Percentage Questions

FACTS  AND  FORMULAE  FOR  PERCENTAGE  QUESTIONS

I.Concept of Percentage : By a certain percent , we mean that many hundredths. Thus x percent means x hundredths, written as x%.

To express x% as a fraction : We have , x% = x/100.

Thus, 20% = 20/100 = 1/5;

48% = 48/100 = 12/25, etc.

To express a/b as a percent : We have, $\frac{a}{b}=\left(\frac{a}{b}×100\right)%$ .

Thus, $\frac{1}{4}=\left(\frac{1}{4}×100\right)%=25%$

II. If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is $\left[\frac{R}{\left(100+R\right)}×100\right]%$

If the price of the commodity decreases by R%,then the increase in consumption so as to decrease the expenditure is $\left[\frac{R}{\left(100-R\right)}×100\right]%$

III. Results on Population : Let the population of the town be P now and suppose it increases at the rate of R% per annum, then :

1. Population after n years = $P{\left(1+\frac{R}{100}\right)}^{n}$

2. Population n years ago =  $\frac{P}{{\left(1+\frac{R}{100}\right)}^{n}}$

IV. Results on Depreciation : Let the present value of a machine be P. Suppose it depreciates at the rate R% per annum. Then,

1. Value of the machine after n years = $P{\left(1-\frac{R}{100}\right)}^{n}$

2. Value of the machine n years ago = $\frac{P}{{\left(1-\frac{R}{100}\right)}^{n}}$

V. If A is R% more than B, then B is less than A by

$\left[\frac{R}{\left(100+R\right)}×100\right]%$

If A is R% less than B , then B is more than A by

$\left[\frac{R}{\left(100-R\right)}×100\right]%$

Q:

A housewife saved Rs. 2.50 in buying an item on sale. If she spent Rs. 25 for the item, approximately how much percent she saved in the transaction ?

 A) 8 B) 9 C) 10 D) 11

Explanation:

Actual price = Rs. (25 + 2.50) = Rs. 27.50.

Saving = 2.50

Percentage Saving = [(2.50/27.50)*100]

= 2500/275

$9111$%

$≈9$%

136 63133
Q:

A number is decreased by 10% and then increased by 10%.  The number so obtained is 10 less than the original number. What was the oiginal number ?

 A) 1000 B) 2000 C) 3000 D) 4000

Explanation:

Let the original number be x.

Final number obtained = 110% of (90% of x) =(110/100 * 90/100 * x) = (99/100)x.

x-(99/100)x=10

=> x =1000

163 61340
Q:

45% of 750  -  25% of 480 = ?

 A) 337.50 B) 217.50 C) 376.21 D) 120

Explanation:

Given expression = (45 x 750/100) - (25 x 480/100) = (337.50 - 120) = 217.50

63 60997
Q:

What percent of 7.2 kg is 18 gms ?

 A) .25% B) .5% C) .75% D) 1%

Explanation:

Required percentage = (18/7200 * 100)% =  1/4% = 0.25%

244 59842
Q:

If A's height is 40% less than that of B, how much percent B's height is more than that of A?

 A) 66.66% B) 76.66% C) 96.66% D) 86.66%

Explanation:

Excess of B's height over A's = [(40/(100 - 40)] x 100% =  66.66%

221 53699
Q:

40% of 75 + 80% of 25 = K% of 250

Find the value of K?

 A) 10 B) 20 C) 30 D) 35

Explanation:

From the given data,

13 53649
Q:

Ram sell his goods 25% cheaper than Shyam and 25% dearer than Bram. How much % is Bram's good cheaper than Shyam ?

 A) 60% B) 40% C) 50% D) 30%

Explanation:

Lets say shyam sells at 100,

Since Ram sells 25% cheaper than Shyam,

Therefor Ram sells at less than 25% of100 or 75 rs.

Ram sells 25% dearer than Bram or 125% of Ram =100% of Bram or

125% of x (say x price of bram )=75rs.
or 100% of x =60rs.
hence price of bram is 60rs.

now Bram's good is cheaper than Shyam's as (100-60)x100/100% or 40%.

Hence Brams Good is 40% cheaper than that of Shyam's good.

101 46162
Q:

If the pice of sugar rises from Rs. 6 per kg to Rs. 7.50 per kg, a person, to have no increase in his expenditure on sugar, will have to reduce his consumpion of sugar by

 A) 15 B) 20 C) 25 D) 30