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School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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16 Nov 2013, 21:53
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School A is 40% girls and school B is 60% girls. The ratio of the number of girls at school A to the number of girls at school B is 4:3. if 20 boys transferred from school A to school B and no other changes took place at the two schools, the new ratio of the number of boys at school A to the number of boys at school B would be 5:3. What would the difference between the number of boys at school A and at school B be after the transfer? A) 20 B) 40 C) 60 D) 80 E) 100
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Re: School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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Assume the common variable for the ratio to be x. As ratio of girls of school A (Ga) to girls of school B (Gb) is 4:3. Therefore, Ga/Gb = 4/3 i.e. there are Ga=4x, Gb=3x Now, as school A has 40% girls, Ga=40/100 * (Total students at school A) Substuting Ga=4x. Total students at school A = 10x Similarly, Total students at school B = 5x From this, we know that boys at school A (Ba) = 6x boys at school B (Bb) = 2x As 20 boys leave school A and join school B new Ba = 6x  20 new Bb = 2x + 20 It is also given that, (new Ba)/(new Bb) = 5/3 i.e (6x  20)/(2x + 20)=5/3 x=20 Hence difference between number of boys after transfer, [6(20)20]  2(20)+20] = 40 Ans : B Hope this helps



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Re: School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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School A...................................................................... School B Boys.............. Girls.......................................Boys................ Girls \(\frac{60a}{100}............... \frac{40a}{100} ........................................... \frac{40b}{100} .................... \frac{60b}{100}\) The ratio of the number of girls at school A to the number of girls at school B is 4:3 \(\frac{40a}{100} / \frac{60b}{100} = \frac{4}{3}\) a = 2b 20 boys transferred from school A to school B \(\frac{120b}{100}  20 = \frac{6b100}{5}\) ............ (1) \(\frac{40b}{100} + 20 = \frac{2b+100}{5}\) ............. (2) Ratio (1) : (2) = 5/3 \(\frac{6b100}{2b+100} = \frac{5}{3}\) Solving b = 100 We require to find \(\frac{6b100}{5}  \frac{2b+100}{5}\) = 100  60 = 40 Answer = B
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Re: School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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registerincog wrote: School A is 40% girls and school B is 60% girls. The ratio of the number of girls at school A to the number of girls at school B is 4:3. if 20 boys transferred from school A to school B and no other changes took place at the two schools, the new ratio of the number of boys at school A to the number of boys at school B would be 5:3. What would the difference between the number of boys at school A and at school B be after the transfer?
A) 20 B) 40 C) 60 D) 80 E) 100 Sometimes, the question becomes very easy with a little bit of reasoning. People often ask  how do we go from Q48 to Q51? The difference often lies in utilization of this reasoning. Let me tell you my thought process when I saw this question. School A  40% girls School B  60% girls "The ratio of the number of girls at school A to the number of girls at school B is 4:3" When I read this line, I went back to the previous line with the % figures. I noticed that school A has more girls than school B (4:3) but its % of girls is lower (only 40% compared to 60% in B). This means that school A has more students than school B. Say something like school A has 200 students while school B has 100. So school A has 80 girls while school B has 60 girls. This gave me a ratio of 4:3. (If I had not got 4:3 on my first try, I might have tweaked the assumed numbers a bit but would have stuck to simple numbers.) Then I verified the rest of the data against these numbers and got my answer easily as 40. Algebra Solution Say, if you do insist on algebra and do not want to rely on reasoning, try to work in a single variable. Girls in A:Girls in B = 4:3 Girls in A = 4n, Girls in B = 3n Since in A, girls are 40% and boys are 60%, number of boys is 6n. Since in B, girls are 60% and boys are 40%, number of boys is 2n. (6n  20)/(2n + 20) = 5/3 You get n = 20 Boys at school A after transfer = 6*20  20 = 100 Boys are school B after transfer = 2*20 + 20 = 60 Difference = 40 Answer (B)
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Re: School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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VeritasPrepKarishma wrote: registerincog wrote: School A is 40% girls and school B is 60% girls. The ratio of the number of girls at school A to the number of girls at school B is 4:3. if 20 boys transferred from school A to school B and no other changes took place at the two schools, the new ratio of the number of boys at school A to the number of boys at school B would be 5:3. What would the difference between the number of boys at school A and at school B be after the transfer?
A) 20 B) 40 C) 60 D) 80 E) 100 Sometimes, the question becomes very easy with a little bit of reasoning. People often ask  how do we go from Q48 to Q51? The difference often lies in utilization of this reasoning. Let me tell you my thought process when I saw this question. School A  40% girls School B  60% girls "The ratio of the number of girls at school A to the number of girls at school B is 4:3" When I read this line, I went back to the previous line with the % figures. I noticed that school A has more girls than school B (4:3) but its % of girls is lower (only 40% compared to 60% in B). This means that school A has more students than school B. Say something like school A has 200 students while school B has 100. So school A has 80 girls while school B has 60 girls. This gave me a ratio of 4:3. (If I had not got 4:3 on my first try, I might have tweaked the assumed numbers a bit but would have stuck to simple numbers.) Then I verified the rest of the data against these numbers and got my answer easily as 40. Algebra Solution Say, if you do insist on algebra and do not want to rely on reasoning, try to work in a single variable. Girls in A:Girls in B = 4:3 Girls in A = 4n, Girls in B = 3n Since in A, girls are 40% and boys are 60%, number of boys is 6n. Since in B, girls are 60% and boys are 40%, number of boys is 2n. (6n  20)/(2n + 20) = 5/3 You get n = 20 Boys at school A after transfer = 6*20  20 = 100 Boys are school B after transfer = 2*20 + 20 = 60 Difference = 40 Answer (B) Great explanation, Karishma!! I mistakenly assumed that both the schools have same number of students. After realizing that I am on the wrong path, changed my approach and finally got the answer after 4.28 mins. School A is 40% girls (means 60% boys) and school B is 60% girls (means 40% boys). The ratio of the number of girls at school A to the number of girls at school B is 4:3 > Number of girls in School A is 4x and in School B, 3x School A statsNumber of girls is 4x and percentage of girls is 40%. We also know that percentage of boys is 60%. So 40% is equivalent to 4x, then 60% is equivalent to 6x Number of girls = 4x Number of boys = 6x School B statsNumber of girls is 3x and percentage of girls is 60%. We also know that percentage of boys is 40%. So 60% is equivalent to 3x, then 40% is equivalent to 2x Number of girls = 3x Number of boys = 2x SO Ratio of boys in School A and School B is 6x/2x After transferring 20 Boys from School A to School B, ratio of boys would be 5/3 > \(\frac{6x  20}{2x + 20} = \frac{5}{3}\) > x = 20 Number of boys in School A after transfer > 6x  20 >100 Number of boys in School B after transfer > 2x + 20 >60 Difference = 40
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Re: School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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07 Jul 2014, 00:02
I am getting number of students in A as 200 so my answer reduces to 60 instead of 40. My approach is as following :
.4A / .6B = 4/3 hence A = 2B where A and B are total number of students in A and B school respectively. .6A  20 / .4B + 20 = 5/3 the new ratio ... when I solve this equation I left with A= 200.
200  100  40 = 60
Can somebody please identify issue with my solution. Thanks



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Re: School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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07 Jul 2014, 04:03
GmatDestroyer2013 wrote: I am getting number of students in A as 200 so my answer reduces to 60 instead of 40. My approach is as following :
.4A / .6B = 4/3 hence A = 2B where A and B are total number of students in A and B school respectively. .6A  20 / .4B + 20 = 5/3 the new ratio ... when I solve this equation I left with A= 200.
Till here, everything is fine. A = 200 and B = 100 (since A = 2B) Number of boys in A = .6A = 120 Number of boys in B = .4B = 40 When 20 are transferred, number of boys in A = 100 and number of boys in B is 60. Difference = 100  60 = 40
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School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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05 May 2015, 23:20
VeritasPrepKarishma wrote: registerincog wrote: School A is 40% girls and school B is 60% girls. The ratio of the number of girls at school A to the number of girls at school B is 4:3. if 20 boys transferred from school A to school B and no other changes took place at the two schools, the new ratio of the number of boys at school A to the number of boys at school B would be 5:3. What would the difference between the number of boys at school A and at school B be after the transfer?
A) 20 B) 40 C) 60 D) 80 E) 100 Sometimes, the question becomes very easy with a little bit of reasoning. People often ask  how do we go from Q48 to Q51? The difference often lies in utilization of this reasoning. Let me tell you my thought process when I saw this question. School A  40% girls School B  60% girls "The ratio of the number of girls at school A to the number of girls at school B is 4:3" When I read this line, I went back to the previous line with the % figures. I noticed that school A has more girls than school B (4:3) but its % of girls is lower (only 40% compared to 60% in B). This means that school A has more students than school B. Say something like school A has 200 students while school B has 100. So school A has 80 girls while school B has 60 girls. This gave me a ratio of 4:3. (If I had not got 4:3 on my first try, I might have tweaked the assumed numbers a bit but would have stuck to simple numbers.) Then I verified the rest of the data against these numbers and got my answer easily as 40. Algebra Solution Say, if you do insist on algebra and do not want to rely on reasoning, try to work in a single variable. Girls in A:Girls in B = 4:3 Girls in A = 4n, Girls in B = 3n Since in A, girls are 40% and boys are 60%, number of boys is 6n. Since in B, girls are 60% and boys are 40%, number of boys is 2n. (6n  20)/(2n + 20) = 5/3 You get n = 20 Boys at school A after transfer = 6*20  20 = 100 Boys are school B after transfer = 2*20 + 20 = 60 Difference = 40 Answer (B) Hi karishma, Thanks for your explanation. Even I started of with the reasoning and using smart numbers. I see the number of students that you initially took were 200 and 100. I started of with 100 and 50 which did not work. So in such cases should one go for the next set of smart numbers or should switch to the equation method??



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Re: School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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06 May 2015, 00:27
believer700 wrote: VeritasPrepKarishma wrote: registerincog wrote: School A is 40% girls and school B is 60% girls. The ratio of the number of girls at school A to the number of girls at school B is 4:3. if 20 boys transferred from school A to school B and no other changes took place at the two schools, the new ratio of the number of boys at school A to the number of boys at school B would be 5:3. What would the difference between the number of boys at school A and at school B be after the transfer?
A) 20 B) 40 C) 60 D) 80 E) 100 Sometimes, the question becomes very easy with a little bit of reasoning. People often ask  how do we go from Q48 to Q51? The difference often lies in utilization of this reasoning. Let me tell you my thought process when I saw this question. School A  40% girls School B  60% girls "The ratio of the number of girls at school A to the number of girls at school B is 4:3" When I read this line, I went back to the previous line with the % figures. I noticed that school A has more girls than school B (4:3) but its % of girls is lower (only 40% compared to 60% in B). This means that school A has more students than school B. Say something like school A has 200 students while school B has 100. So school A has 80 girls while school B has 60 girls. This gave me a ratio of 4:3. (If I had not got 4:3 on my first try, I might have tweaked the assumed numbers a bit but would have stuck to simple numbers.) Then I verified the rest of the data against these numbers and got my answer easily as 40. Algebra Solution Say, if you do insist on algebra and do not want to rely on reasoning, try to work in a single variable. Girls in A:Girls in B = 4:3 Girls in A = 4n, Girls in B = 3n Since in A, girls are 40% and boys are 60%, number of boys is 6n. Since in B, girls are 60% and boys are 40%, number of boys is 2n. (6n  20)/(2n + 20) = 5/3 You get n = 20 Boys at school A after transfer = 6*20  20 = 100 Boys are school B after transfer = 2*20 + 20 = 60 Difference = 40 Answer (B) Hi karishma, Thanks for your explanation. Even I started of with the reasoning and using smart numbers. I see the number of students that you initially took were 200 and 100. I started of with 100 and 50 which did not work. So in such cases should one go for the next set of smart numbers or should switch to the equation method?? During practice, always start with the more intuitive method. Stick to it for a while. Even if you spend a few mins and are unable to get the answer, you can later switch to algebra. With practice, these instances of "not getting the answer" will reduce. You will automatically zero in on the right numbers with some tweaking (for example, is your answer greater than or smaller than what is required? Adjust the assumed numbers accordingly etc). In the actual exam, know where you stand. If, during recent practice, you were able to get most answers without algebra, then go ahead and use the intuitive methods. Even if you have to rely on algebra in some questions after wasting 23 mins on it, you will have plenty and more extra time. If you were almost always forced to resort to algebra during practice, then might as well stick with algebra in the actual exam.
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Re: School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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23 Feb 2016, 07:08
Looking at the answers they are in multiples of 20.
So I will use substitution on this one:
Girls School A to School B 4:3 Multiply y 20 School A= 80 girls School B = 60 girls
From the difference of percentages we can calculate the number of boys School A: 60% is 120 School B: 40% is 40
So if 20 are transferred from A to B then we will have 100 in School A and 60 in School B.
Difference is 10060=40. Choose Answer C



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Re: School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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22 Aug 2016, 04:12
VeritasPrepKarishma wrote: registerincog wrote: School A is 40% girls and school B is 60% girls. The ratio of the number of girls at school A to the number of girls at school B is 4:3. if 20 boys transferred from school A to school B and no other changes took place at the two schools, the new ratio of the number of boys at school A to the number of boys at school B would be 5:3. What would the difference between the number of boys at school A and at school B be after the transfer?
A) 20 B) 40 C) 60 D) 80 E) 100 Algebra Solution Say, if you do insist on algebra and do not want to rely on reasoning, try to work in a single variable. Girls in A:Girls in B = 4:3 Girls in A = 4n, Girls in B = 3n Since in A, girls are 40% and boys are 60%, number of boys is 6n. Since in B, girls are 60% and boys are 40%, number of boys is 2n. (6n  20)/(2n + 20) = 5/3 You get n = 20 Boys at school A after transfer = 6*20  20 = 100 Boys are school B after transfer = 2*20 + 20 = 60 Difference = 40 Answer (B) Responding to a pm: Quote: Can you please explain this part of your solution: Since in A, girls are 40% and boys are 60%, number of boys is 6n. Since in B, girls are 60% and boys are 40%, number of boys is 2n
Why is it that you didn't represent boys in school b as 4n? You simplified it 60/40 to get 3/2. However, in A you left the boys as 6n, you did not reduce the ratio as you did in B.
Your help would be greatly appreeciated.
Girls in A:Girls in B = 4:3 Assume actual numbers with a variable: Girls in A = 4n, Girls in B = 3n Since in school A, girls are 40% and boys are 60%, 40% of Total students in A = 4n Total students in A = 4n *100/40 = 10n So number of boys in school A = 10n  4n = 6n Using the same logic for school B, since in school B, girls are 60%, 60% of Total students in B = 3n Total students in = 3n * 100/60 = 5n So the number of boys in school B = 5n  3n = 2n I hope you understand from where I got 6n and 2n. I did this so that a single variable represents all the relevant quantities.
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Re: School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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09 Sep 2016, 10:30
Hi All,
Why can't we go backwards in such type of questions where two ratios are given and final answer really depends on the final ratio.
Let us start with taking final Boys ratio of 5:3, for school A & B respectively. We know that no. of students in school A is significantly greater than B, so, we can take final numbers of boys in A & B to be 100 & 60 (5:3), also 100 sounds better and is easy for calculation purposes.
now Boys(A original)= 100 +20= 120 & Boys (B original)= 6020=40). These new numbers of 120 & 40 are original percentage of boys in schools A & B respectively (60% & 40%). So, the number of girls in school A & B are 80 & 60 respectively (40% & 60%), which matches with the original ratio 4:3.
I prefer "picking numbers method" in such question more useful as it will eliminate any silly mistakes we may commit while solving algebraic equations and eventually get stuck in getting/marking the right answer.
Experts let me know, if you also suggest to pick numbers in such question types.
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Re: School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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23 Oct 2017, 01:09
registerincog wrote: School A is 40% girls and school B is 60% girls. The ratio of the number of girls at school A to the number of girls at school B is 4:3. if 20 boys transferred from school A to school B and no other changes took place at the two schools, the new ratio of the number of boys at school A to the number of boys at school B would be 5:3. What would the difference between the number of boys at school A and at school B be after the transfer?
A) 20 B) 40 C) 60 D) 80 E) 100 A : Girls = 0.4 A, Boys = 0.6 A B: Girls = 0.6 B, Boys = 0.4 B Now, 0.4A/0.6B = 4/3 > A = 2B So, A= 2B : Girls = 0.8B, Boys = 1.2B B: Girls = 0.6 B, Boys = 0.4 B After transfer of 20 boys from A to B A= 2B : Girls = 0.8B, Boys = 1.2B20 B: Girls = 0.6 B, Boys = 0.4 B+20 (1.2B20)/(0.4B+20) = 5/3 > B = 100 Difference in no. of boys between Class A and B after transfer = (1.2B20)(0.4B+20) = 0.8B 40 = 0.8*100  40 = 40 Answer B
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Re: School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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23 Oct 2017, 02:04
Yash26 wrote: Hi All,
Why can't we go backwards in such type of questions where two ratios are given and final answer really depends on the final ratio.
Let us start with taking final Boys ratio of 5:3, for school A & B respectively. We know that no. of students in school A is significantly greater than B, so, we can take final numbers of boys in A & B to be 100 & 60 (5:3), also 100 sounds better and is easy for calculation purposes.
now Boys(A original)= 100 +20= 120 & Boys (B original)= 6020=40). These new numbers of 120 & 40 are original percentage of boys in schools A & B respectively (60% & 40%). So, the number of girls in school A & B are 80 & 60 respectively (40% & 60%), which matches with the original ratio 4:3.
I prefer "picking numbers method" in such question more useful as it will eliminate any silly mistakes we may commit while solving algebraic equations and eventually get stuck in getting/marking the right answer.
Experts let me know, if you also suggest to pick numbers in such question types.
Regards Yash Yes back solving also works most of the times, unless data is very complex.. For GMAT problems is a very good way to solve the problems.
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Re: School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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26 Oct 2017, 14:27
registerincog wrote: School A is 40% girls and school B is 60% girls. The ratio of the number of girls at school A to the number of girls at school B is 4:3. if 20 boys transferred from school A to school B and no other changes took place at the two schools, the new ratio of the number of boys at school A to the number of boys at school B would be 5:3. What would the difference between the number of boys at school A and at school B be after the transfer?
A) 20 B) 40 C) 60 D) 80 E) 100 If School A has 40% girls, it also has 60% boys, giving us a ratio of 4 : 6. We can let the number of girls and boys in School A be 4x and 6x, respectively. Similarly, we can let the number of girls and boys in School B be 6y and 4y, respectively. Therefore, we can express the ratio of girls at School A to girls at School B as 4x/6y, and we are given that this ratio is also equal to 4 : 3. Thus: 4x/6y = 4/3 For the boys, School A lost 20 boys, giving us (6x  20) boys at School A, and School B gained those 20 boys, so School B now has (4y + 20) boys. The new ratio is given as 5 : 3. Thus: (6x  20)/(4y + 20) = 5/3 Solving the first equation, we have: 12x = 24y x = 2y Solving the second equation, we have: 18x  60 = 20y + 100 18x = 20y + 160 9x = 10y + 80 Since x = 2y, we have: 9(2y) = 10y + 80 18y = 10y + 80 8y = 80 y = 10 Thus, x = 2(10) = 20. After the transfer of 20 boys, School A has 6x  20 = 6(20)  20 = 100 boys and School B has 4(10) + 20 = 60 boys. So, the difference is 100  60 = 40. Answer: B
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Re: School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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11 Feb 2018, 12:34
Hi All, This is a layered algebra question that would take a number of "math steps" to solve. The math involved isn't too hard, but you have to do a lot of work to get the job done…. "School A is 40% girls (thus 60% boys) and School B is 60% girls (thus 40% boys)" We can write these ratios as: School A B:G 3:2 School B B:G 2:3 Since ratios are all about multiples, I'm going to add a "variable" into each ratio (since the schools are different, I need to use a different variable for each). School A B:G 3x:2x School B B:G 2y:3y "The ratio of girls at School A to girls at School B is 4:3" We can write this ratio as… 2x/3y = 4/3 And crossmultiply…. 6x = 12y x = 2y "If 20 boys transferred from School A to School B…the new ratio of boys at School A to boys at School B would be 5:3" We can write this ratio as…. (3x  20)/(2y + 20) = 5/3 And crossmultiply… 9x  60 = 10y + 100 9x = 10y + 160 We now have 2 variables and 2 equations, so we can solve using "System" math to solve for the two variables… x = 2y 9x = 10y + 160 By substituting in, we get… 9(2y) = 10y + 160 18y = 10y + 160 8y = 160 y = 20 Plugging back in, we get… x = 40 The question asks for the DIFFERENCE in the number of boys at School A and at School B AFTER the transfer: Number at School A after the transfer = 3x  20 = 12020 = 100 Number at School B after the transfer = 2y + 20 = 40 + 20 = 60 The difference = 100  60 = 40 Final Answer: GMAT assassins aren't born, they're made, Rich
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School A is 40% girls and school B is 60% girls. The ratio [#permalink]
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11 Feb 2018, 15:45
registerincog wrote: School A is 40% girls and school B is 60% girls. The ratio of the number of girls at school A to the number of girls at school B is 4:3. if 20 boys transferred from school A to school B and no other changes took place at the two schools, the new ratio of the number of boys at school A to the number of boys at school B would be 5:3. What would the difference between the number of boys at school A and at school B be after the transfer?
A) 20 B) 40 C) 60 D) 80 E) 100 let a and b=total students at A and B respectively .4a/.6b=4/3 b=a/2 substituting, (.6a20)/(.4a/2+20)=5/3 a=200; b=100 .6*200=120 boys originally at A 12020=100 boys remaining at A .4*100=40 boys originally at B 40+20=60 boys currently at B 10060=40 boys difference between A and B B




School A is 40% girls and school B is 60% girls. The ratio
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11 Feb 2018, 15:45






