Cambridge IGCSE Mathematics (9-1) 0980 — 2021 Oct/Nov Paper 2 · Variant 1

1 P is a prime number where 60 1 P 1 80 . P is 2 less than a square number. Find the value of P. P = ................................................ [2]

Mark scheme: Question Answer Marks Partial Marks 1 79 2 B1 for 64 or 81 seen or for answer 61, 62, 67, 71 or 73

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2 Hank flies from Los Angeles to Shanghai. (a) The flight departs on Friday 22 July at 21 40. The flight takes 13 hours 35 minutes. The local time in Shanghai is 15 hours ahead of the local time in Los Angeles. Find the day, date and time in Shanghai when Hank’s flight arrives. Day ........................... , Date ........................... , Time ........................... [3] (b) The cost of the flight is $920. The exchange rate is $1 = 6.87 Chinese yuan. Find the cost of the flight in yuan. ......................................... yuan [1]

Mark scheme: 2(a) Sunday 24 [July] 02 15 3 B1 for Sunday 24th [July] as final answer B2 for 02 15 oe as final answer or B1 for sight of any of these 12 40 oe, 11 15 oe, 28h 35min, 50 15, 35 15 or 0215 oe spoilt or M1 for departure time + 13h35min + 15h evaluated as a time with one interval correctly added 2(b) 6320.4[0] 1

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3 Calculate. 4.87 - 2.7 3 - 0.2 + 0.729 ................................................. [1]

Mark scheme: 3 3.1 1

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4 The number of items that each of 22 people buy in a supermarket is shown in the stem-and-leaf diagram. 1 1 3 6 6 2 0 2 2 2 4 8 9 3 1 1 5 8 9 9 4 2 4 6 7 8 Key: 1 | 1 represents 11 items (a) Find the mode. ................................................. [1] (b) Find the median. ................................................. [1]

Mark scheme: 4(a) 22 1 4(b) 30 1

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5 The table shows the relative frequency of the games won by a football team. Result of game won lost drawn Relative frequency 0.1 The number of games lost is twice the number of games drawn. Complete the table. [3]

Mark scheme: 5 3 B2 for 0.6 oe or 0.3 oe lost drawn or M1 for 1 – 0.1 or 0.9 seen 0.6 oe 0.3 oe

6 The scale drawing shows the positions of two towns, P and Q. The scale is 1 cm represents 4 km. North Q North P Scale: 1 cm to 4 km (a) Find the actual distance between town P and town Q. ........................................... km [2] (b) Measure the bearing of town Q from town P. ................................................. [1] (c) Town X is 28 km from town P on a bearing of 140°. On the scale drawing, mark the position of town X. [2]

Mark scheme: 6(a) 32.8 2 M1 for 8[cm] to 8.4[cm] seen or for their measurement [in cm] multiplied by 4 6(b) 065 1 6(c) X correctly placed 7 cm from P 2 M1 for X on bearing of 140 from P on a bearing of 140° or for X 7 cm from P If 0 scored SC1 for X on bearing of 140 from Q and 7 cm from Q

7 Without using a calculator, work out 1 + . 6 5 You must show all your working and give your answer as a mixed number in its simplest form. ................................................. [3]

Mark scheme: 7 25 or 55 12 M1 25k or 55k 12k and Accept and 30 30 30k 30k 7 A2 67 k 37k 230 cao A1 for 30k or 1 30k

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8 Solve the simultaneous equations. You must show all your working. 4x - 2y =- 13 - 3x + 4y = 11 x = ................................................ y = ................................................ [3]

Mark scheme: 8 Correctly eliminates one M1 variable [x =] – 3 , [y =] 0.5 oe A2 A1 for either correct If M0 scored, SC1 for 2 values satisfying one of the original equations If 0 scored, SC1 for correct answers from no working

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9 NOT TO 12 cm SCALE x° 7 cm Calculate the value of x. x = ................................................ [2]

Mark scheme: 9 54.3 or 54.31… 2 7 M1 for cos [x] = oe 12

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10 A regular polygon has an interior angle of 174°. Find the number of sides of this polygon. ................................................. [2]

Mark scheme: 10 60 2 M1 for 360 ÷ (180 – 174) 180 ( n − 2 ) or for = 174 oe n

11 Line L has equation y = 4 - 5x . Find the equation of a line that is perpendicular to line L and passes through the point (0, 6). ................................................. [3]

Mark scheme: 11 1 3 1 1 y = x + 6 oe final answer B2 for y = x + c oe or x + 6 oe or 5 5 5 y = mx + 6 oe 1 or B1 for [gradient =] oe or mx + 6 5

12 Chai invests some money. By the end of the first year, the value of the investment has decreased by 35%. By the end of the second year, the value of the investment has increased by 40% of its value at the end of the first year. Find the overall percentage change in the value of the investment. .............................................. % [3]

Mark scheme: 12 [– ] 9 3  35   40  M2 for [k × ]  1 −  ×  1 +  oe or  100   100  better  35  35  40  or for [k × ]  −  1 −  ×   100  100  100   35  or M1 for [k × ]  1 −  oe  100   40  or [k × ]  1 +  or better  100 

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13 Solve. 6 - x 4 - 3 x H 5 ................................................. [3]

Mark scheme: 13 x  1 final answer 3 6 M1 for 20 – 15x ⩾ 6 – x or 4 − 3x −x 5 5 M1 for correctly isolating terms in x FT their first step of dealing with the 5 x 6 20 – 6 ⩾ – x + 15x or −3x +  − 4 5 5

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14 y is inversely proportional to the square root of ( x - 2 ) . When x = 4.25, y = 12 . Find x when y = 3 . x = ................................................ [3]

Mark scheme: 14 38 3 M2 for 12 × 4.25 − 2 = 3 × x − 2 OR k M1 for y = oe x − 2 their k M1 for 3 = oe x − 2

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15 NOT TO SCALE The diagram shows three shapes that are mathematically similar. The heights of the shapes are in the ratio small : medium : large = 1 : 5 : 8. Find the ratio shaded area : total unshaded area. Give your answer in its simplest form. ..................... : ..................... [4]

Mark scheme: 15 3 : 5 nfww 4 M3 for 52 – 1 oe and 82 – 52 + 1 oe or M2 for 52 – 1 oe or 82 – 52 + 1 oe or M1 for 52 oe or 82 oe seen

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16 Find the nth term of each sequence. (a) 8, 15, 34, 71, 132, .... ................................................. [2] 4 5 6 (b) 2, 3, , , , .... 1 4 16 64 256 ................................................. [3]

Mark scheme: 16(a) n3 + 7 oe final answer 2 B1 for any cubic or for 3rd differences of 6 16(b) n + 1 3 B1 for n + 1 oe final answer n −1 4 B2 for 4n – 1 oe or B1 for 4n – k oe k can be 0 Maximum 2 marks if not correctly combined as a fraction

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17 y = 1 - x Make x the subject of the formula. x = ................................................ [4]

Mark scheme: 17 y + 2 4 M1 y (1 – x) = 3x – 2 or better [x = ] oe final answer y + 3 M1 for correctly isolating x terms on one side FT their first step/bracket expansion M1dep for correctly removing factor of x FT their previous step M1dep for correct division to isolate x Max 3 marks for an incorrect answer

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18 B NOT TO 800 m SCALE 30° A C 2300 m The diagram shows some land in the shape of a triangle ABC. Houses are built on this land. Each house requires 400 m 2 of land. Find the greatest number of houses that can be built on this land. ................................................. [3]

Mark scheme: 18 1150 3  1  M2 for  × 800 × 2300 × sin30  ÷ 400 oe  2  1 or M1 for × 800 × 2300 × sin30 oe 2

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19 Write as a single fraction in its simplest form. 2 x + 2 - x + 3 7 ................................................. [3]

Mark scheme: 19 8 − 5 x − x 2 8 − 5 x − x 2 3 B1 for 7 × 2 – (x + 2)(x + 3) or better seen or 7( x + 3) 7 x + 21 B1 for common denominator 7(x + 3) oe isw final answer

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20 Solve 3 ( 2 + cosx) = 5 for 0° G x G 360° . ................................................. [3]

Mark scheme: 20 109.4 to 109.5 3 B2 for one correct angle and 250.5 to 250.6 5 or M1 for cos x = − 2 or better 3 If 0 scored SC1 for two angles that sum to 360

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21 E NOT TO SCALE 9 cm D C O 5 cm A B 5 cm The diagram shows a pyramid ABCDE. The pyramid has a square horizontal base ABCD with side 5 cm. The vertex E is vertically above the centre O of the base. The height OE of the pyramid is 9 cm. Calculate the angle that EC makes with the base ABCD. ................................................. [4] Question 22 is printed on the next page.

Mark scheme: 21 68.6 or 68.55 to 68.56 4 9 M3 for tan[..] = oe 1 2 2 5 + 5 2 1 2 2 or M2 for 5 + 5 oe 2 or M1 for 52 + 52 oe or 2.52 + 2.52 oe or x2 + x2 = 52 oe or B1 for indicating required angle

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22 (a) Simplify. 2 x 3 8 x 3 ................................................. [1] (b) 16 = 64k Find the value of k. k = ................................................ [1] (c) Solve. 4 - 3 x 3 x 1 3 # = 3 e 9 o x = ................................................ [3]

Mark scheme: 22(a) –2 1 1 x or final answer x 2 22(b) 2 1 3 22(c) 1 nfww 3 M1 for 3–2(4 – 3x) oe or better 3 x 1 or 9 2 × 9 − ( 4 −x3 ) = 9 2 oe or better M1 for 3x + (their– 2)× (4 – 3x)= 1 oe or better 3x 1 or their − ( 4 − 3 x ) = their oe or better 2 2

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