# 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, $\inline \fn_jvn \frac{a}{b}=(\frac{a}{b}\times 100)$% .

Thus, $\inline \fn_jvn \frac{1}{4}=(\frac{1}{4}\times 100)$% = 25%;

II. If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is$\inline \fn_jvn \left [ \frac{R}{(100+R)}\times100 \right]$%

If the price of the commodity decreases by R%,then the increase in consumption so as to decrease the expenditure is$\inline \fn_jvn \left [ \frac{R}{(100-R)}\times100 \right]$%

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 = $\inline \fn_jvn p\left [ 1+\frac{R}{100} \right ]^n$

2. Population n years ago =  $\inline \fn_jvn \frac{P}{\left [ 1+\frac{R}{100} \right ]^n}$

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 = $\inline \fn_jvn P\left [ 1-\frac{R}{100} \right ]^n$

2. Value of the machine n years ago = $\inline \fn_jvn \frac{P}{\left [ 1-\frac{R}{100} \right ]^n}$

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

$\inline \fn_jvn \left [ \frac{R}{(100+R)}\times 100 \right ]$ %

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

$\inline \fn_jvn \left [ \frac{R}{(100-R)}\times 100 \right ]$ %

Q:

In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was :

 A) 2500 B) 2700 C) 2900 D) 3100

Explanation:

Total number of votes = 7500

Given that 20% of Percentage votes were invalid

=> Valid votes = 80%

Total valid votes = $\inline \fn_jvn 7500\times \frac{80}{100}$

1st candidate got 55% of the total valid votes.

Hence the 2nd candidate should have got 45% of the total valid votes

=> Valid votes that 2nd candidate got = total valid votes x $\inline \fn_jvn \frac{45}{100}$

$\inline \fn_jvn =7500\times \frac{80}{100}\times \frac{45}{100}=2700$

234 41157
Q:

If 20% of a = b, then b% of 20 is the same as :

 A) 4% of a B) 6% of a C) 8% of a D) 10% of a

Answer & Explanation Answer: A) 4% of a

Explanation:

20% of a = b  $\inline \fn_jvn \Rightarrow$ $\inline \fn_jvn \frac{20}{100}a=b$

b% of 20 =$\inline \fn_jvn \frac{b}{100} \times 20$ = $\inline \fn_jvn \frac{20a}{100} \times \frac{1}{100}\times 20$=$\inline \fn_jvn \frac{4a}{100}$ = 4% of a.

105 27531
Q:

Fresh fruit contains 68% water and dry fruit contains 20% water. How much dry fruit can be obtained from 100 kg of fresh fruits ?

 A) 20 B) 30 C) 40 D) 50

Explanation:

The fruit content in both the fresh fruit and dry fruit is the same.

Given, fresh fruit has 68% water.so remaining 32% is fruit content. weight of fresh fruits is 100kg

Dry fruit has 20% water.so remaining 80% is fruit content.let weight if dry fruit be y kg.

fruit % in freshfruit = fruit% in dryfruit

$\inline \fn_cm \therefore$(32/100) * 100 = (80/100 )* y

we get, y = 40 kg

207 25925
Q:

A candidate scoring 25% in an examination fails by 30 marks , while another candidate scores 50 % mark, gets 20 marks more than the minimum pass marks . Find the minimum pass marks. Find the minimum pass percentage.

Let x be the maximum marks,

Then (25% of x)+30 = (50% of x)-20

$\inline&space;\Rightarrow$ $\inline&space;\frac{x}{4}+30&space;=&space;\frac{x}{2}-20$

$\inline&space;\Rightarrow&space;2x-x=120+80&space;\Rightarrow&space;x=200$

Hence maximum marks = 200

Minimum pass marks = $\inline&space;\frac{200}{4}+30=80$

Hence, minimum pass marks = $\inline&space;\frac{80}{200}\times&space;100=40$%

23847
Q:

The value of a machine depreciates at the rate of 10% every year. It was purchased 3 years ago. If its present value is Rs. 8748, its purchase price was :

 A) 10000 B) 12000 C) 14000 D) 16000

$\inline \fn_jvn Purchase\; Price = Rs\left [ \frac{8748}{(1-\frac{10}{100})^3} \right ]=Rs. [8748\times \frac{10}{9}\times \frac{10}{9}\times \frac{10}{9}]$