Question 40

The average of first three out of four numbers is 16. The average of last three numbers is 14. The sum of first and last number is 16. The last number is

Solution

let the four numbers be a,b,c,d

the average of first three is 16 

average = $$\frac{sum of numbers}{total count}$$

16 = $$\frac{a+b+c}{3}$$

a+b+c = 48

similarly average of last three

b+c+d = $$14\times{3}$$ = 42

also given a+d = 16 

                a = 16 - d

putting the value of a in equation 1

16 - d + b + c = 48

-d + b + c = 32

now adding this with equation 2 we get

42+32 = b+c+d+b+c-d

74 = 2(b+c)

b+c = 37

we know that

b+c+d = 42

37 + d = 42

therefore d = 5

so the last number is 5 i.e option C


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