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The only people in each of rooms A and B are students, and [#permalink]
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06 Dec 2009, 01:15
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The only people in each of rooms A and B are students, and each student in each of rooms A and B is either a junior or a senior. The ratio of the number of juniors to the number of seniors in room A is 4 to 5, the ratio of the number of juniors to the number of seniors in room B is 3 to 17, and the ratio of the total number of juniors in both rooms A and B to the total number of seniors in both rooms A and B is 5 to 7. What is the ratio of the total number of students in room A to the total number of students in room B ? A. 29/12 B. 59/10 C. 65/8 D. 48/5 E. 29/3
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Re: Kaplan CAT  Ratio problem [#permalink]
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06 Dec 2009, 14:22
Fremontian wrote: The only people in each of rooms A and B are students, and each student in each of rooms A and B is either a junior or a senior. The ratio of the number of juniors to the number of seniors in room A is 4 to 5, the ratio of the number of juniors to the number of seniors in room B is 3 to 17, and the ratio of the total number of juniors in both rooms A and B to the total number of seniors in both rooms A and B is 5 to 7. What is the ratio of the total number of students in room A to the total number of students in room B ? A. 29/12 B. 59/10 C. 65/8 D. 48/5 E. 29/3 Just curious  Is this solvable in < 2mins? We have been given the following information : In room A, ratio of juniors to seniors = 4 : 5 In room B, ratio of juniors to seniors = 3 : 17 Total ratio of juniors to seniors = 5 : 7
Now, from this information we can infer the following : In room A, the if the number of juniors are '4x' then the number of seniors will be '5x'. Also, the total number of students in room A will be 4x + 5x = 9x. Similarly, in room B, if the number of juniors are '3y' then the number of seniors will be '17y' and the total number of students in room B will be 3y + 17y = 20y. And finally, if the total number of juniors are '5z' then the total number of seniors will be '7z' Note : We use different variables (x, y and z) in each case because the value of the multiplying factor can be different in each case. Now, with this information, we can form the following equations : Equation 1 : 4x + 3y = 5z That is, the number of juniors in room A plus the number of juniors in room B = Total number of juniors. Equation 2 : 5x + 17y = 7z That is, the number of seniors in room A plus the number of seniors in room B = Total number of seniors. Now, we can solve these two equations and get the values of 'x' and 'y' in terms of 'z'. Multiplying Eq.(1) by 5 and Eq.(2) by 4 and then subtracting Eq.(1) from Eq.(2), we get :
\(68y  15y = 28z  25z\) OR \(y = \frac{3z}{53}\) Substituting this value of 'y' in Eq.(1) we get : \(x = \frac{64z}{53}\) Now, the ratio of students in room A to students in room B = \(\frac{9x}{20y}\) Substituting values of 'x' and 'y' in terms of 'z' we get : Ratio of students = \(\frac{9*\frac{64z}{53}}{20*\frac{3z}{53}\) = \(9*\frac{64z}{53}*\frac{53}{3z}*\frac{1}{20}\) = \(\frac{48}{5}\) Answer : D
Ps. The amount of time it takes to solve this problem depends on how long it takes you to figure out the approach. Once you know how to go about it, the calculations should be manageable in about 2 minutes.
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Re: Kaplan CAT  Ratio problem [#permalink]
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07 Dec 2009, 11:55
Fremontian wrote: Just curious  Is this solvable in < 2mins?
sriharimurthy has done a pretty good job of explaining the math, but I wanted to address this part of your question specifically. Remember, the GMAT is adaptive. It's designed to give you questions that stretch the limits of your mathematical ability! That means it's important to bank extra time on the easier questions that you can solve quickly and with confidencespecifically so that, when you hit a highdifficulty question like this one, you have some time in the kitty and can afford to spend an extra thirty seconds or minute on the problem. Don't rigidly force yourself to spend 2 minutes per question, that's just the average you are aiming for.
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Re: Kaplan CAT  Ratio problem [#permalink]
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07 Mar 2010, 01:51
KapTeacherEli wrote: Fremontian wrote: Just curious  Is this solvable in < 2mins?
sriharimurthy has done a pretty good job of explaining the math, but I wanted to address this part of your question specifically. Remember, the GMAT is adaptive. It's designed to give you questions that stretch the limits of your mathematical ability! That means it's important to bank extra time on the easier questions that you can solve quickly and with confidencespecifically so that, when you hit a highdifficulty question like this one, you have some time in the kitty and can afford to spend an extra thirty seconds or minute on the problem. Don't rigidly force yourself to spend 2 minutes per question, that's just the average you are aiming for. Class A has ratio 4/5 that gives Juniors around 44% and seniors around 55% Similarly in class B the ratio has 3/17 tha gives Juniors as 15% and seniors as 85% Now for total ..... the ratio is 5/7 that means total juniors are around 43% and seniors around 57% What i see from here that the ratio of total students is very close to that of class A. That means the percentage of students should be maximum (may be even more then 90% because the ratio in class B is very far away) Referring to choices the answer D only provides an option of more than 90% What do you say....am I going the right way???
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Is this a nice shortcut  Tough question from KAPLAN  [#permalink]
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07 Mar 2010, 01:58
The only people in each of rooms A and B are students, and each student in each of rooms A and B is either a junior or a senior. The ratio of the number of juniors to the number of seniors in room A is 4 to 5, the ratio of the number of juniors to the number of seniors in room B is 3 to 17, and the ratio of the total number of juniors in both rooms A and B to the total number of seniors in both rooms A and B is 5 to 7. What is the ratio of the total number of students in room A to the total number of students in room B ? A. 29/12 B. 59/10 C. 65/8 D. 48/5 E. 29/3 My solution: Class A has ratio 4/5 that gives Juniors around 44% and seniors around 55% Similarly in class B the ratio has 3/17 tha gives Juniors as 15% and seniors as 85% Now for total ..... the ratio is 5/7 that means total juniors are around 43% and seniors around 57% What i see from here that the ratio of total students is very close to that of class A. That means the percentage of students in class A should be maximum (may be even more then 90% because the ratio in class B is very far away) Referring to choices the answer D only provides an option of more than 90% What do you say....am I going the right way??? OA is D
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Re: Is this a nice shortcut  Tough question from KAPLAN  [#permalink]
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07 Mar 2010, 03:06
mustdoit wrote: The only people in each of rooms A and B are students, and each student in each of rooms A and B is either a junior or a senior. The ratio of the number of juniors to the number of seniors in room A is 4 to 5, the ratio of the number of juniors to the number of seniors in room B is 3 to 17, and the ratio of the total number of juniors in both rooms A and B to the total number of seniors in both rooms A and B is 5 to 7. What is the ratio of the total number of students in room A to the total number of students in room B ?
A. 29/12 B. 59/10 C. 65/8 D. 48/5 E. 29/3
My solution:
Class A has ratio 4/5 that gives Juniors around 44% and seniors around 55% Similarly in class B the ratio has 3/17 tha gives Juniors as 15% and seniors as 85%
Now for total ..... the ratio is 5/7 that means total juniors are around 43% and seniors around 57%
What i see from here that the ratio of total students is very close to that of class A.
That means the percentage of students in class A should be maximum (may be even more then 90% because the ratio in class B is very far away)
Referring to choices the answer D only provides an option of more than 90%
What do you say....am I going the right way???
OA is D Given: \(\frac{J_a}{S_a}=\frac{4}{5}\) > \(J_a=4x\) and \(S_a=5x\) \(\frac{J_b}{S_b}=\frac{3}{17}\) > \(J_b=3y\) and \(S_b=17y\) \(\frac{J}{S}=\frac{5}{7}\) > \(\frac{J}{S}=\frac{J_a+J_b}{S_a+S_b}=\frac{4x+3y}{5x+17y}=\frac{5}{7}\) > \(28x+21y=25x+85y\) > \(3x=64y\) > \(\frac{x}{y}=\frac{64}{3}\). Q: \(\frac{J_a+S_a}{J_b+S_b}=?\) > \(\frac{J_a+S_a}{J_b+S_b}=\frac{4x+5x}{3y+17y}=\frac{9x}{20y}=\frac{9}{20}*\frac{64}{3}=\frac{48}{5}\). Answer: D. Now the way you solved: A. 29/12 > A=~71% B. 59/10 > A=~86% C. 65/8 > A=~89% D. 48/5 > A=~91% E. 29/3 > A=~91% How can you decide which one from C, D or E is correct?
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Re: Kaplan CAT  Ratio problem [#permalink]
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07 Mar 2010, 06:54
Fremontian wrote: The only people in each of rooms A and B are students, and each student in each of rooms A and B is either a junior or a senior. The ratio of the number of juniors to the number of seniors in room A is 4 to 5, the ratio of the number of juniors to the number of seniors in room B is 3 to 17, and the ratio of the total number of juniors in both rooms A and B to the total number of seniors in both rooms A and B is 5 to 7. What is the ratio of the total number of students in room A to the total number of students in room B ? A. 29/12 B. 59/10 C. 65/8 D. 48/5 E. 29/3 Just curious  Is this solvable in < 2mins?[/quote] Ratio in A: 4/5 so total students in A = 9x Ratio in B: 3/17 so total students in B = 20y (4x+3y)/(5x+17y) = 5/7 solving, x/y = 64/3 required ratio = 9x/20y = (64*9)/(3*20) = 48/5 hence D. I think it is solvable in 2 mins.



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Re: Is this a nice shortcut  Tough question from KAPLAN  [#permalink]
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10 Mar 2010, 23:08
This approach is much simpler. Thanks. Bunuel wrote: mustdoit wrote: The only people in each of rooms A and B are students, and each student in each of rooms A and B is either a junior or a senior. The ratio of the number of juniors to the number of seniors in room A is 4 to 5, the ratio of the number of juniors to the number of seniors in room B is 3 to 17, and the ratio of the total number of juniors in both rooms A and B to the total number of seniors in both rooms A and B is 5 to 7. What is the ratio of the total number of students in room A to the total number of students in room B ?
A. 29/12 B. 59/10 C. 65/8 D. 48/5 E. 29/3
My solution:
Class A has ratio 4/5 that gives Juniors around 44% and seniors around 55% Similarly in class B the ratio has 3/17 tha gives Juniors as 15% and seniors as 85%
Now for total ..... the ratio is 5/7 that means total juniors are around 43% and seniors around 57%
What i see from here that the ratio of total students is very close to that of class A.
That means the percentage of students in class A should be maximum (may be even more then 90% because the ratio in class B is very far away)
Referring to choices the answer D only provides an option of more than 90%
What do you say....am I going the right way???
OA is D Given: \(\frac{J_a}{S_a}=\frac{4}{5}\) > \(J_a=4x\) and \(S_a=5x\) \(\frac{J_b}{S_b}=\frac{3}{17}\) > \(J_b=3y\) and \(S_b=17y\) \(\frac{J}{S}=\frac{5}{7}\) > \(\frac{J}{S}=\frac{J_a+J_b}{S_a+S_b}=\frac{4x+3y}{5x+17y}=\frac{5}{7}\) > \(28x+21y=25x+85y\) > \(3x=64y\) > \(\frac{x}{y}=\frac{64}{3}\). Q: \(\frac{J_a+S_a}{J_b+S_b}=?\) > \(\frac{J_a+S_a}{J_b+S_b}=\frac{4x+5x}{3y+17y}=\frac{9x}{20y}=\frac{9}{20}*\frac{64}{3}=\frac{48}{5}\). Answer: D. Now the way you solved: A. 29/12 > A=~71% B. 59/10 > A=~86% C. 65/8 > A=~89% D. 48/5 > A=~91% E. 29/3 > A=~91% How can you decide which one from C, D or E is correct?



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Re: Kaplan CAT  Ratio problem [#permalink]
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19 Dec 2011, 04:10
I solved this slightly differently, essentially using a weighted average approach.
Let x be the fraction of total students in room A.
(4/9)(x) + (3/20)(1x)=(5/12). Solving for x results in x = 48/53. Therefore the ratio is 48:5.



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Re: The only people in each of rooms A and B are students, and [#permalink]
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26 Oct 2013, 05:25
HOW TO SOLVE IN LESS THAN 1 MIN
In room A, ratio of juniors to seniors = 4 : 5
Therefore, total number people in Room A  4x + 5x = 9x (where x is the unknown multiplier) Similarly students in room B will be 3y + 17y = 20y (where y is a different unknown multiplier)
we are asked to find the ratio of the total number of students, so Room A's figure must contain a factor of 9 and Room b's figure must contain a factor of 20
Ratio should be in the form of 9x:20y
A. 29/12> NO eliminate B. 59/10> 59 NO but 10 yes we need both to work so eliminate C. 65/8>NO eliminate D. 48/5> Works 48 is divisable by 3 and 5 a factor of 20 E. 29/3>NO eliminate
Only option that works is D



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Re: The only people in each of rooms A and B are students, and [#permalink]
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22 Nov 2013, 05:37
Fremontian wrote: The only people in each of rooms A and B are students, and each student in each of rooms A and B is either a junior or a senior. The ratio of the number of juniors to the number of seniors in room A is 4 to 5, the ratio of the number of juniors to the number of seniors in room B is 3 to 17, and the ratio of the total number of juniors in both rooms A and B to the total number of seniors in both rooms A and B is 5 to 7. What is the ratio of the total number of students in room A to the total number of students in room B ?
A. 29/12 B. 59/10 C. 65/8 D. 48/5 E. 29/3 For A> J/S = 4X/5X For B> J/S = 3Y/17Y For A and B > 4X+3Y/5X+17Y = 5/7 (1) We need to find Total A / Total B = 9x/20y (2) Solving for (1)> 3x = 64y, so x/y = 64/3 Replace in (2) The answer is 48/5. Hence D Cheers! J Kudos if it helps



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Re: The only people in each of rooms A and B are students, and [#permalink]
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22 Feb 2016, 20:07
can't say tough, but pretty time consuming one... suppose in room A, we have: 4x juniors, and 5x seniors. total 9x. suppose in room B, we have: 3y juniors, and 17y seniors. total 20y. we know that: the ratio of 4x+3y to the 5x+17y is 5/7. we then are asked to find the ratio of # students in A to the # of students in B or 9x/20y
so...cross multiply: 7*(4x+3y)=5*(5x+17y) 28x+21y=25x+85y 3x=64y multiply by 3: 9x=192y now, 9x/20y = 192y/20y > 96/10 > 48/5
D



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Re: The only people in each of rooms A and B are students, and [#permalink]
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22 Feb 2016, 22:29
mvictor wrote: can't say tough, but pretty time consuming one... suppose in room A, we have: 4x juniors, and 5x seniors. total 9x. suppose in room B, we have: 3y juniors, and 17y seniors. total 20y. we know that: the ratio of 4x+3y to the 5x+17y is 5/7. we then are asked to find the ratio of # students in A to the # of students in B or 9x/20y
so...cross multiply: 7*(4x+3y)=5*(5x+17y) 28x+21y=25x+85y 3x=64y multiply by 3: 9x=192y now, 9x/20y = 192y/20y > 96/10 > 48/5
D This is just a mixtures question which can be quickly solved with our weighted average formula: Focus on one element, say juniors. Room A has 4/9 juniors out of total. Room B has 3/20 Juniors out of total. Combined has 5/12 juniors out of total. \(\frac{wA}{wB} = \frac{(3/20)  (5/12)}{(5/12)  (4/9)}\) \(\frac{wA}{wB} = \frac{48}{5}\) For more on this formula, check: http://www.veritasprep.com/blog/2011/04 ... mixtures/
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Re: The only people in each of rooms A and B are students, and [#permalink]
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22 Feb 2016, 23:01
Fremontian wrote: The only people in each of rooms A and B are students, and each student in each of rooms A and B is either a junior or a senior. The ratio of the number of juniors to the number of seniors in room A is 4 to 5, the ratio of the number of juniors to the number of seniors in room B is 3 to 17, and the ratio of the total number of juniors in both rooms A and B to the total number of seniors in both rooms A and B is 5 to 7. What is the ratio of the total number of students in room A to the total number of students in room B ?
A. 29/12 B. 59/10 C. 65/8 D. 48/5 E. 29/3 HI, another method, although whenever you see Qs on mixture of two Quantities and the average, best approach as mentioned above is weighted average method...lets just work only on strength of juniorsin room A= 4/9 of A.. in room B= 3/20 of B.. in Total= 5/12 of (A+B).. so we can form an equation: 4A/9 + 3B/20= 5(A+B)/12.. (4/9)A+(3/20)B= (5/12)A +(5/12)B.. (4/9)A(5/12)A= (5/12)B(3/20)B.. (1/36)A=(16/60)B.. A/B=36*16/60=48/5 ans
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Re: The only people in each of rooms A and B are students, and [#permalink]
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