Grade 9 : Math : Word Problem

Water is being pumped into a 10-foot-tall cylindrical tank at a constant rate.

• The depth of the water is increasing linearly.
• At 1:30pm, the water depth was 2.4 feet.
• It is now 4:00pm and the depth of the water is 3.9 feet.

What will the depth (in feet) of the water be at 5:00 pm?

Solution

First lets write the given statement in points ( time, water depth in feet)

Given,

At 1:30pm, the water depth was 2.4 feet

which means (1.5 , 2.4 )

 [ 1.30 pm which means 1 1/2 so, we are writing it as 1.5]

also given, 

It is now 4:00pm and the depth of the water is 3.9 feet.

that is ( 4 , 3.9 )

Now we got two points (x1,y1) = (1.5,2.4)

                                  (x2,y2) = (4,3.9)

Slope (m)= y2 - y1 / x2 - x1

               = 3.9 - 2.4 / 4 - 1.5

               = 1.5 / 2.5

                    m  = 0.60

Every hour water level increases 0.60 feet

we have to find the depth (in feet) of the water be at 5:00 pm

we found m=0.60

lets take one of the given point (1.5,2.4) (x1,y1)

we have to find y when x=5(5pm)

y - y1 = m(x-x1)

y - 2.4 = 0.60(5-1.5)

y - 2.4 = 0.60(3.5)

y-2.4 = 2.1

y = 2.4 + 2.1

    = 4.5 feet

Hence the depth (in feet) of the water be at 5:00 pm is 4.5ft