17. Statement I: The factors of 4(xy + 1)2 – 9(x – 1)2 are (2xy + 3y – 1)(2xy – 3x + 5).
Statement II: The factors of 2a5 – 32a are 2a(a – 2)(a + 2)(a2 + 4).
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
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The square and square root of a monomial:
When a monomial is multiplied by itself, it is said to be squared.
*
A monomial which can be written as the square of another monomial is
called a perfect square.
*
If a monomial is a perfect square then it can be expressed as a product of two
equal factors. Those equal factors are called a ‘square root’ of the given
monomial.
*
Every monomial which is a perfect square and has two square roots, of all
the positive square root is called the ‘ principal square root’
Example:
16a 2 = (±4a)2 = ±4a Here +4a is called the principal square root.
Rules for finding the factors of a trinomial which is a perfect square:
*
*
Two of the three terms must always be perfect square with ‘+’ sign.
The remaining term must be twice to the product of the square roots of the
two terms which are perfect squares.
Example: 9x 2 + 12xy + 4y 2 = (3x)2 + 2(3x)(2y) + (2y)2
DESCRIPTIVE TYPE QUESTIONS:
1.
Find the squares of the following monomials:
a, x2, xy, l2m, s2t2, 3ab, xyz, x2yz.
2.
Find the square roots of the following monomials:
(i) 4x2
3.
(ii)
y2
4
(iii)
a4 y2
16
(iv)
144t2 m2
169n2
Determine whether the following trinomials are perfect squares.
If so factorise then:
(i)
(iv)
y2 + y +
4
1
4
2 2
64y – 48x y + 9x
2
(ii)
9p2 – 6p + 1
(v)
2
1
k + k+
3
9
2
(iii) 25d2 – 28d + 4
(vi)
a2
1 2
y+
a
ay +
2
16
2 2
(vii) 12y2 + 60y + 75
LEVEL - 1
SINGLE CORRECT CHOICE TYPE:
4.
The factors of x2 + 8x + 16 are
1) (x + 4)(x + 4)
2) (x + 4)(x – 4)
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4) (x – 2)(x + 4)
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The factors of 4a2 – 4a + 1 are
1) (2a – 1)(2a + 1)
6.
7.
2) (2a – 1)(2a – 1)
3) (2a + 1)(2a + 1)
4) (a – 2)(a – 2)
The factors of x4 – 10x2y2 + 25y4 are
1) (x2 + 5y2)(x2 – 5y2)
2) (x2 + 5y2)(x2 + 5y2)
3) (x2 – 5y2)(x2 – 5y2)
4) (5x2 – y2)(5x2 – y2)
The factors of a4 – 2a2b2 + b4 are
1) (a + b)2(a + b)2
2) (a – b)2(a – b)2
3) (a – b)2(a + b)
4) (a – b)2(a + b)2
3) (2x – y – 3z)
4) (–2x – y + 3z)
MULTI CORRECT CHOICE TYPE:
8.
The factors of 4x2 – 4xy + y2 – 9z2 is/are
1) (2x – y + 3z)
2) (2x + y + 3z)
REASONING TYPE:
9.
Statement I: The factors of 16 – x2 – 2xy – y2 are (4 + x + y)(4 – x – y).
Statement II: The factors of x4 – x – z4 are (2x2 – 2xz + z2)(2x + z)z.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
(x + y)2 = x2 + y2 + 2xy and (x – y)2 = x2 + y2 – 2xy.
10.
The factors of x2 – 2xy + y2 – x + y are
1) (x – y)(x – y + 1)
11.
12.
2) (x – y)(x – y – 1)
2
3) (x + y)(x + y + 1)
4) (x + y)(x – y + 1)
2
The factors of 4a + 12ab + 9b – 8a – 12b are
1) (2a + 3b)(2a + 3b – 4)
2) (2a + 3b)(2a + 3b + 4)
3) (2a – 3b)(2a – 3b – 4)
4) (2a – 3b)(2a + 3b – 4)
The factors of a2 + b2 – 2(ab – ac + bc) are
1) (a + b)(a + b + 2c)
2) (a – b)(a – b – 2c)
3) (a + b)(a – b + 2c)
4) (a – b)(a – b + 2c)
MATRIX MATCH TYPE:
13.
Column-I
Column-II
a) The factors of x2 – y2 – 4xz + 4z2 is/are
2
2
2
1) 5x – 6y – 1
b) The factors of a + 4b – 4ab – 4c is/are
2) x – y + 7
c) The factors of 49 – x2 – y2 + 2xy is/are
3) a – 2b + 2c
2
2
d) The factors of 25x – 10x + 1 – 36y is/are
4) x – y – 2z
INTEGER TYPE:
14.
The value of 29 × 29 + 2 × 29 × 21 + 21 × 21 is ___________
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LEVEL - 2
SINGLE CORRECT CHOICE TYPE:
15. The factors of (a4 – 8a2b2 + 16b4) – 256 are
1) (a2 – 4b2 + 16)(a2 – 4b2 – 16)
2) (a2 + 4b2 + 16)(a2 + 4b2 + 16)
3) (a2 – 4b2 – 16)(a2 – 4b2 – 16)
4) (a2 – 4b2 + 16)(a2 + 4b2 – 16)
2
2
16. The factors of 4(x + y) – 28y(x + y) + 49y are
1) (2x – 5y)2
2) (x + y)(x – 2y)
3) (2x + 5)2
4) x + y(2x – 5y)
2
17. The factors of x + 8x + 15 are
1) (x + 5)(x + 3)
2) (x – 5)(x – 3)
3) (x – 5)(x + 3)
4) (x + 5)(x – 3)
MULTI CORRECT CHOICE TYPE:
18. The factors of 9 – a6 + 2a3b3 – b6 is/are
1) a3 – b3 + 3
2) – a3 + b3 + 3
3) a3 – b3 – 3
4) a3 + b3 + 3
REASONING TYPE:
19. Statement I: The factors of x16 – y16 + x8 + y8 are (x8 + y8)(x8 – y8 + 1).
Statement II: The factors of (p + q)2 – (a – b)2 + p + q – a + b are
(p + q – a + b)(p + q + a – b + 1).
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
Division of multinomials:
*
Arrange the terms of the dividend and divisor in decreasing order of powers
leaving space for missing terms.
*
Divide the first term of the dividend by the first term of the divisor and write
the result as the first term of the quotient.
*
Multiply the entire divisor by this first term of the quotient and put the
product under the dividend, keeping like terms under each other.
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*
*
*
Subtract the product from the dividend and bring down the rest of the dividend.
a)
4x 4 ÷ x = 4x 4 ×
b)
Divide (9p + 9pq + 2q
Step 4 gives us the new dividend, repeat steps 1 to 4.
Continue the process till the degree of the remainder becomes zero or less
than that of the divisor.
Examples:
1
= 4x 3
x
2
2
) by 3p + q
Steps: 1)
3p (3p + q ) = 9p2 + 3pq
3p + q)9p2 + 9pq + 2q 2 (3p + 2q
9p2 + 3pq
(-)
(-)
__________________
9p2
= 3p
3p
2)
6pq
= 2q
3p
2q (3p + q ) = 6pq + 2q 2
6pq + 2q2
6pq + 2q2
(-) (-)
_____________
0
______________
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DESCRIPTIVE TYPE QUESTIONS:
1.
Do the following divisions and verify division rule:
(i)
(–24a b c) ÷ 3ab
3
2
(ii)
(iii) (81y9 – 36y8 + 18y7) ÷ (–9y6)
(v)
2.
(iv)

4
 24p −

(ii)
–6x + 9 + 4x2 by 2x + 3
(iv)
a3 – b3 by a – b
1 p3
6
−
1 p2  ÷ 1 p2

3  2
(r13 − 15r12 − 24r7 ) ÷ 34 r 7
Divide and verify the following:
(i)
x2 + 7x + 12 by x + 3
(iii) –9 – 46x2 + 16x3 + 39x by 8x – 3
3.
−56a 3 b
−14ab2
Find the quotient and remainder:
(b4 – 16) ÷ (b – 2)
(ii) (12x + 4 + 6x3 + x4 + 13x2) ÷ (x2 + 3x + 2)
(iii) (x4 – x2 + 16) ÷ (x2 + 3x + 4)
(i)
LEVEL - 1
SINGLE CORRECT CHOICE TYPE:
4.
The remainder when –72a4b5c8 is divided by –9a2b2c3 is
1) 8a2b3c5
5.
7.
3) 8a3b5c2
4) 8a5b3c2
The remainder when 9m5 +12m4 – 6m2 is divided by 3m2 is
1) 3m3 + 4m2 + 2
6.
2) 8a3b3c5
2) 3m3 + 4m2 – 2
3) 3m3 – 4m2 + 2
4) 3m3 – 4m2 – 2
3a 4 + 2 3a 3 + 3a 2 − 6a is divided by 3a is
The remainder when
1)
1 3
2 2
a +
a +a−2
3
3
2)
1 3
2 2
a −
a +a −2
3
3
3)
1 3
2 2
a +
a −a +2
3
3
4)
1 3
2 2
a −
a −a−2
3
3
The quotient when x3 – 6x2 + 11x – 6 is divided by x2 – 4x + 3 is
1) x + 2
2) x – 1
3) x – 2
4) x + 1
MULTI CORRECT CHOICE TYPE:
8.
The quotient and remainder when 6x3 + 11x2 – 39x – 65 divided by 3x2 + 13x + 13
is
1) 2x – 5
2) 2x + 5
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3) 0
4) 1
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REASONING TYPE:
9.
Statement I: The remainder when 4z3 + 6z2 – z is divided by
−1
z is
2
–8z2 – 12z + 2.
Statement II: The remainder when 3x 3 + 4x 2 + 5x + 18 is divided by x + 2 is
3x2 – 2x + 9.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
dividend = quotient × divisor + remainder
10. The remainder when 14x2 + 13x – 15 is divided by 7x – 4 is
1) –2
2) –3
3) –4
4) –5
2
11. The quotient when 14x + 13x – 15 is divided by 7x – 4 is
1) 2x + 3
2) 2x – 3
3) 3x – 2
4) 3x + 2
3
2
12. The remainder when 15z – 20z + 13z – 12 is divided by 3z – 6 is
1) 52
2) 53
3) 54
4) 55
MATRIX MATCH TYPE:
13. Column-I
Column-II
5
3
2
a) The quotient when 6y – 28y + 3y + 30y – 9 is
1) 0
2
divided by 2y – 6 is
b) The remainder when 6y5 – 28y3 + 3y2 + 30y – 9 is
3
2) 3y − 5y +
3
2
divided by 2y2 – 6 is
4
3
2
c) The quotient when 15y − 16y + 9y −
10
y + 6 is
3
3) 6
divided by 3y – 2 is
4
3
2
d) The remainder when 15y − 16y + 9y −
10
y + 6 is
3
3
2
4) 5y − 2y +
5
y
3
divided by 3y – 2 is
INTEGER ANSWER TYPE:
14. The remainder when 6 + x – 4x2 + x3is divided by x – 3is ___________
LEVEL - 2
SINGLE CORRECT CHOICE TYPE:
15. The remainder when x5 + x4 + x3 + x2 + x + 1 is divided by x3 + 1 is
1) x2 + x + 1
2) x2 + x – 1
3) x2 – x – 1
4) x2 – x + 1
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The remainder when x4 + x2 + 1 is divided by x2 + x + 1 is
1) x2 + x + 1
2) x2 + x – 1
3) x2 – x – 1
REASONING TYPE:
16.
17.
4) x2 – x + 1
2
Statement I: The remainder when –4a3 + 4a2 + a is divided by 2a is −2a + 2a +
Statement II: The remainder when –x 6 + 2x4 + 4x3 + 2x 2 is divided by
−
1)
2)
3)
4)
1
2
1
.
2
2x 2 is
x 4 − 2x 2 + 2 .
Both Statement-I and Statement-II are true.
Both Statement-I and Statement-II are false.
Statement I is true, Statement II is false.
Statement I is false, Statement II is true.
Equation: A statement of equality of two mathematical expressions, is called an
equation.
Example: 3x + y = 8 is an equation.
Formulae: Equations which are used frequently to solve prob-lems are called formulae:
Example: A = S2
Above formula gives the relationship between the variable ‘A’ (area of a square) and
the variable ‘s’ (side of a square)
The subjects of the formulae : Formulae are frequently given with one variable standing
alone on one side of the equality sign and the value of this variable is given in
terms of the others. This variable standing alone on one side is called the subjects
of the formulae.
Example: A = l × b (Area of a rectangular field)
Here the variable ‘A is standing alone on L.H.S and its value is given in the terms of
l (length) and ‘b’ (breadth).
∴ ‘A’ is the subject of this formula.
Starting formulae given symbolically in word form:
*
V =l ×b×h
Volume of a rectangular parallelepiped is equal to the product of its
length, breadth and height.
*
1
bh
2
Area of triangle is equal to half the product of its base and height.
A=
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Expressing the principles stated in words in terms of symbols:
*
The area of a rectangular field is the product of its length, breadth expressed
in symbols as
A=l×b
Where ‘A’ is unit area ‘l’ and ‘b’ are units of length and breadth respectively.
*
Gain in transaction is the difference between selling price and cost price
expressed in symbols as
G = S.P – C.P where,
G
=
Gain
S.P. =
Selling price
C.P. =
Cost Price
Changing the subject of the formula:
Area of a triangle is equal to half the product of its base and height.
A=
where,
In A =
1
bh
2
A = Area b = base h = Corresponding height of the triangle.
1
bh , ‘A’ is called subject of the formula.
2
h=
2A
b
This new formula is deduced from the original formula A =
1
bh. This new formula
2
is called “derived” of “auxiliary” formula.
*
The auxiliary formula depends upon the number of variables.
*
If a formula contain ‘n’ varibles then the number of auxiliary formulae = (n –
1)
The following characteristics of subject in a formula have to be noted:
*
The subject symbol occurs on the left side of the equality sign:
*
It is written independently without being linked with any other quantities
or variables.
*
The coefficient in always one
*
All properties used in solving simple equations are also used in
transforming the subject of a formula.
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DESCRIPTIVE TYPE QUESTIONS:
1.
Write the following statements in symbolic form as formulae:
In a triangle ABC, sum of the three angles is equal to 180 0 .
(ii) The volume of a cuboid with a rectangular base is the product of the length
and breadth of the base and the height of the cuboid.
(iii) The distance covered by an object is given by the product of the average
speed and time taken for the journey.
(iv) The circumference or perimeter of a circle is given by the product of π and
diameter of the circle.
State the following formulae in words:
(i)
When the temperature of a gas is constant, then PV = C. Where P = pressure,
V = volume and C = constant.
(ii) Total curved surface area ‘S’ of a right circular cylinder S = 2 π rh.
(i)
2.
(iii)
∠x + ∠y = 180° .
(iv)
The length of the curve in a sector is given by l =
(v)
Area of trapezium is given by A =
x
× 2ð r ..
360
1
d (a + b ) .
2
LEVEL-1
SINGLE CORRECT CHOICE TYPE:
3.
The formula of Area of rectangle, A = l × b then the subject in the formula is
1) l
2) b
3) l × b
4) A
4.
The coefficient of subject in a formula is
1) 1
2) 2
3) 4
4) 3
5.
“The perimeter of a circle (p) is given by the product of π and its diameter” write it
in symbolic form
1) p = π + d
6.
I=
1)
2) p = π – d
3) p = π × d
4) p =
π
d
PNR
then the formula of p is
100
100I
PR
2)
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100I
NR
3)
NR
100I
4)
100
INR
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7.
In s =
1)
Techno Text book
n
(a + l) then the formula of ‘a’ is
2
2s
+l
n
2)
s
−l
2n
3)
2s
−l
n
4)
2s
n
MULTI CORRECT CHOICE TYPE:
8.
The possible auxiliary formulae from v = lbh are
1) l =
v
bh
2) b =
v
lh
3) h =
v
lb
4) v = l + b + h
REASONING TYPE:
9.
Statement-I: A shopkeeper buys x kgs sugar at Rs y per kg and sells it at Rs z
per kg. Then the formula for profit (p) is p = (xz – xy).
Statement-II: profit = selling price – cost price.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
The Temperature in fahrenheit degrees ‘F’ can be changed to Celsius degrees ‘C’
by the formula c =
10.
Make ‘F’ as the subject
1)
11.
12.
5
(F − 32) then
9
9c
− 32
5
2)
9c
+ 32
5
Find ‘F’ when c = 100° is
1) 148°
2) 212°
Find ‘F’ when c = 0° is
1) 32°
2) 23°
MATRIX MATCH TYPE:
13. Column-I
a) Equation which is used frequently
to solve problems
b) Variable standing alone on the left
side of an equation
c) The formula obtained by transforming
the subject in a given formula
d) Number of auxiliary formulae that
can be derived from p = 2(l + b)
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3)
5c
+ 32
9
4)
5c
− 32
9
3) 221°
4) 184°
3) 42°
4) 24°
Column-II
1) subject
2) Auxiliary formulae
3) Formula
4) 2
5) 3
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INTEGER ANSWER TYPE:
14.
In a triangle, the sum of the three angles is a ___________ right angles.
LEVEL - 2
SINGLE CORRECT CHOICE TYPE:
15.
Write the following statement as an equation.
“If 8 is added to a certain number and sum is multiplied by 4, the result is 56”.
1) 8 + 4x = 56
16.
2) 8 + x + 4 = 56
18.
In M =
2) l =
N (P − R )
40
T
g
2π
3) l =
T 2g
4π2
4) None of these
make ‘R’ as subject
1)
40M
+P =R
N
In
x+a 1
=
then make ‘a’ as subject
a+b c
1) a =
4) 8x + 4 = 56
l
, if ‘l’ is made subject, then
g
In T = 2π
1) l = 2πTg
17.
3) (8 + x)4=56
3) R = P −
2) 40 MNP = R
b + xc
c −1
2) a =
b − xc
c +1
3) a =
40M
N
b − xc
c −1
4) None of these
4) None of these
MULTI CORRECT CHOICE TYPE:
19.
In v2=u2+2as, make ‘u’ as the subject
2) u = ( v 2 − 2as )
1) u = v 2 − 2as
1
2
3) u = v 2 − 2sa
4) u = −2sa + v 2
MATRIX MATCH TYPE:
20.
Column-I
Column-II
a) In S=ut make ‘u’ as subject
1) 3x + y = 8
b) Example of equation is
2)
c) Find ‘x’, if z =
x +3
when z = 3
x −1
d) In P = 2(l + b), if ‘l’ is subject
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P − 2b
2
3) u = s/t
4) 3
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Evalution of the subject of a formula: In a given formula, we substitute the given
values of the variables other than the subject and then find its value.
Examples:
1.
Given Principal (P), Time (T) and rate of interest (R%), the single interest (I) can
be calculated using the formula I =
given P = Rs. 1250; T = 3
Sol: I =
PTR
. Make R the subject of formula. Find R
100
1
years, I = Rs. 175.
2
PTR
100
∴100 × I = 100 ×
PTR
100
Dividing both sides by PT
Hence
(or) 100 I = PTR
=
100 I PTR
=
PT
PT
100 I
100 I
= R (or) R =
PT
PT
Substituting the values of P, T and I
R=
100 × 175 7
2
× = 14 × = 4% .
PT1250 2
7
2.
If P(L + M) = Q(R – M) ; P = 4.5, Q = 13.5, R = 5, L = 2.5, find the value of M ?
Sol: P(L + M) = Q(R – M)
4.5(2.5 + M) = 13.5(5 – M)
11.25 + 4.5M = 67.5 – 13.5M
4.5M + 13.5M = 67.5 – 11.25
18M = 5625
M=
5625
25
=
.
18 × 100
8
DESCRIPTIVE TYPE QUESTIONS:
1.
The total surface area of a solid cylinder is given by S = 2 π r(r+h). Make ‘h’ subject
of the formula. Find its value when S = 220sq.units and r = 3.5 units.
2.
The perimeter of a semi-circular window of a radius ‘r’ meters is given by P = 2r + πr .
Make ‘r’ subject of the formula. Find ‘r’ when P = 2.4m
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If ‘C’ is the cost price, ‘l’ is the percentage of loss, the selling price ‘S’ is given by
 100 − l 
S=
 C . Make ‘l’ subject of the formula. Find ‘l’ when S = Rs. 225 and C =
 100 
Rs. 250.
4.
The length of the curve (l) in a sector is given by l =
x
× 2πr . Make ‘r’ subject of
360
the formula. Find ‘r’ when x = 70° and l = 33 cm.
5.
6.
x
× πr 2 .
360
Make ‘x’ subject of the formula Find x when A = 231 sq.cm and r = 14cm.
Find the missing numbers in the table and finally write the formula between the
variables:
(i)
Changing minutes into seconds.
The area of a sector of radius ‘r’ and included angle x° is given by A =
(ii)
7.
Changing rupees into paise.
How many squares are there altogether in a 2 by 2 square network?
(i)
(ii)
Draw 3 by 3 and 4 by 4 networks and find the number of squares each
contains.
Now draw a K by K network and find the formula that tells the number of
squares contained in it. How many squares does a 15 by 15 network
contain?
LEVEL - 1
SINGLE CORRECT CHOICE TYPE:
8.
The area of a square is A = S2. Find s, when A = 169sq.units
1) 12 units
2) 13 units
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3)
169 units
4) Both (2) & (3)
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Techno Text book
In the formula s =
1)
5 × 10
1 2
gt , then the value of ‘t’ when s = 625, g = 5 is
2
2)
15 × 10
3)
25 × 10
4) 5
10.
Perimeter of a sector is P = l + 2r. If l = 5cm and r = 4 cm then P = _____
1) 13
2) 31
3) 3
4) 9
MULTI CORRECT CHOICE TYPE:
11.
The area of a square is given by A= s 2 . change the subject to ‘s’ . then the value of
‘A’ when s=24 units
1) 546 sq.units
2) 586 sq.units
3) 576 units
4) 576 sq.units
REASONING TYPE:
12.
Statement I: If
r = x 2 + y 2 , express ‘y’ in terms of ‘r’ when r = 17, x = 8 then
y = 14.
Statement II: Make ‘b’ as the subject in x =
a(1 − x 2 )
a −b
b
=
, then
.
(1 + x 2 )
a+b
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
Fahrenheit temperature (F), decreased by 32 is equal to nine-fifths of the
centigrade temperature (c)
13. Frame a formula by making F as subject
1) F − 32 =
9
C
5
2) F =
9
C + 32
5
If C = 30°, then F =
1) 68°
2) 86°
15. If F = 59°, then C =
1) 51°
2) 25°
MATRIX MATCH TYPE:
3) F =
9
C − 32
5
4) F =
5
C + 32
9
14.
16.
3) 78°
4) 87°
3) 15°
4) 52°
The given formula is v 2 − u 2 = 2as
Column-I
Column-II
a) In above formula find v, when u = 4, a = 2, s = 12 is 1) 8
b) Make ‘u’ as a subject
2) 6
c) Find u, when v = 10, a = 2, s = 16
3)
d) The coefficient of ‘as’ in above formula
4) 2
5)
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v 2 − 2as
v 2 + 2as
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LEVEL - 2
SINGLE CORRECT CHOICE TYPE:
17.
A = π (R 2 − r2 ) where R=3.5 cm, r = 1.5cm then A =
1) 31 sq. cm
18.
Let s =
2) 5
3
sq.cm
7
4)
3) 10
3
sq. cm
7
4) 20
The formula of v is π(R 2 − r 2 )h find ‘h’ when R = 2.6, r = 2.3, π =
are given
1) 2.5
REASONING TYPE:
20.
3) 31
n
[2a + (n − 1)d] then the value of ‘a’ when s = 185, n = 10, d = 3 is
2
1) 15
19.
2) 22 sq.cm
2) 25
3) 15
Statement-I: In T = m + mrt, then the value of r is
Statement-II: In m =
22
and v = 115.5
7
4) 35
20
when T = 625, m = 100, t = 5.
21
N(P − R)
40m
, then R is “ p –
”.
40
N
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
If ‘c’ is the cost price, ‘g’ is the percentage of pfofit then the selling price ‘S’ is given
100 + g 
c . Then
by S = 
 100 
21.
22.
The value of ‘c’ when S = Rs. 340, g = 6
1
% are given
4
1) Rs. 320
3) Rs. 380
4) Rs. 300
1
The value of ‘s’ when g = 12 % , c = Rs. 600 are given
2
1) Rs. 675
23.
2) Rs. 340
2) Rs. 650
3) Rs. 600
Percentage of ‘g’ when s = Rs. 720, c = Rs. 600
1) 10%
2) 20%
3) 30%
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4) Rs. 625
4) 40%
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Open Sentences: A mathematical sentence which contains a variable such that it
may be true or false depending on the values of the variable, is called an Open
Sentences.
An open sentence containing the sign of equality is called an equation, we
shall first study the properties of equality.
Linear Equation: An equation involving on variable with highest power 1, is called a
linear equation in that variable.
Properties of equality:
*
Reflexive Property: Every number is equal to itself.
Example: 8 = 8, y = y so on.
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*
Symmetric Property:If in two numbers, the first number is equal to the
second then the second is equal to the first.
If x, y are two numbers and x = y then y = x
Example: (3 + 5) = (2 + 6) hence (2 + 6) = (3 + 5)
*
Transitive Property: In three numbers, if the first number is equal to the
second and second number is equal to the third, then
the first number is equal to the third.
If x = y and y = z then x = z for any three numbers x, y, z
Example: 5 + 3 = 8, 4 × 2 = 8 ∴ 5 + 3 = 4 × 2.
*
If a number is equal to, each of two numbers separately, then these two
numbers are equal to one another.
If x = p and x = q then p = q, for any three numbers x, p, q.
Addition Property: If the same number is added to both sides of an equality, the
equality is no altered.
If x = y, then x + z = y + z where x, y, z are any three numbers.
Subtraction Property: If the same number is subtracted from both sides of an equality, the equality is not altered.
If x = y, then x – z = y – z where x, y, z are any three numbers.
Multiplication Property: If both sides of an equality are multiplied by the same number, the equality is not altered.
If x = y, then xz = yz, where x, y, z are any three numbers.
Division Property: If both sides of an equality are divided by the same non-zero number, then the equality is not altered.
If x = y, then
x y
=
where x, y, z are any three numbers and z ≠ 0 .
z z
*
Whenever a change occurs on one side of an equation, the same change be
made on the other side to maintain the equality (balance) between the sides.
*
To solve a simple equation, the terms containing unknown quantity should
be collected on the left hand side of the equation and constants on the other
side. Then the coefficient of the unknown quantity should be reduced to
unity.
*
If a term is moved (Transposed) from one side of an equation to the other,
then the sign of the term is changed.
*
From E and F in example 3, we observe that, “If a quantity dividing a side is
moved (Transposed) to the other side, it multiplies the other side”.
Similarly from G and H of example 4, we observe that, “if a quantity multiplying a side is moved (Transposed) to the other side, it divides the other side.”
*
From this it is clear that, “If the same term (same both in magnitude and
sign) is present on both sides of an equation then it can be cancelled.
*
“Changing a term from one side of an equation to the other side is called
transposition.”
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DESCRIPTIVE TYPE QUESTIONS:
Solve the following equations and check your result.
1.
x – 10 = –15.
2.
26 = 39y.
3.
x
=5.
3
4.
4p + 12 = 36 – 2p.
5.
7z + 13 = 2z + 4.
LEVEL - 1
SINGLE CORRECT CHOICE TYPE:
6.
Linear equation in one variable is an equation with only one variable and
1) degree 2
7.
If 2x − 1 =
1)
8.
2) degree 1
3) no degree
4) degree 3
x +1
, then x is
3
5
4
2)
6
5
3)
4
5
4)
1
5
If 3x – 5 = 2x + 8, then
1) x = 17
2) x = 11
3) x = 15
4) x = 13
3) 2x = 12
4) x = 3
MULTI CORRECT CHOICE TYPE:
9.
If 5x – 17 = 2x – 8, then
1) x = 6
2) 2x = 6
REASONING TYPE:
10.
Statement-I: If
3
4
=
, then x = –2.
x +8 6−x
Statement-II: If
ax + b m
=
, then n(ax + b) = m(cx + d).
cx + d n
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
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COMPREHENSION TYPE:
Solving the equations of the form
11.
12.
13.
ax + b = cx + d, we get x =
d−b
, then
a −c
If 4x − 9 = 2x + 7 , then
1) x = 8
2) x = 7
3) x = 16
4) x = 6
If 5x + 18 = 11 − 2x , then
1) x = 0
2) x = –2
3) x = –1
4) x = 1
If 21 − 3(x − 7) = x + 20, then x is
1)
11
2
2)
11
4
MATRIX MATCH TYPE:
14. Column-I
3)
6
5
4)
Column-II
a) If 2x − 3 = 7 − 5x , then
1) x = 7
b) If 2(x − 3) = 8 , then
2) x = −
c) If 2x + 5 =
11
7
x
− 2 , then
6
d) If 9x − 6 = 3x + 18 , then
42
11
3) x = 6
4) x =
10
7
5) x = 4
INTEGER ANSWER TYPE:
15. The same quantity can be added from both side of an equation without changing
equality, thus if a = b then a + c = ___________
LEVEL-2
SINGLE CORRECT CHOICE TYPE:
16.
If
2x − 1 3x + 1
=
, then
2x − 3 3x − 1
1) x = 1
17.
3) x = 2
4) x = –2
3) 3 − 2 2
4) 2 − 3 2
If 1 + x = 2(1 − x) , then x is
1) 3 + 2 2
18.
2) x = –1
2) 2 + 3 2
If (5x − 1)(ax + 1) and 3a(3x − 1) are equal when x = 1, find a
1) a = 1
2) a = 2
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4) a = 4
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MULTI CORRECT CHOICE TYPE:
19.
5
3
2
If (x + 2) − (x − 2) = x , then x is
1)
4
2
2)
2
4
3) 2
4)
1
2
4)
7
4
COMPREHENSION TYPE:
If
20.
21.
ax + b m
=
, then n(ax + b) = m(cx + d)
cx + d n
If
4x + 1 2x − 1 3x − 7
+
−
= 6 , then x is
3
2
5
1)
9
2
If
x+7
7
=
, then x is
x + 16 10
2)
1) 19
22.
If
11
4
3)
2) 11
13
2
3) 14
4) 16
3) 5
4) 7
2 + x 10 + x
=
, then x is
7 + x 25 + x
1) 2
2) 3
Equation with fractional and decimal coefficients:
The fractions should be cleared by multiplying both sides of the equation by the
least common multiple (L.C.M) of the denominators of the fractions in the
equation.
2
1
x − x = 4.
3
2
Example:
Solve the equation
Solutions:
L.C.M of denominators 3 and 2 is 6.
Multiplying both sides of the equation by 6.
We get
6×
2
1
x −6× x = 6×4
3
2
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or
4x – 3x = 24
or
x = 24.
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DESCRIPTIVE TYPE QUESTIONS:
1.
2.
3.
4.
5.
6.
7.
8.
x
3
2
3
+4 =
x
2
x +5 =
+3
x
1
.
2
1
−7 .
4
2
(0.3)x – 1.8= x – 1.16.
3 (5 – y) = 4 (3y + 2) + 27.
7x − 3
6
−
2x − 3
4
=
5
4
.
0.2 (2x – 1) – 0.5 (3x – 1 ) = 0.4.
3x − 1
5
5y + 3
4
−
1+ x
−
3y + 2
2
7
=3−
x −1
2
.
+2 = 0 .
LEVEL-1
SINGLE CORRECT CHOICE TYPE:
9.
10.
0.5z + 4 5
= , then z =
1.2z + 6 3
1) 4
2) –4
If
If
x x x −7
+ =
, then x =
2 6
9
1)
−8
5
2) –1.8
3.4x + 8 7.2
=
, then x =
1.6x + 2
5
1) –4.67
2) 4.67
MULTI CORRECT CHOICE TYPE:
11.
12.
3) –8
4) 8
3) –1.4
4) 1.4
3) 46.7
4) –46.7
If
If
z − 1 2z + 5 5
+
= , then z =
6
3
9
1) −
17
15
2) −
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15
3)
71
15
4) –1.13
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REASONING TYPE:
13. Statement I: If 3.5 x +7 = 1.5x + 21, then x = 7.
d−b
.
a −c
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
Statement II: If ax + b = cx + d, then x =
d−b
a −c
If 7.5 x + 4 = 6.5 x + 2, then x =
1) –2
2) 2
If 0.8 x + 7 = 0.5 x + 10.5, then x =
If ax + b = cx + d, then x =
14.
15.
3.5
3.5
2)
0.3
1.3
16. If 0.5 x + 3.5 = 3.5 x + 0.5, then x =
1) 1
2) –1
MATRIX MATCH TYPE:
17. Column-I
a) If 3.6 (0.5x + 10) = 0.2x + 4,then x =
1)
b) If 7.4x – 1 = – 0.6 + 19,then x =
c) If 9.8x + 6 = 4.9 x – 8,then x =
d) If 1.8 x + 5 = – 0.2 x + 7,then x =
3) 4
3)
17.5
1.3
3) 3.5
4) –4
4)
17.5
0.3
4) 0
Column-II
1) 1
20
7
3) –1
2) −
5
2
5) –20
4)
INTEGER ANSWER TYPE:
18.
If
x
x
+ 2 = + 4 , then x = _____________
5
3
LEVEL-2
SINGLE CORRECT CHOICE TYPE:
19.
20.
3
1
(x + 1) − = x − 1 , then x =
4
2
1) 6
2) –5
3) 5
4) –6
If 2 x − 2 (x + 2) = 4 , then x =
5
9
1) 27
2) 25
3) –27
4) –25
If
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3x 4 7
1
x + 5 , then x =
+ − =
7 9 8 21
1) 14.25
2) –14.25
COMPREHENSION TYPE:
21.
If
If Ax + B = Cx + D, then x =
22.
23.
25.
4) –142.5
D−B
where A,B,C,D are non- zero rational numbers.
A −C
5 7x
=
+ 4 , then x =
4
3
1) 3.17
2) –3.17
3) 3.27
4) 3.37
If 7 (x − 1) = 3 (x + 5) , then x =
9
4
1) 163
2) 136
3) 169
4) 166
3) 0.071
4) 0.0071
3) 3.1
4) –2.1
If 3.2x +
12
10
8
6
x−4
= −6 − 24 x , then x =
17
17
11
11
1) –0.017
2) 0.017
MULTI CORRECT CHOICE TYPE:
24.
3) 142.5
If 7
If 0.4(5x − 1) = 0.3(10x + 1) + 2.4 , then x =
1) –1.1
2) –3.1
The method and steps involved in solving a word problem.
*
Read the problem carefully and note down what is given and what is
required.
*
*
Select a latter say a or b or c to represent the unknown quantity asked for.
*
Look for quantities which are equal as per conditions given and form an
equation.
*
*
Solve the equation obtained in step (4).
Represent the word statements of the problem in the symbolic language step
by step.
Check the result for making sure that your answer satisfies the
requirements of the problem.
Examples:
1.
The sum of three consecutive odd numbers is 87. Find the numbers.
Sol: Let numbers be
2x + 1, 2x + 3, 2x + 5
2x + 1 + 2x + 3 + 2x + 5 = 87
6x + 9 = 87
⇒ 6x = 78
⇒ x = 13
Number are
2 × 13 + 1, 2 × 13 + 3, 2x × 13 + 5
Number are
27, 29, 31.
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2.
The sum of two numbers is 100 and their difference is 10. Find the number.
Sol: Let the numbers be x, 100 – x
x – (100 – x) = 10; x – 100 + 10 = 10, 2x = 110, x = 55
Then number are 55, 100 – 55; Then number are 55, 45
3.
The sum of two numbers is 45 and their ratio is 2 : 3 is 2 : 3. Find the numbers.
Sol: Let numbers be 2x and 3x
2x + 3x = 45
⇒ 5x = 45
⇒ x=9
Numbers are 2 × 9 and 3 × 9
Number are
18 and 27.
DESCRIPTIVE TYPE QUESTIONS:
1.
If a number is tripled and 16 is added to it, the result is equal to 40. Find
the number.
2.
If two thirds, one half and one seventh of a number is added to itself, the result
is 37. Find the number.
3.
4 is added to a number and the sum is multiplied by 5. If 20 is subtracted
from the product and the difference is divided by 8, the result is equal to 10.
Find the number.
4.
5.
1
is subtracted from a number and the difference is multiplied by 4. If 25
2
is added to the product and the sum is divided by 3, the result is equal to
10. Find the number.
One number is 3 less than two times the other. If their sum is increased by 7,
the result is 37. Find the number.
LEVEL-1
SINGLE CORRECT CHOICE TYPE:
6.
A number exceeds its four–sevenths by 18. Then the number is _________.
1) 24
2) 42
3) 64
4) 74
7.
Divide 42 into two parts so that one part is twice the other part then two parts are
1) 24, 18
2) 36, 6
3) 26, 16
4) 14, 28
8.
If a number is multiplied by 9 and 10 is subtracted from the product, the result is
equal to 62, then the number is
1) 6
2) 7
3) 8
4) 9
9.
If the sum of two numbers is 240 and their ratio is 5 : 7, then the smallest number
is
1) 140
2) 100
3) 130
4) 110
10. One number is three times the other number. If 15 is added to both the numbers.
Then one of the new number becomes twice that of the other new number, then
greatest number is_________.
1) 15
2) 25
3) 35
4) 45
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MULTIPLE CORRECT CHOICE TYPE:
11. The length of a rectangular park exceeds its breadth by 17 meters. If the perimeter
of the park is 178 meters, then which of the following is correct?
1) l = 53 meters
2) b = 36 meters
3) Area = 1908 sq.m 4) l =56 meters
REASONING TYPE:
12. Statement I: If the sum of two adjacent sides of a parallelogram is 24 cm, then
the perimeter of the parallelogram is 48cm.
Statement II: In a parallelogram opposite sides are equal.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
In a triangle the angles are in the ratio 2 : 3 : 4.
13. The first angle is
1) 200
2) 300
3) 400
4) 500
14. Second angle is
1) 300
2) 600
3) 900
4) 1000
15. Third angle is
1) 200
2) 400
3) 600
4) 800
MATRIX MATCH TYPE:
16. Column - I
Column - II
a) Thrice a number is increased by 6 equals 39,
then the number is_______.
1) 2 : 7
b) The sum of three consecutive odd numbers is 42,
then the least number is______.
2) 6 km/h
c) The present ages of father and son are 30 and 5
the ratio of their ages after 5 years
3) 11
d) A man can row upstream at 7km/hr down stream
at 5km/hr, then the rate of man in still water is
4) 12
5) 7 : 2
INTEGER ANSWER TYPE:
17. If (x + 500) and (x + 700) are linear pair, then the value of x is ___________.
LEVEL-2
SINGLE CORRECT CHOICE TYPE:
18.
If 12 is added to 6 times of a number the resultant is equal to 24, then the number
is
1) 1
2) 2
3) 3
4) 4
19.
If the sum of two number is 24 and their product is 140, then the least number is
20.
1) 10
2) 11
3) 12
Find a number, twice of which decreased by 7 gives 65
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1) 35
2) 36
3) 72
4) 65
Sum of three consecutive odd natural nos is 249. Find the numbers.
1) 81,83,85
2) 83,85,87
3) 71,73,75
4) 75,77,79
MULTI CORRECT CHOICE TYPE:
22. Two number are in the ratio of 3:5. If 5 is subtracted from each of the number,
they become in the ratio of 1:2, the numbers are
1) 15
2) 25
3) 12
4) 20
REASONING TYPE:
23. Statement I: Two numbers are in the ratio 5:8, If the sum of the numbers is
182, then the numbers are 70 and 112 .
Statement II: If the ratio of the two numbers a:b and sum of two numbers m,
21.
a
×m.
a+b
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
MATRIX MATCH TYPE:
24. Column-I
a) A number is decreased by 8% of itself gives 115 is
b) A number is increased by 8% of it self gives 115 is
c) A number is increased by 100% of it self gives 50 is
d) A number is decreased by 20% it self gives 80 is
then first no is
Column-II
1) 106.48
2) 100
3) 125
4) 25
5) 106.58
Introduction: Problems based on ages are generally asked in most of the competitive
examinations. In such problems, there may be three situations:
(i) Age some years ago
(ii) Present age
(iii) Age some years hence
Two of these situations are given and it is required to find the third. The relation
between the age of two persons may also be given. Simple linear equations are
framed and their solutions are obtained.
Illustration: Five years ago the age of a father was twice the age of his son. The sum of
their present ages is 70 years. Find their present ages.
Solution: The age of the father 5 years age = x years
Age of the father 5 years ago = 2x years
∴
Present ages of son and father respectively are
(x + 5) years and (2x + 5) years.
According to problem, sum of their present ages = 70 years
∴ ( x + 5 ) + (2x + 5 ) = 70
∴x =
⇒ 3x + 10 = 70
(or)
3x = 70 − 10 = 60
60
= 20 years .
3
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DESCRIPTIVE TYPE QUESTIONS:
1.
2.
3.
4.
5.
The present age of A is twice that of B. 30 years from now, age of A will be 1
1
2
times that of B. Find the present ages of A and B.
Present ages of A and B are 20 years and 5 years respectively. After how
many years will be the age of A be twice that of B?
Mr. Rao, aged 50 years has a daughter Sowjanya of age 20 years. How
many years ago was the age of Mr. Rao three times that of his daughter?
The present age of a man is three times that of his son. Six years ago, the
age of the man was four times that of his son. Find the ratio of their ages
6 years later.
The present age of Sowmya is one fourth that of her mother Sobhadevi. 20
years later, the age of Sowmya will be half the age of her mother. Find their
present ages?
LEVEL-1
SINGLE CORRECT CHOICE TYPE:
6.
A man is five times as old as his son. after 2 years the man will be four times as
old as his son then present age of father______
1) 35years
2) 30years
3) 20years
4) 40 years
MULTI CORRECT CHOICE TYPE:
7.
The age of father is 4 times the age of his son. If 5 years ago father’s age was 7
times the age of his son at that time, then the present ages of son and father is
1) 40 years
2) 60 years
3) 10 years
4) 15 years
REASONING TYPE:
8.
Statement I: The ages of Ram and Shyam differ by 16 years. Six years ago,
Shyam’s age was thrice as that of Ram’s, then the present ages of
Ram and Shyam is 14 years, 30 years.
Statement II: If 15 years hence, Rohit will be just four times as old as he was 15
years ago, then Rohit present age is 25.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
The age of Mr. Gupta is four times the age of his son. After ten years, the age of
Mr. Gupta will be only twice the age of his son, then
9.
The present age of Gupta is
1) 10 years
2) 20 years
3) 30 years
4) 40 years
10. The present age of Gupta’s Son is
1) 5 years
2) 10 years
3) 15 years
4) 20 years
11. The ratio of the ages of Gupta and his Son is
1) 1 : 4
2) 4 : 1
3) 1 : 2
4) 2 : 1
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MATRIX MATCH TYPE:
12.
The ratio of the age of Father and Son at present is 6 : 1. After 5 years the ratio
will become 7 : 2, then
Column-I
Column-II
a) The present age of Son is
1) 30 years
b) The present age of Father is
2) 5 years
c) 10 years latter the age of Son is
3) 40 years
d) After 10 years the age of Father is
4) 15 years
LEVEL-2
SINGLE CORRECT CHOICE TYPE:
13.
If a man is 35 years old and his son 7 years now, then how many years will the
son be half as old as his father
1) 21 years
14.
15.
2) 22 years
3) 23 years
4) 24 years
If the present age of a boy is one sixth the age of his father. 5 years later the
sum of their ages will be 45 years, then their present ages is
1) 30 years, 5 years
2) 6 years, 36 years
3) 7 years, 42 years
1) 8 years, 48 years
If Shoba’s age 13 years ago is half her age 13 years later, then her present age
is
1) 29 years
2) 39 years
3) 49 years
4) 59 years
COMPREHENSION TYPE:
The sum of the ages of A and B is 42 years. 3 years back, the age of A was 5 times
the age of B, then
16.
The present age of A is
1) 33 years
17.
3) 24 years
4) 23 years
3) 24 years
4) 23 years
The present age of B is
1) 33 years
18.
2) 9 years
2) 9 years
The difference between the present ages of A and B is
1) 33 years
2) 9 years
3) 24 years
4) 23 years
MULTI CORRECT CHOICE TYPE:
19.
The sum of the ages of a Son and Father is 56 years. After 4 years, the age of
the Father will be 3 times that of the Son then their present ages is
1) 12 years
2) 44 years
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4) 52 years
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5
1
of the distance by train,
by bus and the
8
4
remaining 15 km, by boat. Find the total distance travelled by him.
Solutions: Let the total distance travelled by him = x km.
Illustration: A person travelled
Distance travelled by train =
Distance travelled by bus =
5
x km.
8
1
x km.
4
∴ Total distance travelled by train and bus in km. =
5
1
5x + 2x 7x
x+ x =
=
8
4
8
8
7x x
=
8
8
He travelled this distance by boat.
We are given that, this distance = 15 km.
∴
Remaining distance in km. = x −
∴
∴
x
= 15
(or) x = 15 × 8 = 120
8
Total distance travelled = 120 km.
DESCRIPTIVE TYPE QUESTIONS:
1.
1
1
of her property to each of 5 social organizations,
to each of 3
8
15
1
to each of her two sons. The remaining property valued
religious trusts,
24
Sundari gave
Rs. 1,15,000/- was donated for construction of a school building in the name of
her husband. Find the total value of her property.
2.
3.
4.
1
1
of a number of butterflies in a garden are on jasmines and
of them are
5
3
on roses. Three times the difference of the butterflies on jasmines and
roses are on lillys. If the remaining one is flying freely, find the total
number of butterflies in the garden. (modified from Lilavathi Ganitham).
4
of one of its equal sides. If the perimeter of
3
the triangle is 400cm, find the length of its sides.
There are some lotus flowers in a pond and some bees are hovering around.
If one bee lands on each flower, one bee will be left. If two bees land on
each flower, one flower will be left. Find the number of bees and
the number of flowers in the pond.
The base of an Isosceles triangle is
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LEVEL-1
SINGLE CORRECT CHOICE TYPE:
5.
If
1
1
of a flag-pole is black,
th is white and the remaining three metres is
5
4
painted yellow, then the length of the flag-pole is
1) 5
6.
5
m
11
2) 5
6
m
11
3) 5
7
m
11
4) 5
8
m
11
1
1
of his property to his son,
to his daughter and the remaining
3
4
to his wife. If the wife’s share was Rs. 40,000, then the total value of his property is
Murthy left
1) Rs. 76,000
2) Rs. 86,000
3) Rs. 96,000
4) Rs. 32,000
MULTI CORRECT CHOICE TYPE:
7.
A certain amount is distributed among A, B and C. A gets
3
1
and B gets
of the
16
4
whole amount. If C gets Rs. 81, then A, B gets
1) Rs. 30
2) Rs. 27
3) Rs. 33
4) Rs. 36
REASONING TYPE:
8.
Statement I: A tin of oil was
4
full. When 6 bottles of oil were taken out and
5
four bottles of oil were poured into it, it was
3
full, then the tin
4
contain 40 oil bottles.
Statement II: If
1
1
of a pencil is black,
of the remaining is white and the
8
2
remaining 3
1
cm is blue, then the total length of the pencil is 8
2
cm.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
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COMPREHENSION TYPE:
Three prizes are to be distributed in a sportsmen, so that their total cost is
Rs.2,550. The value of the second prize is
third prize is
3
of the first prize and the value of the
4
1
the value of the second prize, then
2
9.
The value of first prizes is
1) Rs. 1200
2) Rs. 900
3) Rs. 450
4) Rs. 600
10. The value of second prizes is
1) Rs. 1200
2) Rs. 900
3) Rs. 450
4) Rs. 600
11. The value of third prizes is
1) Rs. 1200
2) Rs. 900
3) Rs. 450
4) Rs. 600
MATRIX MATCH TYPE:
12. A train starts full of passengers. At the first station, it drops one-third of the
passengers and takes 280 more. At the second station, it drops one-half of the
new total and takes 12 more. One arriving at the third station, it is found to have
248 passengers, then
Column-I
Column-II
a) The number of passengers in the beginning is
1) 184
b) After the first station the number of passengers is
2) 288
c) After the second station the number of passengers is
3) 248
d) How many passengers boarded at first station.
4) 472
LEVEL- 2
SINGLE CORRECT CHOICE TYPE:
13.
A sum of Rs.1360 has been divided among A, B and C such that A gets
B gets and B gets
14.
2
of what
3
1
of what C gets. B’s share is:
4
1) Rs. 120
2) Rs. 160
3) Rs. 240
4) Rs. 300
Three friends had dinner at a restaurant. When the bill was received, Amita
2
1
as much as Veena paid and Veena paid
as much as Tanya paid. What
3
2
fraction of the bill did Veena pay is
paid
1)
1
3
2)
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11
3)
12
31
4)
5
8
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MULTI CORRECT CHOICE TYPE:
15.
1
1
1
1
1
1
of a pole is colored red,
white,
blue,
black,
violet,
yellow
10
20
30
40
50
60
and the rest is green. If the length of the green portion of the pole is 12.08
metres, then the length of the pole is
1) 16 m
2) 18 m
3) 1600 cm
4) 1800 cm
Illustration: In a two digit number, the unit’s digit is two more than tens digit. Sum
of the digits is equal to
1
of the whole number, then the number.
4
Solutions: Let x = ten’s digit
∴ Unit’s digit = x + 2
∴ Number = 10x + (x + 2) = 11x + 2
Sum of the digits = x + x + 2 = 2x + 2
1
(11x + 2)
4
According to problem
= 2x + 2 =
Multiplying both sides by 4
= 8x + 8 = 11x + 2
Transposing 11x and 8
= 8x – 11x = 2 – 8
(or) –3x = –6
−6
= 2 . This is ten’s digit.
−3
∴
x=
∴
Unit’s digit = x + 2 = 4
∴
Number = 24.
DESCRIPTIVE TYPE QUESTIONS:
1.
In a two- digit number, the ten’s digit is 2 more than the unit’s digit. Sum of the
1
of the whole number. Find the digits and number.
7
In a two-digit number, ten’s digit is twice the unit’s digit. The number formed by
interchanging the digits is 36 less than the original number. Find the number.
In a two-digit number, unit’s digit is 3 more than the ten’s digit. The number
formed by interchanging the digits and the original number are in the ratio
7 : 4, then the the number.
In a two - digit number, unit’s digit is twice the ten’s digit. If sum of the digits is
added to the whole number, the result is equal to 30. Find the number.
digits is
2.
3.
4.
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LEVEL -1
SINGLE CORRECT CHOICE TYPE:
5.
In a two-digit number, the unit’s digit is 7 more than the ten’s digit. Sum
of the digits is
1
of the whole number, then the number is
2
1) 16
2) 17
3) 18
4) 19
6.
In a two-digit number, the unit’s digit is 5 less than the ten’s digit. The whole
number is 8 times the sum of the digits, then the number is
1) 69
2) 70
3) 71
4) 72
7.
In a two-digit number, the unit’s digit is 5 less than the ten’s digit. The number
formed by interchanging the digits and the original number are in the ratio 3 : 8,
then the number is
1) 69
2) 70
3) 71
4) 72
MULTI CORRECT CHOICE TYPE:
8.
If the number obtained on interchanging the digits of a two-digit number is 18
more than the original number and the sum of the digits is 8, then the original
number is
1) The units digits is 5
2) The ten’s digits is 5
3) The original number is 35
4) The original number is 53
REASONING TYPE:
9.
Statement I: A two-digit number becomes five-sixth of itself when its digits are
reversed. The two digits differ by one, then the number is 54.
Statement II: A two-digit number is such that the product of the digits is 8. When
18 is added the number, then the digits are reversed, then the
number is 24.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
In a two-digit number, if it is known that its unit’s digit exceeds its ten’s digit by 2
and that the product of the given number and the sum of its digits is equal to 144,
then
10.
The unit’s digit is
1) 2
11.
3) 6
4) 8
2) 4
3) 6
4) 8
2) 24
3) 26
4) 82
The ten’s digit is
1) 2
12.
2) 4
The number is
1) 42
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LEVEL -2
SINGLE CORRECT CHOICE TYPE:
13.
The ratio between a two-digit number and the sum of the digits of that number
is 4 : 1. If the digit in the unit’s place is 3 more than the digit in the ten’s place,
what is the number is
1) 35
14.
2) 36
4) 25
A number consists of two digits. The sum of the digits is 9. If 63 is subtracted
from the number, its digits are interchanged, then the number is
1) 83
15.
3) 26
2) 80
3) 81
4) 82
In a two-digit number, the unit’s digit is 2. If the digits are interchanged, the new
number formed is
3
times the old number, then the number.
8
1) 69
2) 70
3) 71
4) 72
REASONING TYPE:
16.
Statement I: In a two-digit number, the digit in the unit’s place is four times the
digit in ten’s place and sum of the digits is equal to 10, then the
number is 28.
Statement II: A number of two digits has 3 for its unit’s digit, and the sum of
digits is
1
of the number itself. The number is 63.
7
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
MATRIX MATCH TYPE:
17.
If yx be the two-digits number, then
Column-I
Column-II
a) General form two digit number
1) 10y + x=4(x+y)+3
b) 7 times a two digit number is equal to
2) 7(10 y + x ) = 10 x + y
reverse in the order of its digits
c) Difference between two digits is 3
3) y − x = 3
d) Two digit number is 3 more than
4) 10 y + x
4 times the sum of its digits
5) 10 y + x = 3( x + y ) + 4
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Some Important Terms:
Stream: If the water of the river is moving, then it is called stream.
Up Stream: If a boat moves against a stream i.e. in the direction opposite to that of the
stream is called upstream.
Down Stream: If a boat moves with a stream i.e. along the direction of the stream is
called down stream.
Formula:
1.
If the speed of a boat be x km/hr in still water and the speed of the stream or
the current be y km/hr. then
*
*
Speed of boat up stream = (x – y) km/hr
Speed of boat down stream = (x + y) km/hr
2.
Speed of the boat in still water =
3.
Speed of the stream =
1
(Downstream speed + Upstream speed)
2
1
(Down stream speed – up stream speed)
2
Illustration: In a stream the current flows at the rate of 4 km/hr. For a boat the
time taken to cover certain distance upstream is 5 times the time it takes to
cover the same distance downstream. Find the speed of the boat in still water.
Solution: Let the speed of the boat in still water = x km/hr.
Speed of the boat down stream will be greater than its speed in still water, due
to favorable current of water.
∴ Speed of the boat downstream = Speed of the boat + Speed of water current
= (4 + x) km/hr.
Speed of the boat upstream will be less than its speed in still water, due to
opposing current of water.
∴ Speed of the boat upstream
= Speed of the boat – Speed of the current = (4 – x)km/hr.
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Let the time taken to cover the distance down stream = t hrs.
∴ Time taken to cover the distance upstream = 5t hrs.
Now, distance travelled down stream
= Speed × Time
Distance travelled upstream
= (x – 4) 5t km
= (4 + x)t km
Since distances travelled both ways are same, (4 + x)t = (x – 4)5t
Dividing both sides by ‘t’ ( ≠ 0 )
(4 + x ) = 5 ( x − 4 )
Transposing –5x and 4; – x + 5x = 20 + 4(or) 4x = 24
(or) 4 + x = 5x − 20
∴ x = 6 km / hr .
DESCRIPTIVE TYPE QUESTIONS:
1.
Sobha drove her car at a certain speed for the first 4 hours and increased its
speed by 10km./hr. for the next two hours. If the total distance travelled by her
was 500km. find the speeds at which Sobha drove her car at different times.
2.
A postman takes tappals on foot from village P to village Q at a speed of 4km/hr.
If he walks at a speed of 5 km./ hr. then he reaches the village 7 minutes
earlier. Find the distance between the two village.
3.
Raju walked from Tanuku to Undrajavaram at a speed of 3km./hr. He returned by another route which is 5km. longer at a speed of 4 km./hr. If the
total time taken by him for the journey was 4
1
hours. Find the total distance
6
he travelled.
4.
The speed of the motor-boat going down a stream is 50.4km./hr. and going upstream is 23.2km./hr. Determine the speed of the boat in still water and speed
of the current?
LEVEL-1
SINGLE CORRECT CHOICE TYPE:
5.
Rama swamy walked to the town at a speed of 4km./hr. and came back by bicycle
at a speed of 12 km./hr. The total time for to and fro journey was 2
1
hrs, then
2
the times taken by him to travel by foot and bicycle is
6.
1
1
1
1
1
1
1
1
2) 111 m; 37 m
3) 121 m; 37 m
4) 122 m; 36 m
1) 112 m; 37 m
2
2
2
2
2
2
2
2
In a stream whose current flows at the speed of 3 km./hr. a man rowed up
stream and returned to his starting point in 6.25 hours. If the man rows at a
speed of 5km./hr. in still water, how long did he row upstream?
1) 4 hrs
2) 5 hrs
3) 6 hrs
4) 7 hrs
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A man rowed upstream for 4 hours and returned to his starting point in 2
hours. If he rowed at the rate of 5 km./ hr. in still water, what was the rate of
flow of the stream?
4
5
2
7
km/hr
2)
km/hr
3)
km/hr
4)
km/hr
3
3
3
3
MULTI CORRECT CHOICE TYPE:
8.
In a stream running at 2 kmph, a motorboat goes 6 km upstream and back again
to the starting point in 33 minutes, then
1) Speed downstream is 24 kmph.
2) Speed upstream is 20 kmph.
3) Speed of motorboat in still water is 22 kmph.
4) Speed of motorboat in still water is 24 kmph.
REASONING TYPE:
9.
Statement I: In one hour, a boat goes 11 km along the stream and 5 km against
the stream. The speed of the boat in still water (in km/hr) is 8
kmph.
Statement II: A man can row upstream at 8 kmph and downstream at 13 kmph,
then the speed stream is 2.5 kmph.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
A man can row 18 kmph in still water. It takes him thrice as long to row up as
to row down the river, then
10. The rate of upstream is
1) 9 km/hr
2) 18 km/hr
3) 27 km/hr
4) 23 km/hr
11. The rate of downstream is
1) 9 km/hr
2) 18 km/hr
3) 27 km/hr
4) 23 km/hr
12. The rate of stream is
1) 9 km/hr
2) 18 km/hr
3) 27 km/hr
4) 23 km/hr
1)
LEVEL - 2
SINGLE CORRECT CHOICE TYPE:
13. A man takes twice as long to row a distance against the stream as to row the
same distance in favor of the stream. The ratio of the speed of the boat (in still
water) and the stream is
1) 2 : 1
2) 3 : 1
3) 3 : 2
4) 4 : 3
14. A boat running upstream takes 8 hours 48 minutes to cover a certain distance,
while it takes 4 hours to cover the same distance running downstream. What is the
ratio between the speed of the boat and speed of the water current respectively ?
1) 2 : 1
2) 3 : 2
3) 8 : 3
4) Cannot be determined
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15.
A motorboat, whose speed is 15 km/hr in still water goes 30 km downstream and
comes back in a total of 4 hours 30 minutes, then the speed of stream is
1) 4 kmph
2) 5 kmph
3) 6 kmph
4) 10 kmph
MATRIX MATCH TYPE:
16. A sailor goes 8km down stream in 40 minutes and returns in 1 hour.
Column-I
Column-II
a) Speed of sailor in still water
1) 12km\h
b) Speed of current
2) 8 km\h
c) Speed in down stream
3) 2km\hr
d) Speed in up stream
4) 10km\hr
5) 20km\hr
Expressing Decimals as Rational Numbers:
Every Rational Number can be expressed as a terminating or non-terminating
repeating decimal by performing the actual division. How about the reverse
process?
Given a decimal, can we express it in the form p (q ≠ 0 ) . Let us examine case by case.
q
Terminating decimals: In the process of converting a fraction into a decimal by the
division method, if we obtain a zero remainder after a certain number of steps,
then the decimal obtained is a terminating decimal.
Write the number given without decimal in the numerator. Write one
in the denominator and write as many zeros as the number of digits after the
decimal point to the right side of one.
If possible, write the fraction in its lowest terms.
2537
[There are three digits after the decimal point]
1000
Non-terminating recurring decimals: These are of two kinds:
(i) pure recurring decimals
(ii) mixed recurring decimals
Pure Recurring Decimals:
If all the digits after the decimal point recur in a decimal then it is called a pure
recurring decimal.
Example: 2.537 =
(i) 0.333 .............................. = 0.3 .
Mixed Recurring Decimal: A decimal in which some of the digits in the decimal part
are repeated, is called a pure recurring decimal.
Example: 0.447 .
*
Write Pure recurring decimal without integral part in rational number form,
using the formula to remove the decimal point and write the recurring part as
the numerator. Write as many nines in the denominator as the number of
digits in the recurring part. If possible, write the fraction in its lowest terms.
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Example: 0.6 =
*
6 2
= .
9 3
Write Pure recurring decimal with integral part in rational number form,
using the method the given decimal be equal to ‘x’. If there are ‘m’ digits in
the recurring part of the decimal, multiply both sides of the equation by 10m.
From the resulting equation subtract the original equation.
Example: Write 2.34 in the form
Solution: Suppose
p
,q ≠ 0 .
q
x = 2.34 = 2.343434 ...........
100x = 234.34 34 34 .....
x = 2.34 34 34 .....
Subtracting (i) from (ii)
∴
We get x =
.....(i)
.....(ii)
232
.
99
*
Writing Mixed Recurring decimal without integral part in rational number
form, using the method the given decimal be equal to ‘x’. Suppose there are
‘m’ digits in the non-recurring part and ‘n’ digits in the recurring part of the
decimal. Multiply the ‘x’ equation with 10 m+n and write the result. Then
multiply the original ‘x’ equation with 10m and write the result. Subtract the
resulting equations one from the other.
*
Writing Mixed Recurring decimal with integral part in rational number form,
using method the given decimal be equal to ‘x’. Suppose there are m digits in
the non-recurring part and n digits in the recurring part. Multiplying the ‘x’
equation with 10m+n and write the result. Then multiply the original ‘x’ equation with 10 m and write the result. Subtract the resulting equations one
from the other.
*
From some stage we get a digit or more than one digit being repeated in the
same order. The recurring part is called the period and the number of digits
in the recurring part is called the periodicity of the decimal.
DESCRIPTIVE TYPE QUESTIONS:
1.
Write the following decimals as rational number in lowest terms:
(i) 0.52
(ii) 29.2705
(iii) 70.7070
2.
Find the rational numbers in lowest terms represented by the recurring decimals.
3.
4.
(ii) 25.523
(i) 0.4
Finding rational numbers in lowest terms.
(iii) 9.082
(i) 0.0279
(ii) 7.358
(iii) 0.00379
(iv) 29.37486
Write the following terminating decimals in non-terminating recurring form and
prove your results.
(i) 2.3
(ii) 5.2468
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LEVEL-1
SINGLE CORRECT CHOICE TYPE:
1.
0.36 expressed in the form of p/q equals to
1)
2.
2)
4
13
35
90
4)
35
99
3) 100
4)
1
1000
3)
0.213 ÷ 0.00213 =
1)
3.
4
11
1
100
2) 1000
The value of 4.12 is
1) 4
11
90
2) 4
11
99
3)
371
900
4)
407
99
3)
2 .3
40
4)
125
23
MULTI CORRECT CHOICE TYPE:
4.
0.0625 of
1)
23
is equals to
25
23
80
2)
23
400
REASONING TYPE:
5.
Statement I: Fractions in which denominators are powers of 10 are known as
decimal fractions.
Statement II: Fractions in which denominators are except powers of 10 are
known as vulgar fraction.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
If A =
6.
22
47 2
, B=
+
7
99 9
Periodicity of A is
1) 3
7.
3) 5
4) 6
2) 3
3) 1
4) 4
3) 2
4) 3
Periodicity of B is
1) 2
8.
2) 4
Sum of the integer part of A and B is
1) 0
2) 1
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MATRIX MATCH TYPE:
9.
Column - I
Column - II
71
495
a) 5.55
1)
b) 54.56
2) 5
c) 0.143
3)
d) 0.1234
4) 54
11
20
611
4950
14
25
INTEGER ANSWER TYPE:
10.
When 0.232323.......... is converted into a fraction then the result is _________
LEVEL-2
SINGLE CORRECT CHOICE TYPE:
11.
Value of 3.87 − 2.59 =
1)
12.
127
90
Value of
1)
2)
129
90
3)
128
99
4)
127
99
3)
19
10
4) 1
3)
55524
10000
4)
4.2 × 4.2 − 1.9 × 1.9
is
2.3 × 6.1
1
2
2)
21
5
COMPREHENSION TYPE:
If X = 5.55, Y = 0.0024 and Z = 6.005
13.
Fractional value of X + Y is
1)
14.
55024
10000
50524
10000
50024
10000
Sum of integer part of x, y and z is
1) 0.11 x 10-2
15.
2)
The value of
1) 6584
2) 0.11 x 102
3) 0.11 x 10
4) 1.1 x 10-2
3) 6548
4) 6458
x
y
z
+ −4 + −3 is
−2
10
10
10
2) 6854
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Linear Inequations: A statement of inequality between two expression involving
a single variable x with highest power 1, is called a linear inequation.
Example: x < 2, 2x – 3 > 7.
Domain of the Variable or Replacement Set: The set from which the values of
the variable x are replaced in an inequation, is called the domain of the variable
or the replacement set.
Solution Set: It is the subset of the replacement set, consisting of those values of
the variable which satisfy the given inequation.
Example: Find the solution set of x < –3, where replacement set is
B = {–7, –6, –5, –4, –3, –2, –1}
Solution: Solution set =
{x ∈ B : x < −3} = {−7, − 6, − 5, − 4} .
Properties of Inequations:
*
Adding the same number to each side of an inequation does not change the
inequality.
Example: 2x + 3 > 8 ⇒ 2x + 3 + 2 > 8 + 2.
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*
Subtracting the same number from each side of an inequation does not
change the inequality.
Example: 2x + 5 > 12 ⇒ 2x + 5 – 5 > 12 – 5.
*
Multiplying each side of an inequation by the same positive number does not
change the inequality.
Example: x < 5 ⇒ x × 2 < 5 × 2 .
2
2
*
Multiplying each side of an inequation by the same negative number
reverses the inequality.
Example: −
x
< 3 ⇒ x > 3 × ( −2)
2
[Multiplying each side by –2]
*
Dividing each side of an inequation by the same positive number does not
change the inequality.
Example: 3x > 9 ⇒ x > 3
[Dividing each side by 3]
*
Dividing each side of an inequation by the same negative number reverses
the inequality.
Example: –3x > –9 ⇒ x < 3
[Dividing each side by –3]
*
If three numbers are related in such a way that the first is less (greater)
than the second and the second is less (greater) than the third, then the
first is less (greater) than the third.
This is called the transitive property.
*
If x and y are of the same sign and x < y(x > y), then
*
To solve simple inequation, terms containing unknown quantity should
collected on left side and constants on the right side. Then the coefficient of
the unknown quantity should be reduced to unity.
1 11 1
>  <  . If
x yx y
reciprocals are taken to quantities of the same sign on both sides of an
inequality, then the order of the inequality is changed.
DESCRIPTIVE TYPE QUESTIONS:
1.
Solve the following inequations. The values taken by the variable are given
against each:
(i) y + 3 < 9,
y = {0, 1, 2, 3, 4, 5}
(ii) z > 0 and z < 4, z = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
2.
Check whether the value of the variable given against each inequation is solution to it or not.
(i)
2y + 3
3
< 3y - 4 ;
y=2
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(ii)
7t + 5
6
≥
t -1
2
; t = –3
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LEVEL-1
SINGLE CORRECT CHOICE TYPE:
3.
The solution set of x < 4 where replacement set is N
1) {0, 1, 2}
2) {1, 2, 3}
3) {1, 2, 3, 4}
4.
The value of x from the in equation 3x + 4 < 9 is
1) x <
5.
5
3
2) x >
5
3
The value of y from the inequation 7 −
1) 5 < y
2) 5 > y
MULTI CORRECT CHOICE TYPE:
6.
a ≠ b means
1) a > b
2) a = b
REASONING TYPE:
3) x <
13
3
y 5y
>
− 6 is
2
3
3) y < 6
3) a < b
4) {0, 1, 2, 3}
4) x >
13
3
4) y > 6
4) a ≥ b
−2x + 3
> 7 the solution for x is x < -16
5
Statement II: If a > b and c < 0 then a.c < b.c.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
Given x > y, y > z
8.
The relation between x and z is
7.
Statement I: From the in equation
9.
2) x </ z
3) x > z
Given z < o; then the relation between xz and yz
1) x < z
1) xz < yz
10.
2) xz > yz
3) xz = yz
4) x >/ z
xz
4) yz > 1
Given z > 0 then the relation between x + z and y + z
1) x+z < y+z
2) x+z = y+z
MATRIX MATCH TYPE:
11. Column-I
3) x+z > y+z
4) x + z ≤ y + z
Column-II
a b
<
c c
a) y + 3 < 9; y ∈ N
1)
b) a > b and c < 0
c) a < b and c > 0
d) Solution set of z if
2) {1, 2, 3, 4}
3) {1, 2, 3, 4, 5}
4) a.c < bc
0 < z < 5; z ∈ w
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INTEGER ANSWER TYPE:
12.
The greatest whole number that satisfies x − 7 ≤ 11 is ___________
LEVEL - 2
SINGLE CORRECT CHOICE TYPE:
13.
The value of x from the in equation
1) x ≤ 1 20
2) x ≥ 120
13
 1
 x + 4  ≥ ( x − 6 ) is
25
 3
3) x = 120
4) x < 120)
MULTI CORRECT CHOICE TYPE:
14.
Integral value of x satisfying both 2x + 3 > 7 and x + 4 < 10 is
1) 4
2) 6
3) 5
4) 10
REASONING TYPE:
15.
Statement-I: The pairs of consecutive even positive integers both of which are
larger than 5, such that their sum is less than 23 are (6, 8), (8, 10)
and (10, 12).
Statement-II: The sum of two numbers x and y is not more than K is given by
x+y≥k.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
Illustrations:
Solve the inequation 2x – 1 < 9, x ∈ W . Represent the solution set graphically.
Sol: We have
1.
⇒ 2x < 10
2x – 1 < 9
⇒
∴
x<5
Solution Set =
[Adding 1 on each side]
[Dividing each side by 2]
{x ∈ W : x < 5} = {0, 1, 2, 3, 4}
On the number line we mark these points point by dark dots, as shown below.
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Solve the equation
x
+ 3 > 5, x ∈ N . Represent the solution set graphically.
2
Sol: We have
x
+3 >5
2
⇒
x
>2
2
[Subtracting 3 from each side]
⇒x>4
∴ Solution Set =
[Multiplying each side by 2]
{x ∈ N : x > 4} = {5, 6, 7, 8, 9,...} .
On the number line we mark some of these numbers by dark dots and then put
three more dots, as shown below. This indicates the infiniteness of the solutions
set.
DESCRIPTIVE TYPE QUESTIONS:
1.
If the domain of the variable is N, solve the following inequations and check
your result.
(i) 7x + 3 ≤ 17
(iv)
2.
y
3
(ii) 3 + x > 3
(iv) x + 2 ≤ 3x − 4
+1 < 3
Domain of the variable is Q. Solve the following inequations and check your
result.
(i) 2 (x + 3) < 5 (x – 3)
(ii)
2p + 1 3p + 1
≤
5
2
(iii) 7 −
y
2
>
5y
3
−6
LEVEL - 1
SINGLE CORRECT CHOICE TYPE:
3.
3x < 15, x ∈ N the solution set is
1) {1,2,3,4,5}
4.
2) {2,3,4,5}
Solution set of 5x – 9 < 15 where x ∈ w
1) {0,1,2,3,4}
2) {1,2,3,4,5}
5.
3) {1,2,3,4}
4) {0,1,2,3,4}
3) {0,1,2,3,4,5}
4) {2,3,4,5}
represented by the inequation
1) 2x − 1 < 5, x ∈ w
2) 2x − 1 < 3, x ∈ z
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3) 2x − 1 < 5, x ∈ z
4) 2x + 1 < 7, x ∈ N
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MULTI CORRECT CHOICE TYPE:
6.
Value of x from the equation
x
+ 3 < 5, x ∈ N is
2
1) 2
2) 3
3) 4
REASONING TYPE:
7.
Statement I: If x + 2 < 8 then x > 6.
Statement II: a < b and if c is real then a + c < b + c.
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
COMPREHENSION TYPE:
Given −2 ≤ x ≤ 2
8.
First inequation is
9.
10.
4) 1
2) x ≥ 2
1) x > 2
Second inequation is
3) x ≤ −2
4) x ≥ −2
1) x ≤ −2
3) x > −2
4) x < −2
2) x ≤ 2
Solution set of the given inequation if x ∈ z
1) {–2, –1,0,1,2}
2) {–1,0,1}
3) {–2,–1,1,2}
4) {–2, 2}
MATRIX MATCH TYPE:
11.
Column-I
Column-II
a) Inequality
1) 3x – 2 < 4
b) Inequation
2) 3z – 2 ≠ 4
c) Symbolic form of positive Number
3) 8 + 12 > 12
d) If 2 is subtracted from three times
4) x > 0
a number the result is less than 4 is
INTEGER ANSWER TYPE:
12.
Greatest value of x in 0 < x < 7 and x ∈ N is ____________
LEVEL-2
SINGLE CORRECT CHOICE TYPE:
13.
Solution set of 5 - 4x < x - 10, x ∈ {0,1,2,3,4,5,6,7,8} is
1) {3,4,5,6}
14.
2) {4,5,6,7}
3) {4,5,6,7,8}
4) {3,4,5,6,7}
If x > y and y > z then x > z this is called __________ property
1) Symmetric
2) Transitive
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4) Equivalence
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MULTI CORRECT CHOICE TYPE:
15.
If a>0, b>0, x>0 and a>b then which of the following are true
a+x a
<
2)
b+x b
1 1
<
1)
a b
b2 b
3) 2 <
a
a
4) a.x < b.x
REASONING TYPE:
16.
Statement-I: If 30x < 200 then no natural number exists that satisfies the
inequation.
Statement-II: If x > 0 and a > b, then
a b
> .
x x
1)
Both Statement-I and Statement-II are true.
2)
Both Statement-I and Statement-II are false.
3)
Statement I is true, Statement II is false.
4)
Statement I is false, Statement II is true.
LEVEL-3 (TWISTERS)
1.
Find the factors of a4(b2 - c2) + b4(c2 - a2) + c4(a2 - b2).
2.
Find the factors of a(b4 - c4) + b(c4 - a4) + c(a4 - b4).
3.
Factorize: a (b - c)3 + b (c - a)3 + c (a - b)3.
a (b − c )
2
b (c − a )
2
+
c (a − b )
2
+
=a+b+c
4.
Prove that
5.
x
y
z
If (b − c )(b + c − 2a ) = (c − a )(c + a − 2b ) = (a − b )(a + b − 2c ) then find x + y + z.
6.
Find the value of
7.
Find the value of x, if
8.
If x, y, z are positive real numbers , prove that
(c − a )(a − b ) (a − b )(b − c ) (b − c )(c − a )
1
1
1
1 1 1
+ 2
+ 2
+ + = 0.
, if
a − bc b − ca c − ab
a b c
2
x −a x −b x −c
+
+
= 3.
b+c c+a a +b
( x + y + z )2 ( yz + zx + xy )2 ≤ 3( y 2 + yz + z 2 )( z 2 + zx + x 2 )( x 2 + xy + y 2 )
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9.
If (a+b):(b+c):(c+a)=6:7:8, and a+b+c=14, compute the value of c.
(INMO)
10.
Factorize a 3 + b 3 + c 3 − 3abc
(INMO)
11.
Find the value of
12.
Find that x19 in the simplified form of (x + 1)(x + 4) (x + 9) ........... (x + 400).
13.
Prove that x5 - 5x3 + 4x is divisible by 120, ∀ x ∉ N .
14.
Which of the two numbers (1000 )
15.
Compute 12 − 2 2 + 32 − 4 2 + ..... − 19982 + 1999 2
16.
The denominator of a fraction is 9 more than its numerator. If the numerator and
1
1
1
1
+
+
+ ............ +
3.4 4.5 5.6
2007.2008
1000
and 1001999 is greater?
(INMO)
the denominator both are increased by 7, then the new fraction becomes
7
.
10
Find the original fraction.
17.
The sum of the digits of a two-digit number is 8. If the number formed by reversing
the digits is less than the original number by 18, find the original number.
18.
What same number should be added to each of the numbers 2,7, 10, 25 so that
they may be in proportion?
19.
A body covers a distance of 25km in 4 hours partly on foot at the rate of 3.5 kmph
and partly on cycle at 9 kmph. Find the distance covered on foot.
20.
Puneet starts from his home to school on cycle at 10 kmph and reaches 4 minutes
earlier than the scheduled time. However, if he cycles at 8 kmph, he is late by 5
minutes. What is the distance between his residence and the school?
LEVEL-4 (TEASERS)
1.
Factorize 8(a + b + c)3 - (b + c)3 - (c + a)3 - (a + b)3
2.
If
3.
If a, b, c, x, y and z are real and a 2 + b 2 + c 2 = 25, x 2 + y 2 + z 2 = 36 , and
2x + 3
A
B
=
+
then find A and B.
x − 5x + 6 x − 2 x − 3
2
ax + by + cz = 30 , compute
a +b+c
.
x+ y+z
4.
2 3 4
x
If 2 3. . . ...... = 1 then find the value of x + y.
4 5 6
y
5.
−1
−1
−1
−1
−1
−1
Simplify: (a + b ) (a + b − c ) + (b + c ) (b + c − a ) + (c + a ) (c + a − b )
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6.
Simplify:
x + x 2 + x 3 + ...... + x100
x −1 + x −2 + x −3 + ...... + x −100
7.
Solve:
x3 + y3 = 8
xy + x + y = 2
8.
If a = x5 - x3 + x then find the minimum value of x6.
9.
Show that (x + y + z)3 = (y + z - x)3 + (z + x - y)3 + (x + y - z)3 + 24xyz.
10.
If x = 2006, y = 2007 and z =
1
, find the value of
2007
(x + y + z)3 - (x + y -z)3 - (y + z - x)3 - (z + x - y)3 - 23 xyz.
11.
1 
1  2 1
1

 2 1

6
Show that  x +   x −   x + 2 − 1  x + 2 + 1 = x − 6 .
x 
x 
x
x
x



12.
If a + b + c = s, show that
s(s - 2b)(s - 2c) +s(s - 2c)(s- 2a) + s(s - 2a)(s - 2b) = (s - 2a) (s - 2b) (s - 2c) + 8abc.
1
1
1
+ 2
+ 2
= 0.
2
2
2
2
a +b −c
b +c −a
c + a 2 − b2
13.
If a + b + c = 0, show that
14.
If a, b, c > 0 and a + b + c = 1 prove that ab + bc + ca ≤
15.
If (a 2 + b 2 )3 = (a 3 + b 3 ) 2 , and ab ≠ 0 , find the numerical value of
16.
The sum of a certain number of consecutive positive integers is 1000.
2
1
.
3
a b
+
b a
(INMO)
Find these integers.
17.
If the speed of a man with the current is 12 km/hr and the rate of the current is
1
1 km/hr, then his rate against the current is
2
1) 13 km/hr
18.
3) 9 km/hr
4) None of these
A boatman can row 48km downstream in 4 hr. If the speed of the current is 5 km/
hr, then find in what time will he be able to cover 8 km upstream?
1) 6 hr
19.
2) 7 km/hr
2) 4 hr
3) 8 hr
4) None of these
The sum of ages of a father and son is 45 years. Five years ago, the product of
their ages was four times the father’s age at that time. The present age of the
father is
1) 39 years
2) 36 years
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4) None of these
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20.
The ages of A and B are in the ratio of 6 : 5 and sum of their ages is 44 years. The
ratio of their ages after 8 years will be
1) 4 : 5
21.
2) 3 : 4
3) 3 : 7
4) 8 : 7
1
A man’s age is 125% of what it was 10 years ago, but 83 % of what it will be after
3
ten 10 years. What is his present age?
1) 45 years
22.
Techno Text book
2) 50 years
3) 55 years
4) 60 years
A father is twice as old as his son. 20 years back, he was twelve times as old as
the son. What are their present ages?
1) 24, 12
2) 44, 22
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3) 48, 24
4) None of these
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