12 Days of Christmas GMAT Competition - Day 1: If 5x - 2y = 80, how

Question

12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

If xy  ≠  0 and 5x - 2y = 80, how many pairs of (x, y) exist, where x and y are integers with opposite signs?

A. 4
B. 5
C. 6
D. 7
E. 8

Bunuel · Accepted Answer

GMAT Club's Official Explanation:

If x is negative and y is positive, 5x - 2y = negative - positive = negative, which cannot equal the positive number 80. Thus, we are looking for pairs where x is positive and y is negative.

Rearranging, we get 5x = 80 + 2y.

Since the left-hand side is positive, y cannot be -40 or less because that would make 80 + 2y negative. Therefore, y must be more than -40 and less than 0.

Since the left-hand side is a multiple of 5, y must also be a multiple of 5 to ensure that the sum of 80 and 2y is a multiple of 5. There are only seven multiples of 5 between -40 and 0: -35, -30, -25, -20, -15, -10, and -5.

Thus, there are seven pairs of (x, y) possible.

Answer: D.

Oppenheimer1945 · Answer

x=(80+2y)/5=16+(2y/5)

=>x=16+(2y/5)

y=-5, x=14
y=-10, x=12
y=-15, x=10
y=-20, x=8
y=-25, x=6
y=-30, x=4
y=-35, x=2
we have 7 pairs

Krunaal · Answer

Two conditions  1. x and y is an integer not equal to 0  and  2. x and y have opposite signs

5x - 2y = 80

x > 0 and y < 0 is the only case possible as negative - positive gives a negative value.

Let us start with y = - 5, this gives x = 14
 y = -10, => x = 12
 y = -15, => x = 10
(we see a pattern, for every -5 in y there's a -2 in x)
 y = -20, => x = 8
 y = -25, => x = 6
 y = -30, => x = 4
 y = -35, => x = 2

Total no. of possibilities 7

Answer D.

hr1212 · Answer

If we plot line 5x - 2y = 80, it intersects 1st, 3rd and 4th quandrant as can be seen in the attached diagram. Only in the 4th quandrant are the signs for both x and y opposite in the given case.

Here, x-intercept is (16, 0) and y-intercept is (0, -40) --- (1)

So, all the integer points which are between our above intercepts and belong on line 5x - 2y = 80, will be our required set.

Just by looking at the given equation, we can conclude that x cannot be an odd number, as 80 - 5x would be an odd number which won't be divisible by 2. Hence, taking all even values for x between 0 and 16 (from (1)), such that xy != 0, we get,

x 
 y 
 
  2 
 -35 
 
  4 
 -30 
 
  6 
 -25 
 
  8 
 -20 
 
  10 
 -15 
 
  12 
 -10 
 
  14 
 -5

This gives us 7 values in total which have opposite signs for x and y.

Answer : D

charlie1195 · Answer

IMO the answer is 7.

We are given that neither x nor y is equal to 0. 
Simplifying the equation, 
5x - 2y = 80
5x = 80 + 2y
x = (80 + 2y) / 5
x = 16 + (2/5)*y

Now, we know that x is equal to 16 when y = 0; but since y cannot be zero, we need to replace y with negative factors of 5 (because x and y are integers) such that x moves towards 0.

Starting with y = -5, x = 16 + (2/5)*(-5) = 16 - 2 = 14 ------> (x = 14, y =-5)
y = -10, x = 16 + (2/5)*(-10) = 16 - 4 = 12 ------> (x = 12, y = -10)
y = -15, x = 16 + (2/5)*(-15) = 16 - 6 = 10 ------> (x = 10, y = -15)
y = -20, x = 16 + (2/5)*(-20) = 16 - 8 = 8 ------> (x = 8, y = -20)
y = -25, x = 16 + (2/5)*(-25) = 16 - 10 = 6 ------> (x = 6, y = -25)
y = -30, x = 16 + (2/5)*(-30) = 16 - 12 = 4 ------> (x = 4, y = -30)
y = -35, x = 16 + (2/5)*(-35) = 16 - 14 = 2 ------> (x = 2, y = -35)
If we do one more step, x will become zero and post that the signs will be same for x and y.

Hence, 7 pairs of x and y exist where the signs are opposite.

aarati17 · Answer

Ans: D. 7

5x-2y=80
=>x=16+(2y/5) ---------------------(equation (1))
Since x and y are integers, y has to be a multiple of 5.

Since x and y has to be opposite signs, let us consider y as a -ve multiple of 5
Putting values of y as -5, -10,.... in equation (1), we can calculate values of x

x 
 14 
 12 
 10 
 8 
 6 
 4 
 2 
 0 
 
  y 
 -5 
 -10 
 -15 
 -20 
 -25 
 -30 
 -35 
 -40

We cannot consider x=0 and y=-40 because then xy=0 and the condition is given than xy is not equal to zero

Also x cannot be negative because
Let y=45 then x=-2
Putting the values in the equation
5x-2y
= 5(-2)-2(45)
= -10-90
= -100

Hence there are 7 pairs of (x,y)

aarati17 · Answer

Ans: D. 7

5x-2y=80
=>x=16+(2y/5) ---------------------(equation (1))
Since x and y are integers, y has to be a multiple of 5.

Since x and y has to be opposite signs, let us consider y as a -ve multiple of 5
Putting values of y as -5, -10,.... in equation (1), we can calculate values of x

x 
 14 
 12 
 10 
 8 
 6 
 4 
 2 
 0 
 
  y 
 -5 
 -10 
 -15 
 -20 
 -25 
 -30 
 -35 
 -40

We cannot consider x=0 and y=-40 because then xy=0 and the condition is given than xy is not equal to zero

Also x cannot be negative because
Let y=45 then x=-2
Putting the values in the equation
5x-2y
= 5(-2)-2(45)
= -10-90
= -100

Hence there are 7 pairs of (x,y)

Purnank · Answer

I think here as well answer is D.

as it says x and y are integers with opposite signs and xy ≠ 0 means both x and y cant be  "0"

now lets understand the equation. 
5x - 2y = 80 
5x = 2y + 80

here 2y will always be even so RHS will always be even multiple of 5 (because on LHS we have 5x) that means y value vary in multiple of 5s and plus we cant take value of y as +ve because we will always get x as positive and that violates the given condition.

hence taking y as -5, -10, -15, -20, ....
we will see for that value of y we get x as 14, 12, 10, 8, ....

but when value y become -40 or less the value we get for x is 0 or -ve.
so means possible pairs we have for both  (x, y)  are  (14, -5) (12, -10) (10, -15) (8, -20) (6, -25) (4, -30) (2, -35)

So, 7 Such Pairs Possible.

Lizaza · Answer

What opposite sign arrangements can we have here?
1] $5x < 0$ and $2x > 0:$ this is impossible, given we will be subtracting a value out of a negative one, which will only keep decreasing and never be above zero.
2] $5x > 0 $and $2x < 0:$ this is the one we need, as we are adding the absolute value of $ 2x$ to our $5x $ to get to 80.
Therefore, 5x needs to be below 80 for this to work.

As far as exact pairs of X and Y go, it's important that together they make a number divisible by 10: this means that by subtracting 5x we need to be able to fully 'replace' it with 2y, so they need to have the same multiple.
Given 2 and 5 are prime numbers, the only way for 5x and 2y to have a common multiple is when they are mutually divisible - that is, 5x is even and 2y is divisible by 5. Therefore, both 5x and 2y must be divisible by 10, and when we go through our integer 'steps' to get to 0, we must only consider them in tens:

$70-(-10) = 80$
$60-(-20) = 80$
$50-(-30) = 80$
$40-(-40) = 80$
$30-(-50) = 80$
$20-(-60) = 80$
$10-(-70) = 80$

Therefore, there are only 7 non-zero combinations, and the answer is D.

Sujithz001 · Answer

okay let's try inserting the values,

5x - 2y = 80

When

x=1 ; y= -75/2 => Not an integer, so out.
x=2 ; y= -35  (1) 
x=3 ; y= -65/2 => out

Okay now I realize that every odd attribute to x is going to result in fractional y, as y is the factor of 2.

So inserting only even values, till the sign change happens,

x=4 ; y= -30  (2) 
x=6 ; y= -25  (3) 
x=8 ; y= -20  (4) 
x=10 ; y= -15  (5) 
x=12 ; y= -10 (6) 
 x=14 ; y= -5  (7) 
 x=16 ; y= 0 => not possible as xy is not equal to 0.

Now post this inflection point, both of the values are going to be positives.

So 7 is the answer.

Locomotivated · Answer

x and y are integers, and xy ≠ 0, so the equation 5x - 2y = 80 will have limited number of solutions that satisfy the equation.

Find the first pair of (x,y) using x = (80+2y)/5, since x is an int, for y=0,x=16. 
--> (x,y) = (16,0)

For the next pair, reduce (since the signs of x and y have to be opposite), x by coefficient of y (2), and reduce y by coefficient of x (5).
--> (x,y) =

Subsequently, the next set of pairs can be found:
--> (14,-5) (12,-10) ; (10,-15), ...... (2,-35)

Observe that x and y will have opposite signs until x = 0. Therefore, for 7 pairs of (x,y) the signs will be opposite.

Concepts tested:
1. Linear equation with one equation and two unknowns, but with the constraint that x and y are integers. In such a case, we can find out pairs of (x,y) for which the linear equation holds goods.
2. Applying constraints (x and y have opposites signs, and xy ≠ 0) to arrive at the limited set of solutions.

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