Paper - I
S.S.C.
MAHESH TUTORIALS
Batch : SB
Date :
Test - II
Marks : 30
ALGEBRA – Chapter : 1, 2, 3
Time : 1 hr. 15 min.
Q.1. Solve the following :
(i)
The 9th term of an G.P. 3, 6, 12, 24 is ............. .
(ii)
Determine the nature of roots of the following equation from its
3
discriminant : x2 + 3 2 x – 8 = 0
(iii)
Examine whether the point (2, 5) lies on the graph of the equation
3x – y = 1.
Q.2. Solve the following :
(i)
Find S10 if a = 6 and d = 3.
(ii)
Form the quadratic equation if its roots are 0 and – 4.
(iii)
What is the equation of Y - axis? Hence, find the point of intersection
of Y - axis. and the line y = 3x + 2.
Q.3. Solve the following :
(i)
Without plotting the graphs, find the point of intersection of the
lines 2x + 5y = 13 and 4x – 9y = 7.
6
9
Paper - I
... 2 ...
(ii)
Find the 4th and 9th terms of the G.P. with first term 4 and common
ratio 2.
(iii)
Solve the following quadratic equation by completing square :
p2 – 10p + 5 = 0
Q.4. Solve the following :
12
(i)
If S6 = 126 and S3 = 14 then find a and r.
(ii)
If the sum of the roots of the quadratic is 3 and sum of their cubes
is 63, find the quadratic equation.
(iii)
Solve the following simultaneous equations using graphical method :
4x = y – 5; y = 2x + 1
Best Of Luck

Paper - I
S.S.C.
MAHESH TUTORIALS
Batch : SB
Test - II
Marks : 30
ALGEBRA – Chapter : 1, 2, 3
Date :
MODEL ANSWER PAPER
Time : 1 hr. 15 min.
A.1. Solve the following :
(i)
The 9th term of an G.P. 3, 6, 12, 24 is 768.
(ii)
(iii)
1
x2 + 3 2 x – 8 = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = 3 2 , c = – 8
 = b2 – 4ac
= (3 2 )2 – 4 (1) (– 8)
= 9 × 2 + 32
= 18 + 32
= 50
  > 0
Hence roots of the quadratic equation are real and unequal.
1
Substituting x = 2 and y = 5 in the L.H.S. of the equation 3x – y =1
L.H.S. = 3x – y
= 3 (2) – 5
= 6–5
= 1
= R.H.S.
 x = 2 and y = 5 satisfies the equation 3x – y = 1
Hence (2, 5) lies on the graph of the equation 3x – y = 1.
1
A.2. Solve the following :
(i)
For an A.P. a = 6, d = 3
n
[2a + (n – 1) d]
Sn =
2
10
 S 10 =
[2a + (10 – 1) 1]
2
 S 10 = 5 [2 (6) + 9 (3)]
 S 10 = 5 (12 + 27)
 S 10 = 5 (39)
 S 10 =
195
1
1
Paper - I
... 2 ...
(ii)
(iii)
The roots of the quadratic equation are 0 and – 4.
Let  = 0 and  = – 4
  +  = 0 + (– 4) = – 4 and . = 0 × – 4 = 0
We know that,
x2 – ( + )x + . = 0
 x2 – (– 4)x + 0 = 0
 x2 + 4x = 0
1
 The required quadratic equation is x2 + 4x = 0
1
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
A.3. Solve the following :
(i)
2x + 5y = 13
......(i)
4x – 9y = 7
.....(ii)
D
=
2 5
4 –9
= (2 × – 9) – (5 × 4) = – 18 – 20
= – 38
Dx =
13 5
= (13 × – 9) – (5 × 7) = – 117 – 35
7 –9
= – 152
Dy =
2 13
4 7
= – 38
= (2 × 7) – (13 × 4)
= 14 – 52
1
By Cramer’s rule,
x
=
y
=
–152
Dx
= – 38 = 4
D
– 38
= – 38 = 1
D
Dy
 The point for intersection of the lines 2x + 5y = 13 and
4x – 9y = 7 is (4, 1).
1
1
Paper - I
... 3 ...
(ii)








For the G.P.
The first term (a) = 4
Common ratio (r) = 2
t n = arn – 1
t 4 = ar4 – 1
t 4 = ar 3
t 4 = 4 (2)3
t 4 = 4 (8)
t 4 = 32
t 9 = ar9 – 1
t 9 = ar 8
t 9 = 4 (2)8
t 9 = 1024
1
1
 The fourth and ninth term of the G.P. are 32 and
1024 respectively.
(iii)
p2 – 10p + 5 = 0
 p2 – 10p = – 5
Third term
1
..... (i)
1

=  × coefficient of p
2

1

=  × –10
2

2
2
1
= (– 5)2
= 25
Adding 25 to both the sides of equation (i), we get;
p2 – 10p + 25 = – 5 + 25
 (p – 5)2
= 20
 p–5
= +
 p–5
= + 2 5
 p
= 52 5
 p= 52 5
or p = 5 – 2 5
20 [Taking square roots on both sides]
 5  2 5 and 5 – 2 5 are the roots of the given quadratic equation.
1
1
... 4 ...
Paper - I
A.4. Solve the following :
(i)
For a G.P.
Sn
=
a (1 – r n )
1– r
 S6
=
a (1 – r 6 )
1– r
But,
S 6 = 126
[Given]
a (1 – r 6 )

= 126
1– r
.....(i)
1
Similarly,
a (1 – r 3 )
S3 =
1– r
But,
S 3 = 14
[Given]
3

a (1 – r )
= 14
1– r
.....(ii)
1
Dividing (i) by (ii),
 a (1 – r 6 )   a (1 – r3 ) 
126


 =
1
–
r
1
–
r
14

 


a (1 – r 6 )
1– r

1– r
a (1 – r 3 )
= 9

1 – r6
1 – r3
= 9

12 – (r 3 )2
1 – r3
= 9

(1  r 3 ) (1 – r 3 )
1 – r3
= 9



1 + r3 = 9
r3 = 9 – 1
r3 = 8
Taking cube roots on both sides
r = 2
Substituting r = 2 in (ii), we get,
1
... 5 ...
a (1 – 23 )
1– 2
= 14
a (1 – 8)
–1
= 14




 a = 2, r = 2.
(ii)
a (– 7)
–1
7a
a
= 14
= 14
= 2
1
Let  and  be the roots of a quadratic equation.
  +  = 3 and3 +  3 = 63 [Given]
We know that,
.......(i)
x2 – ( + )x + . = 0
Also, 3 +  3
= ( + )3 – 3 . ( + )

63
= (3)3 – 3 . (3)
[  +  = 3 and 3 + 3 = 63]

63
= 27 – 9 .

9 .
= 27 – 63

9 .
= – 36
– 36

.
=
9

.
= –4

x2 – ( + )x + .
= 0
[From (i)]
2

x – 3x + (– 4)
= 0
[  +  = 3 and . = – 4]
2

x – 3x – 4
= 0
 The required quadratic equation is x2 – 3x – 4 = 0.
(iii)
Paper - I
1
1
1
1
4x = y – 5
y = 2x + 1
 4x +5 = y
 y = 4x + 5
x
0
–1
–2
y
5
1
–3
(x, y)
(0, 5) (–1, 1)
(–2, –3)
x
0
1
2
y
1
3
5
(x, y)
(0, 1)
(1, 3)
(2, 5)
1
Paper - I
... 6 ...
Y
(0, 5)
Scale : 1 cm = 1 unit
on both the axes
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
Y
 x = – 2 and y = – 3 is the solution of given simultaneous equations.

1