A and B are friends and their ages differ by 1 yrs. A's father D is twice as old as A and B is twice as old as his sister C. The ages of D and C differ by 40 yrs. Find the age of A and B
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Step-by-step explanation:
let the age of A = x
age of B = x+2
since, A's father D is twice as old as A
So, age of D = 2×x = 2x
and since B is twice as old as his sister
So, age of C = age of B / 2 = (x+2)/2
according to the question,
The ages of D and C differ by 40yrs.
it means,
age of D - age of C = 40
=> 2x - [(x+2)/2] = 40
=> 2x - (x+2)/2 = 40
=> 4x/2 - (x+2)/2 = 40
=> (4x - x + 2)/2 = 40
=> 3x+2/2 = 40
=> 3x + 2 = 40 × 2
=> 3x = 80 - 2
=> x = 78/3
= 26
So, age of A = x = 26yr.s
age of B = x + 2 = 26 + 2 = 28yr.s
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### sakshi
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Step-by-step explanation:
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