# Simplification Questions for Bank CLERK PDF

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## Simplification Questions for Bank CLERK PDF

Download Simplification banking important questions and answers PDF. Practice solved problems provided with detailed explanations for bank clerk exam.

For more questions visit SBI Clerk Previous question papers (solved)

Instructions

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

Question 1: $[(144)^{2}\div48\times18]\div36=\sqrt{?}$

a) 23328

b) 36

c) 216

d) 46656

e) None of these

Question 2: $(27)^{18}\div(27)^{3}=?$

a) $(27)^{54}$

b) $(27)^{21}$

c) $(27)^{15}$

d) $(27)^{6}$

e) None of these

Question 3: $5\frac{1}{4}+6\frac{2}{3}+7\frac{1}{6}=?$

a) 19.5

b) $19\frac{11}{12}$

c) $19\frac{1}{12}$

d) 19

e) None of these

Question 4: $4895+364\times0.75-49=?$

a) 5119

b) 3895

c) 3907

d) 5210

e) None of these

Instructions

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

Question 5: $\frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1, 1\frac{1}{4}, 1\frac{1}{2}, 1\frac{3}{4}, ?$

a) 2

b) 4

c) 61

d) 41

e) 21

Question 6: What will come in place of both the question marks (?) in the following question ?$\frac{(?)^{0.6}}{104}=\frac{26}{(?)^{1.4}}$

a) 58

b) -48

c) -56

d) 42

e) -52

Question 7: Out of the fractions $\frac{1}{2}, \frac{7}{8}, \frac{3}{4}, \frac{5}{6}$, and $\frac{6}{7}$ what is the difference between the largest and smallest fractions ?

a) $\frac{7}{13}$

b) $\frac{3}{8}$

c) $\frac{4}{7}$

d) $\frac{1}{6}$

e) None of these

Question 8: If $(11)^{3}$ is subtracted from $(46)^{2}$ . what will be the remainder ?

a) 787

b) 785

c) 781

d) 783

e) None of these

Question 9: 90.05 + 281 ÷ 4 -151.06 = $\sqrt[3]{?}$

a) 27

b) 343

c) 216

d) 729

e) 176

Question 10: $17.98^{2} + 4.05 \times 90.11 \div 4.98 = ?$

a) 396

b) 336

c) 242

d) 423

e) 816

Lets assume the unknown number to be x.
Now, we will solve in a stepwise manner going by BODMAS rule.
144^2=20736,
20736/48=432
432*18=7776
7776/36=216
Now x=216^2 = 46656 which is option D.

We know that, by the rule of indices,

ab $\div$ ac = a(b-c)

Hence, $(27)^{18}\div(27)^{3}=(27)^{(18-3)}$ = $(27)^{15}$

Hence, Option C is correct option.

First, we convert mixed fractions into proper fractions:
$5\frac{1}{4}=\frac{21}{4}$
$6\frac{2}{3}=\frac{20}{3}$
$7\frac{1}{6}=\frac{43}{6}$
Now, $5\frac{1}{4}+6\frac{2}{3}+7\frac{1}{6}=\frac{21}{4}+\frac{20}{3}+\frac{43}{6}$

= $\frac{229}{12}$

In mixed fraction form 229/12 = $19\frac{1}{12}$
Option C is correct.

The problem can be solved by applying BODMAS rule in a stepwise manner.
$364\times0.75$ = 273
Now, 4895+273= 5168
5168 – 49 = 5119 which is option A.

$\frac{1}{4}$ is added to each number

$\frac{1}{4}$ + $\frac{1}{4}$ = $\frac{1}{2}$

$\frac{1}{2}$ + $\frac{1}{4}$ = $\frac{3}{4}$

$\frac{3}{4}$ + $\frac{1}{4}$ = 1

1 + $\frac{1}{4}$ = $1\frac{1}{4}$

$1\frac{1}{4}$ + $\frac{1}{4}$ = $1\frac{1}{2}$

$1\frac{1}{2}$ + $\frac{1}{4}$ = $1\frac{3}{4}$

$1\frac{3}{4}$ + $\frac{1}{4}$ = 2

$\frac{(x)^{0.6}}{104}=\frac{26}{(x)^{1.4}}$

${(x)^{0.6}} * {(x)^{01.4}}$ = 104*26

${(x)^{2}}$ = 104*26

x = ±52

Given values are ,
$\frac{1}{2}$ = 0.5

$\frac{7}{8}$ = 0.87

$\frac{3}{4}$ = 0.75

$\frac{5}{6}$ = 0.83

$\frac{6}{7}$ = 0.86

∴ Required difference = $\frac{7}{8}$ – $\frac{1}{2}$ = (7-4)/8 = 3/8

Here

$(46)^2$ = 2116

$(11)^3$ = 1331

So, 2116 – 1331 = 785

The given statement can be written as $90 + 60 – 151 = \sqrt[3]{?}$

$9 =\sqrt[3]{?}$

=> ? = 729

The given equation can be written as $18^{2} + \frac{4*90}{5}$
= $324 + 4*18$
= $396$