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# Age problem (m03q13)

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14 Jan 2009, 16:20
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

Last edited by Bunuel on 12 Apr 2014, 03:08, edited 1 time in total.
Edited the question
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14 Jan 2009, 18:41
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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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17 Jan 2009, 00:57
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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

check: daughter 7 yrs , son 5 yrs , father 35 yrs , wife 31 yrs ...sum 78 yrs
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04 May 2009, 07:06
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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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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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12 May 2010, 23:39
I will go with option B)7

S = 5
D = 7
H = 35
W = 31
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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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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

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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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16 May 2011, 10:59

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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18 May 2011, 08:32
Plugging in worked for me a lot quicker. Started with C and worked way backwards
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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

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Last edited by vshrivastava on 18 May 2012, 22:23, edited 1 time in total.
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18 May 2012, 07:29
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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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.

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.

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.

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21 May 2013, 05:29
amit2k9 wrote:
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.

like this one age-problem-m03q13-74810.html#p560685
the try method is always very straight, but I love your method, it seems smarter!
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21 May 2013, 09:45
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.
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21 May 2013, 10:11
Let's say daughter = x
So, Son = (x - 2)
Husband = 7*(x - 2)
Wife = 7*(x - 2) - 4

Add the four: Sum = 16*x - 34 = 78
=> 16*x = 112 => x = 7

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21 May 2013, 17:34
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.
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21 May 2013, 17:59
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Let age of son be x
Father = 7x
Mother =7x - 4
Sister = x + 2

Sum = 16x -2 =78
x=5

Sister's age is x +2 =7
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22 May 2013, 04:54
easy calculation
ans is B
h+w+d+s = 78
hus = 35
wife = 31
dau = 7
son = 5
Re: Age problem (m03q13)   [#permalink] 22 May 2013, 04:54

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# Age problem (m03q13)

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