Question

a fire engine starts pumping water at 10:30 am at the rate af 1000 gados per minute. another fire engine pumping at the rate of 1200 gedoe per mote, stets at 10:40 am at what time will the two engines hove pumped the same number of gal 010.50 011-20 01:00 011:40 01130

Answer 1

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Answer 2

At 11.30 am both engines will pamp same number of water

Given:

A fire engine starts pumping water at 10:30 am at the rate of 1000 gallons per minute.

Another fire engine pumping  started at 10:40 am at the rate of 1200 gallons per minute.

To find:

At what time will the two engines hove pumped the same number of gallons of water

Solution:

Let's assume that after t minutes both engines have pamped same number of water

The amount of water pumped by First engine at rate of 1000 gallons per minute in t minutes = 1000 × t = 1000t gallons

Here, the 2nd engine is started at 10.40 after 10 mins of 1st engine so we take t as (t-10) mins

The amount of water pumped by 2nd engine at rate of 1200 gallons per minute in (t-10) minutes = 1000 × (t-10) = 1200(t-1) gallons  

As we know after t minutes both engines have pamped same number of water  ⇒ 1000t = 1200(t-10)

⇒ 1000t = 1200t - 1200

⇒ 1200t - 1000t = 1200

⇒ 200t = 12000

⇒ 2t = 120

⇒ t = 60

After 60 mins both engines will pamp same number of water

[ which means after 1 hr ]

At 11.30 am both engines will pamp same number of water

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