Quantitative Aptitude - SPLessons

Percentage Problems

Chapter 11

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

Percentage Problems

shape Introduction

Percentage Problems: A Percentage is a dimensionless ratio or number expressed as a fraction of 100. It is often denoted by the character (%).

Example: \(10\)% = \(\frac{10}{100}\) = \(\frac{1}{10}\)


shape Methods

Problems involving percent are called percentage problems. There are three types of percentage problems. They are:

    1. Finding a percent of a given number,

    2. Finding what percent one number is of another,

    3. Finding the number when the percent of the number is given.


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 , \(x\)% = \(\frac{x}{100}\)

Examples:

    1. 20% = \(\frac{20}{100}\) = \(\frac{1}{5}\);

    2. 48% = \(\frac{48}{100}\) = \(\frac{12}{25}\).


To express \(\frac{a}{b}\) as a percent: We have, \(\frac{a}{b}\) = \((\frac{a}{b})\) x 100%

Examples:

    1. \(\frac{1}{4}\) = [\(\frac{1}{4}\) x 100] = 25%

    2. 0.6 = \(\frac{6}{10}\) = \(\frac{3}{5}\) = [\(\frac{3}{4}\) x 100]%= 60%

If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is:

    =[(\(\frac{R}{(100 + R)}\)) x 100]%


If the price of the commodity decreases by R%, then to maintain the same expenditure by increasing the consumption is:

    =[(\(\frac{R}{(100 – R)}\)) x 100]%


Examples 1
In the new budget, the price of kerosene oil rose by 25%. By how much percent must a person reduce his consumption so that his expenditure on it does not increase?

Solution:

    Reduction in consumption = [\(\frac{R}{(100 + R)}\) x 100]% = (\(\frac{25}{125}\) x 100)% = 20%.


Examples 2
The price of wheat falls by 16%. By what percentage a person can increase the consumption in order to maintain the same budget?

Solution:

    Increase in Consumtion = [(\(\frac{R}{(100 – R)}\)) x 100]% = (\(\frac{16}{84}\) x 100)% = \(\frac{400}{21}\)% = 19.04% = 19%.


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 = \(p[1 + (\frac{R}{100})]^{n}\)


    2. Population \(n\) years ago = \(\frac{p}{[1 + (\frac{R}{100})]^{n}}\)


Examples 1
The population of a town is 1,76,400. If it increase at the rate of 5% per annum, what will be its polulation 2 years hence? What was it 2 years ago?

Solution:

    Population after 2 years = 176400 x \((1 + \frac{5}{100})^{2}\) = (176400 x \(\frac{21}{20}\) x \(\frac{21}{40}\)) = 194481

    Population 2 years ago = \(\frac{176400}{(1 + \frac{5}{100})^{2}}\) = (176400 x \(\frac{20}{21}\) x \(\frac{20}{21}\)) = 160000.


Examples 2
The population of a town increases by 5% anually. If its population in 2001 was 1,38,915, what it was in 1998?

Solution:

    Population in 1998 = \(\frac{138915}{(1 + \frac{5}{100})^{3}}\) = (138915 x \(\frac{20}{21}\) x \(\frac{20}{21}\) x \(\frac{20}{21}\)) = 120000.


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 = \(p[1 – (\frac{R}{100})]^{n}\)


    2. Value of the machine \(n\) years ago = \(\frac{p}{[1 – (\frac{R}{100})]^{n}}\)


Examples 1
The value of a machin depreciates at the rate of 10% per annum. If its present value is Rs. 1,62,000, what will be its worth after 2 years? What was the value of he machine 2 years ago?

Solution:

    Value of the machine after 2 years

    = Rs. [162000 x \((1 – \frac{10}{100})^{2}\)] = Rs. (162000 x \(\frac{9}{10}\) x \(\frac{9}{10}\)) = Rs. 131220.

    Value of the machine 2 years ago

    = Rs. \([\frac{162000}{(1 – \frac{10}{100})^{2}}]\) = Rs. (162000 x \(\frac{10}{9}\) x \(\frac{10}{9}\)) = Rs. 200000.


Examples 2
Depreciation applicable to an equipment is 20%. The value of the equipment 3 years from now will be less by:

Solution:

    Let the preent value be Rs. 100.

    Value after 3 years = Rs. [100 x \((1 – \frac{20}{100})^{3}\)] = Rs. (100 x \(\frac{4}{5}\) x \(\frac{4}{5}\) x \(\frac{4}{5}\)) = Rs. 51.20

    Therefore, Reduction in value = (100 – 51.20)% = 48.8%.

shape Formulae

1. To express \(x\)% as a fraction: \(x\)% = \(\frac{x}{100}\)


2. To express \(\frac{a}{b}\) as a percent: \(\frac{a}{b}\) = (\(\frac{a}{b}\) x 100)%


3. If the price rate increases by R%, then decrease in consumption so as not to increase the expenditure is [\(\frac{R}{(100 + R)}\) x 100]%


4. If the price rate decreases by R%, then increase in consumption so as not to decrease the expenditure is [\(\frac{R}{(100 – R)}\) x 100]%


5. Let the population ⇒ P, then

    (a). Increase in rate R% per annum, then

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

      (ii). Population n years ago = \({\frac{p}{(1+ \frac{R}{100})^n}}\)


    (b). Decrease in rate R% per annum, then

      (i). Population after n years = \({P(1 – \frac{R}{100})}^n\)

      (ii). Population n years ago = \({\frac{p}{(1 – \frac{R}{100})^n}}\)

shape Samples

1. Express each of the following as a fraction:
a). 0.8% b). 28% c). 76% d). 0.006% ?

Solution:

    a). Given 0.8%

      0.8% = \(\frac{0.8}{100}\) = \(\frac{8}{1000}\) = \(\frac{1}{125}\)


    b). Given 28%

      28% = \(\frac{28}{100}\) = \(\frac{7}{25}\)


    c). Given 76%

      76% = \(\frac{76}{100}\) = \(\frac{19}{25}\)


    d). Given 0.06%

      0.06% = \(\frac{0.06}{100}\) = \(\frac{6}{10000}\) = \(\frac{3}{5000}\)


    Therefore,

    0.8% = \(\frac{1}{125}\)

    28% = \(\frac{7}{25}\)

    76% = \(\frac{19}{25}\)

    0.06% = \(\frac{3}{5000}\)


2. Determine the value of 14% of 350 + 48% of 250?

Solution:

    Given that

    14% of 350 + 48% of 250

    =\(\frac{14}{100}\) x 350 + \(\frac{48}{100}\) x 250

    =49 + 120

    =169

    Therefore, 14% of 350 + 48% of 250 = 169


3. Find the missing number 10% of ? = 250.

Solution:

    Given that 10% of ? = 250

    Assume the unknown value as \(x\)

    ⇒\(\frac{10}{100}\) x \(x\) = 250

    ⇒\(x\) = 250 x 10

    ⇒\(x\) = 2500

    Therefore, missing value is \(x\) = 2500


4. Difference between two numbers is 122. If 6.8% of one number is 12.4% of the other number, Find the two numbers?

Solution:

    Let the two numbers be \(x\) and \(y\)

    Given that

    6.8% of \(x\) = 12.4% of \(y\)

    ⇒68\(x\) = 124 \(y\)

    ⇒\(x\) = \(\frac{124}{68}y\)

    ⇒\(x\) = \(\frac{31}{17}y\)

    also given \(x\) – \(y\) = 122 ——-(i)

    Now substitute value of \(x\) in equation (i)i.e.

    ⇒\(\frac{31}{17}y\) – \(y\) =122

    ⇒\(\frac{(31-17)y}{17}\) = 122

    ⇒\(14y\) = 122 x 17

    ⇒\(y\) = \(\frac{122 * 17}{14}\)

    ⇒\(y\) = 148.14

    Now substitute the value of y in equation (i)

    \(x\) – \(y\) = 122

    ⇒\(x\) = 122 + 148.4

    ⇒\(x\) = 270.4

    Therefore the two numbers are 270.4 and 148.14


5. The population of a town is 2,00,000. If it increases at the rate of 10% per annum, what will be its population 2 years hence? what was it 2 years ago?

Solution:

    Consider, Population after n years = \({P(1 + \frac{R}{100})}^n\)

    i.e. Population after 2 years = \({200000(1 + \frac{10}{100})}^2\)

    ⇒P = 200000 x \(\frac{121}{100}\)

    ⇒P = 242000

    Now Consider, Population n years ago = \({\frac{p}{(1+ \frac{R}{100})^n}}\)

    i.e. Population 2 years ago = \({\frac{200000}{(1+ \frac{10}{100})^2}}\)

    ⇒P = 200000 x \(\frac{100}{121}\)

    ⇒P = 165289.25

    Therefore, Population after 2 years =242000 and

    Population 2 years ago = 165289.25


6. Express \(5\frac{3}{2}\) as rate percent

Solution:

    Given \(5\frac{3}{2}\)

    ⇒\(\frac{13}{2}\)

    ⇒(\(\frac{13}{2}\) x 100)%

    ⇒(13 X 50)%

    ⇒650 %

    Therefore, \(5\frac{3}{2}\) = 650 %