Question

The ages of the younger and elder brothers are in the ratio of 2:3. If the younger was 6 years older and the elder 5 years younger, the age of the younger would have been thrice the age of the elder. Find their present ages.

Answer 1

Hint: To solve the given question, we will assume that the present age of the younger brother is x and the present age of the elder brother is y. Then, we will take their ratio and equate it to 2:3. Then, we will assume the new age of the younger brother as x’ such that x’ = x + 6 and the new age of the elder brother as y’ such that y’ = y – 5. Again, we will take their ratio and equate it to 3:1. After doing this, we will get a pair of linear equations which we will solve with the help of the substitution method.

Complete step by step solution:
To start with, we will assume that the present age of younger brother is x and the present age of elder brother is y. According to the question, the ratio of their age is 2:3. Thus, we will get the following equation
\[\dfrac{x}{y}=\dfrac{2}{3}\]
\[\Rightarrow 3x=2y\]
\[\Rightarrow 3x-2y=0......\left( i \right)\]
Now, let us assume that the new age of the younger brother is x’. This new age is the age if the younger brother would have been 6 years older. Thus, we get,
\[{{x}^{'}}=x+6....\left( ii \right)\]
Now, let us assume the new age of the elder brother be y’. The new age is the age if the elder brother would have been 5 years younger. Thus, we will get,
\[{{y}^{'}}=y-5.....\left( iii \right)\]
Now, it is given in the question that the ratio of their new age is 3:1. Thus, we will get the following equation
\[\dfrac{{{x}^{'}}}{{{y}^{'}}}=\dfrac{3}{1}\]
\[\Rightarrow {{x}^{'}}=3{{y}^{'}}.....\left( iv \right)\]
Now, we will put the value of x’ and y’ from (ii) and (iii) into (iv). Thus, we will get,
\[\Rightarrow x+6=3\left( y-5 \right)\]
\[\Rightarrow x+6=3y-15\]
\[\Rightarrow x+6+15=3y\]
\[\Rightarrow 3y-x=21....\left( v \right)\]
Now, (i) and (v) are a pair of linear equations in two variables. We will solve it with the substitution method. In the substitution method, we write one variable in terms of others from one equation and put the value of that variable in other equations. Thus, from equation (i), we have,
\[3x-2y=0\]
\[\Rightarrow 3x=2y\]
\[\Rightarrow y=\dfrac{3x}{2}.....\left( vi \right)\]
Now, we will put the value of y in (v). Thus, we will get,
\[\Rightarrow 3\left( \dfrac{3x}{2} \right)-x=21\]
\[\Rightarrow \dfrac{9x}{2}-x=21\]
\[\Rightarrow \dfrac{7x}{2}=21\]
\[\Rightarrow x=\dfrac{2\times 21}{7}\]
\[\Rightarrow x=6\text{ years}\]
Now, we will put the value of x in (vi). Thus, we will get,
\[\Rightarrow y=\dfrac{3}{2}\left( 6\text{ years} \right)\]
\[\Rightarrow y=9\text{ years}\]
Thus, the present age of the younger brother is 6 years and the present age of the elder brother is 9 years.

Note: The pair of linear equations that are formed during the solution can also be solved by the method of elimination. For this method, we will multiply equation (v) with 3 and add in equation (i). Thus, we will get,
\[3\left( 3y-x \right)+3x-2y=3\left( 21 \right)\]
\[\Rightarrow 9y-3x+3x-2y=63\]
\[\Rightarrow 7y=63\]
\[\Rightarrow y=\dfrac{63}{7}\]
\[\Rightarrow y=9\text{ years}\]
On putting this value in (i), we get,
\[3x-2\left( 9 \right)=0\]
\[\Rightarrow 3x=18\]
\[\Rightarrow x=6\text{ years}\]