Question

There are four natural numbers. The average of any three numbers is added in the fourth number and in this way, the numbers $29,23,21$ and $17$ are obtained. One of the numbers is

• A. $11$
• B. $24$
• C. $21$
• D. $10$

Answer 1

Answer: (C)

$21$

We take the numbers as $a,b,c,d.$

Then (the average of a, b & c)+d$=29$,

(the average of b,c & d)+$a=23,$

(the average of c, d & a)+$b=21$&

(the average of d, a & b)+$c=17.4$

So $\frac{a+b+c}{3}+d=29⟹a+b+c+3d=\mathrm{87..........}\left(i\right),\frac{b+c+d}{3}+a=23⟹b+c+d+3a=\mathrm{69..........}\left(ii\right),\frac{c+d+a}{3}+b=21⟹c+d+a+3b=\mathrm{63..........}\left(iii\right)and\frac{d+a+b}{3}+c=17⟹d+a+b+3c=\mathrm{51...........}\left(iv\right).$

Adding the above 4 equations,

$6(a+b+c+d)=270$

i.e $a+b+c+d=45.$.............(v).

Now (v)-(i)$⟶$2d=42

or $d=21.$