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The sum of the ages of a man and his wife is six times the sum of the ages of their children. Two years ago the sum of their ages was ten times the sum of the ages of their children. Six years hence the sum of their ages will be three times the sum of the ages of their children. How many children do they have?
The sum of the ages of a man and his wife is six times the sum of the ages of their children. Two years ago the sum of their ages was ten times the sum of the ages of their children. Six years hence the sum of their ages will be three times the sum of the ages of their children. How many children do they have?
Let present age of man = M
Let present age of wife = W
Let the no. of children = x
Let the sum of the ages of children = C
ATQ
M + W = 6 C … (1)
M – 2 + W – 2 = 10 (C – 2x ) … (2)
M + 6 + W + 6 = 3 (C + 6x) … (3)
From (2) M + W – 4 = 10 C – 20 x
By using (1) 6 C – 4 = 10 C – 20 x
– 4C + 20x = 4
-C + 5x = 1 …. (5)
From (3) M + W + 12 = 3C + 18x
By using (1) 6C + 12 = 3C + 18x
3C – 18x = – 12
C – 6x = – 4 …. (6)
From (5) & (6)
– C + 5x = 1
C – 6x = – 4
∴ – x = – 3
x = 3