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Age problem (m03q13) [#permalink]
14 Jan 2009, 15:20

2

This post received KUDOS

This year, the sum of ages of the Perkins family members is 78. Currently, there are 4 members: husband, wife, daughter, and son. The husband is 4 years older than the wife. The daughter is two years older than the son. If the husband is 7 times as old as the son, how old is the daughter?

This year, the sum of ages of the Perkins family members is 78. Currently, there are 4 members: husband, wife, daughter, and son. The husband is 4 years older than the wife. The daughter is two years older than the son. If the husband is 7 times older than the son, How old is the daughter?

a) 5 b) 7 c) 10 d) 13 e) 14

It is pretty streight.

h+w+d+s = 78 .............................i h = w + 4 ...................................ii d = s + 2 .......................................iii h = 7s ........................................... iv

This year, the sum of ages of the Perkins family members is 78. Currently, there are 4 members: husband, wife, daughter, and son. The husband is 4 years older than the wife. The daughter is two years older than the son. If the husband is 7 times older than the son, How old is the daughter?

a) 5 b) 7 c) 10 d) 13 e) 14

This question can be solved with a strategy as well

let the answer choice c) 10 yrs be daughter , son will be 8 yrs , husband is 56yrs , wife will be 52 yrs ....sum of their ages is 126 ( much higher than 78 ) so eliminate C , D , E

lets work with A) 5 yrs be daughter , son will be 3 yrs , husband is 21 yrs , wife will be 17 yrs ....sum of their ages is 46 ( lower than 78 ) so eliminate A

Answer is B

check: daughter 7 yrs , son 5 yrs , father 35 yrs , wife 31 yrs ...sum 78 yrs

This year, the sum of ages of the Perkins family members is 78. Currently, there are 4 members: husband, wife, daughter, and son. The husband is 4 years older than the wife. The daughter is two years older than the son. If the husband is 7 times older than the son, How old is the daughter?

a) 5 b) 7 c) 10 d) 13 e) 14

It is pretty streight.

h+w+d+s = 78 .............................i h = w + 4 ...................................ii d = s + 2 .......................................iii h = 7s ........................................... iv

7s = w+4 w = 7s - 4

h+w+d+s = 78 .............................i 7s+7s-4+s+2+s = 78 16s = 80 s = 5 d = s+2 = 5+2 = 7

It's correct. Short cut is as follow H= 7S W=7S-4 D=S+2 H+W+D+S=78 Putting the values 7S+(7S-4)+(S+2)+S=78 16S-2=78 16S=80, S=5 D=5+2=7

Re: Age problem (m03q13) [#permalink]
18 May 2012, 06:39

Expert's post

This year, the sum of ages of the Perkins family members is 78. Currently, there are 4 members: husband, wife, daughter, and son. The husband is 4 years older than the wife. The daughter is two years older than the son. If the husband is 7 times as old as the son, how old is the daughter?

A. 5 B. 7 C. 10 D. 13 E. 14

Plug-in method would be the fastest for this question.

Check answer choice C first:

If the daughter is 10 years old then: the son is 10-2=8 years old, the husband is 7*8=56 years old and the wife is 56-4=52 years old. Total: 10+8+56+52=126>78. So, the daughter must be less than 10 years old: eliminate D and E too.

Check answer choice B:

If the daughter is 7 years old then: the son is 7-2=5 years old, the husband is 7*5=35 years old and the wife is 35-4=31 years old. Total: 7+5+35+31=78. Correct answer.

Re: Age problem (m03q13) [#permalink]
21 May 2013, 16:34

RifatSZA wrote:

hello fellas, please refer my quotation, "7 times older than" from the text. if it is said like "7 times", then we can come up with an integer.

The first time I tried 8, which means 7+1, but it did not generate a reasonable answer, so I guess that means the father is older than the son, and the age of the father is 7 times the age of the son. Guess it was not explicitly expressed.