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# Simplification Questions For SSC CPO

SSC CPO Simplification Questions and Answers download PDF, based on previous year question paper of SSC CPO exam. 20 Very important Simplification Questions for SSC CPO.

SSC CPO Free Mock Test (Latest Pattern)

Free 18000 Solved Questions: SSC Study Material

Question 1: What is the value of $10 – [6 – 4 – (2\div 1 – 2)]$?

a) 10

b) 12

c) 14

d) 8

Question 2: What is the value of $\left[22\frac{3}{4}\div\frac{14}{3}of\left(8-\frac{1}{5}+4-2\div\frac{1}{2}\right)\right]$?

a) $\frac{3}{7}$

b) $\frac{91}{213}$

c) $\frac{5}{8}$

d) $\frac{91}{199}$

Question 3: What is the value of:
$8 \times 3 \div 6 + 7 – 4 \times 2 \div 4 + 8 – 10 = ?$

a) 6

b) 5

c) 7

d) 4

Question 4: The value of:
$4 \div 2 + 3 \times 6 – 7$ is:

a) 13

b) 14

c) 11

d) 12

Question 5: The value of $24 \div 4 \times (3 + 3) \div 2$ is:

a) 22

b) 24

c) 20

d) 18

Question 6: The value of $64 \div 2 \times (9 \div 3) \times 3 \div 9 – 32$ is:

a) 1

b) 0

c) -1

d) 2

Question 7: Find the value of:
$(11 + 4) – 9 \times 1 \div 3$ of $4$

a) $\frac{75}{4}$

b) $\frac{57}{4}$

c) $5$

d) $3$

Question 8: Find the value of:
$(7 \times 6 + 5 – 15) \div 4 + 6 \div 3 – 4 + 18 \div 3$

a) 12

b) 14

c) 16

d) 18

Question 9: $11^2 + 11^4 \div 11^3 of 11 – 11$= ?

a) 111

b) 121

c) 131

d) 101

Question 10: $\frac{4}{7} of \frac{8}{9} \div \frac{8}{7} \times 180 of \frac{1}{9} = ?$

a) $\frac{20}{9}$

b) $\frac{9}{20}$

c) $\frac{80}{9}$

d) $\frac{9}{80}$

Question 11: The value of the following expression.
$(47 \times 588) \div (28 \times 120) = ?$

a) 6.284

b) 8.285

c) 7.625

d) 8.225

Question 12: Find the value of the following expression.
$51 \div 3 \times 15 = ?$

a) 140

b) 255

c) 130

d) 100

Question 13: Simplify: $8+3-\left(\frac{5}{2}\times\frac{1}{3}\right) of \frac{12}{5}+\frac{4}{3}\times\frac{3}{8}$

a) 9.5

b) 10

c) 9

d) 19

Question 14: Simplify: $8.65-[4+0.5\ of\ (8.8-2.3\times3.5)]$

a) 4.725

b) 4.275

c) 3.275

d) 4.527

Question 15: The value of $[9.5 \div (0.6 \times 0.75 + 0.8 \div 16) + 0.75] \div (0.03 \div 0.6 of 0.01)$ lies between:

a) 2 and 3

b) 1 and 2

c) 0 and 1

d) 3 and 4

$10 – [6 – 4 – (2\div 1 – 2)] = 10 – [2 – (2 – 2)] = 10 – [2 – 0] = 10-2 = 8$

$\left[22\frac{3}{4}\div\frac{14}{3}of\left(8-\frac{1}{5}+4-2\div\frac{1}{2}\right)\right] = \dfrac{91}{4} \times \dfrac{3}{14} \times \dfrac{5}{39} = \dfrac{5}{8}$

$8 \times 3 \div 6 + 7 – 4 \times 2 \div 4 + 8 – 10 = 8 \times \dfrac{3}{6} + 7 – (4 \times \dfrac{2}{4}) + 8 – 10 = 4+7-2+8-10 = 7$

$4 \div 2 + 3 \times 6 – 7 = 2+18-7 = 13$

$24 \div 4 \times (3 + 3) \div 2 = \dfrac{24}{4} \times \dfrac{6}{2} = 18$

$64 \div 2 \times (9 \div 3) \times 3 \div 9 – 32 = \dfrac{64}{2} \times 3 \times \dfrac{3}{9} – 32 = 32-32 = 0$

$(11 + 4) – 9 \times 1 \div 3$ of $4 = 15-9\times\dfrac{1}{3\times4} = 15-\dfrac{3}{4} = \dfrac{57}{4}$

$(7 \times 6 + 5 – 15) \div 4 + 6 \div 3 – 4 + 18 \div 3 = \dfrac{32}{4} + \dfrac{6}{3} – 4 + \dfrac{18}{3} = 8 + 2 – 4 + 6 = 12$

By simplifying we get 121+1-11

=122-11
=111

By using BODMAS we have
=$(32/63)\div(8/7)\times20$
=$(32/63)\times(7/8)\times20$
=$(4/9)\times20$
=80/9

By Applying BODMAS we have 47*21/120
=47*7/40
=329/40
=8.225

By simplification we get 17*15=255

=$8+3-\left(\frac{5}{2}\times\frac{1}{3}\right) of \frac{12}{5}+\frac{4}{3}\times\frac{3}{8}$
=$8+3-2+\frac{4}{3}\times\frac{3}{8}$
=$8+3-2+\frac{1}{2}$
=9.5

=$8.65-[4+0.5\ of\ (8.8-2.3\times3.5)]$
=$8.65-[4+0.5\ of\ (8.8-8.05)]$
=$8.65-[4+0.5\ of\ (0.75)]$
=$8.65-[35/8]$
=8.65-4.375
=4.275

=$[9.5 \div (0.6 \times 0.75 + 0.8 \div 16) + 0.75] \div (0.03 \div 0.6 of 0.01)$
=$[9.5 \div (0.6 \times 0.75 + 0.05) + 0.75] \div (5)$
=$[9.5 \div (0.45 + 0.05) + 0.75] \div (5)$
=$[19+0.75] \div (5)$