UNIFORM TEST 1 MATHEMATICS Compulsory Part PAPER 1 (Section A)

Size: px

Start display at page:

Download "UNIFORM TEST 1 MATHEMATICS Compulsory Part PAPER 1 (Section A)"

Cassandra Wilkins
5 years ago
Views:

1 SECTION A (70 marks) UNIFORM TEST 1 MATHEMATICS Compulsory Part PAPER 1 (Section A) HKDSE 015 Q ( m n ) Simplify 16 m and express your answer with positive indices. (3 marks) ( m n ) 0 16 m n m m 0 16 m 16 n 4 m 16 n HKDSE 01 Q. Make a the subject of the formula 5a 1 = a + b. (3 marks) 5a 1 = a + b 5a 1 = a + b 5a a = b + 1 3a = b + 1 a = b 1 3 1

2 3. The table below shows the distribution of the heights (in cm) of some students. Height (cm) Number of students (a) What is the modal class of the above distribution? Find the mean of the above distribution. (a) The modal class is 130 cm 139 cm. Class mark (cm) Number of students Mean = cm = cm HKCEE 011 Q6 4. In a training camp, the number of senior secondary students is 4 times that of junior secondary students. If 40 more junior secondary students join the camp, the ratio of the number of junior secondary students to that of senior secondary students becomes 1 :. Find the original number of secondary students in the camp. Let x be the original number of junior secondary students. Then, there are 4x senior secondary students. (x + 40) : 4x = 1 : 4x = (x + 40) x = 80 x = 40 Original number of secondary students in the camp = (40) = 00

3 HKDSE 013 Q5 5. (a) Solve the inequality x 3 x How many integers less than 4 satisfy the inequality in (a)? (a) x 3 x x 1 x 30 5x 9 x 5 9 The integers satisfying both conditions are 1, 0, 1, and 3. The required number is 5. 3

4 6. The stem-and-leaf diagram below shows the distribution of the heights (in cm) of the members in a basketball team. Stem (tens) Leaf (units) (a) Find the mean height of the members. If a member is selected at random, find the probability that the height of the member is greater than the mean height (a) Mean height = cm 15 = 185 cm P(greater than the mean) = 15 9 = 5 3 4

5 7. An ice-cream ball of diameter 8 cm is put into a rectangular container of dimensions 8 cm 8 cm 4 cm. If the ice-cream ball is melted completely, will the ice-cream overflow? Explain your answer. 4 8 Volume of the ice-cream = π cm cm 3 Capacity of the container = (8 8 4) cm 3 = 56 cm 3 < Volume of the ice-cream The ice-cream will overflow. 8. Factorize (a) x y 10xy 5y, 4 y x y 10xy 5. (5 marks) (a) x y 10xy 5 y y ( x 10 x 5) y ( x 5) 4 x y 10xy 5 y 4 ( x y 10xy 5 y ) y ( x 5) [ y ( x 5)][ y( x 5)] ( xy 5y)( xy 5y) 5

6 HKDSE 014 Q8 9. The coordinates of the point P are (5, ). P is reflected about the x-axis to Q. Q is rotated anti-clockwise about the origin O through 90 to R. (a) Find the coordinates of Q and R. Is PQR an isosceles triangle? Explain your answer. (5 marks) (a) Coordinates of Q = ( 5, ) Coordinates of R = (, 5) PQ = ( ) = 4 PR = QR = ( 5 ) ( 5) = 18 ( 5 ) ( 5) = 58 PQ, PR and QR are all different. PQR is not an isosceles triangle. 10. Starting from 1st November, the tram fare is increased by 10% to $3.3. (a) Find the tram fare before the adjustment. The number of passengers decreases by 10% from October to November. Are the total incomes of the tram company the same in October and in November? Explain your answer. (5 marks) (a) Tram fare before the adjustment = $3.3 (1 + 10%) = $3 Let x be the number of passengers in October. Number of passenger in November = x(1 10%) Total income in October = $3x = 0.9x Total income in November = $3.3(0.9x) = $.97x The total incomes of the tram company in October and November are not the same. 6

7 HKCEE 005 Q8 11. In Figure 1, ABCDE is a regular pentagon. AC and BE intersect at F. Find x, y and z. A B y F z E C x D Figure 1 (5 marks) 5x = (5 ) 180 ( sum of polygon) x = 108 AB = AE (given) ABE = AEB (base s, isos. ) ABE + AEB + BAE = 180 ( sum of ) ABE = 180 ABE = 36 y = = 7 ACB = 36 z = y + ACB (ext. of ) = = 108 7

8 HKCEE 1997 Q6 1. In Figure, the bearings of two poles A and B from Peter P are 15 and 15 respectively. A is 300 m apart from P. The bearing of B from A is 45. N P m N A 45 B (a) Find the bearing of P from B. Find the distance of P from B. Figure (5 marks) (a) The required bearing = (alt. s, // lines) = 035 APB = = 90 BAP = (180 15) = 60 BP = tan BAP AP BP = 300 tan 60 m BP = m = 50 m (cor. to 3 sig. fig.) The required distance is 50 m. 8

9 13. In a game, the player throws two fair dice at the same time. If the player gets more than 7 points, he / she wins $0. Otherwise, he / she wins $5. (a) Find the expected amount the player can win. (3 marks) If the player has to pay $10 for playing the game, is it a fair game? If not, is it favourable to the players? Explain your answer. (3 marks) (a) Sum of points: 1st dice nd dice Expected amount = $ = $ 4 Expected amount > $10 It is not a fair game and it is favourable to the players. + 9

10 14. In Figure 3, a trapezoidal prism is formed by removing a triangular prism from a rectangular block, where all the measurements are measured correct to the nearest cm. 5 cm 3 cm 7 cm 10 cm Figure 3 (a) Write down the maximum absolute error of the measurements. (1 mark) Find the least possible volume of the trapezoidal prism. ( marks) (c) The actual volume of the trapezoidal prism is x cm 3. Find the range of values of x. 1cm (a) Maximum absolute error = = 0.5 cm The least possible volume [( ) 9.5](6.5) = (4.5) cm 3 = cm 3 (c) The actual volume of the trapezoidal prism [(10.5.5) 10.5](7.5) < (5.5) cm 3 = cm 3 The required range is x <

11 15. Consider the three points A( 8, 1), B(1, 4) and C(4, 5). (a) Show that there is a straight line passes through the incentre, centroid and orthocentre of ABC. Find the slope of the straight line passing through the incentre, centroid and orthocentre of ABC. ( marks) (a) AB = AC = ( 8 1) [ 1 ( 4)] = 90 ( 8 4) ( 1 5) = 180 BC = ( 1 4) ( 4 5) = 90 ABC is an isosceles triangle. Let M be the mid-point of AC. Then, BM is the angle bisector of ABC, a median and an altitude of ABC. There is a straight line (BM) passes through the incentre, centroid and orthocentre of ABC Slope of AC = = The required slope = = 1 + END OF PAPER 11

1. B (27 9 ) = [3 3 ] = (3 ) = 3 2. D. = c d dy d = cy + c dy cy = d + c. y( d c) 3. D 4. C

1. B (27 9 ) = [3 3 ] = (3 ) = 3 2. D. = c d dy d = cy + c dy cy = d + c. y( d c) 3. D 4. C HKDSE03 Mathematics (Compulsory Part) Paper Full Solution. B (7 9 ) [3 3 ] (3 ) 3 n + 3 3 ( n + ) 3 n + 5 3 6 n + 5. D y y + c d dy d cy + c dy cy d + c y( d c) c + d c + d y d c 3. D hl kl + hm km hn