Question

There are four members. Average age of the first 3 is 16 and last 3 is 15. If the first member is 21. find
the last member's age
​

Answer 1

Answer:

21

Step-by-step explanation:

The average of first 3 = 16 The average of last 3 = 15 The last number =18 Solution:- Sum of Last 3 numbers= 15*3=45 Hence the last number is 18 So, Sum of 2nd & 3rd = 45-18= 27 Again sum of first three numbers =16*3=48 Therefore the 1st number will be: sum of 1st 3 number-sum of 2nd & 3rd number i.e. 48-27=21

So the answer is 21.

Answer 2

We don't know the actual age of all members.

Solution: Let the four age be 21, x, y, z.

Given:

A. The mean of the first 3 is 16.

B. The mean of the last 3 is 15.

C. The first member's age is 21.

By A, we know that $\frac{21+x+y}{3} =16$. ∴x+y=27

By B, we know that $\frac{x+y+z}{3} =15$. ∴x+y+z=45

We have two equations and they have x+y common. Now, all we have to do is to subtract the common parts, then z will be left alone.

→ (x+y+z)-(x+y)=45-27

→ z = 18

Therefore, the last member's age is 18 years old.