GCE 20 Mathematics Past Questions

Published on Sep 1, 2025 • Education

Introduction

Hi guys, welcome to today's tutorial. Today we'll be looking at likely GC mathematics past questions that would be coming out in your forthcoming exams. Now our first question says what is the total surface area of a cuboid whose dimensions are 12 8 and 3 cm. Now looking at a cuboid we have the shape of a cuboid this way.

Key Insight 1

Now we are told that the length is 12, the breadth is eight and the height is three. Now for us to find the total surface area of this total surface area, we'll be using the formula 2 bracket length * bread plus length * height plus bread * height. Now where our length will be 2 brackets 12 * 8 + 12 * 3 + 8 * 3. Okay.

Key Insight 2

So this will give us two brackets. 12 * 8 would give us 96 + 36 + 24. So 96 + 36 + 24 would give us 2 * 1 5 6 which would give us 312 cm squared. So the area of our cuboid total total surface area of our cuboid will be 312 cm squared.

Key Insight 3

If sin 30° is 0.5, what is the cosine of its complement? Now when we talk about complimentary angles, we talk about sum of two angles that sum up to 90°. So if we have a right angle triangle this way, so let's call this adjacent. Let's call this sin 30.

Key Insight 4

So this will become opposite and this will be adjacent. All right. Now in this case we are told that our angle is sin 30 sin theta is 0.5 which is 1 / 2. Okay.

Key Insight 5

So if we are looking for the complement of sin 30 which is 60 the complement of 30 would be 90 - 6 - 30 which will give us 60°. So this will be 60°. So if you're looking for this we need to find sin theta is opposite over hypotenus. Okay, so bearing in mind that our opposite is one and our adjacent is two.

Exam Practice Concept 1

Okay, our hypotenus is two. So we need to find our adjacent using Pythagoras theorem that says that hypotenus squ= opposite 2 + adjacent squ. All right. So this gives us sin 30 will be 2 2 = 1 2 + adjacent squared.

Key Insight 7

So this will give us 2 will be 4 = 1. So plus adjacent. So our adjacent squared will be 4 - 1 which is 3. So adjacent will be <unk>3.

Key Insight 8

Okay. So adjacent being <unk>3. Now the complement of this cosine theta cos 60 in this case which is adjacent over hypotenuse would be here will be our adjacent and here will be our hypotenus. to cos 60 opposite here will be root.

Key Insight 9

Oh sorry considering cos 60 here cos 60 here will be our adjacent and this will be our opposite. Okay and why this is our hypotenus. So considering cos 60 cos 60 here will become 1 / 2. All right.

Key Insight 10

So cos 60 which is a complement of sin 30 will be still will still be 0.5. All right. And next question says a patrol attendant recorded the number of vehicles that bought patrol from him for 3 hours and recorded 12 cars and 27 buses. What is the probability that the next vehicle will be a bus?

Key Insight 11

Now the probability of car equals 12 over our total number of vehicles here will be 39. Now the probability of buses of bus will be 27 over 39. Now when broken down to it lowest 10 we have this as three here 9 three here 13. So the probability that the next vehicle would be a boss will be 9 / 13.

Exam Practice Concept 2

Two pars of a circle are respectively 46 cm and 40 cm long. If the radius of the circle is 25 cm and the center of the circle lies between the two cuts, what is the distance between the cuts? Now let's represent this. We have our cycle right and this is our center of our cycle.

Key Insight 13

We have our first code. They say the cuts are parallel. All right. That means they are straight towards each other.

Key Insight 14

And we have our radius from the center here 25 and this 25. Okay. Now if our first cord which is 46 cm long and this is a 40 cm long cord. All right from our circle we asked to find the distance between these two codes.

Key Insight 15

So we asked to find this. All right. Now for the purpose of this we know that in our circle theorem that when a line from the center of our circle b touches our cord it forms a 90° which means it's perpendicular to the cord. So bringing out this we have this as the first part.

Key Insight 16

Okay, let's call this y 25 and 46 / 2 will give us 23. So this becomes 23. Okay. And for the second one, the lower one, we have this as x 25 / 2 will give us 20.

Key Insight 17

Now in both of these triangles we have it as a a right angle triangle. So we can apply a pythagoras theorem. Now in application of a pythagoras theorem here we have this as 25^ 2 = 23 2 + y 2. So this gives us y^2 = 25 2 - 23 2.

Exam Practice Concept 3

So y = square<unk> of 25 2 - 23 2. Now y will give us uh 25 squar will be 625 - 23 2 which is 529. So 25^ 2 - 23 2 would give us y =<unk> 96. Now roo<unk> over 96 roo<unk> 96 would be 9 8 approximately 9.8 cm.

Key Insight 19

Now that's the value for y. Now working on x still theorem where we have 25^2 = 20^2 + x^2. So this becomes x^2 = 25^ 2 - 20 2. So x =<unk> 25 2 - 20 2.

Key Insight 20

Now roo<unk> 25 is 625 and 20 is 400. So this gives us 225. So square root of 225 will give us 15 cm. So for the distance between the two codes all right they say we should find what is the distance between the two cuts.

Key Insight 21

Distance between the two cuts will become x + y. Now x is 15 + y which is 9.8 would give us 15 + 9.8 8 would give us 24.8 cm. So the distance between the two CS would be 24.8 cm. The angle of elevation of the top t of a palm tree from a point R on the ground is 30°.

Key Insight 22

The distance from R to the base B of the palm tree is 120 m. What is the distance between T and R? Now we have our palm tree and we have our ground. Now this is B base and this is our T.

Key Insight 23

So now and these are points R. All right. Now they say the distance between B and R is 120 m and the point of elevation is 30°. Now remember that this forms what a 90° to the ground forming a right angle triangle.

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Exam Practice Concept 4

All right. Now that I adjust so we asked to find this all right in order to get this we need to use our since we have our adent this is adjacent and we are looking for hypotenus so we are using cos 30 equals was adjacent over hypotenus. All right. Where our adjacent is what?

Key Insight 25

120 and the hypotenus is unknown. Okay. So our hypotenuse x will become 120 / cos 30 and cos 30 is giving us x = 120 / <unk>3 / 2. All right.

Key Insight 26

Inverse of this will give us x = 120 / 1 * 2 /<unk>3. Okay. Now if we rationalize this we have this as 240 / <unk>3 which is going to give us when rationalized will give us 240 * <unk>3 over <unk>3. Okay, multiplying this out, our x will become 240 * <unk>3 all over 3.

Key Insight 27

So 240 / 3 will give us 80 <unk>3 m. So our answer would be 80 <unk>3 m. The probability that a student will pass a mathematics test is 3 over 5. If if three students are chosen at random, what is the probability that at least two of them will pass the test?

Key Insight 28

Now let's look at it. We have probability of pass is given as 3 over 5. That means the probability of fail will be 1 - 3 over 5 which is 2 over 5. Now if three students are chosen at random.

Key Insight 29

Okay. The probability that at least two passes out of the three, okay, will be probability of pass plus probability of fail. So the probability that at least two passes will be the probability of pass plus okay sorry the probability that at least two passes two pass plus the probability that three of them would pass. All right now if we have let's label this as a b c the three chosen candidates.

Exam Practice Concept 5

Now the first probability of two passing will be 3 over 5 3 over 5 and the fail here will be 2 over 5. Probability that A fails will be 2 over 5 3 over 5 3 over 5 and probability that B fails will be 3 over 5 2 over 5 3 over 5. So in each of this probability we have this All right, we have this as 3 * 3 9 * 2 18 over 125 18 over 125 and 18 over 125. All right, that's the probability that two would pass.

Key Insight 31

And probability that three of them would pass will be 3 over 5 * 3 over 5 * 3 over 5 which gives us 27 over 125. Now bringing all this together, let's look at the total sum of this would be 18 / 125 + 18 / 125 + 18 / 125 which gives us 125 all over 18 + 18 + 18 will give us 54 over 125. Now knowing that the probability that two passes will give us 5425 we have it as 54 over 125 + 27 over 125. Okay our LCM will still remain 125.

Key Insight 32

So 54 + 27 would give us 81 over 125. So the probability that at least two of them passes the test would be 81 over 125. What is the total surface area of a cylinder with radius 2 cm? We are given radius as 2 cm and height as 1.5 cm.

Key Insight 33

Now we say correct to two significant figures. Now we have our total surface area of a cylinder as 2 pi r bracket r + h. All right, where r is our radius, pi is our constant and h is the height of our cylinder. Now using this formula we have it as 2 * 3.142 * our radius 2 * 2 + 1.5.

Key Insight 34

All right. Now this gives us 2 * 3.142 * 2 brackets 3.5. Okay. So multiplying through we have this as 2 * 342 * 2 * 3.5 which gives us 43.988 cm m squared.

Key Insight 35

Now since we asked to round off to two significant figures to two significant figures means we should just have just two figures out of this entire numbers. So this gives us 44 cm squared as our answer. A wire is shaped to form a square of area 81 cm squared. We have a square.

Exam Practice Concept 6

All right. Now an equilateral triangle was formed from the same wire. What is the length of the sides of the triangle form? Now we know that for area of a square area of square equ= length * length or it's called L²= 81.

Key Insight 37

So for L will be square root of 81 which will give us 9 cm. So that means for each of these sides is 9 cm 9 cm 9 cm and 9 cm. So for us to know the entire length of the wire we'll calculate the perimeter of the square. So perimeter of square would be 4 L which is 4 * 9 which gives us 36 c.

Key Insight 38

So the perimeter of this square will be 36. 9 + 9 + 9 + 9 would give us 36 cm. Now since I said this wire was not shaped to form an equilateral triangle whose all sides are equal. Now the three sides are equal.

Key Insight 39

So you can have this as L L. Now if the three sides of our equilateral triangle are equal. Now for us to find the length of a triangle. Perimeter of triangle of triangle equals 36 cm which is the perimeter of our square.

Key Insight 40

So length of length of triangle length of one side of triangle triangle side length of triangle side equals 36 / 3 which gives us 12 cm. So one length is 12 cm. So which is our option D. One bag contains four white balls and two black balls.

Key Insight 41

Another bag contains three white balls and five black balls. If one ball is dropped from each bag, what is the probability that both are white? Now we have first bag four whites and two black. Now under this first bag the probability of white probability of white would be 4 / 6 which is 2 over 3 when broken down and probability of black would be 2 / 6 which is 1 / 3.

Exam Practice Concept 7

Now for the second bag we have three white balls, three whites and five blacks. All right. Now, probability of white here would be 3 over 8 and probability of black would be 5 / 8. Now, it says what the probability that if one ball is drawn from each bag.

Key Insight 43

So, this will be probability of white from bag one times probability of white from bag two. All right. So this will give us 2 / 3 * 3 / 8 which gives us what? 6 over 24 or we can still break it down here as two here one two here four three here one three here one which will give us 1 / 4.

Key Insight 44

So the probability that both will be white will be 1 / 4. A quadratic equation with roots x = -1 1 / 2 and x = 1 / 4 is so we looking for the quadratic equation that give rise to this roots. Now let's first of all convert this to improper fraction. This becomes 2 * 1 2 + 3 - 3 / 2.

Key Insight 45

So now having x = -3 / 2 cross multiply this becomes 2x = -3. All right. Moving over the other side becomes 2x + 3 = 0. Now for this other one we have it as x = 1 / 4.

Key Insight 46

Cross multiply we have 4x = 1. So 4x - 4x - 1 = 0. Now bringing these two terms together we have it as 2x + 3 * 4x - 1. Now this time this will give us okay let's do it this way.

Key Insight 47

2 x bracket 4x -1 + 3 bracket 4x -1 = 0. Now 2 * 4 will give us 8 x^ 2 - 2x + 12 x - 3 = 0. Now this will give us 8 x^2 + 10 x - 3 = 0. So this is the equation that gave rise to this root.

Exam Practice Concept 8

So our answer will be option D. Now what are the factors of 2x^2 - xy - 3 y^ 2? Now in order to get the factors of this we have to look for factors that can give us 2x - xy. So minus xy xy we know we will get have as + 2xy - 3x y.

Key Insight 49

All right this will give us minus x y. So computing this for this here in order to have a common factoriable factor we have it as 2x^2 + 2 x y - 3x y - 3 y 2. Now we can now factoriize this. Now what's common here will give us 2x all right x + y minus what's common here is 3 y will leave us with x + y.

Key Insight 50

So here we have it as 2 x - 3 y and x + y. So these are the factors of this. We have 2x - 3 y and x + y. Three sets of rice xy z x y and z are sold for 15 naira, 35 naira and 16 naira per kg respectively.

Key Insight 51

If they are mixed in the ratio of 1 is to 3 is to 2, what is the cost per kg of the mixture? Now if they are mixed in this ratio okay this will become 15 * 1 15 plus 30 35 * 3 would give us 105 + 60 * this will give us 120. Okay our total ratio here is what 6. So 105 + 120 + 15 would give 105 + 15 + 120 would give us 240.

Key Insight 52

So our total mix is 240. So for the cost per package of the mixture will become 240 divided by 6 which will give us 40. So the cost per package would be 40 NRA. When a shirt is sold for 12,000 NRA, a loss of 20% is recorded.

Key Insight 53

How much should it be sold for to gain 20%. Now our selling price is 12,000 NRA and we have loss of 20%. Now we know that for loss, loss is cost price minus selling price. All right?

Exam Practice Concept 9

And percentage loss is cost price minus selling price which is loss over cost price * 100 / 1. Now in this case our selling price is 12,000. So and our loss percent is 20. So imputing those data here we have this as 20 = cost price - 12,000 all over cost price * 100 / 1.

Key Insight 55

All right, cross multiplying we have this as 20 cost price equals 100 cost price minus 1200 0. So we having it as 1.2. Okay. Now collecting electrons this will become 1.2 equals 100 minus 20 100 cost price - 20 cost price.

Key Insight 56

So this gives us 12 1.2 equals 80 cost price. So our cost price will become 12 1.2 divided by 80. So we have 12 0 0 0 divided by 80 which gives us our cost price will be 15,000. Now that our cost price is 15,000 how much should it be sold to make 80% 20% gain?

Key Insight 57

Now our cost price is 15,000. All right. And our gain gain percentage gain is 20%. Now for us to gain we know we are having selling price minus cost price.

Key Insight 58

Okay. And our gain percentage gain would be selling price minus cost price all over cost price * 100 / 1. Now our gain percent is 20 equals selling price which we are looking for minus our cost price which is 15,000 * 100 all over 15,000. Now cross multiplying this gives us 300 300,000 equals 100 selling price minus 1.5.

Key Insight 59

Okay. Now moving our 1.5 over to this side, we have it as 300,000 + 1.5 = 100 selling price. So adding these two together, we have this as 1.8 equals 100 selling price. So selling price would be 1.8 divided by 100 0 cancel 0 leaving us with selling price would be 18,000.

Exam Practice Concept 10

So we we we have to sell the shirt for 18,000 to make 20% gain. The ratio of the circumference to the diameter of a circle is approximate is approximately equal to what? Now if we look at the circumference of a circle circumference of circle is given as 2 pi r and we know that diameter is 2 radius. All right.

Key Insight 61

Now they're asking for the ratio of our circumference to our diameter. Now if our circumference is 2 pi r and our diameter is 2 r, we can divide both sides by 2 r. So this divided by 2 r / 2 r. So this will cancel leaving us with pi is 21.

Key Insight 62

So the ratio of our diameter of our circumference to our diameter would be the value of our pi which is 3.142. Twice a number x is added to five. The result is at least 11. What is the range of the values of x?

Key Insight 63

So if we have a number to be x and they say twice a number. So 2 * x is added to 5 and the result is at least. At least means the least number should be 11. It could be greater than 11.

Key Insight 64

So we are using our sign greater than or equal to 11. All right. Now what is the value range of x? So that means from here we can solve this out as our inequality having this to be 2x + 5 is greater than or = 11.

Key Insight 65

So 2x is greater than or = 11 - 5 2x is greater than or = 6. Dividing both sides by 2 2x / 2 6 / 2. So our x will be greater than or equals to plus or minus 3. Okay, which is three.

Exam Practice Concept 11

So the value of x will ranges from 3 which is x is greater than or equals to 3. The mean height of three group of students consisting of 20, 16 and 14 students are 1.67 m, 1.50 m and 1.40 4 m respectively. What is the mean height of all the students? Now let's look at let's call this group A, group B and group C.

Key Insight 67

Now group A consists of 20 students, group B 16 and group C 14 students. Now the average height here is 1.67 m, 1.50 50 m and 1.4 m. Now we know to calculate our mean is our EFX over N. All right.

Key Insight 68

Now or our EF / N whichever. So for each and every of them we calculate their mean. Now we are given that the mean height is 1.67 equals E FX / N which is 20. All right.

Key Insight 69

And for group B we have it as 1.5 = E FX over 16. And for group C we have it as 1.4 time equals E FX over 14. So our EFX here EFX for group A would be 1.67 * 20 which gives us 33.4. All right, our EFX for group B would be 1.5 * 16 which gives us 24 and the EFX for group C would be 1.4 * 14 which gives us 19.6.

Key Insight 70

Now since we are looking for the mean height of all the students, we are going to add up all our EFX. So we have it as 33.4. 4 + 24 + 19.6 which will give us which will give us 77. Now the total summation for all those height of all the students will be 77.

Key Insight 71

Now this is for the EFS total EFX. Let's call this total EFX. All right. So the total number of students total number would be 20 + 16 + 14.

Exam Practice Concept 12

All right. This will be 30. 30 + 20 will give us 50. So our mean mean will become total efx total e fx over total number.

Key Insight 73

All right. So our mean would become 77 over 50. 77 divided by 50 would give us 1.54 m. So the mean height of all the students would be 1.54 m.

Key Insight 74

All right guys, with this we've come to the end of today's class. Don't forget to subscribe and like. Thank you. Hi, I'm sure you enjoyed the content you just watched.

Key Insight 75

To see more episodes as that preparing you for your oncoming exams, smash the subscribe button on this channel right now and visit the playlist that we've made for you covering over 50 type of different exams including jam wa Power.

Exam Tips for Success

- Always read each mathematics problem carefully before solving.

- Write down the formula clearly before substituting values.

- Cross-check your calculations to avoid simple mistakes.

- Understand trigonometric relationships like sine, cosine, and complementary angles.

- Practice geometry and mensuration regularly to build speed and confidence.

Conclusion

Practicing past questions is one of the best ways to prepare for your GCE mathematics exam.

It helps you familiarize yourself with the question style, improve accuracy, and build confidence.

Stay consistent, revise formulas daily, and you’ll excel in your forthcoming examination.

Keywords

- GCE mathematics past questions

- GCE exam tips

- How to solve cuboid surface area problems

- Trigonometry complementary angles

- Exam preparation for GCE mathematics

- GCE 2025 past questions and answers

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