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

2024
Views:

300

Pages:

5
Author:

Shahbaz A.

IGCSE EXTENDED MATHEMATICS (0580) TOPICAL PAST PAPER QUESTIONS - 2016-2017 ALGEBRA Q5 and Q6 shahbaz ahmed July 2024 1 Q5. (0580-W 2016-Paper 4/3-Q3) (a)Solve. 8x − 5 = 22 − 4x (b)Solve. 6x ≥ 2x + 14 (c)Factorise. x2 − 4x − 21 (d) Expand the bracket and simplify. (3x − 2y)(4x + 3y) Solution 1 (a) 8x − 5 = 22 − 4x 8x − 5 + 5 = 22 − 4x + 5 8x = 27 − 4x 8x + 4x = 27 − 4x + 4x 12x = 27 x= 27 12 (b) 6x ≥ 2x + 14 6x − 2x ≥ 2x + 14 − 2x 4x ≥ 14 x≥ 14 4 x≥ 7 2 = 7 2 = 3.5 x ≥ 3.5 c) x2 − 4x − 21 =x2 − (7 − 3)x − 21 =x2 − 7x + 3x − 21 =(x2 − 7x) + (3x − 21) 2 =x(x − 7) + 3(x − 7) =(x + 3)(x − 7) (d) (3x − 2y)(4x + 3y) =3x(4x + 3y) − 2y(4x + 3y) =12x2 + 9xy − 8xy − 6y 2 =12x2 + xy − 6y 2 6. (0580-W 2016-Paper 4/2-Q7) (a)$1 = 3.67 dirhams Calculate the value,in dollars, of 200 dirhams. Give yours answers correct to 2 decimal places. (b)(i)Write as a single fraction,in its simplest form. 1000 x − 1000 x+1 (ii) One day in 2014, 1 euro was worth x rand. One year later,1 euro was worth x + 1 rand. Winston changed 1000 rand into euros in both years. In 2014,he received 4.50 euros more than in 2015. Write an equation in terms of x and show that it simplifies to 3 9x2 + 9x − 2000 = 0. Solution (a) $1 = 3.67 dirhams $1 × 200 = 3.67 × 200 dirhams $200 = 734 dirhams (b)(i) 1000 x − 1000 x+1 =1000[ x1 − 1 x+1 ] =1000[ x+1−x x(x+1) ] 1000 = x(x+1) (ii) In 2014, 1euro = x rand. 1 x euro = 1 rand 1000 × 1 x euro = 1000 rand One year later 1 euro = x + 1 rand. 1 x+1 euro = 1 rand. 1000 × 1 x+1 euro = 1000 rand. 4 According to the condition 1000 × 1 x − 1000 × 1000[ x1 − 1 x+1 ] 1 x+1 =4.50 euros =4.50 euros 1000[ x+1−x x(x+1) ] =4.50 euros 1 ] =4.50 euros 1000[ x(x+1) 1 x(x+1) = 4.50 1000 euros 1 x(x+1) = 9 2000 euros 2000 = 9x(x + 1) 9x2 + 9x = 2000 9x+ 9x − 2000 = 0 5