Paper - II
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 7th term of G.P. 2, – 6, 18 is ............. .
(ii)
Determine the nature of roots of the following equation from its
discriminant 2y2 – 7y – 3 = 0.
(iii)
Write the co-ordinates of the point of intersection of X-axis and
Y-axis.
Q.2. Solve the following :
(i)
Write down the first five terms of the geometric progression which
has first term 1 and common ratio 4.
(ii)
Form the quadratic equation whose roots are 3 and 10.
(iii)
If the point (3, 2)lies on the graph of the equation 5x + ay = 19, then
find a.
Q.3. Solve the following :
(i)
Without drawing the graphs, show that the following equations are
of concurrent lines : y = 5x – 3; y = 4 – 2x; 2x + 3y = 8
3
6
9
Paper - II
... 2 ...
(ii)
If in a G.P. r = 2 and t8 = 64 then find a and S6.
(iii)
Solve the following quadratic equation by using formula :
7p2 – 5p – 2 = 0
Q.4. Solve the following :
12
(i)
Babubhai borrows Rs. 4000 and agrees to repay with a total interest
of Rs. 500. in 10 instalments, each instalment being less that the
preceding instalment by Rs. 10. What should be the first and the
last instalment ?
(ii)
If  +  = 5 and 3 +  3 = 35, find a quadratic equation whose roots
are  and .
(iii)
Solve the following simultaneous equations :
1
1
1 1
1
1
+
=
+
=
;
3x 5y 15 2x 3y 12
Best Of Luck

Paper - II
MAHESH TUTORIALS
S.S.C.
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 7th term of G.P. 2, – 6, 18 is 1458.
(ii)
(iii)
2y2 – 7y – 3 = 0
Comparing with ay2 + by + c = 0 we have a = 2, b = – 7, c = – 3
 = b2 – 4ac
= (– 7)2 – 4 (2) (– 3)
= 49 + 24
= 73
  > 0
Hence roots of the quadratic equation are real and unequal.
1
The co-ordinates of point of intersection of X-axis and Y-axis is
(0, 0).
1
A.2. Solve the following :
(i)
For the G.P.
The first term (a) = 1
Common ratio (r) = 4
t1 = a
= 1
t 2 = ar = 1 × 4
t 3 = ar 2 = 1 × (4)2
t 4 = ar 3 = 1 × (4)3
t 5 = ar 4 = 1 × (4)4
=
=
=
=
4
16
64
256
 The first five terms of the G.P. are 1, 4, 16, 64 and 256.
(ii)
1
The roots of the quadratic equation are 3 and 10
Let  = 3 and  = 10
  +  = 3 + 10 = 13
 .  = 3 × 10 = 30
We know that,
x2 – ( + ) x + . = 0
 x2 – 13x + 30 = 0
 The required quadratic equation is x2 – 13x + 30 = 0
1
1
1
1
Paper - II
... 2 ...
(iii)
 (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
 x
=
1
= 2
A.3. Solve the following :
(i)
y
= 5x – 3
– 5x + y = – 3
y
= 4 – 2x
 2x + y
= 4
2x + 3y = 8
Solving (i) and (ii),
Substracting (ii) from (i),
– 5x + y = – 3
2x + y = 4
(–) (–)
(–)
–7x
= –7
1
........(i)
........(ii)
.......(iii)
–7
–7
 x
= 1
Substituting x = 1 in (ii),
2 (1) + y = 4
 2+y
= 4
 y
=
4–2
 y
= 2
Substituting x = 1 and y = 2 in L.H.S. of (iii),
L.H.S.
= 2x + 3y
= 2 (1) + 3 (2)
= 2+6
= 8
= R.H.S.
 (1, 2) is the solution for (iii)also.
 (1, 2) is a common solution for all the three equations
 (1, 2) is the common point for all the three lines.
1
 The equations of are concurrent lines.
1
1
... 3 ...
(ii)
For a
r
t8
tn
t8
64
64
64
64
128
 a
Sn
 S6





(iii)
Paper - II
G.P.
= 2
= 64
= arn – 1
= ar8 – 1
= ar 7
= a (2)7
= a (128)
1
= a
1
2
a (1 – r n )
=
1– r
=
=

a 1 – r6

1– r
1
(1 – 26 )
S6
= 2
1– 2
1
(1 – 64)
S6
= 2
–1
1
(– 63)
S6
= 2
–1
63
S6
=
2
1
63
a=
and S6 =
.
2
2
1
1
 7p2 – 5p – 2 = 0
Comparing with ap2 + bp + c = 0 we have a = 7, b = – 5, c = – 2
= (– 5)2 – 4 (7) (– 2)
b2 – 4ac
= 25 + 56
= 81
p
=
=
=
–b 
1
2
b – 4ac
2a
– (– 5)  81
2 (7)
59
14
1
... 4 ...
 p=
59
14
or p =
 p=
14
14
or p =
5–9
14
–4
14
–2
or p =
7
 p=1
 1 and
–2
are the roots of the given quadratic equation.
7
A.4. Solve the following :
(i)
Total money repaid by Babubhai in 10 instalments
= (S 10 )
= 4000 + 500
= Rs. 4500
No. of instalments (n) = 10
Difference between two consecutive instalments (d) = – 10
First instalment = (a) = ?
Last instalment (t10) = ?
n
Sn
=
[2a + (n – 1) d]
2
10

S 10
=
[2a + (10 – 1) d]
2

4500
= 5 [2a + 9 (– 10)]









4500
5
900
900 + 90
990
990
2
a
tn
t 10
t 10
t 10
t 10
1
1
1
= 2a – 90
= 2a – 90
= 2a
= 2a
= a
=
=
=
=
=
=
495
a + (n – 1) d
a + (10 – 1) d
495 + 9 (– 10)
495 – 90
405
 First instalment is Rs. 495 and last instalment is Rs.405.
(ii)
Paper - II
 and  are the roots of a quadratic equation
 + =5
[Given]
3
3
 +  = 35
1
1
Paper - II
... 5 ...
We know that,
x2 – ( + ) x +  . 
Also, 3 +  3

35



35
15 .
15 .

 . 



 . 
x2 – ( + ) x + .
x2 – 5x + 6
= 0
......(i)
3
= ( + ) – 3  .  ( + )
= (5)3 – 3 . (5)
[  +  = 5 and 3 +  3 = 35]
= 125 – 15 .
= 125 – 35
= 90
90
=
15
= 6
= 0
[From (i)]
= 0
 The required quadratic equation is x2 – 5x + 6 = 0.
1
1
1
1
1
1
1
+
=
3x 5y 15
(iii)
Multiplying throught by 15 we get,
 1 
 1 
 1 
15 
 + 15  5y  = 15 

 3x 
 15 



3
5
+ y
x
= 1
........(i)
1
1
1
+ 3y
=
2x
12
Multipying throught by 12,
 1 
 1 
 1 
12 
 + 12  3y  = 12  
 2x 
 12 


4
6

+ y
= 1
.........(ii)
x
1
1
Substituting
= a and y = b in (i) and (ii),
x
5a + 3b
= 1
........(iii)
6a + 4b
= 1
........(iv)
Multiplying (iii) bt 4,
20a + 12b = 4
.........(v)
Multiplying (iv) by 3
18a + 12b = 3
........(vi)
1
... 6 ...
Subtracting
20a + 12b=
18a + 12b=
(–)
(–)
2a
=
 a
=
(vi) from (v)
4
3
(–)
1
1
2
1
Substituting a =
1
6   + 4b
2

3 + 4b

4b

4b


= 1
= 1– 3
= –2
b =
–2
4
–1
2
Resubstituting the values of a and b,
b =
1
1
x
1
1
=
2
x
x = 2
b =

1
in (iv),
2
= 1
a =

Paper - II
1
y
1
–1
= y
2
y = –2
 x = 2 and y = –2 is the solution of given simultaneous equations.

1