Question

two racer run at a speed of 100m/min and 120m/n.lf second takes 10 min less than first. find total distance they hve to cuvr? ​

Answer 1

Given

Speed of 1st racer = 100 m/minute

Speed of 2nd racer = 120 m/minute

2nd racer takes 10 minute less than 1st racer.

To Find

Total distance they have to cover

Solution

As here, both racers will cover same distance. So distance covered by 1st racer = distance covered by 2nd racer.

Here, let the time taken by 1st racer be y m = 60y second & time taken by 2nd racer be (y - 10) m = (60y - 10 × 60) second = (60y - 600) seconds.

Again, speed of 1st racer = 100 m/minute

➻ Speed of 1st racer = 100/60 m/s

➻ Speed of 1st racer = 5/3 m/s

Speed of 2nd racer = 120 m/minute

➻ Speed of 2nd racer = 120/60 m/s

➻ Speed of 2nd racer = 2 m/s

We know that,

➻ Distance = Speed × Time

According to Question :

➻ 5/3(60y) = 2(60y - 600)

➻ 100y = 120y - 1200

➻ 100y - 120y = -1200

➻ -20y = -1200

➻ y = -1200/-20

➻ y = 60 seconds.

Now putting this value in any equation :

➻ Distance = 5/3[60(60)]

➻ Distance = 5/3(3600)

➻ Distance = 5 × 1200

➻ Distance = 6000 m

Tʜᴇʀᴇꜰᴏʀᴇ,

Distance covered by both racers = 6000 m

Answer 2

Answer:

Let the Fixed Distance be D metre.

$\underline{\bigstar\:\textbf{According to the Question :}}$

$:\implies\sf Racer_2=Racer_1-10\:min\\\\\\:\implies\sf Time_1-Time_2=10\:min\\\\\\:\implies\sf \dfrac{Distance}{Speed_1}-\dfrac{Distance}{Speed_2}=10\:min\\\\\\:\implies\sf \dfrac{D}{100\:m/min}-\dfrac{D}{120\:m/min}=10 \:min\\\\\\:\implies\sf \dfrac{6D - 5D}{600 \:m/min} = 10 \:min\\\\\\:\implies\sf \dfrac{D}{600\:m/min} = 10\:min\\\\\\:\implies\sf D = 600 \:m/min \times 10\:min\\\\\\:\implies\sf D = 600\:m \times 10\\\\\\:\implies\underline{\boxed{\sf D = 6000\:m}}$

⠀

$\therefore\:\underline{\textsf{They've to cover distance of \textbf{6000 m or, 6 km}}}.$