Problems on Ages
- 15 years later, A will be twice as old as B but five years ago A was 4 times as old as B. Find the difference of their present ages ?
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Let A's age = x years and B's age = y years
As per the first condition,
∴ (x + 15) = 2(y + 15)
⇒ x - 2 y = 15 ....(i)
As the per second condition,
∴ (x - 5) = 4(y - 5)
⇒ x - 4 y = -15 ....(ii)Correct Option: C
Let A's age = x years and B's age = y years
As per the first condition,
∴ (x + 15) = 2(y + 15)
⇒ x - 2 y = 15 ....(i)
As the per second condition,
∴ (x - 5) = 4(y - 5)
⇒ x - 4 y = -15 ....(ii)
Solving (i) and (ii) one get's, x = 45, y = 15
∴ A's age = 45 years
B's age = 15 years
∴ Difference of their ages = 45 - 15 = 30 years
- The ratio of the present ages of a son and his father is 1 : 5 and that of his mother and father is 4 : 5. After 2 years the ratio of the age of the son to that of his mother becomes 3 : 10. What is the present age of the father ?
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∵ S / F = 1 / 5
⇒ F = 5 S, .....(i)
∵ M / F = 4 / 5
⇒ M = 4 / 5 F ....(ii)
∵ (S + 2) / (M + 2) = 3 / 10
⇒ 10S + 20 = 3M + 6 ....(iii)Correct Option: D
∵ S / F = 1 / 5
⇒ F = 5 S, .....(i)
∵ M / F = 4 / 5
⇒ M = 4 / 5 F ....(ii)
∵ (S + 2) / (M + 2) = 3 / 10
⇒ 10S + 20 = 3M + 6 ....(iii)
From (i), (ii) and (iii)
⇒ (12 - 10)S = 20 - 6
⇒ 2S = 14
∴ S = 7 years
∴ F = 5S = 5 x 7 = 35 years.
