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A bus was travelling from Mumbai to Pune was delayed by 16 minutes and made up for the delay on a section of 80 km travelling with a speed 10 km per hour higher than its normal speed. Find the original speed of the bus?
- a)60 km/h
- b)66.66 km/h
- c)50 km/h
- d)40 km/h
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A bus was travelling from Mumbai to Pune was delayed by 16 minutes and...
80/x – 80/(x+10) = 16/60= 4/15
300/x -300/x+10 =1
X2 + 10x – 3000 = 0
X = -60, +50
So 50km/hr
300/x -300/x+10 =1
X2 + 10x – 3000 = 0
X = -60, +50
So 50km/hr
Most Upvoted Answer
A bus was travelling from Mumbai to Pune was delayed by 16 minutes and...
Given:
- The bus was delayed by 16 minutes.
- The bus made up for the delay on a section of 80 km.
- The bus traveled on this section at a speed 10 km/h higher than its normal speed.
To find:
The original speed of the bus.
Let's assume the original speed of the bus is 'x' km/h.
Finding the time taken to cover the section of 80 km:
- With the original speed, the time taken to cover 80 km is given by the formula: time = distance/speed.
- So, the time taken to cover 80 km at the original speed 'x' km/h is 80/x hours.
Finding the time taken to cover the same section at the increased speed:
- With the increased speed of (x + 10) km/h, the time taken to cover 80 km is given by the formula: time = distance/speed.
- So, the time taken to cover 80 km at the increased speed of (x + 10) km/h is 80/(x + 10) hours.
Since the bus made up for the delay on this section, the time taken at the increased speed should be equal to the original time plus the delay time of 16 minutes.
Converting 16 minutes to hours:
- 16 minutes is equal to 16/60 hours, which simplifies to 4/15 hours.
Equating the two time expressions:
80/x = 80/(x + 10) + 4/15
Solving the equation:
Multiplying through by 15x(x + 10) to eliminate the fractions, we get:
15(80)(x + 10) = 15(80x) + 4x(x + 10)
Simplifying the equation:
1200x + 12000 = 1200x + 4x^2 + 40x
Rearranging the equation:
4x^2 + 40x - 1200x - 40x - 12000 = 0
Simplifying further:
4x^2 - 12000 = 0
Dividing through by 4:
x^2 - 3000 = 0
Factoring:
(x - 50)(x + 60) = 0
So, x = 50 or x = -60.
Since the speed of the bus cannot be negative, the original speed of the bus is 50 km/h.
Therefore, the correct answer is option C) 50 km/h.
- The bus was delayed by 16 minutes.
- The bus made up for the delay on a section of 80 km.
- The bus traveled on this section at a speed 10 km/h higher than its normal speed.
To find:
The original speed of the bus.
Let's assume the original speed of the bus is 'x' km/h.
Finding the time taken to cover the section of 80 km:
- With the original speed, the time taken to cover 80 km is given by the formula: time = distance/speed.
- So, the time taken to cover 80 km at the original speed 'x' km/h is 80/x hours.
Finding the time taken to cover the same section at the increased speed:
- With the increased speed of (x + 10) km/h, the time taken to cover 80 km is given by the formula: time = distance/speed.
- So, the time taken to cover 80 km at the increased speed of (x + 10) km/h is 80/(x + 10) hours.
Since the bus made up for the delay on this section, the time taken at the increased speed should be equal to the original time plus the delay time of 16 minutes.
Converting 16 minutes to hours:
- 16 minutes is equal to 16/60 hours, which simplifies to 4/15 hours.
Equating the two time expressions:
80/x = 80/(x + 10) + 4/15
Solving the equation:
Multiplying through by 15x(x + 10) to eliminate the fractions, we get:
15(80)(x + 10) = 15(80x) + 4x(x + 10)
Simplifying the equation:
1200x + 12000 = 1200x + 4x^2 + 40x
Rearranging the equation:
4x^2 + 40x - 1200x - 40x - 12000 = 0
Simplifying further:
4x^2 - 12000 = 0
Dividing through by 4:
x^2 - 3000 = 0
Factoring:
(x - 50)(x + 60) = 0
So, x = 50 or x = -60.
Since the speed of the bus cannot be negative, the original speed of the bus is 50 km/h.
Therefore, the correct answer is option C) 50 km/h.
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A bus was travelling from Mumbai to Pune was delayed by 16 minutes and made up for the delay on a section of 80 km travelling with a speed 10 km per hour higher than its normal speed. Find the original speed of the bus?a)60 km/hb)66.66 km/hc)50 km/hd)40 km/he)None of theseCorrect answer is option 'C'. Can you explain this answer?
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