Average age of a family of 5 members at present is 35. If After some y

Question

Average age of a family of 5 members at present is 35. If After some years, the eldest member of the family will die, and twins are born just one year before death, such that 6 years after death, the average age of family will become 25.5 years. If the eldest member will die at the age of 85 years, find after how many years from present, will he die?

A 10 years
B 2 years
C 5 years
D 15 years
E 12 years

A 10 years
B 2 years
C 5 years
D 15 years
E 12 years

philipssonicare · Accepted Answer

What is the question source?

Zero chance that it is a sub-600 level. Too many changes in time. I initially missed the 6 years post-death and got 12.x as an answer.

ORIGINALLY
5 members of 35
5(35)=175 total age

AT TIME OF DEATH
The 5 original members minus the death plus the birth of the twins a year ago
 \frac{5(x+35)-85+2}{6}=y

6 YEARS POST DEATH
Same as above +6·6 for the extra years of all the people
 \frac{5(x+35)-85+2+36}{6}=25.5 
5x+175-47=153
5x+128=153
5x=25
x=5

GMATWhizTeam · Answer

Solution
 •  At present:  
 o Average age of the family  = \frac{The\space sum \space of \space ages\space of \space all \space5 \space family \space members}{5 }= 35 
 ⟹ The\space sum \space of \space ages\space of \space all \space5 \space family \space members = 35*5 = 175   
 • Let us assume that the eldest member of the family die after x years from present.
•  Just before the death of the eldest member of the family: 
 o The number of family members  = 7 
o The sum of ages of 5 members of the family(except twins)  = 175 + 5x 
o The sum of the ages of the twins  = 2*1 
o Total age of the family  = (175 + 5x + 2)= (177 +5x)   
 •  6 years after the death: 
 o The total number of family members  = 7 -1 =6 
o And their total ages  = (177 + 5x) – 85 + 6*6 = 128 +5x 
o Average age of the family  = \frac{128 +5x }{6} = 25.5 
  ⟹128 + 5x = 153 
 ⟹5x = 25  
 ⟹ x = 5    
Thus, the correct answer is  Option C.

madlol77777 · Answer

t => Average age of 4 members now (except eldest). 
x => Years before the death

\frac{(85-x+t)}{5} =35

In the numerator, 85-x is the edge of eldest ("death age" - x years from this date = age of the eldest now).

t-x=90  (1)

\frac{(t+4x+24)+14}{6} =25,5

In the numerator, t+4x is the increase in the age of the 4 remaining members before the death of the eldest. 14 is the number of years two twins will leave before the second average.

t+4x=115  (2)

Next, we perform (2)-(1):

5x=25
x=5

finisher009 · Answer

Sol:

Now: 5 people, avg 35 → total age = 175.

Twins born 1 year before death: from 0→t−1 yrs, +5 per year; from t−1→t, +7 (because twins are present); at death subtract the elder’s age (85).

Just after death: total = 175 + 5(t−1) + 7 − 85 = 92 + 5t.

6 years after death: add 6 people × 6 years = +36 → total = 128 + 5t.

Given avg then = 25.5 for 6 people → total 153 → 128 + 5t = 153 → t = 5.

Average age of a family of 5 members at present is 35. If After some y

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Average age of a family of 5 members at present is 35. If After some y

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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