4H June 2017 Model Answers Level Subject Exam Board Paper Booklet IGCSE Maths Edexcel June 2017 4H Model Answers Time Allowed: Score: Percentage: 120 minutes / 100 /100 Grade Boundaries: 9 8 7 6 5 4 3 >90% 80% 70% 60% 50% 40% 30% 2 1 20% 10%
Edexcel IGCSE Maths 4 H June 2017 Question 1. Question 2. P = {p, o, r, t, u, g, a, l} I = {I, t, a, l, y} (a) I) P I = {t, a, l} The intersection of 2 Sets represents the elements that can be seen both in Set P and in Set I, in other words, their common elements. ii) P I= (p, o, r, t, u, g, a, l, I, y} The reunion of 2 Sets represents all the elements in the 2 Sets considered only one time, no repetitions for the common elements of the Sets. (b) F = {f, r, a, n, c, e} Is it true that I F =? No, it is not true because the 2 Sets I and F share a common element, a. Therefore, I F = {a} (a) M = 2t 2 7t Work out the value of M when t = -3. We substitute the value of t and work out the calculations. M = 2 * (-3) 2 7 * (-3) M = 2 * 9 + 21 M = 18 + 21 M = 39 (b) Solve 4(x + 3) = 9x 10 Firstly, we need to expand the bracket on the first side of the equation: 4x + 12 = 9x 10 x = Secondly, we move all the terms that contain the unknown x to one side and the numbers to the other side: 22 = 5x x =4.4 y is an integer -2 < y 3 (c) Write down all the possible values of y. y = {-1, 0, 1, 2, 3}
-2 is not a possible value of y because we know that y is strictly greater than -2. We consider 3 as value of y because it is said that y is lower or equal to 3. Question 3. (a) Exchange rate: 1 = 97 rupees Total amount: 250. From the exchange rate we can work out the proportion necessary for converting pounds into rupees. We note the amout of rupees we need to work out with the unknown x. 1. 97 rupees 250. X x = x = 24250 rupees (b) New exchange rate 1 = 93.5 rupees Total amount: 4 notes of 500 rupees. Total amount of rupees = 4 * 500 rupees Total amount of rupees = 20000 rupees. We note the amount of pounds we need to work out with the unknown y. Using the new exchange rate we can deduce the following proportion necessary for turning rupees into pounds. 1. 93.5 rupees Y. 20000 rupees y = y = 21.39 Question 4. Point A has coordinates (-4, 9) Point B has coordinates (1, 5) Find the coordinates of the midpoint of AB. We note with M (x, y) the midpoint of AB. We know that the coordinates of the midpoint are equal to the coordinates of the 2 points defining line divided by 2. X = and y = X = y = X = -1.5 y = 7
Midpoint: M ( -1.5, 7) Question 5. We know that: probability = (a) The number of total cases is 20. (20 games played on the computer) Number of favourable cases = number of times she wins the game. = 0.3 Number of favourable cases = 0.3 * 20 Number of favourable cases = 6 Answer: 6 (b) We note: The probability that she will win with x; The probability that she will draw with y; And the probability that she will lose with z. x = 0.3 z = 3y The sum of these three probabilities is 1. (the case in which the number of favourable cases = the number of total cases) x + y + z = 1 We now substitute in this equation the values that we do know. 0.3 + y + 3y = 1 => 4y = 1 0.3 => 4y = 0.7 => y = => y = 0.175 z = 3y => z = 3 * 0.175 z = 0.525 (the probability that she will lose for each game of chess) When she plays one game there is a probability of 0.525 that she will lose. Question 6. Small packet 6 batteries
large packet 9 batteries T = total number of batteries The number of batteries from small packets are the number of batteries in a small packet, 6, multiplied by the number of small packets bought, m. The number of batteries from large packets are the number of batteries in a large packet, 9, multiplied by the number of large packets bought, g. the formula for T: T = 6 * m + 9 * g T = 3(2 * m + 3 * g) Question 7. (a) Show that: We need to the have the same denominator for all the fractions, the highest denominator in this case being 24. We notice that: 24 = 12 * 2 and 24 = 8 * 3 Therefore, we need to multiply the first fraction with 2 and the second one with 3. + = + = = (b) Show that: We need to the have the same denominator for all the fractions, the lowest common denominator possible in this case being 45. Question 8. Each interior angle of a regular polygon is 156. Work out the total number of sides of the polygon. We know that a regular polygon has all sides equal and all angles equal. Each angle of a regular polygon = ( ) where n represents the number of sides.
In our case: 156 = ( ) We need to work out the value of n. 156 *n = ( ) We expand the bracket on the one side of the equation: 156 * n = 180 * n - 360 We move all the terms containing the unknown, n, to one side and the numbers to the other side. 360 = n * 24 n = n = 15 sides Question 9. 420 divided by the ratio: 4 : 5 : 3. We note the amount of money each of the 3 boys get with x, y and z. x = the amount of money Manu has y = the amount of money Liam has z = the amount of money Ned has We now that the ratio represents the amount of money each boy gets simplified by a common number, which in this case we note with a. 4a + 5a + 3a = 420 12a = 420 a = a = 35 Using this amount we can now work out the amount of money each boy gets. x = 4a x = 4 * 35 x = 140 y = 5a y = 5 * 35 y = 175 z = 3a z = 3 * 35 z = 105
We know that after this division, Liam gives Ned 75. The final amount of money Liam has is: y - 75 = 100 And the final amount of money Ned has is: z + 75 = 180 We now need to work out the what 180 means as a percentage from 420. We note the percentage with p. = 180 p * 420 = 18000 p = p = % As the decimal 8 is greater than 5 we need to add one to the last digit before the comma so we can correct this answer to the nearest whole number. Answer: p = 43 % Question 10. (a) Simplify e 8 * e 7. When we multiply 2 powers with the same base we just need to keep the base and add up the 2 powers. e 8 * e 7 = e 15 (b) Simplify fully We note that 12 divided by 3 = 4 and g 10 divided by g 2 = g 8 After we simplify: 4g 8 (c) Write down the value of m 0. Any number raised to the power of 0 equals 1. Answer: m 0 = 1 (d) Simplify fully (27x 6 ) 2/3 When we raise a power to another power we simply need to multiply one by the other. 27x 12/3 = 27x 4 Answer: 27x 4 Question 11. ABCD is a square of side 7 cm. Area of the shaded region = Area of the circle Area of the square Area of the square = AB 2 Area of the square = 7 2 = 49 cm
Area of the circle = 2 O To work out the area of the circle we need to find the radius, r. We notice that the length of the radius will be equivalent with half of one diagonal of the square. We draw the 2 diagonals of the square which intersect in the centre of the circle noted with O. We consider at random one of the 4 triangles form, each of them being a right-angled. For example, the triangle AOB is a right-angled triangle with the right angle. The side of the square, AB represents the hypotenuse and the 2 sides of the triangles represents the radius of the circle. AO = OB = r Using Pythagoras Theorem: AB 2 = AO 2 + OB 2 AB 2 = 2r 2 r 2 = cm 2 Area of the circle = Area of the circle = * cm 2 Area of the circle = 76.69 cm 2 Area of the shaded region = 76.96 cm 2 49 cm 2 Area of the shaded region = 27.96 cm 2 Corrected to the nearest whole number: 28 cm 2 2 Question 12. The heights, in millimetres, of 11 seedlings: 16 12 19 17 24 27 19 15 23 27 10 Work out the interquartile range of these heights. The interquartiles range represents the range of the middle 50% of all of our data.
In order to work it out we first need to order our data from lowest to highest. We need to organise it this way so we can find the median of the lower half and the median of the upper half. When we subtract the 2 medians we find the interquartile range. Order from lowest to highest: 10 12 15 16 17 19 19 23 24 27 27 Since we have an odd number of heights we first need to find the middle point. In this case, the middle point is 19: 10 12 15 16 17 19 19 23 24 27 27 (upper half) (lower half) By knowing this, we work out the median of the lower and upper halves without including 19. For the upper half, the median we see that is 15, while for the lower half, the median we see that is 24. Now, by subtracting these 2 medians we work out that the interquartile range is: 24 15 = 9 Question 13. (a) Equations of 4 straight lines: Line A: y = 2x + 3 Line B: 2y = 6 3x Line C: 4x 2y = 3 Line D: y = 3 2x Which 2 of these lines are parallel? If 2 lines are parallel then they should have the same gradient, m. We can work out m by rewriting the equations in the form: y = m * x + n, where m is the gradient. Line A: y = 2x + 3 => m = 2 Line B: y = y = 3 => m = Line C: 2y = 3 4x <=> y = => m = 2 Line D: y = 3 2x => m = 2 The parallel lines are Line A and Line C as they have the same gradient, m = 2. (b) Line L has the gradient and passes through the point with coordinates (1, 3) Find an equation of L. Give your answer in the form ax + by = c where a, b and c are integers.
The coordinates of the point through which line L is passing are (1, 3). These coordinates represent x and y in the equation of the line: y = mx + n We substitute these values and the value of m in the equation. 3 = x + n n = 3 + n = We rewrite the equation in the required form: x + y = We multiply each side by 2 to simplify the equation, given that a, b and c need to be integers: 5x + 2y = 11 Question 14. A, B, C and D are points on a circle, centre O. Angle = 52 (a) i) Write down the size of angle. = 52 ii) Both are angles subtended by the same arc => they are congruent (b) Write down the size of angle. i) = 52 ii) The angle is the angle at center => is twice the size of the angle subtended by the arc at the circumference,. Question 15. (a) We note each vertex of the trapezium with a letter as follows:
A B D C The area of the trapezium is 60 cm 2. We know that the formula for the area (A) of a trapezium is: A = (AB + DC) * AD Where AD represents the height of the trapezium as is a right angle. Substituting the values that we know: A = (2x 4 + x + 5) * (x + 3) We add up the like terms in the first bracket: A = (3x + 1) * (x + 3) And now we expand the 2 brackets: A = (3x 2 + 9x + x + 3) Simplify the terms in the brackets: A = (3x 2 + 10x + 3) We know that the area is 60 cm 2 : 60 cm 2 = (3x 2 + 10x + 3) We multiply each side of the equation by 2: 120 cm 2 = 3x 2 + 10x + 3 We now rewrite the equation in the required form: 3x 2 + 10x 117 = 0 The exact equation that we needed to show. (b) Work out the value of x.
3x 2 + 10x 117 = 0 This is a second order equation in the form: ax 2 + bx + c = 0 We can now work out the value of x using the formula: To solve it we first need to work out the value of b 2 4ac. In our case: a = 3, b = 10, c = -117 We substitute these values in b 2 4ac. 10 2 4* 3 * (-117) We do the calculations: b 2 4ac = 100 + 1404 b 2 4ac = 1504 We substitute the values that we know in the formula We do the calculations: x 1 x 2 x 1 = 4.796 x 2 = - 8.130 We know that x + 3 is the length of the side of a trapezium. Therefore, x + 3 needs to be greater than 0. We impose the condition: x + 3 > 0 x > -3 As x 2 = -8.130 is lower than -3. The only possible value for x is x 1 = 4.796 > -3 To correct this to 3 significant figures we look at the third decimal, in this case 6, and see if it is greater or equal to 5. For our number, the decimal is 6 so we correct the number by adding one to the decimal before it. The correct number will be: x = 4.80 Question 16. (a) The probability that it will rain on Saturday is 0.8. If it rains on Saturday, the probability that it will rain on Sunday is 0.65.
If it does not rain on Saturday, the probability that it will rain on Sunday is 0.4. We know that: probability = In general, the maximum probability of an event is 1, the situation in which the number of favourable cases is equal to the number of total cases. For the first statement, we know that the probability that it wil rain is 0.8 and therefore, the probability that it will not rain is 1 0.8 = 0.2. Following the same principle in each of the cases we fill in the Tree diagram as follows: 0.65 0.35 0.2 0.4 0.6 (b) Work out the probability that it rains only on one of these 2 days. 2 possibilities: 1 st : rains on Saturday and does not rain on Sunday 2 nd : does not rain on Saturday and rains on Sunday These 2 are independent events so the total probability will be their sum. Using the Tree diagram: Probability of the first event = 0.8 * 0.35 = 0.28 Probability of the second event = 0.2 * 0.4 = 0.08 Probability that it rains only on one of these days: P = 0.28 + 0.08 = 0.36 Question 17. Curve C has the equation y = 8x 3 3x 2 25x. (a) Find.
We need to work out the derivation of the equation for curve C. f(x) = 8x 3 3x 2 25x The derivation for any x n is: (x n ) = n * x n 1 Therefore: (8x 3 ) = 3 * 8x 2 = 24x 2 => = 24x 2-6x -25 (-3x 2 ) = 2 * (-3)x = -6x (-25x) = -25 (b) Find the x coordinates of the points on C where the gradient is 5. We consider the equation of the curve as the function: f(x) = 8x 3 3x 2 25x We know that the derivative of the function f (x) = 24x 2-6x -25 The first derivative of the function is equivalent with the gradient. The gradient is 5 f (x) = 24x 2-6x -25 = 5 In order to work out the value of x we solve the second order equation. 24x 2-6x -25 = 5 Where: a = 24, b = -6, c = -25 Using the formula: We substitute the values that we know in the formula. ( ) ( ) We do the calculations: ( ) ( ) x 1 x 2 = X 1 = 1. 153 x 2 = -0.903 Question 18. Frequency table: Time (t minutes) Frequency 0 < t 30 60 30 < t 90 270 90 < t 120 150 120 < t 240 156 240 < t 360 24 On the grid, draw a histogram for this information:
A histogram is a special type of bar chart which shows quantitative information, in this case the frequencies. The horizontal axis of the grid is already showing the intervals of time, in minutes. Therefore, the vertical axis is the one which will show the frequencies. We see that the highest frequency on the table is 270 while the lowest is 24. The histogram will then be drawn following the exact quantities from the table for each interval of time. (Keep in mind that the bars drawn need to be right next to each other) Frequency Question 19. ABCD is a kite. Work out the area of the kite. Give your answer correct to 3 significant figures. In the kite ABCD, we draw the diagonal AC. We notice that AC divides the kite in 2 congruent triangles: ADC and ABC.
Therefore, the area of the kite will be equal to the sum of the areas of these 2 triangles. In the triangle ADC, we know the length of 2 sides and the value of the length between them. We know that in any triangle: A = ab sin C In our case, the sides a and b are AD and DC and the angle is the angle. A = * AD * DC * sin We substitute the values we know in the formula: A = * 6.4 cm * 9.7 cm * sin 110 Using a calculator, we work out that sin 110 = 0.939. We do the calculations: A = 29.168 cm 2 This value represents the area of one triangle. To work out the area of the kite we need to multiply this by 2. A kite = 2 * A triangle A kite = 2 * 29.168 cm 2 A kite = 58.336 cm 2 The corrected answer: A = 58.3 cm 2 Question 20. A car travels a distance of 63.5 km, correct to the nearest 0.5 km. The car takes 45.8 minutes correct to 1 decimal place. Work out the lower bound for the average speed of the car. Give your answer in km/h correct to 1 decimal place. 63.5 km correct to the nearest 0.5 km => 63. 25 km distance < 63.5 km 45.8 minutes correct to one decimal place => 45.8 minutes < time 45.85 minutes To calculate the lower bound we consider 63.25km and 45.85 minutes: Speed =
Speed = = 1.379 km/minute To convert it into km/h we multiply the result by 60 minutes (1 h) Speed = 82.76 km/h Correct the result to 1 decimal place: Speed = 82.8 km/h Question 21. LMNP is a quadrilateral. Work out the size of angle. We draw the diagonal LN in the quadrilateral LMNP. In the triangle NPL we can apply the Cosine rule to work out the length of LN Cosine rule: a 2 = b 2 + c 2-2bc cosa where a = LN, b = LP and c = NP We substitute the values that we know: LN 2 = 15.6 2 cm + 4.3 2 cm - 2 * 15.6 cm * 4.3 cm * cos 72 We do the calculations: LN 2 = 243.36 cm 2 + 18.49 cm 2-134.16 cm 2 *cos 72 Using a calculator, we work out that: cos72 = 0.309 LN 2 = 220.392 cm 2 LN = cm LN = 14.845 cm Knowing the length of the side LN we can now apply the Sine rule in the triangle MLN to work out the size of the angle. Sine rule: = = Where a = LN, sin A = sin 58 ; b = MN and sin B = sin We substitute in the Sine rule the values that we know: =
sin = Using a calculator, we work out that: cos 58 = 0.529 sin = sin = 0.782 = 51 We can also apply the Sine rule in the triangle NLP to work out the angle = We substitute the values we know: = sin = sin = 0.275 = 15.99 = 15.99 + 51 = 67.46 Correct to 3 significant figures: = 67.5 Question 22. m = 8 * 10 9n where n is an integer. Express m -1/3 in standard form. Give your answer, in terms of n, as simply as possible. We raise m to the power of : m -1/3 = (8 * 10 9n ) -1/3 To raise the bracket to the power of bracket with the one outside the bracket for each of the terms. we need to multiply the power inside the The required answer is in the standard form so we keep this form throughout the calculations.
m -1/3 = 8-1/3 * 10-9n/3 Simplified: = 3n m -1/3 = 0.5 * 10-3n m -1/3 = 5 * 10-1 * 10-3n When multiplying 2 powers with the same base we keep the base and add up the 2 powers. m -1/3 = 5 * 10-4n Question 23. The solid hemisphere below has a total surface area of cm 2. The hemisphere has a volume of k cm 3. Find the value of k. Surface area of a sphere: A = 4 r 2, where r is the value of the radidus. The surface area of a hemisphere is half the area of a sphere + the area of the circle inside. (red) A (hemisphere) = 2 r 2 + r 2 = 3 r 2 Volume of a sphere: V = cm 3 The volume of a hemisphere is half the volume of a sphere. V (hemisphere) = cm 3 In our case, the surface area is cm 2. We need to work out the value of r, radius. Surface area: cm 2 = 3 r 2 cm 2 We divide both sides by. cm 2 = 2r 2 cm 2 r 2 = cm 2 r = cm r = cm
Volume of a hemispehere: V (hemisphere) = r 3 cm 3 We substitute the value of r. V (hemisphere) = ( ) 3 cm 3 = k cm 3 We divide by on both sides to work out the value of k. K = ( ) 3 K = * K =