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Q:

A number X is mistakenly divided by 10 instead of being multiplied by 10. What is the percentage error in the result?

 A) -99% B) +99% C) -100% D) +100%

Answer:   A) -99%

Explanation:

By mistake = $\inline&space;\frac{x}{10}$

Actual value = $\inline&space;x\times&space;10$

% Change = $\inline&space;\frac{10x-\frac{x}{10}}{10x}\times&space;100&space;=&space;\frac{99}{100}\times&space;100$

= 99% (negative)

Since actual value is greater than the wrong value.

Q:

96% of the population of a village is 23040. The total population of the village is ?

 A) 24000 B) 32256 C) 25478 D) 26587

Answer & Explanation Answer: A) 24000

Explanation:

Let the total population be 'P'

P x (96/100) = 23040

P = 240 * 100

P = 24000.

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1 17
Q:

If three eighth of a number is 141. What will be the approximately value of 32.08% of this number ?

 A) 101 B) 112 C) 104 D) 120

Answer & Explanation Answer: D) 120

Explanation:

K x 3/8 = 141 => K = 376

376 x 32.08/100 = 120

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0 46
Q:

If L exceeds K by 30%, then K is less than L by ?

 A) 14 2/5 % B) 23 1/3 % C) 20 % D) 16 %

Answer & Explanation Answer: B) 23 1/3 %

Explanation:

K = 100 L = 130
130------ 30
100------- ? => 23 1/13 %

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2 119
Q:

A user type a content on 20 sheets, each sheet consist of 55 lines, and each line can have 66 characters. The content is again retyped on another set of sheets, these new sheets have 65 lines and each line can take 70 characters. Calculate the reduction in sheets ?

 A) 2 or 60% B) 4 or 20% C) 6 or 10% D) 8 or 5%

Answer & Explanation Answer: B) 4 or 20%

Explanation:

20 x 55 x 66 = L x 65 x 70
Here L = 72,600/4550 = 15.95 =~ 16
Reduction is 20-16 = 4

which is 20% reduction.

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2 138
Q:

Kalyan purchased a Maruti van for Rs. 1,96,000 and the rate of depreciation is 14(2/7) % per annum. Find the value of the van after two years ?

 A) Rs. 1,24,000 B) Rs. 1,44,000 C) Rs. 1,32,000 D) Rs. 1,62,000

Answer & Explanation Answer: B) Rs. 1,44,000

Explanation:

Value of Maruti Van, V1 = Rs. 1,96,000
Rate of depreciation, r = 14(2/7)% = 100/7%; Time, t = 2 years
Let V2 is the value after depreciation.
V2 = $\inline \fn_jvn \small V1[1-(r/100)]^{t}$
V2 = $\inline&space;\fn_jvn&space;\small&space;196000[1-((100/7)/100)]^{2}$
V2 = 196000(6/7)(6/7)
V2 = (196000x36)/49 = Rs. 1,44,000.

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