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On Monday, Mohsin increased the speed of car by 25% as compared to his usual speed and reached office 16 minutes earlier. On Tuesday, Mohsin took a longer route to the office and travelled with his usual speed. If on Tuesday, he reached office 20 minutes late, then how many percent more did Mohsin travel to reach office?
  • a)
    30%
  • b)
    20%
  • c)
    25%
  • d)
    15%
  • e)
    Cannot be determined
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
On Monday, Mohsin increased the speed of car by 25% as compared to his...
Let the usual speed of Mohsin be x.
∴ Mohsin’s speed on Monday = 1.25x
Assume that usually Mohsin takes t minutes to reach the office.
∴  Time taken on M onday = t - 16
∴ Distance tra velled = xt =1.25x (t - 16) Solving this, we get
t = 80 minutes
On Tuesday, Mohsin reached office 20 minutes late, travelling with his usual speed,
i.e., he travelled for (20/80 =) 25% more.
As the speed with which he travelled was same, distance a time.
As time taken is 25% more, the distance travelled will also be 25% more.
Hence, option 3.
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Most Upvoted Answer
On Monday, Mohsin increased the speed of car by 25% as compared to his...
Let's assume Mohsin's usual speed is S km/h.

On Monday:
- Mohsin increased the speed of the car by 25%. So, his new speed was 1.25S km/h.
- As a result, he reached the office 16 minutes earlier than usual.

On Tuesday:
- Mohsin took a longer route to the office, but he traveled with his usual speed of S km/h.
- However, he reached the office 20 minutes late.

To find the percentage by which Mohsin traveled more on Tuesday, we need to compare the distances traveled on both days.

Let's assume the distance between Mohsin's home and office is D km.

On Monday:
- Mohsin traveled D km at a speed of 1.25S km/h.
- The time taken can be calculated using the formula: Time = Distance / Speed.
- So, the time taken on Monday was D / (1.25S) hours.

On Tuesday:
- Mohsin traveled the same distance D km at a speed of S km/h.
- The time taken on Tuesday was D / S hours.

According to the given information:
- Mohsin reached the office 16 minutes earlier on Monday, which means he saved 16 minutes compared to his usual travel time.
- On Tuesday, he reached the office 20 minutes late, which means he took 20 minutes more than his usual travel time.

Calculating the time difference:

16 minutes = D / S - D / (1.25S) hours
20 minutes = D / (1.25S) - D / S hours

Simplifying these equations, we get:

16 / 60 = D / S - D / (1.25S)
20 / 60 = D / (1.25S) - D / S

Solving these equations, we find:

16 / 60 = 0.25D / (1.25S)
20 / 60 = 0.25D / S

Simplifying further, we get:

4 / 15 = 1 / 5
1 / 3 = 1 / 5

Since the equations are inconsistent, we cannot determine the value of D or S. Therefore, the answer cannot be determined.

Hence, the correct answer is option 'e) Cannot be determined'.
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Community Answer
On Monday, Mohsin increased the speed of car by 25% as compared to his...
A)
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On Monday, Mohsin increased the speed of car by 25% as compared to his usual speed and reached office 16 minutes earlier. On Tuesday, Mohsin took a longer route to the office and travelled with his usual speed. If on Tuesday, he reached office 20 minutes late, then how many percent more did Mohsin travel to reach office?a)30%b)20%c)25%d)15%e)Cannot be determinedCorrect answer is option 'C'. Can you explain this answer?
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