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Joined: 02 Aug 2017
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Concentration: Strategy, Nonprofit
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A fair coin has 2 distinct flat sidesone of which bears the image of
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Updated on: 12 Jun 2024, 01:02
Updated on: 12 Jun 2024, 01:02
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A fair coin has 2 distinct flat sides—one of which bears the image of a face and the other of which does not—and when the coin is tossed, the probability that the coin will land face up is 1/2. For certain values of M, N, p, and q, when M fair coins are tossed simultaneously, the probability is p that all M coins land face up, and when N fair coins are tossed simultaneously, the probability is q that all N coins land face up. Furthermore, M < N and \(\frac{1}{p} + \frac{1}{q} = 72\).
In the table, select a value for M and a value for N that are jointly consistent with the given information. Make only two selections, one in each column.
In the table, select a value for M and a value for N that are jointly consistent with the given information. Make only two selections, one in each column.
ID: 100444
| M | N | |
| 2 | ||
| 3 | ||
| 4 | ||
| 5 | ||
| 6 | ||
| 7 |
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Official Answer
M: 3
N: 6
Originally posted by Blackishmamba on 25 Aug 2018, 05:32.
Last edited by Bunuel on 12 Jun 2024, 01:02, edited 5 times in total.
Last edited by Bunuel on 12 Jun 2024, 01:02, edited 5 times in total.
Re: A fair coin has 2 distinct flat sidesone of which bears the image of
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15 Mar 2024, 05:03
15 Mar 2024, 05:03
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Expert Reply
This is a Quant Probability question in the garb of a TPA question. So, as long as you have solid quant skills, you should be able to solve this question with utmost ease.
Here is the video solution for you!
If you falter in this question, you should implement the stated corrective actions and once you do that not only will your ability in TPA improve but also in quant Arithmetic. 😊
In addition, this question can be solved using two methods:
Watch the solution to learn how these two methods can be used.
Here is the video solution for you!
If you falter in this question, you should implement the stated corrective actions and once you do that not only will your ability in TPA improve but also in quant Arithmetic. 😊
- If you were not able to translate the scenario into probability equation, then the issue is your lack of conceptual understanding related to probability.
- Corrective Action: Build your probability conceptual understanding.
- If you got scared of this question because you saw 4 different variables in a probability question, then the issue is that you are not comfortable using variables in word problems type questions.
- Corrective Action: Build your process skills in the realm of word problems.
In addition, this question can be solved using two methods:
- Using the answer choices
- Using inferences
Watch the solution to learn how these two methods can be used.
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Re: A fair coin has 2 distinct flat sidesone of which bears the image of
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10 Sep 2018, 20:19
10 Sep 2018, 20:19
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Trial and error would be quicker.
Probability for each event irrespective of the coin ) M or N is 1/2
i.e. if M =1 coin
> p (probability of face) =1/2
Similarly for N
Now for M coins
> p= (1/2)^M
For M coins
> p= (1/2)^N
Now to confuse us , the reciprocal of sums is given so that we get confused.
But one clue is that the sum is 72, an integer, so you need not fear.
1/p +1/q =72
1/(1/2)^M + 1/(1/2)^N = 72
>> 2^M +2^N = 72.
All you have to do list down some powers of two.not much just few.
2
4
8
16
32
64
Given M<N
Trial and error gives us
8+64 =72
2^3 + 2^6 = 72
M=3 N=6
Posted from my mobile device
Probability for each event irrespective of the coin ) M or N is 1/2
i.e. if M =1 coin
> p (probability of face) =1/2
Similarly for N
Now for M coins
> p= (1/2)^M
For M coins
> p= (1/2)^N
Now to confuse us , the reciprocal of sums is given so that we get confused.
But one clue is that the sum is 72, an integer, so you need not fear.
1/p +1/q =72
1/(1/2)^M + 1/(1/2)^N = 72
>> 2^M +2^N = 72.
All you have to do list down some powers of two.not much just few.
2
4
8
16
32
64
Given M<N
Trial and error gives us
8+64 =72
2^3 + 2^6 = 72
M=3 N=6
Posted from my mobile device
