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10 Nov 2021, 08:48
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
86% (02:14) correct
14%
(02:31)
wrong
based on 222
sessions
History
Date
Time
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Not Attempted Yet
Mr. Nadela goes for a run every morning and he runs 6 km in one hour after which he comes back to the starting point along the same route at a particular constant speed. If his average speed during the whole trip was 10 km/hr, what was his speed during the return journey?
A. 30 km/hr
B. 32 km/hr
C. 34 km/hr
D. 35 km/hr
E. 36 km/hr
A. 30 km/hr
B. 32 km/hr
C. 34 km/hr
D. 35 km/hr
E. 36 km/hr
11 Nov 2021, 10:09
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Here is the OE
Solution:
Step 1: Understand Question Statement
Nadela runs a distance of 6 km in 1 hour.
He returns along the same route.
Average speed = 10 kmph
We need to find his speed while returning.
Step 2: Define Methodology
Speed of Nadela while running in morning = \({D}/{S} \) = \({6}/{1}\) = 6 kmph
Let’s assume, speed while returning = “R” kmph.
Since the to and fro distance in the same.
Average speed can be found directly by using the formula: \({2ab}/{a+b} \) , where a and b are the two speeds at which he is running.
Step 3: Calculate the final answer
Average Speed = \({(2*6*R)}/{(6+R)} \)
⟹ 10 = \({12R}/{(6+R)} \)
⟹ 60 + 10R = 12R
⟹ R = 30 kmph
Thus, the correct answer is Option A
Solution:
Step 1: Understand Question Statement
Nadela runs a distance of 6 km in 1 hour.
He returns along the same route.
Average speed = 10 kmph
We need to find his speed while returning.
Step 2: Define Methodology
Speed of Nadela while running in morning = \({D}/{S} \) = \({6}/{1}\) = 6 kmph
Let’s assume, speed while returning = “R” kmph.
Since the to and fro distance in the same.
Average speed can be found directly by using the formula: \({2ab}/{a+b} \) , where a and b are the two speeds at which he is running.
Step 3: Calculate the final answer
Average Speed = \({(2*6*R)}/{(6+R)} \)
⟹ 10 = \({12R}/{(6+R)} \)
⟹ 60 + 10R = 12R
⟹ R = 30 kmph
Thus, the correct answer is Option A
11 Nov 2021, 17:26
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ktzsikka
In problems involving average speeds, the general average-speed formula tends to be surprisingly effective. The formula for average speed is
\(average speed = \frac{total distance}{total time}\)
In some average speed problems no distance is specified and it's a good idea to make up a distance. In this case, however, the total distance is specified in a roundabout way: because Mr. Nadela runs at 6 km/hr for 1 hour before turning around and returning along the same route, his one-way distance is 6 km and his total distance is 12 km.
We're not given Mr. Nadela's total time, though we know that the time for his outbound journey was 1 hour. We also know that his average speed for the entire trip was 10 km/hr. As such, we can put some values into the formula and start to isolate the unknown:
\(10 = \frac{12}{1+t_r}\), where \(t_r\) is the time for Mr. Nadela's return journey
\(10+10t_r = 12\)
\(10t_r = 2\)
\(t_r = \frac{2}{10}\), or \(\frac{1}{5}\)
Recall that \(t_r\) is the time in hours of Mr. Nadela's return journey. We can now use this in a straightforward application of the \(d = r*t\) formula to find his return speed:
\(6 = r*\frac{1}{5}\)
Multiplying both sides by 5, we're left with \(30 = r\), or answer A.
As a general note, the advantages of using the average-speed formula in problems involving average speed are threefold:
1) it breaks the problem into smaller, more targeted steps (finding the individual distances and individual times to populate the formula), and
2) it is suited to handle round-trip journeys, three-leg journeys, and more (that is, it's more versatile than the shortcut formula).
Finally, remember that average speed is rarely going to be calculated as the arithmetic mean of any speeds given in a problem. Such answers are almost always traps!
