Question

22. reversing the digits of father's age we get son's age. one year ago father was twice in age of that of his son? find their current ages?

Answer 1

Hi there!
Here's the answer:

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Let present age of father = 10x+y
where y- units digit of his age
& x - tens digit of his age


Reversing digits gives son's present age
=> Son's present age = 10y+x

Given that,
One year Ago,
Father's age = (10x+y) - 1
Son's age = (10y+x) - 1

Father's Age = 2 × (Son's age)
=> (10x+y) - 1 = 2 × [ (10y+x) - 1 ]
=> 10x+y-1 = 20y+2x-2
=> 8x-19y = -1
=> 8x = 19y-1
=> x = (19/8)y - (1/8)

To make satisfy the above condition satisfied, replace the values of x & y

• When y = 0
x - doesn't satisfy the condition
• When y = 1
x - doesn't satisfy the condition
• When y = 2
x - doesn't satisfy the condition


The only set of digits that satisfies this is
y = 3 & x = 7.


Father's present age = 10x+y = 73
& Son's present age = 10y+x = 37


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Verification:

F - Father's Age ; S - Son's age

F = 73
S = 37

One year ago,
F = 72
S = 36 = (1/2) × F = (1/2) × 72

•°• Given Data Matched.


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Hope it helps