Video: CBSE Class X • Pack 1 • 2018 • Question 18

CBSE Class X • Pack 1 • 2018 • Question 18

06:09

Video Transcript

A plane left 30 minutes after the scheduled departure time. And in order to reach its destination 1500 kilometers away on time, it had to increase its speed by 100 kilometers per hour from the usual speed. Find its usual speed.

So essentially, we have this plane that left late. So it had to increase its speed so that it would still arrive on time. So we know a few things: the plane left 30 minutes late, the destination is 1500 kilometers away, and in order to reach that destination, the plane had to increase its speed by 100 kilometers per hour and that’s from the usual speed. So we need to find the usual speed, what it normally would have went.

We can use the formula that says time is equal to the distance divided by the speed. So out of those three things, let’s see what we know. We know that the total distance is 1500 kilometers. We don’t know the usual speed of the plane. So we’ll let it be called 𝑥 kilometers per hour. But we know that the new speed needed to be quicker than the usual speed because it left late. It increased its speed by 100 kilometers per hour. So we will let it be represented by 𝑥 plus 100.

So now, let’s talk about time. We know that time is equal to the distance divided by the speed. So if we want to know the time where the plane used its usual speed, it would be the distance of 1500 kilometers divided by its usual speed, which we called 𝑥, which is in kilometers per hour. So the kilometers would cancel and our time using the usual speed would be 1500 divided by 𝑥 hours.

So the time it would take if we use the new speed — the speed that had been increased — would be the same distance. But the speed was increased by 100. So it’s 1500 divided by 𝑥 plus 100. And that will also be in hours.

So there’s another detail that we know about time. We know that it arrived on time, but it left 30 minutes late. So the distances were the same, but the time and speed had to be a little bit different.

Our goal is to solve for 𝑥 to find the usual speed. So if this is the usual time, 1500 divided by 𝑥, in order to make the other side of the equation be the exact same amount of time, we would have to take our time with the new speed, we would need to add one-half.

Now, it may seem, “why wouldn’t we subtract if it left 30 minutes late?” Well, the left-hand side of the equation is the amount of time it normally takes — that was with the usual speed — and over here was with the new speed: it went faster. So it left 30 minutes later. So we would need to take that fraction and add 30 minutes — add the one-half hour. And it would be the same amount of time that would have normally taken.

So now, we’ll need to solve for 𝑥, the usual speed. And in order to solve, let’s go ahead and get the 𝑥s all on one side of the equation. So let’s subtract the 1500 divided by 𝑥 plus 100 to both sides of the equation so it’s on the left-hand side.

Now in order to solve, we need to get a common denominator and move all of the 𝑥s up to the top. So our denominator will need an 𝑥, a two, and an 𝑥 plus 100. Now, this isn’t the only way to solve this equation, but it eliminates the denominators.

So before we begin multiplying, we can cancel a little bit. So the 𝑥s cancel with the first fraction, the 𝑥 plus 100 cancel with this fraction, and the twos cancel with this fraction. So now, we can multiply what’s left: 1500 times two is 3000. And now, we need to take 3000 times 𝑥 and then times 100. And we will move our work up here.

So 3000 times 𝑥 is 3000𝑥. 3000 times 100 is 300000. And we do need to be careful with this minus sign. So we have 1500 times two which is 3000 times 𝑥. So we have 3000𝑥. But we need to make sure that we subtract 3000𝑥. And on the right-hand side of the equation, we have 𝑥 times 𝑥 which is 𝑥 squared and 𝑥 times 100 which is 100𝑥. So now to solve, the 3000𝑥s cancel. So let’s go ahead and subtract the 300000 to both sides of the equation so everything is on the right-hand side.

So we have zero equals 𝑥 squared plus 100𝑥 plus 300000. So we need to find two numbers that add to be 100, but multiply to be 300000. That would be 600 and 500. Those multiply to be 300000, but they don’t add to be 100. So we need to make one of them negative. If we need positive 100, we would need the 600 to be positive. So we need positive 600 and negative 500.

So now to solve for 𝑥, we should set each factor equal to zero. So we set 𝑥 plus 600 equal to zero, which when we subtract 600 from both sides of the equation, we get 𝑥 equals 600. And we also set 𝑥 minus 500 equal to zero. So we add 500 to both sides of the equation and we get 500.

So would the usual speed be negative 600 or positive 500? Well, a speed is never negative. So this means the usual speed of the plane would be 500 kilometers per hour. So if that plane had left 30 minutes after the scheduled departure time, it must have increased its speed by 100, which would be 600 kilometers per hour. So once again, the usual speed would be 500 kilometers per hour.

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