If x and y are positive integers and x^2 – y^2 = 101, what is the valu

Question

If x and y are positive integers and x^2 – y^2 = 101, what is the value of x^2 + y^2 ?

A. 5050
B. 5101
C. 5150
D. 5201
E. 5301

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Bunuel · Accepted Answer

x^2 - y^2 = (x-y)(x+y)=101 .

101 is a prime number, so it can be broken into the product of two positive multiples only in one way 101=1*101. Now, since x and y are positive integers, then x - y < x + y, thus x - y = 1 and x + y = 101. Solving for x and y gives x = 51 and y = 50. Thus  x^2 + y^2=51^2+50^2=5,101 .

Answer: B.

EgmatQuantExpert · Answer

Solution

Given  
In this question, we are given that
 • The numbers x and y are positive integers
• Also,  x^2 – y^2 = 101

To find  
We need to determine
 • The value of  x^2 + y^2

Approach and Working out  
The given equation can be written as  x^2 – y^2 = 101 
 • Or, (x + y) (x – y) = 101

Now, 101 is a prime number and can be written as a product of two integers only as 101 * 1
Hence, we can say x + y = 101 and x – y = 1

Solving, we get x = 51 and y = 50
 • Therefore,  x^2 + y^2 = 51^2 + 50^2 = 2601 + 2500 = 5101

Thus, option B is the correct answer.

Correct Answer: Option B

GMATGuruNY · Answer

We can PLUG IN THE ANSWERS, which represent the value of  x^2 + y^2 .
Since  (x^2 + y^2) + (x^2 - y^2) = 2x^2 , the correct answer must yield an EVEN INTEGER when added to  x^2 – y^2 = 101 .
Implication:
The correct answer must be ODD.
Eliminate A and C.

B:  x^2 + y^2 = 5101 
Adding together  x^2 + y^2 = 5101  and  x^2 – y^2 = 101 , we get:
 2x^2 = 5202 
 x^2 = 2601

Since  50^2 = 2500  and  60^2 = 3600 ,  x  in this case must be between 50 and 60.
For its square to yield a units digit of 1,  x  must have a units digit of 1 or 9.
Test  x=51 :
 51^2 = (50+1)^2 = 50^2 + 1^2 + 2*50*1 = 2500 + 1 + 100 = 2601 
Success!

B

MathRevolution · Answer

x^2 - y^2 = 101

=> (x + y) (x - y) = 101 * 1

=> x + y = 101 and x - y = 1

Solving linear equations we get, x = 51 and y = 50.

=>  x^2 + y^2 = 51^2 + 50^2 = 2601+ 2500 = 5101

Answer B

BrushMyQuant · Answer

Given that x and y are positive integers and x^2 - y^2 = 101. And we need to find the value of x^2 + y^2

x^2 - y^2 = 101
=> (x-y) * (x+y) = 1*101
=> x - y = 1 ...(1)
=> x+y = 101 ...(2)

(1) + (2) we get 
2x = 1 + 101 = 102
=> x = 51
=> y = 101 - 51 = 50

=> x^2 + y^2 = 50^2 + 51^2 = 5101

So,  Answer will be B 
Hope it helps!

If x and y are positive integers and x^2 – y^2 = 101, what is the valu

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GMAT Problem Solving (PS) Questions

If x and y are positive integers and x^2 – y^2 = 101, what is the valu

Prep Toolkit

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