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A restaurant serves only tacos and enchiladas. If 60% of all

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A restaurant serves only tacos and enchiladas. If 60% of all

by Anaira Mitch » Sun Dec 16, 2018 10:00 am

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A restaurant serves only tacos and enchiladas. If 60% of all who ate at the restaurant on a certain day ordered tacos, what percentage of those who ate at the restaurant ordered enchiladas that day?

(1) Half of those who ordered tacos also ordered enchiladas.
(2) 3/4 as many people ordered both enchiladas and tacos as ordered only enchiladas.
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Source: — Data Sufficiency |

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by Jay@ManhattanReview » Mon Dec 17, 2018 4:11 am

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Anaira Mitch wrote:A restaurant serves only tacos and enchiladas. If 60% of all who ate at the restaurant on a certain day ordered tacos, what percentage of those who ate at the restaurant ordered enchiladas that day?

(1) Half of those who ordered tacos also ordered enchiladas.
(2) 3/4 as many people ordered both enchiladas and tacos as ordered only enchiladas.
Say those who ordered tacos = T = 60%, those who ordered enchiladas = E and those who ordered both tacos and enchiladas = B

Thus, we have

100 = T + E - B
100 = 60 + E - B
E = 40 + B.

We have to get the value of E.

Let's take each statement one by one.

(1) Half of those who ordered tacos also ordered enchiladas.

=> B = 1/2 of T = 1/2 of 60 = 30.

Thus, E = 40 + 30 = 70%. Sufficient.

(2) 3/4 as many people ordered both enchiladas and tacos as ordered only enchiladas.

=> B = 3/4 (E - B); here E - B = Only enchiladas

B = 3/7E

Thus, from E = 40 + B, we have E = 40 + 3/7E => E = 70%. Sufficient.

The correct answer: D

Hope this helps!

-Jay
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by fskilnik@GMATH » Tue Dec 18, 2018 12:12 pm

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Anaira Mitch wrote:A restaurant serves only tacos and enchiladas. If 60% of all who ate at the restaurant on a certain day ordered tacos, what percentage of those who ate at the restaurant ordered enchiladas that day?

(1) Half of those who ordered tacos also ordered enchiladas.
(2) 3/4 as many people ordered both enchiladas and tacos as ordered only enchiladas.
$$? = x\,\,\,\,\left( {{\rm{see}}\,\,{\rm{figure}}} \right)$$

Image

$$\left( 1 \right)\,\,\,\left[ {{\rm{Blue}}} \right]\,\,\,100 = 60 + x - 30\,\,\,\left( {{\rm{simplifier}}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,x\,\,{\rm{unique}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.$$

$$\left( 2 \right)\,\,\,\left[ {{\rm{Red}}} \right]\,\,\,\,\left\{ \matrix{
{3 \over 4}\left( {x - {\rm{both}}} \right) = {\rm{both}}\,\,\,\,\,\left( {\rm{I}} \right) \hfill \cr
100 = 60 + \left( {x - {\rm{both}}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,40 = x - {\rm{both}}\,\,\,\,\,\left( {{\rm{II}}} \right)\,\,\, \hfill \cr} \right.$$
$$\left( {\rm{I}} \right)\,\,{\rm{in}}\,\,\left( {{\rm{II}}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,40 = {4 \over 3}\left( {{\rm{both}}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{both}} = 30\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {{\rm{II}}} \right)} \,\,\,\,\,\,x\,\,{\rm{unique}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.$$


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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