age problem

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age problem

by coolhabhi » Sat Nov 16, 2013 2:03 am
Five years ago, the age of a person was 3 years more than four times his son's age. Three years hence, the age of the person will be six years less than thrice the age of the son. After how many years from now will their combined age be 80 years?
A) 22
B) 32
C) 16
D) 11
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by theCodeToGMAT » Sat Nov 16, 2013 3:41 am
Person's current age = P
Son's current age = S

(P-5) = 4(S-5) + 3

P - 5 = 4S - 20 + 3

P = 4S - 12 --(1)

(P+3) = 3(S+3) -6
P + 3 = 3S + 9 -6
P = 3S
USING IN (1)
S = 12
P = 36
P + S = 48
To be 80 = 80-48/2 = 32/2 = 16

Answer [spoiler]{C}[/spoiler]?
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by GMATGuruNY » Sat Nov 16, 2013 5:06 am
coolhabhi wrote:Five years ago, the age of a person was 3 years more than four times his son's age. Three years hence, the age of the person will be six years less than thrice the age of the son. After how many years from now will their combined age be 80 years?
A) 22
B) 32
C) 16
D) 11
Algebra is probably the best approach here.
But for those who would struggle with the algebra, an alternate approach is to GUESS AND CHECK.

Case 1: 5 years ago, let S=5
Five years ago, the age of a person was 3 years more than four times his son's age.
P = 4*5 + 3 = 23.

Today:
Since each is 5 years older than the values above, we get:
S = 5+5 = 10.
P = 23+5 = 28.

3 years hence, the age of the person will be six years less than thrice the age of the son.
Since the son is 3 years older, we get:
S = 10+3 = 13.
Since the father is 6 years less than 3 times the son's age, we get:
P = 3*13 - 6 = 33.
Doesn't work:
If today P=28, then 3 years from now P=28+3 = 31.

Case 2: 5 years ago, let S=6
Five years ago, the age of a person was 3 years more than four times his son's age.
P = 4*6 + 3 = 27.

Today:
Since each is 5 years older than the values above, we get:
S = 6+5 = 11.
P = 27+5 = 32.

3 years hence, the age of the person will be six years less than thrice the age of the son.
Since the son is 3 years older, we get:
S = 11+3 = 14.
Since the father is 6 years less than 3 times the son's age, we get:
P = 3*14 - 6 = 36.
Doesn't work:
If today P=32, then 3 years from now P=32+3 = 35.

Case 3: 5 years ago, let S=7
Five years ago, the age of a person was 3 years more than four times his son's age.
P = 4*7 + 3 = 31.

Today:
Since each is 5 years older than the values above, we get:
S = 7+5 = 12.
P = 31+5 = 36.

3 years hence, the age of the person will be six years less than thrice the age of the son.
Since the son is 3 years older, we get:
S = 12+3 = 15.
Since the father is 6 years less than 3 times the son's age, we get:
P = 3*15 - 6 = 39.
Success!
If today P=36, then 3 years from now P=36+3 = 39.

Thus:
Sum of the ages today = 12+36 = 48.
For the sum to increase to 80 -- by a total of 32 years -- each age must increase by 16 years.

The correct answer is C.

Not the most efficient approach, but doable in a reasonable amount of time if we get started right away.
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by Mathsbuddy » Sat Nov 16, 2013 10:45 am
(A) p-5=4(s-5)+3 -> p=4s-12

(B) p+3=3(s+3)-6 -> p=3s

(C) p+x + s+x = 80 -> 2x=80-p-s

Subtract (A)-(B):

0=s-12 -> s=12

Subsitute this into (B): p=3*12=36

Subsitute p and s into (C):

2x=80-36-12=32

x=16