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

2024
Views:

239

Pages:

8
Author:

Shahbaz A.

IGCSE EXTENDED MATHEMATICS (0580) TOPICAL PAST PAPER QUESTIONS - 2016/2017 ALGEBRA (Paper 4) shahbaz ahmed July 2024 1. (0580-S 2016-Paper 4/2-Q2) (a) Work out the value of x in each of the following. (i)3x = 243 Solution 3x = 35 =⇒ x=5 (ii)16x = 4 Solution (42 )x = 41 1 42x = 41 =⇒ 2x = 1 x= 1 2 (iii)8x = 32 Solution (23 )x = 25 23x = 25 =⇒ 3x = 5 x= 5 3 (iv)27x = 1 9 Solution (33 )x = ( 13 )2 33x = ( 13 )2 33x = (3−1 )2 33x = 3−2 =⇒ 3x = −2 2 x = − 23 (b) Solve by factorisation. y 2 − 7y − 30 = 0 y 2 − (10 − 3)y − 30 = 0 y 2 − 10y + 3y − 30 = 0 (y 2 − 10y) + (3y − 30) = 0 y(y − 10) + 3(y − 10) = 0 (y − 10)(y + 3) = 0 Either y − 10 = 0 Or y+3=0 y − 10 = 0 =⇒ y = 10 Similarly y + 3 = 0 =⇒ y = −3 2. (0580-S 2016-Paper 4/3-Q2) (a)Solve the inequality 5x − 3 > 9 5x − 3 + 3 > 9 + 3 5x > 12 3 x> 12 5 (b) Factorise completely. (i)xy − 18 + 3y − 6x (ii)8x2 − 72y 2 Solution xy − 18 + 3y − 6x xy + 3y − 6x − 18 (xy + 3y) − (6x + 18) y(x + 3) − 6(x + 3) (x + 3)(y − 6) (ii)8x2 − 72y 2 8(x2 − 9y 2 ) 8[x2 − 9y 2 ] 8[x2 − (3y)2 ] 8[(x − 3y)(x + 3y)] (c) Make r the subject of the formula. p+5= 1−2r r r(p + s) = 1 − 2r r(p + s) + 2r = 1 − 2r + 2r 4 r(p + s) + 2r = 1 r[p + s + 2] = 1 r= 1 p+s+2 3. (0580-S 2016-Paper 4/1-Q6) (a) The perimeter of the rectangle is 80cm. The area of the rectangle is Acm2 (i) Show that x2 − 40x + A = 0 5 (ii) When A = 300 , solve, by factorising , the equation x2 −40x+A = 0 (iii) When A = 200 , solve, by using the quadratic formula , the equation x2 − 40x + A = 0 Give your answer correct to 2 decimal places. Solution Width=x cm Let length=y cm Perimeter=2(length+width) Perimeter=2(y + x)cm Also Perimeter=80cm =⇒ 2(y + x) = 80 y+x= 80 2 y + x = 40 y = 40 − x Since Area=A=xy 6 A+x(40 − x) A+40x − x2 x2 − 40x + A = 0 (ii) Since x2 − 40x + A = 0 Putting A = 300 x2 − 40x + 300 = 0 x2 − (10 + 30)x + 300 = 0 x2 − 10x − 30x + 300 = 0 x(x − 10) − 30(x − 10) = 0 (x − 10)(x − 30) = 0 =⇒ Either x − 10 = 0 Or x − 30 = 0 Or both expression are zero If x − 10 = 0 =⇒ x=10 cm Or if x − 30 = 0 =⇒ x=30 cm (iii) Since x2 − 40x + A = 0 7 Putting A = 200 x2 − 40x + 200 = 0..............eq(1) ax2 + bx + c = 0..................eq(2) Comparing equations (1) and (2) a = 1, b = −40, c = 200 Putting in the formula x= x= x= x= x= √ −b± b2 −4ac 2a −(−40)± √ (−40)2 −4×1×200 2×1 √ 40± 1600−800 2 √ 40± 800 2 √ 40±20 2 2 √ x = 20 ± 10 2 x = 34.14 ans x = 5.86 8