### Practice Aptitude Questions (Algebra) With Solutions

Practice Aptitude Questions (Algebra) With Solutions Set-61:
Dear Readers, Important Aptitude Questions for SSC/FCI Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

1).If a + [1 / (a – 2) ] = 4, then the value of (a – 2)2 + [ 1 / (a – 2) ]2 is
a)    0
b)    2
c)    - 2
d)    4

2).If a + b + c = 2s, then
[ (s – a)2 + (s – b)2 + (s – c)2 + s2 ] / (a2 + b2 + c2 ) is equal to
a)    a2 + a2 + a2
b)    0
c)    1
d)    2

3).If xy (x + y) = 1 the, the value of [1 / (x3 y3) ] - (x3 - y3) is
a)    3
b)    -3
c)    1
d)    -1

4).If (a3 – b3 – c3 ) = 0 then the value of a9 – b9 – c9 -  3a3  b3  c3
a)    1
b)    2
c)    0
d)    -1

5).The minimum value of (x – 2)(x – 9) is
a)    – (11 / 4)
b)    49 / 4
c)    0
d)    – (49 / 4)

6).If x + y + z = 6 and (x2 + y2 + z2 ) = 20 then the value of x3 + y3 + z3 – 3xyz is
a)    64
b)    70
c)    72
d)    76

7).The third proportional to [ (x / y) + (x / y) ] and √(x2 + y2) is
a)    xy
b)    √xy
c)    3√xy
d)    4√xy

8).If a  : b = 2 : 3, b : c = 4 : 5 and c : d = 6 : 7, then a : d =
a)    12 : 35
b)    24 : 35
c)    16 : 35
d)    24 : 25

9).The value of √(3√0.000729) is
a)    0.03
b)    0.09
c)    0.9
d)    0.3

10).If the average of x and 1 / x be 1, then the value of x10 + (1 / x10)
a)    -2
b)    2
c)    0
d)    1

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

Solution:
1). a + 1 ( a – 2) = 4
= ( a – 2) + (1 / ( a – 2) ) = 4 – 2 = 2
On squaring
= (a – 2)2 + (1 / ( a – 2)2 ) + 2 = 4
= (a – 2)2 + (1 / ( a – 2)2 ) = 2

2). [ (s – a)2 + (s – b)2 + (s – c)2 + s2 ] / (a2 + b2 + c2 )
= [ s2 + 2sa + a2 + s2 + b2 – 2sb + s2 – 2sc + c2 + s2 ] / (a2 + b2 + c2 )
= [ 4s2 + a2 + b2 + c2 – 2s (a + b + c) ] / (a2 + b2 + c2 )
= [ 4s2 + a2 + b2 + c2 – 4s2 ] / (a2 + b2 + c2 ) = 1

3). xy (x + y) = 1
=> x + y = 1 / xy
Cubing both sides,
x3 +  y3 + 3xy (x + y) = 1 / x3  y3
= x3 +  y3 + 3 × 1 = 1 / x3  y3
= (1 / x3  y3 ) - x3 - x3 = 3

4). If a + b + c =0, then a3 + b3 + c3 – 3abc = 0

5). Expression = (x – 2)(x – 9)
= x2 – 11x + 18 = ax2 + bx + c
Minimum Value = (4ac – b2) / 4a
= (4 × 1 × 18 – 121) / 4 = -(49 / 4)

6). x + y + z = 6
On squaring,
x2 + y2 + z2 + 2xy + 2zy + 2zx = 36
ð   20 + 2 (xy + yz + zx) = 36
ð  xy  + yz + zx = 8
∴x3 + y3 + z3 – 3xyz
= (x + y + z) (x2 + y2 + z2 - xy - zy – zx)
= 6 (20 – 8) = 72

7). Third proportional of a and b
= b2 / a = {[ ( √x2 + x2 ) ]2 } / [ (x / y) + (y / x) ]
= (x2 + y2) / [(x2 + y2) / xy ] = xy

8). (a / b) = 2 / 3
(b / c) = 4 / 5
(c / d) = 6 / 7
⇒(a / b) ×(b / c) × (c / d)  = (2 / 3) × (4 / 5) × (6 / 7)
= a / d = 24 / 35

9).3√0.000729
= √3√(0.09)3  = √0.3

10). [ x + (1 / x) ] / 2 = 1
x + (1 / x) = 2
x2 + 1 = 2x
x2 - 2x + 1 = 0
(x – 1)2 = 0
x = 1
x10 + (1 / x10) = 1 + 1 = 2