If x and y are positive integers such that y is a multiple of 5 and 3x

If y is a multiple of 5, 4y must be a multiple of 5 too.

3x = 200 - 4y = Multiple of 5 - Multiple of 5 = Multiple of 5

Since 3 is not a multiple of 5, x MUST be a multiple of 5.

Also, 4y is even. So 3x = Even - Even = Even
Since 3 is not even, x must be even.

Hence x must be a multiple of 10.

Answer (E)

Check out this discussion on numbers and their factors: https://youtu.be/DxIH8rjhpKY

Plug in values and check -

Let y = 5 ; 3x + 4y = 200

So, 3x + 20 = 200

Or, x = 60 ( Options C & D rejected , Left with options A, B & E )


Let y = 20 ; 3x + 4y = 200

So, 3x + 80 = 200

Or, x = 40 ( Options A &B rejected , Left with option E )

Hence, answer will be optin (E)
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If y is a multiple of 5, then we know that y = 5k for some integer k.
So, let's take 3x + 4y = 200 and replace y with 5k to get: 3x + 4(5k) = 200
Simplify to get: 3x + 20k = 200 (where k is some integer)
Subtract 20k from both sides to get: 3x = 200 - 20k
Divide both sides by 3 to get: x = (200 - 20k)/3
Now factor the numerator to get: x = (20)(10 - k)/3
Since x must be an integer, and since 20/3 is not an integer, it must be the case that (10 - k)/3 evaluates to be some integer.
In other words, x = (20)(some integer)
So, x must be a multiple of 20. However, 20 is not one of the answer choices.
Then again, we can rewrite x to get: x = (2)(10)(some integer)
This tells us that x must also be a multiple of 10

Answer:

We are given that x and y are positive integers such that y is a multiple of 5 and 3x + 4y = 200. We need to determine which answer choice MUST be a multiple of x.

Because y is a multiple of 5, we can represent y as 5k, in which k is an integer, and we have:

3x + 4(5k) = 200

3x + 20k = 200

Since the sum of 3x and 20k is 200 (or a number that has a units digit of zero), and since 20k has a units digit of zero, 3x must also contain a units digit of zero. The only way for 3x to contain a units digit of zero is if it’s multiplied by a multiple of 10. Thus, x must be a multiple of 10.

Answer: E
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its given y is multiple of 5 therefore 4y is even and will end in zero as even multiple of 5 ends in zero(zero at units place)
now equation 3x + 4y =200, as we know 4y ends in zero (zero at units place ) we need to add something that ends in zero (zero at units place) to get to 200,3x can only end in zero if x is multiple of 10

If x and y are positive integers such that y is a multiple of 5 and 3x + 4y = 200, then x must be a multiple of which of the following?
A) 3
B) 6
C) 7
D) 8
E) 10

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I drowned in these numbers and failed on my first attempt at this question, but I've got it now.

The most simple approach I can think of is to plug in values of y that are multiples of 5.

1) Plug in 5, 10, 15, & 20.

2) These yield equations of:
3x = 180
3x= 160
3x = 140
3x = 120

3) Of our answer choices, (E) 10 is the only # that divides evenly in all these situations, therefore it must be our answer.

Responding to a pm:


You are given that y is a multiple of 5. So 4y is a multiple of 10.

Hence 3x = 200 - 4y = Multiple of 10 - Multiple of 10

So 3x would be a multiple of 10 too.

So the answer cannot be (B) or (D)

I approached this in a different style I think than the other posters - is this way correct as well?

First, 200 factors down to \(2^3 * 5^2\)

Looking at the information we know that from the equation \(3x + 4y\), where y is a multiple of 5, we have the factors of \(2^2\) (from the 4), 3, and 5. Given this, we can then determine that we need one more factor of both 2 and 5. E - 10 is the only answer choice that provides those two factors.

Can someone pls verify if this method will work for all similar question types? This is the way I approached as well

Since y is a factor of 5, 4y will end in 0.
So get the sum of 200, 3x should end in 0 as well. So x should be a multiple of 10.

i. We see that 4y is always a multiple of 10 when y is a multiple of 5 . So 3x is a multiple of 10 or x is a multiple of 3.33.
ii. However x is an integer . So x has to be a multiple of 3.33*3=10 since that is the minimum value that makes x an integer.
iii. So a multiple of 10.
Hence E.
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Bunuel, VeritasKarishma anyway to do it using number testing ? Thanks.

You can use number testing here though it could end up eating too much of your time.

y is a multiple of 5
3x + 4y = 200

Put y = 5, you get x = 60
Two options (C) and (D) are eliminated because 60 is not a multiple of 7 and 8.

Put y = 10 or 15, x is not an integer.

Put y = 20, you get x = 40
Options (A) and (B) are eliminated because 40 is not divisible by 3 and 6.

Answer must be (E)

Note: For x to be an integer, y must increase in multiples of 3 and y is already a multiple of 5 so you know that y must increase in multiples of 15 for x to be an integer. This can help skipping the y = 10 and y = 15 steps. If you are not sure why, think of integer solutions to equations in 2 variables.

but my doubut is :
but my doubt is :
x = (20)(10 - k)/3
so this can b integer only if k =1 or 4 or 7 ,so 10-k will be =9 or 6 or 3
al;l of which are multiple of 3 then it must be multiple of 3 too .where m i going wrong?

y = 5a
3x + 4y = 200
3x + 4*5a = 200

3x + 20a = 200

x = 20(10 + a)/3

x has to be an integer. So (10 + a) must be divisible by 3 and x must be a multiple of 20. This implies that x must be a multiple of 10.

Note that x may not be a multiple of 3. Say a = 2, x = 20*12/3 = 80 (not a multiple of 3)
On the other hand, if a= 8, x = 20*18/3 = 120 (a multiple of 3)

Whether x is a multiple of 3, depends on the value of a. But it definitely is a multiple of 20 because x = 20 * (10 + a)/3 = Integer
So x = 20 * integer = Integer

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You are given that x and y are positive integers and they satisfy 3x + 4y = 200.
Not all values of x and y will work because they will not satisfy all these constraints.
In your example, x is not an integer.

All examples which do satisfy all constraints will have x as a multiple of 10.
e.g. x = 40, y = 20

Hi All,

This prompt gives us lots of information to work with. We’re told:

X and Y are POSITIVE INTEGERS
Y is a multiple of 5
3X + 4Y = 200

We’re then asked what X MUST be a multiple of. This question can be solved with a bit of Arithmetic and TESTing VALUES. There’s also a subtle Number Property Rule built into this question that can save you some time if you recognize it. If you don’t immediately spot the Number Property, then you can still solve this problem with a bit of ‘brute force.’

Let’s start by TESTing the simplest multiple of 5…

IF…. Y = 5…..
3X + 20 = 200
3X = 180
X = 60
60 is a multiple of both 6 and 10, so the correct answer has to be either B or E.

Next, let’s work up through the next few multiples of 5….

IF…. Y = 10
3X + 40 = 200
3X = 160
160 is NOT evenly divisible by 3 though (meaning that X would be a non-integer, which is not allowed)

IF…. Y = 15
3X + 60 = 200
3X = 140
140 is NOT evenly divisible by 3 though (meaning that X would be a non-integer, which is not allowed)

IF…. Y = 20
3X + 80 = 200
3X = 120
X = 40
Between the two remaining answers, 40 is only a multiple of 10.

Final Answer:

The Number Property in this question is if you add a multiple of 5 to another multiple of 5, then the sum will be a multiple of 5.

Since Y is a MULTIPLE of 5, then 4Y will also be a multiple of 5. The sum of the two terms (200) is ALSO a multiple of 5, so the remaining term (the 3X) must ALSO be a multiple of 5. The answers are written in such a way that there’s only one multiple of 5 among them (re: the correct answer).

GMAT Assassins aren’t born, they’re made,
Rich

Solution:

Given 3x = 200 -4y

=> 3x = Difference of even numbers = even

=> As 3 is odd x must be even (So A & C eliminated)

Also it is a difference of multiples of 5 => 3x is also a multiple of 5

=> x is a multiple of 5 as 3 is not =>So 10

(option e)

D.S
GMAT SME
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