Olympiad Test : Problems On Ages


20 Questions MCQ Test Maths Olympiad Class 6 | Olympiad Test : Problems On Ages


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QUESTION: 1

Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand’s present age in years?

Solution:

Let the present ages of Sameer and Anand be
5x years and 4x years respectively.

⇒ 9(5x + 3) = 11(4x + 3)
⇒ 45x + 27 = 44x + 33
⇒ 45x – 44x = 33 – 27
⇒ x = 6
∴ Anand’s present age = 4x = 4 × 6 = 24 years.

QUESTION: 2

A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, then how old is B?

Solution:

Let C’s age be x years. Then, B’s age
= 2x years. A’s age = (2x + 2) years.
∴ (2x + 2) + 2x + x = 27
⇒ 5x = 25
⇒ x = 5 Hence, B’s age = 2x = 2 × 5 = 10 years.

QUESTION: 3

A father said to his son, “I was as old as you are at the present at the time of your birth”. If the father’s age is 38 years now, the son’s age five years back was:

Solution:

Let the son’s present age be x years. Then,
(38 – x) = x
⇒ 2x = 38
⇒ x = 19
∴ Son’s age 5 years back = (19 – 5) = 14 years

QUESTION: 4

The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?

Solution:

Let the ages of children be x, (x + 3), (x + 6),
(x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
⇒ 5x = 20
⇒ x = 4
∴ Age of the youngest child = x = 4 years.

QUESTION: 5

Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit’s age. After further 8 years, how many times would he be of Ronit’s age?

Solution:

Let Ronit’s present age be x years.
Then, father’s present age = (x + 3x) years = 4x years.
∴ (4x + 8) = 5/2(x + 8)
⇒ 8x + 16 = 5x + 40
⇒ 3x = 24
⇒ x = 8
Hence, required ratio = (4x + 16)/ (x + 16) = 48/24 = 2

QUESTION: 6

The age of father 10 years ago was thrice the age of his son. Ten years hence, father’s age will be twice that of his son. The ratio of their present ages is:

Solution:

Let the ages of father and son 10 years ago be 3x and years respectively.
Then, (3x + 10) + 10 = 2[(x + 10) + 10]
⇒ 3x + 20 = 2x + 40
⇒ x = 20
∴ Required ratio = (3x + 10) : (x + 10)
= 70 : 30 = 7 : 3.

QUESTION: 7

Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q’s age?

Solution:

Given that:
1.  The difference of age b/w R and Q =
The difference of age b/w Q and T.
2.  Sum of age of R and T is 50 i.e. (R + T) = 50
Question: R – Q = ?
Explanation:
R – Q = Q – T
⇒ ( R + T) = 2Q
Now given that (R+ T) = 50
So, 50 = 2Q and therefore Q = 25
Question is (R – Q) = ?
Here we know the value(age) of Q (25), but we don’t know the age of R.
Therefore, (R – Q) cannot be determined.

QUESTION: 8

A person’s present age is two-fifth of the age of his mother. After 8 years, he will be onehalf of the age of his mother. How old is the mother at present?

Solution:

Let the mother’s present age be x years.
Then, the person’s present age = 2x/5 years.
∴ (2x/5 + 8) = 1/2 (x +8)
⇒ 2(2x + 40) = 5(x + 8)
⇒ x = 40

QUESTION: 9

Ayesha’s father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents?

Solution:

Mother’s age when Ayesha’s brother was born = 36 years.
Father’s age when Ayesha’s brother was born = (38 + 4) years = 42 years.
∴ Required difference = (42 – 36) years  = 6 years

QUESTION: 10

The present ages of three persons are in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).

Solution:

Let their present ages be 4x, 7x and 9x years respectively.
Then, (4x – 8) + (7x – 8) + (9x – 8) = 56
⇒ 20x = 80
⇒ x = 4
∴ Their present ages are 4x = 16 years,
7x = 28 years and 9x = 36 years respectively.

QUESTION: 11

A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:

Solution:

Let the son’s present age be x years.
Then, man’s present age = (x + 24) years.
∴ (x + 24) + 2 = 2(x + 2)
⇒ x + 26 = 2x + 4
⇒ x = 22

QUESTION: 12

Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar’s age at present?

Solution:

Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.
Then, (6x + 6) + 4 = 11
(5x + 6) + 4 = 10
⇒ 10(6x + 10) = 11(5x + 10)
⇒ 5x = 10
⇒ x = 2
∴ Sagar’s present age = (5x + 6) = 16 years.

QUESTION: 13

The sum of the present ages of a father and his son is 60 years. Six years ago, father’s age was five times the age of the son. After 6 years, son’s age will be:

Solution:

Let the present ages of son and father be x and (60 – x) years respectively.
Then, (60 – x) – 6 = 5(x – 6)
⇒ 54 – x = 5x – 30
⇒ 6x = 84
⇒ x = 14
∴ Son’s age after 6 years = (x + 6) = 20 years.

QUESTION: 14

At present, the ratio between the ages of Arun and Deepak is 4 : 3. After 6 years, Arun’s age will be 26 years. What is the age of Deepak at present?

Solution:

Let the present ages of Arun and Deepak be
4x years and 3x years respectively. Then,
4x + 6 = 26  4x = 20
⇒ x = 5
∴ Deepak’s age = 3x = 15 years.

QUESTION: 15

Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio of 7 : 9, how old is Sachin?

Solution:

Let Rahul’s age be x years.
Then, Sachin’s age = (x – 7) years.
∴ (x – 7) / x = 7/9
⇒ 9x – 63 = 7x
⇒ 2x = 63
⇒ x = 31.5
Hence, Sachin’s age = (x – 7) = 24.5 years.

QUESTION: 16

The ratio of the father’s age to the son’s age is 4 : 1. The product of their ages is 196. What will be the ratio of their ages after 5 years?

Solution:

Let the ratio of their proportionality be x.
Then,
4x × x = 196
⇒ 4x2 = 196
⇒ x = 7
Thus, father’s age = 28 years and son’s age = 7 years
After 5 years, father’s age = 33 years and son’s age = 12 years
∴ Ratio =33 : 12 = 11 : 4

QUESTION: 17

The ratio of Rita’s age to the age of her mother is 3 : 11. The difference of their ages is 24 years. What will be the ratio of their ages after 3 years?

Solution:

Difference in ratios = 8
Then 8 ≡ 24
∴ 1 ≡ 3
Thus, value of 1 in ratio is equivalent to 3 years.
Thus, Rita’s age = 3 × 3 = 9 years
Mother’s age = 11 × 3 = 33 years
After 3 years, the ratio = 12 : 36 = 1 : 3

QUESTION: 18

The sum of ages of a mother and her daughter is 50 years. Also 5 years ago, the mother’s age was 7 times the age of the daughter. What are the present ages of the mother and the daughter?

Solution:

Let the age of the daughter be x years.
Then, the age of the mother is (50 – x) years
5 years ago, 7(x – 5) = 50 – x – 5
or, 8x = 50 – 5 + 35 = 80
∴ x = 10
Therefore, daughter’s age = 10 years and mother’s age = 40 years.

QUESTION: 19

The sum of the ages of a son and a father is 56 years. After 4 years, the age of the father will be three times that of the son. What is the age of the son?

Solution:

Let the age of the son be x years.
Then, the age of the father is (56 – x) years
After 4 years, 3(x + 4) = 56 – x + 4 or,
4x = 56 + 4 – 12 = 48
∴ x = 12 years Thus, son’s age = 12 years

QUESTION: 20

10 years ago, Sita’s mother was 4 times older than her daughter. After 10 years, the mother will be two times older than the daughter. What is the present age of Sita?

Solution:

Daughter’s age

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