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The ratio of boys to girls in Class A is 3 to 4. The ratio [#permalink]
03 Apr 2010, 11:57

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Question Stats:

70% (02:00) correct
30% (01:54) wrong based on 203 sessions

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?

Re: Help with a ratio question. [#permalink]
03 Apr 2010, 12:42

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IMO E -12

Let for Class A boys = x and girls = y

then for class B boys = x-1 and girls = y-1

x/y = 3/4 and (x-1)/(y-2) = 4/5

put the value of x in second equation we get y =12

another way.. since x = 3/4 * y that means y must be multiple of 4 ( because x must be an integer) so the only possibilities are 8 and 12...for y = 8 and 12 we get x as 6 and 9 substitute in 2nd equation to satisfy..... (x-1)/(y-2) = (6-1)/(8-2) is not equal to 4/5 thus ans is 12. _________________

Re: Help with a ratio question. [#permalink]
10 Apr 2010, 03:24

good question... first I got stuck with big equation ...then applied the short used by gurpreet also...once you figure out it has to be between 8 and 12, it becomes easier.

Re: Help with a ratio question. [#permalink]
09 Oct 2010, 18:04

I understand gurpreetsingh's algebraic explanation, but does that mean that the 17/22 ratio is superfluous? I'm not used to seeing extra info on GMAT questions.

Re: Help with a ratio question. [#permalink]
28 Mar 2012, 06:10

Hey Bunuel, While giving test, I have been able to solve it correctly using back solving. However, when I revisited this question, I was unable to solve it at the first go. Request you to help me out..

Let the multiplier be x for Class A . Therefore, number of boys: 3x and number of girls 4x Let the multiplier be y for Class B. Therefor, number of boys: 4y and number of girls : 5y It is given that - 3x= 4y+1 ----(A) and 4x=5y+2 -----(B) Also As per question:- 3x+4y/4x+5y = 17/22 (combined boys: combined girls) ---(C) Putting the value from A and B in Equation C giving me weird results - y is coming out to be 6/13

I have also gone through the replies posted above and I am not too sure that I would be able to ignore the superfluous statement in the real time exam... so hoping to see the solution utilizing the info. Thanks H _________________

Re: Help with a ratio question. [#permalink]
28 Mar 2012, 09:09

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This post received KUDOS

Expert's post

imhimanshu wrote:

Hey Bunuel, While giving test, I have been able to solve it correctly using back solving. However, when I revisited this question, I was unable to solve it at the first go. Request you to help me out..

Let the multiplier be x for Class A . Therefore, number of boys: 3x and number of girls 4x Let the multiplier be y for Class B. Therefor, number of boys: 4y and number of girls : 5y It is given that - 3x= 4y+1 ----(A) and 4x=5y+2 -----(B) Also As per question:- 3x+4y/4x+5y = 17/22 (combined boys: combined girls) ---(C) Putting the value from A and B in Equation C giving me weird results - y is coming out to be 6/13

I have also gone through the replies posted above and I am not too sure that I would be able to ignore the superfluous statement in the real time exam... so hoping to see the solution utilizing the info. Thanks H

This problem is a rare example of a question which provides redundant information: we CAN solve the question without knowing that the ratio of boys to girls in the combined class is 17 to 22.

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? A. 8 B. 9 C. 10 D. 11 E. 12

The ration of boys to girls in Class A is 3 to 4: # of boys is 3x and # of girls is 4x, for some positive integer multiple x; The ration of boys to girls in Class B is 4 to 5: # of boys is 4y and # of girls is 5y, for some positive integer multiple y;

Class A has one more boy and two more girls than class B: 3x=4y+1 and 4x=5y+2. Now, we have the system of two distinct linear equations with two unknowns, which means that we can solve it. Solving for x we get x=3. Since # of girls in Class A is 4x the there are 4*3=12 girls.

Answer: E.

So, as you can see you've done everything right, you just should have solved the system of equations rather than applying some kind of substitution.

Re: Help with a ratio question. [#permalink]
29 Mar 2012, 00:21

1

This post received KUDOS

Expert's post

imhimanshu wrote:

Hey Bunuel, While giving test, I have been able to solve it correctly using back solving. However, when I revisited this question, I was unable to solve it at the first go. Request you to help me out..

Let the multiplier be x for Class A . Therefore, number of boys: 3x and number of girls 4x Let the multiplier be y for Class B. Therefor, number of boys: 4y and number of girls : 5y It is given that - 3x= 4y+1 ----(A) and 4x=5y+2 -----(B) Also As per question:- 3x+4y/4x+5y = 17/22 (combined boys: combined girls) ---(C) Putting the value from A and B in Equation C giving me weird results - y is coming out to be 6/13

I have also gone through the replies posted above and I am not too sure that I would be able to ignore the superfluous statement in the real time exam... so hoping to see the solution utilizing the info. Thanks H

Also, if you do use this equation, 3x+4y/4x+5y = 17/22, when you solve it, you get x = 3y/2. Put x = 3y/2 is one of the equations above, say 3x= 4y+1. 3*3y/2 = 4y + 1 y = 2 (you probably have a calculation error somewhere) So x = 3 Number of boys and girls in Class A is 3*3 and 4*3.

Mind you, since you have already got two distinct equations (3x= 4y+1 and 4x=5y+2) with two variables, you can easily solve them to get the values of x and y. This question lets you make 3 equations but there are only 2 variables so you can use any 2 of those equations to get the value of the variables.

Actually speaking, you should not be making any equations. You have the ratio 3:4 and 4:5 You know that the number of boys is class A is 1 more so try and multiply the ratios to figure out where you get a difference of 1. I should multiply 3:4 by a bigger number since the number of boys is more in class A.

(3:4)*3 = 9:12 (4:5)*2 = 8:10

The numbers are very simple so you can arrive at the answer very quickly. _________________

Re: The ratio of boys to girls in Class A is 3 to 4. The ratio [#permalink]
19 Apr 2012, 04:08

I will just plug in answers to arrive at decision.

I start with 10. If 10 girls in class A, the boys have to be some integer. With ratio of 3:4, it is not possible to have integer value.

I take 12 as girls, the boys will be 9. This is the answer but to ensure if I did correctly or not, I just add additional information to check ratio of class B. If I reduce 1 boy from 9, I have 8 boys in class B and 10 girls after reducing 2 girls from 12 girls. Thus the ratio of class B is coming out to be 8:10 or 4:5. This is given in the question stem.

Re: The ratio of boys to girls in Class A is 3 to 4. The ratio [#permalink]
13 May 2012, 04:12

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This post received KUDOS

changhiskhan 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?

A. 8 B. 9 C. 10 D. 11 E. 12

Since it is given that the ratio of boys to girls in Class A is 3 to 4 - so we can quickly eliminate the answer choices which are not multiples of 4.

so ans can either be 8 0r 12.

With 8 as answer - no of girls = 8 With 12 as answer - no of girls = 12

Now Class A has one more boy and two more girls than class B and also the ratio of boys to girls in Class B is 4 to 5.

So we are looking for number of girls in B to be divisible by 5. With 8 as answer -> Girls = 8-2 = 6. (Not possible) With 12 as answer -> Girls = 12-2 = 5.

Re: The ratio of boys to girls in Class A is 3 to 4. The ratio [#permalink]
17 May 2012, 03:01

I picked E after about 2:34, but I didn't finish working any algebra I started. As I was getting to about 2 minutes on this problem, I decided to focus just on the 3:4 ratio for class A. Since the number of Boys and Girls each have to be integers, I realized only A or E were possible solutions. At that point, given the time I had spent already, I settled for the 50/50 odds and picked E.

If I had taken that approach immediately, it would have taken only a short amount of time to further test the solutions for fitting into the combined ratio. But I figure if this question came up at the end when I was running out of time to finish, that I could live with the 50/50 chance. _________________

Re: The ratio of boys to girls in Class A is 3 to 4. The ratio [#permalink]
22 Aug 2012, 10:08

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This post received KUDOS

total # of girls = 22k (22, 44, 66 ..) if A has 2 more girls then # of girls in (a,b) = (12,10), (21,23) .. at this point, you know that it can only be 12 because 21 is not an option. _________________

Re: The ratio of boys to girls in Class A is 3 to 4. The ratio [#permalink]
24 Aug 2012, 11:13

changhiskhan 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?

A. 8 B. 9 C. 10 D. 11 E. 12

The information the ratio of boys to girls in the combined class would be 17 to 22 is superfluous. Too much info just to try to confuse you. If one counts the unknowns, we have 4: number of boys in class A, girls in A, boys in B and girls in B. Without the info of the combined rate, we already have in fact four equations: 1) the ratio in class A 2) the ratio in class B 3) one more boy in class A in comparison with class B 4) two more girls in class A in comparison with class B So, we don't need a fifth equation.

If we denote by B and G the number of boys and girls respectively in class A, we can write the following set of equations: B/G=3/4 (B-1)/(G-2)=4/5 Two equations, two unknowns, solve and get G=12.

Or, go with divisibility criteria, G must be a multiple of 4, so choose between A and E. For A, B=6, G=8 but B-1=5, G-2=6, so 5/6\neq4/5. For E, B=9, G=12 and B-1=8, G-2=10, so 8/10=4/5, OK. Also the additional condition holds, as (9+8)/(12+10)=17/22. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: The ratio of boys to girls in Class A is 3 to 4. The ratio [#permalink]
03 Jan 2014, 05:27

changhiskhan 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?

A. 8 B. 9 C. 10 D. 11 E. 12

Quick way to solve

Class A B/G =3X/4X

Class B B/G=4Y/5Y

Also 3X = 4Y+1 (2) 4X = 5Y+2

We need to find 4x

So we also get that

3x + 4y = 17 (1) 4x + 5y = 22

Now, those are quite a number of equations, but let's make the work easy ok?