2007 H2 Math Paper 2 Variant

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Questions

Section A: Pure Mathematics

1
Emily creates custom colored paints by mixing quantities (measured in litres, L) of cyan, magenta and yellow paint. Four of her custom paints are named Aurora, Burgundy, Ceylon and Daisy. The amount of individual paint used to make each custom paint and the cost of producing them are shown in the following table.
Aurora Burgundy Ceylon Daisy
Cyan (L) 1.75 1.50 1.85 2.70
Magenta (L) 3.30 1.10 1.15 0.90
Yellow (L) 3.95 1.75 3.95
Cost ($) 9.54 3.72 4.85 3.90
Assuming that the cost per litre is the same for the cyan, magenta and yellow paints, calculate the cost per litre for each of these three paints and the quantity of yellow paint used to make Daisy.
[6]
2
A sequence u1,u2,u3,…{u_1, u_2, u_3, \ldots} is such that un=1n{\displaystyle u_n = \frac{1}{n}}.
(a)
Show that un−un+1=1n(n+1){\displaystyle u_{ n } - u_{ n + 1 } = \frac{ 1 }{ n ( n + 1 ) }}.
[2]
(b)
Hence find ∑n=1N1n(n+1){\displaystyle \sum_{ n = 1 }^N \frac{ 1 }{ n ( n + 1 ) }}.
[4]
(c)
Give a reason why the series in part (ii) is convergent and state the sum to infinity.
[2]
(d)
Use your answer to part (ii) to find ∑n=2N1n(n−1){\displaystyle \sum_{ n = 2 }^N \frac{ 1 }{ n ( n - 1 ) }}.
[2]
3
(a)
By successively differentiating (1+x)n{( 1 + x )^n}, find the Maclaurin's series for (1+x)n{( 1 + x )^n}, up to and including the term in x3.{x^3.}
[4]
(b)
Obtain the expansion of (25−x)12(1+5x2)12{( 25 - x )^{\frac{1}{2}} ( 1 + 5 x^2 )^{\frac{1}{2}}} up to and including the term in x3.{x^3.}
[5]
(c)
Find the set of values of x{x} for which the expansion in part (ii) is valid.
[2]
4
(a)
Find the exact value of ∫034πsin⁡24x dx.{\displaystyle \int_0^{\frac{3}{4} \pi} \sin^2 4 x \, \mathrm{d}x.} Hence find the exact value of ∫034πcos⁡24x dx.{\displaystyle \int_0^{\frac{3}{4} \pi} \cos^2 4 x \, \mathrm{d}x.}
[6]
(b)
The region R{R} is bounded by the curve y=x2cos⁡x{y=x^2 \cos x}, the line x=13π{x = \frac{1}{3} \pi} and part of the x-axis{x\textrm{-axis}} between 0{0} and 13π{\frac{1}{3} \pi}. Find
(i)
the exact area of R,{R,}
[5]
(ii)
the numerical value of the volume of revolution formed when R{R} is rotated completely about the x-axis,{x\textrm{-axis,}} giving your answer correct to 3 decimal places.
[2]

Section B: Statistics

5
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6
In a lucky draw with a large number of entries, 61% of entries win a small prize.
(a)
Find the probability that, in a random sample of 10 entries, at most more than 4 win a small prize.
[2]
(b)
1000 samples of 10 entries are taken. Find the probability that
(i)
exactly 850 of these samples have at most more than 4 entries winning a small prize,
[3]
(ii)
between 845 and 855 (both inclusive) of these samples have at most more than 4 entries winning a small prize.
[2]
7
A large number of students prepare for a national examination. The time, x{x} hours, taken by a student to prepare for the examination is noted for a random sample of 190 students. The results are summarised by
∑x=8476,∑x2=379747.\sum x = 8476, \quad \sum x^2 = 379747.
(a)
Find unbiased estimates of the population mean and variance.
[2]
(b)
Test, at the 5% significance level, whether the population mean time for a student to prepare for the examination is less than 45 hours.
[4]
(c)
State, giving a reason, whether any assumptions about the population are needed in order for the test to be valid.
[1]
8
In a supermarket, apples can be bought in bags and oranges can be bought in boxes and are sold by weight. The masses, in kg, of a bag of apples and a box of oranges are modelled as having independent normal distribution with means and standard deviation as shown in the table.
Mean mass Standard deviation
Bag of apples 1.8 0.2
Box of oranges 10 2.2
Apples are sold at $1.5 per kg and oranges at $4 per kg.
(a)
Find the probability that a randomly chosen bag of apples has a selling price exceeding $2.9
[2]
(b)
Find the probability that a randomly chosen bag of apples has a selling price exceeding $2.9 and a randomly chosen box of oranges has a selling price exceeding $42.5
[3]
(c)
Find the probability that the total selling price of a randomly chosen bag of apples and a randomly chosen box of oranges is more than $45.4
[4]
(d)
Explain why the answer to part (iii) is greater than the answer to part (ii).
[1]
9
A group of 12 people consists of 6 married couples.
(a)
The group stand in a line.
(i)
Find the number of different possible arrangements.
[1]
(ii)
Find the number of different possible arrangements in which each man stands next to his wife.
[3]
(b)
The group stand in a circle.
(i)
Find the number of different possible arrangements.
[1]
(ii)
Find the number of different possible arrangements if men and women alternate.
[2]
(iii)
Find the number of different possible arrangements if each man stands next to his wife and men and women alternate.
[2]
10
To predict the weather for three days (Monday, Tuesday and Wednesday), a simple computer simulation uses the following model to determine the likelihood of rain.
The probability that it rains on Monday is 0.4.{0.4.} For Tuesday and Wednesday,
  • the probability that it rains, given that it rained the previous day, is 0.5,
  • the probability that it does not rain, given that it did not rain the previous day, is 0.7.
(a)
Construct a probability tree showing this information
[3]
(b)
Find the probability that it rains on all three days.
[2]
(c)
Find the probability that it rains on at least two days.
[2]
(d)
Find the probability that it rains on Wednesday, given that it rained on exactly two days.
[4]
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Answers

1
Cyan: $0.40/L.
Magenta: $2.20/L.
Yellow: $0.40/L.
Quantity of yellow paint used to make Daisy: 2.10L.
2
(b)
1−1N+1{1 - \frac{1}{ N + 1 }}.
(c)
As N→∞,{N \to \infty,} 1N+1→0{\frac{1}{ N + 1 } \to 0} so 1−1N+1→1.{1 - \frac{1}{ N + 1 }\to1.}
Hence the series is convergent and the sum to infinity is 1.{1.}
(d)
1−1N{1 - \frac{1}{ N }}.
3
(a)
1+nx+n(n−1)2x2+n(n−1)(n−2)6x3+…{1 + n x + \frac{n(n-1)}{2} x^2 + \frac{n(n-1)(n-2)}{6} x^3 + \ldots}
(b)
5−110x+124991000x2−1250150000x3+…{5 - \frac{1}{10} x + \frac{12499}{1000} x^2 - \frac{12501}{50000} x^3 + \ldots}
(c)
−155<x<155.{- \frac{1}{5} \sqrt{5} < x < \frac{1}{5} \sqrt{5}.}
4
(a)
∫034πsin⁡24x dx=38π{\displaystyle \int_0^{\frac{3}{4} \pi} \sin^2 4 x \, \mathrm{d}x = \frac{3}{8} \pi}.
∫034πcos⁡24x dx=38π{\displaystyle \int_0^{\frac{3}{4} \pi} \cos^2 4 x \, \mathrm{d}x = \frac{3}{8} \pi}.
(b)
(i)
(1183(π)2+13π−3) units2.{\left(\frac{1}{18} \sqrt{3} \left( \pi \right)^{ 2 } + \frac{1}{3} \pi - \sqrt{3}\right) \textrm{ units}^2.}
(ii)
0.328 units3.{0.328\textrm{ units}^3.}
5
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6
(a)
0.850.
(b)
(i)
0.0353.
(ii)
0.341.
7
(a)
x‾=44.6,s2=8.61.{\overline{x} = 44.6, s^2 = 8.61.}
(b)
(i)
p−value=0.0337≤0.05 ⇒ H0 rejected.{p-\textrm{value} = 0.0337 \leq 0.05} \allowbreak \, \Rightarrow \, \allowbreak {H_0 \textrm{ rejected}.}
Hence there is sufficient evidence at the 5% level of significance to conclude that the population mean time for a student to prepare for the test is less than 45 hours.
(ii)
No assumptions are needed about the population. This is because, as n=190{n=190} is large, by the Central Limit Theorem, X‾{\overline{X}} is normally distributed approximately so no further assumptions on the population are needed.
8
(a)
0.252.
(b)
0.0980.
(c)
0.380.
(d)
The event in part (ii) is a subset of the event in part (iii).
9
(a)
(i)
479,001,600.
(ii)
46,080.
(b)
(i)
39,916,800.
(ii)
86,400.
(iii)
240.
10
(b)
0.1.{0.1.}
(c)
0.35.{0.35.}
(d)
35.{\frac{3}{5}.}
11
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