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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
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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?
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:
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
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31 Oct 2019, 20:26
1
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
Re: If x and y are positive integers such that y is a multiple of 5 and 3x
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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
Re: If x and y are positive integers such that y is a multiple of 5 and 3x
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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.