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Rhonda runs at an average speed of 12 kilometers per hour, a
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02 Jul 2019, 12:32
02 Jul 2019, 12:32
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84% (02:36) correct
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Rhonda runs at an average speed of 12 kilometers per hour, and she bicycles at an average speed of 30 kilometers per hour. When she bicycles to work, her travel time is 2.25 minutes less than when she runs to work. The distance to work, in kilometers, is
A) 27/40
B) 3/4
C) 7/8
D) 11/12
E) 6/5
A) 27/40
B) 3/4
C) 7/8
D) 11/12
E) 6/5
Re: Rhonda runs at an average speed of 12 kilometers per hour, a
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02 Jul 2019, 16:21
02 Jul 2019, 16:21
2
we need to convert the speed from km/h to km/min
therefore, 12km/h to km/min =12/60=1/5km/min
similarly, 30km/h =1/2km/min
speed(km/min) * time(min)=distance(km)
distance covered when running= t/5
distance covered when on bicycle=(t-2.25)/2
distance covered when rhoda runs=distance covered when on bicycle
therefore, t/5=t-2.25/2 ;after simplification t=15/4
using distance covered by run and substituting for t, we have
15/4/5=15/20=3/4
distance from work =3/4km
therefore, 12km/h to km/min =12/60=1/5km/min
similarly, 30km/h =1/2km/min
speed(km/min) * time(min)=distance(km)
distance covered when running= t/5
distance covered when on bicycle=(t-2.25)/2
distance covered when rhoda runs=distance covered when on bicycle
therefore, t/5=t-2.25/2 ;after simplification t=15/4
using distance covered by run and substituting for t, we have
15/4/5=15/20=3/4
distance from work =3/4km
Re: Rhonda runs at an average speed of 12 kilometers per hour, a
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02 Jul 2019, 23:42
02 Jul 2019, 23:42
1
GreenlightTestPrep wrote:
Rhonda runs at an average speed of 12 kilometers per hour, and she bicycles at an average speed of 30 kilometers per hour. When she bicycles to work, her travel time is 2.25 minutes less than when she runs to work. The distance to work, in kilometers, is
A) 27/40
B) 3/4
C) 7/8
D) 11/12
E) 6/5
A) 27/40
B) 3/4
C) 7/8
D) 11/12
E) 6/5
Here,
Let the total distance be = D
Let's convert 2.25 minutes in hours = \(\frac{225}{6000}\)
Since the difference between two travel times = 2.25 minutes
Hence it can be written as,
\(\frac{D}{12} - \frac{D}{30} = \frac{225}{6000}\) ( \(Time =\frac{Distance}{speed}\))
or \(30D -12D = \frac{{225 * 12 * 30}}{{6000}}\)
or D = \(\frac{{225 * 12 * 30}}{{6000 * 18}}\)
or D = \(\frac{3}{4}\)
