Question

The ratio of two numbers is \[2:3\]. If 2 is subtracted from first and 8 from second, the ratio becomes reciprocal of the original ratio. Find the Numbers.

Answer 1

Hint:Here in the question, we have to find the two numbers by using a given two ratio condition. For this, we will consider the two numbers as \[x\] and \[y\] respectively and solve the two ratio conditions and further simplify by the elimination method of the system of linear equations to get the required solution.

Complete step by step answer:
Consider the question: Given, the ratio of two numbers is \[2:3\]. 2 is subtracted from first and 8 from second, the ratio becomes reciprocal of the original ratio. We need to find the two numbers. Let us consider the two numbers as \[x\] and \[y\] respectively.Consider the given first condition:
The ratio of two numbers is \[2:3\]. As we know that the ratio will be written in the form of fraction, then
\[ \Rightarrow \,\,\,\dfrac{x}{y} = \dfrac{2}{3}\]
On cross multiplication, we get
\[ \Rightarrow \,\,\,3x = 2y\]
Take all the RHS term to LHS, then
\[ \Rightarrow \,\,\,3x - 2y = 0\] -----(1)

Now consider, the second condition: 2 is subtracted from first and 8 from second, the ratio becomes reciprocal of the original ratio.Here, \[x\] is the first term and \[y\] is the second term, then by the condition we can written as
\[ \Rightarrow \,\,\,\dfrac{{x - 2}}{{y - 8}} = \dfrac{3}{2}\]
On cross multiplication, we get
\[ \Rightarrow \,\,\,2\left( {x - 2} \right) = 3\left( {y - 8} \right)\]
\[ \Rightarrow \,\,\,2x - 4 = 3y - 24\]
Take all the RHS term to LHS, then
\[ \Rightarrow \,\,\,2x - 4 - 3y + 24 = 0\]
\[ \Rightarrow \,\,\,2x - 3y + 20 = 0\]
Subtract 20 on both side, then we have
\[ \Rightarrow \,\,\,2x - 3y = - 20\] ------(2)
Solve equation (1) and (2) to get the value of \[x\] and \[y\]
\[3x - 2y = 0\]
\[\Rightarrow 2x - 3y = - 20\]

Now we have to solve these two equations to find the unknown
Multiply (1) by 2 and (2) by 3, then we get
\[6x - 4y = 0\]
\[\Rightarrow 6x - 9y = - 60\]
Since the coefficients of \[x\] and \[y\] are the same, we change the sign by the alternate sign.
\[\underline {
   + 6x\,\,\,\, - 4y\,\,\,\, = 0 \\
  \mathop + \limits_{( - )} 6x\mathop - \limits_{( + )} 9y = \mathop - \limits_{( + )} 60 } \\
  \]
Now we cancel the \[x\] term and simplify other terms, so we have
\[
  \underline {
   + 6x\,\,\,\, - 4y\,\,\,\, = 0 \\
  \mathop + \limits_{( - )} 6x\mathop - \limits_{( + )} 9y = \mathop - \limits_{( + )} 60 } \\
    \\
  5y = 60 \\
 \]
\[ \Rightarrow \,\,5y = 60\]
divide 5 on both side, then
\[ \Rightarrow \,\,y = \dfrac{{60}}{5}\]
On simplification, we get
\[\therefore \,\,y = 12\]

We have found the value of \[y\], now we can find the value of \[x\] by substituting the value \[y\]to any one of the equations (1) or (2) . we will substitute the value of \[y\]to equation (1).
Therefore, we have \[3x - 2y = 0\]
\[ \Rightarrow 3x - 2\left( {12} \right) = 0\]
\[ \Rightarrow 3x - 24 = 0\]
Add 24 on both side, we have
\[ \Rightarrow 3x = 24\]
Divide 3 on both side, then
\[ \Rightarrow x = \dfrac{{24}}{3}\]
On simplification, we get
\[\therefore \,\,x = 8\]

Hence, the required two numbers are 8 and 12.

Note:Ratio problems are word problems that use ratios to relate or compare the different items or numbers in the question. It can be simplified by using a proper arithmetic operation. In the elimination method to eliminate the term we must be aware of the sign where we change the sign by the alternate sign and we have made the one variable term have the same coefficient such that it will be easy to solve the equation.