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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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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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 confidence--specifically so that, when you hit a high-difficulty 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???

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

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.

This approach is much simpler. Thanks.

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)(1-x)=(5/12).
Solving for x results in x = 48/53. Therefore the ratio is 48:5.

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

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

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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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 juniors
in 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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