Informative line

Calculations Involving Percents

Finding Percent by Replacing Keyword 

We will now learn to calculate the amount in numbers, expressed in terms of percent by replacing the keyword 'of', with the multiplication operation and then solving it.

Let's solve the following example:

What is \(20\text{%}\) of \(12\) ?

Here, the word 'of' is a keyword for multiplication operation,

\(20\text {%}×12\)

Follow the given steps:

Step 1: Convert the percent to a decimal.

\(20\text {%}=0.20=0.2\)

Step 2: Write the problem again.

\(20\text {%}\) of \(12=0.2\) of \(12=0.2×12\)

\( \begin{array}[b]{r} 12 \\ ×0.2 \\ \hline 24 \\ \hline \end{array}\)

Since, we have a tenth place decimal,

\(\therefore\) \(0.2×12=2.4\)

Thus, \(20\text{%}\) of \(12\) is \(2.4\).

Illustration Questions

At a party, \(60\text{%}\) of the total guests are females. If the total guests are \(200\), how many females are there at the party?

A \(24\)

B \(84\)

C \(120\)

D \(100\)

×

Total number of guests \(=200\)

Number of females \(=60\text{%}\) of \(200\)

'Of' means  multiply

\(\therefore\) \(60\text{%}\) of \(200=60\text{%}×200 \)

Converting the percent to a decimal by moving the decimal point two places to the left. 

\(60\text{%}=0.6\)

Multiplying \(0.6\) and \(200\)

\(0.6×200\)

\( \begin{array}[b]{r} 200 \\ ×0.6 \\ \hline 1200 \\ \hline \end{array}\)

Since, we have a tenth place decimal,

\(\therefore\) \(0.6×200=120.0=120\)

Thus, there are \(120\) females at the party.

Hence, option (C) is correct.

At a party, \(60\text{%}\) of the total guests are females. If the total guests are \(200\), how many females are there at the party?

A

\(24\)

.

B

\(84\)

C

\(120\)

D

\(100\)

Option C is Correct

Percent as a Part of a Whole 

A percent is a part of a whole, where whole is the total quantity.

Let's consider the following example:

\(12\) is what percent of \(60?\)

Here, \(60\) represents the whole and \(12\) represents a part of \(60\).

To solve the above problem, consider the following steps:

Step 1 : Assume the percent as a variable.

Let's assume that \(12\) is \(x\text{%}\) of \(60\).

Step 2: Write the problem as an equation.

 \(x\text{%}\) of \(60=12\)

Step 3: Solve the problem.

 \(x\text{%}\) of \(60=12\)

 \(\Rightarrow\,\dfrac {x}{100}×60=12\) (Converting  \(x\text{%}\) into a fraction and 'of' means multiply)

\(\Rightarrow\,\dfrac {x}{100}×\dfrac {60}{1}=12 \) (Converting \(60\) into a fraction by putting it over \(1\))

\(\Rightarrow\,\dfrac {x×60}{100×1}=12\) (Multiplying both the fractions)

\(\Rightarrow\,\dfrac {60x}{100}=12\)

\(\Rightarrow\,\dfrac {60x}{100}=\dfrac {12}{1}\) ( Converting \(12\) into a fraction by putting it over \(1\))

Applying cross-multiplication method:

\(60x×1=12×100\)

\(60x=1200\)

Dividing both sides of the equation by \(60\) to calculate \(x\).

\(\dfrac {60x}{60}=\dfrac {1200}{60}\)

\(\Rightarrow x=\dfrac {1200}{60}\)

\(x=20\)

Step 4: Write the answer in the required form.

\(x=20\)

Thus, \(12\) is \(20\text {%}\) of \(60\).

Illustration Questions

Jose took out \(36\) colored pencils from a box containing a total of  \(120\) pencils.  What percent of the total number of pencils did Jose take out?

A \(35\text{%}\)

B \(30\text{%}\)

C \(84\text{%}\)

D \(36\text{%}\)

×

Given: Number of pencils Jose took out \(=36\)

Assuming the percent as a variable.

Let Jose took out \(x\text{%}\) of \(120\) pencils.

 

Writing the problem as an equation and solving it.

\(x\text{%}\) of \(120=36\)

'of' means multiply 

\(\therefore\) \(x\text{%}×120=36\)

Converting  \(x\text{%}\) into a fraction.

\(\dfrac {x}{100}×120=36\)

Converting \(120\) and \(36\) into fractions by putting them over \(1\).

\(\dfrac {x}{100}×\dfrac {120}{1}=\dfrac {36}{1}\)

Multiplying both the fractions.

\(\dfrac {120x}{100}=\dfrac {36}{1}\)

Applying cross-multiplication method,

image

\(120x×1=36×100\)

\(120x=3600\)

Dividing both sides of the equation by \(120\) to calculate \(x\).

\(\dfrac {120x}{120}=\dfrac {3600}{120}\)

\(\Rightarrow x=\dfrac {3600}{120}=30\)

Thus, Jose took out \(30\text{%}\) of the total number of pencils.

Hence, option (B) is correct.

Jose took out \(36\) colored pencils from a box containing a total of  \(120\) pencils.  What percent of the total number of pencils did Jose take out?

A

\(35\text{%}\)

.

B

\(30\text{%}\)

C

\(84\text{%}\)

D

\(36\text{%}\)

Option B is Correct

Percent through Figures

  • We can represent percents through figures.

Case I: When the number of figures are given

Five figures are given, out of which two are shaded.

Now the question is,

"What percent of the figures are shaded?"

Here, 

Total number of figures \(=5\)

Number of shaded figures \(=2\)

We can say that \(2\) out of \(5\) figures are shaded.

\(=2\) out of \(5\)

\(=\dfrac {2}{5}\)

To calculate the percent, we should follow the given steps:

\(\dfrac {2}{5}=\dfrac {?}{100}\)

To make the denominator \(100\) , \(2\) and \(5\), both should be multiplied with \(20\).

\(\dfrac {2×20}{5×20}=\dfrac {40}{100}\)

\(\dfrac {2}{5}=\dfrac {40}{100}=40\text{%}\)

Thus, \(40\text{%}\) figures are shaded.

Case II: When only one figure is given (Grid)

  • Consider a grid which has \(100\) small cells, some cells are shaded and some are left unshaded.
  • What percent of the grid is shaded?

    Here,

    Total number of cells \(=100\)

    Number of cells shaded \(=40\)

    There are \(40\) cells out of \(100\) which are shaded.

    \(\therefore\) Shaded cells \(=\dfrac {40}{100}=40\text{%}\)

    Here, we observe that

    \(40\) shaded cells represent \(40\text{%}\).

  • Thus, we can say that  \(1\) shaded cell will represent \(1\text{%}\).
  • Hence, each cell of this grid \((10×10)\) represents \(1\text{%}\).

Case II: When only one figure is given (Grid)

  • Consider a grid which has \(100\) small cells, some cells are shaded and some are left unshaded.

What percent of the grid is shaded?

Here,

Total number of cells \(=100\)

Number of cells shaded \(=40\)

\(\because\)  There are \(40\) cells out of \(100\) which are shaded.

\(\therefore\) Shaded cells \(=\dfrac {40}{100}=40\text{%}\)

Here, we observe that

\(40\) shaded cells represent \(40\text{%}\).

  • so, we can say that  \(1\) shaded cell will represent \(1\text{%}\).
  • Thus, each cell of this grid \((10×10)\) represents \(1\text{%}\).

Illustration Questions

Different number of blocks are given in each option. Some blocks out of the total blocks are shaded. Which one of the following options represents that \(30\text{%}\) blocks are shaded?

A

B

C

D

×

Option (A):

image

Total number of blocks \(=6\)

Number of shaded blocks \(=3\)

\(=3\text{ out of }6\)

\(=\dfrac {3}{6}\)

Simplifying \(\dfrac {3}{6}\)

\(\dfrac {3}{6}=\dfrac {1}{2}\)

Calculating the percent.

\(\dfrac {1}{2}=\dfrac {?}{100}\)

To make the denominator \(100\)\(1\) and \(2\), both should be multiplied with \(50\).

\(\dfrac {1×50}{2×50}=\dfrac {50}{100}=50\text{%}\)

In this option, \(50\text{%}\) blocks are shaded.

Hence, option (A) is incorrect.

Option (B):

image

Total number of blocks \(=5\)

Number of shaded blocks \(=1\)

\(=1\text{ out of }5\)

\(=\dfrac {1}{5}\)

\(\dfrac {1}{5}\) is already in its simplest form.

\(\therefore\) Calculating the percent. 

\(\dfrac {1}{5}=\dfrac {?}{100}\)

To make the denominator \(100\), \(1\) and \(5\), both should be multiplied with \(20\).

\(\dfrac {1×20}{5×20}=\dfrac {20}{100}\)

\(\dfrac {1}{5}=\dfrac {20}{100}=20\text{%}\)

In this option, \(20\text{%}\) blocks are shaded.

Hence, option (B) is incorrect.

Option (C):

image

Total number of blocks \(=10\)

Number of shaded blocks \(=7\)

\(=7\text{ out of }10\)

\(=\dfrac {7}{10}\)

\(\dfrac {7}{10}\) is already in its simplest form.

\(\therefore\) Calculating the percent.

\(\dfrac {7}{10}=\dfrac {?}{100}\)

To make the denominator \(100\), \(7\) and \(10\), both should be multiplied with \(10\).

\(\dfrac {7}{10}=\dfrac {70}{100}=70\text{%}\)

In this option, \(70\text{%}\) blocks are shaded.

Hence, option (C) is incorrect.

Option (D):

image

Total number of blocks \(=10\)

Number of shaded blocks \(=3\)

\(=3\text{ out of }10\)

\(=\dfrac {3}{10}\)

\(\dfrac {3}{10}\) is already in a simplified form.

\(\therefore\) Calculating the percent

\(\dfrac {3}{10}=\dfrac {?}{100}\)

To make the denominator \(100\)\(3\) and \(10\), both should be multiplied with \(10\).

\(\dfrac {3}{10}=\dfrac {30}{100}=30\text{%}\)

In this option, \(30\text{%}\) blocks are shaded.

Hence, option (D) is correct.

Different number of blocks are given in each option. Some blocks out of the total blocks are shaded. Which one of the following options represents that \(30\text{%}\) blocks are shaded?

A image
B image
C image
D image

Option D is Correct

Word Problems

  • Since percents, decimals and fractions are all parts of the whole, so we can easily compare and arrange them in the required order.
  • To understand it more clearly, let's consider the following example:
  • Mrs.Thomson prepares her household budget in the first week of each month.
  • In January, she decides to spend \($1255\).
  • She would spend \($62.77\) on groceries, \(35\text {%}\) on rent, and three-fifths on miscellaneous bills.
  • On which of the items listed above would she spend the most?

To determine this answer, we should compare these expenditures.

The amount she would spend on:

Groceries = \($62.77\)

Rent \(=35\text {%}\) of \($1255\)

\(=\dfrac {35}{100}×1255\)

\(=\dfrac {7}{20}×1255\)

\(=\dfrac {7}{4}×251\) \(=\dfrac {1757}{4}=$439.25\)

Miscellaneous bills = Three - fifth of \(1255\)

\(=\dfrac {3}{5}×1255=$753.00\)

Now, on comparing \($62.77,\;$439.25\) and \($753.00\), we get that

\($753.00>$439.25>$62.77\)

\(\therefore\)  She would spend maximum on her bills.

Illustration Questions

At a dairy farm, three workers, Kara, Larry and Cooper, get the same amount of work to be completed in maximum of 12 hours. If Kara completes the task in \(8\dfrac {1}{2}\) hours, Larry in two fifth of the given hours and Cooper takes \(75\text{%}\) of the allotted time, who is the most efficient?

A Kara is the most efficient.

B Larry is the most efficient.

C Cooper is more efficient than Larry.

D Cooper is the most efficient.

×

By observing the given problem, we get to know that we should compare  who completes the task in minimum amount of time.

For Kara,

Time taken by Kara \(=8\dfrac {1}{2}\) hours

\(8\dfrac {1}{2}=8+\dfrac {1}{2}\)

and  \(\dfrac {1}{2}=0.5\)

\(\therefore\;8\dfrac {1}{2}=8.5\) hours

For Larry,

Time taken by Larry

= Two-fifth of the given hours

\(=\dfrac {2}{5}\) of \(12\)

 \(=\dfrac {2}{5}×12\)

\(=\dfrac {24}{5}=4.8\) hours

For Cooper,

Time taken by Cooper

\(=75\text {%}\) of \(12\)

\(=.75×12\)

\(=9\) hours

Since Larry takes the least amount of time to complete the task, therefore, he is the most efficient.

Hence, option (B) is correct.

At a dairy farm, three workers, Kara, Larry and Cooper, get the same amount of work to be completed in maximum of 12 hours. If Kara completes the task in \(8\dfrac {1}{2}\) hours, Larry in two fifth of the given hours and Cooper takes \(75\text{%}\) of the allotted time, who is the most efficient?

A

Kara is the most efficient.

.

B

Larry is the most efficient.

C

Cooper is more efficient than Larry.

D

Cooper is the most efficient.

Option B is Correct

Percent through Circle Graph-II 

  • A circle graph is a way of displaying data.
  • A full circle represents the \(100\text{%}\).
  • It is divided into a number of sections, known as pie shaped wedges.
  • Each wedge represents a percent of the whole.

Let us consider the following example:

Here, the circle represents the percentage of students that play different sports, at a school.

The total number of students is \(200\).

Now the question is....... "How many students play baseball, cricket and basketball?"

We can calculate it by using the concept of "Percent of a number".

Thus, the number of students who play

Baseball = \(50\text {% of }200\)

\(=.50×200=100\)

Cricket = \(20\text {% of }200\)

\(=.20×200=40\)

Basketball = \(30\text {% of }200\)

\(=.30×200=60\)

This means,  \(100\) students play Baseball, \(40\) students play Cricket, and \(60\) students play Basketball.

Illustration Questions

James starts a business of fruit selling. The circle graph represents the percentage of different types of fruits he has for selling. If there are a total of \(600\) fruits to sell, find how many apples are there?

A \(146\)

B \(132\)

C \(360\)

D \(115\)

×

Total number of fruits \(=600\)

From the circle graph,

Apples \(=22\text {%}\)

Number of apples \(=22\text {% of }\,600\)

\(=.22×600\)

\(=132\)

Hence, option (B) is correct.

James starts a business of fruit selling. The circle graph represents the percentage of different types of fruits he has for selling. If there are a total of \(600\) fruits to sell, find how many apples are there?

image
A

\(146\)

.

B

\(132\)

C

\(360\)

D

\(115\)

Option B is Correct

Calculating Percent using Proportion

We will now learn to calculate the amount in numbers, expressed in terms of percent by using proportion.

For example: \(13\text{% of }30\)

Step 1: Write the percent as a fraction.

\(13\text{%}=\dfrac {13}{100}\)

Step 2: Write the problem as a proportion.

(Let \(13\text{% of }30\) is \(x\))

\(\dfrac {13}{100}=\dfrac {x}{30}\)

Step 3: Apply cross-multiplication method.

\(13×30=x×100\)

\(390=100x\)

Divide both sides by \(100\) to calculate \(x\).

\(\dfrac {390}{100}=x\)

\(\Rightarrow x=3.9\)

Thus, \(13\text{% of } 30\) is \(3.9\).

\(13×30=x×100\)

\(390=100x\)

Divide both the sides by \(100\) to calculate \(x\).

\(\dfrac {390}{100}=x\)

\(\Rightarrow x=3.9\)

Thus, \(13\text{% of } 30\) is \(3.9\)

Illustration Questions

In a class of \(40\) students, \(45\text{%}\) students secured A+ grade. Find the number of students who got A+ grade.

A \(32\)

B \(21\)

C \(8\)

D \(18\)

×

Total number of students \(=40\)

Number of students getting A+

\(=45\text{% of } 40\)

Writing \(45\text{%}\) as a fraction.

\(45\text{%}=\dfrac {45}{100}\)

Writing the problem as a proportion. 

\(\dfrac {45}{100}=\dfrac {x}{40}\) (Let \(45\text{%}\) of \(40\) be \(x\))

Applying cross-multiplication method:

image

\(45×40=x×100\)

\(1800=100x\)

Dividing both sides by \(100\).

\(\dfrac {1800}{100}=\dfrac {100}{100}x\)

\(18=x\)

\(\Rightarrow x=18\)

Thus, \(45\text{% of } 40 \text{ is } 18\).

This means \(18\) students secured A+ grade.

Hence, option (D) is correct.

In a class of \(40\) students, \(45\text{%}\) students secured A+ grade. Find the number of students who got A+ grade.

A

\(32\)

.

B

\(21\)

C

\(8\)

D

\(18\)

Option D is Correct

Practice Now