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# The ratio of blue pens to red pens is 5:7

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Veritas Prep GMAT Instructor
Joined: 01 Jul 2017
Posts: 46
The ratio of blue pens to red pens is 5:7  [#permalink]

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22 Dec 2017, 09:29
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Difficulty:

15% (low)

Question Stats:

90% (01:52) correct 10% (00:56) wrong based on 31 sessions

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The ratio of blue pens to red pens is 5:7. When 3 blue pens are added to the group and 9 red pens are removed, the ratio of blue pens to red pens becomes 3:2. How many red pens are there after the changes?

(A) 3
(B) 5
(C) 9
(D) 12
(E) 18

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Aaron J. Pond
Veritas Prep Elite-Level Instructor

Hit "+1 Kudos" if my post helped you understand the GMAT better.
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Veritas Prep GMAT Instructor
Joined: 01 Jul 2017
Posts: 46
The ratio of blue pens to red pens is 5:7  [#permalink]

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Updated on: 15 Aug 2018, 19:15
There are a handful of ways to solve this problem. Let's look at two of them.

Do the Dang Math
First, we could solve this with just the raw algebra. The question tells us that the original ratio of blue to red pens is 5: 7. Thus, $$B = 5x$$ and $$R = 7x$$ (where $$x$$ is the scaling factor of the ratio.) We don't know what "$$x$$" is yet, but thinking of the ratio in these terms allows us to quickly simplify down to one variable.

The problem then tells us that if we were to add 3 to $$B$$ (in other words, $$5x +3$$) and subtract 9 from $$R$$ (in other words, $$7x-9$$), the new ratio would be 3:2. Mathematically, it would look like this:

$$\frac{5x+3}{7x-9} = \frac{3}{2}$$

Cross-multiplying the fractions gives us:

$$2(5x+3) = 3(7x-9)$$
$$10x+6 = 21x - 27$$
$$11x = 33$$
$$x = 3$$

Note how "3" is a possible answer. However, this is a classic trap of the GMAT: including the "right answer to the wrong question" as one of the answer choices. We were not asked to solve for the original scaling factor prior to the change, $$x$$. (Incidentally, if we solve for the number of blue pens after the change, we arrive at 15, another trap answer!) Make sure you focus on the actual target of the question: the number of red pens after the change. Thus, we need $$7x - 9$$. Plugging $$x$$ into our equation gives us 12.

There is a totally different way to solve this problem as well, using the answer choices as leverage. (I call this tactic "Look Out Below!" in my classes.) The nature of ratios sometimes allows us to compare the answer choices against each other and eliminate any answer choice that doesn't "follow the rules." For example, the problem tells us that after the change, ratio of blue pens to red pens should be 3:2 (or 3x:2x). Since you can't have fractional pens, this means that the number of red pens must be a multiple of 2. And, since answer choices (A), (B), and (C) are all odd, we can eliminate these. We are down to only (D) and (E).

Before the change (in other words, before we added 3 blue pens and subtracted 9 pens), the ratio was 5x:7x.Thus, adding +9 back to answer choices (D) and (E) would give us the theoretical "original" value for $$R$$, which should be a multiple of 7.

Adding 9 to answer choice (D) gives us 21 -- a multiple of 7 and therefore a possible candidate.
Adding 9 to answer choice (E) gives us 27 -- not a multiple of 7. We can eliminate this, leaving only one answer, (D).

Any way you solve it, the answer is still (D).
_________________

Aaron J. Pond
Veritas Prep Elite-Level Instructor

Hit "+1 Kudos" if my post helped you understand the GMAT better.
Look me up at https://www.veritasprep.com/gmat/aaron-pond/ if you want to learn more GMAT Jujitsu.

Originally posted by AaronPond on 22 Dec 2017, 10:16.
Last edited by AaronPond on 15 Aug 2018, 19:15, edited 3 times in total.
##### General Discussion
Intern
Joined: 02 Apr 2017
Posts: 11
Re: The ratio of blue pens to red pens is 5:7  [#permalink]

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22 Dec 2017, 09:40
Given that :

B/R = 5/7

B+3/R-9 = 3/2

It follows that:
5X+3/7x-9 = 3/2
=> After simplification x = 3

Hence Blue pens = 15 and Red Pens = 21
After removing 9 pends, Red Pends = 12

And hence D
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Joined: 22 May 2016
Posts: 1903
The ratio of blue pens to red pens is 5:7  [#permalink]

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22 Dec 2017, 11:06
1
AaronPond wrote:
The ratio of blue pens to red pens is 5:7. When 3 blue pens are added to the group and 9 red pens are removed, the ratio of blue pens to red pens becomes 3:2. How many red pens are there after the changes?

(A) 3
(B) 5
(C) 9
(D) 12
(E) 18

With changed ratio questions like this one:

1) Set up the original ratio with a multiplier, x
$$\frac{Blue}{Red}=\frac{5x}{7x}$$

2) Add or subtract as directed, and set LHS equal to the new ratio*
$$\frac{(5x + 3)}{(7x -9)}=\frac{3}{2}$$

3) Solve for the multiplier, x, which yields quantities for the original ratio
$$2(5x + 3) = 3(7x - 9)$$
$$10x + 6 = 21x - 27$$
$$33 = 11x$$, $$x = 3$$

4) Use the multiplier
Before change, number of red pens is 7x
x = 3: (7)(3) = 21 red pens before
After change: (21 - 9) = 12 red pens

*After setting up the original ratio with x, the numbers in operations AND the new ratio are "just" numbers; i.e., no multipliers attached to them. And almost always you can perform operations on top and bottom simultaneously.
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The ratio of blue pens to red pens is 5:7 &nbs [#permalink] 22 Dec 2017, 11:06
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