Question

Some billiard balls were arranged in an equilateral triangle. And 5 balls were extra. When the same set of billiard balls were arranged into triangle each side has one more ball than in the first arrangement there were 11 balls shortage.How many balls?

Answer 1

`125`

Explanation:

If billiard balls are arranged in rows to form an equilateral triangle, then the first row consists of 1 ball, the second row of 2 balls, the third row of 3 balls, .... , and the `n^(th)` row of `n` balls,
then the total number `t` of balls that forms the equilateral triangle with `n` row is : `t=1+2+3+....+n=(1+n)/2*n`
let `t_1, and t_2` be the total number of balls in the first arrangement and the second arrangement, respectively.
given that there were 5 extra balls in the first arrangement,
`=> t_1=(1+n)/2*n+5`
given that there were 11 balls shortage in the second arrangement,
`=> t_2=(1+(n+1))/2*(n+1)-11=(n+2)/2*(n+1)-11`
As `t_1=t_2`,
`=> (1+n)/2xxn+5=(n+2)/2xx(n+1)-11`
`=> (1+n)*n+10=(n+2)(n+1)-22`
`=> n+n^2=n^2+n+2n+2-22-10`
`=> 2n=22+10-2=30`
`=> n=15`
`=> t_1=(1+15)/2*15+5=125`
check :
`t_2=(15+2)/2*(15+1)-11=125=t_1` -----`(ok)`