Download Algebra - New Age International

Algebra
Algebra is generous: she often gives more than is asked of her
– D’Alembert
„
„
„
„
„
Numbers, H.C.F and L.C.M
Equation
Progression
Matrices & Determinants
Permutation & Combination
CHAPTER
1
Numbers, HCF and LCM
1.1 THEORY OF NUMBERS
The numbers 1, 2, 3, 4... which are used for counting are called counting numbers or
natural numbers. The set of natural numbers is denoted by N.
∴
N = {1, 2, 3, 4, ...}
If we add 2 natural numbers, we get again a natural number. eg.: 2 + 3 = 5 ∈ N. But
if we subtract a bigger number from the smaller number or if we subtract a number from
itself, we won’t get a natural number.
For e.g.: 7 − 8 = −1 ∉ N
or
6 − 6 = 0 ∉N
A set containing natural numbers, zero and negative numbers is called set of integers.
It is denoted by I or Z.
∴
I
or
l
q
Z = 0, ± 1, ± 2, ± 3, ...
or
l ........ − 3, − 2, − 1, 0, 1, 2, 3, ....q
If we add, subtract or multiply 2 integers, we get again an integer. But if we divide, we
won’t get an integer always.
For example:
2÷ 3 =
2
∉I
3
or 1 ÷ 7 =
1
∉Z
7
The numbers which can be expressed in the form
q ≠ 0 are called rational numbers.
Example:
16 3
, , 5,
8 4
FG∵
H
5=
5
1
p
, where p and q are integers and
q
IJ
K
The set of rational numbers is denoted by Q.
The rational numbers, when they are expressed in decimal form are either terminating or non-terminating recurring decimals.
4
Comprehensive Business Mathematics
4
For example:
3 19
=
= 4.75
4 4
(Terminating)
1
= 0.333……
3
(non terminating recurring)
If the decimal form are unending non-repeating or non-terminating for a given fraction
(number), then that fraction (number) is irrational. Also if there is no rational number y
such that yn = x when n is whole number, then y is an irrational number i.e., y = n x is
irrational.
Example:
π=
22
.
...,
= 31415
7
e = 2.718 ...,
2 = 1.41423...
The union of set of rational and irrational numbers is called set of real numbers. It
is denoted by R.
∴
R = Rational numbers U Irrational numbers.
To find square root of a negative number we use ι whose value is −1 i.e. ι 2 = −1
∴
−36 = ± 6ι
The numbers of the form x + ιy where x and y are real and ι = −1 are called Complex numbers. The set of complex numbers is denoted by C.
RS
T
∴ C = x + ιy
UV
W
x, y, ∈R
ι = −1
In the complex number x + iy, if y = 0, then the complex number ‘x’ is just a real
number. The set of complex numbers is the biggest numbers set. It contains every number
set as its subset.
Hence
Set of Natural numbers is a subset of integers. The set of integers is a subset of
rational numbers. The set of rational numbers is a subset of Real Numbers. The set of real
numbers is a subset of complex numbers.
∴ N ⊂ I or Z ⊂ Q ⊂ R ⊂ C
1.2 EVEN AND ODD NUMBERS
The numbers which are divisible by 2 are called even numbers.
Example: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, ...
Clearly the even numbers end with 0 or 2 or 4 or 6 or 8.
Numbers, HCF and LCM 5
The numbers which are not divisible by 2 or the numbers which ends with 1 or 3 or 5
or 7 or 9 are called odd numbers.
Example: 687, 899, 331, etc.
1.3 FACTOR OR DIVISOR
Let a and b be any 2 integers. a ≠ 0, is said to be a factor of b if there exists an integer k
such that b = ka i.e., when b is divided by a, remainder is zero.
Ex. 4 is a factor of 12. ∵ 12 = 3 × 4.
6 is a factor of 60 ∵ 60 = 6 × 10.
1.4 PRIME NUMBERS
A number p > 1 is called a prime number if it is divisible by ± 1 and ± p only i.e. if there
is no divisors for p other than ± 1 and ± itself.
The prime numbers upto 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53,
59, 61, 67, 71, 73, 79, 83, 89, 97.
To test, whether a given number is prime or not, take an integer larger than the
approximate square root of the given number. Let it be x. Test the divisibility of the given
number by every prime number less than x. If it is not divisible by any of them, then it is
prime otherwise it is not a prime number.
Examples:
1. 331.
331
It is approximately 19.
Prime numbers less than 19 are 2, 3, 5, 7, 11, 13, 17.
Clearly 331 is not divisible by any of them. ∴ 331 is a prime number.
2. 481
Consider
481
It is approximately 22.
Prime numbers less than 22 are 2, 3, 5, 7, 11, 13, 17, 19.
Out of these 481 is divisible by 13. (∵ 13 × 37 = 481)
∴ 481 is not a prime number.
Consider
1.5 COMPOSITE NUMBERS
A number which is not a prime number is called a composite number. In other words a
number which has a divisor other than 1 and itself is called a composite number.
Example: 4, 9, 10, 12, 14, 15, 16, ... are composite numbers.
Cannonical Representation: Every composite number can be expressed as a product of
prime factors. This representation is unique except for the different arrangement of factors. This representation is called cannonical representation.
6
Comprehensive Business Mathematics
For Example:
1. 63
63 = 3 × 3 × 7
63 = 32 × 7.
2. 160
160 = 2 × 2 × 2 × 2 × 2 × 5.
= 25 × 5.
3 63
3 21
7 7
1
2 160
2 80
2 40
2 20
2 10
5
5
1
1.6 HIGHEST COMMON FACTOR (HCF) OR GREATEST COMMON
DIVISOR (GCD)
Consider the numbers 12 and 16.
Positive factors of 12: 1, 2, 3, 4, 6, 12
Positive factors of 16: 1, 2, 4, 8, 16.
Common factors of 12 and 16 = 1, 2, 4
Highest Common factor: 4
Incidentally we observe that other common factors of 12 and 16 i.e. 1 and 2 divides 4.
Definition
A positive integer d is called highest common factor of 2 integers a and b (atleast one of
them being non zero) if
(i) d is a factor of both a and b.
(ii) Every other common factor of a and b divides d.
The HCF of a and b is denoted by (a, b) = d.
Relatively prime numbers or co-prime numbers
Two integers a and b are said to be relatively prime numbers if their HCF or GCD = 1
Methods of finding HCF
The method of finding HCF by listing the factors is tedious when the numbers are big. So
we use either factorisation method or division method to find HCF.
(i) Factorisation method: In this method, the given numbers are expressed as the
product of prime factors [canonical representation]. Then the product of the common
factors taken as many times as they appear gives the HCF. If there is no common
factor then HCF = 1.
Example: (1) To find HCF of 120 and 216
Numbers, HCF and LCM 7
120 = 2 × 2 × 2 × 3 × 5
2
2
2
3
3
3
2 120
2 60
2 30
3 15
5
5
1
216 = 2 × 2 × 2 × 3 × 3 × 3
HCF = 2 × 2 × 2 × 3
HCF = 24
216
108
54
27
9
3
1
(ii) To find HCF of 88, 27 and 40.
2
2
2
11
88 = 2 × 2 × 2 × 11
27 = 3 × 3 × 3
40 = 2 × 2 × 2 × 5
88
44
22
11
1
3
3
3
27
9
3
1
2
2
2
5
40
20
10
5
1
There is no common factor in all the 3.
∴ HCF = 1
2. Division Method
In this method bigger number is divided by the smaller one. If the remainder is zero, then
the smaller number is G.C.D. or H.C.F. Otherwise divide the smaller number by the
remainder. If remainder in this division is zero then HCF is first remainder. Otherwise
continue the process till the zero remainder is obtained. The last non zero remainder is
GCD or HCF.
Examples: To find HCF of 120 and 216 by division method:
g
b
120 216 1
120
96 120 1
96
24 96 4
96
00
g
b
g b
Last non zero remainder = 24 = GCD
∴
(120, 216) = 24
Note: To find HCF of 3 or more numbers by division method, first find GCD of 2 numbers
then find the HCF of the first GCD and the remaining number.
Example: To find HCF of 88, 27 and 40.
GCD of 27 and 40.
8
Comprehensive Business Mathematics
g b
g b
27 40 1
27
13 27 2
26
1 13 13
13
00
g b
Last non-zero remainder = 1
Now G.C.D. of 1 and 88.
=
g b
1 88 88
88
00
G.C.D. = 1
∴
HCF of 88, 27 and 40 = 1.
1.7 LEAST COMMON MULTIPLE (LCM)
Consider the numbers 2 and 3:
Positive Multiples of 2 = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 ...
Positive Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33 ...
Common multiples of 2 and 3 = 6, 12, 18, 24 ...
Least Common multiple = 6.
Incidentally we observe that 6 divides other common multiples 12, 18, 24, ... etc. of 2 and
3.
Definition
A positive number is called least common multiple of 2 positive integers a and b if
(i) Both a and b divide .
(ii) divides every other common multiples of a and b.
Methods of Finding LCM
The method of finding LCM by listing the multiples is tedious when the numbers are big.
So we use either factorisation method or Division method to find LCM.
Factorisation Method
The given integers are first expressed in cannonical form i.e. as the product of prime
factors. The LCM is the product of highest powers of each factor.
Numbers, HCF and LCM 9
Example:
(i) To find LCM of 48 and 40.
2
2
2
2
3
48 = 2 × 2 × 2 × 2 × 3 = 24 × 3
40 = 2 × 2 × 2 × 5 = 23 × 5
LCM = 24 × 3 × 5
48
24
12
6
3
1
2
2
2
5
40
20
10
5
1
= 240.
(ii) To find LCM of 20, 72 and 125.
20 = 22 × 5
72 =
23
×32
125 = 53.
2
2
5
LCM = 23 × 32 × 53
20
10
5
1
2
2
2
3
3
= 8 × 9 × 125
72
36
18
9
3
1
5
5
5
125
25
5
1
= 9000
Division Method
Write the given numbers in a line. Divide the given integers by any prime number which
divides atleast 2 of them. Write the quotients and non-multiple in order. Repeat the same
till relatively prime numbers are obtained (i.e. no common divisor except 1 is obtained).
The product of all these gives the LCM.
2 48, 40
Example: (i) To find LCM of 48 and 40.
2 24, 20
LCM = 2 × 2 × 2 × 6 × 5
2
12, 10
= 240
6,
(ii) To find LCM of 20, 72 and 125
LCM = 2 × 2 × 5 × 18 × 25
= 9000
1.8 RELATION BETWEEN LCM AND HCF
2
2
5
5
6 and 5 are
relatively prime.
20, 72, 125
10, 36, 125
5, 18, 125
1, 18, 25 18 and 25 are
relatively prime.
If a and b be any 2 integer whose LCM and HCF be and d respectively. Then
a×b= ×d
i.e., Product of 2 numbers = Product of their LCM and HCF.
Example: (i) If the HCF and LCM of 2 numbers are 8 and 80 respectively. If one number
is 40 then to find another number, we use
10 Comprehensive Business Mathematics
a×b=
×d
40 × b = 80 × 8
b=
80 × 8
40
b = 16
∴
The other number is 16.
1.9 H.C.F AND L.C.M OF FRACTIONS
H.C.F of fractions =
H.C.F of numerators
L.C.M of denominators
L.C.M of fractions =
L.C.M of numerators
H.C.F of denominators
Example:
Find the HCF and LCM of
8 32 2 10
, , ,
9 18 3 27
Solution: HCF of fractions =
=
and LCM of fractions =
H.C.F of numerators
L.C.M of denominators
HCF of 8, 32, 2, 10
LCM of 9, 18, 3, 27
LCM of numerators
HCF of denominators
Now
2 8, 32, 2, 10
4 4, 16, 1, 5
1, 4, 5
Since 4, 16 and 5 have the HCF 1,
HCF of 8, 32, 2, 10 = 2, their LCM = 2 × 16 × 5 = 160
Again
3 9, 18, 3, 27
3 3, 6, 1, 9
1, 2, 1, 3
Numbers, HCF and LCM 11
Since 2 and 3 are relatively prime, LCM of 9, 18, 3, 27 = 3 × 3 × 2 × 3 = 54 and
their HCF = 3 × 3 = 9.
∴ HCF of given fractions =
LCM of given fractions =
2
54
160
9
WORKED EXAMPLES
1. Which of the following are rational numbers?
(a) 3.141314131413 ... (b) 68.2314 ... (c) Every prime number
(d) Every natural number (e) Every Real number.
Ans. a>, c>, d> are rational numbers.
2. Prove that 2 is an irrational number.
Proof: If
2 is rational then
2 can be expressed as
p
where p and q are integers and
q
q ≠ 0.
2=
∴
Let p and q have no common divisors, i.e.,
p
q
p
is in the simplest form.
q
Squaring,
e 2j = FGH qp IJK
2
2=
Cross multiplying
p2 = 2q 2
⇒
p2 is even
⇒
p is even.
Let p = 2k where k ∈ I
∴
b2kg
2
= 2q 2
p2
q2
2
12 Comprehensive Business Mathematics
4k2 = 2q 2
⇒
⇒
q 2 = 2k 2
q 2 is even ⇒ q is even.
So we got both p and q are even. Hence they have a common factor 2, which contradicts
the fact that p and q have no common divisor.
∴
2 is irrational.
3. Express 1.666 ... as a rational fraction.
Solution: 1.666 ... = 1 + 0.6 + 0.06 + 0.006 +...
1+
6
6
+
+ ...
10 100
1+
LM
N
LM OP
N Q
∴ 1.666… = 1 +
= 1+
OP
Q
6
1
1
1+
+
+ ...
10
10 100
1
1
+
+… is an infinite G.P.
10 100
6 10
10 9
1+
6
9
with a = 1, r =
15 5
=
= .
9 3
OR
Let x = 1.666......
S∞ =
S∞ =
...(1)
Multiplying by 10
10x = 16.666
...(2)
Subtracting (1) from (2)
10x − x = 16.666...
9x = 15
x=
15
9
x=
5
3
−1.666...
a
1− r
10
.
9
=
1
10
1
1−
1
10
Numbers, HCF and LCM 13
∴
1.666... =
5
3
4. Express 52323
as rational fraction
.
5.232323 = 5 + 0.23 + 0.0023 + ...
5+
23
23
+
+…
100 10000
5+
23
1
1+
+…
100
100
LM
N
OP
Q
1
+… is an infinite G.P
100
1
with a = 1, r =
100
a
S∞ =
1− r
∵ 1+
LM OP
N Q
23 100
52323
.
= 5+
100 99
= 5+
=
23
99
S∞ =
1
1−
1
100
S∞ =
100
99
518
99
OR
Let x = 52323
.
Multiplying by 100
(1)
100x = 5232323
.
(2)
Subtracting (1) from (2)
− 52323
100x − x = 5232323
.
.
99x = 518
x=
518
99
14 Comprehensive Business Mathematics
5. Find the LCM of 48, 96 and 74.
2
2
2
2
3
LCM = 2 × 2 × 2 × 2 × 3 × 2 × 37
= 25 × 3 × 37
= 3552.
48, 96, 74
24, 48, 37
12, 24, 37
6, 12, 37
3, 6, 37
1, 2, 37
6. Find the greatest number less than 150 and divisible by 18 and 54.
Solution: Greatest number divisible by 18 and 54 is a multiple of LCM of 18 and 54.
Now LCM of 18 and 54 = 2 × 33
= 54
∴
2
3
3
Multiples of 54 = 54 × 1, 54 × 2, 54 × 3, ...
18, 54
9, 27
3, 9
1, 3
∵ 54 × 3 = 162
Number less than 150 = 54 × 2 = 108
7. Find the least number divisible by 32, 76 and 84.
Solution: Least number divisible by 32, 76 and 84 is the least common multiple of 32, 76
and 84.
LCM = 2 × 2 × 8 × 19 × 21
2 32, 76, 84
2
16, 38, 42
8, 19, 21
8, 19 and 21 are relatively prime
= 12768
8. Find the largest integer which divides 48 and 160
Solution: The largest integer which divides 48 and 160 is HCF of 48 and 160.
g
b
48 160 3
144
16 48 3
48
00
g b
Last non zero remainder = 16
∴ HCF = 16
9. Find the greatest number which divides 39, 48 and 90 leaving the remainder 6, 4 and
2 respectively.
Solution: Let d be the greatest number
Numbers, HCF and LCM 15
Given: When d divides 39, 6 is the remainder
⇒
d divides 39 − 6 = 33.
When d divides 48, 4 is the remainder
∴
d divides 48 − 4 = 44.
When d divides 90, 2 is the remainder.
∴
d divides 90 − 2 = 88.
Hence d is the greatest number which divides
33, 44 and 88.
i.e., d is the HCF of 33, 44 and 88.
33 = 3 × 11
3
11
44 = 2 × 2 × 11
88 = 2 × 2 × 2 × 11
2
2
11
33
11
1
44
22
11
1
HCF = 11
2
2
2
11
88
44
22
11
1
∴
d = 11.
Hence the required number = 11.
10. A rectangular field is 32 mts wide 48 mts long. What is the greatest length of the tape
by which the length and breadth can be measured exact number of times?
Solution: The greatest length of the tape required to measure 32 mts and 48 mts is GCD
of 32 and 48.
g b
g b
32 48 1
32
16 32 2
32
00
Last non zero remainder = 16.
Greatest length of the tape = 16 mts.
11. Three heaps of one rupee coin contain 70, 175 and 140 coins. Find the largest number
of coins which can be taken exact number of times from each heap.
Solution: The largest number of coins that can be taken from each heap = HCF of 70, 175,
140.
70 = 2 × 5 × 7
175 = 5 × 5 × 7
140 = 2 × 2 × 5 × 7
HCF = 5 × 7 = 35.
2
5
7
70
35
7
1
5
5
7
175
35
7
1
2 140
2 70
5 35
7
7
1
16 Comprehensive Business Mathematics
∴ From the first heap containing 70 coins, 35 coins can be taken twice.
From the 2nd heap having 175 coins, 35 coins can be taken 5 times and from the 3rd
heap having 140 coins, 35 coins can be taken 4 times.
12. Three bells toll at intervals of 9, 24 and 39 seconds. If they begin to toll together, what
length of time will elapse before they toll together again?
Solution: The least time that should elapse before they toll together = LCM of 9, 24 and
39.
Now
3
9, 24, 39
3, 8, 13
3, 8 and 13 are relatively prime.
∴ LCM of 9, 24 and 39 = 3 × 3 × 8 × 13
= 936.
They toll together after 936 seconds or 15 min. 36 seconds.
13. The cost of a 100 pages note book is Rs. 12, cost of 200 pages note book is 16 and the
cost of 500 pages note book is 24. Find the least amount of money that a person
should possess to enable him to purchase a whole number of 100 pages or 200 pages
or 500 pages note books:
Solution: The least amount is LCM of 12, 16 and 24.
LCM = 2 × 2 × 3 × 2 × 2
= 48
2
2
3
2
12, 16, 24
6, 8, 12
3, 4, 6
1, 4, 2
2
1, 2, 1
1, 1, 1
Minimum amount a person should possess = Rs. 48
So that he can purchase either four 100 pages books (4 × 12 = 48) or three 200 pages
books (3 × 16 = 48) or two 500 pages books (2 × 24 = 48).
14. The H.C.F of 2 numbers is 16 and their LCM is 160. If one of the numbers is 32 then
find the other.
Solution: If d and are HCF and LCM of 2 numbers x and y then we have
×d=x×y
Given:
d = 16,
= 160, x = 32
∴ 16 × 160 = 32 × y
⇒
y=
16 × 160
= 80.
32
Numbers, HCF and LCM 17
So the required number is 80.
15. If the product of 2 two-digit numbers is 2160 and their HCF is 12 then find the
numbers.
Solution: Given: HCF = 12.
So the numbers are 12a and 12b. Where a and b are co-prime numbers.
Given their product = 2160
∴ 12a . 12b = 2160
ab =
2160
= 15.
144
By inspection, (a, b) = (1, 15) or (3, 5)
If we take (1, 15) we get the numbers as 12(1) and 12(15) = 12 and 180.
Since we require 2 two digit numbers, we have to take (3, 5).
∴ The required numbers are 12(3) and 12(5) = 36 and 60.
16. Find any 2 numbers whose HCF is 12 and whose difference is also 12.
Solution: Given 2 numbers have HCF = 12.
∴ The numbers are 12a and 12b where a and b are co-prime.
Also given their difference = 12.
∴
12a − 12b = 12
12 (a − b) = 12
(a − b) = 1
a = 1 + b.
∴ The numbers are 12a and 12b
i.e., 12 (1 + b) and 12b.
∴ Any consecutive multiples of 12 will have HCF 12 and difference also 12.
(12, 24), (24, 36) (36, 48) and so on.
17. The HCF and LCM of 2 numbers are 44 and 264 respectively. If the first number is
divided by 2, the quotient is 44. Find the numbers.
Let the 2 numbers be x and y.
We have
×d=x×y
Given
x
= 44
2
⇒ x = 88
18 Comprehensive Business Mathematics
d = 44
= 264
∴ 264 × 44 = 88 × y
⇒ y=
264 × 44
= 132
88
So the numbers are 88 and 132.
18. Find the number of prime factors in the expression 620 × 717 × 1160
Solution:
6=3×2
7 and 11 are prime numbers
∴ 620 × 717 × 1160 = 320 × 220 × 717 × 1160 .
∴ Total number of prime factors
= 20 + 20 + 17 + 60 = 117.
REMEMBER
• Natural numbers, N = {1, 2, 3, ...}
• Integers = I or Z = {0, ±1, ±2, ±3, ...}
• Rational numbers: Q =
RS p
Tq
p, q ∈ I
q≠0
UV
W
˜ Irrational numbers: If the decimal form of rational number is unending, non-repeating or non-terminating then it is irrational number. Also if there is no rational number
y such that yn = x when n is whole number then y is an irrational.
• Real numbers = R = {Rational numbers} ∪ {Irrational numbers}
• Complex numbers: C =
RSx + iy
T
x, y ∈ R
ι = −1
UV
W
• Even numbers: Multiples of 2
• Odd numbers: The numbers which are not divisible by 2.
• Prime numbers: A number p > 1 is called a prime number if it is divisible by ±1 and
± p only.
• Composite number: A number which is not prime is called composite number.
• HCF or GCD of 2 integers a and b is d if (i) d is a factor of both a and b (ii) Every other
common factor divides ‘d’. It is denoted by (a, b) = d.
• Relatively prime: If GCD or HCF of 2 integers a and b is 1, then they are said to be
relatively prime or co-prime.
Numbers, HCF and LCM 19
• LCM: is LCM of 2 integers a and b if (i) both a and b divide (ii) divides every other
common multiple of a and b.
• If and d are LCM and HCF of 2 integers a and b then × d = a × b.
• H.C.F of fractions =
HCF of numerators
LCM of denominators
• L.C.M of fractions =
LCM of numerators
HCF of denominators
EXERCISES
1. Which is the only even prime number?
2. Express 2.555... as a rational fraction.
3. Express 3.3232 as a rational fraction.
4. Find the HCF of
1 2 3 4
, , , .
2 3 4 5
5. Find the LCM of
3 6 9
, . .
4 7 8
6. Examine whether (i) 101 (ii) 447 is prime number or not.
7. Find the total number of prime factors of the product 2222 × 1524 × 731
8. Find the HCF of the following by factorisation method:
(a) 136, 153
(b) 111, 74 and 185
(c) 234, 288, 270
(d) 75, 255, 345
9. Find the HCF of the following by division method:
(a) 1702 and 1998
(b) 350, 600 and 550
(c) 45, 120 and 180
(d) 88 and 116
10. Find the LCM of the following:
(a) 12, 18, 30, 48 and 60
(b) 18, 46 and 120
(c) 165, 231 and 550
(d) 12, 18 and 40
11. The HCF of 2 numbers is 34 and their LCM is 4284. If one number is 204. Find the
other.
12. If the HCF of 32 and 48 is 16. Find the LCM.
13. Find the least number divisible by 28, 42 and 64.
14. Find the smallest number divisible by 34, 72 and 68.
15. Find the greatest number which divides 44, 57 and 91 leaving remainders 2, 3 and 1
respectively.
16. Find the size of largest square tile to floor a room of 6 m 24 cm length and 9 m 10 cm
breadth.
20 Comprehensive Business Mathematics
17. Three containers can hold 9, 3 and 15 litres of milk. What is the least quantity of milk
which can be filled by these containers as exact number of times.
18. 4 men can walk 105, 112, 126 and 168 kilometres in a week respectively. What is the
least distance they can walk in an exact number of weeks?
19. Find the number which when divided by 14, 49, 84 and 63 leaves remainder 5 in each
case.
20. 6 bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds
respectively. In 30 minutes how many times they toll together?
ANSWERS
1. 2
2.
23
9
3.
329
99
4.
1
60
18
1
8. (a) 17
(b) 37
(c) 18
(d) 15
9. (a) 74
(b) 50
(c) 15
(d) 4
10. (a) 720
(b) 2760
(c) 11550
(d) 360
11. 714
12. 96
13. 1344
14. 1224
15. 6
16. 26 cm
17. 45
18. 5040 kms.
19. 1769
20. 16 times.
5.
6. 101 is prime 447 is not a prime
7. 301