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A car travelling with of ( 5 ) /(7 ) it...

A car travelling with of ` ( 5 ) /(7 ) `its usual speed covers 42 km in 1 hour 40 min 48 sec. What is the usual speed of the car?

A

` 17 ( 6)/(7) `km/hr

B

` 25 km//hr`

C

` 30 km//hr`

D

` 35 km//hr`

Text Solution

AI Generated Solution

The correct Answer is:
To find the usual speed of the car, we will follow these steps: ### Step 1: Convert the time from hours, minutes, and seconds to seconds. The time given is 1 hour, 40 minutes, and 48 seconds. 1 hour = 3600 seconds (since 1 hour = 60 minutes and 1 minute = 60 seconds) 40 minutes = 2400 seconds (40 minutes * 60 seconds) 48 seconds = 48 seconds Now, we add these together: \[ \text{Total time in seconds} = 3600 + 2400 + 48 = 6048 \text{ seconds} \] ### Step 2: Set up the equation for speed. The speed of the car is given as \( \frac{5}{7} \) of its usual speed (let's denote the usual speed as \( x \)). The formula for speed is: \[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \] The distance covered by the car is 42 km, which is equal to 42000 meters (since 1 km = 1000 meters). Thus, we can write: \[ \frac{5}{7} x = \frac{42000 \text{ meters}}{6048 \text{ seconds}} \] ### Step 3: Solve for \( x \). To find \( x \), we can rearrange the equation: \[ x = \frac{42000 \times 7}{6048 \times 5} \] Now, let's calculate the right-hand side: 1. Calculate \( 42000 \times 7 = 294000 \). 2. Calculate \( 6048 \times 5 = 30240 \). Now substitute these values back into the equation: \[ x = \frac{294000}{30240} \] ### Step 4: Simplify the fraction. Now, we simplify \( \frac{294000}{30240} \): \[ x = 9.7 \text{ (approximately)} \] ### Step 5: Convert the speed from meters per second to kilometers per hour. To convert from meters per second to kilometers per hour, we multiply by \( \frac{18}{5} \): \[ \text{Usual speed in km/h} = 9.7 \times \frac{18}{5} \] Calculating this gives: \[ \text{Usual speed in km/h} = 9.7 \times 3.6 = 35 \text{ km/h} \] ### Conclusion: The usual speed of the car is **35 km/h**. ---
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S CHAND IIT JEE FOUNDATION-DISTANCE TIME AND SPEED -QUESTION BANK -14(a)
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