Quantitative Aptitude - SPLessons

Percentage Problems

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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: [latex]10[/latex]% = [latex]\frac{10}{100}[/latex] = [latex]\frac{1}{10}[/latex]

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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 [latex]x[/latex] percent means [latex]x[/latex] hundredths, written as [latex]x[/latex]%
To express [latex]x[/latex]% as a fraction: We have , [latex]x[/latex]% = [latex]\frac{x}{100}[/latex]
Examples:
    1. 20% = [latex]\frac{20}{100}[/latex] = [latex]\frac{1}{5}[/latex];
    2. 48% = [latex]\frac{48}{100}[/latex] = [latex]\frac{12}{25}[/latex].

To express [latex]\frac{a}{b}[/latex] as a percent: We have, [latex]\frac{a}{b}[/latex] = [latex](\frac{a}{b})[/latex] x 100%
Examples:
    1. [latex]\frac{1}{4}[/latex] = [[latex]\frac{1}{4}[/latex] x 100] = 25%
    2. 0.6 = [latex]\frac{6}{10}[/latex] = [latex]\frac{3}{5}[/latex] = [[latex]\frac{3}{4}[/latex] 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:
    =[([latex]\frac{R}{(100 + R)}[/latex]) x 100]%

If the price of the commodity decreases by R%, then to maintain the same expenditure by increasing the consumption is:
    =[([latex]\frac{R}{(100 - R)}[/latex]) 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 = [[latex]\frac{R}{(100 + R)}[/latex] x 100]% = ([latex]\frac{25}{125}[/latex] 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 = [([latex]\frac{R}{(100 - R)}[/latex]) x 100]% = ([latex]\frac{16}{84}[/latex] x 100)% = [latex]\frac{400}{21}[/latex]% = 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 [latex]n[/latex] years = [latex]p[1 + (\frac{R}{100})]^{n}[/latex]
    2. Population [latex]n[/latex] years ago = [latex]\frac{p}{[1 + (\frac{R}{100})]^{n}}[/latex]

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 [latex](1 + \frac{5}{100})^{2}[/latex] = (176400 x [latex]\frac{21}{20}[/latex] x [latex]\frac{21}{40}[/latex]) = 194481
    Population 2 years ago = [latex]\frac{176400}{(1 + \frac{5}{100})^{2}}[/latex] = (176400 x [latex]\frac{20}{21}[/latex] x [latex]\frac{20}{21}[/latex]) = 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 = [latex]\frac{138915}{(1 + \frac{5}{100})^{3}}[/latex] = (138915 x [latex]\frac{20}{21}[/latex] x [latex]\frac{20}{21}[/latex] x [latex]\frac{20}{21}[/latex]) = 120000.

Let the present value of a machine be [latex]P[/latex]. Suppose it depreciates at the rate [latex]R[/latex]% per annum. Then:

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

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 [latex](1 - \frac{10}{100})^{2}[/latex]] = Rs. (162000 x [latex]\frac{9}{10}[/latex] x [latex]\frac{9}{10}[/latex]) = Rs. 131220.
    Value of the machine 2 years ago
    = Rs. [latex][\frac{162000}{(1 - \frac{10}{100})^{2}}][/latex] = Rs. (162000 x [latex]\frac{10}{9}[/latex] x [latex]\frac{10}{9}[/latex]) = 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 [latex](1 - \frac{20}{100})^{3}[/latex]] = Rs. (100 x [latex]\frac{4}{5}[/latex] x [latex]\frac{4}{5}[/latex] x [latex]\frac{4}{5}[/latex]) = Rs. 51.20
    Therefore, Reduction in value = (100 - 51.20)% = 48.8%.

shape Formulae

1. To express [latex]x[/latex]% as a fraction: [latex]x[/latex]% = [latex]\frac{x}{100}[/latex]
2. To express [latex]\frac{a}{b}[/latex] as a percent: [latex]\frac{a}{b}[/latex] = ([latex]\frac{a}{b}[/latex] x 100)%
3. If the price rate increases by R%, then decrease in consumption so as not to increase the expenditure is [[latex]\frac{R}{(100 + R)}[/latex] x 100]%
4. If the price rate decreases by R%, then increase in consumption so as not to decrease the expenditure is [[latex]\frac{R}{(100 - R)}[/latex] x 100]%
5. Let the population ⇒ P, then
    (a). Increase in rate R% per annum, then
      (i). Population after n years = [latex]{P(1 + \frac{R}{100})}^n[/latex]
      (ii). Population n years ago = [latex]{\frac{p}{(1+ \frac{R}{100})^n}}[/latex]

    (b). Decrease in rate R% per annum, then
      (i). Population after n years = [latex]{P(1 - \frac{R}{100})}^n[/latex]
      (ii). Population n years ago = [latex]{\frac{p}{(1 - \frac{R}{100})^n}}[/latex]

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% = [latex]\frac{0.8}{100}[/latex] = [latex]\frac{8}{1000}[/latex] = [latex]\frac{1}{125}[/latex]

    b). Given 28%
      28% = [latex]\frac{28}{100}[/latex] = [latex]\frac{7}{25}[/latex]

    c). Given 76%
      76% = [latex]\frac{76}{100}[/latex] = [latex]\frac{19}{25}[/latex]

    d). Given 0.06%
      0.06% = [latex]\frac{0.06}{100}[/latex] = [latex]\frac{6}{10000}[/latex] = [latex]\frac{3}{5000}[/latex]

    Therefore,
    0.8% = [latex]\frac{1}{125}[/latex]
    28% = [latex]\frac{7}{25}[/latex]
    76% = [latex]\frac{19}{25}[/latex]
    0.06% = [latex]\frac{3}{5000}[/latex]

2. Determine the value of 14% of 350 + 48% of 250?
Solution:
    Given that
    14% of 350 + 48% of 250
    =[latex]\frac{14}{100}[/latex] x 350 + [latex]\frac{48}{100}[/latex] 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 [latex]x[/latex]
    ⇒[latex]\frac{10}{100}[/latex] x [latex]x[/latex] = 250
    ⇒[latex]x[/latex] = 250 x 10
    ⇒[latex]x[/latex] = 2500
    Therefore, missing value is [latex]x[/latex] = 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 [latex]x[/latex] and [latex]y[/latex]
    Given that
    6.8% of [latex]x[/latex] = 12.4% of [latex]y[/latex]
    ⇒68[latex]x[/latex] = 124 [latex]y[/latex]
    ⇒[latex]x[/latex] = [latex]\frac{124}{68}y[/latex]
    ⇒[latex]x[/latex] = [latex]\frac{31}{17}y[/latex]
    also given [latex]x[/latex] - [latex]y[/latex] = 122 -------(i)
    Now substitute value of [latex]x[/latex] in equation (i)i.e.
    ⇒[latex]\frac{31}{17}y[/latex] - [latex]y[/latex] =122
    ⇒[latex]\frac{(31-17)y}{17}[/latex] = 122
    ⇒[latex]14y[/latex] = 122 x 17
    ⇒[latex]y[/latex] = [latex]\frac{122 * 17}{14}[/latex]
    ⇒[latex]y[/latex] = 148.14
    Now substitute the value of y in equation (i)
    [latex]x[/latex] - [latex]y[/latex] = 122
    ⇒[latex]x[/latex] = 122 + 148.4
    ⇒[latex]x[/latex] = 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 = [latex]{P(1 + \frac{R}{100})}^n[/latex]
    i.e. Population after 2 years = [latex]{200000(1 + \frac{10}{100})}^2[/latex]
    ⇒P = 200000 x [latex]\frac{121}{100}[/latex]
    ⇒P = 242000
    Now Consider, Population n years ago = [latex]{\frac{p}{(1+ \frac{R}{100})^n}}[/latex]
    i.e. Population 2 years ago = [latex]{\frac{200000}{(1+ \frac{10}{100})^2}}[/latex]
    ⇒P = 200000 x [latex]\frac{100}{121}[/latex]
    ⇒P = 165289.25
    Therefore, Population after 2 years =242000 and
    Population 2 years ago = 165289.25

6. Express [latex]5\frac{3}{2}[/latex] as rate percent
Solution:
    Given [latex]5\frac{3}{2}[/latex]
    ⇒[latex]\frac{13}{2}[/latex]
    ⇒([latex]\frac{13}{2}[/latex] x 100)%
    ⇒(13 X 50)%
    ⇒650 %
    Therefore, [latex]5\frac{3}{2}[/latex] = 650 %
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