Question

A Bag contains white and black balls. Half the number of white balls is equal to one third of the number of black balls. Twice the total number of balls exceeds three times the number of black balls by$4$. How many balls of each colour does the bag contain?

Answer 1

Step 1: Represent the unknown quantities by two variables

Let the white ball be $A$ and the black ball be $B$

Step 2: Two equations using the assumed variables according to the question

$$\begin{gathered} \frac{1}{2}A=\frac{1}{3}B..............\left(1\right) \\ 2(A+B)=3B+4......\left(2\right)\Rightarrow 2A+2B-3B=4.........\left(3\right) \end{gathered}$$

Step 3: Using elimination method

$$\begin{gathered} \frac{1}{2}A-\frac{1}{3}B=0..........\left(1\right) \\ 2A-B=4...............\left(4\right) \end{gathered}$$

Multiplying equation $\left(1\right)$ by $4$

we get

$$\begin{gathered} \left(1\right)x4\Rightarrow 2A-\frac{4}{3}B=0........\left(5\right) \\ 2A=B=4........\left(4\right) \end{gathered}$$

Subtracting equations $\left(5\right)-\left(4\right)$

$$\begin{gathered} 0-\frac{4}{3}B-B=-4 \\ -\frac{4}{3}B+B=-4 \\ \frac{-4B+3B}{3}=-4 \\ -B=-4\times 3 \\ -B=-12 \\ B=12 \end{gathered}$$

Substitute the value of $B=12$ in equation$\left(4\right)$ by substitution method

we get

$$\begin{gathered} 2A-12=4 \\ 2A=4+12 \\ =16 \\ A=\frac{16}{2}=8 \end{gathered}$$

Hence, Number of white balls $\left(A\right)=8$

Number of Black Balls $\left(B\right)=12$