# Video: CBSE Class X • Pack 2 • 2017 • Question 18

CBSE Class X • Pack 2 • 2017 • Question 18

05:45

### Video Transcript

Water in a canal, 5.4 meters wide and 1.8 meters deep, is flowing with the speed of 25 kilometers per hour. How much area can it irrigate in 40 minutes if 10 centimeters of standing water is required for irrigation?

Let’s start out by sketching this canal. Looking at a cross section of the water as it flows in this canal, we know the canal is 5.4 meters wide and 1.8 meters deep. And along with that, the water flows with a speed we’ll call 𝑆, where 𝑆 is given as 25 kilometers every hour.

Knowing all this, we want to figure out how much area the water in this canal can irrigate in a time of 40 minutes. We can label this irrigated area 𝐴. And the problem statement tells us that for this area there is 10 centimeters of standing water required.

That means that our area sits on top of a depth of water of 10 centimeters. In other words, there’s some total volume of water 𝑉 which is equal to this area multiplied by a depth of 10 centimeters.

This takes us back to our canal, we’re given a cross-sectional area of the canal along with a flow rate and a time during which the water is flowing. We’re also able to calculate a total volume of water. We can consider it this way: in a time of 40 minutes with this flow rate of 25 kilometers an hour, some length of water in the canal — we’ve called it capital 𝐿 — will flow past this point.

We want to solve for 𝐿 because if we can, then we’ll be able to calculate the total volume of water that flows through the canal, which will help us solve for the area that’s irrigated. To start solving for 𝐿, let’s recall that an average speed 𝑆 is equal to a distance travelled divided by the time it took to travel that distance.

In our case, we can write that 𝑆, the flow rate of the water in the canal, is equal to 𝐿 divided by 𝑡, the time that the water is flowing. If we multiply both sides of this expression by the time 𝑡, we see that on the right-hand side that term cancels out of numerator and denominator.

This means that the 𝐿 that we want to calculate to solve for the total volume of water that flows 𝑉 is equal to our flow rate times the time that the water is flowing. That’s equal to 25 kilometers per hour multiplied by the time of 40 minutes. We see though that there’s an issue with combining these two values: the units aren’t consistent. One has time units of hours, but the other has time units of minutes.

To match them up, we can recall that one hour is equal to 60 minutes. When we make that substitution, we see that the units of time cancel out and the fraction 40 divided by 60 reduces to two-thirds. In other words, the length of water that flows along the canal in 40 minutes is equal to 25 kilometers times two-thirds.

There is one last change we’ll want to make to this number though before inserting it into our expression for the volume 𝑉. We can see that the other dimensions of this volume are given in units of meters. But our length currently is in units of kilometers. We can recall that one kilometer is equal to 1000 meters, which means that 𝐿 is equal to 25 kilometers times two-thirds or 25000 meters times two-thirds, which is equal to 50000 meters divided by three.

We can now take this value for 𝐿 and plug it into our equation for the volume 𝑉. We now have an expression for the total volume of water that flows through the canal in 40 minutes. And this is equal to the area that we’re able to irrigate multiplied by a depth of 10 centimeters.

Our next task then is to set these two equations equal to one another and solve for the area 𝐴. Looking at our combined equation, we may notice that there’s a units mismatch between the left side, which has units of meters, and the right side, which we see has units of centimeters. To correct this mismatch, we can recall that one meter is equal to 100 centimeters. This means we can replace 10 centimeters with one over 10 meters.

And now, in order to isolate the area 𝐴 that we want to solve for on one side of this equation, we’ll multiply both sides of the equation by 10 inverse meters. On the right side of our equation, this cancels out everything, except the area 𝐴. And on the left side of our equation, if we multiply by the factor of 10 and group all the units of meters, we find a result of 18 times 5.4 times 50000 over three square meters.

Now, all we need to do is calculate this left side and that will give us our area 𝐴 in units of square meters. To simplify this calculation, we can multiply 5.4 by 10, removing one of the zeros from the 50000, so that the value in parentheses is now 18 times 54 times 5000 over three.

One way to proceed in simplifying this number further is to realize that three divides evenly into 54. It goes into 54 18 times. Now, our value is 18 times 18 times 5000. We can now calculate what 18 squared is by hand. When we carry out this multiplication, we find a final result of 324. Next, we can expand the numbers in the parentheses from 324 times 5000 to 324 times five times 1000. Multiplying 324 by five, we find the result of 1620.

Looking at our expression in parentheses now, we count one, two, three, four zeros, which means we can we rewrite it as 162 times 10 to the fourth and that’s meters squared. This is the area that the canal could irrigate in a time of 40 minutes.