Tricky Math Problems of the Week
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Each Week Dreamz4uonline Team provides you with a tricky Math question to help you improve your analytical thinking abilities.
It does not matter which competitive exam you are preparing for, the tricky maths questions are prepared carefully, trying to keep the flavor of all the competitive exams.
11 Dec, 2006 Solution ( Party in LA )
4 Dec, 2006 Solution ( Pipes working with variable Efficiency )
27 Nov, 2006 Solution ( Trains Meeting Second Time )
20 Nov, 2006 Solution (Woman at Sea Beach)
13 Nov, 2006 Solution (Cuboid immersed in a cylinder)
6 Nov, 2006 Solution (Smoking Cigarette)
30 Oct, 2006 Solution (Number Game (Division by 3))
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23 Oct, 2006 Solution (Late Chauffer)
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16 Oct, 2006 Solution (Bee flying between two trains)
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9 Oct, 2006 Solution (Monkey Climbing a Rope)
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2 Oct, 2006 Solution (Bouncing Ball)
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25 Sep, 2006 Solution (Probability - Three Friends for the same job)
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18 Sep, 2006 Solution (Time and Work- Fencing Field)
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11 Sep, 2006 Solution (Pipes And Cistern)
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4 Sep, 2006 Solution (Triangle Property)
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Solution to the 11 Dec, 2006 Problem ( Party in LA )
Solution to the problem will be posted on 18 Dec. Take your time to solve the problem and check back next Monday about how you did.
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Solution to the 4 Dec, 2006 Solution ( Pipes working with variable Efficiency )
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Pipes working with variable Efficiency
A tap "A" can fill a container in 10 minutes when running alone, another tap "B" can fill the same container in 15 minutes when running alone. A tap "C" can empty the same container in 12 minutes when running alone. Initially the container is empty. Tap A is opened for 1 min and then closed, then tap B is opened for 1 min and then closed. Now tap "C" is opened for 1 min and then closed. This goes on till the container is half filled. after that all the three taps are opened together. What is the total time, from the starting, in which the container gets completely filled. (Assume that the tap "C" works with half the efficiency when running with other tap(s), but with full efficiency when running alone. Also "A" and "B" run with full efficiency whether running alone or with other tap(s). Ignore the time in opening and closing the taps)
... www.dreamz4uonline.com tricky math problem.
18 min
20 min
22 min
24 min
None of the above.
Cannot be determined.
Answer :18 min
Solution (Fast Approach)
Tap "A" can fill 1/10 part of the container in 1 min. same way "B" can fill 1/15 part in 1 min. and "C" can empty 1/12 part in 1 min.
As they are opened individually for 1 min each, in 3 minutes 1/10+1/15 - 1/12 = (6+4-5)/60 = 1/12 part is filled. If we calculate using the above equation, the container can be half filled in 6*3 = 18 minutes, but it's a possibility that after some minutes, there may be no use of opening "C", as the container may be half filled, much before 18 minutes time.
Let us consider opening Taps 4 times, in 4*3 = 12 min, 4/12 = 1/3 part of container will be filled.
Now, 1/2 - 1/3 = 1/6 part is left to be filled, to let the container half filled
Tap "A" and "B" in 2 minutes fill : 1/10 + 1/15 = 5/30 = 1/6 part of the container.
As we had seen above, we conclude that we do not need to open Tap "C" as container is half filled.
Total time taken to fill the first half = 12 + 2 = 14 min
For remaining half, all the taps are opened simultaneously.
But efficiency of "C" is half when opened with other taps.
Hence "C" has an efficiency of emptying the container in 12*2 = 24 minutes
All together "A", "B" and "C" fills 1/10+1/15-1/24 = 15/120 = 1/8 part in 1 minute.
Hence 4*1/8 = 1/2 part will be filled in 1*4 = 4 minutes
So, Total time taken to fill the container = 14 + 4 = 18 minutes
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