Cambridge IGCSE Mathematics 0580 — 2007 May/June Paper 3 · Variant 1

1 (a) Find the value of For Examiner's (i) 50, Use Answer(a)(i) [1] (ii) the square root of 64, Answer(a)(ii) [1] (iii) the cube root of 64, Answer(a)(iii) [1] (iv) the integer closest in value to (1.8)3. Answer(a)(iv) [1] (b) Write down (i) a common factor of 15 and 27, which is greater than 1, Answer(b)(i) [1] (ii) a common multiple of 10 and 12. Answer(b)(ii) [1] (c) (i) Two of the factors of 2007 are square numbers. One of these is 1. Find the other square number. Answer(c)(i) [1] (ii) Write down the two factors of 2007 which are prime. Answer(c)(ii) and [2]

Mark scheme: 1 (a) (i) 1 B1 (ii) 8 or −8 or ±8 B1 (iii) 4 B1 Not −4 (iv) 6 B1 (b) (i) 3 B1 (ii) Multiple of 60 B1 (c) (i) 9 B1 (ii) 3 and 223 B1,B1 [9] 2 5

More questions on Powers and roots

2 Marguerite earns $336 per month. For She divides her earnings between bills, food, savings and personal spending. Examiner's Use (a) Her bills take 2 of her earnings. 7 Show that $240 is left for her other items. Answer(a) [2] (b) She divides the $240 between food, savings and personal spending in the ratio 5 : 3 : 4. Calculate how much she spends on food. Answer(b) $ [2] (c) She saves the same amount each month. Show that she saves $720 in one year. Answer(c) [2] (d) She invests the $720 in a bank which pays 6% per year compound interest. How much will this be worth after 2 years? Answer(d) $ [3]

Mark scheme: 2 5 2 (a) 336 − × 336 or × 336 M1 7 7 (=) 240 E1 240 must be seen for this mark (b) 5 ÷ their(5 + 4 + 3) × 240 M1 100 A1cao www 2 1 3 (c) 3 ÷ their(5 + 4 + 3) × 240 × 12 M1 Allow 2880 for 240 × 12 and for . 4 12 (=) 720 E1 720 must be seen for this mark (d) 720 × 1.06 2 oe M2 Implied by 88.99(2) or 89(total interest)seen M1 for 720 × 1.06 (implied by 763.2 seen) 808.99(2) or 809 A1 SC1 for 806.(4) (Simple Interest) www 3 for 808.99(2) or 809 [9] IGCSE – May/June 2007 0580/0581 03 1

More questions on Ratio and proportion

3 (a) Kinetic energy, E, is related to mass, m, and velocity, v, by the formula For Examiner's 1 Use E = mv2. 2 (i) Calculate E when m = 5 and v = 12. Answer(a)(i) E= [2] (ii) Calculate v when m = 8 and E = 225. Answer(a)(ii) v = [2] (iii) Make m the subject of the formula. Answer(a)(iii) m = [2] (b) Factorise completely xy2 – x2y. Answer(b) [2] (c) Solve the equation 3(x – 5) + 2(14 – 3x) = 7. Answer(c) x = [3] (d) Solve the simultaneous equations 4x + y = 13, 2x + 3y = 9. Answer(d) x = y = [3]

Mark scheme: 1 3 (a) (i) 360 B2 M1 for × 5 × 122oe 2 (ii) 7.5oe B2 M1 for 225 /4 oe (implied by 56.25) 2 E 1 2 1 (iii) or E v B2 B1 for 2E or E or division by v2 2 v 2 2 (b) xy( y – x) final answer B2 B1 for x(y2 – xy) or y(xy – x2) SC1 for xy(y + x) (c) 3x – 15 + 28 – 6x (= 7) MA1 13 – 3x (= 7) M1ft Independent ax + b (=7) from their expansion x= 2 A1cao www 3 (d) Equating coefficients of x or y, or M1 equivalent method. or a correctly substituted substitution. E.g. 5y = 5 oe or 10x = 30 oe A1 y = 13 – 4x ⇒ 2x + 3(13 – 4x) = 9 x = 3, y = 1 A1 www 3 [14]

More questions on Rates

4 (a) The table shows corresponding values of x and y for the function For Examiner's 60 Use y = (x ≠ 0). x x −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 y −12 −15 −30 60 12 10 [2] (i) Fill in the missing values of y in the table above. (ii) Plot the points on the grid below and draw the graph for −6 x −1 and 1 x 6. y 60 50 40 30 20 10 x –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 –10 –20 –30 –40 –50 –60 [4] (b) Write down the order of rotational symmetry of the graph. Answer(b) [1] (c) Draw the lines of symmetry of the graph on the grid. [2] (d) One line of symmetry intersects the graph at two points. (i) Write down the co-ordinates of these two points. Answer(d)(i) ( , ) and ( , ) [2] (ii) Write down the equation of this line of symmetry. Answer(d)(ii) [1] (e) Find the gradient of the other line of symmetry. Answer(e) [1]

Mark scheme: 4 (a) (i) −10, −20, −60, 30, 20, 15 B2 B1 for –20 (x = –3) or 20 (x = 3) (ii) Their 12 points plotted correctly. P3ft P2ft for 10 or 11 points correct. P1ft for 8 or 9 points or 1 quadrant correct. Smooth curves through all points. C1 Two distinct curves; no part of curves between x = –1 and x = 1 (b) 2 B1 (c) Correct lines ruled B1,B1 Minimum length from x = –3 to x = 3. (d) (i) (2.4 to 2.5, 24 to 25) B1ft ft their points of intersection (−2.4 to −2.5, −24 to −25) B1ft ft their points of intersection (ii) y = 10x oe B1 cao (e) −10 B1 cao [13] IGCSE – May/June 2007 0580/0581 03

More questions on Fractions, decimals and percentages

5 A bag contains 24 discs. For 10 discs are red, 9 discs are green and 5 discs are yellow. Examiner's Use (a) The number of discs of each colour can be shown by three sectors on a pie chart. The sector angle for the red discs is 150°. Work out the sector angle for (i) the green discs, Answer(a)(i) [1] (ii) the yellow discs. Answer(a)(ii) [1] (iii) Complete the pie chart below and label the sectors. [2] (b) A disc is chosen at random. For Examiner's Find, as a fraction, the probability of each of the following events. Use (i) Event A: the disc is red. Answer(b)(i) [1] (ii) Event B: the disc is red or yellow. Answer(b)(ii) [1] (iii) Event C: the disc is not yellow. Answer(b)(iii) [1] (c) Probability Scale Impossible Certain (c)(i) ........... (c)(ii) ........... The diagram shows a horizontal probability scale. Write on the dotted lines in the diagram, the probability of (i) an impossible event, [1] (ii) a certain event. [1] (d) Using the notation, A, B and C, mark the positions of your three answers in part (b) on the Probability Scale diagram in part (c). [3]

Mark scheme: 5 (a) (i) 135 (green) B1 (ii) 75 (yellow) B1 (iii) Ruled lines correct to 2° B1ft Only if (a)(i) + (a)(ii) = 210°. 3 correctly labelled sectors B1 Independent of previous marks 10 (b) (i) oe B1 Accept decimals, percentages 24 15 (ii) oe B1 24 19 (iii) oe B1 24 0 12 0 24 (c) (i) 0 B1 SC1 for and or and 12 12 24 24 (ii) 1 B1 (d) Labelled arrows correctly B3ft 1 mark for each. positioned by eye ft their probabilities from (b). [12]

More questions on Circles, arcs and sectors

6 For E Examiner's Use D C 6 m 6 m NOT TO SCALE 56° A B O ABCED is the cross-section of a tunnel. ABCD is a rectangle and DEC is a semi-circle. O is the mid-point of AB. OD = OC = 6 m and angle DOC = 56°. (a) (i) Show that angle COB = 62°. Answer(a)(i) [1] (ii) Calculate the length of OB. Answer(a)(ii) OB = m [2] (iii) Write down the width of the tunnel, AB. Answer(a)(iii) AB= m [1] (iv) Calculate the length of BC. Answer(a)(iv) BC = m [2] (b) Calculate the area of For Examiner's (i) the rectangle ABCD, Use Answer(b)(i) m2 [2] (ii) the semi-circle DEC, Answer(b)(ii) m2 [2] (iii) the cross-section of the tunnel. Answer(b)(iii) m2 [1] (c) The tunnel is 500 metres long. (i) Calculate the volume of the tunnel. Answer(c)(i) m3 [2] (ii) A car travels through the tunnel at a constant speed of 60 kilometres per hour. How many seconds does it take to go through the tunnel? Answer(c)(ii) s [3]

Mark scheme: 6 (a) (i) (180 – 56)/2 B1 Alt. 90 − (56 ÷ 2) (ii) art 2.82 B2 M1 for 6cos 62° (implied by 2.8) Long method must be complete. (iii) 5.63 to 5.64 B1ft 2 × their (a)(ii) (iv) 5.3 or art 5.30 B2 M1 for 6sin 62°oe Long method must be complete. (b)(i) 29.8 to 29.9 B2ft M1 for their (a)(iii) × (a)(iv) (ii) art 12.5 B2ft M1 for 0.5 × π × (their (a)(ii)2) (iii) 42.3 to 42.4 B1ft ft is their (b)(i) + (b)(ii) (c)(i) 21100 to 21200 B2ft M1 for their (b)(iii) × 500  500 3600  (ii)  ×  oe M2 M1 for figs 500 ÷ figs 60  60 1000  1 30 A1 SC2 for answer of min 2 or SC1 for1km per minute seen. www B3 [16] IGCSE – May/June 2007 0580/0581 03

More questions on Right-angled triangles

7 For l Examiner's Use DD CC AA BB O A quadrilateral ABCD, a line l and a point O are shown on the grid above. (a) Write down the mathematical name for the quadrilateral ABCD. Answer(a) [1] (b) On the grid above, draw the images of the quadrilateral ABCD under the following transformations.  9  (i) Translation by the vector   . Label this image P. [2]  − 3    (ii) Reflection in the line l. Label this image Q. [2] (iii) Rotation, centre A, through 90° anti-clockwise. Label this image R. [2] (iv) Enlargement, centre O and scale factor 3. Label this image S. [3]

Mark scheme: 7 (a) Trapezium B1 − 3  (b) (i) Translation 9 across, 3 down B2 B1 for 9 across or 3 down or    9  (ii) Correct reflection B2 B1 any reflection of ABCD in a line parallel to l. (iii) Correct rotation B2 B1 90° clockwise rotation of ABCD about A (iv) Correct enlargement B3 B1 any enlargement of ABCD and B1 any enlargement of ABCD SF 3 or B1 any enlargement of ABCD centre O (not penalise lack of labelling provided intention clear) [10]

More questions on Transformations

8 For Examiner's Use O A R P B The diagram shows a circular garden, centre O. A straight path AB touches the circle at P. (a) (i) Draw on the diagram the diameter PQ and label the point Q. [1] (ii) Without measuring, write down the size of angle APQ. Answer(a)(ii) Angle APQ= [1] (iii) The point R is marked on the circumference of the circle. Draw the lines PR and QR. [1] (iv) Write down the reason why the angle PRQ is 90°. Answer(a)(iv) [1] (b) Showing all your construction lines, use a straight edge and compasses only to construct (i) the perpendicular bisector of QR, [2] (ii) the bisector of angle PRQ. [2] (c) Shade the region of the garden between PQ and QR which is closer to R than to Q and closer to RQ than to RP. [2] Question 9 is on the next page.

Mark scheme: 8 (a) (i) Diameter from P through O to Q B1 (ii) 90 B1cao (iii) P to R and Q to R ruled. B1 (iv) (angle in a ) semi-circle B1 Angle on a diameter. Half the angle at the centre. (b) (i) Bisector of QR with arcs. B2 SC1 if accurate without arcs. Maximum errors 2mm from mid-point and 2° from perpendicular. (ii) Bisector of PRQ with arcs. B2 SC1 if accurate without arcs. Maximum error 2° in line from R. If wrong line and/or angle used treat as misread each time. (c) Correct Shading 2 Dep. on B2 in (b)(i) and (b)(ii). SC1 for ‘correct’ shading but dependent on at least SC1 in (b)(i) and (b)(ii). [10] IGCSE – May/June 2007 0580/0581 03

More questions on Geometrical terms

9 In the pattern below each diagram shows a letter E formed by joining dots. For Examiner's Diagram 1 Diagram 2 Diagram 3 Diagram 4 Use (a) Draw the next letter E in the pattern. [1] (b) Complete the table showing the number of dots in each letter E. Diagram 1 2 3 4 5 Dots 8 15 [3] (c) How many dots make up the letter E in (i) Diagram 10, Answer(c)(i) [2] (ii) Diagram n? Answer(c)(ii) [2] (d) The letter E in Diagram n has 113 dots. Write down an equation in n and use it to find the value of n. Answer(d) n = [3]

Mark scheme: 9 (a) Letter E correctly drawn B1 (b) 22, 29, 36 B3 B1 for each correct number. (c) (i) 71 B2 B1 for 7 × 10 + 1 or 8 + 9 × 7 seen. (ii) 7n + 1 or 8 + (n – 1) × 7 oe B2 SC1 for 7n + k seen. (k is an integer) oe (d) Their (c)(ii) = 113 B1ft ft any expression involving n. Full method of solution of their M1ft ft only a linear equation. equation. (113 – k)/ ‘7’ 16 A1cao www B2 [11]

More questions on Sequences