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A truck after covering 1 / 6th of the total distance decreases its speed by 2 / 3 of its usual speed and reaches destination 1 hr 40 minutes late. Find the usual time taken (in hours) by the truck to cover the complete distance?
  • a)
    4
  • b)
    3
  • c)
    2
  • d)
    1
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A truck after covering 1 / 6th of the total distance decreases its sp...
Let's assume the total distance to be D km and the usual speed of the truck to be S km/hr.

**Distance covered in the first 1/6th of the total distance:**
The distance covered in the first 1/6th of the total distance = (1/6) * D = D/6 km.

**Time taken to cover the first 1/6th of the total distance:**
As the truck is traveling at its usual speed, the time taken to cover the first 1/6th of the total distance = (D/6) / S = D/6S hr.

**Distance remaining after covering the first 1/6th of the total distance:**
The distance remaining after covering the first 1/6th of the total distance = D - D/6 = (5/6)D km.

**Decreased speed of the truck:**
The truck decreases its speed by 2/3 of its usual speed. Therefore, the decreased speed of the truck = S - (2/3)S = (1/3)S km/hr.

**Time taken to cover the remaining distance with decreased speed:**
The time taken to cover the remaining distance with decreased speed = (5/6)D / ((1/3)S) = (5/6) * (3/1) * D/S = (5/2) * D/S hr.

**Total time taken to cover the complete distance:**
The total time taken to cover the complete distance = Time taken to cover the first 1/6th of the total distance + Time taken to cover the remaining distance with decreased speed
= D/6S + (5/2) * D/S
= (D/6S) + (5D/2S)
= (D + (15D/2)) / (6S)
= (17D/2) / (6S)
= (17D/12S) hr.

Given that the truck reaches the destination 1 hr 40 minutes late, which is equal to 1 + (40/60) = 1.67 hours.

Therefore, (17D/12S) = (17D/12S) + 1.67
=> 17D/12S - 17D/12S = 1.67
=> 0 = 1.67

This implies that the equation does not hold true, which means our assumption about the usual speed of the truck is incorrect.

Since the equation does not hold true for any value of D and S, it means that there is no valid solution to this problem. Hence, the given information is inconsistent, and we cannot determine the usual time taken by the truck to cover the complete distance.
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Community Answer
A truck after covering 1 / 6th of the total distance decreases its sp...
Let the total distance be 6x and the usual speed is v.
The actual time taken by the truck ⇒ (x / v) + [5x / (v / 3)] = 16x / v
The usual time the truck would have taken ⇒ 6x / v
⇒ Difference in time = 16x / v - 6x / v = 5 / 3 [1 hr 40 minutes = 5 / 3 hour]
⇒ x / v = 1 / 6
The usual time is taken by the truck to cover the complete distance = 6x / v = 6 × 1 / 6 = 1
∴ the usual time taken by the truck to cover the complete distance is 1 hour.
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A truck after covering 1 / 6th of the total distance decreases its speed by 2 / 3 of its usual speed and reaches destination 1 hr 40 minutes late. Find the usual time taken (in hours) by the truck to cover the complete distance?a)4b)3c)2d)1Correct answer is option 'D'. Can you explain this answer?
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