O-LevelMathematicsAlgebra and graphs4024/21 · Oct/Nov 2014 · Q712 Marks
Solve 4x/3 = 1,
- (a)(i)Solve 4x/3 = 1,[1]
- (a)(ii)4y - 3 (2y + 1) = 5 .[2]
- (b)Simplify 15w² - 30w / 5w² - 20[3]
- (c)(i)The diagram shows the plan of a patio made from rectangular paving slabs. The width of each paving slab is pcm. The length of each paving slab is 20 cm longer than its width. Find an expression, in terms of p, for the area, in cm², of one paving slab.[1]
- (c)(ii)Given that the area of the patio is 12.25 m², show that p satisfies the equation p² + 20p - 3500 = 0 .[2]
- (c)(iii)(a)Solve by factorisation p² + 20p – 3500 = 0 .[2]
- (c)(iii)(b)Hence state the length of each paving slab.[1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2014). The mark or mark type is shown in brackets.
- [1] 3/4 or 0.75
- [2] (y =)-4
- [2] M1 for 4y – 6y – 3 = 5 or correctly rearranges their linear equation
- [3] 3w / w+2 final answer
- [3] B1 for 15w(w - 2) and B1 for 5(w + 2)(w – 2)
- [1] p(p+20) or p² + 20p
- [2] Correct equation and the given form correctly derived.
- [2] M1 for 35( p² + 20p)
- [2] and A1 for 35( p² + 20p ) = 122500 And the given form established.
- [2] p = 50 and p = - 70
- [2] M1 for (p + h)(p + k) where hk = 3500
- [1ft] 70
- [1ft] Accept their positive p + 20
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →