# SBI Clerk Prelims 2016- Practice Aptitude Questions (Simplification)

SBI Clerk Prelims 2016- Practice Aptitude Questions (Simplification) Set-49:
Dear Readers, Important Practice Aptitude Questions with solution for Upcoming SBI Clerk Exam, candidates those who are preparing for those exams can use this practice questions.

1).The index form of9โ(4/5)3 is
a)  (4/5)1/3
b)  (4/5)3
c)  (4/5)1/2
d)  (4/5)1/27
e)  (4/5)

2).The radical form of (13/25)3/4 is
a)  3โ(13/25)4
b)  4โ(13/25)3
c)  4โ(25/13)3
d)  3โ(25/13)4
e)  None of these

3).The value of 5/(121)-1/2 is
a)  -55
b)  1/55
c)  โ(1/55)
d)  55
e)  60

4).The value of (512)-3/9 is
a)  1/4
b)  8
c)  1/8
d)  -(1/8)
e)  โ(3/7)

5).The value of 3 ร 9-(3/2) ร 91/2 is
a)  1/3
b)  3
c)  27
d)  -(1/3)
e)  None of these

6).The value of(2162/3)1/2 is
a)  3
b)  9
c)  12
d)  6
e)  None of these

7).The value of 27-1/3 ร [(27)2/3/ (27)1/3] is
a)  4
b)  3
c)  2
d)  1
e)  None of these

8).The value of (6.25)-1/2 is
a)  0.25
b)  25
c)  1/2.5
d)  2.5
e)  1.5

9).The value of (โ63 ร โ7) / (3โ27) is
a)  7
b)  9
c)  21
d)  18
e)  None of these

10).If โ3 = 1.732, then the value of (โ3 + 1) / (โ3 โ 1) is
a)  3.732
b)  1/3.732
c)  0.732
d)  2.732
e)  None of these

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

Solution:

1).9โ(4/5)3 = (4/5)3/9 = (4/5)1/3

2).The radical form of (13/25)3/4 =4โ(13/25)3

3).5 / (121)-1/2 = 5 ร (121)1/2= 5 ร (112)1/2
= 5 ร 11 = 55

4).(512)-(3/9) = (29)-(3 / 9) = 29 ร (- 3 / 9)
= 2-3 = (1 / 2)3 = (1/2) ร (1/2) ร (1/2) = (1/8)

5).3 ร 9-(3 / 2) ร 91/2 = 3 ร [32 ร (- 3 / 2)] ร ( 32 ร(1 / 2))
= 3 ร (3)-3 ร 3 = 3 ร (1/3)3 ร 3
= 3 ร (1 / 27) ร 3 = 1/3

6).(2162 / 3 )1/2 = (63 ร (2 / 3))1 / 2= 62 ร (1 / 2 ) = 61= 6

7).271 / 3 ร (272/3/ 271/3) = 33 ร – (1/3)ร [(33)2/3 / (33)1/3]
= 3-1 ร (32/ 3)
= (1/3) ร ( 9 / 3 ) = (1 / 3) ร 3 = 1

8).(625)-(1/2) = (625 / 100)-1/2 = (25 / 4)-1/2
= (4 / 25)1/2 = (22 / 52)1/2 = (2 / 5)2 ร ยฝ
= (2 / 5) or (1 / 2.5)

9).(โ63 ร โ7) /3โ27 = โ(63 ร 7) /3โ33 = โ(441) / 33 ร1/3
= (21 / 3) = 7

10).(โ3 + 1) / (โ3 – 1)
Rationalising, we have
= [(โ3 + 1) / (โ3 – 1)] ร [(โ3 + 1) / (โ3 + 1)] = (โ3 + 1)2 / [(โ3)2โ (1)2 ]
[(x + y)2 = x2 + y2 + 2 xy and (x – y) (x + y) = (x)2โ (y)2 ]
= (3 + 1 + 2โ3) / (3 – 1) = ( 4 + 2โ3) / 2 = [2 (2 + โ3)] / 2 = 2 + โ3
= 2 + 1.732 = 3.732