Question

$n$ similsr balls each of weight $w$ when weighed in pairs, the sim of the weights of all the possible pairs is $120$ is $120$ and when they are weighed in triplets, the sum of the weights comes out to be $480$ for all possible triplets. Then $n$ is equal to

• A. $5$
• B. $10$
• C. $15$
• D. $20$

Answer 1

The correct option is B $10$

Given,

n = similar balls, each of weight W

when,

Weight in pair = 120

Weight in triplets = 180

Now,

Total number of selection in pairs

$=\frac{n(n-1)}{2}$

A/Q

$120=\frac{n(n-1)}{2}\cdot 2w$ - (i)

Total number of selection in triplets

$\begin{array}{cc}& =\frac{n(n-1)(n-2)}{6}\\ 480& =\frac{n(n-1)(n-2)}{6}-3w\end{array}$ - (ii)

Dividing equation (i) and (ii)

We get,

$\begin{array}{cc}\frac{4480}{120}& =\frac{\frac{3(n-1)(n-2)\cdot 3w}{6}}{\frac{n(n-1)}{2}\cdot 2w}\\ 4& =\frac{n-2}{n}\end{array}$

$8+2=n$ (n = 10 option B)