University:
University of Cambridge
Course:
0580 IGCSE | Mathematics
Academic year:

2024
Views:

317

Pages:

6
Author:

Shahbaz A.

IGCSE EXTENDED MATHEMATICS (0580) TOPICAL PAST PAPER QUESTIONS - 2016-2017 ALGEBRA shahbaz ahmed July 2024 Q 3. (b)(0580-S 2016-Paper 4/1-Q6) A car completes 200 Km journey with an average speed of x Km h . The car complete the return journey of 200 Km with an average speed of (x + 10) Km h . (i)Show that the difference between the time taken for each of the two journeys is 2000 x(x+10 hours. (ii)Find the difference between the time taken for each of the two journeys when x = 80. Give yours answer in minutes and seconds. Solution Average speed to complete journey =x km h 1 Since speed= distance time x= 200 Km t1 h Or t1 = 200 x hours Where t1 is the time to complete journey Similarly t2 = 200 x+10 hours Where t2 is the time to complete return journey Now Difference=t1 − t2 = 200 x − 200 x+10 Difference=t1 − t2 = 200[ x1 − hours 1 x+10 ] hours x+10−x Difference=t1 − t2 = 200[ x(x+10) ] hours Difference=t1 − t2 = 2000 x(x+10) hours (ii) Putting x = 80 in the equation Difference=t1 − t2 = 2000 80(80+10) hours Difference=t1 − t2 = 2000 80(90) = 5 18 Difference=t1 − t2 = 2000 80(90) = 16 minutes and 40 seconds hours 2 Q 4. (0580-S 2016-Paper 4/3-Q7) Alfonso runs 10 Km at an average speed of x Km h . The next day he runs 12 Km at an average speed of x-1 Km h . The time taken for 10 Km run is 30 minutes less than the time taken for the 12 Km run. (a)(i) Write an equation in x and show that it simplifies to x2 − 5x − 20 = 0 (ii) Use the quadratic formula to solve the equation x2 −5x−20 = 0. Give answer correct to 2 decimal places. (iii)Find the time that Alfonso takes to complete the 12 Km runs. Give your answer in hours and minutes correct to the nearest minutes. (b) A cheetah runs for 60 seconds. The diagram shows the speed-time graph. (i)Work out the acceleration of the cheetah during the first 10 sec3 onds. (ii) Calculate the distance travelled by the cheetah. Solution(a)(i) x= 10 km t1 h Where t1 is the time taken in covering 10km. t1 = 10 x hours Similarly t2 = 12 x−1 hours Where t2 is the time taken in covering 12km. According to the condition t2 − t1 = 30 minutes t2 − t1 = 30 60 t2 − t1 = 1 2 t2 = 1 2 + t1 t2 = 1 2 + t2 = x+20 2x hours. hours. 10 x Putting in the equation t2 = 12 x−1 x+20 2x = hours 12 x−1 hours 4 (x + 20)(x − 1) = 24x x2 − x + 20x − 20 = 24x x2 + 19x − 24x − 20 = 0 x2 − 5x − 20 = 0 (ii) x2 − 5x − 20 = 0 ax2 + bx + c = 0 =⇒ a=1,b=-5,c=-20 Putting in the quadratic formula x= x= x= √ −b± b2 −4ac 2a 5± √ (−5)2 −4(1)(−20) 2×1 √ 5± 105 2 x = 7.62 Km, x = −2.62 Km(which is impossible) Hence x = 7.62 Km (iii) Since t2 = 12 x−1 hours Putting x = 7.62 Km 5 t2 = 12 7.62−1 t2 = 12 6.62 hours = 1.81 = 1 hour 48 minutes and 45.68 seconds (b)(i) acceleration= 25−0 10 = 2.5 m s2 (ii) Distance travelled in first 10 seconds s1 = 0 × 10 + 21 (2.5)(100) = 125 m Distance travelled in 45 seconds s2 = 25 × 45 = 1125 m Distance travelled in the last 5 seconds Here a= 25−0 5 =5 m s2 s3 = 25 × 5 + 12 × 5 × 25 = 187.5 m Total distance=s1 + s2 + s3 = 125 + 1125 + 187.5 = 1437.5m 6