B is 18 years younger than A. The ratio of B's age six years hence to C's Present age is 3 : 2. If at present A's age is twice the age of C, then what was B's age four years ago?
Answers
B's age four years ago was 26 years.
Step-by-step explanation:
B is 18 years younger than A so, the eq. can be written as,
B = A – 18
⇒ A – B = 18 ……. (i)
The ratio of B’s age six years hence to C’s present age is 3:2 so, the eq. will be,
[B+6]/C = 3/2 …… (ii)
At present, A's age is twice the age of C so, the eq. will be,
A = 2C
⇒ C = A/2 …… (iii)
Substituting (iii) in (ii), we get
[B+6]/[A/2] = 3/2
⇒ 2[B + 6] = A (3/2)
⇒ 4[B + 6] = 3A
⇒ 4B + 24 = 3A
⇒ 3A – 4B = 24 ….. (iv)
Now, by multiplying eq. (i) by 3 and subtracting eq. (ii) from it, we get
3A – 3B = 54
3A – 4B = 24
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B = 30 ← present age of B
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Thus,
B’s age 4 years ago is given by,
= B – 4
= 30 – 4
= 26 years
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