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Problems on Ages  01
01.
The ratio of a man's age and his son's age is 7 : 3 and the product of their ages is 756. The ratio of their ages after 6 years will be ?
1
5 : 2
2
2 : 1
3
11 : 7
4
13 : 9
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Answer:
1
Explanation:
Solution:
Let the man's age be 7x years ,
Then son's age = 3x years
∴
7
x
×
3
x
=
756
⇒
21
x
2
=
756
⇒
x
2
=
756
⇒
x
2
=
36
⇒
x
=
6
$$\begin{array}{rl}& \therefore \text{7}x\times 3x=756\\ & \Rightarrow 21{x}^{2}=756\\ & \Rightarrow {x}^{2}=756\\ & \Rightarrow {x}^{2}=36\\ & \Rightarrow x=6\end{array}$$
The ratio of their ages after 6 years
=
(
7
x
+
6
)
:
(
3
x
+
6
)
=
(
7
×
6
+
6
)
:
(
3
×
6
+
6
)
=
48
:
24
=
2
:
1
$$\begin{array}{rl}& \text{=}\left(7x+6\right):\left(3x+6\right)\\ & =\left(7\times 6+6\right):\left(3\times 6+6\right)\\ & =48:24\\ & =2:1\end{array}$$
02.
Reenu's father was 38 years of age when she was born while her mother was 36 years old when her brother 4 years younger to her was born. What is the difference between the ages of her parents ?
1
2 years
2
4 years
3
6 years
4
8 years
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Answer:
1
Explanation:
Solution:
Mother's age when Reenu's brother was born = 36 year
Father's age when Reenu's brother was born = (38 + 4) years = 42 year
Required difference = (42  36) years = 6 years
03.
A couple has a son and a daughter. The age of the father is four times that of the son and the age of the daughter is onethird of that of her mother. The wife is 6 years younger to her husband and the sister is 3 years older then her brother. The mother's age is = ?
1
42 years
2
48 years
3
54 years
4
63 years
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Answer:
1
Explanation:
Solution:
M
→
Mother,
F
→
Father,
S
→
Son and
D
→
Daughter
F
=
4
S
,
D
=
1
3
M
,
M
=
F
−
6
a
n
d
S
=
D
−
3
∴
M
=
3
D
=
3
(
S
+
3
)
=
3
S
+
9
=
3
4
F
+
9
=
3
4
(
M
+
6
)
+
9
=
3
4
M
+
3
4
×
6
+
9
⇒
(
M
−
3
4
M
)
=
(
9
2
+
9
)
⇒
1
4
M
=
27
2
⇒
M
=
(
27
2
×
4
)
=
54
years
$$\begin{array}{rl}& M\to \text{Mother,}F\to \text{Father,}S\to \text{Son and}D\to \text{Daughter}\\ & F=\text{}4S,\text{}D\text{}=\frac{1}{3}M,\text{}M=\text{}F6\text{}and\text{}S=\text{}D3\\ & \therefore M=3D=3\left(S+3\right)\\ & =3S+9=\frac{3}{4}F+9=\frac{3}{4}\left(M+6\right)+9\\ & =\frac{3}{4}M+\frac{3}{4}\times 6+9\\ & \Rightarrow \left(M\frac{3}{4}M\right)=\left(\frac{9}{2}+9\right)\\ & \Rightarrow \frac{1}{4}M=\frac{27}{2}\\ & \Rightarrow M=\left(\frac{27}{2}\times 4\right)\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=54\phantom{\rule{thinmathspace}{0ex}}\text{years}\end{array}$$
∴ The mother is 54 years old.
04.
Rajan got married 8 years ago. His present age is times his age at the time of his marriage. Rajan's sister was 10 years younger to him at the time of his marrige. The age of Rajan's sister is = ?
1
32 years
2
36 years
3
38 years
4
40 years
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Answer:
1
Explanation:
Solution:
Let Rajan's age 8 years ago be x years,
Then, present age = (x + 8) years
∴
x
+
8
=
6
5
x
⇒
5
x
+
40
=
6
x
⇒
x
=
40
$$\begin{array}{rl}& \therefore x+8=\frac{6}{5}x\\ & \Rightarrow 5x+40=6x\\ & \Rightarrow x=40\end{array}$$
Rajan's sister's age 8 years ago
= (40  10) years
= 30 years
∴ His sister's age now
= (30+8) years
= 38 years
05.
Ram's son's age is
1
3
$$\frac{1}{3}$$
of Ram's wife's age. Ram's wife's age is
4
5
$$\frac{4}{5}$$
of Ram's age and Ram's age is
3
5
$$\frac{3}{5}$$
of Ram's father's age. Find the age of Ram's son, if Ram's father is 50 years old ?
1
6 years
2
8 years
3
10 years
4
12 years
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Answer:
1
Explanation:
Solution:
Ram's father's age
= 50 years,
Ram's age
=
(
3
5
×
50
)
years
= 30 years
Ram's wife's age
=
(
4
5
×
30
)
years
= 24 years
Ram's son's age
=
(
1
3
×
24
)
years
= 8 years
$$\begin{array}{rl}& \text{Ram's father's age}\\ & \text{= 50 years,}\\ & \text{Ram's age}\\ & \text{=}\left(\frac{3}{5}\times 50\right)\text{years}\\ & \text{= 30 years}\\ & \text{Ram's wife's age}\\ & \text{=}\left(\frac{4}{5}\times 30\right)\text{years}\\ & \text{= 24 years}\\ & \text{Ram's son's age}\\ & \text{=}\left(\frac{1}{3}\times 24\right)\text{years}\\ & \text{= 8 years}\end{array}$$
06.
The sum of the ages of a daughter and her mother is 56 years. After 4 years, the age of the mother will be three times that of the daughter. At present their ages are ?
1
10 years, 46 years
2
12 years, 44 years
3
11 years, 45 years
4
13 years, 43 years
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Answer:
1
Explanation:
Solution:
Let daughter's age be x years.
Then, mother's age = 3 years
∴
(
3
x
+
12
)
=
2
(
x
+
12
)
⇒
3
x
+
12
=
2
x
+
24
⇒
x
=
12
$$\begin{array}{rl}& \therefore \left(3x+12\right)=2\left(x+12\right)\\ & \Rightarrow 3x+12=2x+24\\ & \Rightarrow x=12\end{array}$$
∴ Daughter's age today = 12 years
07.
The ages of A and B are presently in the ratio of 5 : 6 respectively. Six years hence, this ratio became 6 : 7 respectively. What was B's age 5 years ago ?
1
25 years
2
30 years
3
31 years
4
36 years
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Answer:
1
Explanation:
Solution:
Let A's age be 5x years.
Then, B's age = 6 years
∴
5
x
+
6
6
x
+
6
⇒
7
(
5
x
+
6
)
=
6
(
6
x
+
6
)
⇒
35
x
+
42
=
36
x
+
36
⇒
x
=
6
$$\begin{array}{rl}& \therefore \frac{5x+6}{6x+6}\\ & \Rightarrow 7\left(5x+6\right)=6\left(6x+6\right)\\ & \Rightarrow 35x+42=36x+36\\ & \Rightarrow x=6\end{array}$$
B's age 5 years ago = (6x  5) years
= (6 × 6  5) years
= 31 years
08.
The ages of Samina and Suhana are in the ratio of 7 : 3 respectively. After 6 years, the ratio of their ages will be 5 : 3. What is the difference in their ages?
1
6 years
2
8 years
3
10 years
4
12 years
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Answer:
1
Explanation:
Solution:
Let Samina's age be 7x years.
Then, Suhana's age = 3x years
∴
7
x
+
6
3
x
+
6
⇒
3
(
7
x
+
6
)
=
5
(
3
x
+
6
)
⇒
21
x
+
18
=
15
x
+
30
⇒
6
x
+
12
⇒
x
=
2
$$\begin{array}{rl}& \therefore \frac{7x+6}{3x+6}\\ & \Rightarrow 3\left(7x+6\right)=5\left(3x+6\right)\\ & \Rightarrow 21x+18=15x+30\\ & \Rightarrow 6x+12\\ & \Rightarrow x=2\end{array}$$
Difference in their ages = (7x  3x) years
= 4x years
= (4 × 2) years
= 8 years
09.
Ten years ago, a man was seven times as old as his son. Two years hence, twice his age will be equal to five times the age of his son. What is the present age of the son ?
1
12 years
2
13 years
3
14 years
4
15 years
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Answer:
1
Explanation:
Solution:
Let son's age 10 years ago be x years.
Then, man's age 10 years ago = 7 years
Son's present age = (x + 10) years,
Man's present age = (7x + 10) years
∴
2
[
(
7
x
+
10
)
+
2
]
= 5
[
(
x
+
10
)
+
2
]
⇒
2
(
7
x
+
12
)
=
5
(
x
+
12
)
⇒
14
x
+
24
=
5
x
+
60
⇒
9
x
=
36
⇒
x
=
4
$$\begin{array}{rl}& \therefore \text{2}\left[\left(7x+10\right)+2\right]\text{= 5}\left[\left(x+10\right)+2\right]\\ & \Rightarrow 2\left(7x+12\right)=5\left(x+12\right)\\ & \Rightarrow 14x+24=5x+60\\ & \Rightarrow 9x=36\\ & \Rightarrow x=4\end{array}$$
Son's present age = (x + 10) years
= (4 + 10) years
= 14 years
10.
The age of a father 10 years ago was thrice the age of his son. 10 years hence , the father's age will be twice that of his son. The ratio of their present age is = ?
1
8 : 5
2
7 : 3
3
9 : 5
4
5 : 2
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Answer:
1
Explanation:
Solution:
Let son's age 10 years ago be x years
Then, father age 10 years ago = 3 years
Son's age now = (x + 10) years,
Father age now = (3x + 10) years
(
3
x
+
10
)
+
10
=
2
[
(
x
+
10
)
+
10
]
⇒
3
x
+
20
=
2
(
x
+
20
)
⇒
3
x
+
20
=
2
x
+
40
⇒
x
=
20
$$\begin{array}{rl}& \left(3x+10\right)+10=2\left[\left(x+10\right)+10\right]\\ & \Rightarrow 3x+20=2\left(x+20\right)\\ & \Rightarrow 3x+20=2x+40\\ & \Rightarrow x=20\end{array}$$
Ratio of present ages of father and son
=
3
x
+
10
x
+
10
=
3
×
20
+
10
20
+
10
=
70
30
=
7
3
=
7
:
3
$$\begin{array}{rl}& \text{=}\frac{3x+10}{x+10}\\ & =\frac{3\times 20+10}{20+10}\\ & =\frac{70}{30}\\ & =\frac{7}{3}\\ & =7:3\end{array}$$
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