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If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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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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Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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AbdurRakib wrote:
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

OG 2017 New Question

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.

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Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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AbdurRakib wrote:
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

OG 2017 New Question

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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Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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AbdurRakib wrote:
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

OG 2017 New Question

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

Here are two videos that cover the skills needed to answer this question:

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Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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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
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GRE 1: Q164 V161 Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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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

----------------------
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.
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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VeritasPrepKarishma wrote:
AbdurRakib wrote:
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

OG 2017 New Question

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.

Responding to a pm:
Quote:
If 3x has to be even, why can't the answer be B oder D, since 3*6 = even and 3*8 = even?

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)
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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AbdurRakib wrote:
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

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.

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Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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We know: 1. x and y are positive integers
2. y is a multiple of 5
3. 3x + 4y = 200

3x + 4y = 200

=> 3x = 200 - 4y

It's known that if we add or subtract multiples of N, the result is a multiple of N. Here, both 200 and 4y are multiple of 10(20 actually). So 3x is a multiple of 10.
And therefore x is a multiple of 10.

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Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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AbdurRakib wrote:
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

let y=35
x=20
only E works
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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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.
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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GMATPrepNow wrote:
AbdurRakib wrote:
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

OG 2017 New Question

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

Here we say that (10-k)/3 = (some-int) => (10-k) = 3*(some-int)
and x = 20*(10-k)/3
x = 20*3*(some-int)
so x is a multiple of 3 as well.

Where am I going wrong?
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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this is hard

3x+4y=200
y=(200-3x)/4

= 50-3x/4

because y is divisible by 5, 3x/5 is divisible by 5

3x/4=5a
x=20a/3

a must divisible by 3, so that x can be interger, so x= multiple of 20

x divisible by 10

hard one
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If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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x and y are positive integers

This is the most important information that will help us find the answer.

y=5k (k is a positive constant)

3x+4*5k= 200
3x+20k= 200
x= (200-20k)/3

x= 20(10-k)/3

As x is an integer, the above equation has to be an integer. As 20 is not divisible by 3, (10-k) must be divisible by 3 to make RHS an integer.

Looking at the options, E works.
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If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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AbdurRakib wrote:
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

OG 2017 New Question

3x+4y=200➡3x=4(50-y)
because y is a multiple of 5, and 4 is even,
4(50-y) will always have a units digit of 0
10
E
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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3x + 4y = 200
y is a multiple of 5 so go for the trail and error
1. 3x+4*5 = 200
3x = 180, it is mutiple of both 3 and 10
2. 3x+4*(5*2) = 200
3x = 160, 3 can be eliminated so answer should be 10
E
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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VeritasPrepKarishma wrote:
AbdurRakib wrote:
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

OG 2017 New Question

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.

hi mam

The way I approached the problem is as under:

3x + 4y = 200
Now, as, y = 5k

3x + 20k = 200

x = 20(10 - k)/3
Now, (10 - k) must be 3n

So, x = 60n

Thus answer choice E must be true, and other choices could be true ....

please say to me whether it is okay...

thanks in advance, mam
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If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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AbdurRakib wrote:
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

OG 2017 New Question

3x/4+y=50
x=either 20 or 40
10
E
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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gmatcracker2017 wrote:
VeritasPrepKarishma wrote:
AbdurRakib wrote:
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

OG 2017 New Question

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.

hi mam

The way I approached the problem is as under:

3x + 4y = 200
Now, as, y = 5k

3x + 20k = 200

x = 20(10 - k)/3
Now, (10 - k) must be 3n

So, x = 60n

Thus answer choice E must be true, and other choices could be true ....

please say to me whether it is okay...

thanks in advance, mam

Yes, perfectly fine.
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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I think the trick here is to use unit digit concept.
3X+20K=200 means 3X must have 0 as unit digit.
from the answer choices, only E works.

Thanks
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