Question

Some amount of marbles were arranged in an equilateral triangle. And 2 marbles were extra. When the same set of marbles were arranged into triangle in which each side has one more marble than in the first arrangement there were 13 marbles shortage. How many marbles were in the set?

Answer 1

Answer:

The total number of marbles in the set is 107 marbles

Step-by-step explanation:

The parameters given are;

Let the size of the marbles be 1 unit

The marbles are arranged to form the equilateral triangle as follows;

1st row = 1 marble

2nd row = 2 marbles and so on so that the total number of marbles for an arithmetic series as follows;

Total number of marbles, t = 1 + 2 + 3 + ... + n

Therefore;

$t = \dfrac{1 + n}{2} \times n$

Since there are 2 marbles left in forming the first equilateral triangle, we have the total number of marbles in the set, t₁, is given as follows;

$t_1 = \dfrac{1 + n}{2} \times n + 2$

When the same marble set are arranged into a triangle in which each side has one more marble than in the first arrangement, there where 13 marble shortage, hence, the total number of marbles is given as follows;

$t_1 = \dfrac{1 + (n+1)}{2} \times (n+1) -13$

We therefore have;

$t_1 = \dfrac{1 + n}{2} \times n + 2 = \dfrac{1 + (n+1)}{2} \times (n+1) -13$

Which gives;

$\dfrac{n^{2}+n+4}{2}=\dfrac{n^{2}+3\cdot n-24}{2}$

Therefore;

n² + n + 4 = n² + 3·n - 24

2·n = 24 + 4 = 28

n = 14

From which we have;

$t_1 = \dfrac{1 + 14}{2} \times 14 + 2 = 107$

Therefore, the total number of marbles in the set, t₁ = 107 marbles.