Centre Number 71 Candidate Number
General Certificate of Secondary Education 2011
Mathematics Unit T4
(With calculator) Higher Tier
TUESDAY 31 MAY
For Examiner’s use only Question Number Marks
9.15 am–11.15 am
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
INSTRUCTIONS TO CANDIDATES
Write your Centre Number and Candidate Number in the spaces provided at the top of this page. Write your answers in the spaces provided in this question paper. Answer all nineteen questions. Any working should be clearly shown in the spaces provided since marks may be awarded for partially correct solutions. You may use a calculator for this paper. INFORMATION FOR CANDIDATES
The total mark for this paper is 100. Figures in brackets printed down the right-hand side of pages indicate the marks awarded to each question or part question. Functional Elements will be assessed in this paper. Quality of written communication will be assessed in question 11. You should have a calculator, ruler, compasses and a protractor. The Formula Sheet is overleaf.
Volume of prism area of cross section length
– Area of trapezium = 1 (a + b)ha + b)h 2 Area of trapezium 1 ( 2
b length – Area of trapezium = 1 (a + b)h 2 a
– Area of trapezium = 1 (a + b)h 2
Volume of prism = area of cross section × length Volume of cone 1 πr 2 h 3 In anyCurved surface area of cone πrl triangle ABC – Area of triangle = 1 ab sin C 2
In any triangle ABC
Volume of prism = area of cross section × length a = b = c Sine rule : sin A sin B sinCross C section b of c Volume of a Sine Rule: prism = area cross section × length sin A In any triangle ABC sin B sin C Cosine rule: a2 = b2 + c2 – 2bc cos A Cross section c A c B b a l r r
Cosine Rule: a2 Area of triangle = 1 ab sin C b2 c2 2bc cos – A 2 In any triangle ABC a b c Sine rule : = = Area of triangle 1 ab sin C Area of triangle = – ab sin C sin A sin B sin C 2 Sine rule :
b Surface ar
2 2 a2 a Cosine rule: c = b + c – 2bc cos A b = = B – sin A sin B sin C Volume of cone = 1 πr 2h 3 2 2 2
Cosine rule: a = b + c – 2bc cos A Quadratic Equation The solutions of ax2 bx c 0 where a ≠ 0, are given by x
Curved surface area of cone = πrl C Volume of sphere 4 πr 3 Volume of sphere = – 3 Surface area of sphere r 4πr 2 = Surface area of sphere 4πr 2 Quadratic– equation: Volume of sphere = 4 πr 3 3
– b ± b 2 – 4 ac Surface area of sphere = 4πr22 + bx + c = 0, where a ≠ 0, are g The solutions of ax 2a – Volume of cone = 1 πr 2h 3 h – b ± l b 2 – 4 ac x= Curved surface area of cone = πrl 2a r 1 2 Volume of cone = – πr h 3 l h 5414
Curved surface area of coneequation: Quadratic = πrl
The solutions of ax2 + bx + c = 0, where a ≠ 0, are given by Quadratic equation: – b ± b 2 – 4 ac x2= bx + c = 0, where a ≠ 0, 2 given by The solutions of ax + are 6528.06RR 2a 2
Answer all questions. 1 The angle of elevation of the top of a telephone mast is 23° from a point 60 metres from the base of the mast on horizontal ground. Calculate the height of the mast.
Examiner Only Marks Remark
Answer __________ m 
2x – 4 (a) Solve the equation 5 Show your working.
x 11 2
A solution by trial and improvement will not be accepted.
Answer x _________  (b)
Write down the equation of this line in the form y mx c. Answer ___________________  6528.06RR
The box plots show the distribution of test results for two different classes. Class P
Examiner Only Marks Remark
Comment on two differences between the classes. 1. ___________________________________________________________...