2019 Mathematics paper 1 internal fully answered (Q1-Q23)

Added: 2026-05-25

[00:00:00]Hello, good day viewers. I welcome you to this mathematics presentation. So in this presentation I want us to answer mathematics paper one for 201 19. So this is paper one. So we are going to answer the entire paper from question one up to question 23. So let's look at the questions now. All right. So here is our question one. So we have evaluate. So in brackets we have -1^ 0 * 2^ 3. So we are going to write the question the way it is -1^ 0 * 2^ 3. So here we know that any number raised to the^ 0 it's a one of course not 0^ 0 this is not equal to 1.

[00:00:49]Okay. So this one to be a one times. So 2^ 3 it means the best two is being multiplied by itself three times. So you write 2 * 2 * 2 and this will give us 1 * 2 * 2 it's a 4. 4 * 2 it's 8. So 8 * 1 you still get a 8. So this is our answer for question one. We can now move on to question two. So this is our question two right here. So the question reads solve the equation x -1 and 2x + 5. So these are in brackets = 0. So I'm going to copy the question which is x - 1.

[00:01:30]Again we have 2x + 5 is = 0. So the way we should solve this equation just equate. So we going to equate x -1 = 0.

[00:01:44]x -1 = 0 or we can say 2x + 5 = 0. So this is a quadratic equation. If you expand, you get a quadratic equation. Okay? So there's no need for you to start expanding. Just x - 1 = 0 or 2x + 5 = 0. Then you solve them separately. So x is = 0 + 1 because this -1 go to the right hand side or here we have 2 x. So this positive 5 also go to the right hand side. You're going to have 0 - 5. So from there we can say x = 0 + 1 it's a 1 or now here we are going to have 2x = -15.

[00:02:34]So on the left hand side we have already calculated the value of x. So we say x = 1 or so here we can divide by 2 so that we find we find the exact value of what of x. So you can divide by two even this side by two. So this two and this two will cancel. X = -5 / 2. So these are the answers. Okay. So these are the answers. You can leave your answers at that's the job. You can write this one as a mixed fraction. Still you're going to be marked. So you can now move on to question three. So this is our this is our question three. So here we are told to simplify. We have 4 x^ 2 + 2 y^ 2 - x y - 6 x^ 2 + 2 x y. So we are told to simplify this expression. This is an algebraic expression. So when you are simplifying an algebraic expression, the first thing that you must do is group the like terms together. So we have 4x each like is -6x^2.

[00:03:52]So here you move this term with each sign. So I'm going to write 4 x^ 2 - 6 x^ 2. Then we have + 2 y^2. So when you look at these terms, we don't have any other term which is having a y^2. So this one just write it the way it is + 2 y^ 2. Then we have x y which is like to positive 2x y. So I can group them by starting with 2xy then - x y. So even if you start with x y and then plus 2x y the answer will be the same. So 4x^2 - 6 x^ 2. So here it's like you have 4 - 6 and you know that a smaller number four is subtracting a bigger number 6. So your answer will always have a negative.

[00:04:43]So 4x^2 - 6 x 2 x^ 2 then + 2 y^ 2 then 2x y - x y this will give us x y. So this is our answer four. So the question reads shed A union B A union C in brackets complement intersection B in the vin diagram in the answer space. So here we need to shade in brackets we have A union C complement intersection B. So we know that A union C when we combine set A and C. Okay that's the union. Now there is this complement. So complement complement means the solution in the solution set we should not have any element coming from set A and set C. So we are just going to consider set B only. So when I say set B only it's only this part this is the part that we're going to do what to shade this part only. That's all. So so this is our answer and you get your two marks there. Okay. So we we can now move on. So we come to question five. So under question five, we are told to factoriize completely. So we have 4 a b + 6 a c - 6 b - 9 c d. So this is factorization by grouping of terms. So here you need to observe the first term and the second term if they have a similar or they have common terms. Okay. So here we can observe that there is a even here there is a then third term and fourth term there is a d even here there is a d. So what you can do is you can just separate them after observing those things like this in brackets. So in the first bracket here we can look at the numbers we have a four and a six. So we look for the highest common factor of 4 and six which is what a two. Then a and a is common. So we are going to factor out 2 a in the first bracket. So we open the bracket here. Then we start dividing this two into four. So here we dividing this two into this four. It goes there many times two times. So we write a two.

[00:07:12]This a and this a will cancel. It means you remain with a b. Okay. Then plus 2 into 6 it's a three. This a and this a will cancel remain with what? A c we close minus. So whatever that you have you have in the first bracket you should get the same things when you when you simplify in the other bracket.

[00:07:34]Okay. When you do the factoring out. So here we have 6 and 9. We look for the highest common factor of 6 and 9 which is a three. Then we have B D and C D. So you can notice that letter G is common.

[00:07:48]So you factor it out open bracket. So 3 into 6 3 into 6 it's a two. Okay. Then d and d will cancel you remain with a b.

[00:08:00]So you remain with a 2 b. Then this negative time this negative we will get a positive. So you have to be careful here. Negative * negative is a positive.

[00:08:10]Then 3 into 9 goes there many times three times. D and D will cancel. You remain with a C. Okay. So you can notice that the terms in the brackets they are the same. So what you do? Just pick one term which is 2 B + 3 C. You close open bracket. You get this 2 A minus 3 D. So 2 A - 3 D. Then we close. So this is our answer. Okay.

[00:08:40]So this is our answer. I hope it's clear. You have understood this part. We can now move on to the next question which is our question six. So our question six reads a company declared a dividend of one quarter 15 per share.

[00:08:58]Malala has 600 shares in the company.

[00:09:01]How much will she get? So we are looking at a company which has got which is which has declared a dividend of one quarter 15 per share. So this person Malara has 600 shares. So we want or they want us to find out how much she's going to get at the end of the year once they share the profit. So here what we do is we will get the number of shares which is 600 shares times the amount per share which is one quarter 15 way. Okay.

[00:09:35]So once we do the multiplication of these two we are going to get 900 quatcher. So we get 900 quatcher. Now let's see how we have come up with this answer. So we have multiplied 600 * 1 15 we ignored the point. So I just write 150. So now how have we gotten this answer? So we going to multiply here 0 * 0 it's a 0. 0 * this 0 it's a 0. 0 * 6 it's a 0. So I can add the 0 here. 5 * 0 it's a 0. 5 * 0 it's a 0. 5 * 6 it's a 30. Like this. Okay. So this is a zero.

[00:10:22]Then one add two zeros here. Then you say 1 * 0 it's 0. 1 * 0 again it's a 0.

[00:10:30]1 * 6 we get a six here. Then we can close. We say plus. So I can add my zeros on top there. So 0 + 0 get a 0 get a zero. A zero. Even here you get a zero. 6 + 3 will get a 9. Now this 1 5 0 it was a decimal number which was one 15. So it had two decimal places here.

[00:10:55]So you count from the right one two. So the point will be. So this is the 900 quatcher that we got. That's how it came about. All right. So we have now come to question seven. So under question seven we have a and b. So we are starting with question 7 a. Given that 17 + m + 27 + consecutive terms of an arithmetic progression. Find the arithmetic mean.

[00:11:23]So this is an arithmetic progression where we have 17 + m + 27.

[00:11:31]Okay. Now we need to find the arithmetic mean. Okay. So how do we find the arithmetic mean in an AP? So if you're finding the arithmetic mean arithmetic mean arithmetic mean in an AP. This is like a simple formula you must use. So here we have one, we have two, then we have three. So what we normally do is you get one then plus 3 then you divide by two.

[00:12:03]This is a simple formula that you must know. So from here what are we going to do? We just going to substitute our one it's a 17 plus our two it's a 27 over a two. Okay. So here now when we add 7 + 7 is a 14 we write a 4 1 + 2 is a 3 + that one 4 over a 2. So here 2 1 2 2 into 42 again 2 into 42 1 into 2 22 we get a 22. So the arithmetic mean was a 22. So we have answered our question A.

[00:12:43]I hope this question is clear. Okay we now come to question B. So question A is clear we have answered that one. Let us now move on to question B. So I'm rubbing this part. So under question B they are asking us to find the formula for the anything term. So how do we find the formula for the n theme? So here we have this sequence 11 13 and 15. So to find the formula for the n term we going to come up with the formula t subscript n= a + n -1 in brackets then time z the common difference. So how do we find the common difference in an a we say 2us 1 or you can say 3 -2 this will give us 13 - 11. So this will be a 2. So the common difference is a two. Then a is the first term from this sequence. The first term is 11. So we just going to substitute on g and a. Okay. So we can now substitute. So I just substituting on d and a. So a is 11 plus n. We leave it because we don't have the number of the term minus one. Then d it's a two.

[00:14:07]So 11 plus these two will expand to multiply throughout each and every term inside the brackets. So 2 * n will get to n 2 * -1 we get -2. So this is what we will get. So from here we can say we are going to have 11 - 2 + 2 n. Okay. So 11 - 2 we get 9 + 2 n. So this is our answer. So in this answer space here the first one we said it's a 22. Now here we have 9 + 2 n. Okay. So we have answered our question seven. We can now move on to question eight. Okay. We can now move on to question 8. So this is our this is our question eight right here. So under question eight we have a and b. So we're going to start with question a. So question A is saying the transpose of a matrix A is -1 4 and 5. Okay. So A transpose is equal to -1 4 and 5. This is the transpose of the matrix that have been given. This is the transpose matrix that they have given us. Now they want us to find the original matrix. So to find the original matrix here, we know that if you're looking at a transpose of the matrix, we just swap. Okay, we just swap or we just change the rows becomes the columns, the columns becomes the rows and vice versa. So this is a a row matrix. So the original matrix now here matrix A it was a column matrix which was like 1 -1 4 and 5. So this will be our matrix what our matrix A. So that is the answer. I hope it's clear. We can now move on to question B. So we have answered this question. Now we are moving on to question B. Okay.

[00:16:09]All right. So here is our question B. So question B reads given that we have this matrix which is 1 2 3 and it's been multiplied by this matrix 1 x and 5 which is equal to 24. Find the value of X. So to find the value of X, we are just going to multiply this row matrix time the column matrix. So row by column. So what we get here is we do this. We say one. So this 1 * this one 1 * 1 + 2 * X 2 * X + 3 * 5 3 * 5. Okay.

[00:16:56]then is equal to 24.

[00:16:59]Then from this stage we are just going to 1 * 1 is 1 + 2 * x we get 2x + 3 * 5 get 15 is equal to 24. So here we can say 2x is = 24. We are just correcting the like terms. So 1 + 15 it's a 16. So it will be 24 - 16. I hope you are following here. Okay. So from this stage we just get 2x = 24 minus 16. This will give us 8. So I want to find the value of x. It means we're going to divide both side of the equation by the coefficient of x which is a two. So we can cancel 2 and 2. x is equal to 2 a 1 2 8 it's a 4. So we get a four here. So x= 4. That is our answer for this question. So we are done answering question eight. We can now move on to the next question. So all right. So this is our question n. So this question n it has two questions A and B. So we're going to start with question A. So question A reads, a number is chosen at random from the set.

[00:18:19]So we have this set from 1 2 3 4 5 6 7 8 9 up to 10. Find the probability that it is a perfect square number. So here we are going to write we are going to write the formula we say probability of picking probability of picking perfect probability of picking perfect square is equal to so we are going to have Number number of favorable outcome meaning how many perfect square number we going to to pick number of favorable outcome over total number of outcomes.

[00:19:22]So here what we we mean we have a perfect square here. So a perfect square we are looking at a number which has got it perfect square root without a decimal number. So here we have one. One is a perfect square number.

[00:19:42]Also four is a perfect square number. Meaning that when you square one you get a one.

[00:19:50]When you square two, you get a four.

[00:19:53]Another perfect square number is a 9.

[00:19:55]When we square a three, we get a 9. So the perfect square numbers that we have picked from this list, we have 1 4 and 9. So here we will say number of favorable outcomes. How many have we picked?

[00:20:11]There are three. 1 2 3 over total number of outcomes. How many are they? 1 2 3 4 5 6 7 8 9 10. So we just divide by 10.

[00:20:23]So that means this is the probability of picking a perfect square number which is 3 / 10 from the list that you have been given. So this is our answer here. I hope it's clear. This is our answer.

[00:20:38]Okay. So we have answered our question A. We can now move on to our next question which is question 9b. Okay. So our question 9b reads solve the equation. So we have the equation which is 8 rais^ x is equal to 128. So here the first thing you observe that the base here is 8. Then here we have 128. So we can't express 128 in index form with a base of 8. No because if we write 8 squ this will give us 64.

[00:21:15]If we write 8^ 3 this will give us 512.

[00:21:20]So you see if the power of 8 the exponent increases we going to have a very big number. So I can't write 128 in index form the base of what? 8. But we can do this. Here we have 8 we have 128 here. So this 8 can be written in index form as 2^ 3 which is the same as 8 not.

[00:21:43]So this one we can write it like this. Then we write x here. Even this one we can write two power 7. So 2^ 7 it means you multiply the base two by itself seven times and you're going to get 128. Okay. So from here we can now apply one of the rules of indices which is power of the power we say 3 * x. So write 3 then we have 3x because 3 * x is 3x is equal to 2^ 7. So from here we can finish it here. The bases are the bases are the same. We can equate the powers which is 3x = 7. So we write 3 x = 7. Want to find the value of x we divide by 3. Even here we divide by 3. So this and this will cancel. x = 7 / 3 or you can say 3 into 7 is a 2 or number remainder 1 / 3. So this is our answer. Okay. So we have answered our question n. I believe we can now move on to the next question. Okay. So here is our question 10. So under question 10 we have a and b. We are going to start with question a. So question a says question a says set a is equal to prime numbers less than 12. So we need we need just to know the meaning of prime numbers. Set a. So a prime number is a number which has only two factors. So if we write the numbers from 1 to 12, these are the numbers we're considering from 1 to 12. 0 1 2 3 4 5 6 7 8 9 10 11 12. I' I'm not saying all these are prime numbers. No, I'm just following the statement prime numbers less than 12. So I've just considered the numbers which are less than 12.

[00:23:51]Okay, which are less than 12. Which are less than 12. And these are positive prime numbers. That's why I'm not including the negative prime numbers.

[00:24:00]Okay. So here zero is not a prime number. I've just put it because of the condition of the of the equation. Okay. Then we pick my prime numbers. So prime numbers are numbers which have only two factors. Zero not a prime number. One is not a prime number.

[00:24:17]It it only has one factor. Even zero only has one factor. Two is a prime number. So when you are listing set a here, you get two. It's a prime number.

[00:24:29]Three is also a prime number. Four is not a prime number because it has more than two factors. Five is a prime number.

[00:24:38]Then six is not a prime number. Seven is a prime number. So you write seven. Then 8 is not a prime number. 9 is not a prime number. 10 is not a prime number.

[00:24:49]11 is a prime number. So you end there because they said less than what? 12.

[00:24:56]And 12 is not a prime number. Now they said less than 12. That's why we ending on what? On 11. So this is our answer for question uh 10 a. Okay. This is the answer for question 10 a which is set A is 2 3 5 7 11. And this is a set of prime numbers less than 12. Okay, we can now come to question 10b. So question 10 B reads, the diagram below shows a sector A O B.

[00:25:25]So this is the sector A O B ar this is the ark A subains an angle of 21°. So here there is an angle of 21° at the center O. Okay. So we now look at this question. Given that the area of the sector is 14.85 85 square cm. Calculate the radius and take pi to be 22. So here we should come up with the formula for finding the area of the sector which is area is equal to theta / 360°*<unk> r².

[00:26:04]So here we need to come up with our data. Area we have been given in the question which is this one 14.85.

[00:26:14]Then radius is what they want us to find. We don't have. Then theta is an angle. Okay. We have an angle here which is this one 21°.

[00:26:26]So the angle there is 21°. Then the pi we have been given which is 22 over 7.

[00:26:35]So we just need to substitute on this formula. Where there is area we put the area which is 14 85 where there is theta we put an angle 21 / 3°* where there's pi we put 22 / 7 * r² that is what we want not so so I can rub this part here.

[00:27:02]So from here we can try to reduce the numbers. 7 can go into 21. 7 1 7 into 21 is a 3. So this is step one. You need to know these tricks. Then we have 22 and 3. You can divide by 2. 2 into 22 is 11.

[00:27:20]Uh 2 into 3 which will give us 180. So we are having something like this. 14.85 85 is equal to we'll get 3 * we have 3 here 3 over 180 then time 11 * r² okay so I hope we are moving together here I hope this part is it's very clear to you so now here this is paper one.

[00:27:58]This is paper one. So you need to work out this question step by step. So you can reduce 31 3 into 18 is a or 3 into 18 and 0.

[00:28:11]So you can now say 1 * 11 * r². You're going to have you're going to get 14.85 is = 11 * r 2. I've just multiplied 1 * 11 * r² over so from this stage now what we can do is we can divide this by 1 then we cross multiply. So when we cross multiply we are going to get something like this 11 * r² = 14.85* 85 time that is what you're going to get. So what you only need to do is you need to know 14.85 what is it going to give you? In fact let me just yeah anyway finish it. Yeah.

[00:29:08]So 14 85 * I'm multiplying it here. So I've ignored the decimal point on 14.85.

[00:29:18]Okay. So 0 * 5 is a 0. 0 * 8 say 0. 0 * 4 is a 0. 0 * 1 it's a 0. So I'll put a zero here. 6 * 5 is a 30. Write zero. We carry three. 6 * 8 we are going to get a 48. 48 + 3 we are going to get 51.

[00:29:41]So I'll write a one here. Then I'll carry a 5. 6 * 4 it's a 24 plus a 5 29 I'll write a 9 I'll carry a 2. 6 * 1 6 + 2 I'll get 8. So write eight here. Okay write 8 here. Let me just write it nicely.

[00:30:05]So we get eight here. Then here there is nine. We can close. Then we add. So we get 0 0 1 9 8. Now this number on top here it had a decimal point which was two decimal places 14.85 this one. So you put the one2 here. So meaning that when we multiply 14.85 * we get 891 this is the number that we get. Okay. So we come back here we finish our question. So we have 11 * r² which is equal to this we have said this giving us 891.

[00:30:47]Okay. So you want to find r² you divide by 11 we divide by 11 we cancel. So here we remain with r² = 8 91 / 11. So we can now do the actual division here. R² is equal to 11 into 8 9 it's 8 because 11 * 8 you normally get 88 okay so 11 into 89 it's 8 remainder one 11 into this 11 it's a one that is what we get so we have r² is equal to 81 we want the value of r not r² hence we introduce a square root so R will be equal to plus or minus 9. So therefore radius is equal to 9. So this is our answer. Okay. So we have answered our question 10b. Uh this question right here. We have answered it. Question 10 B. We have found the value of what of radius which is to be a nine. Okay. So we can now move on to our next question.

[00:32:05]All right. So this is our question 11.

[00:32:08]So the question reads the diagram below shows the positions of towns A, B and C on the A surface. So this is the diagram showing the A surface where we have positions A, B, okay, and C. So we are going to look at question A. The question reads, if it is 0820 at A. So at this point A, it is 0820 hours. What time is it at C? So what is the time at C? So here how are we going to find the time at C? Then we have the time at A. So this is very very easy for you. So what you need to do is the first thing we need to find what we call time difference.

[00:32:55]We need to find the time difference.

[00:32:57]Okay, we need to find the time difference between point A and point C.

[00:33:03]So how do we find the time difference?

[00:33:06]So TD meaning time difference. So the time difference is equal to difference in longitude. So we're going to find the difference difference in longitude between point A and B over a 15°.

[00:33:27]Okay. So we need to find the difference in longitudes between point A and B over 15°. Okay, that is how we we are going to do it. So from here point A is lying on which longitude?

[00:33:42]Is lying on 20°?

[00:33:45]A is on longitude 20° east. Okay. Then C is on longitude 95° east. So you can see that both of these two longitudes they are lying on the same hemisphere east east. So to find the difference in longitude between point A and B we say 95°us 20°. So we subtract. Okay. So 5 - 0 it's a 5 9 - 2 to give us 75°. So the difference in longitude will be a 75°.

[00:34:27]So what we are going to do now is we are going to say time difference is equal to difference in longitudes between A and B. We have found that it's a 75° then we divide by 15°. Okay. Then we divide by 15°.

[00:34:46]So here we just need to know that how many times can 15 go into 75. So 15 here 1 15 into 75 it's a five. So the time difference we having 05 hours. Now in case someone has a question where where is the 15° coming from? So we know that in 24 hours, okay, in 24 hours, in 24 hours the earth rotates at an angle 360°.

[00:35:18]Then in 1 hour, in 1 hour, how at what degrees is the earth going to to rotate?

[00:35:25]So we put x here so that we find out in 1 hour how many degrees the earth was going to rotate. So what we do we divide three° divided by the 24 and this gives us a 15 that is the 15 you are seeing okay that is the 15 you are seen I'm I'm doing this because I get a lot of questions from students where the 15 is coming from so that is where the 15 comes from so after we have found our difference in wrong our time difference between point A and B then We can find the time at C.

[00:36:03]We can say therefore time at Calal.

[00:36:08]So we get the time at A. At the time at A they said is 0820. The time at A they said it's 0820.

[00:36:17]Then the time at C the time difference we have found that is 05 hours. So what are we going to do? Are we going to add or subtract? Let's look at the longitude for C. C is on longitude 95° east. So because it's on east, east gain time. So we are going to add because it's on east. So that's why you're going to add here. We say plus. So we get zero. Here we get a two. 8 + 5 it's a 13. We write a 3 carry 1. 0 + 0 is 0 + 1 13. So this will be the time at at C. So the time at C will be uh 1320 hours. This is the time at C. So we have answered our question A. I hope it's clear. We can now move on to our next question which is our question B. Okay, you can always pause the video and see the calculations. If you have a question, you are free to ask. So now to question B right here. So question B reads, a plane, a plane flies from A to B at a speed of 100 knots. How long does the journey take if AB is equal to 4,800 N hours? A plane flies from A to B at a speed of 400 knots. How long does the journey take if A is equal to 400 knots?

[00:37:46]So what you can notice here is uh this question they saying how long if they say how long they are talking about time. So because they are talking about time we going to use this formula speed is equal to distance over time. So in our question when we organize our data the speed you have been given which is 400 knots. Then distance we have been given which is 4,800.

[00:38:16]Okay, then time is what they are asking us to calculate. So to calculate time, we are just going to substitute where there is s. Okay, where there is s speed, we put 400.

[00:38:30]Where there is distance, we put our distance which is 4,800 divided by time which we don't know. So from here we are going to divide this by one. And then we cross multiply 400 * t we will get 400 t okay = to 4, 800. So from here uh we can divide this by 400 so that we just remain with t.

[00:39:02]Okay. So here we can now say uh we can cancel my z. I'm sure you have seen this 400 and this 400 will cancel. We say t is equal to so these my zeros can cancel here. Then 4 here 1 4 into this 4 it's a 1. 4 into 8 it's a 2. So 400 into 4,800 we are going to get a 12. So say 12 hours. So this will be our time. Okay.

[00:39:33]So we have also answered our question B for question for question 11. Okay, we have answered both question A and B.

[00:39:44]Okay, so we can move on. So we said the time here is 1320 hours then the time is 12 hours.

[00:39:58]Very simple. We are now moving on to the next question which is question 12. So under question 12 they are they asking us to find the length of a piece of wire. Okay sorry they're asking us to find the tolerance. So the question reads the length of a piece of wire is measured as 4.5 cm. Calculate the tolerance.

[00:40:20]So how do we calculate or how do we find tolerance? So tolerance is equal to absolute error * 2. Yeah. So now when you look at the question, we are given the true value.

[00:40:38]So the true value is 4.5.

[00:40:41]Then we do not have the recorded value.

[00:40:44]So to find the absolute error if you do not have the recorded value we can just find the smallest unit of measurement for this true value because it's a decimal number. So the absolute error or the least unit of measurement will be 0.05.

[00:41:01]So here where there is absolute error we put 0.05 that is absolute error* 2.

[00:41:08]Okay.

[00:41:10]So to understand this topic approximation in detail make sure you look for my for the videos on my channel for approximation. I've explained fully how to find the absolute error. So here we multiply by two. So 2 * 5 it's a 10 you write a 0 carry 1. 2 * 0 it's a 0 plus that one write 1 2 * 0 it's a 0. So this means we have 0.1. So the tolerance here was a 0.1 that was the tolerance. Okay. Then we can come to question B. They are saying can we find the relative error? Okay.

[00:41:50]Can we find the relative error? So to find the relative error what you are going to do is this. We are going to say relative error. So this is question B is equal to it has a formula absolute error divided by true value. Okay. So remember our absolute error we have said it's a 0.05 over the true value. This is the true value 4.5.

[00:42:26]So the the numerator is a decimal number and the dometer is a decimal number. So to get rid of the points on both the numerator and the denominator we are going to multiply by 100 even multiply by 100. So we are following the place values of the of the numerator. Okay. So here 0.05 * 100 you get a 5 over 4.5 * 100 we get a 450.

[00:42:55]Okay. So we can just finish it here. We can finish it this side where we say relative error is equal to. So we can now divide five here 1 then 5 into 45.

[00:43:11]We know that it goes there many times 9. I hope we are moving together. It goes there nine times. So 9 and zero we have a 90. So we have 1 / 90. So this is our relative error. Okay. So we have answered this question right here both question A and B. So on the answer space here we have said the tolerance was a 0.1 and the relative error it's a one over a 90.

[00:43:44]So we are we are done with this question. We are done with this question which was question 12. We can now move on to the next question. All right. So this is our question 12 and this is a question coming from grade 11 work under theorem. So the question reads in the diagram below a b c and d are points on the circumference of the circle center o. So we have this diagram where we have the points a b c d which are points on this circumference of the circle at the center o. Uh we have d a t we have d a t the angle which is which is 40° and we have b d c b d c which is a 30° then we have a t is a tangent to the circle at a. So a t this is a tangent of the circle at the point a. Okay. So before we before we even look at the questions these are the questions before we pay attention to them it's it's advisable for you to come up with the answers here when you look at the circle that you have just using the circle theorems you can come up with some angles here so the first part I can notice is here there is a 30 so if there is a 30 there it means even here I'm going to get a 30°. Okay, I'm going to get 30°. Why angles in the same segment? That point is very very important. Then here we can notice that from this tangent and from this center there is this radius and you have this tangent. So at the point of contact the radius and the tangent this forms this will form a 90° this entire angle. So if here you have a 40 it means for this angle to be it means there will be a 50.

[00:45:46]So that when you add 50 + 40 you get an I I I hope this part is I hope this part is is clear. I hope this part is clear.

[00:45:58]So now if there is a 50 here again even here we are going to have a 50.

[00:46:05]Okay. 50° angles in the same segment.

[00:46:11]Angles in the same segment. So this is very very important for you to to understand. Okay. Very very important for you to to understand. Then let's look at other other things here.

[00:46:29]Remember I've said here there is a 40.

[00:46:32]So this 40 again even here there will be a 40 and you cannot see that 40 50 + 40 this will give us a and when you look at this B GC this will be angles in a semi semicircle of which here it will be an okay so this again and this it will have a 40 because of the angles in the in the same segment okay So if we have a 50 here, we have a 40 here, we have a 30 meaning this and this 90, 90 + 30 uh gives us what 120.

[00:47:16]So it means to find the sum of these angles this will be degrees. Why?

[00:47:22]Because if you say 90 + 30, this gives us 120. Then we say 180us 120. This gives us a 50. Okay, this is how I found this here. I've considered this triangle because here I've got and there is a 30.

[00:47:43]Okay, so it's very easy for me to calculate the angle which will be right here that it will be a I hope you you are getting me. Okay, I hope it's clear. Now I've said here this and this angles in the alternate segment angle is in the same segment. Even this and this to be the same angles in the same segment. So 40° here. Okay. We have 40° here. We have a 40° here. Then uh we can notice that from this point here from this point a this radius here. Okay. Again from this point here the distance of this radius from A to O and from A to B it means that here if there is a 40 even here there will be a 40 because the rad we have two radius which are equal. So this entire angle because there is a 30 it means even here there will be a 10. So this 30 + 10 it will make this entire angle to make this entire angle to also be a 40°.

[00:49:06]I I hope you are getting me there. It will make that entire angle to be a 40°.

[00:49:12]So if this entire angle if a 40 and here is a 40, it means we can find the angle which will be uh we can find the angle which will be right here. So what we can we can do is since we have this triangle there is a 40 here again there is a 40 here then we wanted to find the angle let's say x so normally say x + 40 + 40° then is equal to 180. So this will mean x = 180° - 80° because 40 + 40 is 80. So x is = 100°. So it means here we are going to have 100°.

[00:49:57]We are going to have 100°.

[00:50:00]I I I hope we are we are moving together. I hope we are moving together there. Okay. So now uh from there let's try to let's try to look at the questions. Calculate CBD. So CB D C B D C B D this angle here we have said this angle it's a ski° then B A C we have B A C B A C B A C this angle here this is a 30 we said it's a 30 don't get that 40 is the entire angle Okay. Equal to a b.

[00:50:51]A or b. We have already found the angle which is 100. So here we get 100°.

[00:51:00]That means we have answered our question 13. We can now move on to question.

[00:51:07]>> So now we have come to question 14 right here. So under question 14 we have a and b. We are starting with question 14 a.

[00:51:14]The question reads in the diagram below B C D is a straight line. So this is the diagram where we have BC D which is a straight line. Then A= 12 cm and A is equal to 20 cm and angle A B C. So the angle which is on B is 90°. Find the value of cos A C. So they want us to find the value of cos A C D.

[00:51:42]So this is the angle they are looking for. Okay, that is the co. So to find cos AC, we know that looking at we know that for co it's going to be adjacent over hypotenus. So what you going to do is in order for you to find cos AC first we must find the co which is inside here which will be cos a cb. So we must find cos a cb the co which is inside a cb here. Again it's also adjacent over hypotenus. So if I want to find the co which is inside this will be the opposite. So the adjacent is here then the hypotenus it's a 20 we have. So how do we find adjacent? We are just going to use the pythogas theorem to find this bc which is the adjacent. So since we have the hypotenus and we have one of the opposite side to find this adjacent side we are going to subtract from the hypotenus. So we going to say maybe this is y. So you say y = the square roo<unk> of 20 2 - 12 2. So here y is = square roo<unk> of 20 is 400 - 12 2 it's a 144.

[00:53:15]So y is = So here when we have 400 - 144 so 10 - 4 it's a 6. Here we have a 9. 9 - 4 it's a 5. Then 3 - 1 it's a 2. So 400 - 144 it's 256 square root of 256 this gives us a 16.

[00:53:39]So here there was a 16 cm. Okay. So I've just used the pythograph theorem to find BC. Okay. Now cos ACB cos AC is equal to our adjacent is 16. Then the hypotenus is a 20. So I can simplify this by looking for the highest common factor of 20 which is a four. Okay. 4 into 16 it's a 4. 4 into 20 it's a five.

[00:54:09]So this is what we have 4. This is cosb.

[00:54:14]Since we have cos there we can now say therefore we can now say therefore therefore cos will be equal. So cos a cd the one which is here is a 4 over 5. So the one which will be outside here it will have a 4 over 5. So cos a cd is equal to -4 / 5. So meaning that the one we have found inside is in the first quadrant where all angles are positive.

[00:54:46]So probably the one which is here it's outside. So look at those quadrants. So the one which is in the first quadrant where positive then the one that you found which will be outside here will be in the second quadant where the co is negative that's why we have put4 so that is the answer there cose a c=4 5 we now move on we come to question B so this is question B so the question reads A and B are points with coordinates -3 and 5a 9 respectively. Find the length AB. So the length AB is the distance. So first of all we have the coordinates for A is -3A 3. Then for B we have 5A 9. So we can name our coordinates. This is X1 Y1. This is X2 Y2. So the distance. So I can say distance distance of AB is equal to. So this is the formula for distance square root of x2 - x1 in brackets squar + y2 - y1 in brackets squared. So we just going to substitute.

[00:56:09]Okay. Where there is x2, we going to put a five. Where there is x1, we put a -3.

[00:56:16]So this this will change. It will be + 3. Because we going to have 5 - -3. So you say 5 negative it will be positive 3. Okay? That's why I've written a 5 + 3 in brackets plus y 2 it's a 9us y1 it's a 3 squared. So from here can just say you're going to have square root of 5 + 3 it's 8 squar plus 9 - 3 9 - 3 it's a 6 2. So from here we are going to get square root of 8 2 is 64 + 6 2 it's a 36. So when we add 64 + 36 this gives us a square root of 100 which is equal to square root of 100 is a 10 can even put 10 units. So that's our answer here. So in our answer space the first one we said we have -4 over 5. The second one the distance we have found 10 units. I hope this is clear. We have answered our question 14. We can now move on to our next question. So this is our question 15. So under question 15, we're dealing with the variation coming from driven work. The question reads, A varies directly as B and as the square of C and A= 30 when B= 2.5 and C= 2. Find the A the value of K the constant of variation. So here to answer question A we must come up with the variation equation. A varies directly. So because it's varying directly to B. A varies directly as B. So as B. So first we are going to write the constant and then we write B and as the square of C and as the square of C. So this is the variation that we're going to come up with. So it will be a is equal to k b c c².

[00:58:29]So you should find the value of k here when we have the value of a which is 30.

[00:58:36]So a = 30 b is = 2.5 and c is = 2. So throughout question a, b and c you'll be using this equation.

[00:58:47]Okay? So if you get it wrong, it means everything is going to be wrong. So there is a we put a 30 k is what we looking for b it's a 2.5 * c it's a 2 c is a 2 so it will be 2^ 2. So this will be 30 = k * 2.5 * 4. So this will be 30 = 2.5 * 4 2.5 * 4 let me just multiply rather than saying the answer. So 2 * 5 is a 20 you write a zero carry 2 then there's a point there 2 * 4 it's 8 + 2 it's a 10. So meaning that 2.5 * 4 it will give us a 10. So I'm going to have k * 10. Okay. So you divide by 10. Even this side we divide by 10 and we going to get the value of K here. K is equal to. So this will cancel. This 0 and this Z will cancel. 1 into 2 3 it's a three. So K is = 3.

[00:59:58]Don't forget this answer. K is = 3. So we can now come to question B. So under question B they saying can we calculate we find the value of A when B= 2 and C= 3. So we are going to use this same equation remember. So we write it here.

[01:00:19]This is B. A = K * B * C² if you want.

[01:00:27]So we want to find A. K we know it's a three time B it's a 2.

[01:00:35]* C it's a 3 which is 3 * 3 because the C² so A is equal to 3 * 2 uh 3 * 3 it's 9 * 2 it's 18 okay then 18 * 3 that's what we have 18 * 3 I believe this gives us a 54 okay 18 * a 3 this is uh 24.

[01:01:06]Yeah, he say 54. So, A is equal to 54.

[01:01:11]So, we have answered the question A and question B. We now move on to question C. So, question C they are saying find the values of C when A is equal to 300 and B is equal to 4. So, they are saying values of C meaning that C is going to have two possible values. So we can write the equation a = k * b * c². So they have given us to say a is equal to this k we already know it's a three and c is what we are looking for. So let's look at the value of a. They have said a is equal to 300. So a is equal to 300 here. So where there is a you put a 300.

[01:01:55]Where there is K you put a three then times where there is a B I'm sure they have given us a B is equal to 4. So A B is equal to 4. So there is a B you put a four then times C².

[01:02:12]Okay. So here we can divide by the same 3 * 4. Even here you divide by 3 * 4. So this is paper one. You must have this.

[01:02:23]You must know simple tricks to use to to to work out the numbers. Okay, you cancel this.

[01:02:30]So we have C² is equal to here 3 1 3 into 2 3 it's 1. So I'm remaining with 100 / 4. Okay 100 / 4. Again I can simplify further. We can say C² is equal to 4 year 1 4 into 10 is a 2 remainder a 2 4 into 20 is a 5 4 into 100 goes there 25 times. So introduce a square root even here. So we going to have C is equal to plus or minus 5. So here K we said this three. Then A we said this 54.

[01:03:10]Then for C since we have C is equal to plus or minus 5. So meaning that C is equal to positive 5 or C is equal to -5.

[01:03:20]Okay, that is our answer. I hope it's it's clear. So we have answered this question which is very very easy to to deal with. Okay, we can now move on. We go to the next question. Uh this is our next question. Question 16 which is having A and B. So we are going to start with question A. We should find the integral of 3x^2 + 8 x - 5. So we are finding the integral of 3 x² + 8x - 5 dx. So this is very easy to work out because we are just integrating. So when you integrating what you do is the power we add by 1 then we divide by the same thing. So this will be 3 x^ 2 + 1. So here the power is 2 + 1 even said over 2 + 1 + 8 x. So this x has a power of 1 then + 1 again over 1 + 1. Okay, I hope you are following here. Then I'll say minus 5 x okay 5x meaning this x is having a zero because x^0 it's one so that's why I've put a zero here then plus a one okay then after you are done doing this then you are not done you should say plus a c the bit constant so this gives us 3 x^ 3 even here it will be over 3 + 8 x^ 2 again / 2 - 5 x^ 1 because 0 + 1 is 1 then plus a c.

[01:05:14]So from there we can now cancel the 3 is here we have x^ 3 + 2 into 8 it's a 4. So we have 4x^2 - 5x + c. So this is our answer for our question A. Very simple to integrate.

[01:05:36]Okay. Very simple. So we can now move on to question B. So this is our question B right here. Okay. So for question B, uh they are saying uh the diagram below shows two triangles A and B. So we have triangle A and triangle B. So this triangle A and we have triangle B. So the question says describe fully a single transformation which maps triangle A and to triangle B.

[01:06:06]So triangle A was mapped onto triangle B. So it means this triangle A this triangle A right here. This is uh this is the object and this triangle B is the image. So from this stage now before we describe this transformation before we describe this transformation there are certain things that we must know. We have seen that triangle A was mapped to triangle B triangle A was mapped to triangle B. So you can see that the shape is the is the same this shape of triangle A is the same with triangle B. It's the same not so then we can notice that it was not rotated. It's not rotating because if it was rotating, it would have been something like this.

[01:06:54]If it was rotating again, it's not a reflection because if it's a reflection, it would have been it would have come down here or the other side. Okay.

[01:07:04]Again, we can notice that the shape does not change the the size the size of the shape was not changed. It's the same thing. So this will tell you that again this same shape which is not being enlarged. So this will tell you that here it was just the shape this shape triangle A it was just moved from one position to this other position without changing its size without changing the shape or the direction. So it means that this shape was translated. So we are looking at translation. We know the type of transformation we're looking at. is a translation. So translation is um is just a type of trans transformation which involve the movement of a shape.

[01:07:53]Okay, like this shape is moving from one position to the other position without changing the size, the shape or the direction. So how can we describe fully this transformation? So because it's a translation, we are going to need the translation vector.

[01:08:11]we are going to need the translation vector. So how do we come up with the translation vector? It's very simple.

[01:08:17]There are two ways. I'll show you the first way. Okay. So from this point here from this point which will correspond to this point here. Okay. I'll move in the x-axis I'll move I'll move 1 2 3 4 5 6. meaning here because I'm moving in the x ais which is a negative so I'm going to have a -6 then I'll go up in the y ais from here 1 2 to come here so it's a two this will be my translation my translation vector so this transformation the way we are going to describe it you're just going to say it's a it's a translation We say it's a it's a translation. Then then we say the translation vector which is uh 2 sorry -6 over pos2. So this is your answer and you get your full maxi. If you don't want this method, you can use the other method of saying translation vector is equal to coordinates of the image the coordinates of the of this image minus the coordinates of the what of the object of which this you say transition vector is equal to I've said of the image. So here the coordness of the image I look at what I have this image for this first point here. Okay, I'll get a -2.

[01:09:57]Okay. Then there is a three here. -2 and a three. I'll marry them. Then minus the coordinates of the image for this point here. Okay. We have a four in the x axis over a one in the y ais for this point over a one. That's what I'll do. Then I'll say -2 -2 - 4 I'll get -6. 3 - 1 I'll get a 2.

[01:10:23]Again my translation vector will be the same. You can even do that. It's fine.

[01:10:27]Okay. So, we are done answering this question. We can now move on to to the other question. So, this is the other question. So, this is a um a bearing coming from grade 10 work. So, the question reads in the diagram below A, B and C are three points on level ground.

[01:10:45]So, A, B and C. These are three points on the level ground. The bearing of A the bearing of B from A B from A is 62 and angle A B C A B C is 128 C is due east of A find the bearing of C from B C from B so where they have said from that's where going to draw the we going to draw the north on B because C from B so on this We will draw a north like this. Okay, we will draw a north like this. So when we draw the north, when we draw the north, uh this is what we are going to do. Uh let me indicate the north. Here we have a north. So the bearing they talking about is coming from the north in the clockwise direction until we meet this line for BC. So we need this entire angle right here. Now when you extend this, you're going to have a 180 because you said this is a straight angle. This gives us a 180°. So to find this entire angle here, I must subtract 180 minus the angle which will be down here. The angle which will be here. So I need to find this angle. So to find this angle, we know that this entire angle is 128. So we should know the angle which will be right here. Now you can notice that if you have a north and you have a north so it means this angle which is alternate to the angle which is so it means even what okay this angle alternate angles they are equal.

[01:12:35]So to know the angle it will be here because this entire angle is 128. So I'll say 128°us 62° then 8 - 2 is a 6 um 12 2 - 6 it can't borrow a one here remain with a zero 12 - 6 I get a 6. So even here there is 6° so remember I've said to find this bearing we know that the entire angle here it's 180. So to find just this angle here we say 180°us 66° this angle which is making it 180.

[01:13:17]So what we are going to do is we just subtract where we say 180 - 6. So 0 - 6 it can't we borrow here remain with 7.

[01:13:26]10 - 6 it's a 4. 7 - 6 it's a 1. Then 1 - 0 it's a 1. we get 114°.

[01:13:35]So meaning that the bearing of C from B will give us 114°.

[01:13:42]Very very easy. Then we come to the other one. The second one says find the bearing of A from C. A from C. So it's saying from C. That's where we draw the the north. We draw the the north on C.

[01:14:02]Okay. Let me just draw it exactly on on only C.

[01:14:10]We will draw the north.

[01:14:18]Mhm.

[01:14:22]Okay. So, we have drawn the north on C. Yeah. Let me just do extension. So we have the north here. So the bearing you're talking about is all the way from here up to this line. This is the bearing we are talking about of a from C. So we need this entire what angle?

[01:14:45]We need this entire angle. Very very important. We need this entire angle. So how do we find this angle? It's very easy to find this angle.

[01:14:59]Okay, it's very easy to find this entire angle. So what you can do is uh to find this entire angle we know that here we have a 90 not so we have a 90 even here we have a 90. So this 90° here and this angle which can be here this also will be a 90° we know that again you can have 180 just this part here it's 180 okay I hope you are getting me this is the first method so you can say 180° plus 90° this 18 here and this angle here which is And this can give us a 270°.

[01:15:51]Okay. Again, if you don't want this method because we have drawn a north here. So this will be an okay. So if I'm having here and this entire angle, let me show it. This entire angle if it makes a complete 10. If it makes a a complete 10 like this, if it makes a complete 10, just a minute. If it makes a complete 10 from here up to here, okay, it gives us 300.

[01:16:20]We know that. Now, if I've got here and I want this just this entire bearing ending up to here, I will say three minus this here. So, it also give us a 270°.

[01:16:41]So this bearing is very easy to be found. So this is a 270° very easy. So we have answered our we have answered our question.

[01:16:53]We have answered this question. I I hope it has it has helped you. I hope it has helped you. I've used two different formulas here. Okay. I've used this one.

[01:17:07]I've said if there is a 90 even here there is a 90 and this angle is 180 from here to here you can add them 180° plus 90° and this will give you 270° or if you don't want you know that if there is a north this will be 90 and this entire angle gives us a three so you can say 3°us this angle which is here so that you remain with with this angle which will also give you a 270. Okay, we move on to the next question and I believe this is our next question. So this question says the diagram below shows a plane figure made up of congrent semicircles. So congrent semic looking at circles which are equal they have the same length.

[01:17:56]Okay. So congrent angle Ale O here this is a 90. So it will mean that even angle A even it will be even so that we have three. So the question says describe fully the symmetry of the figure. So how do we describe the symmetry of the figure? So here we are looking at the order of the rotational symmetry. So the order of the rotational symmetry the way we find it the order of rotational symmetry. First we must find the order of rotational symmetry order is equal to so this entire angle this entire angle is 360°.

[01:18:46]So you get 360°.

[01:18:48]Okay. Then you divide by one angle which is 90° here. Okay. And when you divide you're going to get a four. So the order of rotational symmetry here is going to be a four about the center. So to describe this food we say we say the order we say the order of the order of rotational symmetry the order of rotational symmetry is four. This four that you have found here is four about is four about the center about the center about the center at 90°. So this is your answer you have answered. That's all.

[01:19:52]Very very easy to answer this question.

[01:19:55]We now move on to we now move on to the next question which is this question right here. So we have answered this one. We now come to question B. So question B says in the answer space below is an incomplete program written in pseudo code for calculating volume of a cuboid. So we are we are calculating the volume of the cuboid given the length L base B and the height H. Complete the program by filling in the blank spaces with appropriate statement. So this is the the statement here. So we are talking about volume.

[01:20:34]We are talking about volume of a cuboid.

[01:20:40]So the volume of the cuboid this is the formula L * B * H. So if we begin the program what are we going to enter? We going to enter L. Then we also enter B.

[01:20:53]Then we also enter H. Here we have we are going to display volume is equal to L * B * H. That's all. This is the answer and you get your two marks here without wasting your without wasting too much time. Okay, I believe this is clear also relations and functions. So the question reads the functions f and g are defined by fx= 2x - 3 and g of x = 3x find function inverse of x. So question a we are finding the function inverse of x.

[01:21:36]So this function inverse of x we are looking at the function of f which is this one. So we are going to say f ofx is = 2x - 3. So there is f of x we put a y = 2x - 3. Then here we can make x the subject of the formula. So we have y + 3 is = 2x. So this three has gone to the left hand side. It was a negative has become a positive. Then we divide by x.

[01:22:11]Uh we divide by 2 because we are making x the subject of the formula.

[01:22:17]So I can cancel going to have x = y + 3 / 2. So the final answer now we will say therefore function inverse of x. So we are saying function inverse of x where there is y replace x + 3 / 2. So this is the answer for question one. That's the function inverse of x. We can now move on to the next question.

[01:22:49]Okay. So the next question is question B right here. Under question B, we should find G f of X. So this means G sorry F is going on G. So you should write the G function which is this one which is 3X. Okay.

[01:23:12]So I've said f will go on g. So the function for f is 2x - 3. So it will go on g where there is x. So 3 on this x we put the function of f which is 2x - a 3 like this. Then we can expand 3 * 2x this will be 6 x - 3 * 3 it's a 9. So this is the the answer. This is the g f of x. After finding the GF of X, we can now look at uh question C. Right here they are saying GF of two. So GF of two, we already have GF of X which is this one. So we use the same function. Now we'll be saying G F of 2. So this two is representing X.

[01:24:02]So this function 6 X - 9 where there is X we put a two. So it will be 6 then a 2 - 9.

[01:24:11]So this will give us uh 6 - 2 we get a 12. 6 - uh sorry 6 * 2 we get a 12 - a 9. Then 12 - 9 we get a 3. So this is the gf of 2 it's a three. That is our answer there.

[01:24:34]We we are done answering question 19 which is also very very easy. Okay. Very very easy. This is our question 20. So under question 20 starting with question 20. In the diagram below A is the point 0a 4. This is A 0A 4. B is the point 2 0. So B is the point 2 0 and O. O is the origin. Find the equation of the straight line through O which is parallel to the line A. So we are finding the equation of the straight line through point O which is parallel to the line A. So the equation of the straight line which is parallel to the line OB. Now one thing you must know is that we are looking at parallel lines.

[01:25:21]So this line A does another line which is parallel to this. So parallel lines parallel lines have same gradient.

[01:25:35]This is one thing you must know.

[01:25:37]Parallel lines have same gradient. So if we find the gradient of this line AB, we can even use it now to find the equation of the the straight line through or which is parallel to the line A. So here to find the gradient we use the points A which is 0 4 then you also have B which is 2 0. So we use these two coordinates.

[01:26:03]Okay. So what we going to do is we come up with the formula that we normally use where we say m this is the formula for gradient y2 - y1 over x 2 - x1. So m stands for grad. So this is x1 and this is y1. This is x2 and this is y2. So we just going to substitute here. y2 it's a 0 minus y1 it's a 4 over x2 it's a 2 or x1 is a zero. So this will definitely give us m = -4 because 0 - 4 is -4 over a 2. So 2 1 2 into 4 it's a 2. So the gradient will be a -2. So if you have found our gradient -2, we can now find the equation of the line of a straight line through which is paral to a. So this equation we are going to find it. We are going to use this formula y - y1 is = m open bracket x - x1. So here we have two points a and b. We can pick one. So for me I like picking the point where y1 will be zero. So I'll pick b so that it becomes very easy for me to substitute.

[01:27:34]So I've picked these points 2 0. So this will be x1 and this will be y1. Then m we have found that is -2. So we just substitute on m and y1 and x1. That's all. So y -0. This zero is not going to have any impact. So you can even remove it just y= your m is a -2 your x will maintain your x1 it's a two like this.

[01:28:05]So y = -2 * x you get -2x * you get positive 2 * 2 you get a 4. So this is the equation of the straight line. Okay this is the answer. Now even if someone picks the points for a you are going to get the same answer. Okay you get the same answer. So we have answered our question.

[01:28:31]We have answered our question A. We can now answer our question B. So we now come to question B right here. So question B reads the heights of two similar cylinders are 4 cm and 6 cm. If the volume of the smaller cylinder is 48 cm cubed, find the volume of the larger cylinder. So here we have we have we have smaller we have smaller cylinder then we also have larger cylinder.

[01:29:09]That's the that's the information that we have. So what we are going to notice is we have the height. So four is the height of the smaller cylinder. Then six is the height of the bigger cylinder.

[01:29:21]Okay. So what we going to do is we take them back to the original heights. Half of four it's a two. Then half of six it's a three. So since we talking we want to compare to the volume. So we are going to write them 2^ 3 3^ 3 which will give us 2^ 3 it's 8 and 3^ 3 this gives us a 27. Now we can equate this to their volume because at least we have equated to the cubes. Okay. So the height of the smaller the volume of the smaller cylinder it's a 48.

[01:29:58]Then the volume of the larger cylinder we put x because we don't know.

[01:30:04]So we are going to write this as a fraction because these are ratios. 8 / 27 is equal to 48 / x.

[01:30:15]So from this stage we can now cross multiply. We have 8 x is = 27 * 48. This is paper one. So you need to know how to play with the numbers. So divide by 8.

[01:30:31]Divide by 8. So this and this will cancel. Then we are going to continue on this other side. X is equal to. So here what we have? We have 27 * 8. 8 can go into 48 without giving us a remainder.

[01:30:56]So 8 here 1 8 into 48 is 6. So just multiply 27 * 6 we see what you're going to get. 6 * uh 6 * 7 gives us a 42. We carry a 4. 6 * 2 is 12. 12 + uh 12 + 4. So 12 + 4 it gives us a 16.

[01:31:20]So meaning that 27 * 6 gives us 162.

[01:31:26]Remember x was representing the volume of the larger cylinder. So you write 162 cm cubed. So this was the answer. So meaning that volume volume of the volume of the larger cylinder is equal to 162 cm cubed. Very easy to answer this question. So we are done answering this question. All right. So we now come to question 21. So question 21 reads, write down the three inquities. So they want three inequalities. Okay. They want three inequalities that define the unshaded region error on the diagram below. So we have inequality 1 2 3. So we say this will be our inequality one. Then this will be inequality two. This will be our inquiry three. So the first inequality it is cutting the yaxis here only 6 because we have 0 6. So this is y = 6. Okay. So we start with the easiest one y = 6. So this is our first one. Now they saying we should do it defines the unshaded region. So we should do here we are defining the unshaded region. So the shaded region is greater. The upper part is greater. Then the unshaded region is less. Okay. So this line is a solid line. So we're going to write y is less than or equal to 6. This is our first one. Then for this line here, the second one which is cutting both the x ais and the y ais. So the x ais is cutting at 8 6 then uh sorry the y ais. Then the x ais cutting at 2 0. So we pick these coordinates 8 6 and 2 0. Then we find the the gradient.

[01:33:35]When we find the gradient, we're going to use it to find the equation of the straight line here. So what we going to do is the formula for gradient, we know first we write x1 y1 x2 y2. So we can now write the formula for gradient which is m = y2 - y1 over x2 - x1. So this will give us let's just substitute our y2 it's a 0 y1 it's a 6 our x2 it's a 2 and then our our x1 it's 8. So this will give us 0 - 6 to give us -6 over 2 - 8 to also give us -6. So we have -6 / -6 which would definitely give us a positive 1. So gradient there is a one. When you find the gradient, now we can use it to find the equation of the straight line by using this formula. Y - Y1 is equal to M open bracket X - C X1.

[01:34:45]This is the formula. So remember we have two coordinates. We have ha 6 and 2 0 for this line. I'll pick this one. 2 0.

[01:34:55]You can even pick the other coordinates is fine. You'll get the same answer. So this will be x1 and this will be y1. So I'll just substitute y - 0 = where there is m we put the gradient which is which is a one because we have found a positive one. Then xus our x1 is a 2.

[01:35:19]This is how it's going to be. So this zero won't have any impact. So just write y = 1 * x is x. 1 * 2 is a 2. This is the equation that you have. Now we need to find we need to use the inequality. So the equation we have for this line is y = xus 2. Now we need to use the inquality sign to describe or to define the unshaded region here. So the upper region is greater the lower region is less. So we're describing the upper region which is greater which is unshaded. So this type of the line it's a it's a broken line. So the inequality is not going to have equal. So just write y is greater than x - 2 because we are describing the unshaded region which is upper region which is greater. So this is the second one. We are now remaining with our third line. So the third line we'll answer it from here. So this is the third line which is which is having these two coordinates 0 6 and 2 0. So don't forget 0a 6 and the 2 0. Again find the gradient here. So this is x1 y1 x2 y2.

[01:36:41]We are doing the same things. M = y2 it's a 0us y1 is a 6. Then our x2 it's a 2. Our x1 is a 0. So 0 - 6 is a - 6. 2 - 0 is a 2.

[01:36:57]So 2 a 1 2 into 2 6 is a 3. So the gradient here is a -3. This question carries five marks. So just by finding the gradient you'll be given a mark.

[01:37:08]Okay. Then from there we can find the equation of the straight line by using our formula which we know y - y1 is equal to m in bracket x - x1 we close.

[01:37:22]So we pick two coordinates again we are one coordinate either this one or this one. I'll pick this one x1 y1. The reason why I'm picking this one because I'm looking at y1 will be a zero. So I I I'm going to have less work here. I'll just write y. I know it will be y. Now this zero won't have any impact. So I'll remove it. Okay. So I'll just write y= my m is a -3 open bracket x - x1 here it's a 2.

[01:37:54]Okay that is what you do. So from there we are going to have y is equal to -3 * x is -3x.

[01:38:04]Uh * positive 3 * 2 it's a 6. This is the equation that you're going to have.

[01:38:10]So for this line, this line here, this line here, this line was y = -3x + 6. Now we want to describe or we want to describe the unshaded region which is the upper region. So the upper region is greater. This lower region is less. So this line is a solid line. So it's going to be greater or equal to. So we say therefore y is greater than or = -3x + 6. So this is our answer. So these are the three inquiries that we are needed.

[01:38:50]This one, this one and the first one. So these are the answers. Okay, we are done answering this question. If you have any question, okay, or any contributions, you are free to leave a comment in the comment section. and I'll be able to respond to you. We go to the next question guys. We have come to question 22. So under question 22, we have a and b. So we going to start with question a.

[01:39:17]So for question a normally m is the point 0a 5 and mn is equal to this 3 and 4. Find n. This is a vector a position vector question. So to answer this question, normally what I do is I I like drawing a triangle like this. So I'll put the origin here. Then I'm having point M and N. Okay, like this. Even put the arrows. So they saying M is the point 0a 5. So this M it's a point from O M. It's a point which is 0 and five. I've written it in vector form. Then MN we have -3 and four. So they want us to find O N. So to find O N O N you just say to find O N you say O M plus MN.

[01:40:26]So what we are going to do is we know what OM is. OM 0 and 5 plus MN we have -3 and 4. So once we add these two okay we say O N is = 0 + -3. This will give us -2.

[01:40:51]Okay. Uh this will give us sorry to give us -3. Sorry for that. 0 + -3 this will give us -3. 5 + 4 this will give us a 9.

[01:41:05]So this is the answer. This is O N. Very very simple to answer this one. Okay. So we can now come to question B. So look at the answer here. Very clear. Okay. We come to the other question now. So the other question here the other question it's question B that we are going to answer now. Okay. So for this question B we can look at question B. The question reads the question reads the sketch shown below represents the graph of the c y = x^2 - 4x + 3. So this is the parabola.

[01:41:51]Okay. So we can now look at find the coordinates of the points B and C. So we should find the coordinates of the points B and what and C. So what we are going to do is what we are going to do is look at our parabola here. So we have this parabola which has points A, B and C. So this line this is the X-axis.

[01:42:16]Again this is the line Y=0.

[01:42:18]So meaning that the coordinates for B and the coordinates for C here the Y coordinate will be zero here. So we just need to find the X coordinate here and the X coordinate here because B and C are lying in the X ais where Y=0 where the Y value is equal to Z. So how do we find how do we find the X coordinate? We look at the equation this equation of the parabola which is y= x^2 - 4x + 3. So what you do is where there is y here we going to put a zero. So going to have x^2 - 4x + 3 is = 0. Then want to find the value of x. So what we do is we can do this.

[01:43:09]We can do this equal to z. So I'll write x here. I want to factoriize this. So here I've got three. The product is a3.

[01:43:19]The sum is -4. So I'll look for two numbers such that when I multiply them, I'll get 3. When I add them, I'll get a -4. So those two numbers are -1 and -3.

[01:43:30]So I'll write 1. Then I'll write -3 here. So meaning that the first one I will say x -1 = 0 which will mean x = 1.

[01:43:41]Then the other one will be x - 3 = 0 which will be x = 3. So these are my points. So here where there is b it was a 1. Here where there was a 3. So the coordinates here for b it will be 1 comma 0.

[01:44:00]Then for c it will be 3 comma 0. So these are the answers very simple.

[01:44:09]I hope this is clear. Then we come to question two where they are saying find the minimum value of y. So the minimum value of y we are looking at the the turning point right here.

[01:44:26]Here what value what minimum value of y is presenting this turning point? What value of y do we have here? So to find this value of y, you can use different methods. Okay, you can use different methods and these methods are going to give you the the same answer. The first one is we are going to find the x coordinate of the turning point which can be found by x=b / 2 a. So here what we have is what is our equation? We have this is what we have. We have x^ 2 - 4 x + 3.

[01:45:13]This is what we have. Okay. So what are we going to do? We will say x = So from this we have a = 1. We look at the x² which is a 1. B = X on the middle is -4 then C = 3. So you are going to X = where there is B you put -4 over 2 where there is A you put a one because A it's a 1. So from here we say X = A4 2 * 1 is a 2. Then two here 1 2 into 4 it's a 2. So the x value is a two.

[01:45:56]Since we have found our x value, we can now find our y value. So to find the y value, we'll get this equation here, which is y = x^2 - 4x + 3 when x = a 2. So we just going to substitute here where there is x in this equation, we put a 2. y = 2^ 2 - 4 then 2 then + 3. So this will be y = So here 2^ 2 it's a 4. This will be 4 - 4 * 2 it's 8 + 3 and the final answer you're going to have 4 - 8 we get a -4 -4 + 3 it will give us a1.

[01:46:47]So this is the answer here.

[01:46:50]So to write in our answer space uh the first one we found that we are getting a -3 and 9. This was for question a for b1 the coordinates for b we have found that is 1 0. Then for c we have found that is 3 0. Then the minimum value of y we have found that is a -1. So we have answered our question 22.

[01:47:17]Okay, this was question 22.

[01:47:20]So we are just remaining with our last question which is question 23. All right, so this is our question 23 now.

[01:47:28]So the question reads, the diagram below is the speed time graph of a car.

[01:47:34]The car starts from rest and accelerates uniformly at 2 m/s squared. So the car starts from rest here and it accelerates at 2 m/s squared. So the acceleration here it was 2 m/s square 40 seconds 40 seconds here. So the acceleration was 2 m/s² for seconds until it reaches a speed of 16 m. So this is a speed 16 m/s.

[01:48:06]It then travels at 16 m/s for 40 seconds. So from here it travels for 40 seconds. Okay. at 16 m/s.

[01:48:17]It then travels 16 m/s for 40 seconds after which it comes to rest in a further 10 seconds. Okay. Find the value of t. So you should find the value of this t. So for question a to find the value of t. To find the value of t, this is what we are going to do.

[01:48:40]We are going to say acceleration is equal to final velocity minus initial velocity over time. So the acceleration here we said it was 2 m/s. So 2 = final velocity is 16 as you can see the initial velocity 0 here - 0 / t. So going to have 2 = 16 / t. We divide by 1. So we are going to have 2 * t is 2 t = 16. So over 2 / 2 that means our time there t is equal to 8 seconds. Okay.

[01:49:28]So time is equal to 8 seconds. All right. So we now come to question B. So we have answered our question A. We have said the time here is 80 seconds. Okay.

[01:49:41]So if the time here is 8, what does this mean? If you read the the question again, they said uh the car starts from rest and accelerates from 2 m/s 40 seconds. This is where the eight is until it it reaches a speed of m/s. It then travels at 16 m/s for 40 seconds. So from here from 16 m/s it traveled for 40 seconds. So from here to here it was 40 seconds. Now when you look at the 8. So here it means you're going to have a 48.

[01:50:20]Okay? Because 48 - 8 that's when you're going to have the 40 seconds that the car traveled. Then they have said after which it comes to rest in a further 10 seconds. So it comes to rest in the further 10 seconds. So meaning that where it was resting okay in the further 10 seconds when you add this 48 plus the 10 that is when you are going to get the 58 seconds. So you need to understand these principles. So find the distance traveled in the last 50 seconds. So we should find now the distance that was traveled in the last 50 seconds. So the distance that was traveled in the last 50 seconds we just consider uh we consider this.

[01:51:15]Okay we consider this. So remember we have said here it's 8 and here it's a 48 and in the last 10 seconds we should so consider this. Okay in the last 10 seconds. So in the last 10 seconds mean 40 + 10 that's when you get what a 50 here. So please pay attention this 58 is coming from because you have added 48 + a 10. There's a difference with what I've indicated. So what you going to do is we know that distance under a graph this is a trapezium is area under the graph is the same as distance. So this is a and this is b. So what we do is we are going to say our distance is equal to a + b in brackets h over 2 / 2 okay over 2. So we are going to substitute here. So distance is equal to so what you going to get here is we are going to substute a is the difference between 48 - 8 so 48 - 8 you're getting a 40 so a is a 40 plus b is from uh b we are looking at okay it's fine even if you maintain the 50 the 58 where we say 58 minus 8 you still get what you still get a 50 here. Okay, I hope we are moving together here. You still get a 50. Okay, just maybe for the sake of those who are confused, let's just maintain a 58 here.

[01:53:04]Remember we for B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B we consider this entire uh distance the difference between the zero and the 58. Now we have cut here.

[01:53:16]So we are considering this 8 and the 58 say 58 - 8 which is a 50. Okay. Then times h is the height here. So the height you look at the difference which is here which is a 16 from this. So the height is a 16 over a two. So what you are going to have here is distance is equal to 2 here 1 2 here it's 8 40 + 50 we get 90 * 8 and this will give us 90 * 8 8 * 0 it's 0 8 * 9 it's 72. So the distance is 72 m.

[01:54:01]So this was the distance 72 what? M very very easy there. Okay. So we we now come to the last question. So under the under the last question here under the last question they are saying find the speed of the car when t is equal to 53 seconds. So we should find the speed of the car when time is equal to 53 seconds. So this means the 53 seconds was somewhere here. We can say this was where the 53 second was. So the time here it means 58 - 53. Okay let me just write 58 minus 53 we are getting a five. So meaning that time was 5 seconds. We should know this. So don't get this actual time. So time here is 5 seconds. So what you do now? What the way going to find the the going to find the speed? We going to say since we know this we have this formula is equal to final velocus initial over time. We can make v the subject of the formula. We can make v the subject of the formula.

[01:55:19]We going to come up with a * t. We have a t is equal to v minus u. So this will be uh this will be a t this u go here it's going to be plus u is equal to v.

[01:55:34]So we have made v the subject of the formula which is becoming as v = a t + u. So we are going to use this. Now here we need to find acceleration. Remember we are looking at the last 10 seconds in the last 10 seconds. So acceleration we can find acceleration in the last 10 seconds. So acceleration in the last 10 seconds when we look at the diagram here. Acceleration in the last 10 seconds.

[01:56:06]Acceleration in the last 10 seconds what are we going to what are we going to to consider? So in the last 10 seconds normally we are just looking at deceleration there. Okay. of which the of which the the we are going to say okay here we are going to write 0 - 16 / 10 that's the time okay because we are saying final veloc minus initial velocity. So the final velocity when you look at the diagram here the final velocity is going to change the final velocity will be a zero the initial velocity will be a 16 okay that's why I've written 0 - 16 / 10 t will be the last 10 seconds so here we are going to have a = -16 / 10 and this will give us we can simplify uh 2 into 16 it's 8 so this is8 / 2 into 10 it's a 5. This will be our acceleration. So you say v = a acceleration you have said is -8 over a 5 * time remember time I said this is the five we close then plus what is the initial we have said this is 16. So v = this and this will cancel 8 + 16. So from here we going to have v = 8 - 8 + 16 you're going to have 80 m/ second. This was the final velocity. So I can just write our answer here. The first one we said the acceleration was it acceleration? it was time where we said t is equal to uh I think we came up with 8 seconds. Okay. Then the second one we are finding the distance we have said this is 720 m. The last one we are finding velocity which is equal to 80 m/ second. So thank you so much everybody. We have come to the end of our presentation. This has been your presenter Mr. Bye-bye.