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Please refer to the graph regarding the relation between spee 15 Feb 2020, 19:11
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C
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GMATBusters’ Quant Quiz Question -3





Please refer to the graph regarding the relation between speed and distance for a journey undertaken by Bob. The average speed (km/hr) for the first 150 km is:
A 12
B 20.5
C 22.5
D 25
E None of the above
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C
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Please refer to the graph regarding the relation between spee 15 Feb 2020, 19:21
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Average speed=150/(100/20)+(50/30)
=150/(5+5/3)
=150×3/20
=45/2
=22.5

C is the answer

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Please refer to the graph regarding the relation between speed and distance for a journey undertaken by Bob. The average speed (km/hr) for the first 150 km is:
A 12
B 20.5
C 22.5
D 25
E None of the above

For the first 100 kms Rate is 20 kmph -> Time taken for first 100 kms is 5 hrs

For the next 50 kms Rate is 30 kmph -> Time taken for next 50 kms is \(\frac{5}{3}\) hrs

Average for 150 kms is
\(\frac{150}{(5+[m][fraction]5/3}\))[/fraction][/m] = 22.5 hrs -> Answer - C
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Please refer to the graph regarding the relation between spee Updated on: 16 Feb 2020, 06:42
Originally posted by KhulanE on 15 Feb 2020, 20:54.
Last edited by KhulanE on 16 Feb 2020, 06:42, edited 2 times in total.
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Total time to travel 150miles 100/20+50/30=6.66
Average speed 150/6.66=22.5
so C is right
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Please refer to the graph regarding the relation between spee 15 Feb 2020, 20:56
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Please refer to the graph regarding the relation between speed and distance for a journey undertaken by Bob. The average speed (km/hr) for the first 150 km is:
A 12
B 20.5
C 22.5
D 25
E None of the above

0 to 100 km, t1 = 100/20 = 5
100 to 150 km, t2 = 50/30 = 1.66

Average speed = total distance/total time = 150 / 6.66 = 22.5

So, C is correct.
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Please refer to the graph regarding the relation between spee 15 Feb 2020, 20:59
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The equation for average speed is (total distance)/(total time)
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Please refer to the graph regarding the relation between spee 15 Feb 2020, 21:29
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time taken for first 100 km =\( \frac{100}{20 }\)= 5hrs
time taken to cover next 50 km = \(\frac{50}{30} = \frac{5}{3}\) hrs
avg speed
total distance /total time
= \(\frac{150}{(5+5/3)}\)
\(\frac{45}{2}\)
= 22.5
C
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total distance = 150 km

bob covered first 100 km with speed = 20 km/hr
time taken to cover 100 km = \(\frac{100}{20 }\)= 5 hours

bob covered next 50 km with speed = 30 km/hr
time taken to cover next 50 km = \(\frac{50}{30}\) = 5/3

time taken to cover 150 km = \(\frac{20}{3}\) hours

average speed = \(\frac{total-distance-covered}{total-time-taken}\) = \(\frac{150}{20/3 }\) = \(\frac{45}{2}\) = 22.5

C is the answer
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T= 100/20 + 50/30 = 20/3
S=150
V=s/t = 450/20=22.5

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Average speed = total distance / total time

total distance = 150

total time time = time to cover (100km with speed of 20kmph + 50km with 30kmph)
= 5+ (5/3)

Average speed = 150/(5+ (5/3)) = 90/4 = 22.5kmph
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Solution:

Question 3: Please refer to the graph regarding the relation between speed and distance for a journey undertaken by Bob. The average speed (km/hr) for the first 150 km is:

A 12
B 20.5
C 22.5
D 25
E None of the above

In this question, we need to find the average speed for first 150 km shown on X-axis. The formula for average speed is \frac{{Total distance} }{ {Total time}}.

Here, total distance = 150 Km and total time can be calculated by considering the two intervals i.e., distance of 100 km at speed of 20 Km/hour and interval of 100-150 km at speed of 30 Km/hour. Thus, total time = \(\frac{100}{20} + \frac{50}{30}\) = \(5 + \frac{5}{3}\) = \(\frac{20}{3}\)

Therefore, Average speed = \(\frac{150 }{ (20/3)}\) = 22.5 Km/hour, which is option C.

Option C is the correct answer.
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Solution:

Let's see: Avg speed for a particular distance is : total distance/total time taken.

So lets break the speed to required times till 150 kms is completed.

speed during the 1st 50kms = 20km/hr, so time taken : 5/2
speed during the next 50kms = 20km/hr, so time taken : 5/2
speed during the last 50kms= 30km/hr, so time taken is : 5/3

so total time taken: 20/3hr.

So the avg speed for the 1st 150 kms will be : 150/(20/3) = 22.5km/hr
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