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06 May 2021, 01:02
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
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Question Stats:
61% (02:37) correct
39%
(02:52)
wrong
based on 118
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If x^2 + y^3 = 593 and y^2 = x + 55, then what is the value of x - y ?
A. 1
B. 2
C. 3
D. 4
E. 7
A. 1
B. 2
C. 3
D. 4
E. 7
When the square of a number and the cube of a smaller number are added the result is 593. If the square of the smaller number exceeds the bigger number by 55, find the difference of the two numbers.
06 May 2021, 01:44
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As I see it, the answer is A (1).
Let the two numbers be x and y such that x>y.
x 2 +y 3 =593 y 2 =x+55
We know that y+55 is a perfect square. Therefore x=9 and y=8 is one possible combination. x=8 and y=9 fit the bill for the first equation too because 8\sqrt{3}+9\sqrt{2} =593 As a result x−y=1
Let the two numbers be x and y such that x>y.
x 2 +y 3 =593 y 2 =x+55
We know that y+55 is a perfect square. Therefore x=9 and y=8 is one possible combination. x=8 and y=9 fit the bill for the first equation too because 8\sqrt{3}+9\sqrt{2} =593 As a result x−y=1
06 May 2021, 02:01
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Bunuel
Assuming x and y are positive integers (the options for x-y are integers), we have:
y² = x + 55
Looking at squares of numbers greater than 55:
y² = 64 or 81 or 100 etc.
x = 64-55=9 or 81-55=26 or 100-55=45 etc.
We have:
x² + y³ = 593
If x = 9 => y³ = 593-81=512 => y = 8
If x = 26 => y³ < 0 i.e. y<0 - violates condition that y is positive.
Thus, we have: x=9 and y=8 => x - y = 1
Answer A
Note: If we try to solve by substituting x = y² - 55 in the first equation, we will get a 4th degree equation which is too complicated to solve (beyond the scope of the GMAT)
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