GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 09 Apr 2020, 20:16

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 62676
If x and y are positive integers and x^2 – y^2 = 101, what is the valu  [#permalink]

### Show Tags

21 Feb 2020, 04:47
00:00

Difficulty:

25% (medium)

Question Stats:

78% (02:10) correct 22% (01:48) wrong based on 36 sessions

### HideShow timer Statistics

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

Are You Up For the Challenge: 700 Level Questions

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 62676
Re: If x and y are positive integers and x^2 – y^2 = 101, what is the valu  [#permalink]

### Show Tags

21 Feb 2020, 04:48
Bunuel wrote:
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

$$x^2 - y^2 = (x-y)(x+y)=101$$. 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 $$x^2 + y^2=51^2+50^2=5,101$$.

_________________
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3348
Re: If x and y are positive integers and x^2 – y^2 = 101, what is the valu  [#permalink]

### Show Tags

21 Feb 2020, 09:36

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.

_________________
Director
Joined: 04 Aug 2010
Posts: 558
Schools: Dartmouth College
Re: If x and y are positive integers and x^2 – y^2 = 101, what is the valu  [#permalink]

### Show Tags

21 Feb 2020, 14:27
Bunuel wrote:
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

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!

_________________
GMAT and GRE Tutor
New York, NY

Available for tutoring in NYC and long-distance.
Re: If x and y are positive integers and x^2 – y^2 = 101, what is the valu   [#permalink] 21 Feb 2020, 14:27
Display posts from previous: Sort by