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IGCSE Past Papers Maths
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0580_m24_qp_22 Question 8 Solution
Part (a):
The n-th term of the sequence is given as n2 - 3. Substituting the values for n = 1, n = 2, and n = 3:
For n = 1: 12 - 3 = -2
For n = 2: 22 - 3 = 1
For n = 3: 32 - 3 = 6
Therefore, the first three terms are:
-2, 1, 6
Part (b):
The given sequence is: 1, 3, 9, 27, 81. Let's analyze the differences:
From 1 to 3: Multiply by 3
From 3 to 9: Multiply by 3
From 9 to 27: Multiply by 3
From 27 to 81: Multiply by 3
This is a geometric sequence where each term is obtained by multiplying the previous term by 3. The general form of a geometric sequence is:
nth term = first term × (common ratio)n-1
Here:
First term = 1
Common ratio = 3
Therefore, the n-th term of this sequence is:
3n-1
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0580_m24_qp_42 Question 4 Solution
Part (a):
(i) Calculate the volume of the pyramid:
The formula for the volume of a pyramid is:
V = (1/3) × A × h
Where:
- A is the area of the base
- h is the perpendicular height of the pyramid
The base is a square with side length 12 cm:
A = 12 × 12 = 144 cm2
The perpendicular height of the pyramid is 9 cm.
Substituting the values into the formula:
V = (1/3) × 144 × 9 = 432 cm3
Final Answer: 432 cm3
(ii) Calculate the total surface area of the pyramid:
The total surface area consists of:
- The area of the base: 144 cm2
- The area of the 4 triangular faces
Each triangular face has a base of 12 cm. The slant height (l) can be calculated using Pythagoras' theorem:
l = √(62 + 92) = √(36 + 81) = √117 ≈ 10.8 cm
The area of one triangular face:
Atriangle = (1/2) × base × slant height = (1/2) × 12 × 10.8 = 64.8 cm2
Total area of the 4 triangular faces:
4 × 64.8 = 259.2 cm2
Adding the base area:
Total surface area = 144 + 259.2 = 403.2 cm2
Final Answer: 403.2 cm2
Part (b):
The total surface area of the toy is the sum of:
- The curved surface area of the cone: Acone = πr × l
- The curved surface area of the hemisphere: Ahemisphere = 2πr2
The slant height of the cone is 3r. Therefore:
Acone = πr × (3r) = 3πr2
The total surface area is given as 304 cm2:
304 = 3πr2 + 2πr2
304 = 5πr2
Solve for r2:
r2 = 304 / (5π) ≈ 304 / 15.707 ≈ 19.36
Solve for r:
r = √19.36 ≈ 4.4 cm
Final Answer: 4.4 cm
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0580_m23_qp_22 Question 7 Solution
Part (a):
Solve 15t + 8 = 4 - t
- Combine like terms:
15t + t = 4 - 8
16t = -4
- Solve for t :
t = -4 ÷ 16 = -1/4
Final Answer:t = -1/4
Part (b):
Solve (25 - 2u) ÷ 3 = 2
- Eliminate the denominator by multiplying both sides by 3:
25 - 2u = 6
- Rearrange to isolate u :
-2u = 6 - 25
-2u = -19
- Solve for u :
u = 19 ÷ 2 = 9.5
Final Answer: u = 9.5
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0580_s23_qp_41 Question 3 Solution
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0580_s23_qp_41 Question 3 Solution
Part (a):
The table shows information about the mass of each of 1000 eggs:
| Mass (m grams) | Frequency |
|---|---|
| 40 < m ≤ 50 | 126 |
| 50 < m ≤ 56 | 520 |
| 56 < m ≤ 64 | 154 |
| 64 < m ≤ 70 | 200 |
(i) Calculate an estimate of the mean:
Step 1: Calculate the midpoint for each group:
- 40 < m ≤ 50: midpoint = (40 + 50) ÷ 2 = 45
- 50 < m ≤ 56: midpoint = (50 + 56) ÷ 2 = 53
- 56 < m ≤ 64: midpoint = (56 + 64) ÷ 2 = 60
- 64 < m ≤ 70: midpoint = (64 + 70) ÷ 2 = 67
Step 2: Multiply midpoint by frequency:
- 126 × 45 = 5670
- 520 × 53 = 27560
- 154 × 60 = 9240
- 200 × 67 = 13400
Step 3: Total frequency = 126 + 520 + 154 + 200 = 1000
Step 4: Total of midpoint × frequency = 5670 + 27560 + 9240 + 13400 = 55870
Step 5: Mean = Total of midpoint × frequency ÷ Total frequency:
Mean = 55870 ÷ 1000 = 55.87 g
Final Answer: 55.87 g
(ii) Find the probability that an egg has a mass greater than 56 g:
Step 1: Eggs with mass greater than 56 g = 154 + 200 = 354
Step 2: Probability = (Eggs with mass > 56 g) ÷ Total eggs:
Probability = 354 ÷ 1000 = 177 ÷ 500
Final Answer: 177 ÷ 500
Part (b):
A farmer makes a profit of $24,730 selling eggs.
(i) Write this profit correct to 2 significant figures:
Profit = $25,000
Final Answer: $25,000
(ii) Write this profit in standard form:
Profit = 2.473 × 104
Final Answer: 2.473 × 104
Part (c):
On a farm, there are 500 hens, correct to the nearest 10.
(i) Calculate the upper bound for the total number of eggs all the hens lay in that year:
Step 1: Upper bound of hens = 500 + 5 = 505
Step 2: Upper bound of eggs per hen = 320 + 10 = 330
Step 3: Total upper bound = 505 × 330 = 166,650
Final Answer: 166,650
(ii) Calculate the lower bound for the difference between the number of hens on two farms:
Step 1: Lower bound for 500 hens = 500 - 5 = 495
Step 2: Lower bound for 800 hens = 800 - 10 = 790
Step 3: Difference = 790 - 495 = 285
Final Answer: 285
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0580_s23_qp_41 Question 3 Solution
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0580_m22_qp_22 Question 3 Solution
Question:
A journey starts at 21:15 one day and ends at 04:33 the next day. Calculate the time taken, in hours and minutes.
Solution:
- Calculate the time from 21:15 to midnight (00:00):
21:15 to 00:00 = 2 hours and 45 minutes
- Calculate the time from midnight (00:00) to 04:33:
00:00 to 04:33 = 4 hours and 33 minutes
- Add the times:
- Hours: 2 + 4 = 6
- Minutes: 45 + 33 = 78
- Convert 78 minutes into hours and minutes:
78 minutes = 1 hour and 18 minutes
- Add the additional 1 hour to the total hours:
- Total hours: 6 + 1 = 7
- Remaining minutes:18
Final Answer: 7 hours and 18 minutes
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0580_s22_qp_42 Question 3 Solution
Given:f(x) = 1 + 4x , g(x) = x^2
(a) Find:
(i) g(f(3)) :
- Substitute x = 3 into f(x) :
f(3) = 1 + 4(3) = 13
- Substitute f(3) = 13 into g(x) :
g(13) = 13^2 = 169
Final Answer: g(f(3)) = 169
(ii) f(g(x):
- Substitute g(x) = x^2 into f(x) :
f(g(x) = 1 + 4(x^2)
Final Answer: f(g(x) = 1 + 4x^2
(iii) f^{-1}(f(x)) :
- \( f^{-1}(f(x)) returns x , since f^{-1} is the inverse of f :
f^{-1}(f(x)) = x
Final Answer: f^{-1}(f(x) = x
(b) Find the value of x when f(x) = 15 :
- Substitute f(x) = 1 + 4x into the equation:
1 + 4x = 15
- Solve for x :
4x = 15 - 1 = 14
x = 14 ÷ 4 = 3.5
Final Answer: x = 3.5 or x = 7 ÷ 2
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0580_s22_qp_21 Question 25 Solution
Question:
w is proportional to the square root of y: w = jsqrt{y}.
y is inversely proportional to x: y = c/x.
When x = 4, y = 16, and w = 8.
Find w in terms of x.
Solution:
- From w = jsqrt{y}, substitute w = 8 and y = 16:
8 = jsqrt{16}
8 = j x 4
j = 2
- From y = c/x, substitute y = 16 and x = 4:
16 = c/4
c = 64
- Substitute y = 64/x into w = jsqrt{y}:
w = 2sqrt{64/x}
w = 16/sqrt{x}
Final Answer: w = 16/sqrt{x}
Solution Coming Soon
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Solution Coming Soon
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Solution Coming Soon
IGCSE Past Papers Math
IGCSE Past Papers Math
Examiner report :
March 2015 0580_m15_er.pdf
March 2016 0580_m16_er.pdf
March 2017 0580_m17_er.pdf
March 2018 0580_m18_er.pdf
March 2019 0580_m19_er.pdf
Grade Threshold:
March 2015 question paper & Mark Scheme Paper 1: [Core]
0580_m15_qp_12.pdf / 0580_m15_ms_12.pdf
May 2015 question paper & Mark Scheme Paper 1: [Core]
0580_s15_qp_11.pdf / 0580_s15_ms_11.pdf [Core]
0580_s15_qp_12.pdf / 0580_s15_ms_12.pdf [Core]
0580_s15_qp_13.pdf / 0580_s15_ms_13.pdf [Core]
October 2015 question paper & Mark Scheme Paper 1: [Core]
0580_w15_qp_11.pdf / 0580_w15_ms_11.pdf [Core]
0580_w15_qp_12.pdf / 0580_w15_ms_12.pdf [Core]
0580_w15_qp_13.pdf / 0580_w15_ms_13.pdf [Core]
March 2016 question paper & Mark Scheme Paper 1: [Core]
0580_m16_qp_12.pdf / 0580_m16_ms_12.pdf
May 2016 question paper & Mark Scheme Paper 1: [Core]
0580_s16_qp_11.pdf / 0580_s16_ms_11.pdf [Core]
0580_s16_qp_12.pdf / 0580_s16_ms_12.pdf [Core]
0580_s16_qp_13.pdf / 0580_s16_ms_13.pdf [Core]
October 2016 question paper & Mark Scheme Paper 1: [Core]
0580_w16_qp_11.pdf / 0580_w16_ms_11.pdf [Core]
0580_w16_qp_12.pdf / 0580_w16_ms_12.pdf [Core]
0580_w16_qp_13.pdf / 0580_w16_ms_13.pdf [Core]
March 2017 question paper & Mark Scheme Paper 1: [Core]
0580_m17_qp_12.pdf / 0580_m17_ms_12.pdf
May 2017 question paper & Mark Scheme Paper 1: [Core]
0580_s17_qp_11.pdf / 0580_s17_ms_11.pdf [Core]
0580_s17_qp_12.pdf / 0580_s17_ms_12.pdf [Core]
0580_s17_qp_13.pdf / 0580_s17_ms_13.pdf [Core]
October 2017 question paper & Mark Scheme Paper 1: [Core]
0580_w17_qp_11.pdf / 0580_w17_ms_11.pdf [Core]
0580_w17_qp_12.pdf / 0580_w17_ms_12.pdf [Core]
0580_w17_qp_13.pdf / 0580_w17_ms_13.pdf [Core]
March 2018 question paper & Mark Scheme Paper 1: [Core]
0580_m18_qp_12.pdf / 0580_m18_ms_12.pdf [Core]
May 2018 question paper & Mark Scheme Paper 1: [Core]
0580_s18_qp_11.pdf / 0580_s18_ms_11.pdf [Core]
0580_s18_qp_12.pdf / 0580_s18_ms_12.pdf [Core]
0580_s18_qp_13.pdf / 0580_s18_ms_13.pdf [Core]
October 2018 question paper & Mark Scheme Paper 1: [Core]
0580_w18_qp_11.pdf / 0580_w18_ms_11.pdf [Core]
0580_w18_qp_12.pdf / 0580_w18_ms_12.pdf [Core]
0580_w18_qp_13.pdf / 0580_w18_ms_13.pdf [Core]
March 2019 question paper & Mark Scheme Paper 1: [Core]
0580_m19_qp_12.pdf / 0580_m19_ms_12.pdf [Core]
May 2019 question paper & Mark Scheme Paper 1: [Core]
0580_s19_qp_11.pdf / 0580_s19_ms_11.pdf [Core]
0580_s19_qp_12.pdf / 0580_s19_ms_12.pdf [Core]
0580_s19_qp_13.pdf / 0580_s19_ms_13.pdf [Core]
October 2019 question paper & Mark Scheme Paper 1: [Core]
0580_w19_qp_11.pdf / 0580_w19_ms_11.pdf [Core]
0580_w19_qp_12.pdf / 0580_w19_ms_12.pdf [Core]
0580_w19_qp_13.pdf / 0580_w19_ms_13.pdf [Core]
March 2020 question paper & Mark Scheme Paper 1: [Core]
0580_m20_qp_12.pdf / 0580_m20_ms_12.pdf [Core]
May 2020 question paper & Mark Scheme Paper 1: [Core]
0580_s20_qp_11.pdf / 0580_s20_ms_11.pdf [Core]
0580_s20_qp_12.pdf / 0580_s20_ms_12.pdf [Core]
0580_s20_qp_13.pdf / 0580_s20_ms_13.pdf [Core]
October 2020 question paper & Mark Scheme Paper 1: [Core]
0580_w20_qp_11.pdf / 0580_w20_ms_11.pdf [Core]
0580_w20_qp_12.pdf / 0580_w20_ms_12.pdf [Core]
0580_w20_qp_13.pdf / 0580_w20_ms_13.pdf [Core]
0580 IGCSE Math Paper 2 [ Extended]
March 2015 question paper & Mark Scheme Paper 2 Extended
0580_m15_qp_22.pdf / 0580_m15_ms_22.pdf [ Extended]
October 2015 question paper & Mark Scheme Paper 2:[ Extended]
0580_w15_qp_21.pdf / 0580_w15_ms_21.pdf [ Extended]
0580_w15_qp_22.pdf / 0580_w15_ms_22.pdf [ Extended]
0580_w15_qp_23.pdf / 0580_w15_ms_23.pdf [ Extended]
March 2016 question paper & Mark Scheme paper 2 [ Extended]
0580_m16_qp_22.pdf / 0580_m16_ms_22.pdf [Extended]
March 2017 question paper & Mark Scheme paper 2 [ Extended]0580_m17_qp_22.pdf / 0580_m17_ms_22.pdf [ Extended]
May 2017 question paper & Mark Scheme Paper 2:[ Extended]
0580_s17_qp_21.pdf / 0580_s17_ms_21.pdf [ Extended]
0580_s17_qp_22.pdf / 0580_s17_ms_22.pdf [ Extended]
0580_s17_qp_23.pdf / 0580_s17_ms_23.pdf [ Extended]
October 2017 question paper & Mark Scheme Paper 2:[ Extended]0580_w17_qp_21.pdf / 0580_w17_ms_21.pdf [ Extended]
0580_w17_qp_22.pdf / 0580_w17_ms_22.pdf [ Extended]
0580_w17_qp_23.pdf / 0580_w17_ms_23.pdf [ Extended]
March 2018 question paper & Mark Scheme paper 2 [ Extended]
May 2018 question paper & Mark Scheme Paper 2:[ Extended]
0580_s18_qp_21.pdf / 0580_s18_ms_21.pdf [ Extended]
0580_s18_qp_22.pdf / 0580_s18_ms_22.pdf [ Extended]
0580_s18_qp_23.pdf / 0580_s18_ms_23.pdf [ Extended]
October 2018 question paper & Mark Scheme Paper 2:[ Extended]0580_w18_qp_21.pdf / 0580_w18_ms_21.pdf [ Extended]
0580_w18_qp_22.pdf / 0580_w18_ms_22.pdf [ Extended]
0580_w18_qp_23.pdf / 0580_w18_ms_23.pdf [ Extended]
March 2019 question paper & Mark Scheme Paper 2:[ Extended]
May 2019 question paper & Mark Scheme Paper 2:[ Extended]
[ Extended]
[ Extended]
[ Extended]
0580 IGCSE Math Paper 3 Core
October 2018 question paper & Mark Scheme Paper 3: [Core]
0580_s18_qp_31.pdf / 0580_s18_ms_31.pdf
0580_s18_qp_32.pdf / 0580_s18_ms_32.pdf
0580_s18_qp_33.pdf / 0580_s18_ms_33.pdf
0580 IGCSE Math Paper 4 Extended
May 2015 question paper & Mark Scheme paper 4 [ Extended]
0580_s15_qp_41.pdf / 0580_s15_ms_41.pdf
0580_s15_qp_42.pdf / 0580_s15_ms_42.pdf
0580_s15_qp_43.pdf / 0580_s15_ms_43.pdf
October 2015 question paper & Mark Scheme paper 4 [ Extended]
0580_w15_qp_41.pdf / 0580_w15_ms_41.pdf
0580_w15_qp_42.pdf / 0580_w15_ms_42.pdf
0580_w15_qp_43.pdf / 0580_w15_ms_43.pdf
May 2016 question paper & Mark Scheme paper 4 [ Extended]
0580_s16_qp_41.pdf / 0580_s16_ms_41.pdf
0580_s16_qp_42.pdf / 0580_s16_ms_42.pdf
0580_s16_qp_43.pdf / 0580_s16_ms_43.pdf
October 2016 question paper & Mark Scheme paper 4 [ Extended]
0580_w16_qp_41.pdf / 0580_w16_ms_41.pdf
0580_w16_qp_42.pdf / 0580_w16_ms_42.pdf
0580_w16_qp_43.pdf / 0580_w16_ms_43.pdf
May 2017 question paper & Mark Scheme paper 4 [ Extended]
0580_s17_qp_41.pdf / 0580_s17_ms_41.pdf
0580_s17_qp_42.pdf / 0580_s17_ms_42.pdf
0580_s17_qp_43.pdf / 0580_s17_ms_43.pdf
October 2017 question paper & Mark Scheme paper 4 [ Extended]
0580_w17_qp_41.pdf / 0580_w17_ms_41.pdf
0580_w17_qp_42.pdf / 0580_w17_ms_42.pdf
0580_w17_qp_43.pdf / 0580_w17_ms_43.pdf
May 2018 question paper & Mark Scheme paper 4 [ Extended]
0580_s18_qp_41.pdf / 0580_s18_ms_41.pdf
0580_s18_qp_42.pdf / 0580_s18_ms_42.pdf
0580_s18_qp_43.pdf / 0580_s18_ms_43.pdf
October 2018 question paper & Mark Scheme paper 4 [ Extended]
0580_w18_qp_41.pdf / 0580_w18_ms_41.pdf
0580_w18_qp_42.pdf / 0580_w18_ms_42.pdf
0580_w18_qp_43.pdf / 0580_w18_ms_43.pdf