Alex 發問於 科學及數學數學 · 6 年前

# mock paper LQ

Below are the weights of beef steak bought by Mary and Peter.

330g, 255g, 330g, 315g, 290g, 280g, 370g, 310g

(a) Find the mean, mode and median value.

(b) Mary claims that by adding 3 more pieces of steak, she can keep the median value, decrease the mode and increase the mean, suggest the weight of these three steaks.

(c) Peter claims that by adding more pieces of steaks, he can change the inter quartile range value to zero, do you agree with him? Explain your answer.

Let H(x) =x4 + 5x3 +10x2－x－5.

(a) Find the quotient and remainder when H (x) is divided by x2 + 3x + 3.

(b) If H (x) + Ax+ B is divisible by x2 + 3x + 3, find the values of A and B.

(c) If H (x) + Cx2 + Dx is divisible by x2 + 3x + 3, find the values of C and D by using the result of (b).

There are 15 boys and 25 girls in a class. If 4 students are randomly selected one after another, find

the probabilities that

(a) the second student selected is a boy,

(b) the first student selected is a boy given that the second student selected is a boy

(c) there is at least one boy and one girl in the group.

(My answer for (a) and (b) is 3/8 and 14/39, and (c) is 1 - P(no boy) - P(no girl) = 0.8466 = 0.847, not sure)

A restaurant has 90 tables. Fig. II shows its floor plan where each circle represents a table. Each table is assigned a number between 01 and 90. A rectangular coordinate system is introduced to the floor plan such that the table numbered (10x + y) is located at (x , y) where x and y are the tens digit and

the units digit of the table number respectively (notice that the tens digit here may be zero).

The table numbered 54 has been marked in the figure as an example.

https://www.flickr.com/photos/130230543@N06/155561...

(a) In Fig. II, shade all the circles satisfies the following constraints:

0 <= x <= 9

0 <= y <= 9

4x + 3y - 24 < 0

(b) If 3 different tables are randomly selected one after another from shaded table, find the probability that the sum of the table numbers will less than 130.

### 1 個解答

• 6 年前
最愛解答

1. Rearrange the weights :255, 280, 290, 310, 315, 330, 330, 370(a) mean = sum/8 = 310 g mode = 330 g median = (310 + 315)/2 = 312.5 g

(b) In order to increase the mean by adding 3 more pieces of steak, the sum of these 3 pieces must greater than 3*310 g, ie 930 g.In order to keep the median value, one of the steak must be weighted 312.5 g.In order to decrease the mode, the count of that steak is 3.So, the possible weight of these 3 steaks must be 312.5 g, 312.5 g, 312.5 g.

(c) I agree with Peter's claim. If the inter-quartile range value is zero, then, Q1 = Q3.For example, add 7 pieces of steak with weight 311 g, as there are totally 15 pieces, Q1 is the 4th one, which is weighted 311 g, and, Q3 is the 12th one, which is also 311 g.

2a. As x⁴+ 5x³ + 10x² - x - 5 = (x² + 3x + 3)(x² + 2x + 1) - 10x - 8So, the quotient is (x² + 2x + 1), and, the remainder is (-10x - 8).

2b. H(x) + Ax + B = (x² + 3x + 3)(x² + 2x + 1) - 10x - 8 + Ax + BIf it is divisible by (x² + 3x + 3), thenA = 10, B = 8

2c. H(x) + Cx² + Dx= (x² + 3x + 3)(x² + 2x + 1) - 10x - 8 + Cx² + Dx= (x² + 3x + 3)(x² + 2x + 1) + C(x² + 3x + 3) + (D - 3C - 10)x - (3C + 8)So, D - 3C - 10 = 0 and 3C + 8 = 0, that is,C = -8/3, D = 2

3a. P(boy, boy) + P(girl, boy)= (15/40)*(14/39) + (25/40)*(15/39)= 3/8

3b. (15/40)*(14/39) / (3/8)= 14/39

3c. 1 - P(all boys) - P(all girls)= 1 - 15C4/40C4 - 25C4/40C4= 15475/18278= 0.8467 (corr. 3 sf) ⋯⋯⋯⋯ (you are right)

4a. Shade 01, 02, 03, 04, 05, 06, 07, 10, 11, 12, 13, 14, 15, 16,20, 21, 22, 23, 24, 25, 30, 31, 32, 33, 40, 41, 42, 50, 51

(Total 29 tables)

4b. Number of possible outcomes = 29C3 = 3654The sum of 3 tables >= 130 are :(30,50,51), (31,50,51), (32,50,51), (33,50,51), (40,41,50), (40,41,51), (40,42,50), (40,42,51), (40,50,51), (41,42,50), (41,42,51), (41,50,51), (42,50,51) totally 13 possible outcomes.So, the required prob. is 1 - 13/3654= 3641/3654