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26 Mar 2020, 03:22
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
87% (02:01) correct
13%
(02:21)
wrong
based on 269
sessions
History
Date
Time
Result
Not Attempted Yet
Two cyclists start from the same place to ride in the same direction. A starts at noon at 8 kmph while B starts at 2 pm at the rate of 10 kmph. How far will A have ridden before he is overtaken by B?
A. 75 km
B. 76 km
C. 78 km
D. 80 km
E. 84 km
Are You Up For the Challenge: 700 Level Questions
A. 75 km
B. 76 km
C. 78 km
D. 80 km
E. 84 km
Are You Up For the Challenge: 700 Level Questions
26 Mar 2020, 03:34
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Bunuel
Let the time taken by A = \(t\) hours
--> Time taken by B = \(t - 2\) hours
When A meets B, Distance traveled by both is same.
--> \(D_A = D_B\)
--> \(8*t = 10*(t - 2)\)
--> \(8t = 10t - 20\)
--> \(2t = 20\)
--> \(t = 10\) hours
Distance traveled by A, \(D_A = 8*10 = 80\) kms
Option D
26 Mar 2020, 10:44
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Solution
Given
In this question, we are given that
- • Two cyclists start from the same place to ride in the same direction.
• A starts at noon at 8 kmph while B starts at 2 pm at the rate of 10 kmph
To find
We need to determine
- • The distance A has travelled before he is overtaken by B
Approach and Working out
When B starts journey, A has already travelled for 2 hours, and a distance of 8 x 2 = 16 km
- • Hence, from 2 pm, the time taken by B to catch A = \(\frac{16}{10 – 8}\) hrs = 8 hrs
Therefore, A travels a total of 2 + 8 = 10 hours before B catches and overtakes A
- • So, total distance travelled by A till that time = 10 x 8 = 80 km
Thus, option D is the correct answer.
Correct Answer: Option D
