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

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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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New post 09 Feb 2018, 17:09
Can someone pls verify if this method will work for all similar question types? This is the way I approached as well

wvu wrote:
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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New post 10 Feb 2018, 01:37
mrdlee23 wrote:
Can someone pls verify if this method will work for all similar question types? This is the way I approached as well

wvu wrote:
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.


We know that -
\(3x + 4y = 200, y = 5m\)
\(3x + 20m = 200\)
\(x = \frac{10(20 - m)}{3}\)

So, x has to be 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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New post 12 Feb 2018, 04:38
mrdlee23 wrote:
Can someone pls verify if this method will work for all similar question types? This is the way I approached as well

wvu wrote:
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.


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.
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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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New post 16 Apr 2018, 00:59
Abhishek009 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

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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)


Hey, Abhishek009! 40 is also a multiple of 8, option D. So we still have D and E remaining.
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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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New post 16 Apr 2018, 01:19
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advait92 wrote:
Hey, Abhishek009! 40 is also a multiple of 8, option D. So we still have D and E remaining.


Hey advait92 ,

As Abhishek009 has explained, we have already rejected option D when we substituted y = 5 in the equation.

Remember for a "MUST" be true question, it should work for all the scenarios but we are not getting x a multiple of 8 whenever we have y = 5.

Hence, D cannot be the answer.

Does that make sense?
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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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New post 16 Apr 2018, 02:00
abhimahna wrote:
advait92 wrote:
Hey, Abhishek009! 40 is also a multiple of 8, option D. So we still have D and E remaining.


Hey advait92 ,

As Abhishek009 has explained, we have already rejected option D when we substituted y = 5 in the equation.

Remember for a "MUST" be true question, it should work for all the scenarios but we are not getting x a multiple of 8 whenever we have y = 5.

Hence, D cannot be the answer.

Does that make sense?


Hey abhimahna! Thank you for bringing this to my notice.
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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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New post 26 Jul 2018, 08:12
X,Y >0 i.e.X,Y both positive
Y=mul(5) i.e. Y is multiple of 5

3X+4Y=200
Because Y=mul(5),4Y will be multiple of 20
200 is also multiple of 20.
Hence 200-4Y will also be multiple of 20
I.e. 3X= multiple of 20
Hence, X will be multiple of 20, 10×2, and hence of 10.

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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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New post 28 Jul 2018, 01:58
niks18 chetan2u pushpitkc KarishmaB gmatbusters

Quote:
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?


How about this approach?
Given: y is a multiple of 5, so unit digits of product with y must end with 0 or 5.

4*y = (Unit digit will always end with 0, since 4*5 = 20 and 4*0 =0 )

Now look RHS, I need unit digit as 0, __ + 0 = 0
Hence 3 must be some multiple an integer ending with 0.
Ans: 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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New post 28 Jul 2018, 02:41
This is the perfect approach. :thumbup:

adkikani wrote:
niks18 chetan2u pushpitkc KarishmaB gmatbusters

Quote:
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?


How about this approach?
Given: y is a multiple of 5, so unit digits of product with y must end with 0 or 5.

4*y = (Unit digit will always end with 0, since 4*5 = 20 and 4*0 =0 )

Now look RHS, I need unit digit as 0, __ + 0 = 0
Hence 3 must be some multiple an integer ending with 0.
Ans: 10

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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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New post 06 Aug 2018, 18:09
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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If x and y are positive integers such that y is a multiple of 5 and 3x  [#permalink]

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New post 28 Aug 2018, 01:38
Clearly we can assume y = 5k since y is a multiple of 5. Now the equation becomes
3x + 4(5k) = 200
So 3x + 20k = 200
Now for any integer value of k , we will have unit digit ending with 0 (e.g 20*1 = 20 , 20*2 = 40 etc)

Applying the unit digit concept, we know that
sum of both terms of LHS is 200 and since 20k has unit digit 0 , therefore 3x must have unit digit 0.
So 3x can end with unit digit 0 only when x is a multiple of 10

Hope i could clarify in a simpler way.
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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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New post 01 Oct 2018, 09:43
Bunuel, VeritasKarishma anyway to do it using number testing ? Thanks.
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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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New post 02 Oct 2018, 04:24
sadikabid27 wrote:
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. For more, check: https://www.veritasprep.com/blog/2011/0 ... -of-thumb/
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x &nbs [#permalink] 02 Oct 2018, 04:24

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