x and y are positive integers and 5x + 4y = 240. If x is a multiple of

\(5x + 4y = 240\)...let \(x=3z\)..
\(4y=240-5*3z....y=\frac{15(16-z)}{4}\)


I. 6.......\(y=\frac{15(16-z)}{4}\), Possible whenever z is a multiple of 8, since (16-z)/4 will be even.. z=8---\(y=\frac{15(16-8)}{4}=30=6*5\)

II. 9.......\(y=\frac{15(16-z)}{4}\), Possible whenever 16-z is a multiple of 3, n.. z=4---\(y=\frac{15(16-4)}{4}=45=9*5\)

III. 15.......\(y=\frac{15(16-z)}{4}\), TRUE always

All can be true

E
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Can someone please explain how y will be a multiple of 6?

when x is 24. 5x = 120 and 4y = 120. Y = 30. 30 is a multiple of 6.

Hope this helps.

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This is a copy of the following OG question: https://gmatclub.com/forum/if-x-and-y-a ... 20282.html
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The best way to solve this question is to write the equation on this way:
y = (240 - 5x)/4
Now, we know x is a multiple of 3, so we can write x = 3k (note k cannot be 0, since x is a positive integer) and substitute it in the above equation and get:
y = (240 -15k)/4
y = 15* (16 - k)/4
Now, (16 -k)/4 can be 1,2, and 3 only.. thus, the possible values of y are 15, 30, and 45 and y can be a multiple of 6, 9, and 15.
Hence, correct answer is Option E. fyi

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Lets try to express 240 as 5x+ 4y, where x is a multiple of 3. So, 5x can be 15, 30, 45, 60, 75, 90, 105, 120, 135,150. When 5x =120, 4y = 120, y= 30, which is a multiple of 6 and 15. When 5x =60, 4y = 180, so y = 45, which is a multiple of 9.
Answer is option E.

Given 5x +4y= 240

So, (x,y) = (48,0), (44,5), (40,10), (36,15), (32,20), (28,25), (24,30), (20,35), (16,40), (12,45), (8,50), (4,55) and (0,60)

If x is divisible by 3 then, (x,y) = (48,0), (36,15), (24,30), (12,45), and (0,60)

So, I think E.

5X + 4Y = 240

Divide each side by the coefficient 4 and take the remainder:

(1X / 4) Rem of = (240 / 4) Rem of


4 divided evenly into 240 / so we are looking for a positive integer value of X that will leave a remainder of 0 when divided by 4

X = 4 ———> Y = 55

Since the linear equation is given by:

5x + 4y = 240

the positive consecutive values of X that are integers and will satisfy the given equation will increase by the Coefficient In front of the Y variable ——- (+4)

Similarly, the corresponding Y integer values will change by ——-(-5)

X values———- corresponding Y value
4 —————— 55

12 ——— ——-45

24 ———- ———30

In all 3 cases, X is a value that satisfies the equation and is a multiple of 3.

You can see the corresponding Y value that works with that X can be a multiple of all three factors

15
9
Or
6

(E) all 3

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Start with x=0 then y=60
Now for integral solutions in such a scenario, always, the next value of x that satisfies the equation will be the previous value plus the coefficient of y in the equation.
Hence the next value of x that satisfies will be 4. But it says x must be a multiple of 3. Hence next value of x will be 12 and that time y=45. Now x will keep increasing by 12 and y will keep decreasing by 15. So next x=24 y=30 next x=36 y=15 finally x=48 y=0

Hence y COULD be a multiple of 6 (y=30) , multiple of 9 (y=45) and multiple of 15 (all the y values listed)
Hence, E is the answer

Since we know x is a multipole of 3, we can say x = 3k. Thus, we have:

5(3k) + 4y = 240

15k + 4y = 240

4y = 240 - 15k

4y = 15(16 - k)

y = 15(16 - k)/4

We see that k can be 4, 8, or 12.

When k is 4, y is 45.

When k is 8, y is 30.

When k is 12, y is 15.

Thus, we see that y can be a mutlipole of 6, 9, and 15.

Answer: E
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