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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 08 Oct 2019, 22:59
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

Answer:




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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?
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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 Oct 2019, 21:23
vanam52923 wrote:
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

Answer:




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


and

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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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 31 Oct 2019, 20:26
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StrugglingGmat2910 wrote:
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


can we apply this approach to other questions also
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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 31 Oct 2019, 20:35
Rexnaster wrote:
StrugglingGmat2910 wrote:
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


can we apply this approach to other questions also


Gmat is all about seeing the pattern : either you can apply some logical pattern or just test the answer to prove the question
in order to score well in efficient time one must leverage these small logical steps that will save time
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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 Dec 2019, 11:56
Its Simple.

3x + 4y = 200

But y = 5k( Since y is a multiple of 5)

3x + 5k(4) = 200

Now 5k(4)= 20 k and it must end with a zero(0)

Sum of 3x + 20k equals 200

So 3x must end with a zero.

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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New post 14 Jan 2020, 14:21
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


Plug in the options given

for example option (A) 3(3) + 4y = 200
9 + 4y = 200

4y = 200 - 9

4y = 191


Since it is given that y is a multiple of 5. Plugging in 3 does not satisfy the equation

Option (B) 3(6) + 4y = 200

4y = 200 - 18

4y = 182
Does not satisfy the given that statement y is a multiple of 5

Try plugging in for all option.

Clearly only option (E) satisfies the given condition
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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 Jan 2020, 11:29
okaypompeii 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

----------------------
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.


If y = 10, then 3x+4(10)=200
3x+40=200
3x=160
x=53.333333

does this not mean X is not a multiple of 10?
What am I missing here?
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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 17 Feb 2020, 12:20

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Re: If x and y are positive integers such that y is a multiple of 5 and 3x   [#permalink] 17 Feb 2020, 12:20

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