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

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 12PM and 11.30PM 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 \(\frac{x}{x+1}=\frac{3}{5}\) --> \(5x=3x+3\) --> \(x=\frac{3}{2}\neq{integer}\), 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 \(\frac{x-2}{x+1}=\frac{3}{5}\) --> \(5x-10=3x+3\) --> \(x=\frac{13}{2}\neq{integer}\), 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.