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Age problem (m03q13)

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Age problem (m03q13) [#permalink] New post 14 Jan 2009, 15:20
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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 as old as the son, how old is the daughter?

(A) 5
(B) 7
(C) 10
(D) 13
(E) 14

[Reveal] Spoiler: OA
B

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Last edited by Bunuel on 12 Apr 2014, 02:08, edited 1 time in total.
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Re: Age problem [#permalink] New post 14 Jan 2009, 17:41
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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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Re: Age problem [#permalink] New post 16 Jan 2009, 23:57
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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

Answer is B

check: daughter 7 yrs , son 5 yrs , father 35 yrs , wife 31 yrs ...sum 78 yrs
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Re: Age problem [#permalink] New post 04 May 2009, 06:06
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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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Re: Age problem (m03q13) [#permalink] New post 12 May 2010, 19:13
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] New post 12 May 2010, 22:39
I will go with option B)7

S = 5
D = 7
H = 35
W = 31
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Re: Age problem (m03q13) [#permalink] New post 27 Nov 2010, 22:30
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] New post 16 May 2011, 03:52
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] New post 16 May 2011, 06:42
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] New post 16 May 2011, 09:59
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] New post 18 May 2011, 07:32
Plugging in worked for me a lot quicker. Started with C and worked way backwards
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Re: Age problem (m03q13) [#permalink] New post 18 May 2012, 05:12
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.

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Last edited by vshrivastava on 18 May 2012, 21:23, edited 1 time in total.
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Re: Age problem [#permalink] New post 18 May 2012, 06:29
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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Re: Age problem (m03q13) [#permalink] New post 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.

Answer: B.
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Re: Age problem (m03q13) [#permalink] New post 21 May 2013, 04:29
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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Re: Age problem (m03q13) [#permalink] New post 21 May 2013, 08:45
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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Re: Age problem (m03q13) [#permalink] New post 21 May 2013, 09:11
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

Correct answer is B.
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Re: Age problem (m03q13) [#permalink] New post 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.
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Re: Age problem (m03q13) [#permalink] New post 21 May 2013, 16:59
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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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Re: Age problem (m03q13) [#permalink] New post 22 May 2013, 03:54
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, 03:54
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