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# Approximation Questions For IBPS PO PDF

Download Approximation Questions For IBPS PO PDF. Practice ApproximationÂ  Questions with Solutions for Banking exams based on asked questions in previous papers.

Instructions

What should come in place of the question mark (?) in the following questions ?

Question 1:Â ${2^{0.2}} \times 64 \times {8^{1.3}} \times {4^{0.2}} = {8^?}$

a)Â 2.7

b)Â 2.5

c)Â 3.7

d)Â 3.2

e)Â None of these

Question 2:Â 83% of 6242 x 12% of 225 = ?

a)Â 146286.42

b)Â 134263.18

c)Â 139883.22

d)Â 1562218.23

e)Â None of these

Question 3:Â $1{1 \over 8} + 1{6 \over 7} + 3{3 \over 5}$ = ?

a)Â $8{{121} \over {140}}$

b)Â $6{{163} \over {280}}$

c)Â $9{{197} \over {280}}$

d)Â $7{{117} \over {140}}$

e)Â None of these

Question 4:Â ? Ă· 25 Ă· 12 = 248.76

a)Â 74628

b)Â 497.52

c)Â 62452

d)Â 870.66

e)Â None of these

Question 5:Â ${(73)^3}$ = ?

a)Â 365127

b)Â 298627

c)Â 305867

d)Â 389017

e)Â None of these

Instructions

What should come in place of the question mark (?) in the following questions?

Question 6:Â $\sqrt[3]{1331}$ =

a)Â 27

b)Â 21

c)Â 17

d)Â 9

e)Â None of these

Question 7:Â 188.21 – 27.54 – 11.93 = ?

a)Â 139.74

b)Â 126.64

c)Â 148.74

d)Â 184.64

e)Â None of these

Question 8:Â 1268 Ă· 8 Ă· 2 = ?

a)Â 71.75

b)Â 317

c)Â 268

d)Â 79.25

e)Â None of these

Question 9:Â ${8^{1.1}} \times {4^2}^{.7} \times {2^{3.3}} = {2^?}$

a)Â 7.1

b)Â 14

c)Â 0.5

d)Â 9

e)Â None of these

Instructions

What approximate value will come in place of question mark (?) in the given questions? (You are not expected to calculate the exact value).

Question 10:Â 344 Ă· 4.99 + 144.08 Ă· 8.89 =?

a)Â 119

b)Â 85

c)Â 43

d)Â 54

e)Â 132

Question 11:Â 43.99 x 20.001 – 1439 Ă· 6 = ?

a)Â 500

b)Â 640

c)Â 540

d)Â 600

e)Â 680

Question 12:Â 459.85 + 519.82 = ?% of 1399.92

a)Â 90

b)Â 70

c)Â 75

d)Â 50

e)Â 80

Question 13:Â 40 % of 309 Ă· 4 + ? = 6.9992

a)Â 24

b)Â 18

c)Â 42

d)Â 56

e)Â 34

Expression : ${2^{0.2}} \times 64 \times {8^{1.3}} \times {4^{0.2}} = {8^?}$

=> $(2)^{0.2} \times (2)^6 \times (2)^{3 \times 1.3} \times (2)^{2 \times 0.2} = (2)^{3 \times x}$

=> $(2)^{0.2 + 6 + 3.9 + 0.4} = (2)^{3 x}$

=> $(2)^{10.5} = (2)^{3 x}$

Comparing powers of 2, we get :

=> $3x = 10.5$

=> $x = \frac{10.5}{3} = 3.5$

ExpressionÂ :Â 83% of 6242 x 12% of 225 = ?

= $(\frac{83}{100} \times 6242) \times (\frac{12}{100} \times 225)$

= $5180.86 \times 27$

= $139883.22$

ExpressionÂ :Â $1{1 \over 8} + 1{6 \over 7} + 3{3 \over 5}$ = ?

= $(1 + 1 + 3) + (\frac{1}{8} + \frac{6}{7} + \frac{3}{5})$

= $5 + (\frac{35 + 240 + 168}{280})$

= $5 + \frac{443}{280} = 5 + \frac{280 + 163}{280}$

= $5 + (1 + \frac{163}{280}) = 6{{163} \over {280}}$

ExpressionÂ :Â ? Ă· 25 Ă· 12 = 248.76

=> $\frac{x}{25} \times \frac{1}{12} = 248.76$

=> $x = 248.76 \times 300$

=> $x = 74628$

ExpressionÂ :Â ${(73)^3}$

= $73 \times 73 \times 73$

= $5329 \times 73 = 389017$

Shortcut MethodÂ : We can find the last two digits of $73^3$, which is ’17’

Thus, ans – (D)

Expression : $\sqrt[3]{1331}$

= $\sqrt[3]{11 \times 11 \times 11}$

= $11$

ExpressionÂ :Â 188.21 – 27.54 – 11.93 = ?

= 188.21 – 39.47

= 148.74

ExpressionÂ :Â 1268 Ă· 8 Ă· 2 = ?

= $\frac{1268}{8} \div 2$

= $\frac{158.5}{2} = 79.25$

Expression : ${8^{1.1}} \times {4^2}^{.7} \times {2^{3.3}} = {2^?}$

=> $(2)^{3 \times 1.1} \times (2)^{2 \times 0.7} \times (2)^{3.3} = (2)^x$

=> $(2)^{3.3} \times (2)^{1.4} \times (2)^{3.3} = (2)^x$

=> $(2)^{3.3 + 1.4 + 3.3} = (2)^x$

Comparing both sides, we get :

=> $x = 3.3 + 1.4 + 3.3 = 8$

ExpressionÂ :Â 344 Ă· 4.99 + 144.08 Ă· 8.89 =?

= $(\frac{345}{5}) + (\frac{144}{9})$

= $69 + 16 = 85$

ExpressionÂ :Â 43.99 x 20.001 – 1439 Ă· 6 = ?

= $(44 \times 20) – (\frac{1440}{6})$

= $880 – 240 = 640$

ExpressionÂ :Â 459.85 + 519.82 = ?% of 1399.92

=> $460 + 520 = \frac{x}{100} \times 1400$

=> $980 = 14x$

=> $x = \frac{980}{14} = 70$