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Two years ago a man was five times as old as his son. Two years later his age will be 8 more than |
Answer» Given:-
To FIND out:-Find the present ages of the man and his son. Solution:-Let the present age of his son be a. Then, the present age of the father = b According to the question, ★ Two years ago,
⇒( b - 2 ) = 5 ( a - 2 ) ⇒ b - 2 = 5a - 10
⇒ b = 5a - 8 .......( 1 ) ★ Two years later, b + 2 = 8 + 3 ( a + 2 ) ⇒ b + 2 = 8 + 3a + 6 ⇒ b + 2 = 3a + 14 ⇒ b = 3a + 12.......( 2 ) From ( 1 ) and ( 2 ) 5a - 8 = 3a + 12 ⇒ 5a - 3a = 12 + 8 ⇒ 2a = 20 ⇒ a = 10 Now, Substituting the value of a in EQUATION ( 1 ) b = 5a - 8 b = 5 × 10 - 8 b = 50 - 8 b = 42 Hence,the present age of man and his son are 42 and 10 respectively. |
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