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Question
Present age of mother of Manish is 1 year more than 5 times the present age of Manish. Four years before, if the product of their ages was 22, then find the present age of Manish and his mother
Solution
Let the present age of Manish be x years.
The present age of the mother of Manish is 1 year more than 5 times the present age of Manish.
∴ The present age of the mother of Manish will be (5x + 1) years.
Four years before,
Manish’s age was (x – 4) years and
Mother’s age was (5x + 1 – 4) years i.e., (5x – 3) years
According to the given condition,
Four years before, the product of their ages was 22.
∴ (x – 4)(5x – 3) = 22
∴ 5x2 – 3x – 20x + 12 – 22 = 0
∴ 5x2 – 23x – 10 = 0
| 5 × (-10) = | -50 |
| +25 +2 | |
| (-25) × 2 = -50 | |
| -25 + 2 = -23 |
∴ 5x2 – 25x + 2x – 10 = 0
∴ 5x(x – 5) + 2(x – 5) = 0
∴ (x – 5)(5x + 2) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
x – 5 = 0 or 5x + 2 = 0
∴ x = 5 or x = `(-2)/5`
But, age cannot be negative.
∴ x = 5
∴ Manish’s present age = x = 5 years,
Mother’s present age = 5x + 1
= 5(5) + 1
= 25 + 1
= 26 years
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