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?
“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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