Real World Problems With Rational Numbers

Rational numbers are the 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. Now 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' 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.

Now to solve this problem suppose that the required number is n.
given: The sum of half of the number and its reciprocal is equal to 51divided by the number itself.
$\frac{n}{2}$ + $\frac{1}{n}$ = $\frac{51}{n}$
$\frac{n}{2}$  + $\frac{1}{n}$  =  $\frac{51}{n}$
or
$\frac{n^{2}}{2n}$ + $\frac{2}{2n}$ = $\frac{51}{n}$
or
n² + 2 = 102,

On simplifying we obtain n² = 100
or 
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?

We will proceed in this way the first guy can do $\frac{1}{6}$ part of job per hour and 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  that is why $\frac{1}{t}$ job will be completed per hour,

Equating the two expressions we get:
$\frac{1}{6}$ + $\frac{1}{8}$ = $\frac{1}{t}$
or 
$\frac{7}{24}$ = $\frac{1}{t}$,

As they work for 2 hours so 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?

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,

$\frac{1}{10}$ + $\frac{1}{x}$ = $\frac{1}{6.5}$
or
$\frac{x+10}{10x}$ = $\frac{1}{6.5}$
or
6.5(x +10) = 10x
or
13(x+10) = (10x).2
or
7x = 130
or 
x = 18.57 hours,

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