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GRE Math Challenge #1- Below are some tough word problems

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GRE Math Challenge #1- Below are some tough word problems [#permalink] New post 29 May 2014, 03:32
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Below are some tough word problems that require both close reading and sharp problem solving skills. To make these already difficult questions even more fiendish, I’ve made each a Numeric Entry question (sorry, no cool boxes). In some of the questions below, there are a few possible solution paths, some more elegant than others. If you are feeling brave, share your answer(s) and how you got there.
Good luck!

Numeric Entry Questions



1. Set S consists of the integers from -1 to 5, inclusive. If N is the product of three distinct members of Set S, how many unique values of N are there?
Numeric Entry ____________

2. A combination lock has three dials, each consisting of a single digit, 0-9. The lock can be opened only if each dial is in the correct position. If the code for the lock consists of three prime numbers, what is the probability that 252 is the code? (Provide answer in fraction form).
Numeric Entry ___________

3. Twelve doctors are to be selected by hospital staff to sit on a six-person committee. Within that committee, an additional subcommittee of three doctors will be formed. How many unique sub-committees are possible?
Numeric Entry ____________
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Re: GRE Math Challenge #1- Below are some tough word problems [#permalink] New post 26 May 2016, 07:51
1)210
2)1/64
3)120

plz provide the QA'z with explaination
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Re: GRE Math Challenge #1- Below are some tough word problems [#permalink] New post 27 May 2016, 00:53
1. Order doesn't matter, hence this is a combination problem. Let us exclude zero from the list as any number multiplied by zero is zero. So there are 6C3 ways choosing 3 numbers out of 6 numbers that is 20. Adding a case were product is zero, the answer is 20+1 = 21
2. There are four single digit prime numbers: 2,3,5,7. So 4*4*4 ways to arrange the prime numbers. and there is only to form the number 252, hence the probability is 1/64
3. 12C3 = 220. The intermediate step of choosing 3 people from six-person committee should not matter after all you choose 3 people out of 12 with all having the same chance of being selected
Re: GRE Math Challenge #1- Below are some tough word problems   [#permalink] 27 May 2016, 00:53
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GRE Math Challenge #1- Below are some tough word problems

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