# Video: GCSE Mathematics Foundation Tier Pack 4 • Paper 1 • Question 10

GCSE Mathematics Foundation Tier Pack 4 • Paper 1 • Question 10

07:45

### Video Transcript

Alaine is tiling a T-shaped section of wall in her kitchen. The section of wall is shown in the picture along with its dimensions in millimeters. Each tile is a rectangle with dimensions of 30 centimeters by 25 centimeters. Alaine uses exactly 12 whole tiles to tile the T-shaped section of wall. Show in a labelled sketch exactly how Alaine tiled the wall.

Now the first thing that we notice in this question is that the dimensions of the wall are given in millimeters, but the dimensions of the tiles are given in centimeters. And so our first step in solving this problem can be to convert these dimensions so that they’re in the same units. We know that one centimeter is equal to 10 millimeters. If we divide both sides here by 10, we get that one over 10 centimeters is equal to one millimeter. And we can use this conversion in order to convert the dimensions of the wall, which are in millimeters, to centimeters. So we simply divide each length by 10. The length of 1200 millimeters becomes 120 centimeters. The length of 500 millimeters become 50 centimeters. The length of 300 millimeters become 30 centimeters. And the length of 600 millimeters becomes 60 centimeters.

Now that we’ve got the dimensions of the wall in centimeters and the question gives us the dimensions of the tiles in centimeters, we have one standard unit for all of the measurements in this question. And that unit is centimeters. So we’re ready to answer the question. Let’s consider our tiles with dimensions of 30 centimeters by 25 centimeters. Let’s look at the top left of the wall and consider the different ways in which Alaine could have placed a tile here. She could have placed a tile with the 30- centimeter side vertical, which would’ve looked like this. We can see that the height of this section of the wall on the left is 50 centimeters. And in this orientation, the height of the tile is 30 centimeters. So this means that the height of the gap which is left below the tile is 50 minus 30, which is also equal to 20 centimeters. And since this is smaller than any of the dimensions on the tiles, Alaine would not be able to fit a tile in this gap. And the question tells us that Alaine used exactly 12 whole tiles. So this means that Alaine didn’t cut up any of the tiles.

So we can conclude the Alaine did not place the tile in this orientation in this corner. Let’s see what happens if we put the tile with the 30- centimeter side horizontal. Now we can calculate the height of the gap below the tile with the tile in this orientation. The height of the wall on the left is still 50 centimeters. The height of the tile is 25 centimeters. And so the height of the bit of wall left below the tile will be 50 minus 25, which is also equal to 25 centimeters. And now we can see that one of the dimensions of our tile is also 25 centimeters. So we’ll be able to fit another tile below this tile, and this tile will be in the same orientation as the one above it. The 30- centimeter side will be horizontal, and the 25-centimeter side will be vertical. And since we have tried both possible orientations of the tiles, we know that this is the only possible way in which we can fit tiles in this left-hand side of the T. Since the T-shape is symmetrical down the horizontal axis, we know that the tiles on the opposite side of the T must be arranged the same way.

Now we have fitted four tiles into our T-shaped wall. Let’s now consider the section of wall which we have remaining. This section of wall has got a width of 60 centimeters. And its height is 50 centimeters plus 50 centimeters, which is also equal to 100 centimeters. Now let’s try and fit the tiles in this rectangular section so that the 30-centimeter side is vertical. What this means is at the height of the tiles is going to be a multiple of 30. The height of the wall in this rectangular section is 100 centimeters. So let’s see how many times 30 centimeters goes into 100 centimeters. So we divide 100 by 30, and we can cancel a multiple of 10 from the top and bottom. So we simply cross the zero out, leaving us with 10 over three. Now we notice that we have an improper or top-heavy fraction. And we know that three times three is equal to nine. So we can write this improper fraction as nine plus one all over three. Then we can separate the addition in the fraction into two separate fractions to give us nine over three plus one over three. Then nine divided by three is simply three. So we’re left with three plus one-third, which we can write as a mixed number. And that mixed number is three and a third.

Now what this tells us is it three and a third tiles will fit vertically in this rectangle when the 30- centimeter side is vertical. So we can draw three tiles with the 30-centimeter side vertical on our rectangle in the middle. However, this leaves us with a gap at the bottom. And the length of this gap is one-third the height of the tile in this orientation, or one- third times 30, which is also equal to 30 over three or 10 centimeters. And since 10 centimeters is too small to fit another tile in, this orientation of the tiles will not work. Next we can try to fit the tiles with the 30-centimeter side horizontal. What this means is that the 25-centimeter side will be the vertical side. Let’s see how many times this height of 25 centimeters of the tile will fit in the height of the rectangle of the remaining wall, so that’s the height of 100 centimeters.

In order to do this, we divide the height of the wall, so that’s 100 centimeters, by the height of the tile in this orientation, so that’s 25 centimeters. So we have 100 over 25. And first we notice that both the top and bottom of the fraction are multiples of five. So we can divide both the top and bottom by five, leaving us with 20 divided by five. And again both the top and bottom of the fraction are multiples of five. So we will divide both numerator and denominator by five. This gives us an answer of four over one, which is also equal to four since anything divided by one is simply itself. What this tells us is that in this orientation, we can fit four tiles above one another inside this rectangle. So let’s draw these four tiles in.

Now we’ve figured out how many of the tiles we can fit vertically in this orientation, let’s quickly calculate and check that we can fit another tile next to this one. We have that the width of our tile in this orientation is 30 centimeters, and the width of the rectangle is 60 centimeters. So therefore, the width of the gap remaining will be 60 minus 30, which is also equal to 30 centimeters. And since this matches the width of the tile in this orientation, this tells us that we can fit another four tiles in the same orientation in the remaining space. So let’s draw them in. And now we can see the all 12 tiles fit perfectly on the T-shaped section of wall. And so this must be exactly how Alaine tiled the wall. And therefore, this is the solution to this question.