Three years ago,the sum of the ages of A, B and C was 52 years. Four years ago…
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Three years ago,the sum of the ages of A, B and C was 52 years. Four years ago B, who is a year younger than C, was 1(1/2) times as old as A. The sum of the ages of B and C three years hence will be -
- A.
45 years
- B.
55 years
- C.
47 years
- D.
51 years
Attempted by 8 students.
Show answer & explanation
Correct answer: D
Answer: 51 years
Reasoning and steps:
Let present ages of A, B, C be a, b, c respectively. Three years ago their sum was 52, so (a-3)+(b-3)+(c-3)=52 ⇒ a+b+c=61.
B is 1 year younger than C, so b = c - 1.
Four years ago, B was 1½ times as old as A: (b-4) = (3/2)(a-4). Multiply both sides by 2: 2(b-4)=3(a-4).
Use b = c-1 in a+b+c=61 ⇒ a+2b=60 ⇒ a=60−2b. Substitute into 2(b-4)=3(a-4):
2b−8 = 3(60−2b)−12 = 168−6b ⇒ 8b = 176 ⇒ b = 22. Then a = 60−2·22 = 16 and c = b+1 = 23.
Three years hence B + C = (22+3)+(23+3) = 25+26 = 51 years.