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Byju's Answer
Standard VIII
Mathematics
Reducing Equations to Simpler Form
six years hen...
Question
six years hence a mans age will be three times his sons age and three years ago he was nine times as old as his son . find their present ages
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Solution
Dear student
Let
the
present
age
of
man
be
x
years
and
the
present
age
of
his
son
be
y
years
Three
years
ago
Man
'
s
age
=
(
x
-
3
)
years
Son
'
s
age
=
(
y
-
3
)
years
It
is
given
that
three
years
ago
man
was
nine
times
as
old
as
his
son
i
.
e
x
-
3
=
9
(
y
-
3
)
⇒
x
-
3
=
9
y
-
27
⇒
x
=
9
y
-
24
.
.
.
.
(
1
)
Six
years
hence
Man
'
s
age
=
(
x
+
6
)
years
Son
'
s
age
=
(
y
+
6
)
years
It
is
given
that
six
years
hence
man
will
three
times
as
old
as
his
son
i
.
e
x
+
6
=
3
(
y
+
6
)
⇒
x
+
6
=
3
y
+
18
⇒
x
=
3
y
+
12
.
.
.
.
.
(
2
)
Equate
(
1
)
and
(
2
)
,
we
get
9
y
-
24
=
3
y
+
12
⇒
6
y
=
36
⇒
y
=
6
Put
the
value
of
y
in
(
2
)
,
we
get
x
=
3
(
6
)
+
12
⇒
x
=
18
+
12
⇒
x
=
30
So
,
Man
'
s
present
age
=
30
years
Son
'
s
present
age
=
6
years
Regards
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Q.
Six years hence a man's age will be three times the age of his son and three years ago he was nine times as old as his son. Find their present ages.
