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
E
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Must be a multiple of what number?
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AbeNeedsAnswers
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RULE:AbeNeedsAnswers 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
(multiple of k) ± (multiple of k) = multiple of k.
Since y is a multiple of 5, let y=5a.
Substituting y=5a into 3x + 4y = 200, we get:
3x + 4(5a) = 200.
3x + 20a = 200
3x = 200 - 20a.
3x = (multiple of 20) - (multiple of 20).
In accordance with the rule above:
(multiple of 20) - (multiple of 20) = multiple of 20.
Thus:
3x = multiple of 20.
For 3x to be a multiple of 20, x must be a multiple of 20.
Thus, x must also be a multiple of 10.
The correct answer is E.
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Jay@ManhattanReview
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We have 3x + 4y = 200, such that y is a multiple of 5 and x and y are positive integers.AbeNeedsAnswers 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
E
Say y = 5k; where k is a positive integer
=> 3x + 20k = 200
=> 3x = 200 - 20k
=> x = 20(10 - k)/3
=> x is a multiple of 20.
=> x is a multiple of all the factors of 20: 1, 2, 4, 5, 10, and 20.
Only one option qualifies, i.e. option E, the correct answer!
The correct answer: E
Hope this helps!
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ceilidh.erickson
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I would recommend relying on strategy whenever possible; on this problem, using the rule that Mitch pointed out is probably the fastest approach. However, you could also TEST NUMBERS here.
Since y is a multiple of 5, just test any multiple of 5:
y = 5
4y = 20
3x + 4y = 200
3x + 20 = 200
3x = 180
x = 60
Eliminate C and D.
Test another value:
y = 10
4y = 40
3x + 40 = 200
x = 160
x = 160/3 --> not an integer. We have to test another value.
y = 15
4y = 60
3x + 60 = 200
3x = 140
x = 140/3 --> not an integer
y = 20
4y = 80
3x + 80 = 200
3x = 120
x = 40
Eliminate A and B.
The only answer left is E.
This strategy is more cumbersome, but in a pinch it would get you there!
Since y is a multiple of 5, just test any multiple of 5:
y = 5
4y = 20
3x + 4y = 200
3x + 20 = 200
3x = 180
x = 60
Eliminate C and D.
Test another value:
y = 10
4y = 40
3x + 40 = 200
x = 160
x = 160/3 --> not an integer. We have to test another value.
y = 15
4y = 60
3x + 60 = 200
3x = 140
x = 140/3 --> not an integer
y = 20
4y = 80
3x + 80 = 200
3x = 120
x = 40
Eliminate A and B.
The only answer left is E.
This strategy is more cumbersome, but in a pinch it would get you there!
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
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We can write "y is a multiple of 5" as y = 5m, where m is some integer whose value we don't need.
From there:
3x + 4y =>
3x + 4*5m =>
3x + 20m
We're told this equals 200, so
3x + 20m = 200
3x = 200 - 20m
3x = 20 * (10 - m)
So x must be a multiple of 20. 20 is itself a multiple of 10, so answer E fits.
From there:
3x + 4y =>
3x + 4*5m =>
3x + 20m
We're told this equals 200, so
3x + 20m = 200
3x = 200 - 20m
3x = 20 * (10 - m)
So x must be a multiple of 20. 20 is itself a multiple of 10, so answer E fits.
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Alternatively, start with
3x + 4y = 200
3x = 200 - 4y
3x = 4 * (50 - y)
From this, we know x is a multiple of 4. We're told in the prompt that it's a multiple of 5. Since it's a multiple of 4 and of 5, it must also be a multiple of 4*5, or 20. That tells us x is also a multiple of ANY factor of 20 and E fits the bill.
3x + 4y = 200
3x = 200 - 4y
3x = 4 * (50 - y)
From this, we know x is a multiple of 4. We're told in the prompt that it's a multiple of 5. Since it's a multiple of 4 and of 5, it must also be a multiple of 4*5, or 20. That tells us x is also a multiple of ANY factor of 20 and E fits the bill.
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Scott@TargetTestPrep
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Since y is a multiple of 5, we can let y = 5k and we have:AbeNeedsAnswers 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
3x + 4(5k) = 200
3x = 200 - 20k
3x = 20(10 - k)
x = [20(10 - k)]/3
Since x is an integer, (10 - k) must be divisible by 3. Then, x is the product of 20 times some integer [which is the quotient of (10 - k)/3]. Thus, x must be a multiple of 10. For instance, when k = 1, x = 90, when k = 4, x = 40, and when k = 7, k = 20.
Answer: E
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Hi All,
This prompt gives us lots of information to work with. We’re told:
X and Y are POSITIVE INTEGERS
Y is a multiple of 5
3X + 4Y = 200
We’re then asked what X MUST be a multiple of. This question can be solved with a bit of Arithmetic and TESTing VALUES. There’s also a subtle Number Property Rule built into this question that can save you some time if you recognize it. If you don’t immediately spot the Number Property, then you can still solve this problem with a bit of ‘brute force.’
Let’s start by TESTing the simplest multiple of 5…
IF…. Y = 5…..
3X + 20 = 200
3X = 180
X = 60
60 is a multiple of both 6 and 10, so the correct answer has to be either B or E.
Next, let’s work up through the next few multiples of 5….
IF…. Y = 10
3X + 40 = 200
3X = 160
160 is NOT evenly divisible by 3 though (meaning that X would be a non-integer, which is not allowed)
IF…. Y = 15
3X + 60 = 200
3X = 140
140 is NOT evenly divisible by 3 though (meaning that X would be a non-integer, which is not allowed)
IF…. Y = 20
3X + 80 = 200
3X = 120
X = 40
Between the two remaining answers, 40 is only a multiple of 10.
Final Answer: E
The Number Property in this question is if you add a multiple of 5 to another multiple of 5, then the sum will be a multiple of 5.
Since Y is a MULTIPLE of 5, then 4Y will also be a multiple of 5. The sum of the two terms (200) is ALSO a multiple of 5, so the remaining term (the 3X) must ALSO be a multiple of 5. The answers are written in such a way that there’s only one multiple of 5 among them (re: the correct answer).
GMAT Assassins aren’t born, they’re made,
Rich
This prompt gives us lots of information to work with. We’re told:
X and Y are POSITIVE INTEGERS
Y is a multiple of 5
3X + 4Y = 200
We’re then asked what X MUST be a multiple of. This question can be solved with a bit of Arithmetic and TESTing VALUES. There’s also a subtle Number Property Rule built into this question that can save you some time if you recognize it. If you don’t immediately spot the Number Property, then you can still solve this problem with a bit of ‘brute force.’
Let’s start by TESTing the simplest multiple of 5…
IF…. Y = 5…..
3X + 20 = 200
3X = 180
X = 60
60 is a multiple of both 6 and 10, so the correct answer has to be either B or E.
Next, let’s work up through the next few multiples of 5….
IF…. Y = 10
3X + 40 = 200
3X = 160
160 is NOT evenly divisible by 3 though (meaning that X would be a non-integer, which is not allowed)
IF…. Y = 15
3X + 60 = 200
3X = 140
140 is NOT evenly divisible by 3 though (meaning that X would be a non-integer, which is not allowed)
IF…. Y = 20
3X + 80 = 200
3X = 120
X = 40
Between the two remaining answers, 40 is only a multiple of 10.
Final Answer: E
The Number Property in this question is if you add a multiple of 5 to another multiple of 5, then the sum will be a multiple of 5.
Since Y is a MULTIPLE of 5, then 4Y will also be a multiple of 5. The sum of the two terms (200) is ALSO a multiple of 5, so the remaining term (the 3X) must ALSO be a multiple of 5. The answers are written in such a way that there’s only one multiple of 5 among them (re: the correct answer).
GMAT Assassins aren’t born, they’re made,
Rich
