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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

A cyclist travels the length of a bike path that is 225 [#permalink]

Show Tags

12 Jul 2011, 05:32

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

30% (02:33) correct
70% (01:58) wrong based on 421 sessions

HideShow timer Statistics

A cyclist travels the length of a bike path that is 225 miles long, rounded to the nearest mile. If the trip took him 5 hrs, rounded to the nearest hour, then his average speed must be between:

A. 38 and 50 mph B. 40 and 50 mph C. 40 and 51 mph D. 41 and 50 mph E. 41 and 51 mph

A cyclist travels the length of a bike path that is 225 miles long, rounded to the nearest mile. If the trip took him 5 hrs, rounded to the nearest hour, then his average speed must be between: A. 38 and 50 mph B. 40 and 50 mph C. 40 and 51 mph D. 41 and 50 mph E. 41 and 51 mph

Length of a path is 225 miles long, rounded to the nearest mile --> \(224.5\leq{distance}<225.5\); The trip took him 5 hrs, rounded to the nearest hour --> \(4.5\leq{time}<5.5\);

Lowest average rate is \(\frac{224.5}{5.5}\approx{40.8}\) (take the lowest value of nominator and highest value of denominator); Highest average rate is \(\frac{225.5}{4.5}\approx{50.1}\) (take the highest value of nominator and lowest value of denominator);

\(40.8<rate<50.1\).

Now, the question is: "the average speed must be between..." hence the range from correct answer choice MUST cover all possible values of rate, so must cover all the range: \(40.8<rate<50.1\). Only C does that: \((40)<40.8<rate<50.1<(51)\). D can not be the answer as if \(rate=40.9\) or if \(rate=50.01\) then these possible values of the average rate are not covered by the range from this answer choice, which is (41-50).

Answer: C.

Alternative approach. It's based on observing the answer choices. On the PS section always look at the answer choices before you start to solve a problem. They might often give you a clue on how to approach the question. A. 38 and 50 mph B. 40 and 50 mph C. 40 and 51 mph D. 41 and 50 mph E. 41 and 51 mph

Notice that since the range from A covers entire range from B and D, then B and D are out (if B or D is correct so is A and we cannot have two correct answer, leave the bigger range). Similarly since the range from C covers entire range from E, then E is out too (if E is correct so is C and we cannot have two correct answer, leave the bigger range).

Thus we are left only with two answer choices A (38, 50) and C (40, 51). From here it's much easier to get the correct answer.

Re: A cyclist travels the length of a bike path that is 225 [#permalink]

Show Tags

17 Apr 2012, 17:02

Bunuel,

Thanks for the explanations and the alternative approach. Great as always.

I am having a problem in selecting the denominator range. In some problems, I noticed that the lowest denominator value is usually taken as <0.5, i.e. 0.4, e.g. in the above problem I thought it should be 4.4. How do you choose between .4 and .5?

Hope my question makes sense. :D
_________________

Thanks for the explanations and the alternative approach. Great as always.

I am having a problem in selecting the denominator range. In some problems, I noticed that the lowest denominator value is usually taken as <0.5, i.e. 0.4, e.g. in the above problem I thought it should be 4.4. How do you choose between .4 and .5?

Hope my question makes sense. :D

It's because of the rounding rules: math-number-theory-88376.html (check chapter for Rounding). 4.5 rounded equals to 5, while 4.4 rounded equals to 4.

So, "the trip took him 5 hrs, rounded to the nearest hour" means \(4.5\leq{time}<5.5\).

Re: A cyclist travels the length of a bike path that is 225 [#permalink]

Show Tags

18 Apr 2012, 06:09

Bunuel, I understand the rounding concept but sometimes it is just confusing to apply it in problems.

e.g. On a recent trip Cindy drove her car 290 miles, rounded to the nearest 10 miles and used 12 gallons of gasoline rounded to the nearest gallon. The actual number of miles per gallon that Cindy's car got on this trip must have been between:

A) 290/12.5 and 290/11.4 B) 294/12 and 284/11.4 C) 284/12 and 295/12 D) 284/12.5 and 295/11.4 E) 295/12.5 and 284/11.4

In this is the correct choice is D 12 rounded to the nearest gasoline should be 11.5<=Value<12.5. But why is it 11.4 in the correct choice? Also why is it 284? Is it not supposed to be 285? I thought the answer will be 285/12.5 to 295/11.5?
_________________

Bunuel, I understand the rounding concept but sometimes it is just confusing to apply it in problems.

e.g. On a recent trip Cindy drove her car 290 miles, rounded to the nearest 10 miles and used 12 gallons of gasoline rounded to the nearest gallon. The actual number of miles per gallon that Cindy's car got on this trip must have been between:

A) 290/12.5 and 290/11.4 B) 294/12 and 284/11.4 C) 284/12 and 295/12 D) 284/12.5 and 295/11.4 E) 295/12.5 and 284/11.4

In this is the correct choice is D 12 rounded to the nearest gasoline should be 11.5<=Value<12.5. But why is it 11.4 in the correct choice? Also why is it 284? Is it not supposed to be 285? I thought the answer will be 285/12.5 to 295/11.5?

Question should read: On a recent trip, Cindy drove her car 290 miles, rounded to the nearest 10 miles, and used 12 gallons of gasoline, rounded to the nearest gallon. The actual number of miles per gallon that Cindy's car got on this trip must have been between A. 290/12.5 and 290/11.5 B. 295/12 and 285/11.5 C. 285/12 and 295/12 D. 285/12.5 and 295/11.5 E. 295/12.5 and 285/11.5

Cindy drove her car 290 miles, rounded to the nearest 10 miles --> \(285\leq{m}<295\); Used 12 gallons of gasoline, rounded to the nearest gallon --> \(11.5\leq{g}<12.5\);

Minimum Miles per gallon, m/g --> \(\frac{285}{12.5}<\frac{m}{g}<\frac{295}{11.5}\) (to get lower limit take min possible for nominator and max possible for denominator, and for upper limit take max possible for nominator and min possible for denominator).

A cyclist travels the length of a bike path that is 225 miles long, rounded to the nearest mile. If the trip took him 5 hrs, rounded to the nearest hour, then his average speed must be between: A. 38 and 50 mph B. 40 and 50 mph C. 40 and 51 mph D. 41 and 50 mph E. 41 and 51 mph

Length of a path is 225 miles long, rounded to the nearest mile --> \(224.5\leq{distance}<225.5\); The trip took him 5 hrs, rounded to the nearest hour --> \(4.5\leq{time}<5.5\);

Lowest average rate is \(\frac{225.5}{4.5}\approx{50.1}\) (take the highest value of nominator and lowest value of denominator); Highest average rate is \(\frac{224.5}{5.5}\approx{40.8}\) (take the lowest value of nominator and highest value of denominator);

\(40.8<rate<50.1\).

Now, the question is: "the average speed must be between..." hence the range from correct answer choice MUST cover all possible values of rate, so must cover all the range: \(40.8<rate<50.1\). Only C does that: \((40)<40.8<rate<50.1<(51)\). D can not be the answer as if \(rate=40.9\) or if \(rate=50.01\) then these possible values of the average rate are not covered by the range from this answer choice, which is (41-50).

Answer: C.

Alternative approach. It's based on observing the answer choices. On the PS section always look at the answer choices before you start to solve a problem. They might often give you a clue on how to approach the question. A. 38 and 50 mph B. 40 and 50 mph C. 40 and 51 mph D. 41 and 50 mph E. 41 and 51 mph

Notice that since the range from A covers entire range from B and D, then B and D are out (if B or D is correct so is A and we can not have two correct answer, leave the bigger range). Similarly since the range from C covers entire range from E, then E is out too (if E is correct so is C and we can not have two correct answer, leave the bigger range).

Thus we are left only with two answer choices A (38, 50) and C (40, 51). From here it's much easier to get the correct answer.

Hope it's clear.

Hi Bunuel, Shouldnt it be vice versa??

Lowest average rate is \(\frac{225.5}{4.5}\approx{50.1}\) (take the highest value of nominator and lowest value of denominator);- Highest Avg Highest average rate is \(\frac{224.5}{5.5}\approx{40.8}\) (take the lowest value of nominator and highest value of denominator);-Lowest Avg.

A cyclist travels the length of a bike path that is 225 miles long, rounded to the nearest mile. If the trip took him 5 hrs, rounded to the nearest hour, then his average speed must be between: A. 38 and 50 mph B. 40 and 50 mph C. 40 and 51 mph D. 41 and 50 mph E. 41 and 51 mph

Length of a path is 225 miles long, rounded to the nearest mile --> \(224.5\leq{distance}<225.5\); The trip took him 5 hrs, rounded to the nearest hour --> \(4.5\leq{time}<5.5\);

Lowest average rate is \(\frac{225.5}{4.5}\approx{50.1}\) (take the highest value of nominator and lowest value of denominator); Highest average rate is \(\frac{224.5}{5.5}\approx{40.8}\) (take the lowest value of nominator and highest value of denominator);

\(40.8<rate<50.1\).

Now, the question is: "the average speed must be between..." hence the range from correct answer choice MUST cover all possible values of rate, so must cover all the range: \(40.8<rate<50.1\). Only C does that: \((40)<40.8<rate<50.1<(51)\). D can not be the answer as if \(rate=40.9\) or if \(rate=50.01\) then these possible values of the average rate are not covered by the range from this answer choice, which is (41-50).

Answer: C.

Alternative approach. It's based on observing the answer choices. On the PS section always look at the answer choices before you start to solve a problem. They might often give you a clue on how to approach the question. A. 38 and 50 mph B. 40 and 50 mph C. 40 and 51 mph D. 41 and 50 mph E. 41 and 51 mph

Notice that since the range from A covers entire range from B and D, then B and D are out (if B or D is correct so is A and we can not have two correct answer, leave the bigger range). Similarly since the range from C covers entire range from E, then E is out too (if E is correct so is C and we can not have two correct answer, leave the bigger range).

Thus we are left only with two answer choices A (38, 50) and C (40, 51). From here it's much easier to get the correct answer.

Hope it's clear.

Hi Bunuel, Shouldnt it be vice versa??

Lowest average rate is \(\frac{225.5}{4.5}\approx{50.1}\) (take the highest value of nominator and lowest value of denominator);- Highest Avg Highest average rate is \(\frac{224.5}{5.5}\approx{40.8}\) (take the lowest value of nominator and highest value of denominator);-Lowest Avg.

Lowest and highest were mixed. Edited. Thank you. +1.
_________________

A cyclist travels the length of a bike path that is 225 miles long, rounded to the nearest mile. If the trip took him 5 hrs, rounded to the nearest hour, then his average speed must be between: A. 38 and 50 mph B. 40 and 50 mph C. 40 and 51 mph D. 41 and 50 mph E. 41 and 51 mph

Length of a path is 225 miles long, rounded to the nearest mile --> \(224.5\leq{distance}<225.5\); The trip took him 5 hrs, rounded to the nearest hour --> \(4.5\leq{time}<5.5\);

Lowest average rate is \(\frac{224.5}{5.5}\approx{40.8}\) (take the lowest value of nominator and highest value of denominator); Highest average rate is \(\frac{225.5}{4.5}\approx{50.1}\) (take the highest value of nominator and lowest value of denominator);

\(40.8<rate<50.1\).

Now, the question is: "the average speed must be between..." hence the range from correct answer choice MUST cover all possible values of rate, so must cover all the range: \(40.8<rate<50.1\). Only C does that: \((40)<40.8<rate<50.1<(51)\). D can not be the answer as if \(rate=40.9\) or if \(rate=50.01\) then these possible values of the average rate are not covered by the range from this answer choice, which is (41-50).

Answer: C.

Hi Bunuel,

Refering to the rounding chapter in Number Theory examples

Rounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.

Example: 5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5. 5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5. 5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.

Shouldn't the range for distance be 224.5 < D< 225.4 ------> If we take the higher value of distance i.e 225.4 and drop the last digit (4) then it gets rounded off to 225 but if we keep 225.5 then after dropping (5) it should be rounded off to 226.

A cyclist travels the length of a bike path that is 225 miles long, rounded to the nearest mile. If the trip took him 5 hrs, rounded to the nearest hour, then his average speed must be between: A. 38 and 50 mph B. 40 and 50 mph C. 40 and 51 mph D. 41 and 50 mph E. 41 and 51 mph

Length of a path is 225 miles long, rounded to the nearest mile --> \(224.5\leq{distance}<225.5\); The trip took him 5 hrs, rounded to the nearest hour --> \(4.5\leq{time}<5.5\);

Lowest average rate is \(\frac{224.5}{5.5}\approx{40.8}\) (take the lowest value of nominator and highest value of denominator); Highest average rate is \(\frac{225.5}{4.5}\approx{50.1}\) (take the highest value of nominator and lowest value of denominator);

\(40.8<rate<50.1\).

Now, the question is: "the average speed must be between..." hence the range from correct answer choice MUST cover all possible values of rate, so must cover all the range: \(40.8<rate<50.1\). Only C does that: \((40)<40.8<rate<50.1<(51)\). D can not be the answer as if \(rate=40.9\) or if \(rate=50.01\) then these possible values of the average rate are not covered by the range from this answer choice, which is (41-50).

Answer: C.

Hi Bunuel,

Refering to the rounding chapter in Number Theory examples

Rounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.

Example: 5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5. 5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5. 5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.

Shouldn't the range for distance be 224.5 < D< 225.4 ------> If we take the higher value of distance i.e 225.4 and drop the last digit (4) then it gets rounded off to 225 but if we keep 225.5 then after dropping (5) it should be rounded off to 226.

A cyclist travels the length of a bike path that is 225 miles long, rounded to the nearest mile. If the trip took him 5 hrs, rounded to the nearest hour, then his average speed must be between: A. 38 and 50 mph B. 40 and 50 mph C. 40 and 51 mph D. 41 and 50 mph E. 41 and 51 mph

Length of a path is 225 miles long, rounded to the nearest mile --> \(224.5\leq{distance}<225.5\); The trip took him 5 hrs, rounded to the nearest hour --> \(4.5\leq{time}<5.5\);

Lowest average rate is \(\frac{224.5}{5.5}\approx{40.8}\) (take the lowest value of nominator and highest value of denominator); Highest average rate is \(\frac{225.5}{4.5}\approx{50.1}\) (take the highest value of nominator and lowest value of denominator);

\(40.8<rate<50.1\).

Now, the question is: "the average speed must be between..." hence the range from correct answer choice MUST cover all possible values of rate, so must cover all the range: \(40.8<rate<50.1\). Only C does that: \((40)<40.8<rate<50.1<(51)\). D can not be the answer as if \(rate=40.9\) or if \(rate=50.01\) then these possible values of the average rate are not covered by the range from this answer choice, which is (41-50).

Answer: C.

Hi Bunuel,

Refering to the rounding chapter in Number Theory examples

Rounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.

Example: 5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5. 5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5. 5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.

Shouldn't the range for distance be 224.5 < D< 225.4 ------> If we take the higher value of distance i.e 225.4 and drop the last digit (4) then it gets rounded off to 225 but if we keep 225.5 then after dropping (5) it should be rounded off to 226.

Refering to the rounding chapter in Number Theory examples

Rounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.

Example: 5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5. 5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5. 5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.

Shouldn't the range for distance be 224.5 < D< 225.4 ------> If we take the higher value of distance i.e 225.4 and drop the last digit (4) then it gets rounded off to 225 but if we keep 225.5 then after dropping (5) it should be rounded off to 226.

The Answer remains C. Please confirm the range selection

Thanks

Should be as written. What about 225.499999? So, take 225.5 but exclude the endpoints, by writing < and > instead of <= and >=.

Hope it's clear.

Hi Bunuel,

Can you elaborate a bit more. I am not able to get the message.

Thanks

When you take maximum distance as 225.4 you exclude distance from 225.4 to 225.5. For example d could be 225.47, 225.43, 225.4111111. So, take 225.5 as max distance but write rate<225.5/4.5 instead of rate<=225.5/4.5.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...