All the students in a school are made to stand in rows so as to form an equilateral triangle. The first row consists of one student, the second row 2 students, third row 3 students and so on. Had there been 90 more students, the total number of students could have been arranged in the shape of a square, so that each side of the square has 5 students less than the number of students in the side of the triangle. Find the total number of students initially present in the school?

as first row consist 1 student, 2nd row consist 2 students and so on...

let the no of rows in triangle be "n" now the total no of students in triangle will be 1+2+3+....+n=n(n+1)/2
given if 90 more students are added that is n(n+1)/2+90 is equal to no of students in square arranged in such a manner that has 5 less students each side than the triangle.
so now n(n+1)/2+90=(n-5)^2 (area of the square )
n^2-21n-130=0
n=26 substituting in n(n+1)/2 we get 351.
answer is 351

You are dealing with the triangular numbers whose sums of consecutive rows would be

1 3 6 10 15 21 28 ...

these numbers are generated by (n^2+n)/2 for n equal to the set of natural numbers.

So I added 90 to each of these to see if I get a perfect square.
The first such n value was at n=4
the sum of students would have been 10, and adding 90 gave me 100.
But that did not satisfy the second condition you gave, namely that the side of the square had to be 5 less than the side of the triangle.

I kept going for a while, almost gave up, but came to n = 26

the sum of the students in the triangle is (26^2 + 26)/2 = 351
adding 90 got me 441 which is 21^2
and the difference between these is indeed 5

so there were initially 351 students in the school

Ah, the twisted tale of triangular rows and square sides. Let's see if I can solve this riddle without getting tangled up in shapes.

Let's assume the number of rows in the equilateral triangle is "n." So, the number of students in each row can be represented as 1, 2, 3, ..., n. The total number of students in the equilateral triangle can be found using the formula for the sum of an arithmetic series: n(n+1)/2.

Now, if we add 90 more students, we can arrange them into a square shape. The number of students in each side of the square will be (n - 5). So, the total number of students in the square can be found using the formula for the sum of the first n - 1 natural numbers: (n - 1)(n - 5)/2.

But wait! The total number of students in the square is the same as the total number of students in the equilateral triangle plus 90. Setting up an equation, we get:

n(n+1)/2 + 90 = (n - 1)(n - 5)/2

Now, let me put my math hat on and solve this equation. After some calculations, we find that n^2 - 10n - 279 = 0. Solving this quadratic equation gives us two possible values for n: 23 and -13.

Now, we know that the number of rows can't be negative (unless we're dealing with an alternate reality school). So, we discard -13 and settle with n = 23.

Plugging n = 23 into the equation for the total number of students in the equilateral triangle, we get:
23(23+1)/2 = 276 students.

So, the total number of students initially present in the school is 276. Ta-da!

Now, wasn't that a triangularly square solution?

Let's solve this problem step-by-step:

Step 1: Determine the number of rows in the equilateral triangle.
Let's assume the number of rows in the equilateral triangle is "n".
The first row consists of 1 student, the second row 2 students, the third row 3 students, and so on.
So, the total number of students in the equilateral triangle is given by the sum of the first n natural numbers, which is n*(n+1)/2.

Step 2: Determine the total number of students in the square.
If there were 90 more students, the total number of students could have been arranged in the shape of a square.
Let's assume the number of students in each side of the square is "m" (m-5).

The total number of students in the square is given by m^2.

Step 3: Set up the equation.
Based on the information given, we can set up the following equation:
n*(n+1)/2 + 90 = (m(m-5))^2

Step 4: Solve the equation.
To solve the equation, we need to find values of "n" and "m" that satisfy the equation.

By trial and error, we can find that n = 9 and m = 9.

Step 5: Calculate the total number of students initially present in the school.
Using the values of "n" and "m" we found in step 4:
Total number of students in the equilateral triangle = n*(n+1)/2 = 9*(9+1)/2 = 45
Total number of students in the square = m^2 = 9^2 = 81

Therefore, the total number of students initially present in the school is 45.

To solve this problem, let's break it down step by step.

Step 1: Define the variables
Let's assume the number of rows in the equilateral triangle is n. Therefore, the number of students in the first row is 1, the second row is 2, the third row is 3, and so on. We can represent the number of students in the nth row as n.

Let's also assume that the total number of students initially present in the school is S.

Step 2: Find the number of students in the equilateral triangle
The number of students in the equilateral triangle can be found by summing up the first n natural numbers. This can be represented as the sum of an arithmetic series:

Sum = (n/2) * (first term + last term)
= (n/2) * (1 + n)

Step 3: Find the number of students in the square
According to the problem, if there were 90 more students, the total number of students (S + 90) could be arranged in the shape of a square. Each side of the square has 5 students less than the number of students in the side of the triangle. Therefore, if the triangle has n rows, the square would have n-5 rows.

The number of students in the square can be found using the formula to find the sum of the first n-5 natural numbers:

Square_sum = ((n-5)/2) * (first term + last term)
= ((n-5)/2) * (1 + (n-5))

Step 4: Set up the equation
We know that the sum of the students in the triangle is equal to the sum of the students in the square, plus the additional 90 students:

(n/2) * (1 + n) = ((n-5)/2) * (1 + (n-5)) + 90

Step 5: Solve the equation
Now we can solve the equation to find the value of n, which represents the number of rows in the equilateral triangle.

(n/2) * (1 + n) = ((n-5)/2) * (1 + (n-5)) + 90

Simplifying the equation:

(n^2 + n) = ((n-5)^2 - 5(n-5)) + 180

Simplifying further:

n^2 + n = (n^2 - 10n + 25 - 5n + 25) + 180

n^2 + n = n^2 - 15n + 230

Rearranging the terms:

15n = 230

n = 230/15

n = 15.33

Since n represents the number of rows, it must be a whole number. Therefore, n = 15.

Step 6: Find the total number of students initially present in the school
To find the total number of students initially present in the school (S), we substitute the value of n (15) into the formula for the sum of the first n natural numbers:

S = (n/2) * (1 + n)
= (15/2) * (1 + 15)
= 7.5 * 16
= 120

Therefore, the total number of students initially present in the school is 120.