Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Efrida and Frazer who live 10 miles apart, meet at a restaur [#permalink]
06 Jan 2013, 08:16

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

66% (02:30) correct
33% (01:23) wrong based on 45 sessions

Efrida and Frazer who live 10 miles apart, meet at a restaurant that is directly north of Efrida's home and directly east of Frazer's home. If the restaurant is two miles closer to Efrida's home, than to Frazer's home, how many miles is the restaurant from Frazer's home?

Re: efrida and frazer who live 10 miles apart meet at a restaur [#permalink]
07 Jan 2013, 01:32

1

This post received KUDOS

Expert's post

chiccufrazer1 wrote:

efrida and frazer who live 10 miles apart,meet at a restaurant that is directly north of efrida's home and directly east of frazer's home.if the restaurant is too miles closer to efrida's home,than to frazer's home,how many miles is the restaurant from frazer's home A.6 B.7 C.8 D.10 E.11

Posted from my mobile device

Please check the questions when posting and format them properly.

Efrida and Frazer who live 10 miles apart, meet at a restaurant that is directly north of Efrida's home and directly east of Frazer's home. If the restaurant is two miles closer to Efrida's home, than to Frazer's home, how many miles is the restaurant from Frazer's home?

A. 6 B. 7 C. 8 D. 10 E. 11

It's a simple geometry problem. Look at the diagram below:

Attachment:

Restraunt.PNG [ 4.84 KiB | Viewed 879 times ]

From above it follows that the restaurant is 8 miles from Frazer's home and 6 miles from Efrida's home.

Re: efrida and frazer who live 10 miles apart meet at a restaur [#permalink]
07 Jan 2013, 03:01

1

This post received KUDOS

thanks bunuel..i really appreciate on how you have solved the problem..i also used the same terms except for x +2 which i ended up using x-2 instead..i did use the pythagorean theorem but i ended up getting something like wrong and getting more confused

Re: Efrida and Frazer who live 10 miles apart, meet at a restaur [#permalink]
07 Jan 2013, 11:02

bunuel i have noticed one thing on how your have illustrated the information given in the diagram..the diagram indicates frazer's home as directly west of the restaurant rather than as directly east..does the direction change anything at all?

Re: Efrida and Frazer who live 10 miles apart, meet at a restaur [#permalink]
08 Jan 2013, 02:38

Expert's post

chiccufrazer1 wrote:

bunuel i have noticed one thing on how your have illustrated the information given in the diagram..the diagram indicates frazer's home as directly west of the restaurant rather than as directly east..does the direction change anything at all?

Posted from my mobile device

The stem says that the restaurant is directly east of Frazer's home, hence Frazer's home is directly west of the restaurant. Though it wouldn't change anything even if it were vise-versa.
_________________

Re: Efrida and Frazer who live 10 miles apart, meet at a restaur [#permalink]
08 Jan 2013, 05:45

Bunuel,

did you apply the pythagorean theorem to the diagram above? I tried to apply it, but I get 6 and -8 as a result.

A second way that I was thinking about is the 90-60-30 degree triangular rule where the sides of the triangle are 3, 4, and 5. However, we are not certain that this triangle has this property.

Re: Efrida and Frazer who live 10 miles apart, meet at a restaur [#permalink]
08 Jan 2013, 13:02

marcoppe wrote:

Bunuel,

did you apply the pythagorean theorem to the diagram above? I tried to apply it, but I get 6 and -8 as a result.

A second way that I was thinking about is the 90-60-30 degree triangular rule where the sides of the triangle are 3, 4, and 5. However, we are not certain that this triangle has this property.

@marcoppe most definitely bunuel used the pythagorean theorem..it is as it is because of the word DIRECTLY NORTH OR DIRECTLY EAST in the stem question signifying that we have a 90 degrees angle somewhere..well marcoppe am sorry awil ask you a dumb question but please bear with me..what does the 90-60-30 triangular rule state?and how do you apply it?

Re: Efrida and Frazer who live 10 miles apart, meet at a restaur [#permalink]
09 Jan 2013, 00:50

@chiccufrazer1

The 90-60-30 rule, sometimes called "Pythagorean triple" or "3-4-5 triangle", applies to those triangles that hold this property: 90°, 60°, and 30° angles. The sides of any 90-60-30 degree triangle will have its sides bearing specific length proportions: 3, 4, 5. There would be no need, therefore, to use the Pythagorean theorem in such cases.

The triangle formed in the excercise above will certainly be a 3-4-5 triangle as we have a 3-4-5 proportion in the length of its sides.

Re: Efrida and Frazer who live 10 miles apart, meet at a restaur [#permalink]
09 Jan 2013, 01:13

marcoppe wrote:

@chiccufrazer1

The 90-60-30 rule, sometimes called "Pythagorean triple" or "3-4-5 triangle", applies to those triangles that hold this property: 90°, 60°, and 30° angles. The sides of any 90-60-30 degree triangle will have its sides bearing specific length proportions: 3, 4, 5. There would be no need, therefore, to use the Pythagorean theorem in such cases.

The triangle formed in the excercise above will certainly be a 3-4-5 triangle as we have a 3-4-5 proportion in the length of its sides.

@marcoppe. in other words what you trying to say is that we might be given a triangle that only has one side given but holds the 90-60-30 property and can still be able to find the other sides by using the pythagorean triple..if saw then how would test for the 90-60-30 property?

Re: Efrida and Frazer who live 10 miles apart, meet at a restaur [#permalink]
09 Jan 2013, 02:14

Expert's post

marcoppe wrote:

Bunuel,

did you apply the pythagorean theorem to the diagram above? I tried to apply it, but I get 6 and -8 as a result.

A second way that I was thinking about is the 90-60-30 degree triangular rule where the sides of the triangle are 3, 4, and 5. However, we are not certain that this triangle has this property.

90-60-30 (angles) triangle and 3-4-5 (sides) triangle are not the same. For more check here: math-triangles-87197.html _________________

Re: Efrida and Frazer who live 10 miles apart, meet at a restaur [#permalink]
09 Jan 2013, 02:38

Thanks Bunuel, I realized now that I got confused with the two principles. To know that we are dealing with a 3-4-5 right triangle there is no need to know its angles.

Sorry chiccufrazer1.

Anyways, I was curious to know how Bunuel got that result. I am pretty sure he used the pythagorean theorem, but I got -8 and 6 using Phythagora. I am not sure how to decide which one to pick.

Re: Efrida and Frazer who live 10 miles apart, meet at a restaur [#permalink]
09 Jan 2013, 02:54

Expert's post

marcoppe wrote:

Thanks Bunuel, I realized now that I got confused with the two principles. To know that we are dealing with a 3-4-5 right triangle there is no need to know its angles.

Sorry chiccufrazer1.

Anyways, I was curious to know how Bunuel got that result. I am pretty sure he used the pythagorean theorem, but I got -8 and 6 using Phythagora. I am not sure how to decide which one to pick.

Actually I didn't use Pythagorean theorem.

We know that the hypotenuse is 10 and one leg is 2 more than the other, which means that it should be 6-8-10 right triangle.

If we use the theorem then we'll have x^2+(x+2)^2=10^2 --> 2x^2+4x-96=0 --> x^2+2x-48=0 --> (x-6)(x+8)=0 --> x=6 (discard x=-8 since the distance cannot be negative).

Re: efrida and frazer who live 10 miles apart meet at a restaur [#permalink]
20 Nov 2013, 02:21

Bunuel wrote:

chiccufrazer1 wrote:

efrida and frazer who live 10 miles apart,meet at a restaurant that is directly north of efrida's home and directly east of frazer's home.if the restaurant is too miles closer to efrida's home,than to frazer's home,how many miles is the restaurant from frazer's home A.6 B.7 C.8 D.10 E.11

Posted from my mobile device

Please check the questions when posting and format them properly.

Efrida and Frazer who live 10 miles apart, meet at a restaurant that is directly north of Efrida's home and directly east of Frazer's home. If the restaurant is two miles closer to Efrida's home, than to Frazer's home, how many miles is the restaurant from Frazer's home?

A. 6 B. 7 C. 8 D. 10 E. 11

It's a simple geometry problem. Look at the diagram below:

Attachment:

Restraunt.PNG

From above it follows that the restaurant is 8 miles from Frazer's home and 6 miles from Efrida's home.

Answer: C.

Solved using pure math, but took me more than 3 minutes... the draw is reaaly a shortcut.

Thanks Bunuel for this.

gmatclubot

Re: efrida and frazer who live 10 miles apart meet at a restaur
[#permalink]
20 Nov 2013, 02:21