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22 Dec 2017, 09:29
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
82% (02:01) correct
18%
(02:27)
wrong
based on 252
sessions
History
Date
Time
Result
Not Attempted Yet
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
(A) 3
(B) 5
(C) 9
(D) 12
(E) 18
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.
The answer is (D).
Leverage the Answer Choices
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).
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.
The answer is (D).
Leverage the Answer Choices
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).
General Discussion
22 Dec 2017, 09:40
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Bookmarks
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
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
