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Q1. | Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit’s age. After further 8 years, how many times would he be of Ronit’s age? |

A. | 2 times [Correct Answer] |

B. | 2(1/2) times [Wrong Answer] |

C. | 2(3/4) times [Wrong Answer] |

D. | 3 times [Wrong Answer] |

View Answer
Explanation:-
Answer : ADiscuss it below :Let the age of father be ‘x’ years and son be ‘y’ yearsGiven : Father age is 3 times more than his son Ronit. X = 3y x- 3y = 0 ------------------(1) Given : after 8 years, father would be 2 (1/2) of Ronit age X + 8 = 5/2 (y +8) ⇒ 2x +16 = 5y +40 ⇒ 2x – 5y = 24 ----------(2) To find: After further 8 years, how many times would be of Ronit age If (x + 16) = z (y +16) ; z = ? Multiply (1) by 2 and subtract with (2) 2x- 6y = 0 (-) 2x – 5y = 24 From this solving of above equations we get y = 24 and x = 72 Therefore (x + 16) = z (y +16) (72 +16) = z (24 +16) 88 = z (40) 88/40 = z ⇒ 2.2 ; z = 2 !! OOPS Login [Click here] is required to post your answer/resultHelp other students, write article, leave your comments |

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