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
Answers:
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
Answer: a)
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)
Answer:
b)
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.
Answer:
c)
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)
Answer:
a)
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)
Answer:
a)
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
Answer:
b)
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)
Answer:
c)
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)]
Answer:
d)
9). Evaluate (125)1
/ 3
(125)1
/ 3 = (53)1 / 3
= 53×1 / 3
= 51
= 5
Answer:
b)
10). (81)-1
/ 4 = ?
(81)-1
/ 4 = (34)-1 /
4
= 34×-1 / 4
= 3-1
= 1 / 3
Answer:
d)
