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Age problem (m03q13) [#permalink]
 14 Jan 2009, 16:20
Question Stats:
85% (02:17) correct
14% (02:11) wrong based on 118 sessions
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 Source: GMAT Club Tests - hardest GMAT questions
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vksunder wrote: 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
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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
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graduatetutor wrote: 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
I agree with this strategy....it's much quicker then algebra
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Let H=Husband, W=Wife, D=Daughter, S=Son.
Express all of them in terms of S so that you have only one unknown.
S = S
D = S+2 [2 y. older than the son]
H = 7S [7 times older than the son]
W = 7S-4 [wife 4 years younger than the hubby]
S+(S+2)+7S+(7S-4) = 78
16S - 2 = 78
16S = 80
S=5
D=S+2
D=5+2=7
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Re: Age problem (m03q13) [#permalink]
12 May 2010, 09:12
easy calculation... Option B=7 yrs
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Re: Age problem (m03q13) [#permalink]
12 May 2010, 11:31
relatively easy question; agree, answer is B (7).
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Re: Age problem (m03q13) [#permalink]
12 May 2010, 11:37
Option B. quite easy question. 
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Re: Age problem (m03q13) [#permalink]
12 May 2010, 20:13
Simpal one , Ans:B h+w+d+s=78........1 h=4+w................2 d=2+s..................3 h=7s.....................4 re-arrange above equation 7s+7s-4+2+s+s=78 16s=80 s=5 so,d=5+2=7
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Re: Age problem (m03q13) [#permalink]
12 May 2010, 23:39
I will go with option B)7
S = 5
D = 7
H = 35
W = 31
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Re: Age problem (m03q13) [#permalink]
27 Nov 2010, 23:30
Shouldn't the wording of the question be "The father is 7 times as old as his son"?I got the question wrong because I
wrote the equation as f=s+7
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Re: Age problem (m03q13) [#permalink]
16 May 2011, 04:52
H + W + D + S = 78 D = S + 2 H = W + 4 H = 7S H + H - 4 + H/7 + H/7 + 2 = 78 => 2H + 2H/7 = 80 => H = 80/16 * 7 = 35 => S = H/7 = 5 years => D = 5 + 2 = 7 years Answer - B
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Re: Age problem (m03q13) [#permalink]
16 May 2011, 07:42
since father is 7* son's age. son's age cannot be more than 10. options C,D and E are gone. check for daughters age = 5, son's age = 3 father age = 21 and wife's age = 17. total does not add up to 78. Hence B an obvious choice.
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Re: Age problem (m03q13) [#permalink]
16 May 2011, 10:10
I really have to learn to recognize those easy strategies more quickly. Algebra was fairly quick, but the other way is a huge time saver!
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Re: Age problem (m03q13) [#permalink]
16 May 2011, 10:59
Answer is B)7
Explanation:Assuming wife's age to be x,husband age is x+4.
Assuming son's age to be y,daughter's age is y+2.
Given,
x+(x+4)+y+(y+2) = 78---(1)
7y=x+4------------------(2)
Solving (1) and (2) gives
y=5
So,daughter's age is 7.
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Re: Age problem (m03q13) [#permalink]
18 May 2011, 08:32
Plugging in worked for me a lot quicker. Started with C and worked way backwards
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graduatetutor wrote: 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 Agree...much faster approach. I guess approaches like this in questions like these will help us accumulate time for the difficult ones.
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Re: Age problem (m03q13) [#permalink]
18 May 2012, 06:12
Suppose Daughter's age = x Son = (x - 2) Husband = 7*(x-2) = (7x - 14) Wife = 7*(x-2) - 4 = (7x - 18) So, x + (x - 2) + (7x - 14) + (7x - 18) = 78 => 16x = 112 => x =7 Answer: Daughter's age is 7. B is the correct answer. If you like my submission, please do not forget to click on kudos. Cheers!!
Last edited by vshrivastava on 18 May 2012, 22:23, edited 1 time in total.
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GMAT TIGER wrote: vksunder wrote: 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
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Re: Age problem (m03q13) [#permalink]
18 May 2012, 07:39
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. Answer: B.
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Re: Age problem (m03q13)
[#permalink]
18 May 2012, 07:39
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