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A cyclist travels the length of a bike path that is 225 [#permalink]
12 Jul 2011, 05:32

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (medium)

Question Stats:

29% (02:34) correct
71% (02:01) wrong based on 188 sessions

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

Re: Inequalities word problem [#permalink]
17 Apr 2012, 05:18

8

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

ENAFEX wrote:

Isn't the answer supposed to be D?

224.5/5.4 = 41.57 AND 225.5/4.5 = 50.1

41 and 50 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]
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
_________________

Re: A cyclist travels the length of a bike path that is 225 [#permalink]
18 Apr 2012, 03:25

Expert's post

ENAFEX wrote:

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

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]
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?
_________________

Re: A cyclist travels the length of a bike path that is 225 [#permalink]
18 Apr 2012, 07:34

Expert's post

ENAFEX wrote:

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).

Re: Inequalities word problem [#permalink]
24 Jul 2013, 09:53

1

This post received KUDOS

Bunuel wrote:

ENAFEX wrote:

Isn't the answer supposed to be D?

224.5/5.4 = 41.57 AND 225.5/4.5 = 50.1

41 and 50 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{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.

Re: Inequalities word problem [#permalink]
24 Jul 2013, 09:57

Expert's post

up4gmat wrote:

Bunuel wrote:

ENAFEX wrote:

Isn't the answer supposed to be D?

224.5/5.4 = 41.57 AND 225.5/4.5 = 50.1

41 and 50 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{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.
_________________

Re: Inequalities word problem [#permalink]
24 Jul 2013, 21:00

Bunuel wrote:

ENAFEX wrote:

Isn't the answer supposed to be D?

224.5/5.4 = 41.57 AND 225.5/4.5 = 50.1

41 and 50 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.

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.

Re: Inequalities word problem [#permalink]
25 Jul 2013, 01:21

Expert's post

mridulparashar1 wrote:

Bunuel wrote:

ENAFEX wrote:

Isn't the answer supposed to be D?

224.5/5.4 = 41.57 AND 225.5/4.5 = 50.1

41 and 50 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.

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.

Re: Inequalities word problem [#permalink]
25 Jul 2013, 01:46

Bunuel wrote:

mridulparashar1 wrote:

Bunuel wrote:

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.

Re: Inequalities word problem [#permalink]
25 Jul 2013, 02:06

1

This post received KUDOS

Expert's post

mridulparashar1 wrote:

Bunuel wrote:

mridulparashar1 wrote:

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.

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.
_________________