Problems on Ages
- If C's age is twice the average age of A, B and C, A's age is one half the average of A, B and C. If B is 5 years old, the average age of A, B and C is ?
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Let the average age of A, B and C be N years.
∴ Total age of A, B and C = 3 x N = 3N years
Now, according to the question,
∵ 3N - (2N + N/2) = 5Correct Option: A
Let the average age of A, B and C be N years.
∴ Total age of A, B and C = 3 x N = 3N years
Now, according to the question,
∵ 3N - (2N + N/2) = 5
∴ N = 10 years.
- Two years ago, a mother was four times as old as her daughter. 8 years later, mother's age will exceed her daughter's age by 12 years. The ratio of the present ages of mother and daughter is ?
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Let the mother's age 2 years ago be 4x and daughter's age 2 years ago be be x.
∵ (4x + 8) - (x + 8) = 12
⇒ 3x = 12
⇒ x = 4
∴ Mother's present age = 4x + 2 = 18 years
and daughter's present age = x + 2 = 6 yearsCorrect Option: A
Let the mother's age 2 years ago be 4x and daughter's age 2 years ago be be x.
∵ (4x + 8) - (x + 8) = 12
⇒ 3x = 12
⇒ x = 4
∴ Mother's present age = 4x + 2 = 18 years
and daughter's present age = x + 2 = 6 years
∴ Required ratio = 3 : 1
- Five years ago, the total of the ages of father and son was 60 years. The ratio of their present ages is 4 : 1. Then the present age of the father is ?
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Let the present age of the son be x and that of the father be 4x years.
∵ (x - 5) + (4x - 5) = 60Correct Option: C
Let the present age of the son be x and that of the father be 4x years.
∵ (x - 5) + (4x - 5) = 60
⇒ 5x = 70
∴ x = 14 years
∴ Father's present age = 4x = 56 years
- A is as much younger than B as he is older then C. If the sum of B's and C's age is 40 years. Find the age of A. ?
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Age of C < Age of A < Age of B
From question,
A = C + x ....(i)
B = A + x ....(ii)
B + C = 40.Correct Option: A
Age of C < Age of A < Age of B
From question,
A = C + x ....(i)
B = A + x ....(ii)
From equation (i) and (ii)
A - B = C - A
⇒ 2A = B + C
⇒ A = (B + C) / 2
Given that sum of the ages of B and C is 40 years.
So, A = (B + C) / 2 = 40/2 = 20 Years
- A says to B "I am twice as old as you were when I was as old as you are." The sum of their ages is 63 years. Find the difference of their ages ?
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Let the present age of A is x and present age of B is y.
Therefore, x + y = 63 ....(i)
Difference of their ages is = (x - y) years.
When A was as old as B then, A's age was 'y; years and B's age was [y - (x - y)] = (2y - x) years.
Given that present age of A is twice the past age of B.
∴ x = 2(2y - x)
⇒ 3x = 4y .....(ii)
From (i) and (ii)
x = 36 and y = 27
So the difference in age of A and B is 36 - 27 = 9 years.Correct Option: C
Let the present age of A is x and present age of B is y.
Therefore, x + y = 63 ....(i)
Difference of their ages is = (x - y) years.
When A was as old as B then, A's age was 'y; years and B's age was [y - (x - y)] = (2y - x) years.
Given that present age of A is twice the past age of B.
∴ x = 2(2y - x)
⇒ 3x = 4y .....(ii)
From (i) and (ii)
x = 36 and y = 27
So the difference in age of A and B is 36 - 27 = 9 years.
