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1. A is the point (5,7) and B is the point (1,5). What are the coordinates of the midpoint of line AB?
(1,2)(2,1)(3,1)(2,3)You calculate this by adding the x coordinates together and dividing by two to get x = 2, then doing the same with the y coordinates to get y = 1.

2. What is the gradient of AB?
234Subtract one y coordinate from the other, do the same with the x coordinates, then divide the first result by the second. Voila!

3. Which plot represents y=9Ë£?
Via wolframalpha.comVia wolframalpha.comThe other plot represents y = 2Ë£.

4. Which of the following is equivalent to sin(x)  cos(x) = 6 cos(x) / tan(x) ?
tanÂ²(x)  tan(x)  6 = 0cos(x) + tanÂ²(x)  6 = 06  tanÂ²(x) = 0Multiply the whole thing by tan(x) then divide it by cos(x). Simplify using trigonometric identities, and rearrange.

5. A curve has equation y = (2x + 4)/(xÂ² + 5). What is dy/dx?
Hint: Try the quotient rule.
dy/dx = (xÂ² + 10x  8) / (xÂ² + 5)Â²dy/dx = (2xÂ²  8x + 10) / (xÂ² + 5)Â²You need to use the quotient rule for this one. Treat the top and bottom of the fraction as different functions, use the rule, then simplify what you get.

6. Now you have dy/dx, what are the coordinates of one of the stationary points on the above curve?
Type the coordinates like this: x,y.
The two stationary points are (1,1) and (5,1/5). Solve 2xÂ²  8x + 10 to find x = 1 and x = 5, then substitute those back into the original curve equation to find y.

7. A particle is speeding towards the ground at 3.5 metres per second vertically downwards, from a height of 5 metres above the ground. What equation would you use to work out its speed just before hitting the ground?
v = u + atvÂ² = uÂ² + 2asYou don't have the time, so you have to use the second equation where all you need are the starting velocity (u) the acceleration (a) and the distance (s).

8. What is the speed of the particle in the previous question, just before it hits the ground?
510.515.1Just substitute in u = 3.5, a = 9.8 and s = 5 to vÂ² = uÂ² + 2as to find v.

9. This is a stem and leaf diagram showing the heights of a random sample of trees. What is the median tree height?
Type your answer to two decimal places.
There are 20 tree heights in the diagram, so the median height is in between the 10th and 11th values, which are 7.4 and 7.5 respectively.

10. For the same sample of trees, what is the interquartile range?
Type your answer to two decimal places.
The IQR = 7.75  6.7 = 1.05.
Questions adapted from OCR past exam papers for all four core modules as well as mechanics and statistics.