# Video: CBSE Class X • Pack 4 • 2015 • Question 23

CBSE Class X • Pack 4 • 2015 • Question 23

05:05

### Video Transcript

A train travelled at a certain average speed for a distance of 54 kilometres and then travelled a distance of 63 kilometres at an average speed of six kilometres per hour more than the initial average speed. If the train took three hours to complete the total journey, what was the initial average speed?

Let’s start by calling the initial average speed of the train 𝑢. And now, we can use the equation for average speed, which is that average speed is equal to total distance over total time and then rearrange this for the total time to get that the total time is equal to total distance over average speed.

And now, we have that for the first part of the journey the train travelled a distance of 54 kilometres with an average speed of 𝑢. Now, if we call the time for the first part of the journey 𝑇 one, then we have that 𝑇 one is equal to the distance travelled which is 54 kilometres over the initial average speed which is 𝑢 kilometres per hour.

Now, we can also find 𝑇 two, which is the time for the second part of the journey. And this will again be equal to the distance over the average speed. And we have in the question that the distance was 63 kilometres and that the average speed were six kilometres per hour more than the initial average speed. And so its average speed was 𝑢 plus six kilometres per hour.

And now, the time for the total journey, which we can call 𝑇, is simply the sum for the time of the first part of the journey 𝑇 one plus the time for the second part of the journey 𝑇 two. And we can substitute in what we found for 𝑇 one and 𝑇 two to obtain that 𝑇 is equal to 54 over 𝑢 plus 63 over 𝑢 plus six.

However, we’re also told in the question that it took the train three hours to complete the whole journey. And so we have that 𝑇 is also equal to three. And so now, we have formed an equation involving only 𝑢, which we can solve in order to find 𝑢. Now, in this equation, we have 𝑢 and 𝑢 plus six in the denominators of the fractions on the left.

In order to move these out of the denominators, we need to multiply both sides of the equation by 𝑢 and 𝑢 plus six. So let’s multiply term by term. If we multiply the first terms, that’s 54 over 𝑢 by 𝑢, we simply get 54 and then we have to multiply it by 𝑢 plus six. And so we get 54 lots of 𝑢 plus six. And then, the second term 63 over 𝑢 plus six: if we multiply that by 𝑢 plus six, we simply get 63 and then we also have to multiply it by 𝑢. And so we obtain 63𝑢. Then, on the right-hand side of the equation, we have three which we’re multiplying by 𝑢 and 𝑢 plus six. And so we obtain three 𝑢 lots of 𝑢 plus six.

Now, let’s expand the brackets. Expanding the first bracket, we get 54 𝑢 plus six times 54, which is simply 324. Then, we add the 63𝑢. And this is all equal to three 𝑢 times 𝑢 which is three 𝑢 squared plus three 𝑢 times six which gives us 18𝑢. Then, on the left-hand side of the equation, we can combine the 63𝑢 and the 54𝑢 to obtain 117𝑢.

Next, we want to move all of the terms onto one side of the equation. And so we can subtract 117𝑢 plus 324 from both sides of the equation. This leaves us with zero is equal to three 𝑢 squared plus 18𝑢 minus 117𝑢 minus 324. And now, we can combine the 𝑢 terms and flip the equation to the other side to leave us with the equation three 𝑢 squared minus 99𝑢 minus 324 equals zero.

And now, we can spot that each term here is a multiple of three. So we can divide the whole equation by three, leaving us with the equation 𝑢 squared minus 33𝑢 minus 108 is equal to zero. And now that we have the equation in this form, we simply need to factorize it.

In order to do this, we need to find factor pairs of negative 108 that sum to give negative 33. And we find that the factor pair of three and negative 36 sum to give negative 33. And so now, we can factorize our equation. And we obtain that 𝑢 plus three times 𝑢 minus 36 is equal to zero. And this means that either 𝑢 plus three equals zero which gives us a speed of 𝑢 equals negative three kilometres per hour or 𝑢 minus 36 equals zero which gives a speed of 𝑢 equals 36 kilometres per hour.

Now, since an average speed has to be positive, the answer of 𝑢 equals negative three kilometres per hour must be ignored. And so therefore, we find that the initial average speed of the train was 36 kilometres per hour.