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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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09 Feb 2018, 16:09
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
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10 Feb 2018, 00:37
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
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12 Feb 2018, 03:38
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
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15 Apr 2018, 23:59
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 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)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
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16 Apr 2018, 00:19
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
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16 Apr 2018, 01:00
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
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26 Jul 2018, 07:12
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 2004Y 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. Posted from my mobile device
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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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28 Jul 2018, 00:58
niks18 chetan2u pushpitkc KarishmaB gmatbustersQuote: 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
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28 Jul 2018, 01:41
This is the perfect approach. adkikani wrote: niks18 chetan2u pushpitkc KarishmaB gmatbustersQuote: 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
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06 Aug 2018, 17:09
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
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28 Aug 2018, 00:38
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
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01 Oct 2018, 08:43
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
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02 Oct 2018, 03:24
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 ... ofthumb/
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