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Real World Problems with Rational Numbers

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Rational numbers are a part of Real Numbers but the only necessary condition is that, they need to be represented in the form of $\frac{p}{q}$, where p and q are integers and q $\neq$ 0.

Real world problems on Rational Numbers are those problems which represent a real life situation and we need to solve them. It basically involves some equations and expressions involving rational numbers.

To solve these problems, we have to apply different operations and properties of the rational numbers and then, we need to further simplify our solution to obtain the required answer.

Few examples of real world problems on rational numbers which will help you to understand the topic are given below:

Example 1:

A box contains some blue and yellow balls. The Ratio of number of blue balls to the number of yellow balls is 4:5. If the bag contains 100 balls, then how many blue balls are there in the bag?

Now, let 'y' be the number of blue balls. By expressing the ratio as a fraction, we get

$\frac{4}{5}$ = $\frac{y}{100}$

Now, this is an equation which can be solved by using cross multiplication technique,

4 $\times$ 100 = 5y
or
y = $\frac{400}{5}$

y = 80
$\therefore$ 80 blue balls are there in the bag.

Example 2:

The sum of half of a number and its reciprocal is the same as 51 divided by the number itself. Find the number.

Solution:

To solve this problem, let the required number be n.
Given: The sum of half of the number and its reciprocal is equal to 51 divided by the number itself.

$\frac{n}{2}$ + $\frac{1}{n}$ = $\frac{51}{n}$

$\frac{n}{2}$
+ $\frac{1}{n}$ = $\frac{51}{n}$

$\frac{n^{2}}{2n}$ + $\frac{2}{2n}$ = $\frac{51}{n}$

n2 + 2 = 102,

On simplifying, we obtain n2 = 100
n = 10
The required number is 10.

Example 3:

Two mechanics were working on a car. One can complete a given job in 6 hours. But, the new guy takes 8 hours. They work together for first two hours. But then, the first guy left to help another mechanic on a different job. How long will it take for the new guy to finish the car work?

Solution:

The first guy can do $\frac{1}{6}$ part of job per hour and the second guy can do $\frac{1}{8}$ part of job per hour and together they can do $\frac{1}{6}$ + $\frac{1}{8}$ part of job per hour.

Now, let ’t’ hours is the time to complete the car job. So, $\frac{1}{t}$ job will be completed per hour,

Equating the two expressions, we get:

$\frac{1}{6}$ + $\frac{1}{8}$ = $\frac{1}{t}$

$\frac{7}{24}$ = $\frac{1}{t}$

As they work for 2 hours, 2 $\frac{7}{24}$ = $\frac{14}{24}$ part of job will be done.

The work remaining is 1 - $\frac{1}{t}$ = (1 - $\frac{14}{24}$)

= $\frac{10}{24}$

$\therefore$ $\frac{10}{24}$ job is left which has to be completed by the second guy, who will take $\frac{10}{24}$ $\div$ $\frac{1}{8}$

= $\frac{40}{12}$

= $\frac{10}{3}$

= 3.33 hours to complete the car job.

Example 4:

Brian can lay a foundation for a house in 10 hours. Together Brian and Robin can lay a foundation in 6.5 hours. How long will it take Robin to lay a foundation when working alone?

Solution:

Now to solve this problem, let's say it will take x hours for Robin to lay a foundation.

Brian’s work per hour = $\frac{1}{10}$ and

Robin’s work per hour = $\frac{1}{x}$.

If they work together, then they will complete $\frac{1}{10}$ + $\frac{1}{x}$ work per hour which equals $\frac{1}{6.5}$. Now, equating the two expressions we obtain,

$\frac{1}{10}$ + $\frac{1}{x}$ = $\frac{1}{6.5}$

$\frac{x+10}{10x}$ = $\frac{1}{6.5}$

6.5(x +10) = 10x
13(x+10) = (10x).2
7x = 130
x = 18.57 hours,

It will take 18.57 hours for Robin to lay a foundation.

Real Word Problems

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Rational numbers are terminating or recurring decimal numbers written in the form of fraction $\frac{a}{b}$ in which 'a' and 'b' are integers and the denominator 'b' not equal to zero.

Rational numbers can include operations like addition, subtraction, multiplication and division.

Some of the solved examples of word problems with Rational Numbers are given below:

Solved Examples

Question 1: If Sachin covers $\frac{3}{4}$ km distance by foot and then, he covers $\frac{11}{3}$ km distance by bus, while he was going to school, how far is his house from the school?
Solution:
Distance covered by Sachin by foot = $\frac{3}{4}$ km

Distance covered by him by Bus = $\frac{11}{3}$ km

Total distance covered to reach school = $\frac{3}{4}$ + $\frac{11}{3}$ = $\frac{9}{12}$ + $\frac{44}{12}$ = $\frac{53}{12}$ km

So, the house is 4.42km far from Sachin's school.

Question 2: If Shobha has a red ribbon with length $\frac{25}{9}$ m and a piece of $\frac{5}{7}$ m is cut from it and gave to Seema. Find how much long ribbon is left with her?
Solution:
Shobha has a ribbon of length = $\frac{25}{9}$ m

Part of ribbon given to seema = $\frac{5}{7}$ m

So, the length of the ribbon left =$\frac{25}{9}$ - $\frac{5}{7}$ = $\frac{175}{63}$ - $\frac{45}{63}$ = $\frac{130}{63}$ m

So, the ribbon left with Shobha is 2.06m

Question 3: What should be multiplied with -$\frac{3}{5}$ to make 1?
Solution:
Let x be the number multiplied with -$\frac{3}{5}$ to get 1.

So, -$\frac{3}{5}$ * x = 1

x = -$\frac{5}{3}$

Question 4: If a basket has $\frac{35}{6}$ kg of grapes and if it is to be distributed among 7 children equally, how much grapes will the each child get?
Solution:
Total amount of grapes in the basket = $\frac{35}{6}$ kg

Number of students to which it is to be distributed equally = 7

So, each child gets grapes = $\frac{35}{6}$ divided by 7 = $\frac{35}{6}$ $\times$ $\frac{1}{7}$ = $\frac{5}{6}$ kg

So, each child gets 0.83 kg of grapes.