### Practice Aptitude Questions With Solutions for SSC CGL / Railway Exam

Practice Aptitude Questions With Solutions Railway / SSC Exams Set-11:
Dear Readers, Important Aptitude Questions for Railway / SSC / FCI Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

1).The factors of x2 – 64 are
a)    ∓ 8
b)    ∓ 6
c)    ∓ 4
d)    ∓ 2

2).One of the factors of x2 + 2x – 35 is
a)    (x – 7)
b)    (x – 5)
c)    (x + 5)
d)    None

3).If 5 – x is a one factor of -3x2 + 13x + 10, the other factor is
a)    5 + x
b)    2x + 3
c)    3x + 2
d)    -5 + x

4).If (x2 – 10x + 21) / (x2 – 7x + 12) = (x – 7) / ?
a)    x - 4
b)    x - 3
c)    x + 3
d)    x + 4

5).The factors of x2 + x - 56
a)    (x + 8), (x – 7)
b)    (x - 8), (x – 7)
c)    (x + 8), (x + 7)
d)    (x - 8), (x + 7)

6).For what value of a 6x2 + ax + 12 will have one factor 3x + 4.
a)    9
b)    17
c)    8
d)    1

7).If (x – 4) is a common factor to both Nr and Dr of (x2 – 16) / [(x – 4) (x + 2)] the other factor is
a)    (x + 4) / (x – 2)
b)    (x - 4) / (x – 2)
c)    (x + 4) / (x + 2)
d)    (x - 2) / (x + 4)

8).Simplify : (2x2 + 3xy + y2) / (2x2 - 3xy + y2)
a)    (x + y) / (x – y)
b)    (2x + y) / (2x – y)
c)    [(2x – y) (x – y)] / [(x + y) (2x – y)]
d)    [(2x + y) (x + y)] / [(x - y) (2x – y)]

9).For what value of a 2x2 + x – 3 becomes zero
a)    x = 2
b)    x = 3
c)    x = 1
d)    x = 1 / 4

10).If (x + 2) is one factor of 4x2 + 13x + 10, the other factor is
a)    4x + 5
b)    5x + 4
c)    2x + 5
d)    5x + 2

1). a) 2). b) 3). c) 4). a) 5). a) 6). b) 7). c) 8). d) 9). b) 10). d)

Solution:
1). x2 – 64  = (x + 8) (x – 8)
Hence the factors are +8, - 8
2). Factorise : x2 + 2x – 35
= x2 + 7x – 5x – 35
= x(x + 7) - 5(x + 7)
=(x – 5) (x + 7)
Factros are (x – 5) (x + 7)
3). Divide – 3x2 + 13x + 10 by (5 – x)
5 – x ) – 3x2 + 13x + 10 ( 3x + 2
– 3x2 + 15x
-2x + 10
-2x + 10
The other factor is 3x + 2.
4). Simplify : (x2 – 10x + 21) / (x2 – 7x + 12) = (x – 7) / ?
Nr = (x2 – 10x + 21) = (x – 7) (x – 3)
Dr = (x2 – 7x + 12) = (x – 4) (x – 3)
(x2 – 10x + 21) / (x2 – 7x + 12) = [(x – 7) (x – 3)] / [(x – 4) (x – 3)]
= (x – 7) / (x – 4)
5). Factors of x2 + x – 56
By factorizing x2 + x – 56, we get
=  x2 + 8x – 7x – 56
= x (x + 8) – 7(x + 8)
= (x + 8) (x – 7)
The factors are (x + 8) and (x – 7)
6). Divide 6x2 + ax + 12 by 3x + 4
3x + 4 ) 6x2 + ax + 12 ( 2x + 3
6x2 + 8x
(a – 8)x + 12
9x + 12
∴ a – 8 = 9
∴ a = 9 + 8 = 17
7). Simplify : (x2 – 16) / [(x – 4) (x + 2)]
Nr : (x2 – 16) = (x – 4) (x + 4)
∴(x2 – 16) / [(x – 4) (x + 2)] = [(x – 4) (x + 4)] / [(x – 4) (x + 2)]
= (x + 4) / (x + 2)
8). Simplify : (2x2 + 3xy + y2) / (2x2 - 3xy + y2)
(2x2 + 3xy + y2) = 2x2 + 2xy + xy + y2
= 2x (x + y) + y (x + y)
= (2x + y) (x + y)
(2x2 - 3xy + y2) = 2x2 - 2xy - xy + y2
= 2x (x - y) + y (x - y)
= (2x - y) (x - y)
∴(2x2 + 3xy + y2) / (2x2 - 3xy + y2) = [(2x + y) (x + y)] / [(2x - y) (x - y)]
9). Evaluate (125)1 / 3
(125)1 / 3 = (53)1 / 3
= 53×1 / 3
= 51 = 5