Question

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

Answer 1

Lets say billiard balls are arranged in rows to form an equilateral triangle, then the first row consists of 1 ball, second row consists of 2 balls, and third row consists of 3 balls,  and so on. So there must be $n$ balls in the $n^{th}$ row.  

So, the total number of balls that forms the equilateral triangle with $n$ rows is:  

$1+2+3+4+5+....+n=\frac{n(n+1)}{2}$

Let $x_1$ and $x_2$ be the total number of balls in the first and second arrangements respectively.  

Then,

$x_1=\frac{n(n+1)}{2} +5$

It has been said that there were 11 lesser balls in the second arrangement:  

$x_2=\frac{1+(n+1)}{2} \times (n+1)-11=(n+1) \times \frac{(n+2)}{2} -11$

Since, $x_1=x_2$

$\frac{(n+1)}{2} \times n+5=\frac{(n+2)}{2} \times(n+1)-11$

multiplying both the sides by 2

$(n+1)\times n+10=(n+2)(n+1)-22$

$n+n^2=n^2+n+2n+2-22-10$

$2n=22+10-2$

$2n=30$

$n=15$

Therefore,

$x_1=\frac{(n+1)}{2}\times n+5=\frac{15+1}{2}  \times 15+5=125$

So, there were 125 balls at the set.