**:**

__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 of

_{9}√(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)

_{3}√(13/25)^{4}b)

_{4}√(13/25)^{3}c)

^{4}√(25/13)^{3}d)

^{3}√(25/13)^{4}e) None of these

**3).**The value of 5/(121)

^{-1/2}is

a) -55

b) 1/55

c) –(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) –(3/7)

**5).**The value of 3 × 9

^{-(3/2) }× 9

^{1/2}is

a) 1/3

b) 3

c) 27

d) -(1/3)

e) None of these

**6).**The value of(216

^{2/3})

^{1/2}is

a) 3

b) 9

c) 12

d) 6

e) None of these

**7).**The value of 27

^{-1/3 }× [(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 (√63 × √7) / (

_{3}√27) is

a) 7

b) 9

c) 21

d) 18

e) None of these

**10).**If √3 = 1.732, then the value of (√3 + 1) / (√3 – 1) is

a) 3.732

b) 1/3.732

c) 0.732

d) 2.732

e) None of these

__Answers__**:**

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

__Solution:__

**1).**

_{9}√(4/5)

^{3}= (4/5)

^{3/9}= (4/5)

^{1/3}

**Answer: a)**

**2).**The radical form of (13/25)

^{3/4}=

_{4}√(13/25)

^{3}

**Answer: b)**

**3).**5 / (121)

^{-1/2}= 5 × (121)

^{1/2}= 5 × (11

^{2})

^{1/2}

= 5 × 11 = 55

**Answer: d)**

**4).**(512)

^{-(3/9)}= (2

^{9})

^{-(3 / 9)}= 2

^{9 × (- 3 / 9)}

= 2

^{-3}= (1 / 2)^{3}= (1/2) × (1/2) × (1/2) = (1/8)**Answer: c)**

**5).**3 × 9

^{-(3 / 2)}× 9

^{1/2}= 3 × [3

^{2 × (- 3 / 2)}] × ( 3

^{2 ×(1 / 2)})

= 3 × (3)

^{-3}× 3 = 3 × (1/3)^{3}× 3= 3 × (1 / 27) × 3 = 1/3

**Answer: a)**

**6).**(216

^{2 / 3})

^{1/2}= (6

^{3 × (2 / 3)})

^{1 / 2}= 6

^{2 × (1 / 2 )}= 6

^{1}= 6

**Answer: d)**

**7).**27

^{1 / 3}× (27

^{2/3}/ 27

^{1/3}) = 3

^{3 × – (1/3)}× [(3

^{3})

^{2/3}/ (3

^{3})

^{1/3}]

= 3

^{-1}× (3^{2}/ 3)= (1/3) × ( 9 / 3 ) = (1 / 3) × 3 = 1

**Answer: d)**

**8).**(625)

^{-(1/2)}= (625 / 100)

^{-1/2}= (25 / 4)

^{-1/2}

= (4 / 25)

^{1/2}= (2^{2}/ 5^{2})^{1/2}= (2 / 5)^{2 × ½ }= (2 / 5) or (1 / 2.5)

**Answer: c)**

**9).**(√63 × √7) /

_{3}√27 = √(63 × 7) /

_{3}√3

^{3}= √(441) / 3

^{3 ×1/3}

= (21 / 3) = 7

**Answer: a)**

**10).**(√3 + 1) / (√3 – 1)

Rationalising, we have

= [(√3 + 1) / (√3 – 1)] × [(√3 + 1) / (√3 + 1)] = (√3 + 1)

^{2}/ [(√3)^{2}– (1)^{2}][(x + y)

^{2}= x^{2}+ y^{2}+ 2 xy and (x – y) (x + y) = (x)^{2}– (y)^{2}]^{}= (3 + 1 + 2√3) / (3 – 1) = ( 4 + 2√3) / 2 = [2 (2 + √3)] / 2 = 2 + √3

= 2 + 1.732 = 3.732

**Answer: a)**