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a, b, and c are multiples of 15 and a < b < c

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GRE Math Challenge #8- a, b and c are multiple of 15 [#permalink] New post   05 Aug 2014, 03:42
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a, b and c are multiples of 15, and 0 < a < b < c

Quantity A
Quantity B
The remainder when b is divided by c
The remainder when (b + c) is divided by a


A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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A
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Re: GRE Math Challenge #8- a, b and c are multiple of 15 [#permalink] New post   21 Sep 2016, 04:57
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Given: a < b < c

Quantity A: The remainder when b is divided by c
Notice that c is greater than b
So, b divided by c will equal 0 with remainder b
If this isn't 100% clear, here are a few examples.
4 divided by 6 equals 0 with remainder 4.
Likewise, 2 divided by 7 equals 0 with remainder 2.
So, we get.....
Quantity A: b


Quantity B: The remainder when (b + c) is divided by a
IMPORTANT CONCEPT
When positive integer N is divided by positive integer D, the remainder R is such that 0 < R < D
For example, if we divide some positive integer by 7, the remainder will be 6, 5, 4, 3, 2, 1, or 0
So, we can conclude that....
Quantity B: some integer that's less than a

Since it's given that a < b, we can be certain that quantity A is greater.

Answer:
Show: ::
A


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Re: GRE Math Challenge #8- a, b and c are multiple of 15 [#permalink] New post   19 Sep 2016, 10:03
hi

Is the answer 'A'.
can anyone please confirm.

Thanks
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Re: GRE Math Challenge #8- a, b and c are multiple of 15 [#permalink] New post   19 Sep 2016, 10:41
I think the answer is D. I took 15,30 and 45 at first and here A is greater. But then I took 15, 30 and 150 30/150 leaves a quotient of 0.2 and a remainder of 0. Hence D
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a, b, and c are multiples of 15 and a < b < c [#permalink] New post   29 Oct 2016, 04:40
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a, b, and c are multiples of 15 and a < b < c

Quantity A
Quantity B
The remainder when b is divided by c
The remainder when (b + c) is divided by a


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Drill 2
Question: 2
Page: 290
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Re: a, b, and c are multiples of 15 and a < b < c [#permalink] New post   29 Oct 2016, 04:43
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Explanation

Try Plugging In. If a = 15, b = 30, and c = 60, Quantity A is 30 because c cannot divide into b even one time. Quantity B is 0 because 90 divided by 15 has no remainder. Eliminate choices (B) and (C).

Try a new set of numbers to further narrow your choices. If a = 30, b = 45, and c = 120, Quantity A is 45, and Quantity B is 15. The answer is choice A.
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Re: GRE Math Challenge #8- a, b and c are multiple of 15 [#permalink] New post   19 Dec 2016, 01:17
i think the answer will be D, when I take 15, 30, 45 then for the A quantity remainder will be 15, and for B is 0, then A is BIG, but if I take 15, 30, 150 ( i can take this, cause You didnt tell me that it will be consecutive multiples) then the remainder will be 0 for the quantity A and B, then the answer ll be D! Can you explain it?
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Re: a, b, and c are multiples of 15 and a < b < c [#permalink] New post   26 Mar 2018, 02:20
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sandy wrote:
Explanation

Try Plugging In. If a = 15, b = 30, and c = 60, Quantity A is 30 because c cannot divide into b even one time. Quantity B is 0 because 90 divided by 15 has no remainder. Eliminate choices (B) and (C).

Try a new set of numbers to further narrow your choices. If a = 30, b = 45, and c = 120, Quantity A is 45, and Quantity B is 15. The answer is choice A.



When it comes to this question, don't we need to consider negative numbers?

If we do, what the result will be?
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Re: a, b, and c are multiples of 15 and a < b < c [#permalink] New post   26 Mar 2018, 08:05
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This is a good question..

All the questions I have encountered have positive numbers when asking for remainder.

Remainder is a positive number. So Say \(-11 \div 5\) has a remainder of 4 not -1.

Let me look around and get back to you on this one.
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Re: a, b, and c are multiples of 15 and a < b < c [#permalink] New post   27 Aug 2018, 07:45
when dealing with these types of questions we usually consider negative numbers as well. So why we are not considering negative numbers in this question?
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Re: a, b, and c are multiples of 15 and a < b < c [#permalink] New post   27 Aug 2018, 10:16
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sandy wrote:
a, b, and c are multiples of 15 and a < b < c

Quantity A
Quantity B
The remainder when b is divided by c
The remainder when (b + c) is divided by a


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Drill 2
Question: 2
Page: 290


Remainder rule:
When positive integer N is divided by positive integer D, the remainder R is such that 0 ≤ R < D
For example, if we divide some positive integer by 7, the remainder will be 6, 5, 4, 3, 2, 1, or 0

QUANTITY A: The remainder when b is divided by c
Since c is greater than b, we know that: b divided by c equals zero with remainder b
For example, 30 divided by 75 equals 0 with remainder 30
So, QUANTITY A = b

QUANTITY B: The remainder when (b + c) is divided by a
According to the above rule, the remainder must be LESS THAN a
So, QUANTITY B = some number less than a

We have:
QUANTITY A: b
QUANTITY B: some number less than a
Since we are told that a < b < c, we can see that Quantity A is greater.

Answer: A

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Re: GRE Math Challenge #8- a, b and c are multiple of 15 [#permalink] New post   15 Sep 2019, 14:04
Given we use 15, 30 and 45 for our sample numbers we could have

A

\(\frac{30}{45}\) which, although not conventionally done, could be expressed as mixed fraction, like so:

\(0\frac{30}{45}\) which shows that 45 goes into 30 zero times and leaves a remainder of 30.

B

We have \(\frac{75}{15}\), which becomes

\(5\frac{0}{15}\), which shows that 15 goes into 75 five times and leaves 0 remainder.

We can see that the remainder for A(30) is greater than the remainder for B(0).
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Re: GRE Math Challenge #8- a, b and c are multiple of 15 [#permalink] New post   16 Oct 2019, 06:53
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IlCreatore wrote:

This is right instead. 0 divided by any number is 0 with remainder of zero and the definition of multiple does not forbid to look for negative multiples, unless stated we are using positive multiples. So the Answer should be D.

Does anybody have the OA/OE?


Good point. The question does not state that a, b and c are POSITIVE multiples.
I never noticed that the first time I read the question.

When it comes to remainder/divisibility questions on the GRE, there will always be some proviso that states the values are positive integers.
In my opinion, it's a bad question.

I'm going to edit the question so that it has the required provisos.

Cheers,
Brent
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Re: a, b, and c are multiples of 15 and a < b < c [#permalink] New post   10 Jun 2020, 06:30
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consider a=15, b=30, c=45

Quantity A = b/c = 30/45, Remainder = 30

Quantity B = (b+c)/a = 75/15, Remainder = 0

Quantity A is greater

option A should be the answer
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Re: a, b, and c are multiples of 15 and a < b < c [#permalink] New post   12 Jun 2020, 16:57
We have to remember that the remainder of a/c when a is less than c is equals to a.

With this condition. A = b, because b is less than c. And B = 0, because the remainder of the division of multiples of the same number is 0.

Since b is greater than 0, the correct answer is A.
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Re: a, b, and c are multiples of 15 and a < b < c [#permalink] New post   12 Oct 2023, 19:02
mocafaj wrote:
hi

Is the answer 'A'.
can anyone please confirm.

Thanks


yes correct
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Re: a, b, and c are multiples of 15 and a < b < c [#permalink] New post   07 Mar 2025, 14:22
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