Edexcel 2026 GCSE Maths Paper 2 Higher Predicted Paper

Added: 2026-05-27

[00:00:00]This is the ED XL June 2026 practice paper 2 calculator paper higher tier.

[00:00:12]Question one. The frequency table shows the temperature in degrees of 30 days and we to draw a frequency polygon to show this information.

[00:00:24]For a frequency polygon, we are plotting the midpoint against the frequency. So the middle of 10 and 14 is 12. The frequency is 1. The first point is at 12 1.

[00:00:40]So it's 11 12 1. So here 164 2016 247 and 282.

[00:01:06]So we plot all the midpoints against the frequencies and we join them up with a ruler.

[00:01:14]It's a frequency polygon. So, we're plotting midpoint against frequency.

[00:01:19]Don't confuse it with cumulative frequency where you're going to be plotting the top point against the cumulative frequency. So, frequency polygon is midpoint against frequency joined up with a ruler.

[00:01:35]Part B, work out an estimate for the mean temperature.

[00:01:39]So, we've got 30 days But we don't know the temperature on each of the days. We only know what group they're in. So one day between 10 and 14, we're going to estimate that it was 12°, the midpoint. 4 days, we're going to say we're all 16°, the midpoint. 16 days of 20°, 7 days at 24°, and 2 days at 28°.

[00:02:10]So we're estimating that every day on every day the temperature was the midpoint of the group. So we need to add them up. 112 416s 1620s 7 24s and two 28s.

[00:02:26]We'll do that on the calculator.

[00:02:31]So, I've got one 12, four 16s, 1620s, 7 24s, and 28s.

[00:02:49]It adds up to 620.

[00:02:54]It adds up to 620.

[00:02:57]and there were 30 days. So 620 over 30. It's a good idea to show your working. So So show where the 620 came from. So you could do this on the table. 112 is 12.

[00:03:15]46 are 64 and so on. Fill out them there. Just put a sum at the bottom.

[00:03:24]620 over 30.

[00:03:27]Again, type it in the calculator.

[00:03:32]It's 20.6 recurring.

[00:03:36]20.6 recurring.

[00:03:41]Question two. Hannah is going to play one game of chess and one game of back gammon. The probability she will win the game of chess is 0.6.

[00:03:53]So winning chess is 0.6. Not winning is 1 minus 0.6 0.4. The probability she will win the game of bat gammon is 0.7.

[00:04:07]So that's the probability of her winning regardless of whether she wins or loses the chess. And losing is 1 minus 0.7.

[00:04:17]That's 0.3.

[00:04:20]So that is the completed probability tree diagram. Work out the probability that Hannah will win both games. So when we go across the branches, we times 0.6* 0.7 is 0.42.

[00:04:39]So you could work out the probability of any of these happening. So win then does not win is 0.6 * 0.3.

[00:04:48]does not win then win 0.4 * 0.7 and losing both or not winning both 0.4 * 0.3.

[00:05:03]Question three, write 7.3 * 10 ^ of -3 as an ordinary number. So it's 7.3 divided by 10 three times. It's going to be 073.

[00:05:22]You can check on the calculator. So the calculator will always either change the question into the answer or the answer back into the question. If you write in 7.3 * 10 to the^ of -3 and then press your S to D or format or F toD button, it doesn't want to change it. It wants to keep it as it is. But if you type the answer in 0.73, it will change it back into standard form. So it's always going to change the question to the answer or the answer back into the question.

[00:06:04]Part B, calculate the value of 8.5 * 10 ^ of 5 + 2.84 * 10 ^ of 6 give your answer in standard form. So we can just type this into the calculator.

[00:06:23]8.5 * 10 ^ 5 + 2.84 84 * 10 to the power of 6 we get 3 million690,000.

[00:06:42]The question said give your answer in standard form. So it's got to be a number between 1 and 10 which is going to be 3.69 * 10 to a power. How many times have you got to multiply 3.69 69 by 10 to get 3.69 million.

[00:07:02]Millions are always the power of six.

[00:07:05]Thousands are always the power of three.

[00:07:07]So it is the power of six. And of course you can check your answer.

[00:07:14]So if you type in 3.69 * 10 ^ of 6, it will take you back to 3,690,000.

[00:07:29]Question four. Sarah purchases a smartphone. She receives a discount of 15% off the regular price. Sarah pays £612 for the smartphone. Calculate the regular price of the smartphone. So if we took 15% off, you start with 100%.

[00:07:51]You take 15% off, that means she paid 85%.

[00:07:56]So 612 is 85% of the original of X.

[00:08:08]So we know 85%. We need to find 100%.

[00:08:14]I'm going to change this is I'm going to change to equals of I'm going to change to times. So 85% time x is 612 and I can change 85% into a decimal as well. 0.85 x= 612 to get x by itself. At the moment it's multiplied by 0.85.

[00:08:45]I'm going to do the opposite of tsing divide. So 612 divided by 0.85 will give me x which is 720.

[00:09:04]If you start with 720 and take 15% away from it, you get 612.

[00:09:14]So 15% of 720 is 108. 720 minus 108 is 612.

[00:09:28]Question five. The diagram shows a solid cylinder with base radius three and height of 12. The cylinder is made from steel. It's got a mass of 2,650 g. Work out the density of the steel.

[00:09:45]Give your answer correct to three significant figures.

[00:09:50]So density is equal to mass divided by volume.

[00:09:55]We know the mass.

[00:09:58]We don't know the volume, but we can work it out. So, the volume of a cylinder is the area of the base, the area of the circle, p<unk> r 2* the height h, p<unk> r 2 h. So that's<unk> * 3^ 2 * 12<unk> * 3^ 2 * 12 is 108 pi. I'll leave my answer in terms of pi because I'm going to use it again in the next calculation. And then we can use density mass over volume.

[00:10:41]So, density is the mass 2,650 g over the volume 108 pi cm cubed.

[00:10:57]2650 over answer is 7.81 7.81 81 g per cm cubed.

[00:11:13]And question five, a number x is written correct to two significant figures. The result is 5.4. Complete the error interval for x.

[00:11:24]So we have 5.4.

[00:11:27]The one up would be 5.5.

[00:11:29]The one down 5.3.

[00:11:32]So to round to 5.4, four. It's got to be between the two midpoints which are at 5.35 and 5.45.

[00:11:44]You can see it can equal 5.35, but it's just less than 5.45 because of the inequality symbols that have been used.

[00:11:58]Sophie, Thomas, and Umar share some money in the ratio 3 1:2. Umar gets £110.

[00:12:07]Sophie then gives some of her share to Thomas and some of her share to Umar.

[00:12:12]The money that Sophie, Thomas, and Umar each have is now in the ratio 26-7.

[00:12:18]How much money did Sophie give to Umar?

[00:12:23]Let's work out how much they started with. It was the ratio 3 1:2. Sophie to Thomas to Umar.

[00:12:31]and Umar got £110.

[00:12:34]So, two parts for Umar and that's worth £110.

[00:12:43]Which means one part must be half of £110 55.

[00:12:49]3 1 and two. That's six parts in total which is well 6 110s. No three 110s £330.

[00:13:05]So that's how much money they're sharing. They're sharing £330.

[00:13:12]And it's going to be the same amount of money that's in the ratio 2-6 to 7. 2 + 6 + 7 is 15. So there's now 15 parts.

[00:13:28]So the 330 gets shared out between 15 parts.

[00:13:36]That's 22.

[00:13:40]So each part is worth 22. Sophie, Thomas, Umar, it was 2 to 6 to 7. So Sophie gets two lots of 22, which is 44.

[00:13:53]Thomas gets six lots of 22, which is 132. and Umar gets 7 22s 154.

[00:14:07]How much did Sophie give to Umar? If Umar started with 110, he's now got 154.

[00:14:17]How much money did he get? He got 44 extra pounds.

[00:14:28]Question eight, write 90 as a product of its prime factors.

[00:14:35]So, we're going to break 90 down until it's just prime numbers multiplied together.

[00:14:41]So, I'll start with 90. I'm going to change 90 into two lots of 45.

[00:14:49]Two is prime, 45 isn't. I can break it down again.

[00:14:53]So 45 is in the three times table. It's three 15s.

[00:14:59]15 isn't prime, but 3 is. 15 can be broken down again into 3 * 5. And now they're all prime. So 90 can be written as 2 * 3 * 3 * 5 or 2 * 3^ 2 * 5.

[00:15:20]Question 9. and a B and BC are two sides of a regular polygon with n sides.

[00:15:27]Work out the value of n. So work out how many sides it's got. So we've got this polygon. So it's got loads of sides.

[00:15:35]Keeps going around.

[00:15:37]Loads and loads of sides. We need to know how many sides it's got. We can work out the interior angle because angles around the point add to 360.

[00:15:47]So 360 take away the 109 and the 86 and we will find the interior angle.

[00:15:58]So 360us 109 and minus 86 is 165°.

[00:16:06]That's the interior angle. So this polygon has got interior angles of 165°.

[00:16:14]We can use that to find the exterior angles.

[00:16:19]So 180 take away 165 is 15. So each exterior angle for this polygon is going to be 15°.

[00:16:32]And that's useful because all the exterior angles add up to 360.

[00:16:39]So imagine you started at one point You walk around the shape. In total, to get back to the same point, you have to turn around 360° to go the whole way back to where you started. So, all the exterior angles add to 360. They're 15 each. So, the question is, how many 15s make 360?

[00:17:06]360 / 15.

[00:17:12]360 over 15 is 24. It must have 24 sides.

[00:17:25]Question 10. There are 30 students in a class. A teacher is going to choose at random two of the students. Work out the number of different pairs of students the teacher can choose.

[00:17:41]So for the first student, for the first pick, there are 30 options. There are 30 options for the first pick.

[00:17:49]And then for each of those picks, there's 29 people they can get matched with. So it's 30 * 29.

[00:17:58]Now, the only problem is we're going to be double counting the people here. So we're going to have person A and person B and also person B with person A. So we're going to have twice as many possible pairs here. So we have to half our answer.

[00:18:17]Let's say we had just four people A, B, C, and and D in the class. If I pick A first, I could get three options for second. I could get B, I could get C, and I could get D. But then if I got B first, I could get B A B C B D. If I got C, I could get C A C B C D. If I got D, I could get D A D B D C. So I could say there's 12 different options here, which is four different people for the first pick and then three for the second.

[00:18:52]That' be four * 3, which is 12. But you can see I double counted them all. I've got A and B and B and A. I've got A and C and C and A. I've got A and D and D and A. I've got B and C and C and B.

[00:19:13]B and D and D and B and C and D and D and C. So you can see I've double counted them all. So the number of different options is actually six. 12 over 2. So it's 30 * 29 over 2 for this question and that is 435.

[00:19:43]Question 11. Megan invests £2,600 in a savings account for 3 years. She was paid 4.8% peranom compound interest.

[00:19:56]for year one and paid something r interest for each of the second and third years. Megan had £2,846 at the end of 3 years. What is the value of r?

[00:20:13]So what what's happened? We got £2,600.

[00:20:17]We got £4.8% in year 1. So to add on 4.8% 8% in one calculation. We want 104.8%.

[00:20:27]That's times by 1.048.

[00:20:32]And then we're going to times by some of a multiplier for two years for the second and third year. And then it's going to equal 2,846.

[00:20:47]So we can work out what the multiplier is, what X is, and then from that we'll be able to see what the interest rate was.

[00:20:58]So after year 1, there would be 2,724.80 in the account. 2724.8 2724.8x 8 x^2 is equal to 2846.

[00:21:21]We want to get x^2 by itself. So divide by our last answer.

[00:21:29]2846 over answer. So this is x squared is 1.0 444.

[00:21:45]And then to get x by itself, the opposite of squaring is square rooting.

[00:21:51]Square root answer 1.02 2. 1.022.

[00:22:01]So that's what you times by 1.022.

[00:22:05]What is the percentage increase from what? From multiplying by 1.022.

[00:22:12]That would be 102.2%. 2% each year. So what's the increase? Well, it's 2.2%.

[00:22:21]100% plus 2.2%.

[00:22:26]Is 102.2%.

[00:22:31]Question 12, box plots. So we've got box plots showing the time it took year 7 and year 11 students to travel to school. Compare the distribution.

[00:22:42]So when we're comparing distributions, we have to compare the median, the average.

[00:22:49]So let's make a comment on that. The year 11 times were longer on average.

[00:22:55]And we have to say in context, but just saying times is absolutely fine for that. So the year 11 median times were longer.

[00:23:13]So that's the comment on the median.

[00:23:15]We've written the word median and it's in context because we put times.

[00:23:20]Now let's compare the spread which can be the range or the interquartile range.

[00:23:28]The inter quartile range is the difference between the lower quartile and the upper quartile.

[00:23:36]And you can see it's longer, it's bigger, it's more spread out for year 11.

[00:23:44]So year 11 interquartile range again in context of times was greater was bigger.

[00:24:01]So one comparison for the median one comparison for the range or interquartile range and that is all we need to do.

[00:24:12]Question 13. Here are the first five terms of a sequence. Find an expression in terms of n for the nth term.

[00:24:22]So what kind of sequence do we have here? Let's look at the difference.

[00:24:26]-5 to -1 goes up by four. -1 to 7 is 8.

[00:24:34]7 to 19 is 12. And 19 to 35 is 16. The difference isn't the same every time, but the second difference from 4 to 8 is 4. From 8 to 12 is 4. From 12 to 16 is 4. The second difference is the same.

[00:24:53]And that makes it a quadratic sequence.

[00:24:58]So it's in the form a n 2 + b n + c.

[00:25:06]So to find the m term for a quadratic sequence you can remember that this number here the second difference is 2 a.

[00:25:16]So that one's 2 a.

[00:25:19]This one is 3 a + b and the first term is a + b + c.

[00:25:32]And you can just solve the equations. So 2 a is 4. So a is 2.

[00:25:40]3 a + b is 4. a is 2. So 3 a's are six.

[00:25:47]6 + b is 4. b must be -2.

[00:25:52]And a + b + c is equal to -5. a is 2. B is -2.

[00:26:02]C must be -5.

[00:26:06]So the mth term is 2 n^ 2 - 2 n - 5. And you can check by substituting numbers in. If we substitute in five for the fifth term, we should get 35 out. So 2 * 5^ 2 - 2 5s - 5 is 35. So we know it is correct.

[00:26:38]Question 14. A O is a sector of a circle. Center O radius 7.6 cm. The area is 47.9 cm squared. Calculate the size of the angle X. Give your answer to three significant figures.

[00:26:59]So a sector area is a fraction of the area of a circle.

[00:27:07]So what fraction is it? It's we've got x degrees out of 360 in total. So the fraction is x 360s.

[00:27:20]If if it was 180°, it would be half. If it was 90°, it' be a quarter. It's x 360s.

[00:27:28]We don't know what that fraction is.

[00:27:31]That's what we're going to work out.

[00:27:33]It's some fraction of so times the area of the circle pi r 2.

[00:27:45]So we know the sector area, the area of the sector is 47.9.

[00:27:52]That's equal to x 360 some fraction of p<unk> r 2<unk> * 7.6^ 2.

[00:28:04]So what is p<unk> r 2? P<unk> * 7.6 squared. The area of the whole circle is 181.45. Four. Five.

[00:28:17]Eight.

[00:28:28]So that's the area of the whole circle.

[00:28:30]If we do the sector area 47.9 divided by that, that will tell us what fraction we've got, what proportion of the whole circle. We've got like 26.4% of it. 0.2639 and so on. 0.2639.

[00:28:57]So we've got like 26.39% of it. But how many degrees is that?

[00:29:03]Well, times it by 360 and we will have our answer.

[00:29:10]So 95 0 degrees.

[00:29:25]Question 15. Enlarge the shaded shape by scale factor -2 with center of enlargement at the origin. So the center of enlargement is at 0 0.

[00:29:39]Let's look at each point and see how far we've got to go to get there. So to get to this green point, it was back two down two.

[00:29:51]As a vector, that's -2 - 2. If we times that by -2, what do we get? We get 44. A negative * a negative is a positive.

[00:30:04]It's going to be over there.

[00:30:09]For this blue point, it's back five down two. So min - 5 - 2 times it by the scale factor. That would be 10 4. So right 10 up four back three down four times by -2 is 68.

[00:30:37]and back six down four times by -2 will be 128. So right 12 up 8.

[00:30:54]And we'll just join these up with a ruler.

[00:31:00]And that is our new shape.

[00:31:08]Question 16. A solid frostm is made by removing a small cone from a large cone as shown on the diagram. The slanted height of the small cone is 6 cm. The slanted height of the large is 10 cm and the base has a radius of 4 cm. Calculate the total surface area of the frustr.

[00:31:37]So when we have a frost drumm, when we cut off a small cone from the top of a large cone, we have similar shapes.

[00:31:47]So we can work out the radius of the small cone using a scale factor.

[00:31:59]So for the large cone we had a slant height of 10. For the small one it's slant height of six. So what's the scale factor? Well it's 60% 6/10. The small one is 610 of the big one. So times 0.6 is the scale factor to go from 10 to 6 10 * 0.6 is 6.

[00:32:26]So from large to small we times by 0.6.

[00:32:31]So four 0.6's will give us the radius of the small cone and that is 2.4.

[00:32:42]So the radius of small cone is 2.4 cm.

[00:32:50]We can then work out the curved surface area.

[00:32:56]And we're given a formula. It's pi RL.

[00:33:01]And we know the radius and we know the slant height of both of them. So we can just type it in. So for the frustr I'm going to do large minus small. So pl for the large one.

[00:33:24]* 4 * 10 take away<unk> * 2.4 * 6. So the big one take away the small one will be the curved part we've got left.

[00:33:45]* 4 * 10 -<unk> * 2.4 * 6 80.4 I'm going to store that in the memory.

[00:34:03]80.4 80.4 cm squared.

[00:34:12]So that is just the curved part. We've also got two circles. We've got this small circle on the top and the larger circle on the bottom.

[00:34:26]So the area of a circle is p<unk> r^ 2.

[00:34:31]So p<unk> * 2.4 4^ 2 and<unk> * 4^ 2<unk>i * 2.4^ 4 squared is 18.1 and I'll store that in the memory as well.

[00:35:02]And pi * 4^ 2 16 pi or 50.3 So, we're going to add them together.

[00:35:19]We've got 80.4 for the curved part, 18.1 for the small circle, and 50.3 for the large circle.

[00:35:30]So, add them all together. So, I've got answer plus I save them as a and b.

[00:35:41]What answer am I expecting here? 80, 100, 150ish.

[00:35:48]Let's see.

[00:35:51]149 to three significant figures. 149 cm squared.

[00:36:03]Question 17. Show that x /x - 3 - 3x -1 over x + 3 + 1 can be written in the form ax + b over x^2 - 9 where a and b are integers.

[00:36:24]So to add fractions or take away fractions, we need the denominators to be the same. So we're going to have to make the denominators the same. And they're all going to have to be x - 3 * x + 3. So for our first fraction, we're going to times top and bottom by x + 3.

[00:36:46]So 2x * x + 3 over x + 3 * x - 3.

[00:36:56]For the second one, times top and bottom by x - 3.

[00:37:02]3x -1 * x - 3 over x + 3 * x - 3.

[00:37:12]And for the one, we're going to times top and bottom by x + 3 and times top and bottom by x - 3. We're going to change it into x + 3 * x - 3 over x + 3 * x - 3. So now all the denominators are the same. We can make it into one fraction with a denominator of x + 3 * x - 3. I don't know why I've just written that the other way around.

[00:37:46]When we expand and simplify that denominator, it's going to give us x^ 2 - 9 because it is the difference of two squares.

[00:37:55]I'm going to just write it out. I'm not going to expand it this time.

[00:38:11]And now I'm going to start expanding these brackets. I've got to be really careful this negative sign here. We want to take away all of the expansion, not just the first term.

[00:38:26]But apart from that we can just start expanding. So 2x * x 2x^ 2 x * 3 is 6 x to keep this negative sign. I'm going to keep a bracket around the expansion.

[00:38:43]3x * x 3x^ 2 3x * -3 - 9x -1 * x is -x -1 * -3 is a positive 3 and x + 3 * x - 3 that expands to give x^2 - 9.

[00:39:05]The 3x - 3x that we get cancels out. The middle bit disappears.

[00:39:13]And then I'm just going to expand the bracket again, changing all the signs in there.

[00:39:20]So it's going to be - 3x^2 + 9 x + x - 3 + x^2 - 9 / x^2 - 9.

[00:39:34]So then let's collect the like terms.

[00:39:36]We've got these x squar terms which add up to nothing. 2 - 3 + 1 is zero.

[00:39:46]For the x terms, we've got well 16 6 + 10 16 x's and we've got - 3 - 9 - 12.

[00:40:05]So 16x - 12 / x^2 - 9.

[00:40:17]Question 18. Here is a velocity time graph for a car. Calculate an estimate the acceleration of the car when t is five.

[00:40:28]So it's a curve and we want the acceleration. The acceleration on a velocity time graph is the gradient. So we need to know the gradient when t is five. We do that by drawing a tangent.

[00:40:46]So t is five here. We want to know the gradient at that point.

[00:40:52]So we draw a tangent. I'm going to slide that into place.

[00:41:00]A tangent that's got the same gradient at that point.

[00:41:05]So something like that.

[00:41:10]And then we got to work out the gradient of the tangent.

[00:41:14]So let's find some good points on this tangent.

[00:41:18]Let's say this is 222.

[00:41:23]That's quite a nice point.

[00:41:26]and it's close to being at 70. Let's just say it's at 70. So, let's go for 222 and 70. When we're working out the gradient, it's the change in y over the change in x. So the y's have gone from 22 to zero.

[00:41:58]Actually I'm going I'm going to show you a formula. Let's have x1 y1 x2 y2.

[00:42:06]So for the change in y I'm going to do y 2 - y1. And for the change in x2 - x1.

[00:42:16]So that makes it 0 - 22.

[00:42:22]over 7 - 2, which is -22 over 5.

[00:42:34]-22 over 5, it's -4.4.

[00:42:44]There'll be a range of possible answers as long as you've drawn a tangent.

[00:42:49]That's at the right point and it looks like a tangent, you should be absolutely fine.

[00:42:59]Work out an estimate for the distance the car travels in the first 6 seconds.

[00:43:03]Use three strips of equal width. So in a velocity time graph, when we have a velocity time graph, the gradient is the acceleration. The area under the graph is the distance traveled.

[00:43:17]So, it's the first six seconds in three strips.

[00:43:22]We're going to split it into 0 to two seconds and we're going to have a trapezium.

[00:43:30]Then from 2 to 4 seconds will be another trapezium.

[00:43:37]And from 4 to 6 seconds will be another trapium.

[00:43:47]So the area of a trapezium is half the sum of the parallel sides times the distance between them. So for 0 to 2 seconds, the distance between them is two. The parallel sides, we've got 23 and 17.

[00:44:14]So it's half of 23 + 17 * 2 a for two to four we've got 17 and 12.

[00:44:31]So half of 17 + 12.5 * 2 2 is the distance between them. And for four to six, we've got 12 1/2 and 5 and 1/2.

[00:44:47]So half of 12 1/2 + 5 1/2 times the distance between them.

[00:44:56]And well, we're halfing and tsing by two for all of them. So half of 40 * 2 is going to be 40.

[00:45:06]Half of 29 1/2 * 2 is going to be 29 1/2 and half of 18 * 2 is 18.

[00:45:19]So add them all up.

[00:45:24]40 29.5 and 18 add to make 87.5.

[00:45:31]So our estimate is 87.5 m.

[00:45:42]Question 19. Y is inversely proportional to the square of X. Let's write that as a formula. So if it's inversely proportional, it's something over the square of X X^ 2.

[00:45:57]So Y is inversely proportional to the square of X as a formula is Y = KX^2.

[00:46:05]Y is 28 when X is 1 and a half.

[00:46:10]So we can use that to find this K. K is just a number.

[00:46:16]And K is going to be 28* 1.5^ 2 63.

[00:46:26]So the formula for this question is y is 63x^2.

[00:46:34]That is the formula. Find the negative value of x when y is 7. So y is 7.

[00:46:42]What's the negative value of x?

[00:46:45]So 7 is equal to 63 /x^2 times both sides by x^2.

[00:46:53]Divide both sides by 7 and square root. So x is plus or minus the square root of 9.

[00:47:04]which means it's plus or minus 3. We wanted the negative value. So -3.

[00:47:17]Question 20. ABC and a DC are triangles.

[00:47:23]The area of triangle A is 58 m squared.

[00:47:30]Work out the length of BC.

[00:47:34]Give your answer correct to one decimal place.

[00:47:39]So, we've got the area of one of the triangles. We can use the formula. Area equals half a b sin c which will give us this length c to d. So half a b sin c.

[00:47:58]To use that formula we use two lengths and the angle between them. So we can find C to D using half A B sin C. Then we can use the cosine rule.

[00:48:13]The cosine rule will give us the length A to C which is shared by both triangles. So that's a 2 = b^ 2 + c^ 2 - 2 b c cos a.

[00:48:28]But then finally we'll be able to use the sign rule with opposites and we know this missing angle here because angles in a triangle add up to 180. So 180 take away the 112 and the 51 it's 17°.

[00:48:54]So we're going to use half a b sin c the cosine rule and the sine rule. So a over sin a equals b over sin b.

[00:49:08]So, we're going to use all of our different formulas for non-right angle triangles all in the same question. So, let's get started. Half a b sin c half of 12 * y sin 102 a b sin c is 58.

[00:49:35]So y is going to be equal to 58 / half of 12 * sin 102.

[00:49:44]So 6 sin 102 and I'll type that in the calculator. 58 over 6 sin 102 and it is 9.88.

[00:50:00]So that's 9.88.

[00:50:02]That's this one here. And then we're using the cosine rule. So we're going to find a to c. Now let's call it zed.

[00:50:16]So a 2 = b 2 + c^ 2 - 2 b c cos a the cosine rule. So z^ 2 = 12^ 2 + 9.88 88^ 2 - 2 * 12 * 9.88 cos 102.

[00:50:38]So again for the cosine rule we're using two lengths and the angle between them to find the other length to find the length opposite the angle.

[00:50:51]And we can just type this all in the calculator and it will give us z^ squ.

[00:51:00]So 12^ 2 + answer squar - 2 * 12 * answer cos 102 98 that's z squed so square root and we will find that length 17.06.

[00:51:39]So we've got 17.06 here. And then finally, it's the sign rule. So we've got the opposites a over sin a. So x over sin 17. They go together. x and sin 17.

[00:52:02]equals 17.06 over sin 112. So the opposites go together. I'm putting the lengths on top because we're calculating a length. If I was working out an angle, I'd put the angles in the top just to make the rearranging easier. So it would be sin a / a= sin b over b. So 17.06 06 over sin 112.

[00:52:32]So to get x by itself, I'm going to times both sides by sin 17.

[00:52:42]So I've got my answer over sin 112 times sin 17 and 5.3 8 5.38 is my answer at one decimal place 5.4 four.

[00:53:16]Question 21.

[00:53:19]f ofx is cube roo<unk>x. g of x is 3x + 5. h of x is f g ofx. Find the inverse of h.

[00:53:33]So h of x is fgx.

[00:53:37]So that's putting g into f. So g is 3x + 5. So if we put that into f, f of 3x + 5, we're just changing x into 3x + 5 is the cube root of 3x + 5.

[00:53:58]So that's h of x. We need to know the inverse of this. What's the function that does the opposite thing? I'm going to rewrite it as y equals. So y equals the cube root of 3x + 5. And then it's a to change the subject of the formula question.

[00:54:18]So when we have an inverse function, the input becomes the output and the output becomes the input. They have to switch around, which means the x becomes the y and the y becomes the x. So if we rearrange this to get x by itself to make it x equals that will tell us what the inverse function is. So we're going to get rid of rid of a cube root. So we're going to cube both sides.

[00:54:48]We're going to take five away from both sides and then we're going to divide by three.

[00:54:59]So the inverse of our function is cubing then taking away five and then dividing by three.

[00:55:09]But we don't want it in this x and y format. We want it in h the inverse function h minus one of x and it's going to be x cubed - 5 over 3. That is our answer. That is the function does the opposite thing to our original. And you can check that. You can check it's the opposite. If I put a number into the original one into h of x, let's put 10 in. It's going to give us a decimal answer, but that's fine. So, if we substitute 10 into the original, we get out 3.27.

[00:55:57]So if we substitute 3.27 into our inverse function, it should give us back 10.

[00:56:05]So let's put this 3.27 into the inverse function. Answer cubed - 5 over 3. It should give us 10 out.

[00:56:19]And it does. So it does the opposite thing.

[00:56:24]and it's x cub - 5 / 3.

[00:56:30]Question 22. T is equal to v minus u / a. v is 38.2 correct to one decimal place. U is 10.9 to one decimal place and a is 9.81 to three significant figures. By considering bounds, it's a bounds question. Calculate the value of t to a suitable degree of accuracy.

[00:56:57]So let's start with V.

[00:57:00]It's been rounded to 38.2.

[00:57:04]The one up is 38.3.

[00:57:07]The one down is 38.1.

[00:57:10]So what are the bounds?

[00:57:12]It's got to be between the two midpoints.

[00:57:15]38.15 and 38.25.

[00:57:22]The same with you. It's 10.9.

[00:57:26]The one up is 11.0.

[00:57:28]The one down is 10.8.

[00:57:31]It must be between the midpoints of 10.85 and 10.95.

[00:57:40]And a has been rounded to 9.81.

[00:57:44]The one up is 9.82.

[00:57:46]The one down is 9.80.

[00:57:50]So it must be between the midpoints 9.805 and 9.815.

[00:57:58]We want to know the biggest thing T can be and the smallest thing T can be.

[00:58:04]So the biggest possible answer we can get out. So the biggest thing we can get out from V minus U is going to be the biggest V take away the smallest U.

[00:58:19]And to get the biggest possible answer, we want to divide divide by the smallest possible A.

[00:58:29]For the lower bound for T, we're doing the opposite. So to get the smallest possible number, we're going to start with the smallest V.

[00:58:38]take away the highest u and divide by the biggest possible number.

[00:58:45]We're just going to type these into the calculator. See what the upper bound and the lower bound for t is. So the upper b 38.25 25 take away the lowest U divided by the lowest A and the lowest possible B take away the biggest possible U over the biggest possible A.

[00:59:16]So let's just type them into the calculator and we'll see the biggest and smallest things that T can be.

[00:59:25]38.25US 10.85 over 9.805 2.79449 2.79449 I'm doing five decimal places four or five should be fine and just change these 2.77 1 27.

[01:00:10]They don't look much like sevens. Okay.

[01:00:12]So now we've got an upper bound and a lower bound. We've got to see what they both agree to. So, I can't do three significant figures because I've got 2.79 and 2.77, but I can do two. They're both 2.8.

[01:00:31]So, that's the highest degree of accuracy they both agree to. So, it's definitely 2.8 to two significant figures because it's the highest degree of accuracy.

[01:00:51]They both agree.

[01:00:57]So 2.8 is the answer.

[01:01:08]Question 23. Solve algebraically the simultaneous equations. So, we've got nonlinear. They've got squares in them.

[01:01:19]So, we can't use elimination. We have to use substitution.

[01:01:24]So, we're going to make our linear one, our 5x + 2 y = 6, either x equals or y equals, and then substitute that in to the top one.

[01:01:36]So, I'm going to make it y equals.

[01:01:39]So, I'm going to change it into 2 y = 6 - 5x. Then half both sides 6 - 5x over 2. So, I've rearranged the linear one to make it y equals. And then I can substitute that in to my top equation.

[01:01:59]2x^2 plus So, it's y^2. It's 6 - 5x / 2^ squared = 12. So now I've got rid of the y's.

[01:02:11]I've just got x's and I can solve this equation to find out x or to find the two values for x. So squared means times itself.

[01:02:25]So I'm going to write it as 6 - 5x / 2 * 6 - 5x / 2.

[01:02:35]I'm going to expand those brackets.

[01:02:39]So expand the top like usual and the bottom. So it's times the top times the bottom. 2 * 2 is 4. So 6es are 36.

[01:02:51]Six negative 5 -30 -5 * 6 again -30 and -5x * - 5x will be positive 25x^2 I'm going to get rid of this fraction so divide no times by four I need to times everything by four so I need to times by four times by four times by four tsing by four here just gets rid of the divide by four, but I'm tsing everything by four.

[01:03:29]So that's going to give me 8 x^2 plus the divide by 4 is gone here. So I've got 36 - 60x + 25 x^2 = 48.

[01:03:45]I've got a quadratic. We solve quadratics by making them equal to zero.

[01:03:50]So, I'm going to get everything over to the left side. I've got 8 x^2 and 25 x^2, which is 33 x^2.

[01:04:00]I've got - 60x and I've got 36 take away 48, which is -12.

[01:04:10]So, I've got a quadratic to solve. We could use the quadratic formula, but I'm going to try and make it easier.

[01:04:18]try and divide through by three and see if I can factoriize it.

[01:04:28]So again, we could use the quadratic formula here. A is 11, B is -20, C is -4. Type it in and find our two answers.

[01:04:41]But this will factoriize.

[01:04:44]I'm going to do the extra step in the middle.

[01:04:48]So a * c 11 fours are 44. Does anything multiply to make 44 and add to make a 20? So 144 22s or 4 11s. Yes. 2 22s.

[01:05:04]So it can be 11 x^2 + 2 x - 22x - 4 = z.

[01:05:16]So then what can I take out the first two terms? That's an x x * something is 11 x^ 2 11 x * 2 is 2x and then -2 * 11 x is - 22x and - 2 * 2 is -4.

[01:05:38]So it factorizes x can be two that's what makes the first bracket zero or x is We have two 11ths.

[01:05:47]That's what makes the second bracket zero. So we've got two values for x.

[01:05:52]There'll be two for y as well. y is 6 - 5x / 2. Remember 6 - 5 2's over 2 or 6 - 5 * -2 11th over two.

[01:06:13]Let's use the calculator.

[01:06:20]6 - 5 2's over 2 is -2.

[01:06:32]So x is 2, y is -2 or x is - 21 and y is 38 11ths and again we can check on the check with the original.

[01:07:03]So 2 x^ 2 * - 2 11th squared + y^ 2 which was was it 38 11th should equal 12. No it doesn't.

[01:07:33]I didn't press squared.

[01:07:36]It does equal 12.

[01:07:38]And x was 2 and y was -2 was the other one.

[01:07:54]And that equals 12 as well.

[01:07:58]Question 24. A circle has the equation x^2 + y^2 = 45. Write down the length of the radius of the circle. Give your answer as a simplified third. The equation of a circle is x^2 + y^2 = r 2.

[01:08:18]The equation of a circle with a center in the middle is x^2 + y^2 = r 2. Which means 45 must be the radius squared.

[01:08:30]So what's the radius? It's square<unk> 45 3<unk> 5. That is a simplified third.

[01:08:40]That is our answer.

[01:08:42]P is the point 3, -6 on the circle x^2 + y^2 = 45. work out the equation of the tangent to the circle at P.

[01:08:56]So we have a circle that goes the center at the origin. There's a point on it 36 which is down here somewhere.

[01:09:10]And there's a tangent, a line that just touches the circle, doesn't go through the circle, just touches it at that point. We want to know the equation of the tangent.

[01:09:26]So the radius meets the tangent at a 90° angle.

[01:09:33]And because they meet at a 90° angle, they're perpendicular lines. So we can work out the gradient of the radius. And then the gradient of the tangent is the negative reciprocal of that.

[01:09:49]So the gradient is the change in y over the change in x.

[01:09:54]y2 - y1 over x2 - x1.

[01:09:59]And if we've got x1 y1 x2 y2, we've got -6 - 0 over 3 - 0, which is -2. That's the gradient of the radius.

[01:10:18]The perpendicular gradient is the negative reciprocal of that. Flip a minus, which is 1/2.

[01:10:27]two perpendicular lines the gradients have to multiply to make negative -1.

[01:10:33]So half * -2 is -1. It's a negative reciprocal flip and minus.

[01:10:40]So that's the gradient of our tangent.

[01:10:44]And the equation of a line is in the form y = mx + c. So y = x + c. We don't know what C is, but we can use these coordinates 3, -6 to find it. X is 3, Y is -6. So, -6 is half of 3 + C. Half of 3 is 1 and a half or 3 over 2.

[01:11:16]So, C is -6 - 1 and a half.

[01:11:23]-6 - 1.5 is -15 /2.

[01:11:32]So the equation of the tangent is half x - 15 / 2