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100 approximation questions with solutions

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

21) \(\displaystyle \sqrt{{8000}}=?\)

(a) 76

(b) 89

(c) 65

(d) 97

(e) 58


Solution: (b)
\(\displaystyle ?\approx \sqrt{{8000}}\approx 89\)
We can orally do it by, approximation, 90 multiplied two time is 8100. Our answer is just 100 short of the 8000, so the answer is 89.

22) \(\displaystyle 99999\div 99\div 9=?\)

(a) 112

(b) 211

(c) 121

(d) 221

(e) 222


Solution: (a)
\(\displaystyle ?=\frac{{99999}}{{99\times 9}}\approx 112\)
Without pen and paper,
Assume all numbers near to their respective nearest tens, hundreds and thousands and start dividing, we will get 100 as answer. So, by approximation our answer should be very near to 100. We see that out of the given options 112 is the least number near to 100. If you get a doubt that it may be option (c) i:e 121. We can check that, like 100000 divided by 100 is 1000, which if again divided by 9 we get 111.11. So by approximation method the answer is (a) only.

23) 7984.986 + 2541.005 + 127.996 = ?

(a) 11280

(b) 15600

(c) 14650

(d) 10654

(e) 16500


Solution: (d)
? = 7985 + 2541 + 128 = 10654

24) \(\displaystyle {{13.001}^{3}}=?\)

(a) 1900

(b) 2200

(c) 2000

(d) 1800

(e) 2100


Solution: (b)
\(\displaystyle ?={{13.001}^{3}}={{13}^{3}}\)
\(\displaystyle 2197\approx 2200\)

25) \(\displaystyle 50.001\%of99.99\div 49.999=?\)

(a) 1

(b) 0.1

(c) 0.01

(d) 0.02

(e) None of these


Solution: (a)
\(\displaystyle ?=\frac{{100\times 50}}{{100}}\div 50=1\)

26) \(\displaystyle \frac{{{{{\left( {10008.99} \right)}}^{2}}}}{{10009.001}}\times \sqrt{{3589}}\times 0.4987=?\)

(a) 3000

(b) 300000

(c) 3000000

(d) 5000

(e) 9000000


Solution: (b)
\(\displaystyle ?=\frac{{{{{\left( {10008.99} \right)}}^{2}}}}{{10009.001}}\times \sqrt{{3589}}\times 0.4987\)
By taking approximate values, we write as
\(\displaystyle \frac{{{{{\left( {10009} \right)}}^{2}}}}{{10009}}\times \sqrt{{3600}}\times 0.5=?\)
\(\displaystyle \left( {10009} \right)\times \sqrt{{3600}}\times 0.5=?\)
\(\displaystyle 10009\times 60\times 0.5\approx 300000\)

27) \(\displaystyle 196.1\times 196.1\times 196.1\times 4.01\times 4.01\times 4.001\times 4.999\times 4.99= {{196.1}^{3}}\times 4\times ?\)

(a) 100

(b) 16

(c) 10

(d) 64

(e) 32


Solution: (a)
\(\displaystyle 196.1\times 196.1\times 196.1\times 4.01\times 4.01\times 4.001\times 4.999\times 4.99={{196.1}^{3}}\times 4\times ?\)
or \(\displaystyle 4\times ?=4.01\times 4.001\times 4.999\times 4.999or?=4\times 5\times 5=100\)

28) \(\displaystyle ?\%of45.999\times 16\%of83.006=116.073\)

(a) 6

(b) 24

(c) 19

(d) 30

(e) 11


Solution: (c)
Let x be there in place of question mark so, \(\displaystyle x\%of45.999\times 16\%of83.006=116.073\)
We take \(\displaystyle \frac{x}{{100}}\times 46\times \frac{{16}}{{100}}\times 89=116\)
By approximation,
\(\displaystyle \frac{x}{{100}}\times 50\times \frac{{16}}{{100}}\times 83=116\)
\(\displaystyle x\times 0.5\times 1.28=116\)
\(\displaystyle x\times 6=116\) (approx)
\(\displaystyle \Rightarrow x=19.33\approx 19\)

29) 9228.789 – 5021.832 + 1496.989 = ?

(a) 6500

(b) 6000

(c) 6300

(d) 5700

(e) 5100


Solution: (d)
Having a glance at the given options one can find out that the two nearest values have a difference of 300. So round off the numbers to the nearest ten’s values.
\(\displaystyle 9228.789\approx 9230;\text{ }5021.832\approx 5020\text{ }and\text{ }1496.989\approx 1500\)
Now the equation will become
9230 – 5020 + 1500 = ?
? = 5710

30) \(\displaystyle 1002\div 49\times 99-1299\text{ }=\text{ }?\)

(a) 700

(b) 600

(c) 900

(d) 250

(e) 400


Solution: (a)
It can be rounded off to the nearest ten’s places.
\(\displaystyle 1002\approx 1000;\text{ }49\approx 50;\text{ }99\approx 100\text{ }and\text{ }1299\approx 1300\)
Now the equation will become
\(\displaystyle 1000\div 50\times 100-1300=?\)
\(\displaystyle 20\times 100-1300=?\)
2000 – 1300 = ?
? = 700