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Probability

by tsmith93 » Sun Feb 21, 2010 3:58 pm
Jason has twice as many red marbles as blue marbles. He puts them in two jars in such a way that the ratio of the number of red marbles to blue marbles in jar I is 2:7 and there are only red marbles in jar II. The number of red marbles in jar II is how many times the number of red marbles in jar I?
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by harsh.champ » Sun Feb 21, 2010 4:16 pm
tsmith93 wrote:Jason has twice as many red marbles as blue marbles. He puts them in two jars in such a way that the ratio of the number of red marbles to blue marbles in jar I is 2:7 and there are only red marbles in jar II. The number of red marbles in jar II is how many times the number of red marbles in jar I?
R=2 x B
In Jar 1,ratio R:Bis 2:7, in jar II there are only red marbles
so in jar 2 we have all the blue marbles i.e. B and some red marbles its no. 2/7 B
Let it be 2x and 7x
and y
2x + y= 14x
Hence,y=12x
so, we have y/2x = 12x/2x =6:1
Last edited by harsh.champ on Sun Feb 21, 2010 4:26 pm, edited 2 times in total.
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by shashank.ism » Sun Feb 21, 2010 4:20 pm
tsmith93 wrote:Jason has twice as many red marbles as blue marbles. He puts them in two jars in such a way that the ratio of the number of red marbles to blue marbles in jar I is 2:7 and there are only red marbles in jar II. The number of red marbles in jar II is how many times the number of red marbles in jar I?
so R = 2 B
ratio R:B in jar I is 2:7, in jar II there are only red marbles
so in jar 2 we have all the blue marbles i.e. B and some red marbles its no. 2/7 B
in jar II red marbles = 2B- 2/7 B = 12/7 B
so jar II /jar I = [12/7 B]/[2/7 B] = [spoiler]6 ans
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by thephoenix » Sun Feb 21, 2010 5:15 pm
tsmith93 wrote:Jason has twice as many red marbles as blue marbles. He puts them in two jars in such a way that the ratio of the number of red marbles to blue marbles in jar I is 2:7 and there are only red marbles in jar II. The number of red marbles in jar II is how many times the number of red marbles in jar I?
let the # of blue marbles=x
so # of red marbles=2x

now let # of red marbles in jar2=z

so # of red marbles in jar1=2x-z

ratio=(2x-z)/x=2/7-------->z=12x/7
so # of red marbles in jar1=2x-(12x/7)=2x/7............say z1

so z=n(z1) i.e n times of # of red marbles in z1
----->n=z/z1=12x/7/2x/7=6 is the ans
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