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The ratio of boys to girls in Class A is 3 to 4.

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The ratio of boys to girls in Class A is 3 to 4.

by melguy » Mon Nov 14, 2016 6:19 am
The ratio of boys to girls in Class A is 3 to 4. The ratio of boys to girls in Class B is 4 to 5. If the two classes were combined, the ratio of boys to girls in the combined class would be 17 to 22. If Class A has one more boy and two more girls than class B, how many girls are in Class A?

8
9
10
11
12
(Ratios, Problem source : Manhattan , Difficulty : 600 - 700)

Please help me with this problem. I used the approach that number of girls in Class A must be a multiple of 4 so the answer must be 8 or 12. Then I plugged back the values to figure out the answer is E. I was really pressed of time in the sample exam and this was the final Q so got this wrong. But during the review I got it correct. So please confirm if there is a better approach /lesson I can take away from this problem / this type of problem.

OA is E

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by GMATGuruNY » Mon Nov 14, 2016 6:28 am
The approach that you took -- PLUGGING IN THE ANSWERS -- is a very efficient way to determine the correct answer.
The ratio of boys to girls in Class A is 3 to 4. The ratio of boys to girls in Class B is 4 to 5. If the two classes were combined, the ratio of boys to girls in the combined class would be 17 to 22. If Class A has one more boy and two more girls than class B, how many girls are in Class A?

8
9
10
11
12
The answers choices represent the number of girls in Class A.
Since b:g = 3:4, the number of girls must be a multiple of 4.
Eliminate B, C and D.

Answer choice A: g = 8.
Since b:g = 3:4 = 6:8, b=6.
Since Class A has one more boy and two more girls than Class B, in Class B, b=5 and g=6.
Doesn't work: the required ratio in Class B is 4:5.
Eliminate A.

The correct answer is E.

Note that we had to try ONLY ONE ANSWER CHOICE and perform only SIMPLE ARITHMETIC to determine the correct answer -- a very efficient way to solve the problem.
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by Brent@GMATPrepNow » Mon Nov 14, 2016 6:40 am
melguy wrote:The ratio of boys to girls in Class A is 3 to 4. The ratio of boys to girls in Class B is 4 to 5. If the two classes were combined, the ratio of boys to girls in the combined class would be 17 to 22. If Class A has one more boy and two more girls than class B, how many girls are in Class A?

8
9
10
11
12
The ratio of boys to girls in Class A is 3 to 4.
Let B = number of boys in class A
Let G = number of girls in class A
We get: B/G = 3/4
Cross multiply to get: 4B = 3G

Class A has one more boy and two more girls than class B
So B - 1 = number of boys in class B
So G - 2 = number of girls in class B

The ratio of boys to girls in Class B is 4 to 5
We get: (B - 1)/(G - 2) = 4/5
Cross multiply to get: 5(B - 1) = 4(G - 2)
Expand: 5B - 5 = 4G - 8

So, we now have the following system to solve for G:
4B = 3G
5B - 5 = 4G - 8

Take 4B = 3G and solve for B to get: B = 3G/4

Take 5B - 5 = 4G - 8 and replace B with 3G/4
We get: 5(3G/4) - 5 = 4G - 8
Expand: 15G/4 - 5 = 4G - 8
Multiply both sides by 4 to get: 15G - 20 = 16G - 32
Solve to get: G = 12
Answer: E

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by MartyMurray » Mon Nov 14, 2016 7:08 am
melguy wrote:The ratio of boys to girls in Class A is 3 to 4. The ratio of boys to girls in Class B is 4 to 5. If the two classes were combined, the ratio of boys to girls in the combined class would be 17 to 22. If Class A has one more boy and two more girls than class B, how many girls are in Class A?

8
9
10
11
12
From the answer choices we know that the number of girls in class A is 12 or less. We also know that there are 2 more girls in A than in B. So the number of girls in B is 10 or less.

So the answer choices indicate that maximum number of girls total is 12 + 10 = 22.

17 is prime. So there are no integers - numbers of boys and girls are always integers - less than 17 and 22 that would be in the ratio 17:22. Therefore the total number of girls in the two classes has to be at least 22.

The maximum according to the answer choices is 22 girls. The minimum according the question is 22.

The total number of girls must therefore be 22, making the maximum answer choice, 12, the correct one.

The correct answer is E.
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by [email protected] » Mon Nov 14, 2016 1:33 pm
Hi melguy,

You made the proper deduction that the number of girls in Class A had to be a multiple of 4 - meaning that either '8' or '12' would be the correct answer to this question. Thus, you clearly understand that 'multiples' is the key concept behind this question. If you just did a little more brute-force work, then you probably could have gotten the correct answer (in the time that you originally had). Here's how:

The number of girls in Class A is a multiple of 4 and the number of girls in Class B is a multiple of 5. When COMBINED, the total is a multiple of 22. So...

Can you quickly find a multiple of 5 and add it to '8' to get a multiple of 22?
Can you quickly find a multiple of 5 and add it to '12' to get a multiple of 22?

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by crackverbal » Wed Nov 16, 2016 1:35 am
Hi Melguy,

One good way of answering this question is to use the answer options i.e. eliminate B, C and D and back solve using A and E. The easiest approach to solve this question algebraically is where we do not use two variables B and G and do not use the the ratio 17 : 22 of the combined class since this ratio is just redundant information.

The ratio of boys to girls in class A is 3 : 4. So let the number of boys in class A be 3x and the number of girls in class A be 4x.

Since the number of boys in class B is one less than the number of boys in class A, the number of boys in class B will be 3x - 1.

Since the number of girls in class B is two less than the number of girls in class A, the number of girls in class B will be 4x - 2.

We are given the ratio of boys to girls in class B as 4 : 5. So 3x - 1 : 4x - 2 = 4 : 5
Solving we get x = 3. Now since the number of girls in class A is given by 4x, the number of girls in class A = 4 * 3 = 12

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by Matt@VeritasPrep » Fri Nov 25, 2016 3:50 pm
If I were really pressed for time, I'd try to find numbers that work, going from the most unusual numbers in the stem. (This is the Lazy Test Writer Principle: if a ratio such as, I don't know, 17 : 22 appears, it's because we've probably got exactly 17 of one and exactly 22 of the other.)

Since the strangest ratio is 17 : 22, I'd start there. Let's say that we have 17 boys and 22 girls in the combined class.

From the last sentence, that gives us 9 boys and 12 girls in A, and 8 boys and 10 girls in B. Hey, what a "coincidence": those perfectly match the ratios in the first two sentences!

Now we're done, and it didn't even take that long.

If that HADN'T worked, I would've taken 17 : 22 to be exactly 34 boys and 44 girls and tried those numbers, etc. By the strong form of the Lazy Test Writer Principle -- the calculations required should never be very great -- it shouldn't take long to find the set that works.
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