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Five years ago, a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages. - Mathematics

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Question

Five years ago, a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.

Sum
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Solution

Five years ago:

Let the age of the son be x years.

Therefore, the age of the father will be 7x years. 

Thus, Present age of the son = (x + 5) years, and Present age of the father = (7x + 5) years,

After five years:

Age of the son = (x + 5 + 5) = (x + 10) years,

Age of the father = (7x + 5 + 5) = (7x + 10) years,

According to the given question,

7x + 10 = 3(x + 10)

7x − 3x = 30 − 10

4x = 20

∴ x = 5

Thus, Calculation of the ages,

(x + 5) years

= (5 + 5)

= 10 years,

(7x + 5) years

= (7 × 5 + 5)

= 40 years,

Hence, the present ages of the son and his father are 10 years and 40 years, respectively.

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Chapter 9: Linear Equation in One Variable - Exercise 9.4 [Page 30]

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RD Sharma Mathematics [English] Class 8
Chapter 9 Linear Equation in One Variable
Exercise 9.4 | Q 13 | Page 30
B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 5 Simultaneous Linear Equations
EXERCISE 5B | Q 8. | Page 59
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