If 55.55% increased speed of car then find how much reduce it’s time


To solve this problem, we'll assume the initial speed of the car is \( v_1 \) and the initial time taken is \( t_1 \). When the speed is increased by 55.55%, the new speed \( v_2 \) becomes:

\[
v_2 = v_1 \times \left(1 + \frac{55.55}{100}\right) = v_1 \times 1.5555
\]

Time taken is inversely proportional to speed when the distance remains constant, so:

\[
t_2 = \frac{t_1 \times v_1}{v_2} = \frac{t_1 \times v_1}{1.5555 \times v_1} = \frac{t_1}{1.5555}
\]

Now, to find the reduction in time:

\[
\text{Reduction in time} = t_1 - t_2 = t_1 - \frac{t_1}{1.5555}
\]

\[
\text{Reduction in time} = t_1 \left(1 - \frac{1}{1.5555}\right)
\]

\[
\text{Reduction in time} = t_1 \times \frac{0.5555}{1.5555}
\]

Calculating the reduction factor:

\[
\frac{0.5555}{1.5555} \approx 0.3571
\]

So, the time is reduced by approximately 35.71%.

### Do you want details or have any questions?

Here are 5 related questions:
1. How would a 75% increase in speed affect the time taken for the journey?
2. What happens to the time if the speed is doubled?
3. If the initial time was 2 hours, what would be the new time after a 55.55% speed increase?
4. How does a 10% reduction in speed affect the time taken for a constant distance?
5. Can you derive the time reduction formula for a general percentage increase in speed?

**Tip:** Always remember that for a constant distance, time is inversely proportional to speed.

Learn how to calculate the reduction in time when a car's speed increases by 55.55%. Understand the relationship between speed, distance, and time, with a detailed mathematical solution. Suitable for students in Grades 9-12.

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