Elitmus
Exam
Numerical Ability
Age Problem
the average of a group of 8 friends increases by 4 years when the youngest of them is not considered and decreases by 3 years when the oldest of them is not considered.find the difference between the ages of oldest and the youngest of them
Read Solution (Total 7)
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- let x be the average age of all friends and total age 8x
when youngest of them not considered total age 7x+7*4=7(x+4) as average is x+4
so the age of youngest 8x-7(x+4)=x-28
when the age of oldest not considered total age 7x-7*3=7(x-3) as average is x-3
so the age of oldest 8x-7(x-3)=x+21
so difference between their ages=x+21-(x-28)=49 - 10 years agoHelpfull: Yes(31) No(1)
- (x+6friend)/7=a+4
(6friend+y)/7=a-3
solve both eq.
we get x-y=49
- 11 years agoHelpfull: Yes(18) No(4)
- Let Average age of the group = x
age of the youngest person = a
age of the oldest person = z
so 1st condition : 8x-a = 7(x+4) => x-a = 28.....(1)
2nd condition : 8x-z = 7(x-3) => x-z = -21 or x = z-21.....(2)
Now put value of eq(2) in eq(1) => z-21-a = 28 => z-a = 49 Ans - 10 years agoHelpfull: Yes(4) No(0)
- let the average age of 6 frnds be a after removng the youngest and oldest both from the 8 frnds..... for 7 frnds including the youngest we have (y+6f)/7 = a-4 (as when the youngest is not there the average increases thus when he was there then the a average decreases by 4) similarly in second case we have (o+6f)/7=a+3..thus subtracting both we get 6f+o-(6f+y)= 7a+21-(7a-28) thus we get o-y=21+28=49.... :)
- 11 years agoHelpfull: Yes(0) No(3)
- (x+6)/7=a+4;
(y+6)=a-3;
by solvng both eq u ll get (x-y)=7 - 10 years agoHelpfull: Yes(0) No(0)
- (a1+a2+a3+a4)/4=a
(a1+a2+a3)/3=a-3
(a2+a3+a4)/3=a+4
substracting 2nd from 3rd we will get
a4-a1=7*3=21 years
- 10 years agoHelpfull: Yes(0) No(2)
- 49
average=(largeV+midV+smallV)/8
case 1: average+4=(largeV+MidV)/7
Case 2:average-3=(MidV+smallV)/7
case1-case2===>4-(-3)=(largeV-smallV)/7
:. largeV-smallV=7x7=49 - 5 years agoHelpfull: Yes(0) No(0)
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