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Paper - I
S.S.C.
MAHESH TUTORIALS
Batch : SB
Date :
Test - III
Marks : 30
ALGEBRA – Chapter : 2, 3
Time : 1 hr. 15 min.
Q.1. Solve the following :
(i)
If the point (3, 2)lies on the graph of the equation 5x + ay = 19, then
find a.
(ii)
Without actually solving the simultaneous equations given below,
decide which simultaneous equations have unique solution, no
solution or infinitely many solutions. 3x + 5y = 16; 4x – y = 6
Q.2. Solve the following :
(i)
Three times the square of a natural numbers is 363. Find the
numbers.
(ii)
Solve the following simultaneous equations using Cramer’s rule :
3x – y = 7; x + 4y = 11
(iii)
Tinu is younger than Pinky by three years. The product of their ages
is 180. Find their ages.
Q.3. Solve the following : (Any Three)
(i)
The length of one diagonal of a rhombus is less than the second
diagonal by 4 cm. The area of the rhombus is 30 sq.cm. Find the
length of the diagonals.
4
9
12
Paper - I
... 2 ...
(ii)
Solve the following simultaneous equations using graphical method :
3x + 4y + 5 = 0; y = x + 4
(iii)
The sum of the squares of five consecutive natural numbers is 1455.
find them.
(iv)
Seg AB is the diameter of a circle. C is the point on the circumference
such that in ABC, B is the less by 10º than A. Find the measures
of all the angles of ABC.
Q.4. Solve the following : (Any One)
5
(i)
A car covers a distance of 240km with some speed, if the speed increased
by 20 km/hr, it will cover the same distance in 2 hours less. find the
speed of the car.
(ii)
Some part of a journey of 555 km was completed by a car with speed
60 km/hr then the speed is increased by 15 km/hr and the journey is
completed. If it takes 8 hours to reach, find the time taken and
distance covered by 60km/hr speed.
Best Of Luck

Paper - I
MAHESH TUTORIALS
S.S.C.
Batch : SB
Test - III
Marks : 30
ALGEBRA – Chapter : 2, 3
Date :
A.1.
(i)
Time : 1 hr. 15 min.
Solve the following :
 (3, 2) lies on the graph of the equation 5x + y = 19.
It satisfies the equation,
 Substituting x = 3 and y = 2 in the equation we get,
5 (3) + a (2)
= 19

15 + 2a
= 19

2a
= 19 – 15

2a
= 4
4

a
=
2

a
= 2
(ii)





A.2.
(i)
MODEL ANSWER PAPER
3x + 5y = 16
Comparing with a1x + b1y = c1 we get, a1 = 3, b1 = 5, c1 = 16
4x – y = 6
Comparing with a2x + b2y = c2 we get, a2 = 4, b2 = – 1, c2 = 6
a1
3
a2 = 4
b1
5
=
b2
–1 = – 5
c1
16
8
=
c2 =
6
3
a1
b1
a 2  b2
The simultaneous equations 3x + 5y = 16 and 4x – y = 6 have
unique solution.
Solve the following :
Let the natural number be ‘x’
From the given condition,
3x2 = 363
363
 x2 =
3
 x2 = 121
 x = + 11
[Taking square roots]
1
1
1
1
1
1
Paper - I
... 2 ...
x  – 11 because x is a natural number
 x = 11
 The natural number is 11.
(ii)
1
3x – y = 7
x + 4y = 11
D
=
3 –1
1
4
Dx =
7
–1
11
4
Dy =
3 7
1 11
= (3 × 4) – (– 1 × 1)
= 12 + 1 = 13
= (7 × 4) – (– 1 × 11)
= 28 + 11 = 39
= (3 × 11) – (7 × 1)
= 33 – 7 = 26
1
By Cramer’s rule,
x
=
y
=
Dx
D
Dy
D
=
39
= 3
13
=
26
= 2
13
1
 x = 3 and y = 2 is the solution of given simultaneous equations.
(iii)









Tinu’s age be ‘x’ years
Pinky’s age is (x + 3) years
As per the given condition,
x (x + 3) =
x2 + 3x – 180 =
x2 – 12x + 15x – 180 =
x (x – 12) + 15 (x – 12) =
(x – 12) (x + 15) =
x – 12 = 0 or
x = 12 or
Age cannot be negative.
x 
x =
And x + 3 = 12 + 3 = 15
180
0
0
0
0
x + 15 = 0
x = – 15
1
1
1
– 15
12
 Tinu’s age is 12 years and Pinky’s is 15 years.
1
... 3 ...
A.3.
(i)
Paper - I
Solve the following : (Any Three)
Let the length of other diagonal of a rhombus be ‘x’ cm.
 The length of first diagonal is (x + 4) cm.
1
Area of rhombus =
× Product of length of diagonals
2
1
Area of rhombus =
× x × (x + 4)
2
As per the given condition,









1
x (x + 4) =
2
x (x + 4) =
x2 + 4x – 60 =
x2 + 10x – 6x – 60 =
x (x + 10) – 6 (x + 10) =
(x + 10) (x – 6) =
x + 10 = 0 or
x = – 10 or
The length of diagonal of the
x  – 10
Hence x = 6
And x + 4 = 6 + 4 = 10
30
60
0
0
0
0
x–6=0
x=6
rhombus cannot be negative.
 The length of smaller diagonal of a rhombus is 6 cm and bigger
diagonal is 10 cm.
(ii)
1
1
1
1
3x + 4y + 5 = 0
 3x = –5 – 4y
– 5 – 4y
 x=
3
x
–3
1
5
y
1
–2
–5
(x, y) (–3, 1) (1, –2) (5, –5)
y=x+4
x
0
1
2
y
4
5
6
(x, y)
(0, 4) (1, 5) (2, 6)
1
Paper - I
... 4 ...
Y
Scale : 1 cm = 1 unit
on both the axes
8
2
7
6
(2, 6)
(1, 5)
5
4
3x
3
+
4y
+
5
2
=
0
1
(–3, 1)
X
-5
y
=
x
(0, 4)
-4
+
4
-3
-2
0
-1
1
2
3
4
5
X
-1
(1, –2)
-2
-3
-4
(5, –5)
-5
Y
 x = – 3 and y = 1 is the solution of given simultaneous equations.
1
Paper - I
... 5 ...
(iii)
Let the five consecutive natural numbers be x, x + 1, x + 2, x + 3
and x + 4 respectively.
As per the given condition,
x2 + (x + 1)2 + (x + 2)2 + (x + 3)2 + (x + 4)2
= 1455
2
2
2
2
2
x + x + 2x + 1 + x + 4x + 4 + x + 6x +9 + x + 8x + 16 – 1455 = 0
5x2 + 20x + 30 – 1455
=0
5x2 + 20x – 1425
=0
Dividing throughout by 5 we get,
x2 + 4x – 285 = 0
2

x – 15x + 19x – 285 = 0

x (x – 15) + 19 (x – 15) = 0

(x – 15) (x + 19) = 0

x – 15 = 0 or x + 19 = 0

x = 15 or x = – 19
 x is a natural number x  – 19
Hence x = 15
 x + 1 = 15 + 1 = 16
x + 2 = 15 + 2 = 17
 x + 3 = 15 + 3 = 18
x + 4 = 15 + 4 = 19
 The required five consecutive natural numbers are
15, 16, 17, 18 and 19 respectively.
(iv)
1
1
1
1
C
0
y0
Let the measures of A be xº
A x
B
and measure of B be yº.
Seg AB is the diameter of a circle
and C is a point on the circumference.

m C
=
900
[ Diameter subtends a right
angle at any point on the circle]
In ABC,
m A + m B + m C = 1800 [ Sum of measures of the angles
of a triangle is 1800]

x + y + 90
=
180

x+y
=
180 – 90

x+y
=
90
........(i)
As the given condition,
y
=
x – 10

–x+y
=
– 10
.......(ii)
Adding (i) and (ii),
x+y
=
90
–x+y
=
– 10
2y
=
80
80

y
=
2

y
=
40
1
1
1
... 6 ...
Paper - I
Substituting y = 40 in (i),
x + 40
=
90

x
=
90 – 40

x
=
50
 The measures of angle of ABC are 50º, 40º and 90º.
A.4.
(i)
1
Solve the following : (Any One)
Let the original speed of car be x km/hr.
 Distance covered is 240 km
Dis tance
Speed =
Time
Dis tance
 Time =
Speed
 Time taken by car =
New speed of car =
 240 

 hrs
x 
(x + 20) km/hr
 240 
hrs
 New time taken by car = 
 x  20 
As per the given condition,
240
240
–
= 2
x
x  20
1
1 
240  –
 = 2

 x x  20 

x  20 – x
x (x  20)
=

20
x  20x
=











1
2
240
1
120
20 (120) = 1 (x2 + 20x)
2400 = x2 + 20x
0 = x2 + 20x – 2400
x2 + 20x – 2400 = 0
2
x + 60x – 40x – 2400 = 0
x (x + 60) – 40 (x + 60) = 0
(x + 60) (x – 40) = 0
x + 60 = 0 or x – 40 = 0
x = – 60 or x = 40
The speed of car can never be negative.
x  – 60
Hence x = 40
2
1
 The original speed of car is 40 km/hr.
1
1
1
... 7 ...
(ii)
Paper - I
Let the time for which the car travels at 60 km/hr be x hours
and at 75 (i.e. 60 + 15) km/hr be y hours .
We know that, Distance = Speed × Time
Distance covered by car with the speed of 60 km/hr = 60 × x = 60x km
Distance covered by car with the speed of 75 km/hr = 75 × y = 75y km
As per first condition,
60x + 75y
=
555
15 (4x + 5y)
=
555
555
4x + 5y
=
15
4x + 5y
=
37
.............(i)
As per second condition,
x + y
= 8
............(ii)
Multiplying (ii) by 5, we get,
5x + 5y
= 40
............(iii)
Subtracting (i) from (iii), we get
5x + 5y
=
40
4x + 5y
=
37
–
–
–
x
=
3
 Time taken by the car at the speed of 60 km/hr = 3 hours and
Distance covered with the speed of 60km/hr = 60 × 3 = 180 km.

1
1
1
1
1
Paper - II
MAHESH TUTORIALS
S.S.C.
Batch : SB
Date :
Q.1.
Test - III
Marks : 30
ALGEBRA – Chapter : 2, 3
Time : 1 hr. 15 min.
Solve the following :
4
1.2
0.57
0.03
– 0.23
(i)
Find the value of the following determinants :
(ii)
What is the equation of X - axis? Hence, find the point of intersection of
the graph of the equation x + y = 3 with the X - axis.
Q.2. Solve the following :
9
(i)
The length of the rectangle is greater than its breadth by 2 cm. The area
of the rectangle is 24 sq.cm, find its length and breadth.
(ii)
Solve the following simultaneous equations using Cramer’s rule :
y=
(iii)
5x – 10
; 4x + 5 = – y
2
The sum of a natural number and its reciprocal is
10
. Find the number.
3
Q.3. Solve the following : (Any Three)
(i)
The sum of the areas of two squares is 400sq.m. If the difference
between their perimeters is 16 m, find the sides of two square.
12
Paper - II
... 2 ...
(ii)
Solve the following simultaneous equations using graphical method :
4x = y – 5; y = 2x + 1
(iii)
The divisor and quotient of the number 6123 are same and the remainder
is half the divisor. Find the divisor.
(iv)
Durga’s mother gave some 10 rupee notes and some 5 rupee notes to
her, which amounts to Rs. 190. Durga said, ‘if the number of 10 rupee
notes and 5 rupee notes would have been interchanged, I would have
Rs. 185 in my hand.’ So how many notes of rupee 10 and rupee 5 were
given to Durga ?
Q.4. Solve the following : (Any One)
5
(i)
The cost of bananas is increased by Re. 1 per dozen, one can get 2 dozen
less for Rs. 840. Find the original cost of one dozen of banana.
(ii)
Students of a school were made to stand in rows for drill. If 3 students
less were standing in each row, 10 more rows were required and if 5
students more were standing in each row then the number of rows was
reduced by 10. Find the number of students participating in the drill.
Best Of Luck

Paper - II
S.S.C.
MAHESH TUTORIALS
Batch : SB
Test - III
Marks : 30
ALGEBRA – Chapter : 2, 3
Date :
A.1.
(i)
MODEL ANSWER PAPER
Time : 1 hr. 15 min.
Solve the following :
1.2
0.03
0.57 – 0.23
= (1.2 × – 0.23) – (0.03 × 0.57)
= – 0.276 – 0.0171
= – 0.2931
(ii)
The equation of X-axis is y = 0
Let the point of intersection of graph x + y = 3 with X-axis be (h, 0)
 (h, 0) lies on the graph, it satisfies the equation
 Substituting x = h and y = 0 in the equation we get,
h + 0= 3

h = 3
1
1
1
1
 The line x + y = 3 intersects the X-axis at (3, 0).
A.2.
(i)
Solve the following :
Let the breadth of the rectangle be ‘x’ cm.
 It’s length is (x + 2) cm.
As per the given condition.

Area of rectangle = Length × Breadth

24 = (x + 2) × x

24 = x2 + 2x

0 = x2 + 2x – 24
2

x + 2x – 24 = 0

x2 – 4x + 6x – 24 = 0

x (x – 4) + 6(x – 4) = 0

(x – 4) (x + 6) = 0

x – 4 = 0 or x + 6 = 0

x = 4 or x = -6
 The breadth of rectangle cannot be negative.
 x  -6
Hence x = 4 and x + 2 = 4 + 2 = 6
 The length of rectangle is 6 cm and its breadth is 4 cm.
1
1
1
Paper - II
... 2 ...
(ii)
5x – 10
2
2y
– 5x + 2y
4x + 5
4x + y
y=



D
=
Dx =
Dy =
=
=
=
=
–5 2
4 1
–10 2
–5
1
– 5 –10
4
–5
5x – 10
– 10
–y
–5
= (– 5 × 1) – (2 × 4)
1
= –5–8
= – 13
= (– 10 × 1) – (2 × – 5) = – 10 + 10 = 0
= (– 5 × – 5) – (– 10 × 4 ) = 25 + 40
= 65
1
By Cramer’s rule,
x
=
y
=
Dx
D
Dy
D
0
= –13 = 0
65
= –13 = – 5
 x = 0 and y = – 5 is the solution of given simultaneous equations.
(iii)









Let natural number be ‘x’
1
Its reciprocal is
x
From the given condition,
1
10
=
x+
x
3
Multiplying throughout by 3x,
3x2 + 3 = 10x
3x2 – 10x + 3 = 0
3x2 – 9x – x + 3 = 0
3x (x – 3) – 1 (x – 3) = 0
(3x – 1) (x – 3) = 0
3x – 1 = 0
or x – 3 = 0
3x = 1
or x = 3
1
x=
or x = 3
3
1
x 
because x is natural number
3
x=3
 The natural number is 3.
1
1
1
1
... 3 ...
A.3.
(i)
Solve the following : (Any Three)
Difference between the perimeters of two squares is 16 m
 The difference between the sides of the same squares is 4 m
Let the side of smaller square be x cm
 The side of bigger square is (x + 4) cm

Area of square = (side) 2
As per the given condition,
x2 + (x + 4)2 = 400
 x2 + x2 + 8x + 16 – 400 = 0

2x2 + 8x – 384 = 0
Dividing throughout by 2 we get,
x2 + 4x – 192 = 0
2

x – 12x + 16x – 192 = 0
 x (x – 12) + 16 (x – 12) = 0

(x – 12) (x + 16) = 0

x – 12 = 0 or x + 16 = 0

x = 12 or x = – 16
 The side of square cannot be negative
 x  – 16
Hence x = 12
 x + 4 = 12 + 4 = 16
 The side of smaller square is 12 m and bigger square is 16 m.
(ii)
Paper - II
1
1
1
1
4x = y – 5
 4x + 5 = y
 y = 4x + 5
x
0
–1
–2
y
5
1
–3
(x, y)
(0, 5) (–1, 1) (–2, –3)
y = 2x + 1
x
0
1
2
y
1
3
5
(x, y)
(0, 1) (1, 3) (2, 5)
1
Paper - II
... 4 ...
Y
Scale : 1 cm = 1 unit
on both the axes
6
(0, 5)
5
2
(2, 5)
4
(1, 3)
3
2
(–1, 1)
-5
-4
-3
-2
(0, 1)
0
-1
1
2
3
4
5
X
-1
-2
(–2, –3)
-3
-4
-5
y=
2x
+ 1
X
1
-6
-7
-8
Y
 x = – 2 and y = – 3 is the solution of given simultaneous equations.
1
... 5 ...
(iii)
Let divisor of 6123 be ‘x’
 Divisor = Quotient
[Given]
 Quotient = x
x
[Given]
Remainder =
2
We know,
Dividend = Divisor × Quotient + Remainder
1
x
6123 = x . x +
2
Multiplying throughout by 2,
2 (6123) = 2x2 + x

12246 = 2x2 + x
2

2x + x – 12246 = 0
 2x2 + 157x – 156x – 12246 = 0
1
 x (2x + 157) – 78 (2x + 157) = 0

(2x + 157) (x – 78) = 0
 2x + 157 = 0
or
(x – 78) = 0
 2x + 157 = 0
or
x – 78 = 0
– 157
 x=
or
x = 78
1
2
– 157
x=
cannot be acceptable because divisor cannot be negative.
2
 x = 78
 The divisor of 6123 is 78.
(iv)
Paper - II
Let the no. of Rs. 10 notes given to Durga be x and the no. of
Rs.5 notes given to her be y.
As per the first condition,
10x + 5y
=
190
.......(i)
As per the second condition,
5x + 10y
=
185
......(ii)
Adding (i) and (ii),
15x + 15y
=
375
Dividing throughout by 15 we get,
375
x+y
=
15

x+y
=
25
......(iii)
Subtracting (ii) from (i),
5x – 5y
=
5
Dividing throughout by 5 we get,
x–y
=
1
.......(iv)
Adding (iii) and (iv),
1
1
1
... 6 ...




x+y
=
25
x–y
=
1
2x
=
26
x
=
13
Substituting x = 13 in (iii),
13 + y
=
25
y
=
25 – 13
y
=
12
 Durga had 13 notes of Rs. 10 rupee and 12 notes of Rs. 5.
A.4.
(i)
Solve the following : (Any One)
Let the cost of banana per dozen be Rs. x.
Amount for which bananas are bought = Rs. 840
840
No. of dozens of bananas for Rs 840 is x
New cost of banana per dozen = Rs. (x + 1)
840
New No. of dozens of bananas for Rs 840 = x  1
As per the given condition,
840
840
–
2
x
x 1 =












1
1 
840  –
x  1 
x
=
x  1 – x
840 
=

 x (x  1) 
 1 
840  2
 =
x  x
840 =
2x2 + 2x – 840 =
Dividing throughout by 2,
2x2 + x – 420 =
2
x – 20x + 21x – 420 =
x (x – 20) + 21 (x – 20) =
x – 20 = 0 or
x = 20 or
The cost of bananas cannot
x  –21
Hence x = 20
Paper - II
1
1
1
0
2
1
2
2 (x2 + x)
0
0
0
0
x + 21 = 0
x = –21
be negative.
 The original cost of one dozen banana is Rs. 20.
1
1
1
... 7 ...
(ii)
Let the no. of students standing in each row be x and let no. of
rows be y.
 Total no. of students participating in the drill = xy
As per the first given condition,
(x – 3) (y + 10)
=
xy
 x (y + 10) – 3 (y + 10) =
xy

xy + 10x – 3y – 30
=
xy

10x – 3y
=
30
.......(i)
As per the second given condition,
(x + 5) (y – 10)
=
xy
 x (y – 10) + 5 (y – 10) =
xy

xy – 10x + 5y – 50
=
xy

– 10x + 5y
=
50
......(ii)
Adding (i) and (ii),
10x – 3y
=
30
– 10x + 5y
=
50
2y
=
80
80

y
=
2

y
=
40
Substituting y = 40 in (i),
10x – 3 (40)
=
30

10x – 120
=
30

10x
=
30 + 120

10x
=
150
150

x
=
10

x
=
15

xy = 15 × 40 = 600
 600 students were participating in the drill.

Paper - II
1
1
1
1
1
Paper - III
S.S.C.
MAHESH TUTORIALS
Batch : SB
Date :
Test - III
Marks : 30
ALGEBRA – Chapter : 2, 3
Time : 1 hr. 15 min.
Q.1. Solve the following :
(i)
What is the equation of Y - axis? Hence, find the point of intersection of
Y - axis. and the line y = 3x + 2.
(ii)
Without actually solving the simultaneous equations given below, decide
which simultaneous equations have unique solution, no solution or
infinitely many solutions. 3y = 2 – x; 3x = 6 – 9y
Q.2. Solve the following :
(i)
The sum of the squares of two consecutive natural numbers is 113. Find
the numbers.
(ii)
Solve the following simultaneous equations using Cramer’s rule :
3x + y = 1; 2x = 11y + 3
(iii)
The length of the rectangle is greater than its breadth by 2 cm. The area
of the rectangle is 24 sq.cm, find its length and breadth.
Q.3. Solve the following : (Any Three)
(i)
Around a square pool there is a footpath of width 2km. if the area of the
5
footpath is
times that of the pool. Find the area of the pool.
4
4
9
12
Paper - III
... 2 ...
(ii)
Solve the following simultaneous equations using graphical method :
x + y = 8, x – y = 2
(iii)
The length of one diagonal of a rhombus is less than the second diagonal
by 4 cm. The area of the rhombus is 30 sq.cm. Find the length of the
diagonals.
(iv)
On the first day of the sale of tickets of a drama. 35 tickets in all were
sold. If the rates of the tickets were Rs.20 and Rs.40 per ticket and the
total collection was Rs. 900. Find the number of tickets sold of each rate.
Q.4. Solve the following : (Any One)
5
(i)
One tank can be filled up by two taps in 6 hours. The smaller tap alone
takes 5 hours more than the bigger tap alone. Find the time required by
each tap to fill the tank separately.
(ii)
A bus covers a certain distance with uniform speed. If the speed of the
bus would have been increased by 15 km/h, it would have taken two
hours less to cover the same distance and if the speed of the bus would
have been decreased by 5 km/h, it would have taken one hour more to
cover the same distance. Find the distance covered by the bus.
Best Of Luck

Paper - III
S.S.C.
MAHESH TUTORIALS
Batch : SB
Test - III
Marks : 30
ALGEBRA – Chapter : 2, 3
Date :
A.1.
(i)
(ii)
A.2.
(i)
MODEL ANSWER PAPER
Time : 1 hr. 15 min.
Solve the following :
The equation of Y-axis is x = 0
Let the point of intersection of the line y = 3x + 2 with Y-axis be (0, k)
 (0, k) lies on the line it satisfies the equation
 Substituting x = 0 and y = k in the equation we get,
k
= 3 (0) + 2

k
= 2
1
 The point of intersection of the line y = 3x + 2 with Y-axis is (0, 2).
1
3y = 2 – x
 x + 3y = 2
Comparing with a1x + b1y = c1 we get, a1 = 1, b1 = 3, c1 = 2
3x
=
6 – 9y
 3x + 9y =
6
Comparing with a2x + b2y = c2 we get, a2 = 3, b2 = 9, c2 = 6
a1
1
 a
=
3
2
b1
3
1
 b =
=
9
3
2
c1
2
1
 c
=
=
6
3
2
a1
b1
c1
= b
= c
 a
2
2
2
 The simultaneous equations 3y = 2 – x and 3x = 6 – 9y have
infinitely many solutions.
Solve the following :
Let the two consecutive natural numbers be x and x + 1
As per the given condition,
x2 + (x + 1)2 = 113
2

x + x2 + 2x + 1 = 113
2

2x + 2x + 1 – 113 = 0

2x2 + 2x – 112 = 0
Dividing throughout by 2 we get,
1
1
1
... 2 ...







x2 + x – 56
x2 + 8x – 7x – 56
x (x + 8) – 7 (x + 8)
(x + 8) (x – 7)
x+8=0
x=–8
x is a natural number
x  -8
Hence, x = 7
And x + 1 = 7 + 1 = 8
=
=
=
=
or
or
Paper - III
0
0
0
0
x–7=0
x=7
 The two consecutive natural numbers are 7 and 8 respectively.
(ii)
1
1
3x + y = 1
2x = 11y + 3
 2x – 11y = 3
D
=
Dx =
3 1
= (3 × – 11) – (1 × 2) = – 33 – 2 = – 35
2 –11
1
1
3 –11
= (1 × – 11) – (1 × 3) = – 11 – 3 = – 14
3 1
= (3 × 3) – (1 × 2) = 9 – 2 = 7
2 3
By Cramer’s rule,
–14
Dx
2
=
=
x =
– 35
D
5
Dy =
y
 x=
(iii)








=
Dy
D
=
7
–1
– 35 = 5
2
–1
and y =
is the solution of given simultaneous equations.
5
5
Let the breadth of the rectangle be ‘x’ cm.
It’s length is (x + 2) cm.
As per the given condition.
Area of rectangle = Length × Breadth
24 = (x + 2) × x
24 = x2 + 2x
0 = x2 + 2x – 24
x2 + 2x – 24 = 0
2
x – 4x + 6x – 24 = 0
x (x – 4) + 6(x – 4) = 0
1
1
1
1
Paper - III
... 3 ...

(x – 4) (x + 6) = 0

x – 4 = 0 or x + 6 = 0

x = 4 or x = -6
 The breadth of rectangle cannot be negative.
 x  -6
Hence x = 4 and x + 2 = 4 + 2 = 6
1
 The length of rectangle is 6 cm and its breadth is 4 cm.
A.3.
(i)
(ii)
1
Solve the following : (Any Three)
Let the side of inner square i.e. pool be x m.
 The width of foot path around the pool is 2 m
 The side of outer square is (x + 2) m

Area of square = side 2
As per the given condition,
(Area of outer square)= (Area of inner square) + (Area of footpath)
5 2
x
(x + 4)2 = x2 +
4
5

x2 + 8x + 16 = x2 + x2
4
5 2

8x + 16 =
x
4
Multiplying throughout by 4 we get,
32x + 64 = 5x2
2

5x – 32x – 64 = 0

5x2 – 40x + 8x – 64 = 0

5x (x – 8) + 8 (x – 8) = 0

(x – 8) (5x + 8) = 0

x – 8 = 0 or 5x + 8 = 0

x = 8 or 5x = – 8
–8

x = 8 or x =
5
–8
x =
is not acceptable because side of pool cannot be negative.
5

x = 8

x2 = 8 2

x2 = 64
 Area of pool is 64 sq. m.
x+y=8
 y=8–x
x
0
1
2
x
2
3
4
y
8
7
6
y
0
1
2
(x, y)
1
1
1
1
x–y =2
 x=2+y
(x, y) (0, 8) (1, 7) (2, 6)
1
(2, 0) (3, 1)
(4, 2)
1
Paper - III
... 4 ...
Y
8
Scale : 1 cm = 1 unit
on both the axes
(0, 8)
2
(1, 7)
7
(2, 6)
6
x
5
+
y
=
4
8
(5, 3)
3
(4, 2)
2
(3, 1)
1
(2, 0)
X -4
-3
-2
0
-1
1
-1
-2
x
-
y
=
2
3
4
5
X
2
-3
Y
 x = 5 and y = 3 is the solution of given simultaneous equations.
(iii)
Let the length of other diagonal of a rhombus be ‘x’ cm.
 The length of first diagonal is (x + 4) cm.
1
× Product of length of diagonals
Area of rhombus =
2
1
× x × (x + 4)
Area of rhombus =
2
As per the given condition,
1
1
... 5 ...









1
x (x + 4) =
2
x (x + 4) =
2
x + 4x – 60 =
x2 + 10x – 6x – 60 =
x (x + 10) – 6 (x + 10) =
(x + 10) (x – 6) =
x + 10 = 0 or
x = – 10 or
The length of diagonal of the
x  – 10
Hence x = 6
And x + 4 = 6 + 4 = 10
Paper - III
30
60
0
0
0
0
x–6=0
x=6
rhombus cannot be negative.
 The length of smaller diagonal of a rhombus is 6 cm and
bigger diagonal is 10 cm.
(iv)
Let the no. of tickets sold at Rs. 20 each be x and Rs. 40 each be y.
As per first given condition,
x+y
=
35
......(i)
As per second given condition,
20x + 40y
=
900
Dividing throughout by 20,
x + 2y
=
45
......(ii)
Subtracting (ii) from (i),
x+y
=
35
x + 2y
=
45
(–) (–)
(–)
–y
=
– 10

y
=
10
Substituting y = 10 in (i),
x + 10
=
35

x
=
35 – 10

x
=
25
 The no. of tickets sold at Rs. 20 each and Rs. 40 each are 25
tickets and 10 tickets respectively.
A.4.
(i)
Solve the following : (Any One)
Let the time taken to fill a tank by a bigger tap alone be x hrs.
 The time taken by smaller tap alone is (x + 5) hrs.
Time taken by both the taps together to fill the same tank is 6 hrs.
1
 Portion of tank filled in 1 hr by bigger tap =
x
1
Portion of tank filled in 1 hr by smaller tap = x  5
1
1
1
1
1
1
1
1
Paper - III
... 6 ...
Portion of tank filled in 1 hr by both taps together =













As per the given condition,
1
1
1
+ x5
=
x
6
x5x
1
=
x (x  5)
6
2x  5
1
=
x 2  5x
6
6 (2x + 5) = 1 (x2 + 5x)
12x + 30 = x2 + 5x
0 = x2 + 5x – 12x – 30
2
x + 3x – 10x – 30 = 0
x (x + 3) – 10 (x + 3) = 0
(x + 3) (x – 10) = 0
x + 3 = 0 or x – 10 = 0
x = – 3 or x = 10
x is the time taken bybigger tap
x  – 3
Hence x = 10
x + 5 = 10 + 5 = 15
1
6
 Time taken by bigger tap alone is 10 hrs and smaller tap alone
is 15 hrs.
(ii)







Let the speed of bus be x km/hr. and time taken be y hrs.
Distance
= Speed × Time
Distance
= xy km
According to the first condition,
(x + 15) (y – 2)
= xy
x (y – 2) + 15 (y – 2)
= xy
xy – 2x + 15y – 30
= xy
– 2x + 15y
= 30 ......(i)
According to the second condition,
(x – 5) (y + 1)
= xy
x (y + 1) – 5 (y + 1)
= xy
xy + x – 5y – 5
= xy
x – 5y
= 5 ......(ii)
Multiplying (ii) by 3 we get,
3x – 15y
=
15
......(iii)
Adding (i) and (iii) we get,
– 2x + 15y
= 30
3x – 15y
= 15
x
= 45
Substituting x = 45 in (ii),
1
1
1
1
1
1
1
... 7 ...



45 – 5y
– 5y
– 5y
=
=
=
5
5 – 45
– 40

y
=
– 40
–5


y
Distance
=
=
=
=
8
xy
45 × 8
360
 Distance covered by bus is 360 km.

Paper - III
1
1
Paper - IV
S.S.C.
MAHESH TUTORIALS
Batch : SB
Date :
Test - III
Marks : 30
ALGEBRA – Chapter : 2, 3
Time : 1 hr. 15 min.
Q.1. Solve the following :
(i)
If (a, 3) is the point lying on the graph of the equation 5x + 2y = – 4, then
find a.
(ii)
Without actually solving the simultaneous equations given below, decide
which simultaneous equations have unique solution, no solution or
infinitely many solutions. 3x – 7y = 15; 6x = 14y + 10
Q.2. Solve the following :
(i)
The sum of the ages of father and his son is 42 years. The product of
their ages is 185, find their ages.
(ii)
Find the value of k for which the given simultaneous equations have
infinitely many solutions :4x + y = 7; 16x + ky = 28.
(iii)
A natural number is greater then three times its square root by 4. Find
the number.
Q.3. Solve the following : (Any Three)
(i)
A car covers a distance of 240km with some speed, if the speed is increased
by 20 km/hr, it will cover the same distance in 2 hours less. find the
speed of the car.
4
9
12
Paper - IV
... 2 ...
(ii)
Solve the following simultaneous equations using graphical method :
x + 2y = 5; y = – 2x – 2
(iii)
For doing some work Ganesh takes 10 days more than John. If both
work together they complete the work in 12 days. Find the number of
days if Ganesh worked alone?
(iv)
AB is a segment. The point P is on the perpendicular bisector of segment
AB such that length of AP exceeds length of AB by 7 cm. If the perimeter
of ABP is 38 cm. Find the sides of ABP.
Q.4. Solve the following : (Any One)
5
(i)
A man riding on a bicycle cover a distance of 60 km in a direction of wind
and comes back to his original position in 8 hours. If the speed of the
wind is 10 km/hr. Find the speed of the bicycle.
(ii)
Some part of a journey of 555 km was completed by a car with speed
60 km/hr then the speed is increased by 15 km/hr and the journey is
completed. If it takes 8 hours to reach, find the time taken and distance
covered by 60km/hr speed.
Best Of Luck

Paper - IV
MAHESH TUTORIALS
S.S.C.
Batch : SB
Test - III
Marks : 30
ALGEBRA – Chapter : 2, 3
Date :
A.1.
(i)
MODEL ANSWER PAPER
Time : 1 hr. 15 min.
Solve the following :
 (a, 3) is a point lying on the graph of the equation 5x + 2y = – 4,
it is satisfies the equation.
 Substituting x = a and y = 3 in the equation we get,
5 (a) + 2 (3)
= –4

5a + 6
= –4

5a
= –4–6

5a
= – 10
–10

a
=
5

a
= –2
(ii)





3x – 7y = 15
Comparing with
6x = 14y + 10
6x – 14y = 10
Comparing with
a1
3
=
a2 = 6
b1
–7
=
=
b2
–14
c1
15
=
c 2 = 10
a1
b1
a 2 = b2 
1
a1x + b1y = c1 we get, a1 = 3, b1 = – 7, c1 = 15
a2x + b2y = c2 we get, a2 = 6, b2 = – 14, c2 = 10
1
2
1
2
3
2
c1
c2
 The simultaneous equations 3x – 7y = 15 and 6x = 14y + 10
have no solution.
A.2.
(i)
1
Solve the following :
Sum of ages of father and son = 42 years
Let father’s age be x years
 Son’s age = (42 – x) years
 The age of his son is (42 – x) years.
As per the given condition,
1
1
... 2 ...









x (42 – x) =
42x – x2 =
0 =
x2 – 42x + 185 =
x2 – 5x – 37x + 185 =
x (x – 5) – 37 (x – 5) =
(x – 5) (x – 37) =
x – 5 = 0 or
x = 5 or
If x = 5, father’s age is less
x 5
Hence x = 37
And 42 – x = 42 – 37 = 5
Paper - IV
185
185
x2 – 42x + 185
0
0
0
0
x – 37 = 0
x = 37
than son’s age.
 The father’s age is 37 years and son’s age 5 years.
(ii)







4x + y = 7
Comparing with a1x + b1y = c1 we get, a1 = 4, b1 = 1, c1 = 7
16x + ky = 28
Comparing with a2x + b2y = c2 we get, a2 = 16, b2 = k, c2 = 28
a1
4
1
=
a 2 = 16
4
b1
1
b2 = k
c1
7
1
c 2 = 28 = 4
The simultaneous equations have infinitely many solutions.
a1
b1
c1
=
=
a2
b2
c2
1
1
1
=
=
4
k
4
1
1
=
4
k
 k
(iii)
=
4
Let the square root of natural number be x
 The natural number = x2
As per the given condition,
x2 = 3x + 4
2

x – 3x – 4 = 0

x2 – 4x + 1x – 4 = 0

(x – 4) (x + 1) = 0

x – 4 = 0 or x + 1 = 0

x = 4 or x = – 1
1
1
1
1
1
1
1
... 3 ...




Square root of natural number cannot be negative
x  –1
x = 4
x2 = 42
x2 = 16
 The natural number is 16.
A.3.
(i)
Solve the following : (Any Three)
Let the original speed of car be x km/hr.
 Distance covered is 240 km
Dis tance
Speed =
Time
Dis tance
 Time =
Speed
 240 
 Time taken by car = 
 hrs
x 
New speed of car = (x + 20) km/hr
 240 
hrs
 New time taken by car = 
 x  20 
As per the given condition,
240
240
–
= 2
x
x  20
1
1 
240  –
 = 2

 x x  20 
x  20 – x
2

=
x (x  20)
240
20
1

=
x 2  20x
120

20 (120) = 1 (x2 + 20x)

2400 = x2 + 20x

0 = x2 + 20x – 2400
2

x + 20x – 2400 = 0

x2 + 60x – 40x – 2400 = 0
 x (x + 60) – 40 (x + 60) = 0

(x + 60) (x – 40) = 0

x + 60 = 0 or x – 40 = 0

x = – 60 or x = 40
 The speed of car can never be negative.
 x  – 60
Hence x = 40
 The original speed of car is 40 km/hr.
Paper - IV
1
1
1
1
1
1
Paper - IV
... 4 ...
(ii)
x + 2y = 5
y = –2x – 2
 x = 5 – 2y
1
x
5
3
1
x
0
1
2
y
0
1
2
y
–2
–4
–6
(x, y)
(5, 0) (3, 1) (1, 2)
(x, y) (0, –2)
Y
–2
–2 x
y=
x+
2y
=5
(1, –4) (2, –6)
2
Scale : 1 cm = 1 unit
on both the axes
5
4
(–3, 4)
3
(1, 2)
2
(3, 1)
1
(5, 0)
X
-5
-4
-3
-2
0
-1
1
2
3
4
5
X
-1
-2
(0, –2)
-3
(1, –4)
-4
-5
-6
(2, –6)
Y
 x = – 3 and y = 4 is the solution of given simultaneous equations.
1
... 5 ...
(iii)
Paper - IV
Let the number of days required by John alone to complete
the work be x days and Ganesh alone is (x + 10) days
 No. of days requred by ganesh along is (x + 10 ) days.
Also number of days required by both to complete the same
work is 12 days
1
 Work done by John in 1 day =
x
1
 Work done by Ganesh in 1 day = x + 10
1
 Work done by both in 1 day =
12
As per the given condition,
1
1
1
+
=
x x + 10
12
x + 10 + x
1

=
x (x + 10)
12
2x + 10
1

=
x 2 + 10x
12

12 (2x + 10) = 1 (x2 + 10x)

24x + 120 = x2 + 10x

0 = x2 + 10x – 24x – 120
2

x – 14x – 120 = 0

x2 – 20x + 6x – 120 = 0

x (x – 20) + 6 (x – 20) = 0

(x – 20) (x + 6) = 0

x – 20 = 0 or x + 6 = 0

x = 20 or x = – 6
 The numbers of days cannot be negative
 x–6
Hence x = 20
 x + 10 = 20 + 10 = 30
1
1
1
 Ganesh alone worked for 30 days.
(iv)
Let the length of seg AB be x cm and that of seg AP be y cm
 l (BP) = l (AP) = y
[By perpendicular bisector theorem]
As per first condition,
y
=
x+7

–x+y
=
7
.......(i)
As per the second condition,
P
x+y+y
=
38

x + 2y
=
38
......(ii)
cm
y
Adding (i) and (ii),
–x+y
=
7
A
T
x +2y
=
38
x cm
3y
=
45
1
1
1
B
... 6 ...





Paper - IV
45
3
y
=
15
Substituting y = 15 in (ii),
x + 2 (15)
=
38
x + 30
=
38
x
=
38 – 30
x
=
8
l (AB) = 8 cm, l (BP) = l (AP) = 15 cm.
y
=
 The length of sides of ABP are 8 cm, 15 cm and 15 cm.
A.4.
(i)
1
1
Solve the following : (Any One)
Let the speed of bicycle be x km / hr
Speed of wind is 10 km /hr
 Speed of bicycle in the direction of wind = (x + 10) km/hr
Speed of the bicycle against the direction of wind = (x – 10) km/hr
Dis tance
Also, Speed =
Time
Dis tance
 Time = Speed
Time taken by man while riding in the direction of wind
=
 60 

 hrs
 x  10 
Time taken by man while riding against the direction of wind
 60 

 hrs
 x – 10 
As per the given condition,
60
60

x  10 x – 10
1
=





=
8
 1
1 
60 


= 8
 x  10 x – 10 
x – 10  x  10
8
=
(x  10) (x – 10)
60
2x
2
=
x 2 – 100
15
Dividing throughout by 2 we get,
x
1
=
2
x – 100
15
x2 – 100 = 15x
x2 – 15x – 100 = 0
1
1
... 7 ...

x2 – 20x + 5x – 100 =

x (x – 20) + 5 (x – 20) =

(x – 20) (x + 5) =

x – 20 = 0 or

x = 20 or
 The speed of bicycle cannot
 x– 5
Hence x = 20
Paper - IV
0
0
0
x+5=0
x=–5
be negative
 The speed of bicycle is 20 km / hr.
(ii)
Let the time for which the car travels at 60 km/hr be x hours
and at 75 (i.e. 60 + 15) km/hr be y hours
We know that, Distance = Speed × Time
Distance covered by car with the speed of 60 km/hr = 60 × x = 60x km
Distance covered by car with the speed of 75 km/hr = 75 × y = 75y km
As per first condition,
60x + 75y
=
555
15 (4x + 5y)
=
555
555
4x + 5y
=
15
4x + 5y
=
37
.............(i)
As per second condition,
x + y
= 8
............(ii)
Multiplying (ii) by 5, we get,
5x + 5y
= 40
............(iii)
Subtracting (i) from (iii), we get
5x + 5y
=
40
4x + 5y
=
37
–
–
–
x
=
1
1
1
1
1
1
3
 Time taken by the car at the speed of 60 km/hr = 3 hours and
Distance covered with the speed of 60km/hr = 60 × 3 = 180 km.

1
Paper - V
S.S.C.
MAHESH TUTORIALS
Batch : SB
Date :
Test - III
Marks : 30
ALGEBRA – Chapter : 2, 3
Time : 1 hr. 15 min.
Q.1. Solve the following :
4
3 6
–4 2
5 3
2
(i)
Find the value of the following determinants :
(ii)
If the point (3, 2)lies on the graph of the equation 5x + ay = 19, then find
a.
Q.2. Solve the following :
(i)
The sum ‘S’ of the first ‘n’ natural numbers is given by S =
Find ‘n’, if the sum (S) is 276.
9
n (n + 1)
.
2
(ii)
Find the value of k for which the given simultaneous equations have
infinitely many solutions : 4y = kx – 10; 3x = 2y + 5
(iii)
Tinu is younger than Pinky by three years. The product of their ages is
180. Find their ages.
Q.3. Solve the following : (Any Three)
(i)
For doing some work Ganesh takes 10 days more than John. If both
work together they complete the work in 12 days. Find the number of
days if Ganesh worked alone?
12
Paper - V
... 2 ...
(ii)
Solve the following simultaneous equations using graphical method :
3x + 4y + 5 = 0; y = x + 4
(iii)
Divide 40 into two parts such that the sum of their reciprocals is
(iv)
Monthly hostel charges in a college comprises of two parts, one fixed
part for the stay in the hostel and the varying part depending on the
number of days one has taken food in the mess. Ram takes food for 20
days and pays Rs. 1700 as hostel charges and Rahim takes food for 24
days and pays Rs.1900 as hostel charges. Find the fixed charges and
the cost of the food per day.
8
.
75
Q.4. Solve the following : (Any One)
5
(i)
One diagonal of a rhombus is greater than other by 4 cm. If the area of
the rhombus is 96 cm2, find the side of the rhombus.
(ii)
A bus covers a certain distance with uniform speed. If the speed of the
bus would have been increased by 15 km/h, it would have taken two
hours less to cover the same distance and if the speed of the bus would
have been decreased by 5 km/h, it would have taken one hour more to
cover the same distance. Find the distance covered by the bus.
Best Of Luck

Paper - V
MAHESH TUTORIALS
S.S.C.
Batch : SB
Test - III
Marks : 30
ALGEBRA – Chapter : 2, 3
Date :
A.1.
(i)
A.2.
(i)
Time : 1 hr. 15 min.
Solve the following :
3 6 –4 2
5 3
2
=
 3 6  2 –  – 4 2  5 3 
 6 6  –  – 20 6 
=
6 6  20 6
=
26 6
=
(ii)
MODEL ANSWER PAPER
 (3, 2) lies on the graph of the equation 5x + y = 19.
It satisfies the equation,
 Substituting x = 3 and y = 2 in the equation we get,
5 (3) + a (2)
= 19

15 + 2a
= 19

2a
= 19 – 15

2a
= 4
4

a
=
2

a
= 2
Solve the following :
n (n + 1)
S=
2
From the given condition,
n (n + 1)
276 =
2
 552 = n2 + n
 n2 + n – 552 = 0
 n2 + 24n – 23n – 552 = 0
 n (n + 24) (n – 23) = 0
 n + 24 = 0
or n – 23 = 0
 n = – 24
or n = 23
n  – 24 because ‘n’ cannot be negative
 n = 23
1
1
1
1
1
1
1
... 2 ...
(ii)
4y = kx – 10
 – kx + 4y = – 10










(iii)









C om parin g w ith a 1x + b1y = c1 we get, a1 = – k, b1 = 4, c1 = – 10
3x = 2y + 5
3x – 2y = 5
Comparing with a2x + b2y = c2 we get, a2 = 3, b2 = – 2, c2 = 5
a1
–k
a2 =
3
b1
4
–2
b2 = – 2 = 1
c1
–10
–2
=
c2 =
5
1
The simultaneous equations have infinitely many solutions.
a1
b1
c1
=
=
a2
b2
c2
–k
–2
–2
=
=
3
1
1
–k
–2
=
3
1
–k = –6
k
=
Tinu’s age be ‘x’ years
Pinky’s age is (x + 3) years
As per the given condition,
x (x + 3) =
x2 + 3x – 180 =
x2 – 12x + 15x – 180 =
x (x – 12) + 15 (x – 12) =
(x – 12) (x + 15) =
x – 12 = 0 or
x = 12 or
Age cannot be negative.
x 
x =
And x + 3 = 12 + 3 = 15
1
1
1
6
180
0
0
0
0
x + 15 = 0
x = – 15
1
1
– 15
12
 Tinu’s age is 12 years and Pinky’s is 15 years.
A.3.
(i)
Paper - V
Solve the following : (Any Three)
Let the number of days required by John alone to complete
the work be x days and Ganesh alone is (x + 10) days
 No. of days requred by ganesh along is (x + 10 ) days.
1
Paper - V
... 3 ...
Also number of days required by both to complete the same
work is 12 days
1
 Work done by John in 1 day =
x
1
 Work done by Ganesh in 1 day = x + 10
1
 Work done by both in 1 day =
12
As per the given condition,
1
1
+
x x + 10
=
1
12

x + 10 + x
x (x + 10)
=
1
12

2x + 10
x 2 + 10x
=












1
1
1
12
12 (2x + 10) = 1 (x2 + 10x)
24x + 120 = x2 + 10x
0 = x2 + 10x – 24x – 120
2
x – 14x – 120 = 0
x2 – 20x + 6x – 120 = 0
x (x – 20) + 6 (x – 20) = 0
(x – 20) (x + 6) = 0
x – 20 = 0 or x + 6 = 0
x = 20 or x = – 6
The numbers of days cannot be negative
x– 6
Hence x = 20
x + 10 = 20 + 10 = 30
1
 Ganesh alone worked for 30 days.
(ii)
1
3x + 4y + 5 = 0
y=x+4
 3x = –5 – 4y
 x=
– 5 – 4y
3
x
–3
1
5
x
0
1
2
y
1
–2
–5
y
4
5
6
(x, y) (–3, 1) (1, –2) (5, –5)
(x, y)
(0, 4) (1, 5)
(2, 6)
1
Paper - V
... 4 ...
Y
Scale : 1 cm = 1 unit
on both the axes
8

7
6
(2, 6)
(1, 5)
5
4
3x
3
+
4y
+
5
2
=
0
1
(–3, 1)
X
-5
y
=
x
(0, 4)
-4
+
-3
-2
0
-1
4
1
2
3
4
5
X
-1
(1, –2)
-2
-3
-4
(5, –5)
-5
Y
x = – 3 and y = 1 is the solution of given simultaneous equations.
(iii)
Sum of two parts is 40
Let one of the part is x
 The other part is 40 – x
1
1
... 5 ...
As per the given condition,
1
1

=
x 40 – x
40 – x + x

=
x (40 – x)
40

=
40x – x 2

40 (75) =









(iv)
8
75
8
75
8
75
8 (40x – x2)
40 × 75
= 40x – x2
8
375 = 40x – x2
2
x – 40x + 375 = 0
x2 – 25x – 15x + 375 = 0
x (x – 25) – 15 (x – 25) = 0
(x – 25) (x – 15) = 0
x – 25 = 0 or x – 15 = 0
x = 25 or x = 15
If
x = 25 or if x = 15
then 40 – x = 40 – 25 = 15
then 40 – x = 40 – 15 = 25
The two parts are 25 and 15.
Let the fixed charge for stay in hostel be Rs. x and cost of food
per day be Rs. y
Ram’s total expenditure = Rs. (x + 20y)
Rahim’s total expenditure = Rs. (x + 24y)
As per the first given condition,
x + 20y
=
1700
......(i)
As per the second given condition,
x + 24y
=
1900
.....(ii)
Substracting (ii) from (i),
x + 20y
=
1700
x + 24y
=
1900
(–) (–)
(–)
– 4y
=
– 200
– 200

y
=
–4

y
=
50
Substituting y = 50 in (i),
x + 20 (50)
=
1700

x + 1000
=
1700

x
=
1700 – 1000

x
=
700
 Fixed charge for stay in the hostel is Rs. 700 and cost of food
per day is Rs. 50.
Paper - V
1
1
1
1
1
1

... 6 ...
A.4.
(i)
Solve the following : (Any One)
Let the length of smaller diagonal of a rhombus be x cm
 The length of bigger diagonal is (x + 4) cm
As per the given condition,
1

Area of rhombus =
× Diagonal 1 × Diagonal 2
2
As per the given condition,
1

× x × (x + 4) =
96
2

x2 + 4x = 192
2

x + 4x – 192 = 0
2

x – 12x + 16x – 192 = 0
 x (x – 12) + 16 (x – 12) = 0

(x – 12) (x + 16) = 0

x – 12 = 0 or x + 16 = 0

x = 12 or x = – 16
 The length of diagonal of the rhombus cannot be negative
 x  – 16
Hence x = 12
 x + 4 = 12 + 4 = 16
Considering ABCD a rhombus
 l (AC) = 12 cm
 l (BD) = 16 cm
O is the intersection point of diagonal AC and BD
B
A
1
1
l (AO) =
l (AC) =
× 12 = 6 cm
2
2
1
1
l (BO) =
l (BD) =
× 16 = 8 cm
O
2
2
AOB is a right angled triangle
C
D
In right angled AOB,
2
2
2
[By Pythagoras theorem]
[l (AB)] = [l (AO)] + [l (BO)]
2
2
2
 [l (AB)] = (6) + (8)
 [l (AB)]2 = 36 + 64
 [l (AB)]2 = 100
Taking square root on both the sides we get,
l (AB) = 10 cm
 The side of a rhombus is 10 cm.
(ii)
Paper - V
Let the speed of bus be x km/hr. and time taken be y hrs.
Distance
= Speed × Time

Distance
= xy km
According to the first condition,
1
1
1
1
1
Paper - V
... 7 ...



(x + 15) (y – 2)
= xy
x (y – 2) + 15 (y – 2)
= xy
xy – 2x + 15y – 30
= xy
– 2x + 15y
= 30
According to the second condition,
(x – 5) (y + 1)
= xy
x (y + 1) – 5 (y + 1)
= xy
xy + x – 5y – 5
= xy
x – 5y
= 5
Multiplying (ii) by 3 we get,
3x – 15y
= 15
Adding (i) and (iii) we get,
– 2x + 15y
= 30
3x – 15y
= 15
x
= 45
Substituting x = 45 in (ii),
45 – 5y
= 5
– 5y
= 5 – 45
– 5y
= – 40

y
=
– 40
–5


y
Distance
=
=
=
=
8
xy
45 × 8
360






 Distance covered by bus is 360 km.

......(i)
1
......(ii)
1
......(iii)
1
1
1
Paper - VI
S.S.C.
MAHESH TUTORIALS
Batch : SB
Date :
Test - III
Marks : 30
Time : 1 hr. 15 min.
ALGEBRA – Chapter : 2, 3
Q.1. Solve the following :
4
(i)
What is the equation of X - axis? Hence, find the point of intersection of
the graph of the equation x + y = 3 with the X - axis.
(ii)
Find the value of the following determinants :
1.2
0.57
0.03
– 0.23
Q.2. Solve the following :
9
(i)
A natural number is greater than twice its square root by 3. Find the
number.
(ii)
Find the value of p for which the given simultaneous equations have unique
solution : 3x + y = 10; 9x + py = 23
(iii)
In garden there are some rows and columns. The number of trees in a
row is greater than that in each column by 10. Find the number of trees
in each row if the total number of trees are 200.
Q.3. Solve the following : (Any Three)
(i)
A rectangular playground is 420 sq.m. If its length is increased by 7 m
and breadth is decreased by 5 metres, the area remains the same. Find
the length and breadth of the playground ?
12
Paper - VI
... 2 ...
(ii)
Solve the following simultaneous equations using graphical method :
4x = y – 5; y = 2x + 1
(iii)
The sum of the areas of two squares is 400sq.m. If the difference between
their perimeters is 16 m, find the sides of two square.
(iv)
A man starts his job with a certain monthly salary and a fixed increment
every year. If his salary will be Rs. 11000 after 2 years and Rs. 14000
after 4 years of his service. What is his starting salary and what is the
annual increment ?
Q.4. Solve the following : (Any One)
5
(i)
The product of four consecutive positive integers is 840. find the
largest number.
(ii)
Students of a school were made to stand in rows for drill. If 3 students less
were standing in each row, 10 more rows were required and if 5 students
more were standing in each row then the number of rows was reduced by
10. Find the number of students participating in the drill.
Best Of Luck

Paper - VI
MAHESH TUTORIALS
S.S.C.
Batch : SB
Test - III
Marks : 30
ALGEBRA – Chapter : 2, 3
Date :
A.1.
(i)
MODEL ANSWER PAPER
Time : 1 hr. 15 min.
Solve the following :
The equation of X-axis is y = 0
Let the point of intersection of graph x + y = 3 with X-axis be (h, 0)
 (h, 0) lies on the graph, it satisfies the equation
 Substituting x = h and y = 0 in the equation we get,
h+0
= 3

h
= 3
 The line x + y = 3 intersects the X-axis at (3, 0).
1.2
(ii)
=
=
=
A.2.
(i)
1
0.03
0.57 – 0.23
(1.2 × – 0.23) – (0.03 × 0.57)
– 0.276 – 0.0171
– 0.2931
Solve the following :
Let square root of the natural number be x
 The natural number = x2
As per the given condition,
x2 = 2x + 3
2

x – 2x – 3 = 0
2

x – 3x + 1x – 3 = 0

x (x – 3) + 1 (x – 3) = 0

(x – 3) (x + 1) = 0

x – 3 = 0 or x + 1 = 0

x = 3 or x = – 1
 Square root of a natural number cannot be negative

x  –1

x = 3

x2 = 9
 The natural number is 9.
(ii)
1
3x + y = 10
Comparing with a1x + b1y = c1 we get, a1 = 3, b1 = 1, c1 = 10
1
1
1
1
1
... 2 ...
9x + py = 23
Comparing with a2x + b2y = c2 we get, a2 = 9, b2 = p, c2 = 23
a1
3
1
=
=
a2
9
3
b1
1
b2 = p
For the equations to have unique solution.
a1
b1

a2
b2
1
1

 p
3
 p
 3
 The simultaneous equations will have unique solution for all
values of p except 3.
(iii)










Let the number of trees in each column be x
The number of trees in each column is x + 10
As per the given condition,
x (x + 10) = 200
x2 + 10x – 200 = 0
x2 + 20x – 10x – 200 = 0
x (x + 20) – 10 (x + 20) = 0
(x + 20) (x – 10) = 0
x + 20 = 0 or x – 10 = 0
x = –20 or x = 10
The number of trees cannot be negative
x  – 20
Hence
x = 10
x + 10 = 10 + 10 = 20
 The number of trees in each row is 20.
A.3.
(i)
Paper - VI
1
1
1
1
1
1
Solve the following : (Any Three)
Let the length of a rectangular playground be ‘x’ m.
 The area of playground is 420 sq.m.
420
m
x
New length = (x + 7) m
 It’s breadth is
 420

– 5 m
New breadth = 

x
As per the given condition,
1
Paper - VI
... 3 ...
 Area of new rectangle
=
Length × Breadth
New area
=
 420

– 5
(x + 7) 

x

420
=
 420

5
(x + 7) × 
 x


420
=
420 – 5x +









2940
– 35
x
Multiplying throughout by x, we get;
0 = – 5x2 + 2940 – 35x
5x2 + 35x – 2940 = 0
x2 + 7x – 588 = 0
[Dividing throughout by 5]
2
x – 21x + 28x – 588 = 0
x (x – 21) + 28 (x – 21) = 0
(x – 21) (x + 28) = 0
x – 21 = 0 or x + 28 = 0
x = 21 or x = -28
The length of playground cannot be negative.
x  -28
Hence x = 21
And
1
420
420
=
= 20
x
21
 The length of a rectangular playground is 21 m and
its breadth is 20 m.
(ii)
1
1
4x = y – 5
 4x +5 = y
 y = 4x + 5
x
0
–1
–2
y
5
1
–3
(x, y)
(0, 5) (–1, 1) (–2, –3)
y = 2x + 1
x
0
1
2
y
1
3
5
(x, y)
(0, 1) (1, 3) (2, 5)
1
Paper - VI
... 4 ...
Y
Scale : 1 cm = 1 unit
on both the axes
6
(0, 5)

5
2
(2, 5)
4
(1, 3)
3
2
(–1, 1)
X
-5
-4
-3
-2
1
(0, 1)
0
-1
1
2
3
4
5
X
-1
-2
(–2, –3)
-3
-4
y=
2x
+ 1
-5
-6
-7
-8
Y
 x = – 2 and y = – 3 is the solution of given simultaneous equations.
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... 5 ...
(iii)
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
Paper - VI
Difference between the perimeters of two squares is 16 m
The difference between the sides of the same squares is 4 m
Let the side of smaller square be x cm
The side of bigger square is (x + 4) cm
Area of square = (side)2
As per the given condition,
x2 + (x + 4)2 = 400
x2 + x2 + 8x + 16 – 400 = 0
2x2 + 8x – 384 = 0
Dividing throughout by 2 we get,
x2 + 4x – 192 = 0
x2 – 12x + 16x – 192 = 0
x (x – 12) + 16 (x – 12) = 0
(x – 12) (x + 16) = 0
x – 12 = 0 or x + 16 = 0
x = 12 or x = – 16
The side of square cannot be negative
x  – 16
Hence x = 12
x + 4 = 12 + 4 = 16
 The side of smaller square is 12 m and bigger square is 16 m.
(iv)
As
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Let starting salary of man be Rs. x and the fixed annual
increment be Rs. y.
per the first given condition,
x + 2y
=
11000
......(i)
As per the second given condition,
x + 4y
=
14000
......(ii)
Subtracting (ii) from (i),
x + 2y
=
11000
x + 4y
=
14000
(–) (–)
(–)
– 2y
= – 3000
– 3000
y
=
–2
y
=
1500
Substituting y = 1500 in (i),
x + 2 (1500)
=
11000
x + 3000
=
11000
x
=
11000 – 3000
x
=
8000
 The starting salary of man is Rs. 8000 and his fixed annual
increment is Rs. 1500.
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... 6 ...
A.4.
(i)
(ii)
Paper - VI
Solve the following : (Any One)
Let the four consecutive positive integers be x, x + 1, x + 2 and x + 3
As per the given condition,
x × (x + 1) × (x + 2) × (x + 3) = 840

x (x + 3) × (x + 1) (x + 2) = 840
2

(x + 3x) × (x2 + 2x + x + 2) = 840

(x2 + 3x) (x2 + 3x + 2) = 840
Substituting x2 + 3x = m we get,
m (m + 2) = 840
2

m + 2m – 840 = 0
2

m + 30m – 28m – 840 = 0

m (m + 30) – 28 (m + 30) = 0

(m + 30) (m – 28) = 0

m + 30 = 0 or m – 28 = 0

m = – 30 or m = 28
Resubstituting m = x2 + 3x we get,
x2 + 3x = – 30 ......(i) or x2 + 3x = 28 ......(ii)
From (i),
x2 + 3x = – 30
2

x + 3x + 30 = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = 3, c = 30
b2 – 4ac = (3)2 – 4 (1) (30)
= 9 – 120
= – 111

b2 – 4ac < 0
 The roots of the above quadratic equation are not real.
Hence not considered.
From (ii),
x2 + 3x = 28
2

x + 3x – 28 = 0

x2 + 7x – 4x – 28 = 0

x (x + 7) – 4 (x + 7) = 0

(x + 7) (x – 4) = 0

x + 7 = 0 or x – 4 = 0

x = – 7 or x = 4
 x is positive integer

x  –7
Hence x = 4
 x+3=4+3=7
 The largest required number is 7.
Let the no. of students standing in each row be x and let no. of rows
be y.
 Total no. of students participating in the drill = xy
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Paper - VI
... 7 ...
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As per the first given condition,
(x – 3) (y + 10)
=
xy
x (y + 10) – 3 (y + 10)
=
xy
xy + 10x – 3y – 30
=
xy
10x – 3y
=
30
As per the second given condition,
(x + 5) (y – 10)
=
xy
x (y – 10) + 5 (y – 10) =
xy
xy – 10x + 5y – 50
=
xy
– 10x + 5y
=
50
Adding (i) and (ii),
10x – 3y
=
30
– 10x + 5y
=
50
2y
=
80
80
y
=
2
y
=
40
Substituting y = 40 in (i),
10x – 3 (40)
=
30
10x – 120
=
30
10x
=
30 + 120
10x
=
150
150
x
=
10
x
=
15
xy = 15 × 40 = 600
.......(i)
1
......(ii)
1
 600 students were participating in the drill.
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1
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