Two different candles are lit. They burn at different rates and one is

Question

Not A GMAT question. Just wanted to know how to solve it.

Two different candles are lit. They burn at different rates and one is 3cm longer than the other. The longer one was lit at 6 PM and the shorter one at 730PM. At 1000 PM they both are of the same length. The longer one burned out at 1200AM and the shorter one at 1130AM.How long was each candle originally?

24cm and 27 cm

there are 5 yellow balls in bag 1. One ball is transferred to bag 2 which contains an unknown number of green balls. Bag 2 is then shaken and a ball is selected at random and transfeered to bag1 without seeing its color. Bag1 is then shaken and a ball is selected at random and transferred to bag2 without seeing its color. If the probability of selecting a green ball from bag 2 is now 3/5, how many green balls were there in bag2 at the begining?

4 green balls

Bunuel · Answer

Two different candles are lit. They burn at different rates and one is 3cm longer than the other. The longer one was lit at 6 PM and the shorter one at 730PM. At 1000 PM they both are of the same length. The longer one burned out at  1200AM  and the shorter one at  1130AM .How long was each candle originally?

I think there should be 12 PM  and 11.30 PM  to get the answers 27 and 24.

Let the rate of longer one be  x  and the shorter  y  cm per hour, their length  a  and  a-3  cm respectively.

From 6PM to 10PM (in 4 hours) longer one decreased in length by  4x  cm and from 7:30PM to 10PM (in 2.5 hours) shorter one decreased in length by  2.5y  cm and at 10PM their length were equal so:  a-4*x=(a-3)-2.5y ;

Next: from 10PM to 12PM (in 2 hours) longer one burned out so  a-4x=2x  (it took 2 hours at rate  x cm per hout to "cover"  a-4x cm) and also from 10PM to 11:30PM (in 1.5 hours) shorter one burned out so  (a-3)-2.5y=1.5y . As from above  a-4*x=(a-3)-2.5y , then  2x=1.5y  --> solving for  a  -->  a=27  and  a-3=24

There are 5 yellow balls in bag 1. One ball is transferred to bag 2 which contains an unknown number of green balls. Bag 2 is then shaken and a ball is selected at random and transfeered to bag1 without seeing its color. Bag1 is then shaken and a ball is selected at random and transferred to bag2 without seeing its color. If the probability of selecting a green ball from bag 2 is now 3/5, how many green balls were there in bag2 at the begining?

I guess there are ONlY yellow balls in bag #1 and ONLY green balls in bag #2.

Let's say initially there were  x  green balls in the bag #2. After all theese transfers # of balls in bag #2 became  x+1  (2 transfers from bag #1 to bag #2 AND one transfer from bag #2 to bag #1). Initially probability of picking green ball from bag #2 was 1 and we are told that the probability of picking green at the end decreased to  \frac{3}{5} , so obviously # of green balls either decreased (we transferred either 1 or 2 green balls to the bag #1, as there were 2 transfers from bag #2) or remained the same (we were moving one yellow ball all the time). So 3 cases:

If we transfered 0 green balls to bag #1, then  \frac{x}{x+1}=\frac{3}{5}  -->  5x=3x+3  -->  x=\frac{3}{2}
eq{integer} , which is not possible as  x  represent # of green balls and it must an integer.

If we transfered 1 green ball to bag #1, then  \frac{x-1}{x+1}=\frac{3}{5}  -->  5x-5=3x+3  -->  x=4

If we transferred 2 green balls to bag #1, then  \frac{x-2}{x+1}=\frac{3}{5}  -->  5x-10=3x+3  -->  x=\frac{13}{2}
eq{integer} , which is not possible as  x  represent # of green balls and it must an integer.

So initially there were 4 green balls in bag #2.

Hope it's clear.

tkarthi4u · Answer

wow thank you.

yes u r right regarding the typo.

TallJTinChina · Answer

I started with a drawing (feel free to resell :-D

) 2bags.JPG At the end we have P(green) = 3/5 As always total probability = 1 So P(yellow) = 1- P(green) = 2/5 Most possible yellow that moved to green is 2 because we only have 2 transfers from Bag 1 to Bag 2. This means at the end of transfers Bag 2 has 2 yellow and 3 green (It can't have more than 5 total balls) So then we know both transfers to Bag 2 moved yellow balls. Bag 1(end) = Bag 1(beginning) - 2 yellow balls It is given Bag 1 (beginning) = YYYYY So Bag 1 (end) = YYY? ?=Green Ball because if not green, then it would be yellow and this would make Bag 2(end) P(green) = less than 2/5 So Bag 1(end) = YYYG This tells us the one transfer Bag 2 to Bag 1 was in fact a green ball. So we work it backward Bag 2(end) = YYGGG Bag 2 (end) - 2Y + 1 G= Bag 2 (beginning) YYGGG-YY+G=GGGG

saynchalk · Answer

First one is harder than a 600 level I think

Two different candles are lit. They burn at different rates and one is

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