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Walking at 6/7th of his usual speed, a man is 12 minutes late. The usual time taken by him to cover that distance is-
- a)1 hr
- b)1 hr 12 min
- c)1 hr 15 min
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Walking at 6/7th of his usual speed, a man is 12 minutes late. The us...
S1 : S2 = 6 : 7
View all questions of this test
T1 : T2 = 7 : 6 (S ∝ 1/T)
Difference between
T1 & T2 = 1(7 - 6 = 1)
If 1 = 12 min.
then 6 = 72 min. 1 hrs 12 min
Most Upvoted Answer
Walking at 6/7th of his usual speed, a man is 12 minutes late. The us...
Problem Analysis:
Let's assume the usual speed of the man is S km/hr and the usual time taken by him to cover the distance is T hours. We need to find the value of T.
Solution:
Given that the man is walking at 6/7th of his usual speed. So, his speed becomes (6/7)S km/hr.
We know that Speed = Distance/Time.
Let the distance be D km.
So, (6/7)S = D/T1 ----(1) (where T1 is the time taken by the man when he is walking at 6/7th of his usual speed)
Given that the man is 12 minutes late. So, his usual time taken to cover the distance is T + 12/60 = T + 1/5 hours.
So, S = D/(T + 1/5) ----(2)
From equation (1), we have D = (6/7)S * T1.
Substituting this value of D in equation (2), we get
S = (6/7)S * T1 / (T + 1/5).
Simplifying the above equation, we get T1 = (5/6) * (T + 1/5).
Expanding and simplifying further, we get
T1 = (5T + 1) / 6.
Now, let's substitute this value of T1 in equation (1).
(6/7)S = D/(5T + 1)/6.
Simplifying the above equation, we get
(6/7)S = 6D/(5T + 1).
Simplifying further, we get
(6/7)S * (5T + 1) = 6D.
Simplifying and cancelling S from both sides, we get
(6/7) * (5T + 1) = 6.
Expanding and simplifying further, we get
(30T + 6)/7 = 6.
Multiplying both sides by 7, we get
30T + 6 = 42.
Subtracting 6 from both sides, we get
30T = 36.
Dividing both sides by 30, we get
T = 36/30.
Simplifying further, we get
T = 6/5 = 1.2 hours.
Therefore, the usual time taken by the man to cover the distance is 1 hour and 12 minutes.
Hence, the correct answer is option B) 1 hr 12 min.
Let's assume the usual speed of the man is S km/hr and the usual time taken by him to cover the distance is T hours. We need to find the value of T.
Solution:
Given that the man is walking at 6/7th of his usual speed. So, his speed becomes (6/7)S km/hr.
We know that Speed = Distance/Time.
Let the distance be D km.
So, (6/7)S = D/T1 ----(1) (where T1 is the time taken by the man when he is walking at 6/7th of his usual speed)
Given that the man is 12 minutes late. So, his usual time taken to cover the distance is T + 12/60 = T + 1/5 hours.
So, S = D/(T + 1/5) ----(2)
From equation (1), we have D = (6/7)S * T1.
Substituting this value of D in equation (2), we get
S = (6/7)S * T1 / (T + 1/5).
Simplifying the above equation, we get T1 = (5/6) * (T + 1/5).
Expanding and simplifying further, we get
T1 = (5T + 1) / 6.
Now, let's substitute this value of T1 in equation (1).
(6/7)S = D/(5T + 1)/6.
Simplifying the above equation, we get
(6/7)S = 6D/(5T + 1).
Simplifying further, we get
(6/7)S * (5T + 1) = 6D.
Simplifying and cancelling S from both sides, we get
(6/7) * (5T + 1) = 6.
Expanding and simplifying further, we get
(30T + 6)/7 = 6.
Multiplying both sides by 7, we get
30T + 6 = 42.
Subtracting 6 from both sides, we get
30T = 36.
Dividing both sides by 30, we get
T = 36/30.
Simplifying further, we get
T = 6/5 = 1.2 hours.
Therefore, the usual time taken by the man to cover the distance is 1 hour and 12 minutes.
Hence, the correct answer is option B) 1 hr 12 min.
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