Aptitude - Average - Discussion

The average age of the mother and her six children is 12.
ie. (sum of ages of 6 children + Age of mother)/7 = 12.
ie Sum of ages of 6 children + age of mother = 12*7 = 84.
Age of mother = 84 - sum of ages of 6 children ----> 1.

The average is reduced by 5 yrs when the age of mother is excluded
ie.the average age of 6 children = 12 -5.
ie the sum of ages of children / 6 = 7.
sum of ages of children = 6*7 =42.
Sum of ages of children = 42.

Substitute this sum in (1) we get,
Age of mother = 84 - 42.
Age of mother = 42 years (Answer).

In the first case,

Let, X be the average weight of 8 persons
W1 be total weight of 8 persons
7sw be the total weight of 7 persons excluding 65 year old person
W1 = 7sw + 65
we have W1/8=X ==> (7sw+65)/8 = X
7sw+65 = 8 X
7sw = 8 X - 65 --------> equation no.(1)

In the second case,
Let, W2 be the total weight of 8 persons including the newcomer
Now be the weight of new comer
We get W2 = 7sw + Nw --------> (2)
We have W2/8 =X+2.5
W2 = (X+2.5)*8
W2 = 8X +20
Therefore (2) becomes
7sw+Nw = 8X + 20 --------> (3)

Substitute (1) in (3)
We get, 8X - 65 +Nw = 8X + 20.
ie: Nw = 8X + 20 - 8X +65,
Nw = 20+65.
= 85,
ie, the weight of the person who joined in place of the person weighing 65 kg is 85 kg.

@Naveen.

Total number of students = 120,
Let X be the number of passed candidates then 120-X is the number of failed candidates.
Average marks of 120 students = 35,
Average of passed candidates =39,
Average of failed candidates =15,

Therefore,
Total marks of 120 students = 35*120 = 4200,
Total marks of passed candidates= 39 * X,
Total marks of failed candidates = 15 *(120 - X) = 1800 -15 X,
Total marks of passed candidates + Total marks of failed candidates = Total marks of 120 sudents.

ie 39X + 1800 - 15X = 4200,
39X - 15X = 4200 - 1800,
24X = 2400,
X = 100,
ie The number of passed candidtes = 100 (Answer).

The average of six consecutive integers in increasing order size is 9.5. What is the average of the last three numbers?

Can anyone solve this?

The average weight of three men 'X', 'Y' and 'Z' is 75 kgs. Another man 'A' joins
the group and the average weight now becomes 80 kgs. If another person 'B' whose
weight is 5 kgs more than 'A' replaces 'X', then the average weight of 'Y', 'Z', 'A' and
'B' will be 85 kgs. What is the weight of 'X'?

Can anyone solve this?

@Niraj.

The answer is 46.

Thanks @Bhanu.

Your method is good, Thanks @Bhanu.

Thanks @Anonymous.

Let the average weight of 8 persons be x and the weight of first 7 persons be A.

The weight of 8 persons is A+ 65 = 8x ---> (1)
After replacing the person having weight 65, the average weight of 8 persons increased by 2.5
Now the average weight of 8 persons will be 8(x+2.5) and;

The weight of 8 persons will be A + Z(replaced person weighing 65) = 8(x+2.5) ---> (2)
Subtract (1) from (2) i.e.,
A+Z-(A+65) = 8(x+2.5) - 8x,
A+Z-A-65 = 8x + 20 -8x,
Z-65 = 20,
Z = 20+65,
Z = 85 Weight of replaced person.