GAC 010 Test 3 Sample Questions 2Word下载.docx

1、Nil$6 001 - $20 00017 cents for each $1 over $6000$20 001 - $50 000$2 380 + 30 cents for each $1 over $20 000$50 001 - $60 000$11 380 + 42 cents for each $1 over $50 000$60 001 and over$15 580 + 47 cents for each $1 over $60 000Source: 2001 ATO Tax GuideSolutions1.a.）i.）a.）ii.） 3 + 16 + 18 = 37a.）ii

2、i.） 18 + 10 + 8 + 5 = 41b.）i.） Taxable Income = gross - deductions = $57 678 - $1 745 = $ 55933b.）ii.） Tax Payable on $57 678 is $57 678 - $1 745 = $ 55933 on $50 000 = $11 380 on $55933 - $50 000 = $5 933 * .42 = $2 491.86 Tax Payable on $ 55933 = $11 380 + $2 491.86 = $13 871.86Question 2 16 marks

3、The stem and leaf diagram displays the scores obtained by 40 students in their final Mathematics exam.Stem Leaf45 73 5 7 862 4 5 6 8 971 1 1 2 3 4 4 5 5 6 6 6 6 8 8 93 5 5 7 8 8 993 3 5 8 8a.） Calculate the mean, median and mode. 5 marksb.） Calculate the range, and the lower and upper quartiles. 5 m

4、arksc.） Construct a box and whisker diagram. 3 marksd.） Hence, comment on the symmetry or otherwise of the data. 1 marke.） Find the variance and standard deviation. 2 marksa.） Mean = S x nMedian = X （n + 1） = X 41/2 = X 20.5 2Mode = 76b.） Range = Highest Value - Lowest value = 98 - 45 = 53Q1 = X （n

5、+ 1） 4 Q1 = X （41） 4 = X 10.25Q3 = X 3（n + 1） Q3 = X 3（41） = X 30.75c.） Box and Whisker diagram _30 40 50 60 70 80 90 45 Q1 Md Q3 98d.） The data is skewed to the left （negatively skewed） due to the greater number of smaller values.e.） Variance is S2 = S （x - x ）2*f n -1 Standard deviation is s =s2Qu

6、estion 3 15 marksa.） Households in a certain town were surveyed to determine whether they would subscribe to a Pay TV Channel. The households were classified according to “high”, “medium”, and “low” income levels. The results of the survey were summarized in the following table.Income LevelWill Subs

7、cribeWill not subscribeHigh16503300Medium160063407940Low520190024203770989013660A household is chosen at random. Find the probability that the household:i.） will subscribe. 1 markii.） has a low income level. 1 markiii.） will subscribe and a has a low-income level. 2 marksiv.） will subscribe or has a

8、 low income level. 2 marks v.） will subscribe given that the household has a low income level. 2 marksb.）i.） 12 people are to be divided into two groups with 8 in one and 4 in the other. In how many ways can this be done? 2 marks ii.） A television firm has in stock 9 television sets with 32 inch tub

9、es and 6 sets with 28 inch tubes. Four sets are removed at random. What is the probability that they are all of the same size? 3 marks iii.） A particular firm has in type of TV tube is almost certain to fail within 3500 hours. If the life of a tube follows a normal distribution with a standard devia

10、tion of 150 hours, find the mean life of these tubes. 2 marks 3.a）i.） （will subscribe） = 3770 = 0.27 13660ii.） P（low level income） = 2420 = 0.17iii.） P（will subscribe and has low level income） = 520 = 0.03iv.） P（will subscribe or has low level income） = 3770 + 1900 = 0.41v.） P（will subscribe given h

11、ousehold has low level income） = 520 = 0.21 2420b.） i.） Number of arrangements n C r = n! r!（n-r）! 12 C 8 * 4 C 4 = 12! * 4! 8!（12 - 8）! 4!（4-4）! = 12! （4!） 4!（0）! = 12 * 11 * 10 * 9 * 8! * 1（4）! = 12 * 11 * 10 * 9 4 * 3 * 2 * 1 = 11880 = 495 24 ii.） P（same size） = P（LLLL） + P（SSSS） = （ 9 * 8 * 7 *

12、6 ） + （ 6 * 5 * 4 * 3 ） 15 14 13 12 15 14 13 12 = 0.10 iii.） Almost certain to fail is 3 standard deviations above the mean. therefore, mean is + 3 = 3500 = 3500 - 3 = 3500 - 3 * 150 = 3500 - 450 = 3050 or, 3500 - 150 = 3350 3350 - 150 = 3200 3200 - 150 = 3050 3050 3200 3350 3500Question 4 13 marksM

13、ost movie trailers are normally distributed with a mean of 2 minutes and a standard deviation of 30 seconds.a.） Find the probability that a movie trailer selected at random will be i.） less than 160 seconds ii.） less than 80 seconds iii.） over 180 seconds iv.） over 100 seconds but less than 140 seco

14、nds.b.） If you watch 100 of these movie trailers, how many will be under 140 seconds?4. a.）i.）Let X = time in seconds = 120, = 30 X: N（120, 302） z = X - = X - 120 30 P（X 160） = P（Z 160 -120） = P（Z 140 ） 30 30 = P（Z 4.66） = .5 + P（0 Z 1.66） = .5 + .4515 = .9515 120 160_ 0 4.66 Z a.）ii.） 80） = P（Z 80

15、-120） = P（Z - 40 ） - 1.33） = .5 - P（-1.33 Z 0） = .5 - P（0 180） = P（Z 180 -120） = P（Z 60 ） = P（Z 2） = .5 - P（0 = .5 - .4772 = .0228 120 180_ 0 2 Za.） iv）P（100 X 140） = P（100 -120 150 -120） = P（ -20 30 ） = P（-.66 1） 0） + P（0 = P（0 .66） + P（0 1） = .2454 + .3413 = .5867 100 120 150_ -.66 0 1 Zb.） 140） =

16、 P（Z 140 - 130 ） 30 .33） = .5 + P（0 about 63 trailers will be under 140 seconds. 120 140_ 0 .33 Question 5 14 marksa.） Frank recently bought a small boat. He put $8000 as down payment and borrowed the rest of the money, with payments of $400 at the end of every 4 months for 4 years. If the interest

17、is 8% p.a. compounded three times a year, what is the current value of the boat? 4 marksUnit 7F, Ex 7.7, p. 235b.） My brother Tony, who lives in Canada, just bought 2 houses. the first costed $60 400 and the second $ 80 600. Later on, he sold the first one at a profit of 15% and the second at a loss

18、 of 12%. How much profit or loss did the agent gain or lose in total?Unit 6D, Ex 6.3, p. 207c.） A newly opened offices purchased equipment - computers, printers, fax machines, etc. costs $25 000. John puts down a $3000 deposit and agrees to pay the balance in 3 years at 7% interest on the balance.i.

19、） If M is the monthly repayment, show that the amount owed at the end of the fourth month is 22 000 * 1.00584 - M（1.00583 + 1.00582 + 1.0058 + 1）2 marksUnit 7Gii.） Find the monthly repayment to repay the loan in 36 months. 4 marks Unit 7G, Ex 7.11, p.238 （ Application of Geometric Series）5 a.）Ordina

20、ry AnnuitiesPresent Value of an Ordinary AnnuityThe present value （A） of an ordinary annuity is given by:A = R 1 - （1 + i）-n i Where: R periodic payment of the annuity i = interest rate per period （as a decimal） n = the number of periodic payments to be made The term 1 - （1 + i）-n can be abbreviated

21、 to nitherefore the present value of an ordinary annuity is A = R ni, where ni = 1 - （1 + i）-n i=Deposit = $ 8000R = $400n = 4 * 3 = 12i = 8% / 3 = .026Present value （A） = $8000 + A = $8000 + 400 1 - （ 1.026） -12 .026 = $8000 + 40010.195931 = $8000 + 4078.37 =$ 12078.37b.）cost of house 1 = 60 400cos

22、t of house 2 = 80 600cost of both houses = 141 000Mark-up on house 1 = .15 * 60 400 = 9060Sale price of house 1 = 60 400 + 9060 = 69460Loss on house 2 = .12 * 80 600 = 9672Sale price on house 2 = 80 600 - 9672 = 70928Total sales price of houses 1 and 2 = 69460 +70928 = 140388Total loss =141 000 - 14

23、0388 = 612$612 is （612/141000） * 100 = 4.340 = about 4.3 % of Total Cost Pricec.）The balance left to pay is 25 000 - 3000 = 22 000The calculation must first be based on monthsTherefore monthly interest rate = 7 % = .58% =.0058 12so 1 + r = 1.0058 100Number of months = 3 * 12 = 36i.）Let Ai be the amo

24、unt still owing after i months and M the monthly repayments.A1 = 22 000 * 1.0058 - MA2 = A1 * 1.0058 - M = （ 22 000 * 1.0058 - M） * 1.0058 - M = 22 000 * 1.00582 - 1.0058M - M = 22 000 * 1.00582 - M（1.0058 + 1）A3 = A2 * 1.0058 - M = （22 000 1.00582 - M（1.00581 + 1） * 1.0058 - M = 22 000 * 1.00583 -

25、（1.00581M - M） * 1.0058 - M = 22 000 * 1.00583 - 1.00582 - 1.0058M - M = 22 000 * 1.00583 - M（1.00582 + 1.0058 + 1）A4 = A3 * 1.0058 - M = . = 22 000 * 1.00584 - M（1.00583 + 1.00582 + 1.0058 + 1）ii.）A36 = 22 000 * 1.005836 - M（1.005835 + 1.005834 + .+1） （ using the geometric series formula for Sn） The Sum of n Terms of a Geometric Progressio