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Standard VII

Mathematics

Solving Equations with Linear Expressions on One Side and Numbers on the Other Side

Five years ag...

Question

Five years ago, a man was seven times older than his son .Five years hence, he is three times older. Find out their current ages

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Solution

Let the past age of son = x
Let the past age of father = 7x
So, Present ages are x+5 and 7x+5
Future age of son = x+10 Future age of father = 7x+10
Given 7x+10 = 3*(x+10)7x+10 = 3x+30 7x-3x = 30-10 4x = 20 x = 20/4x = 5 ,7x = 35
So x=5
x+5 = 10 7x+5 = 40
current age of son 10 and father age is 40

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Five years ago, a man was seven times older than his son .Five years hence, he is three times older. Find out their current ages

Let the past age of son = x Let the past age of father = 7x So, Present ages are x+5 and 7x+5 Future age of son = x+10 Future age of father = 7x+10 Given 7x+10 = 3*(x+10)7x+10 = 3x+30 7x-3x = 30-10 4x = 20 x = 20/4x = 5 ,7x = 35 So x=5 x+5 = 10 7x+5 = 40 current age of son 10 and father age is 40

Let the past age of son = x
Let the past age of father = 7x
So, Present ages are x+5 and 7x+5
Future age of son = x+10 Future age of father = 7x+10
Given 7x+10 = 3*(x+10)7x+10 = 3x+30 7x-3x = 30-10 4x = 20 x = 20/4x = 5 ,7x = 35
So x=5
x+5 = 10 7x+5 = 40
current age of son 10 and father age is 40