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20 Dec 2016, 01:29
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
81% (01:45) correct
19%
(01:40)
wrong
based on 149
sessions
History
Date
Time
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Not Attempted Yet
James is selecting CDs to take with him on his trip. James will select 1 jazz disc, 1 pop, 1 country and western and 1 blues. If he has 12 jazz, 10 pop and 15 each of country and western and blues, then how many possible different combinations of CDs could James take on his trip?
A. 52
B. 1,800
C. 6,750
D. 22,500
E. 27,000
A. 52
B. 1,800
C. 6,750
D. 22,500
E. 27,000
Updated on: 21 Dec 2016, 04:10
Originally posted by EgmatQuantExpert on 21 Dec 2016, 02:30.
Last edited by EgmatQuantExpert on 21 Dec 2016, 04:10, edited 1 time in total.
Last edited by EgmatQuantExpert on 21 Dec 2016, 04:10, edited 1 time in total.
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Hey Everyone,
PFB the solution.
We are given that we need to select –
Notice that this question clearly tells us to find the different “combinations”, and the word “select” is a clear indicator that we need to apply the concept of combination to solve this question.
Thus, the different combination of CDs can be written as –
Therefore, the Correct Answer is Option E
Thanks,
Saquib
Quant Expert
e-GMAT
PFB the solution.
We are given that we need to select –
- • 1 jazz cd out of 12 jazz.
• 1 pop cd out of 10 pop.
• 1 country and western cd out of 15 country and western.
• 1 blues cd out of 15 blues.
Notice that this question clearly tells us to find the different “combinations”, and the word “select” is a clear indicator that we need to apply the concept of combination to solve this question.
Thus, the different combination of CDs can be written as –
- \(= ^{12}C_1\) x \(^{10}C_1\) x \(^{15}C_1\) x \(^{15}C_1\)
\(= 12\) x \(10\) x \(15\) x \(15\)
\(= 120\) x \(225\)
\(=27000\)
Therefore, the Correct Answer is Option E
Thanks,
Saquib
Quant Expert
e-GMAT
To practise ten 700+ Level Number Properties Questions attempt the The E-GMAT Number Properties Knockout
21 Dec 2016, 03:44
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Bunuel
Hi ,
The equation is something which we all will be able to get but effort would be in solving it at the earliest..
So 10 of one kind, 12 of another and 15 of two other kinds will give us 10*12*15*15 combinations..
What will be 10*12*15*15..
Should we get into multiplication or there is some way out..
Look at 0s you will have..
10 will give you one..
Each 15 will give you another 0, when multiplied by 2.. so another 2..
Total 0s .. 1+2=3..
Only E has 3 0s..
Ans E
