For integers x and y, |x 2y| = 5 and |2x y| = 11. What i

Question

For integers x and y, |x − 2y| = 5 and |2x − y| = 11. What is the value of xy?

A. 14
B. 36
C. 48
D. 63
E. 72

RohanSharma15 · Accepted Answer

Step 1: Analyze the first equation

|x − 2y| = 5

Since 5 is odd, the expression  x − 2y must be odd . 
 Note that 2y is always even,  
 So:    x − (even) = odd  → x must be  odd

Step 2: Analyze the second equation

|2x − y| = 11

Since 11 is odd, the expression  2x − y must be odd . 
 Note that 2x is always even. So: 
  (even) − y = odd  → y must be  odd

Step 3: Determine the parity of xy 
 
  x is odd  
  y is odd  
   odd × odd = odd 
   
 So  xy must be odd .

Step 4: Scan the answer choices 
 
  A. 14 → even ✗  
  B. 36 → even ✗  
  C. 48 → even ✗  
  D. 63 →  odd  ✓  
  E. 72 → even ✗
   
 Only one option is odd. 
 Answer: D. 63

Dereno · Answer

If we remove the modulus values, we can get the possible values of (5,11); (5,-11); (-5,11) and (-5,-11).

Case 1: x-2y = 5 , and 2x - y = 11. Solving, we get y = 1/3 , which is not an integer.

Case 2: x-2y = 5, and 2x - y = -11. Solving, we get y = -7 and x = -9. So, xy = (-7)*(-9) = 63.

Case 3: x-2y = -5, and 2x - y = 11. Solving, we get x = 17/3, which is not an integer.

Case 4: x -2y = -5, and 2x -y = -11. Solving, we get y=1/3 , which is not an integer.

Only Case 2 satisfies.

xy = (-7)*(-9) =  63

Option D

gmatcat8 · Answer

Hey Kevin, could you please show another way of solving this by checking cases with absolute values?

egmat · Answer

Hi gmatcat8,

You're asking to see the full absolute-value case method spelled out, so let me lay all four cases side by side. The idea is simple: each absolute value splits into a  plus  and a  minus  version, and since there are two equations, that's  2 × 2 = 4  systems to test.

The two splits are:
- x − 2y = 5  or  x − 2y = −5
- 2x − y = 11  or  2x − y = −11

Solving each pairing

A quick trick for all four: double the first equation to get 2x − 4y, then subtract it from the second to knock out x and leave 3y.

Case 1:  x − 2y = 5, 2x − y = 11
- 3y = 11 − 10 = 1 - y =  1/3 . Not an integer. ✗

Case 2:  x − 2y = 5, 2x − y = −11
- 3y = −11 − 10 = −21 - y =  −7 , x =  −9 .
Check: |−9 + 14| =  5  ✓ and |−18 + 7| =  11  ✓. So xy = (−9)(−7) =  63 . ✓

Case 3:  x − 2y = −5, 2x − y = 11
- 3y = 11 + 10 = 21 - y =  7 , x =  9 . So xy = (9)(7) =  63 . ✓

Case 4:  x − 2y = −5, 2x − y = −11
- 3y = −11 + 10 = −1 - y =  −1/3 . Not an integer. ✗

Notice the pattern: the two  same-sign  pairings (both +, both -) fail the integer test, and the two  opposite-sign  pairings both land on xy =  63 . That's why the answer is locked in as  D , and why you never even needed to worry about which of Cases  2  or  3  is "the" solution - they agree.

The key discipline here is to write out  all four  sign combinations and let the  integer  condition filter them, exactly as Dereno did above.

Answer: D

For integers x and y, |x 2y| = 5 and |2x y| = 11. What i

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For integers x and y, |x 2y| = 5 and |2x y| = 11. What i

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