Informative line

Calculation Of 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

Finding Whole from a Part 

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

Let's consider the following example:

If \(60\text{%}\) of a number is \(42\), find the number.

Here, the number which is to be found out, represents the whole.

To solve the above problem, consider the following steps:

Step 1 : Assume the number as a variable.

Let the number be \(x\).

Step 2: Write the problem as an equation.

\(60\text{%}\) of \(x \) is \(42\)

\(\Rightarrow60\text{%}\) of \(x=42\)

Step 3: Solve the equation.

\(60\text{%}\) of \(x=42\) ( 'of' means multiply)

\(60\text{%}×x=42\)

\(\dfrac {60}{100}×x=42\) (converting \(60\text{%}\) into a fraction)

\(\dfrac {60}{100}×\dfrac {x}{1}=42\) (converting \(x\) into a fraction by putting it over \(1\))

\(\dfrac {60×x}{100}=42\) (multiplying both the fractions)

\(\dfrac {60x}{100}=\dfrac {42}{1}\) ( converting \(42\) into a fraction by putting it over \(1\))

Applying cross-multiplication method.

\(60x×1=42×100\)

\(60x=4200\)

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

\(\dfrac {60x}{60}=\dfrac {4200}{60}\)

\(x=70\)

Step 4: Write the answer.

\(x=70\)

Thus, \(60\text {% of } 70\) is \(42\).

Illustration Questions

\(75\text {%}\) of the total students of grade \(6\) have passed the final examination. If \(24\) students have cleared the exam, find the total number of students in grade \(6\).

A \(30\)

B \(32\)

C \(64\)

D \(51\)

×

We are asked to find out 'the total number of students'.

Let the total number of students \(=x\) 

Writing the problem in the form of an equation.

\(75\text {%}\) of \(x\) is \(24\)

\(\Rightarrow75\text {%}\) of \(x=24\)

Solving the equation.

\(75\text {%}\) of \(x=24\)

\(\because\;\) 'of' means multiply

\(\therefore75\text {%}×x=24\)

Converting \(75\text {%}\) into a fraction.

\(\dfrac {75}{100}×x=24\)

Converting \(x\) and \(24\) into fractions by putting them over \(1\).

\(\dfrac {75}{100}×\dfrac {x}{1}=\dfrac {24}{1}\)

Multiplying both the fractions.

\(\dfrac {75×x}{100×1}=\dfrac {24}{1}\)

\(\Rightarrow \dfrac {75×x}{100}=\dfrac {24}{1}\)

Applying cross-multiplication method.

image

\(75x×1=24×100\)

\(75x=2400\)

Dividing both sides by \(75\) to calculate \(x\) .

\(\dfrac {75x}{75}=\dfrac {2400}{75}\)

\(\Rightarrow x=32\)

There are total \(32\) students in grade \(6\).

Hence, option (B) is correct.

\(75\text {%}\) of the total students of grade \(6\) have passed the final examination. If \(24\) students have cleared the exam, find the total number of students in grade \(6\).

A

\(30\)

.

B

\(32\)

C

\(64\)

D

\(51\)

Option B 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 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

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