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MCQ on Approximation for ssc exam

Approximation for ssc exam like ssc cgl, ssc chsl, ssc mts, ssc gd, ssc je
approximation questions with solutions

11) \(\displaystyle 805.0003\div 34.999=?\)

(a) 23

(b) 9

(c) 14

(d) 18

(e) 27


Solution: (a)
\(\displaystyle \frac{{805}}{{35}}=\frac{{161}}{7}=23\)

12) \(\displaystyle {{13.5}^{2}}=?\)

(a) 170

(b) 200

(c) 150

(d) 215

(e) 180


Solution: (e)
\(\displaystyle {{13.5}^{2}}=182.25\approx 180\)
Alternately without pen and paper,
13 and 14 are the nearest numbers. Two times of 13 is 169 and 14 two times is 196. By elimination and approximation, we can see that the answer should lie in between these so, (e) is the answer.

13) \(\displaystyle 750.0003\div 19.999=?\)

(a) 49

(b) 18

(c) 22

(d) 45

(e) 38


Solution: (e)
\(\displaystyle ?=\frac{{750}}{{20}}=37.5\approx 38\)
Alternately without pen and paper,
750 divided by 10 is 75 and half of it is approximately 38. By approximation we can see that the answer should is, (e) is the answer.

14) \(\displaystyle 16.003\times 27.998-209.010=?\)

(a) 150

(b) 200

(c) 75

(d) 240

(e) 110


Solution: (d)
\(\displaystyle ?=16.003\times 27.998-209.010=16\times 28-209\)
\(\displaystyle 448-209=239\approx 240\)

15) \(\displaystyle 16.928+24.7582\div 5.015=?\)

(a) 20

(b) 24

(c) 22

(d) 26

(e) None of these


Solution: (c)
\(\displaystyle 16.928+24.7582\div 5.015=?\)
\(\displaystyle 16.928+4.93=?\)
\(\displaystyle ?=21.86\approx 22\)

16) \(\displaystyle \sqrt[3]{{7.938}}\times {{6.120}^{2}}-4.9256=?\)

(a) 70

(b) 55

(c) 30

(d) 25

(e) 90


Solution: (a)
\(\displaystyle ?=\sqrt[3]{{7.938}}\times {{6.120}^{2}}-4.9256\)
\(\displaystyle (2\times 37.4)-4.9256\)
= 74.9088 – 4.9256 = which is approximate 70

17) \(\displaystyle \sqrt{{963}}+{{4.895}^{2}}-9.24=?\)

(a) 60

(b) 35

(c) 85

(d) 45

(e) 25


Solution: (d)
\(\displaystyle \sqrt{{963}}+{{4.895}^{2}}-9.24=?\)
31 + 23.9 – 9.24 = ?
54.91 – 9.24 = ?
\(\displaystyle ?=45.6\approx 45\)
Alternate mind calculation,
30 two times is 900 so, 963 should be 31 or 32. So, take lower number 31. Now, 4.8 is approximately 5, two times of 5 is 25. So, 31+25 – 9 = 47. So, by approximation and elimination we see the answer is (d).

18) \(\displaystyle \frac{5}{8}of4011.33+\frac{7}{{10}}of3411.22=?\)

(a) 4810

(b) 4980

(c) 4890

(d) 4930

(e) 4850


Solution: (c)
\(\displaystyle \frac{5}{8}of4011.33+\frac{7}{{10}}of3411.22=?\)
\(\displaystyle \frac{5}{8}\times 4010+\frac{7}{{10}}\times 3410\Rightarrow 2506+2387\)
\(\displaystyle 4893\approx 4890\)

19) 23% of 6783 + 57% of 8431 = ?

(a) 6460

(b) 6420

(c) 6320

(d) 6630

(e) 6360


Solution: (d)
23% of 6783 + 57% of 8431 = ?
? = 1559 + 4805
? = 6364 (Approximately 6360)

20) \(\displaystyle 839.999\div 48.007=?\)

(a) 9.5

(b) 23.5

(c) 11.5

(d) 28.5

(e) 17.5


Solution: (e)
\(\displaystyle ?\approx 840\div 48\approx 17.5\)
Alternately without pen and paper
840 divided by 50 (nearest number to 48) is 16.8 or approximately 17. So, by elimination and approximation we see the answer as (e)