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26 May 2017, 04:41
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There are 2 candles with the same height. It takes t mins for A to burn, and it takes 2t mins for B to burn. When the candles are burnt at the same time, how many minutes did it take for the height of B to be twice the height of A?
A. \(\frac{t}{3}\)
B. \(\frac{2t}{3}\)
C. \(\frac{t}{2}\)
D. \(\frac{3t}{4}\)
E. \(\frac{4t}{5}\)
A. \(\frac{t}{3}\)
B. \(\frac{2t}{3}\)
C. \(\frac{t}{2}\)
D. \(\frac{3t}{4}\)
E. \(\frac{4t}{5}\)
26 May 2017, 04:41
Kudos
Bookmarks
Official Solution:
There are 2 candles with the same height. It takes t mins for A to burn, and it takes 2t mins for B to burn. When the candles are burnt at the same time, how many minutes did it take for the height of B to be twice the height of A?
A. \(\frac{t}{3}\)
B. \(\frac{2t}{3}\)
C. \(\frac{t}{2}\)
D. \(\frac{3t}{4}\)
E. \(\frac{4t}{5}\)
As shown in the diagram above, there are candles A and B with the same height of h. If you assume the speed of burning as v and s, respectively, you get speed*time=distance, so each becomes \(vt=h, s(2t) = h\). If you burn them at the same time and assume the time it took the height of B to become twice the height of A as x minutes, you get \(h-(\frac{h}{2t})x = 2[h-(\frac{h}{t})x]\), and if you take out h from both sides, you get \(1- (\frac{x}{2t}) = 2[1-(\frac{x}{t})]\). If you multiply both sides by \(2t\), you get \(2t-x = 4t-4x\), then \(4x-x = 4t-2t, 3x = 2t\), and then \(x = \frac{2t}{3}\). B is the answer.
Answer: B
There are 2 candles with the same height. It takes t mins for A to burn, and it takes 2t mins for B to burn. When the candles are burnt at the same time, how many minutes did it take for the height of B to be twice the height of A?
A. \(\frac{t}{3}\)
B. \(\frac{2t}{3}\)
C. \(\frac{t}{2}\)
D. \(\frac{3t}{4}\)
E. \(\frac{4t}{5}\)
As shown in the diagram above, there are candles A and B with the same height of h. If you assume the speed of burning as v and s, respectively, you get speed*time=distance, so each becomes \(vt=h, s(2t) = h\). If you burn them at the same time and assume the time it took the height of B to become twice the height of A as x minutes, you get \(h-(\frac{h}{2t})x = 2[h-(\frac{h}{t})x]\), and if you take out h from both sides, you get \(1- (\frac{x}{2t}) = 2[1-(\frac{x}{t})]\). If you multiply both sides by \(2t\), you get \(2t-x = 4t-4x\), then \(4x-x = 4t-2t, 3x = 2t\), and then \(x = \frac{2t}{3}\). B is the answer.
Answer: B
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