
The sum of age of a father and his son is \[45\] years. Five years ago, the product of their ages (in years) was \[124\]. Determine their present ages.
Answer
622.8k+ views
Hint: Form linear equations in two variables based on the given data and then solve them to find the present ages of father and son.
We have some relations regarding the age of father and son. We have to find their exact present ages.
We will find their present ages by forming linear equations in two variables and then solving them.
Let’s assume that the present age of son is \[x\] years and the present age of father is \[y\] years.
We know that the sum of age of father and son is \[45\] years. Thus, we have \[x+y=45\]. \[...\left( 1 \right)\]
Five years ago, the son’s age would be \[x-5\] years and father’s age would be \[y-5\] years.
We know that five years ago, the product of son’s and father’s age was \[124\].
Thus, we have \[\left( x-5 \right)\left( y-5 \right)=124\]. \[...\left( 2 \right)\]
We now have two linear equations in two variables. We will solve them to get the present ages of father and son using elimination methods. In this method, we solve the system of linear equations by using the additional property of equality.
Rearranging the equation \[\left( 1 \right)\] to write variable \[x\] in terms of \[y\], we get \[x=45-y\].
Substituting the above equation in equation \[\left( 2 \right)\], we get \[\left( 45-y-5 \right)\left( y-5 \right)=124\].
Solving this equation, we have \[\left( 40-y \right)\left( y-5 \right)=124\].
\[\begin{align}
& \Rightarrow 40y-200-{{y}^{2}}+5y=124 \\
& \Rightarrow {{y}^{2}}-45y+324=0 \\
& \Rightarrow {{y}^{2}}-9y-36y+324=0 \\
& \Rightarrow y\left( y-9 \right)-36\left( y-9 \right)=0 \\
& \Rightarrow \left( y-9 \right)\left( y-36 \right)=0 \\
& \Rightarrow y=9,36 \\
\end{align}\]
If we substitute \[y=9\] in equation \[\left( 1 \right)\], we get \[x=36\]. But as \[x\] represents the age of the son, it can’t be greater than the age of the father. Thus, we can’t have \[y=9,x=36\].
If we substitute \[y=36\] in equation \[\left( 1 \right)\], we get \[x=9\].
Hence, we have the age of son as \[x=9\] years and the age of father as \[y=36\] years.
When we substitute these values in the given conditions, we observe that they satisfy the conditions.
Thus, we have the age of son as \[x=9\] years and the age of father as \[y=36\] years.
Note: We can also solve this question by using linear equations in one variable. We can assume the age of son as a variable\[x\]and then write the age of father in terms of age of son under the given conditions and solve the equations to get the present age of father and son.
We have some relations regarding the age of father and son. We have to find their exact present ages.
We will find their present ages by forming linear equations in two variables and then solving them.
Let’s assume that the present age of son is \[x\] years and the present age of father is \[y\] years.
We know that the sum of age of father and son is \[45\] years. Thus, we have \[x+y=45\]. \[...\left( 1 \right)\]
Five years ago, the son’s age would be \[x-5\] years and father’s age would be \[y-5\] years.
We know that five years ago, the product of son’s and father’s age was \[124\].
Thus, we have \[\left( x-5 \right)\left( y-5 \right)=124\]. \[...\left( 2 \right)\]
We now have two linear equations in two variables. We will solve them to get the present ages of father and son using elimination methods. In this method, we solve the system of linear equations by using the additional property of equality.
Rearranging the equation \[\left( 1 \right)\] to write variable \[x\] in terms of \[y\], we get \[x=45-y\].
Substituting the above equation in equation \[\left( 2 \right)\], we get \[\left( 45-y-5 \right)\left( y-5 \right)=124\].
Solving this equation, we have \[\left( 40-y \right)\left( y-5 \right)=124\].
\[\begin{align}
& \Rightarrow 40y-200-{{y}^{2}}+5y=124 \\
& \Rightarrow {{y}^{2}}-45y+324=0 \\
& \Rightarrow {{y}^{2}}-9y-36y+324=0 \\
& \Rightarrow y\left( y-9 \right)-36\left( y-9 \right)=0 \\
& \Rightarrow \left( y-9 \right)\left( y-36 \right)=0 \\
& \Rightarrow y=9,36 \\
\end{align}\]
If we substitute \[y=9\] in equation \[\left( 1 \right)\], we get \[x=36\]. But as \[x\] represents the age of the son, it can’t be greater than the age of the father. Thus, we can’t have \[y=9,x=36\].
If we substitute \[y=36\] in equation \[\left( 1 \right)\], we get \[x=9\].
Hence, we have the age of son as \[x=9\] years and the age of father as \[y=36\] years.
When we substitute these values in the given conditions, we observe that they satisfy the conditions.
Thus, we have the age of son as \[x=9\] years and the age of father as \[y=36\] years.
Note: We can also solve this question by using linear equations in one variable. We can assume the age of son as a variable\[x\]and then write the age of father in terms of age of son under the given conditions and solve the equations to get the present age of father and son.
Recently Updated Pages
Master Class 7 English: Engaging Questions & Answers for Success

Master Class 7 Maths: Engaging Questions & Answers for Success

Master Class 7 Science: Engaging Questions & Answers for Success

Class 7 Question and Answer - Your Ultimate Solutions Guide

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 Social Science: Engaging Questions & Answers for Success

Trending doubts
The value of 6 more than 7 is A 1 B 1 C 13 D 13 class 7 maths CBSE

Convert 200 Million dollars in rupees class 7 maths CBSE

How many lines of symmetry does a regular pentagon-class-7-maths-CBSE

List of coprime numbers from 1 to 100 class 7 maths CBSE

AIM To prepare stained temporary mount of onion peel class 7 biology CBSE

The plural of Chief is Chieves A True B False class 7 english CBSE


