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Math Expert V
Joined: 02 Sep 2009
Posts: 58340
At the same time Sue began rolling a wheelbarrow from X to Y, Nancy  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 66% (02:05) correct 34% (02:00) wrong based on 361 sessions

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At the same time Sue began rolling a wheelbarrow from X to Y, Nancy started walking along the same road from Y to X. If Sue's traveling rate was 2 feet per second and Nancy's was 3 feet per second, and if the wheelbarrow wheel rolled without slipping, what is the radius of the wheel on Sue's wheelbarrow?

(1) At the moment Sue and Nancy crossed paths, Sue had traveled 60 feet with the wheelbarrow.
(2) The total number of wheel revolutions during Sue's trip from X to Y was 75.

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Re: At the same time Sue began rolling a wheelbarrow from X to Y, Nancy  [#permalink]

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Bunuel wrote:
At the same time Sue began rolling a wheelbarrow from X to Y, Nancy started walking along the same road from Y to X. If Sue's traveling rate was 2 feet per second and Nancy's was 3 feet per second, and if the wheelbarrow wheel rolled without slipping, what is the radius of the wheel on Sue's wheelbarrow?

(1) At the moment Sue and Nancy crossed paths, Sue had traveled 60 feet with the wheelbarrow.
(2) The total number of wheel revolutions during Sue's trip from X to Y was 75.

CONCEPT: Distance travelled by Wheel in 1 revolution = Circumference of Wheel = 2 *(Pi)* Radius
and Total Distance covered by Wheel = No. of Revolutions made to cover that Distance * (2*(Pi)*r)

Statement 1: At the moment Sue and Nancy crossed paths, Sue had travelled 60 feet with the wheelbarrow.

Speeds of Sue / Speeds of Nancy = 2/3

Speeds of Sue / Speeds of Nancy = Distance of Sue / Distance of Nancy

i.e. Distance travelled by Nancy in same time = (3/2)*60 = 90 feet

i.e. Total Distance between X and Y = 60 + 90 = 150 Feet

But This doesn't provide us the information about how many revolutions does the wheel takes to cover total or any part of distance

Hence NOT SUFFICIENT

Statement 2: The total number of wheel revolutions during Sue's trip from X to Y was 75

But This doesn't provide us the information about Total distance travelled by the wheel in 75 Revolutions

Hence NOT SUFFICIENT

Combining the two statements

Total Distance = 150 feet (Calculated from statement 1)
Total Revolutions made by wheel to cover total distance = 75

i.e. 150 = 75 * (2*pi*r)

i.e. Radius of wheel, r = 1/(pi)

SUFFICIENT

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GMAT 1: 690 Q49 V34 Re: At the same time Sue began rolling a wheelbarrow from X to Y, Nancy  [#permalink]

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Q: At the same time Sue began rolling a wheelbarrow from X to Y, Nancy started walking along the same road from Y to X. If Sue's traveling rate was 2 feet per second and Nancy's was 3 feet per second, and if the wheelbarrow wheel rolled without slipping, what is the radius of the wheel on Sue's wheelbarrow?

we need the total # of revolutions and total distance covered by wheelbarrow.
Sue | 2 feet per sec |
Nancy | 3 feet per sec |

Stmt (1) At the moment Sue and Nancy crossed paths, Sue had traveled 60 feet with the wheelbarrow.
Sue time to cover 60 feet = distance/speed = 60 feet/2 = 30 sec
Nancy took same time to cover = 30 sec * 3 = 90 feet
total Dist = 90 + 60 = 150 feet.
but no info about # of revolutions by wheelbarrow. ----- not sufficient

Stmt (2) The total number of wheel revolutions during Sue's trip from X to Y was 75.
info about # of revolutions by wheelbarrow.
but no info about total distance. ----- not sufficient

Together Stmt (1) and Stmt (2)
total Dist = 150 feet.
# of revolutions by wheelbarrow = 75
$$2 \pi r = 150/75.$$ ----- sufficient

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Math Expert V
Joined: 02 Sep 2009
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Re: At the same time Sue began rolling a wheelbarrow from X to Y, Nancy  [#permalink]

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Bunuel wrote:
At the same time Sue began rolling a wheelbarrow from X to Y, Nancy started walking along the same road from Y to X. If Sue's traveling rate was 2 feet per second and Nancy's was 3 feet per second, and if the wheelbarrow wheel rolled without slipping, what is the radius of the wheel on Sue's wheelbarrow?

(1) At the moment Sue and Nancy crossed paths, Sue had traveled 60 feet with the wheelbarrow.
(2) The total number of wheel revolutions during Sue's trip from X to Y was 75.

MANHATTAN GMAT OFFICIAL SOLUTION:

For the wheelbarrow, we can set up a distance formula:
Distance Traveled = (Wheel Circumference) (Number of Wheel Revolution)
Distance Traveled = (2πr) (Number of Wheel Revolutions)

$$\frac{Distance \ Traveled}{(2\pi{r})(Number \ of \ Wheel \ Revolutions)}=r$$

We can rephrase the question from “What is r?” to “What is the distance traveled and the number of wheel revolutions?”

(1) INSUFFICIENT: This provides the distance traveled, but no information about the number of wheel revolutions.

(2) INSUFFICIENT: This gives the number of wheel revolutions, but no information about the total distance from X to Y.

(1) AND (2) SUFFICIENT: If Sue traveled 60 feet at a rate of 2 feet per second, she traveled for 60/2 = 30 seconds before meeting Nancy. During those 30 seconds, Nancy had traveled at a rate of 3 feet per second, for a distance of (3)(30) = 90 feet. The total distance between X and Y is 60 + 90 = 150 feet.

We also know the total number of wheel revolutions that occurred between X and Y, so we have the answer to the rephrased question.

The correct answer is C.
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Posts: 58340
Re: At the same time Sue began rolling a wheelbarrow from X to Y, Nancy  [#permalink]

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Bunuel wrote:
Bunuel wrote:
At the same time Sue began rolling a wheelbarrow from X to Y, Nancy started walking along the same road from Y to X. If Sue's traveling rate was 2 feet per second and Nancy's was 3 feet per second, and if the wheelbarrow wheel rolled without slipping, what is the radius of the wheel on Sue's wheelbarrow?

(1) At the moment Sue and Nancy crossed paths, Sue had traveled 60 feet with the wheelbarrow.
(2) The total number of wheel revolutions during Sue's trip from X to Y was 75.

MANHATTAN GMAT OFFICIAL SOLUTION:

For the wheelbarrow, we can set up a distance formula:
Distance Traveled = (Wheel Circumference) (Number of Wheel Revolution)
Distance Traveled = (2πr) (Number of Wheel Revolutions)

$$\frac{Distance \ Traveled}{(2\pi{r})(Number \ of \ Wheel \ Revolutions)}=r$$

We can rephrase the question from “What is r?” to “What is the distance traveled and the number of wheel revolutions?”

(1) INSUFFICIENT: This provides the distance traveled, but no information about the number of wheel revolutions.

(2) INSUFFICIENT: This gives the number of wheel revolutions, but no information about the total distance from X to Y.

(1) AND (2) SUFFICIENT: If Sue traveled 60 feet at a rate of 2 feet per second, she traveled for 60/2 = 30 seconds before meeting Nancy. During those 30 seconds, Nancy had traveled at a rate of 3 feet per second, for a distance of (3)(30) = 90 feet. The total distance between X and Y is 60 + 90 = 150 feet.

We also know the total number of wheel revolutions that occurred between X and Y, so we have the answer to the rephrased question.

The correct answer is C.

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At the same time Sue began rolling a wheelbarrow from X to Y, Nancy  [#permalink]

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Bunuel wrote:
At the same time Sue began rolling a wheelbarrow from X to Y, Nancy started walking along the same road from Y to X. If Sue's traveling rate was 2 feet per second and Nancy's was 3 feet per second, and if the wheelbarrow wheel rolled without slipping, what is the radius of the wheel on Sue's wheelbarrow?

(1) At the moment Sue and Nancy crossed paths, Sue had traveled 60 feet with the wheelbarrow.
(2) The total number of wheel revolutions during Sue's trip from X to Y was 75.

Circumference * Rev = Total distance covered.

Distance covered by Sue = 60 feet, Time taken = 60/2 = 30 secs.
Distance covered by Nancy in same time = 30 * 3 = 90 feet.

Total D = 60 + 90 = 150 feet.

2 * pi * r * 75 = 150

using I and II, we can get the value of the wheelbarrow radius, hence C
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Re: At the same time Sue began rolling a wheelbarrow from X to Y, Nancy  [#permalink]

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_________________ Re: At the same time Sue began rolling a wheelbarrow from X to Y, Nancy   [#permalink] 23 Dec 2018, 03:06
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