Mathematics GCSE Higher Tier Taster Pages

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Kelley Cunningham
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Question 14 (June 2011 4306/1H) a) on the cumulative frequency diagram you can work out the lower quartile, median and upper quartile. These have been got by using the dashed red lines.

1 Question 14 (June /1H) a) on the cumulative frequency diagram you can work out the lower quartile, median and upper quartile. These have been got by using the dashed red lines. Draw them across and where they meet the curve drop them down to meet the x axis. Lower quartile = 23 Median = 34 Upper quartile = 42 We can also see that the 40 th piece of data finishes at 58 We are told that the first piece of data starts at 7 b) this is shown on the cumulative frequency diagram as the green dashed line. Draw this up from 45 on the x axis to meet the curve. Then across to meet the y axis at 33. This means that 33 people took less than 45 minutes. So 7 (40 33) took more than 45 minutes Answer = 7 people c) it is better to use the cumulative frequency diagram rather than the box plot as the box plot is just a summary it only gives details about medians and quartiles and minimum and maximum values Page 1

2 Question 15 (June /1H) Substitution method 3x 2y = 5 equation 1 x = y + 3 equation 2 As x is already in terms of y in equation 2 I am going to use the substitution method and just replace the x in the first equation with y + 3 3(y + 3) 2y = 5 Expand 3y + 9 2y = 5 Group terms y + 9 = 5 Subtract 9 from both sides y = -4 Now we have y we can put this back into equation 2 to get x x = y + 3 x = x = -1 Check our answer by putting both x and y back into both equations Equation 1: (3 x -1) (2 x -4) = = = 5 Equation 2: -1 = = -1 x = -1 and y = Page 2

3 alternatively we could solve by elimination Elimination method 3x 2y = 5 equation 1 x = y + 3 equation 2 multiply equation 2 by 3 so that we have the same number of x 3x 2y = 5 equation 1 (unchanged) 3x = 3y + 9 equation 3 Now we can subtract equation 1 from equation 3 (doing it this way around so that end up with a positive number of y) LHS 3x (3x 2y) = 3x 3x + 2y = 2y RHS 3y = 3y + 4 Now put the RHS = LHS 3y + 4 = 2y Subtract 2y from both sides y + 4 = 0 Subtract 4 from both sides y = -4 Now we have y we can put this back into equation 2 to get x x = y + 3 x = x = -1 Check our answer by putting both x and y back into both equations Equation 1: (3 x -1) (2 x -4) = = = 5 Equation 2: -1 = = -1 x = -1 and y = Page 3

Question 19 (June 2011 4306/1H) Looking at just triangle ABC we can work out length AC using basic trigonometry A B x 4 cm adj opp Label each side from the point of view of the angle I haven t

4 Question 19 (June /1H) Looking at just triangle ABC we can work out length AC using basic trigonometry A B x 4 cm adj opp Label each side from the point of view of the angle I haven t labelled the hypotenuse (hyp) as tan doesn t use hyp tan x = = = 0.6 Multiply both sides by 4 AC = 2.4 cm Now we can do a similar thing with triangle ACD C A opp hyp 2.4 cm y C D Label each side from the point of view of the angle y. This time I have not shown the adjacent as sin doesn t use adjacent (adj) sin y = =. = Multiply both sides by AD 2.4 = Multiply both sides by 3 AD = 7.2 cm Page 4

5 Question 6 (June /2H) a) we can draw this line by plotting three points and joining them up or by using gradients and intercepts plotting 3 points x y (3 x -4) - 1 = -13 (3 x 0) - 1 = -1 (3 x 4) - 1 = 11 Plot these 3 coordinates (-4, -13), (0, -1), (4, 11) and join the 3 points with a straight line Gradients and intercepts y = mx + c where m is the gradient and c is the y intercept In this case the intercept is -1 (where the graph crosses the y axis) and the gradient is 3 (level of steepness of the graph) As the gradient is 3 this means we go up 3 for every 1 we go across Page 5

Question 22 (June 2011 4306/2H) a) the graph is y = x 2 + 3x 2 we want to know where 0 = x 2 + 3x 2 so this happens when y = 0 the curve meets the line y = 0 when x = 0.6 x = 0.

6 Question 22 (June /2H) a) the graph is y = x 2 + 3x 2 we want to know where 0 = x 2 + 3x 2 so this happens when y = 0 the curve meets the line y = 0 when x = 0.6 x = 0.6 b) we have to rearrange x 2 + 2x 4 = 0 to look like x 2 + 3x 2 = something x 2 + 2x 4 = 0 add x to both sides x 2 + 3x 4 = x add 2 to both sides x 2 + 3x 2 = x + 2 we can replace the LHS with y y = x + 2 we draw this line on the same graph and where the line and curve meet these will be the approximate solutions to the equation the line meets the graph when x = 1.2 and x = -3.2 (shown with red dashed lines) Page 6

7 If you like these worked answers then why not purchase the complete set of answers for these question papers by visiting It should be noted that Chatterton Tuition is responsible for these solutions. The solutions have not been produced nor approved by Edexcel. In addition these solutions may not necessarily constitute the only possible solutions Page 7

Mathematics IGCSE Higher Tier, June /4H (Paper 4H)

Mathematics IGCSE Higher Tier, June /4H (Paper 4H) Link to examining board: http://www.edexcel.com The question paper associated with these solutions is available to download for free from the Edexcel website. The navigation around the website sometimes