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A father says to his son, "7 years ago, I was 7 times as old as you were and after 3 years ,I will be 3 times as old as you will be". Find their present ages. |
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Answer» Answer: 42 years and 12 years Step-by-step explanation: Let father's present age = x years and SON's present age = y years According to the first condition:x − 7 = 7 (y − 7)⇒ x − 7 = 7y − 49⇒ x − 7y = −49 + 7⇒ x − 7y = −42 ...........(1) According to the second condition:x + 3 = 3 (y + 3)⇒ x + 3 = 3Y + 9⇒ x − 3y = 9 − 3⇒ x − 3y = 6 ...........(2) Subtracting (1) from (2) we get, 4y = 48⇒ y = 48/4 = 12 Putting the VALUE of y in (1) we get,x − 7 × 12 = −42⇒ x − 84 = −42⇒ x = −42 + 84⇒ x = 42 Hence father's age is 42 years and son's present age is 12 years. PLS MARK ME THE BRAINLIEST... |
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