O-LevelMathematicsAlgebra and graphs4024/21 · May/June 2014 · Q1112 Marks
Imran drives a distance of 180 km on a business trip. He drives the first 100 km at an average speed of x km/h. He drives at an average speed 5 km/h slower than this for the remainder of the journey.
- (a)Find, in terms of x, an expression for the time taken, in hours, for the first 100 km.[1]
- (b)Given that the journey takes a total of 2 hours 30 minutes, form an equation in x and show that it simplifies to x²-77x + 200 = 0.[4]
- (c)Solve the equation x²-77x + 200 = 0, giving each answer correct to 2 decimal places.[4]
- (d)Which of the solutions in part (c) represents the speed for the first 100 km of Imran's trip? Give a reason for rejecting the other solution.[1]
- (e)Find the difference between the times taken for the first and second parts of the journey. Give your answer in minutes, correct to the nearest minute.[2]
Mark scheme
Marking points from the official mark scheme (May/June 2014). The mark or mark type is shown in brackets.
- [1] 100/x
- [4] B1 for 80/(x-5) seen
- [4] M1 for 100/x + 80/(x-5) = 2.5 oe
- [4] M1 for 100(x - 5) + 80x = 2.5x(x - 5)
- [4] B3 for one correct root seen
- [4] or for 74 to 74.31 and 2.69 to 2.7
- [4] If in the form P±(or+or−)√9/r
- [4] B1 for p = 77 and r = 2
- [4] And B1 for q = 5129 or √q = 71.6...
- [1] 74.31, because 2.69 would give negative speed for second part
- [2] M1 for 100/74.31 - 80/(74.31-5) or 0.191 [hours]
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