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MGMT ratio problem [#permalink]
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02 Aug 2011, 22:38
Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 30 white marbles. How many red marbles could be in bag A? 1 3 4 6 8 Bag A: R/w = 1/3 and w/b=2/3 Bag B: R/w = 1/4 Number of White marbles in A and B is 30 . i.e A+B =30 so in Bag B Number of white marbles will be 30/5 * 4 = 24 since A+B = 30 => B= 24 then in Bag A number of white marbles will be 6 Now R/w = 1/3 => R/6 = 1/3 => R=2But there is NO 2. what i'm doing wrong. please explain. regards
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Re: MGMT ratio problem [#permalink]
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shrive555 wrote: Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 30 white marbles. How many red marbles could be in bag A? 1 3 4 6 8
Bag A: R:W:BR:W:B 1x:3x:2x:6x: :2x:3x:6x:9x Bag A: R:W:B 2x:6x:9x Bag B: R:W 1y:4y If we just consider white marbles now: 6x+4y=30 y=(306x)/4 So, x=1; y=6 x=2; y=Non integer. Ignore x=3; y=3 x=4; y=Non integer x=5; y=0(Not possible) Thus, x can be either 1 or 3. If x=1; Bag A: R:W:B 2x:6x:9x R=2*1=2 If x=3; R=2*3=6 So, both 2 and 6 are possible numbers of red marbles in Bag A. Ans: "D"
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Re: MGMT ratio problem [#permalink]
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02 Aug 2011, 23:14
fluke wrote: shrive555 wrote: Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 30 white marbles. How many red marbles could be in bag A? 1 3 4 6 8
Bag A: R:W:BR:W:B 1x:3x:2x:6x: :2x:3x:6x:9x Bag A: R:W:B 2x:6x:9xBag B: R:W 1y:4y If we just consider white marbles now: 6x+4y=30 y=(306x)/4 So, x=1; y=6 x=2; y=Non integer. Ignore x=3; y=3 x=4; y=Non integer x=5; y=0(Not possible) Thus, x can be either 1 or 3. If x=1; Bag A: R:W:B 2x:6x:9x R=2*1=2 If x=3; R=2*3=6 So, both 2 and 6 are possible numbers of red marbles in Bag A. Ans: "D" Thank you for your detail explanation. just little confusion. how did you get the red part. because i have to solve it through making equations R/w = 1/3 w/b=2/3 => R/b = 2/9 after this i don't know how to set up R:W:B I assume that you have used some short cut.
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Re: MGMT ratio problem [#permalink]
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02 Aug 2011, 23:22
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shrive555 wrote: Thank you for your detail explanation. just little confusion. how did you get the red part. because i have to solve it through making equations R/w = 1/3 w/b=2/3 => R/b = 2/9 after this i don't know how to set up R:W:B I assume that you have used some short cut. Not really; Just take the LCM of the common term: R:W=1:3 1 W:B=2:3 2 Here W is common; Take the LCM of 2,3=6 So, eq 1 needs to be multiplied by 2: R:W=(1:3)*2=2:63 So, eq 2 needs to be multiplied by 3: W:B=(2:3)*3=6:94 Now; 3 and 4 are both 6 whites; we can join them R:W:B 2:6:9
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Re: MGMT ratio problem [#permalink]
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03 Aug 2011, 09:00
you guys are very much right 2 / 6 are the options possible



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Re: MGMT ratio problem [#permalink]
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03 Aug 2011, 13:03
good question .. Bag A R:W:B = 2x: 6x: 9x and Bag B R:W = y:4y
6x + 4y = 30 => 3x + 2y = 15 ..
the only +ve integers that satisfy the above is 1, 6 and 3, 3 ..
red marbles in Bag A = 2x = 2 or 6 => answer is D



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Re: Ratios [#permalink]
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18 Sep 2011, 16:10
Bag A
R/W = X/(3X)  1 W/B = 2Y/((3Y) 2
from 1 and 2 , we can see W= 3X = 2Y
Bag B R/W = Z/4Z
X=?
the two bags contain 30 white marbles => 3X+4Z = 30
=> Z = (303X)/4
then by plugging in X values , we can see that only D gives an integer value for Z.
Answer is D.



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Re: Ratios [#permalink]
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19 Sep 2011, 03:18
aeros232 wrote: Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 30 white marbles. How many [highlight]red marbles could be in bag A[/highlight]? [/[color=#40FF00]color]
a 1 b 3 c 4 d 6 e 8 Bag A: R:W = 1:3 W:B = 2:3 W is the common one here so make it equal i.e. R:W = 2:6 and W:B = 6:9 (the ratios remain the same). So R:W:B = 2:6:9 Since number of marbles has to be an integer, number of red marbles in this bag must be 2 or a multiple of 2 and number of white marbles must be 6 or a multiple of 6. Bag B: R:W = 1:4 The number of white marbles must be 4 or a multiple of 4. To make 30 white marbles, you could mix white marbles from Bag A and Bag B in many ways. BagA: 6 + BagB: 24 (No. of red marbles in BagA = 2) BagA: 12 + BagB: 18  Not possible because 18 is not a multiple of 4 BagA: 18 + BagB: 12 (No. of red marbles in BagA = 6) BagA: 24 + BagB: 6  Not possible because 6 is not a multiple of 4 No of red marbles in bag A can be both 2 and 6. Answer (D)
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Re: Ratios [#permalink]
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21 Sep 2011, 07:52
please comment is this method wrong :
ratio of balls in bag A: red:white:blue >2:6:9 (2x+6x+9x) ratio of balls in bag B: red:white >1:4 (1y+4y)
so we can derive a equation where white balls in both bags> 6x+4y=30 1 (as the total number of white balls are 30) red balls in bag A > 2x=?2
solving eqn 1 > we get x value as 1 and 3 to get a whole number as balls cannot be in fraction.
substituting the value in 2 > we get the value as 2 and 6 for red balls in bag A. As 2 is not there in the answer , we select 6 (ie answer D)



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Re: Ratios [#permalink]
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21 Sep 2011, 10:12
abhi398 wrote: please comment is this method wrong :
ratio of balls in bag A: red:white:blue >2:6:9 (2x+6x+9x) ratio of balls in bag B: red:white >1:4 (1y+4y)
so we can derive a equation where white balls in both bags> 6x+4y=30 1 (as the total number of white balls are 30) red balls in bag A > 2x=?2
solving eqn 1 > we get x value as 1 and 3 to get a whole number as balls cannot be in fraction.
substituting the value in 2 > we get the value as 2 and 6 for red balls in bag A. As 2 is not there in the answer , we select 6 (ie answer D) Nothing wrong with the method. Essentially, both, this one and the one above, find integral solutions of 6x + 4y = 30.
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Re: MGMAT PS: Bag A contains red, white and blue marbles [#permalink]
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21 Sep 2011, 11:00
D) 6
Let x be the # of Red balls in Bag A Therefore the ratio for R/W = 1/3 = x/3x
Let y be the # of Red balls in Bag B Therefore the ratio for R/W = 1/4 = y/4y
Now we know White (Bag A) + White (Bag B) = 30 3x + 4y = 30
So lets see what value of x solves this equation from the choices given: 1, 3, 4, 6, 8 1: y is not an integer (27/4) 3: y is not an integer (21/4) 4: y is not an integer (14/4) 6: y is an integer 8: y is not an integer
Thus 6.




Re: MGMAT PS: Bag A contains red, white and blue marbles
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