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A local restaurant prepares one of three desserts every day

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by BTGmoderatorLU » Tue Aug 14, 2018 3:06 pm

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A

B

C

D

E

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Difficulty—

Source: Veritas Prep

A local restaurant prepares one of three desserts every day they are open, Monday through Saturday: cake, pastries, or brownies. If the dessert on Wednesday must be brownies and the same dessert is not served on consecutive days, how many different menus for the week are possible?

A. 3*2^5
B. 3^6
C. 3^2*2^4
D. 2^5
E. 2^3*3^3

The OA is D.
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by [email protected] » Tue Aug 14, 2018 5:26 pm
Hi All,

We're told that a local restaurant prepares one of three desserts every day they are open, Monday through Saturday: cake, pastries, or brownies, the dessert on Wednesday MUST be brownies and the same dessert is NOT served on consecutive days. We're asked for the number of different menus that are possible for each week. This question is essentially a Permutation question - although you can beat it with just a bit of logic and some Arithmetic.

Since brownies MUST be served on Wednesday - and the same dessert CANNOT be served on consecutive days - then we know that there will be 2 possible desserts on both Tuesday and Thursday (and those options would be limited to cake or pastries). Regardless of which option is served, there will only be 2 options for Monday and Friday (again - since the same dessert cannot be served on back-to-back days). Whatever option is chosen on Friday, there will only be 2 options for Saturday.

Since Wednesday has one option 'locked' in place, the total number of permutations for the Monday-Saturday period would be:
(2)(2)(1)(2)(2)(2) = 2^5

Final Answer: D

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by swerve » Thu Aug 16, 2018 9:18 am
Monday - Tuesday - Wednesday - Thursday - Friday - Saturday

2c1 2c1 B 2c1 2c1 2c1 = 2 x 2 x 1 x 2 x 2 x 2= 2^5

Treat Wednesday as the center point for solving the problem and move in either direction.

Since no dessert can be served on consecutive days, hence Tuesday and Wednesday have 2c1 possibilities, and so on.

Therefore, D is the correct answer.
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by Scott@TargetTestPrep » Thu Aug 23, 2018 3:53 pm
BTGmoderatorLU wrote:Source: Veritas Prep

A local restaurant prepares one of three desserts every day they are open, Monday through Saturday: cake, pastries, or brownies. If the dessert on Wednesday must be brownies and the same dessert is not served on consecutive days, how many different menus for the week are possible?

A. 3*2^5
B. 3^6
C. 3^2*2^4
D. 2^5
E. 2^3*3^3
We are given that on Wednesday, the dessert must be brownies. Going backward, we have two choices (cake or pastries) for Tuesday. Once a dessert is chosen for Tuesday, we have two choices for Monday since it can't be the one that is same as Tuesday.

Similarly, going forward from Wednesday, we have two choices (cake or pastries) for Thursday. Once a dessert is chosen for Thursday, we have two choices for Friday since it can't be the one that is same as Thursday. Once a dessert is chosen for Friday, we have two choices for Saturday since it can't be the one that is same as Friday.

Therefore, we have 2 choices of desert for every day from Monday to Saturday except for Wednesday, when there is only 1 choice. Therefore, the number of different menus for the week is:

2 x 2 x 1 x 2 x 2 x 2 = 2^5

Answer: D

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