Video: GCSE Mathematics Foundation Tier Pack 3 • Paper 2 • Question 6

GCSE Mathematics Foundation Tier Pack 3 • Paper 2 • Question 6

04:25

Video Transcript

An airline company is selling tickets to fly from London to New York. The cost of two adult tickets and two child tickets is 1620 pounds. The cost of three adult tickets is 1650 pounds. Calculate the cost for one adult ticket and the cost for one child ticket.

So first of all, before we actually try to solve the problem, what I’m gonna say is that actually the cost of one adult ticket is gonna be 𝐴 and the cost of one child ticket is going to be 𝐶. So we can use our 𝐴 and our 𝐶 to set up a couple of equations.

The first equation we’re gonna set up is two 𝐴 plus two 𝐶 equals 1620 and that’s because the cost of two adult tickets and the cost of two child tickets is 1620 pounds. And the other equation we can set up is that three 𝐴 equals 1650 and that’s because the cost of three adult tickets is 1650 pounds.

Well, now, we can actually use this equation to actually work out the value of an adult ticket because that’s the only variable we have in this equation. Well, in order to find out how much one adult ticket is going to cost, what we’re gonna do is actually divide each side of our equation by three. And that’s because we have three adult tickets. So we know the cost of it and we want to find one adult ticket. So therefore, we divided by three.

And when we do that, we’re gonna get 𝐴 is equal to 550. And that’s because three 𝐴 divided by three is just 𝐴. And 1650 divided by three is 550. And we could work that out using the bus stop method or short division.

So if we see how many threes go into 1650, first of all how many threes go into one is zero. Then, we carry the one. So then, we’ve got how many threes is going into 16 which is five, carry the one. So we’ve got how many threes are in 15 which is five with no remainder. So then, finally, how many threes is going to zero? There are none with no remainder. So therefore, that’s how we got our 550.

Okay, great, so now, we know the cost of an adult ticket. So what we’re gonna do now is actually substitute our value for 𝐴, so the cost of an adult ticket, into our original equation. So when we do that, we’re gonna have two multiplied by 550. And that’s because we said that 𝐴 is equal to 550 plus two 𝐶 is equal to 1620.

So therefore, we can say that 1100 plus two 𝐶 is equal to 1620. We can work that out using column multiplication. So we’d have 550 multiplied by two. So we do two multiplied by zero is just zero. Then, we’d have two multiplied by five which would give us 10. So we put a zero in the tens column and carry the one into the hundreds column. And then finally, we’d have two multiplied by the five which will come from the 500. Again, this gives us 10 but plus the one we carried gives us 11. So we put one in the hundreds column and one in the thousands column. So therefore, we’ve got the result 1100.

So now, what we need to do is we’re trying to find out the cost of a child’s ticket. So we want to find out what 𝐶 is. So in that case, we’re gonna subtract 1100 away from each side of our equation. So then, when we do that, we get two 𝐶 is equal to 520.

So then as that’s the amount for two 𝐶, what we want to do is find out one 𝐶. So to do that, we divide each side of the equation by two, which gives us 𝐶 is equal to 260. And that’s because two 𝐶 divided by two is just 𝐶 and 520 divided by two is 260. We can think of that as 52 divided by two is 26. And then we’ve got a zero cause that’s was 520. And therefore, we have a result of 260 or we could use the bus stop method as we showed previously.

So therefore, we can say that if an airline company is selling tickets to fly from London to New York and the cost of two adult tickets and two child tickets is 1620 pounds, the cost of three adult tickets is 1650 pounds, then the cost for one adult ticket is 550 pounds and the cost for one child ticket is 260 pounds.

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