Question

In three coloured boxes - Red, Green and Blue, \[108\] balls are placed. There are twice as many balls in the green and red boxes combined as there are in the blue box and twice as many in the blue box as there are in the red box. How many balls are there in the green box?

Answer 1

Hint: In the given question, we have to find the number of boxes in the green box out of the three given boxes. We are given data about the ratio of balls in the boxes. Hence, we can form an equation and use a substitution method to arrive at the answer.

Complete step-by-step answer:
Let the Red, Green and Blue boxes be \[R,G\] and \[B\] respectively.
We are given that there are total \[108\] balls in the three boxes.
Hence, we can form the following equation:
\[R + G + B = 108\] …(1)
Next, we are given that there are twice as many balls in the green and red boxes combined as there are in the blue box which can be written in form of equation as:
\[G + R = 2B\] …(2)
Next, we are given that there are twice as many in the blue box as there are in the red box which can be written in form of equation as:
\[B = 2R\] …(3)
Comparing equation (2) and (3), we can substitute the value of \[B\]and get-
\[G + R = 2(2R) = 4R\] …(4)
Comparing equation (4) and (1), we can substitute the value of \[G + R = 4R\] and \[B = 2R\]to get-
\[R + G + B = 108\]
\[4R + 2R = 108\]
\[6R = 108\]
\[R = {{108}}{6}\]
\[R = 18\]
Now, we can get the value of \[G\] by substituting \[R = 18\] in equation in equation (4):
\[G + 18 = 4(18)\]
\[G = 72 - 18\]
\[G = 54\].
Hence, the number of balls in the green box are \[54\] balls.

Note: The algebraic method for solving simultaneous linear equations is the substitution method. In this method, a pair of linear equations is converted into a single linear equation with only one variable, which can then be solved quickly.
We have to read the question carefully and form such an equation that it has only one unknown variable by analysing the relationship between the different variables.