Practice
Aptitude Questions with Solution for SSC/FCI Exam Set-10:
The list
of practice Aptitude Questions with Solutions for SSC / FCI Exams were given
here, candidates those who are preparing for the examination can use this
practice questions.
1).If (x2+7x+12) / (x2-9)
= (x+4) / A, then value of A is
a)x+4
b)x+3
c)x+4
d)x-3
Answer: d)
Explanation:
Simplify (x2+7x+12) / (x2-9) = (x+4) / A
Factors of x2+7x+12 are (x+4) (x+3)
Factors of x2-9 are (x+3) (x-3)
∴ (x2+7x+12) / (x2-9) = (x+4)(x+3) / (x+3)(x-3)
= (x+4) / (x-3)
2).For what value of x, (x2+11x+30)
/ (x2-25) becomes zero.
5).If (x+1) is a factor of x3+6x2+11x+6,
the other factors are
a)(x-2)(x-3)
b)(x+3)(x-2)
c)(x+2)(x+3)
d)(x+1)(x+6)
Answer: c)
Explanation:
(x+1) is a factor of x3+6x2+11x+6
∴f(-1) = (-1)3 + 6(-1)2 + 11(-1) + 6
→ -1+6-11+6 =0
∴-1 1 6 11 6
0 -1 -5 -6
-------------------------------------
+1 5 6 |___ 0
∴The quotient is of x2+5x+6
Factorise x2+5x+6
∴x2+5x+6= (x+3)(x+2)
The other factors are (x+2) (x+3)
6).Any one of the factors of x3-6x2+11x-6,
is
a)(x+1)
b)(x+2)
c)(x+3)
d)(x-1)
Answer: d)
Explanation:
Factorise of x3-6x2+11x-6
To find the factors, first we have to see whether (x-1) is a factor
f(x) = x3-6x2+11x-6
f(1)= 13-6 ×12+11 ×1-6
=1-6+11×1-6
=1-6+11-6 =12-12 =0
∴(x-1) is a factor.
By division method
1 1 -6 11 -6
0 1 -5 6
--------------------------------
1 -5 6 0
∴The quotient is x2-5x+6
∴The factor of x2-5x+6 are (x-2)(x-3)
Hence the factors of x3-6x2+11x-6 are (x-1),(x-2) and (x-3)
7).If the difference between the two numbers
is 8 and the sum of their squares is 274 the two numbers are
a)7,15
b)8,16
c)-7,15
d)-15,7
Answer: a)
Explanation:
Let x be the one number and (x+8) be the other number
∴x2+(x+8)2 =274
x2+x2+64+16x =274
2x2+16x+64-274 =0
2x2+16x-210 =0
→x2+8x-105 =0
→x2+15x-7x=105 =0
→x(x+15)-7(x+15) =0
(x+15) (x-7) =0
∴x+15 =0 →x=-15
x-7 =0 →x=7
Let x=7, difference is 8
∴The other Number is 8
∴The Numbers are 7,15
8).For what values of x, x2-x-6 = 0
a)+2
b)-3
c)3
d)4
Answer: c)
Explanation:
x2- x2-x-6 = 0
x2-x-6 = (x-3)(x+2)
∴(x-3)(x+2) = 0
x-3 =0→x= 3
x+2 =0→x= -2
∴When x = 3 and x = -2,
∴ x2-x-6=0
9).For what value of x, (x2-3x+2) /
(x2-5x+6) becomes zero.