A Right Angled Triangle has its Sides in Arithmetic Progression and Being Integers GMAT Problem Solving

Question: A right angled triangle has its sides in Arithmetic progression and being integers. At most one of its sides is a multiple of 10. Its perimeter is less than 115. How many such triangles are possible?

1. 9
2. 8
3. 7
4. 6
5. 5

Correct Answer: A

Solution and Explanation:
Approach Solution 1:
The problem statement states that:
Given:

• A right angled triangle has its sides in Arithmetic progression and being integers.
• At most one of its sides is a multiple of 10.
• Its perimeter is less than 115.

Find out:

• The number of such possible triangles.

Algebraically let's see how many sides are there:
Let sides be a-d, a and a+d
a² − 2ad + d² + a² = a² + 2ad+d²......a²−4ad = 0 

So a=4d or multiple of 4.
So 3:4:5 or 6:8:10 and so on

Otherwise, also we know 3:4:5 is a right angled triangle.

At Most one is a multiple of 10, which means it may not have any multiple of 10 or just one at the max.
Least perimeter = 3+4+5=12
The next bigger perimeter will be 12*2 when sides are 6:8:10
So the total possible perimeter will be the greatest multiple of 12 just less than 115
12*9=108, so 9 such perimeter
All 9 will have at most one multiple of 10.
Then the answer will be 9.
Hence the number of such possible triangles = 9.

Approach Solution 2:
From the question stem, we can deduce the following information

1. The sides of the right-angled triangle are in an AP.
2. At most one of the sides is a multiple of 10
3. The perimeter is less than 115

We need a set of numbers which is both a Pythagorean triplet and in an AP
One such set of numbers that is both in an AP and sides of a right-angled triangle are 3x-4x-5x

The triangle possible are
3,4,5(when x = 1) | 6,8,10(when x = 2) | 9,12,15(when x = 3) |
12,16,20(when x = 4) | 15,20,25(when x = 5) | 18,24,30(when x = 6) |
21,28,35(when x = 7) | 24,32,40(when x = 8) | 27,36,45(when x = 9) |
At x = 10, the perimeter of the triangle is 12x or 120(and this fails condition 3)

Therefore, there are at least 9(Option A) possibilities where right-angled triangles can be formed. Hence, the answer is A.

Approach Solution 3:
The problem statement states that:
Given:

• A right angled triangle has its sides in Arithmetic progression and being integers.
• At most one of its sides is a multiple of 10.
• Its perimeter is less than 115.

Find out:

• The number of such possible triangles.

Let the sides be (a−d), a, (a+d), where d is the common difference for the sides to form an AP.

Given the perimeter (a−d) + a + (a+d) < 115

Hence we have a < 38.3, since a is an integer we get a = 1,2,......38

Also, the given triangle is a right triangle, hence (a−d)²+a²=(a+d)², the longest side as the hypotenuse.

we get a=4d, hence a is a multiple of 4 & has a value between 1 & 38.

Therefore, we get 9 such values for a to satisfy the above criterion.
Hence the number of such possible triangles = 9.

”A right angled triangle has its sides in Arithmetic progression and being integers”- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. This question has been taken from the book “501 GMAT Questions”. To solve the GMAT Problem Solving questions, the candidates need to access data to calculate the sum accurately. The candidates can go through several sorts of questions from the GMAT Quant practice papers that will help them to enhance their mathematical learning.

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