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The sum of husband and his wife is 4 times the sum of ages of their children.
Four years ago, the ratio of their ages to the sum of ages of their children was 18:1.
Two years hence,the ratio will be 3:1.
How many children do the have?
The answer is 4.
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1 solution
Let dad's age be x, mum's age y, the average age of the children be z. The sum of the children's ages would be nz where n is the number of children. so x+y=4nz or n z = 4 x + y .
Now 4 years ago, each child would have been 4 years younger so the sum back then was nz - 4n. The sum of the parents ages is x-4+y-4 = x+y-8. So x+y-8 = 18(nz-4n) = 18nz-72n.
Similarly, 2 years on, every child would age by 2 years, so will both parents hence we have x+2+y+2 = 3(nz+2n) = 3nz+6n.
Substitute n z = 4 x + y into the 2 equations we just formed and upon simplification, we would get 7(x+y) = 144n-16 and x+y = 24n-16. It follows that 7(24n-16) = 144n-16 so 24n=96 and n = 2 4 9 6 = 4 .