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Ratios: Bag A contains red, white and blue marbles such that

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Ratios: Bag A contains red, white and blue marbles such that

by II » Fri Mar 21, 2008 12:35 am
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?

A) 1 B) 3 C) 4 D) 6 E) 8

Interested in finding out the most effective and efficient way of answering these types of questions.

Thanks in advance.
Last edited by II on Mon May 05, 2008 1:53 am, edited 1 time in total.
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by musicdaemon » Fri Mar 21, 2008 2:02 am
For Bag A:

R:W = 1:3 = 2:6
W:B=2:3=6:9
Thus, R:W:B= 2:6:9

If x be the common factor for this proportion , then number of balls in bag A

= 2x+6x+9x …………………..(1)

For bag B:
R:W = 1:4
If y be the common factor for this proportion, then number of balls in bag B

= y+4y…………………………(2)

Give,
Total no. of white balls = 30 ,
6x+4y = 30 ………………………………(putting values of white balls from (1) & (2))
=> 3x+2y=15
By trial and error (x,y)=(3,3) & (1,6)
For x = 3
Total number of Red balls in Bag A = 2x = 2*3 = 6 ………………….. Ans

For x = 1
No of red balls in bag A = 2*1=2 ………………. None of the options match

Thsu 6 is the only answer in this case
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useful trick

by musicdaemon » Fri Mar 21, 2008 10:30 am
A useful Trick:
i use it often to find the range of integer values of a linear equation in two variables

for example, in the solution for this question, the equation

3x+2y=15 , gives

x=(15-2y)/3, now go on putting y=0,1,2,3 .....

since, x is to be an integer, (15-2y) must always be a multiple of 3

thus, for y=1, x = (15-2*1)/3 = 13/3 ...... ( not an integer)
for y=2, x = (15-2*2)/3 = 11/3 ...... ( not an integer)
for y=3, x = (15-2*3)/3 = 9/3=3 ...... (an integer solution)


Hope you get the point
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by mmukher » Fri Mar 21, 2008 5:42 pm
A different way to solve this..

Bag A => Red(R):White(W) = 1:3 Therefore R = W/3
Bag B => R:W = 1:4 Therefore R = W/4

Total Bag A, white (let it be x) + Bag B, whites (let it be y) = 30

We need x and y such that x+y = 30 and x is divisible by 3 and y is divisible by 4.

if you plug in a few numbers you see that x = 18 and y = 12 works

And therefore Red Balls in Bag A are x/3 = 18/3 = 6.
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by ritz » Fri Mar 21, 2008 6:35 pm
musicdaemon wrote:For Bag A:

R:W = 1:3 = 2:6
W:B=2:3=6:9
Thus, R:W:B= 2:6:9

If x be the common factor for this proportion , then number of balls in bag A

= 2x+6x+9x …………………..(1)

For bag B:
R:W = 1:4
If y be the common factor for this proportion, then number of balls in bag B

= y+4y…………………………(2)

Give,
Total no. of white balls = 30 ,
6x+4y = 30 ………………………………(putting values of white balls from (1) & (2))
=> 3x+2y=15
By trial and error (x,y)=(3,3) & (1,6)
For x = 3
Total number of Red balls in Bag A = 2x = 2*3 = 6 ………………….. Ans

For x = 1
No of red balls in bag A = 2*1=2 ………………. None of the options match

Thsu 6 is the only answer in this case
an easier way to get to the running rations when individual ratios are given..
if
a/b = n1/d1
b/c = n2/d2
c/d = n3/d3
then
a:b:c:d = n1n2n3:d1n2n3:d1d2n3:d1d2d3
start with all numerators & replace them one by one by denominators.
you will get the running ratios. hope this helps.
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by II » Sat Mar 22, 2008 3:23 am
Guys ... thanks for the different ways to solve this type of problem !
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by simplyjat » Sat Mar 22, 2008 4:50 am
II wrote:Guys ... thanks for the different ways to solve this type of problem !
Can you tell us, from where you got the question?
simplyjat
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by II » Sun Mar 23, 2008 6:08 am
simplyjat wrote:
II wrote:Guys ... thanks for the different ways to solve this type of problem !
Can you tell us, from where you got the question?
Manhattan GMAT practice test.
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