Question: When a certain perfect square is increased by 148, the result is another perfect square. What is the value of the original perfect square?

1. 1296
2. 1369
3. 1681
4. 1764
5. 2500

“When a certain perfect square is increased by 148, the result is”- is a topic of the GMAT Quantitative reasoning section of GMAT. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:
Approach Solution 1:
Let’s call the two perfect squares x^2 and y^2, respectively. Then the given information translates as x^2+148=y^2. Subtracting x^2 gives 148=y^2−x^2, a difference of squares. This, in turn, factors as (y+x)(y−x)=148.

The next step is tricky. It begins with factoring 148, which breaks down as 2∗2∗37. Since we’re dealing with perfect squares, x and y are positive integers, and (y+x) and (y−x) must be paired integer factors of 148. The options are 148∗1,74∗2, and 37∗4. But our number properties establish that (y+x) and (y−x) must be either both odd or both even, so only 74∗2 is an actual possibility. And because for any positive integers (y+x)>(y−x), we can conclude that y+x=74 and y–x=2. Solving by elimination, 2y=76, y=38, and x=36.

Finally, we just need to square 36. But rather than multiplying it out, note that 36^2 ends in 6

Correct Answer: A

Approach Solution 2:
Let x² = the smaller perfect square and y² = the greater perfect square.
A certain perfect square is increased by 148.
The result is another perfect square.
Translated into maths:

x² + 148 = y².

Thus:
x² - y² = -148
(x+y)(x-y) = (74)(-2)
(x+y)(x-y) = (36+38)(36-38).

Resulting values:
x= 36, y=38.
x² = 36² = integer with a units digit of 6.

Correct Answer: A

Approach Solution 3:
Let the original value (before it’s squared) = x for some positive integer x, and the new value (before it’s squared) = x + k for some positive integer k. We can create the equation:

x^2 + 148 = (x + k)^2
x^2 + 148 = x^2 + 2kx + k^2
148 = 2kx + k^2

Since both k and x are positive, we see that k^2 < 148. Thus k ≤ 12. Also, since 148 and 2kx are even, k must be even also. Thus k can only be 2, 4, 6, 8, 10 or 12. Let’s analyse each of these values until we find a suitable value for x.

If k = 2, then
148 = 2(2)x + 2^2
144 = 4x
36 = x

We see that x can be 36 and 36^2 = 1296

Correct Answer: A

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