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# At noon of a certain day, when 5 pens and 3 pencils were

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At noon of a certain day, when 5 pens and 3 pencils were [#permalink]  10 Sep 2018, 00:47
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At noon of a certain day, when 5 pens and 3 pencils were placed in a drawer, the ratio of the number of pens to the number of pencils in that drawer became 47 to 17.

 Quantity A Quantity B The ratio of the number of pens to the number of pencils in the drawer immediately before noon of that day $$\frac{3}{1}$$

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

Can someone explain this in very very simpler terms?
[Reveal] Spoiler: OA
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Re: At noon of a certain day, when 5 pens and 3 pencils were [#permalink]  10 Sep 2018, 04:59
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Expert's post
rajlal wrote:
At noon of a certain day, when 5 pens and 3 pencils were placed in a drawer, the ratio of the number of pens to the number of pencils in that drawer became 47 to 17.

 Quantity A Quantity B The ratio of the number of pens to the number of pencils in the drawer immediately before noon of that day $$\frac{3}{1}$$

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

Can someone explain this in very very simpler terms?

KEY INFORMATION: After the pens and pencils were added, the ratio of the number of pens to the number of pencils in that drawer became 47 to 17 (aka 47/17).

Here's one possible scenario:
BEFORE NOON, the drawer contained 42 pens and 14 pencils.
So, after the 5 pens and 3 pencils were added, the drawer contained 47 pens and 17 pencils, giving us our ratio of 47/17
In this case, ratio of the number of pens to the number of pencils in the drawer BEFORE NOON = 42/14 = 3/1
We get:
Quantity A: 3/1
Quantity B: 3/1
The two quantities are EQUAL

IMPORTANT: Are there any other scenarios that satisfy the given conditions?
YES! There are infinitely many scenarios, because the key ratio in the question (47/17) has INFINITELY MANY equivalent ration.
For example, 47/17 = 94/34 = 141/51 = 470/170 etc

So, let's examine another possible scenario
BEFORE NOON, the drawer contained 89 pens and 31 pencils.
So, after the 5 pens and 3 pencils were added, the drawer contained 94 pens and 34 pencils, giving us our ratio of 94/34, which is EQUIVALENT to 47/17
In this case, ratio of the number of pens to the number of pencils in the drawer BEFORE NOON = 89/31
We get:
Quantity A: 89/31
Quantity B: 3/1
The two quantities are NOT equal

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com

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At noon of a certain day, when 5 pens and 3 pencils were pla [#permalink]  10 Jul 2019, 05:45
At noon of a certain day, when 5 pens and 3 pencils were placed in a drawer, the ratio of the number of pens to the number of pencils in that drawer became 47 to 17.

 Quantity A Quantity B The ratio of the number of pens to the number of pencils in the drawer immediately before noon of that day 3/1
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Re: At noon of a certain day, when 5 pens and 3 pencils were pla [#permalink]  10 Jul 2019, 06:36
Expert's post
arpitjain wrote:
At noon of a certain day, when 5 pens and 3 pencils were placed in a drawer, the ratio of the number of pens to the number of pencils in that drawer became 47 to 17.

 Quantity A Quantity B The ratio of the number of pens to the number of pencils in the drawer immediately before noon of that day 3/1

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com

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Re: At noon of a certain day, when 5 pens and 3 pencils were [#permalink]  21 May 2020, 13:30
@GreenlightTestPrep

Hi Brent,

In practice, I've come upon similar problems in which equations are set up with a before and after change to solve. I'm not sure if that approach doesn't apply here, but this is what I did. Can you tell me where I went wrong or if that is not the correct approach for this type of problem?

pen + 5/ pencil + 3 = 47/17

I then solved the equation to get pen/pencil = 103/17 which is a little greater than 6 so I said B is greater than A.

Is the above method not correct? If not, when can you apply the above method to create an equation and solve for the proportion?

GreenlightTestPrep wrote:
rajlal wrote:
At noon of a certain day, when 5 pens and 3 pencils were placed in a drawer, the ratio of the number of pens to the number of pencils in that drawer became 47 to 17.

 Quantity A Quantity B The ratio of the number of pens to the number of pencils in the drawer immediately before noon of that day $$\frac{3}{1}$$

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

Can someone explain this in very very simpler terms?

KEY INFORMATION: After the pens and pencils were added, the ratio of the number of pens to the number of pencils in that drawer became 47 to 17 (aka 47/17).

Here's one possible scenario:
BEFORE NOON, the drawer contained 42 pens and 14 pencils.
So, after the 5 pens and 3 pencils were added, the drawer contained 47 pens and 17 pencils, giving us our ratio of 47/17
In this case, ratio of the number of pens to the number of pencils in the drawer BEFORE NOON = 42/14 = 3/1
We get:
Quantity A: 3/1
Quantity B: 3/1
The two quantities are EQUAL

IMPORTANT: Are there any other scenarios that satisfy the given conditions?
YES! There are infinitely many scenarios, because the key ratio in the question (47/17) has INFINITELY MANY equivalent ration.
For example, 47/17 = 94/34 = 141/51 = 470/170 etc

So, let's examine another possible scenario
BEFORE NOON, the drawer contained 89 pens and 31 pencils.
So, after the 5 pens and 3 pencils were added, the drawer contained 94 pens and 34 pencils, giving us our ratio of 94/34, which is EQUIVALENT to 47/17
In this case, ratio of the number of pens to the number of pencils in the drawer BEFORE NOON = 89/31
We get:
Quantity A: 89/31
Quantity B: 3/1
The two quantities are NOT equal

Cheers,
Brent
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Posts: 3907
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Re: At noon of a certain day, when 5 pens and 3 pencils were [#permalink]  21 May 2020, 16:56
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Expert's post
GREStudent2020 wrote:
@GreenlightTestPrep

Hi Brent,

In practice, I've come upon similar problems in which equations are set up with a before and after change to solve. I'm not sure if that approach doesn't apply here, but this is what I did. Can you tell me where I went wrong or if that is not the correct approach for this type of problem?

pen + 5/ pencil + 3 = 47/17

I then solved the equation to get pen/pencil = 103/17 which is a little greater than 6 so I said B is greater than A.

Is the above method not correct? If not, when can you apply the above method to create an equation and solve for the proportion?

If you got pen/pencil = 103/17, then I think you made a mistake.
You should get: 17N - 47P = 56

Can you show me your steps?

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com

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Re: At noon of a certain day, when 5 pens and 3 pencils were [#permalink]  22 May 2020, 08:10
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@GreenlightTestPrep

Hi Brent,

Please see my steps below. (n = pen and p = pencil)

n+5/p+3 = 47/17
17n + 85 = 47p + 141 (subtract 85 from both sides)
17n = 47p + 56 (divide both sides by 17)
n = 47p/17 + 56/17 (divide both sides by p to get n/p since the question asks for the original proportion value)
n/p = 47/17 + 56/17 (add fraction)
n/p = 103/17

I think I might have caught my mistake as I was typing my steps out. When I divide both sides by p, I would also have to divide the 56 by p, correct? I didn't do that above so maybe that's where I made my first mistake?

Additionally, you said I should get 17n = 47p + 56. Which I did in step 3. How would I proceed after that step to get the value of the original proportion, n/p?

[/quote]

Yes you correctly identified your mistake.
If we take: n = 47p/17 + 56/17
And divide both sides by p, we get: n/p = 47/17 + 56/17p
----------------

Once we get to the equation 17n = 47p + 56, we might recognize that there are infinitely many solutions.
If we're able to identify pairs of INTEGER values of n and p that yield different values for Quantity A, then we're done.
However, that could be quite time-consuming.

A faster approach is to do what I did in my solution.
If we know that (n+5)/(p+3) = 47/17, then it's also true that (n+5)/(p+3) = 94/34 (since 47/17 and 94/34 are equivalent fractions)
We could also have (n+5)/(p+3) = 940/340 or (n+5)/(p+3) = 9400/3400
etc

So one possible situation is n+5 = 47 and p+3 = 17, in which case n/p = 42/14 = 3/1
Another possible situation is n+5 = 94 and p+3 = 34, in which case n/p = 89/31
Another possible situation is n+5 = 940 and p+3 = 340, in which case n/p = 935/337
etc

Does that help?
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Re: At noon of a certain day, when 5 pens and 3 pencils were [#permalink]  23 May 2020, 15:03
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Hi Brent,

Thank you for the explanation. That does make sense, I guess I figured that when I set up an equation and get a value for n/p that will only result in one possible value.

How is this question different from say something like the below?

"On Monday, a class has 8 girls and 20 boys. On Tuesday, a certain number of girls joined the class just as twice that number of boys left the class, changing the ratio of girls to boys to 7 to 4. How many boys left the class on Tuesday?"

Because in the the above problem you can simply set up an equation to equal the new proportion to determine what the original proportion was.
girls = boys = 8+x/20-2x = 7/4 to determine what x is and therefore what x, the number of girls, is and use that to solve for what the number of boys is.

Is the reason the girls and boys question is different than the pen and pencil question because in the former there's a relationship identified between girls and boys (boys = 2*girls) whereas in the pen and pencil question there is not relationship identified between the two values other than the proportion which allows the proportion to be any multiple of 47/17.

Apologies for all the back and forth and for so many questions, but I really appreciate you explanations!

Thank you very much again!

GREStudent2020 wrote:
@GreenlightTestPrep

Hi Brent,

Please see my steps below. (n = pen and p = pencil)

n+5/p+3 = 47/17
17n + 85 = 47p + 141 (subtract 85 from both sides)
17n = 47p + 56 (divide both sides by 17)
n = 47p/17 + 56/17 (divide both sides by p to get n/p since the question asks for the original proportion value)
n/p = 47/17 + 56/17 (add fraction)
n/p = 103/17

I think I might have caught my mistake as I was typing my steps out. When I divide both sides by p, I would also have to divide the 56 by p, correct? I didn't do that above so maybe that's where I made my first mistake?

Additionally, you said I should get 17n = 47p + 56. Which I did in step 3. How would I proceed after that step to get the value of the original proportion, n/p?

Yes you correctly identified your mistake.
If we take: n = 47p/17 + 56/17
And divide both sides by p, we get: n/p = 47/17 + 56/17p
----------------

Once we get to the equation 17n = 47p + 56, we might recognize that there are infinitely many solutions.
If we're able to identify pairs of INTEGER values of n and p that yield different values for Quantity A, then we're done.
However, that could be quite time-consuming.

A faster approach is to do what I did in my solution.
If we know that (n+5)/(p+3) = 47/17, then it's also true that (n+5)/(p+3) = 94/34 (since 47/17 and 94/34 are equivalent fractions)
We could also have (n+5)/(p+3) = 940/340 or (n+5)/(p+3) = 9400/3400
etc

So one possible situation is n+5 = 47 and p+3 = 17, in which case n/p = 42/14 = 3/1
Another possible situation is n+5 = 94 and p+3 = 34, in which case n/p = 89/31
Another possible situation is n+5 = 940 and p+3 = 340, in which case n/p = 935/337
etc

Does that help?[/quote]
Re: At noon of a certain day, when 5 pens and 3 pencils were   [#permalink] 23 May 2020, 15:03
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