Trial Stpm Mathematics M 2 (sarawak) Smk Paku 2v5j3z
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CONFIDENTIAL* TRIAL STPM MATHEMATICS M (SARAWAK) –SMK Paku QA
Section A [45 marks] Answer all questions in this section. 1
2
In a box containing 80 light bulbs, 60 are in good condition. Two bulbs are taken randomly, one after the other, from the box. Find the probability that only one of the two bulbs taken is in good condition. (a)
If the first bulb is returned into the box.
[3 marks]
(b)
If the first bulb is not returned into the box.
[3 marks]
The table below shows the prices of fish and the quantities of it bought by a housewife at a market in the year 2011 and 2012. Fish
Parang Tenggiri Bawal Putih Kembung Selar Kuning
2011 Price Quantity (RM per kg) (kg) 11.00 2 12.00 2 10.00 2 8.00 3 4.00 5
2012 Price Quantity (RM per kg) (kg) 12.00 1 13.00 2 a 1 10.00 2 5.00 6
(a) If the simple aggregate price index increases by 20 % from 2011 to 2012, determine the value of a. [3 marks] (b) Calculate the Laspeyres price index, and comment on the housewife’s change in expenditure on fish. [3 marks]
3.
A random sample of 24 people were asked to record the number of kilometres they travelled by car in a given week. The distances, to the nearest kilometres, are shown below. 67 72 70
66 77 70
85 67 41
93 48 108
63 66 58
44 50 74
87 52 62
57 74 68
(a)
Construct a stem-and-leaf diagram to represent these data. [2 marks]
(b)
Determine the median and interquartile range of this distribution. [3 marks] Draw a box-and-wisker plot for this data. [3 marks]
(c) STPM 950/2
[Turn over CONFIDENTIAL* This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
2
4 The probability density function of the time t, in minutes, of a student spends every morning waiting for bus to school is given by 2 75 t 0 ≤ t < 5, 1 1 f (t ) = − t , 5 ≤ t < 15, 5 75 otherwise 0 (a) Find the cumulative distribution function of T. [4 marks] (b)
5.
Calculate the mean waiting time of the student.
The water vapour content, w (in g) per 100 g samples of mud, taken at a depth of d meters from a river mouth is measured. The results are shown in the table below. Depth (d) 0 5 10 Water vapour content (w) 92 80 57 (a) Plot these data on a scatter diagram (b)
15 44
20 32
25 22
30 35 20 17 [2 marks]
Calculate the Pearson’s correlation coefficient for the above data. [4 marks] Comment on the relationship between the depth of the mud and the water vapour content. [2 marks]
(c)
6.
[3 marks]
Unemployed school leavers in the Malaysia (figures in thousands) is tabulated below. Year 2010 2011 2012 (a) (b) (d)
Quarter 1 22 21 50
Quarter 2 12 26 36
Quarter 3 110 150 146
Quarter 4 31 70 110
Find the four-quarter centred moving average for the time series. [3 marks] Calculate the seasonal variation using an additive model. [3 marks] Use this model to predict unemployed school leavers for the first quarter of the year 2013. [4 marks]
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
3
Section B [15 marks] Answer any one question in this section. 7.
Everyday, a bus leaves town A for town B at 0800 hours. The time taken for this journey is recorded for a certain period. The result are shown in the following table. Time (minutes) Frequency 6 25 ≤ x ≤ 28 12 28 ≤ x ≤ 29 27 29 ≤ x ≤ 30 30 30 ≤ x ≤ 31 18 31 ≤ x ≤ 32 14 32 ≤ x ≤ 33 9 33 ≤ x ≤ 34 4 34 ≤ x ≤ 35 5 35 ≤ x ≤ 40 a) Draw a histogram to represent these data. [3 marks] b) c)
d)
8
Calculate the median, mean and standard deviation for this data. [5 marks] Hence, calculate the Pearson coefficient of skewness and comment the skewness of the distribution. [2 marks] Plot a cumulative frequency curve of the above data. Hence, find the number of times a bus arrives at town B between 0830 hours and 0836 hours. [5 marks]
(a) On average, the number of books read by an adult is 5 books per year. Using the Poisson distribution, find the probability that [2 marks] (i) an adult reads exactly 3 books per year, (ii) an adult reads more than 2 books in 5 years. [2 marks] (b) The marks obtained by the candidates for a paper in an examination are distributed normally with mean of 50 and standard deviation 10. (i) If a candidate must obtain 70 marks to score a distinction for the paper, find the percentage of candidates who obtained distinctions for this paper in the examination. [3 marks] (ii) If 70% of the candidates the paper, determine the minimum marks required to get a ing grade for the paper. [3 marks]
(c) In a group of teachers, the expected number of teachers who own Proton cars is 8 and the variance is 1.6. Find the probability that [3 marks] (i) a teacher chosen at random owns a Proton car, (ii) exactly 4 teachers from the group own Proton cars. [2 marks] STPM 950/2 END OF QUESTION PAPER CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
4
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
5
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
6
ANSWER TRIAL STPM MATHEMATICS M (SARAWAK) –SMK Paku QA 1
1. a) P(only one of the two bulbs is in good condition ,replacing the first bulb) 60 20 = × ×2 80 80 = 0.375 b) P(only one of the two bulbs is in good condition, with the first bulb not returned into the box) 60 20 = × ×2 80 79 = 0.3797 [1M] 2. a) Simple aggregate price index =
∑x ∑x
n
[2M] [1M]
[2M] [1M]
× 100
0
12 + 13 + a + 10 + 5 × 100 = 120 11 + 12 + 10 + 8 + 4 a =14 ∑ pn qn × 100 b) Laspeyres Price Index = ∑ p0q0
2M
12(2) + 13(2) + 14(2) + 10(3) + 5(5) × 100 11(2) + 12(2) + 10(2) + 8(3) + 4(5) = 120.9 The housewife’s expenditure on fish in year 2012 increased by 20.9% when compared with 2011.
2M
1M
=
1M
3a) stem 4 5 6 7 8 9 10
leaf 1 0 2 0 5 3 8
4 2 3 0 7
8 7 6 2
8 6 4
7 4
7 7
8 2M Key: 4 1 means 41
b)
1M STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
7
1 Median = (24)th 2 th = 12 = 67
1 Q1 = (24)th 4 th =6 57 + 58 = 2 = 57.5
3 Q3 = (24)th 4 th = 18
1M Interquartile range
= 74
3 Lower boundary = Q1 − (Q3 − Q1 ) 2 3 c. = 57.5 − (74 − 57.5) 2 = 32.75 Outlier = 108
= Q3 − Q1 = 74 -57.5 = 16.5
3 Upper boundary = Q1 − (Q3 − Q1 ) 2 3 = 74 − (74 − 57.5) 2 = 106.75 1M
2M 4a)
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
8
5.
2M
6 a) Graph Year
Quarter
3M Unemployed
4-quarter
Centred 4-quarter
Deviation
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT school leavers 2010
1 2
22 12
3
110
9 moving average
moving average (t)
(y-t)
43.625
66.375
45.25
-14.250
52.00
-31.000
61.875
-35.875
70.375
79.625
75.250
-5.250
76.000
-26.000
80.500
-44.500
43.75 43.50 4
31 47.00
2011
1
21 57.00
2
26 66.75
3
150 74.00
4
70 76.50
2012
1
50 75.50
2
36 85.50
3 4
146 110
b) seasonal variation Year Quarter 1 2010 2011 -31.000 2012 -26.000 Unadjusted -28.500 seasonal variation Correction -1.3595 factor Seasonal -27.14 Variation C) The least square method
∑x
=
2
= 482
∑ x = 66 ∑ y
(66)(784) 12 = 2.664 (66) 2 482 − 12
4629 −
2
Quarter 2
Quarter 3 66.375 79.625
Quarter 4 -14.250 -5.250
73.000
-9.750
-1.3595
-1.3595
-1.3595
-38.83
74.36
-35.875 -44.500 -40.188
= 79418
∑ y = 784
-8.39
[2M]
[1M]
∑ xy = 4627
1M
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
10
784 66 − 2.664 =50.68 12 12 Regression line, y = 50.68+2.664x For the first quarter of the year 2013, Te = 50.68+2.664(13) = 85.312 Unemployed school leavers for the first quarter of the year 2013 =85.312+ (-27.14) = 58.172 ≈ 59
1M 1M
Or
1M
a=
1M
80.5 − 43.625 Average quarterly increment = =5.2679 7 For the first quarter of the year 2013, Te = 80.5 + 3 x 5.2679 = 96.304 Unemployed school leavers for the first quarter of the year 2013 = 96.304 + (-27.14)= 69.164 ≈ 69
2M 1M
7a Histogram
1 1 N = (125) = 62.5 2 2 1 N −F 2 c Median, M = L + fm
b)
1 (125) − 45 2 (1) = 30.58minutes = 30 + 30 x 26.5 28.5 29.5 30.5 31.5 32.5 33.5 34.5 37.5
f 6 12 27 30 18 14 9 4 5
fx
fx 2
∑ fx = 3861.5
∑ fx
2
2M
Cumulative frequency,F 6 18 45 75 93 107 116 120 125
= 119905.25
∑ fx = 3861.5 = 30.89 ∑ f 125 fx fx 119905.25 Standard Deviation, σ = ∑ −∑ = ∑ − (30.89) f f 125 ∑ ∑
1M
Mean, x =
2
c. Pearson coefficient of skewness
2
= 2.220
2M 1M
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
11
3(mean − median) 3(30.89 − 30.58) = = 0.4189 standard deviation 2.220 The distribution is positively skewed or skewing to the right. 7d) =
. The number of times a bus arrives at town B between 0830 hours and 0836 hours = 122-45 = 77 8a) Let X be the number of books read by an adult Then, X ≈[ P0 (5) e −5 (5)3 a) 3! = 0.1404 b) In 5 years, λ = 5 × 5 = 25
1M
2M
P ( X = 3) =
1M 1M
1M STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
12
P ( X > 2) = 1 − [ P( X = 0) + P ( X = 1) + P ( X = 2)]
1M
e − 25 (25)0 e − 25 (25)1 e − 25 (25) 2 =1- + + 0! 1! 2! = 1 .0 b)(i) 70 − 50 P ( X ≥ 70) = P z ≥ 10 = P(z ≥ 2) = 1 - P(z < 2) = 1 - 0.9772 = 0.0228 Hence, percentage of candidates = 0.0228 x 100% = 2.28% (ii) P ( X ≥ x) = 0.70
x − 50 P Z ≥ = 0.70 10 P(Z ≤ z) = 0.70 z = 0.524 x − 50 ∴ = −0.524 10 x = 44.76 Hence, minimum marks required = 45 8c) Expected number, E ( x) = np = 8 Variance, Var (x) = npq = 1.6 npq 1.6 = np 8 q = 0.2 p = 0.8 n = 10 X ≈ (10,0.8) (i) P( a teacher chosen at random owns a proton car) = 0.8 P( exactly 4 teachers from the group own Proton cars) (ii) = P(X = 4) = 10C4 (0.8) 4 (0.2)10 − 4 = 0.005505
2M 1M
2M
1M
2M 1M
2M
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
Section A [45 marks] Answer all questions in this section. 1
2
In a box containing 80 light bulbs, 60 are in good condition. Two bulbs are taken randomly, one after the other, from the box. Find the probability that only one of the two bulbs taken is in good condition. (a)
If the first bulb is returned into the box.
[3 marks]
(b)
If the first bulb is not returned into the box.
[3 marks]
The table below shows the prices of fish and the quantities of it bought by a housewife at a market in the year 2011 and 2012. Fish
Parang Tenggiri Bawal Putih Kembung Selar Kuning
2011 Price Quantity (RM per kg) (kg) 11.00 2 12.00 2 10.00 2 8.00 3 4.00 5
2012 Price Quantity (RM per kg) (kg) 12.00 1 13.00 2 a 1 10.00 2 5.00 6
(a) If the simple aggregate price index increases by 20 % from 2011 to 2012, determine the value of a. [3 marks] (b) Calculate the Laspeyres price index, and comment on the housewife’s change in expenditure on fish. [3 marks]
3.
A random sample of 24 people were asked to record the number of kilometres they travelled by car in a given week. The distances, to the nearest kilometres, are shown below. 67 72 70
66 77 70
85 67 41
93 48 108
63 66 58
44 50 74
87 52 62
57 74 68
(a)
Construct a stem-and-leaf diagram to represent these data. [2 marks]
(b)
Determine the median and interquartile range of this distribution. [3 marks] Draw a box-and-wisker plot for this data. [3 marks]
(c) STPM 950/2
[Turn over CONFIDENTIAL* This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
2
4 The probability density function of the time t, in minutes, of a student spends every morning waiting for bus to school is given by 2 75 t 0 ≤ t < 5, 1 1 f (t ) = − t , 5 ≤ t < 15, 5 75 otherwise 0 (a) Find the cumulative distribution function of T. [4 marks] (b)
5.
Calculate the mean waiting time of the student.
The water vapour content, w (in g) per 100 g samples of mud, taken at a depth of d meters from a river mouth is measured. The results are shown in the table below. Depth (d) 0 5 10 Water vapour content (w) 92 80 57 (a) Plot these data on a scatter diagram (b)
15 44
20 32
25 22
30 35 20 17 [2 marks]
Calculate the Pearson’s correlation coefficient for the above data. [4 marks] Comment on the relationship between the depth of the mud and the water vapour content. [2 marks]
(c)
6.
[3 marks]
Unemployed school leavers in the Malaysia (figures in thousands) is tabulated below. Year 2010 2011 2012 (a) (b) (d)
Quarter 1 22 21 50
Quarter 2 12 26 36
Quarter 3 110 150 146
Quarter 4 31 70 110
Find the four-quarter centred moving average for the time series. [3 marks] Calculate the seasonal variation using an additive model. [3 marks] Use this model to predict unemployed school leavers for the first quarter of the year 2013. [4 marks]
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
3
Section B [15 marks] Answer any one question in this section. 7.
Everyday, a bus leaves town A for town B at 0800 hours. The time taken for this journey is recorded for a certain period. The result are shown in the following table. Time (minutes) Frequency 6 25 ≤ x ≤ 28 12 28 ≤ x ≤ 29 27 29 ≤ x ≤ 30 30 30 ≤ x ≤ 31 18 31 ≤ x ≤ 32 14 32 ≤ x ≤ 33 9 33 ≤ x ≤ 34 4 34 ≤ x ≤ 35 5 35 ≤ x ≤ 40 a) Draw a histogram to represent these data. [3 marks] b) c)
d)
8
Calculate the median, mean and standard deviation for this data. [5 marks] Hence, calculate the Pearson coefficient of skewness and comment the skewness of the distribution. [2 marks] Plot a cumulative frequency curve of the above data. Hence, find the number of times a bus arrives at town B between 0830 hours and 0836 hours. [5 marks]
(a) On average, the number of books read by an adult is 5 books per year. Using the Poisson distribution, find the probability that [2 marks] (i) an adult reads exactly 3 books per year, (ii) an adult reads more than 2 books in 5 years. [2 marks] (b) The marks obtained by the candidates for a paper in an examination are distributed normally with mean of 50 and standard deviation 10. (i) If a candidate must obtain 70 marks to score a distinction for the paper, find the percentage of candidates who obtained distinctions for this paper in the examination. [3 marks] (ii) If 70% of the candidates the paper, determine the minimum marks required to get a ing grade for the paper. [3 marks]
(c) In a group of teachers, the expected number of teachers who own Proton cars is 8 and the variance is 1.6. Find the probability that [3 marks] (i) a teacher chosen at random owns a Proton car, (ii) exactly 4 teachers from the group own Proton cars. [2 marks] STPM 950/2 END OF QUESTION PAPER CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
4
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
5
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
6
ANSWER TRIAL STPM MATHEMATICS M (SARAWAK) –SMK Paku QA 1
1. a) P(only one of the two bulbs is in good condition ,replacing the first bulb) 60 20 = × ×2 80 80 = 0.375 b) P(only one of the two bulbs is in good condition, with the first bulb not returned into the box) 60 20 = × ×2 80 79 = 0.3797 [1M] 2. a) Simple aggregate price index =
∑x ∑x
n
[2M] [1M]
[2M] [1M]
× 100
0
12 + 13 + a + 10 + 5 × 100 = 120 11 + 12 + 10 + 8 + 4 a =14 ∑ pn qn × 100 b) Laspeyres Price Index = ∑ p0q0
2M
12(2) + 13(2) + 14(2) + 10(3) + 5(5) × 100 11(2) + 12(2) + 10(2) + 8(3) + 4(5) = 120.9 The housewife’s expenditure on fish in year 2012 increased by 20.9% when compared with 2011.
2M
1M
=
1M
3a) stem 4 5 6 7 8 9 10
leaf 1 0 2 0 5 3 8
4 2 3 0 7
8 7 6 2
8 6 4
7 4
7 7
8 2M Key: 4 1 means 41
b)
1M STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
7
1 Median = (24)th 2 th = 12 = 67
1 Q1 = (24)th 4 th =6 57 + 58 = 2 = 57.5
3 Q3 = (24)th 4 th = 18
1M Interquartile range
= 74
3 Lower boundary = Q1 − (Q3 − Q1 ) 2 3 c. = 57.5 − (74 − 57.5) 2 = 32.75 Outlier = 108
= Q3 − Q1 = 74 -57.5 = 16.5
3 Upper boundary = Q1 − (Q3 − Q1 ) 2 3 = 74 − (74 − 57.5) 2 = 106.75 1M
2M 4a)
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
8
5.
2M
6 a) Graph Year
Quarter
3M Unemployed
4-quarter
Centred 4-quarter
Deviation
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT school leavers 2010
1 2
22 12
3
110
9 moving average
moving average (t)
(y-t)
43.625
66.375
45.25
-14.250
52.00
-31.000
61.875
-35.875
70.375
79.625
75.250
-5.250
76.000
-26.000
80.500
-44.500
43.75 43.50 4
31 47.00
2011
1
21 57.00
2
26 66.75
3
150 74.00
4
70 76.50
2012
1
50 75.50
2
36 85.50
3 4
146 110
b) seasonal variation Year Quarter 1 2010 2011 -31.000 2012 -26.000 Unadjusted -28.500 seasonal variation Correction -1.3595 factor Seasonal -27.14 Variation C) The least square method
∑x
=
2
= 482
∑ x = 66 ∑ y
(66)(784) 12 = 2.664 (66) 2 482 − 12
4629 −
2
Quarter 2
Quarter 3 66.375 79.625
Quarter 4 -14.250 -5.250
73.000
-9.750
-1.3595
-1.3595
-1.3595
-38.83
74.36
-35.875 -44.500 -40.188
= 79418
∑ y = 784
-8.39
[2M]
[1M]
∑ xy = 4627
1M
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
10
784 66 − 2.664 =50.68 12 12 Regression line, y = 50.68+2.664x For the first quarter of the year 2013, Te = 50.68+2.664(13) = 85.312 Unemployed school leavers for the first quarter of the year 2013 =85.312+ (-27.14) = 58.172 ≈ 59
1M 1M
Or
1M
a=
1M
80.5 − 43.625 Average quarterly increment = =5.2679 7 For the first quarter of the year 2013, Te = 80.5 + 3 x 5.2679 = 96.304 Unemployed school leavers for the first quarter of the year 2013 = 96.304 + (-27.14)= 69.164 ≈ 69
2M 1M
7a Histogram
1 1 N = (125) = 62.5 2 2 1 N −F 2 c Median, M = L + fm
b)
1 (125) − 45 2 (1) = 30.58minutes = 30 + 30 x 26.5 28.5 29.5 30.5 31.5 32.5 33.5 34.5 37.5
f 6 12 27 30 18 14 9 4 5
fx
fx 2
∑ fx = 3861.5
∑ fx
2
2M
Cumulative frequency,F 6 18 45 75 93 107 116 120 125
= 119905.25
∑ fx = 3861.5 = 30.89 ∑ f 125 fx fx 119905.25 Standard Deviation, σ = ∑ −∑ = ∑ − (30.89) f f 125 ∑ ∑
1M
Mean, x =
2
c. Pearson coefficient of skewness
2
= 2.220
2M 1M
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
11
3(mean − median) 3(30.89 − 30.58) = = 0.4189 standard deviation 2.220 The distribution is positively skewed or skewing to the right. 7d) =
. The number of times a bus arrives at town B between 0830 hours and 0836 hours = 122-45 = 77 8a) Let X be the number of books read by an adult Then, X ≈[ P0 (5) e −5 (5)3 a) 3! = 0.1404 b) In 5 years, λ = 5 × 5 = 25
1M
2M
P ( X = 3) =
1M 1M
1M STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
CONFIDENTIAL*/SULIT
12
P ( X > 2) = 1 − [ P( X = 0) + P ( X = 1) + P ( X = 2)]
1M
e − 25 (25)0 e − 25 (25)1 e − 25 (25) 2 =1- + + 0! 1! 2! = 1 .0 b)(i) 70 − 50 P ( X ≥ 70) = P z ≥ 10 = P(z ≥ 2) = 1 - P(z < 2) = 1 - 0.9772 = 0.0228 Hence, percentage of candidates = 0.0228 x 100% = 2.28% (ii) P ( X ≥ x) = 0.70
x − 50 P Z ≥ = 0.70 10 P(Z ≤ z) = 0.70 z = 0.524 x − 50 ∴ = −0.524 10 x = 44.76 Hence, minimum marks required = 45 8c) Expected number, E ( x) = np = 8 Variance, Var (x) = npq = 1.6 npq 1.6 = np 8 q = 0.2 p = 0.8 n = 10 X ≈ (10,0.8) (i) P( a teacher chosen at random owns a proton car) = 0.8 P( exactly 4 teachers from the group own Proton cars) (ii) = P(X = 4) = 10C4 (0.8) 4 (0.2)10 − 4 = 0.005505
2M 1M
2M
1M
2M 1M
2M
STPM 950/2 CONFIDENTIAL* *This question paper is CONFIDENTIAL until the examination is over.
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