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Standard VIII
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Mathematics
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Applications (Word Problem)
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Five years ag...
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.
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Solution
5 years ago,
Let age of son = x years
Then, age of father = 7x years
Present age of son = x + 5 years
and age of father = 7x + 5 years
5 years hence,
age of son = x + 5 + 5 = x + 10
and age of father = 7x + 5 + 5 = 7x + 10
According to the condition :
7
x
+
10
=
3
(
x
+
10
)
⇒
7
x
+
10
=
3
x
+
30
⇒
7
x
−
3
x
=
30
−
10
=
20
⇒
4
x
=
20
⇒
x
=
20
4
=
5
∴
Present age of son
=
x
+
5
=
5
+
5
=
10
years and
Age of father
=
7
x
+
5
=
7
×
5
+
5
=
35
+
5
=
40
years
