Question 11 (June 2011 - 4MA0/3H) a) Weight (w kg)
Frequency
Mid point
0w2 2w4 4w6 6w8 8 w 10 10 w 12 Total
8 14 26 17 10 5 80
1 3 5 7 9 11
Midpoint x frequency 8 42 130 119 90 55 444
Total weight is 444 kg b) Weight (w kg) 0w2 2w4 4w6 6w8 8 w 10 10 w 12
Cumulative Frequency 8 22 48 65 75 80
To get the cumulative frequency you just keep adding the next category to the previous Eg cum freq for 2nd category is 8 + 14 = 22, cum freq for 3rd category is 22 + 26 = 52 The final cumulative frequency should be the same as the total frequency – it is
www.chattertontuition.co.uk 0775 950 1629
Page 1
Mathematics IGCSE Higher Tier 2011 4MA0 Taster Pages c) the points can be joined as straight lines or as a curve as I have shown
d) Draw a vertical line up from 5.2 kg to meet curve then draw a horizontal line to meet the y axis The cumulative frequency is 38. This means that there were less than 38 parcels that weighed less than 5.2 kg 38 parcels
www.chattertontuition.co.uk 0775 950 1629
Page 2
Mathematics IGCSE Higher Tier 2011 4MA0 Taster Pages Question 15 (June 2011 - 4MA0/3H) a) we can work out the area of the rectangle and the semi circle separately and then add the two areas together Area rectangle Length is 7.1 cm Width is 2 x 2.7 = 5.4 cm Area = 7.1 x 5.4 = 38.34 cm2 Area semicircle Area of a full circle is πr2 so we want ½ of this π x 2.72 ÷ 2 = 11.451 cm2 Total area = 38.34 + 11.451 = 49.8 (3 significant figures) b) P = πr + 2L + 2r put all terms involving r on one side and all other terms on the other side subtract 2L from both sides P – 2L = πr + 2r Rewrite the other way around πr + 2r = P – 2L factorise out the r r(π + 2) = P – 2L divide both sides by π + 2 r=
www.chattertontuition.co.uk 0775 950 1629
Page 3
Mathematics IGCSE Higher Tier 2011 4MA0 Taster Pages Question 18 (June 2011 - 4MA0/3H) We have two right angled triangles (triangle ABC and triangle CDB) so we can use SOHCAHTOA on each.First triangle ABC to get length BC then triangle CDB to get DB Label all the sides from the point of view of the angle Triangle ABC
C 47 m
hyp
opp
32⁰ A
adj
B
We have hyp (H) and we are trying to find opp (O). We don’t have Adj (A) and don’t want it. SOHCAHTOA this leaves SOH (sine) sin 32⁰ = =
multiply both sides by 47 47 sin32⁰ = BC BC = 47 sin32⁰ = 24.9062 Now we can work on triangle CDB to get BD Label all the sides from the point of view of the angle Triangle CDB
C hyp
opp 24.9062 m
51⁰ D
adj
B
We have opp (O) and we are trying to find adj (A). We don’t have hyp (H) and don’t want it. SOHCAHTOA this leaves TOA (tan) . tan 51⁰ = = multiply both sides by BD BD x tan51⁰ = 24.9062 Divide both sides by tan51⁰ BC = 24.9062 ÷ tan 51⁰ = 20.2 m (3 significant figures)
www.chattertontuition.co.uk 0775 950 1629
Page 4
Mathematics IGCSE Higher Tier 2011 4MA0 Taster Pages Question 17 (June 2011 - 4MA0/4H) a) the probability of being late or not late will be the same on both days (as not been told any different). If probability of being late is then probability of not being late is 1 - = Monday meeting
Tuesday meeting
1 8
late
late
1 8
7 8 1 8
7 8
7 8
Not late
Not late
late
Not late
b) The probability that Alan is not late for either meeting is x = So the probability that Alan is late for at least one meeting is
1 - = - = We could have obtained the same answer by adding up the three probabilities that he is Late Monday, late Tuesday ( x = )
)
Late Monday, not late Tuesday ( x = ) Not late Monday, late Tuesday ( x = +
+ =
www.chattertontuition.co.uk 0775 950 1629
Page 5
Mathematics IGCSE Higher Tier 2011 4MA0 Taster Pages Question 23 (June 2011 - 4MA0/4H) We want our answer in terms of k√6 so divide 48 and 108 by 6 √48 = √8 x 6 = √8 x √6 = √4 x 2 x√6 = √4 x √2 x √6 = 2√2 x √6 √108 = √18 x 6 = √18 x √6 = √9 x 2 x√6 = √9 x √2 x √6 = 3√2 x √6 Now add the two surds to get 5√2 x √6 So k = 5√2
If you like these worked answers then why not purchase the complete set of answers for these question papers by visiting www.chattertontuition.co.uk/IGCSE-maths-Edexcel-2011 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.
