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经典24题 Test


NUMERICAL TEST 2
Answer Booklet

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Please note – the correct answers are shown in bold

Example Questions

City Trading last 6 months (number of trades made, in 000s)
30 27 28 Phone 20 22 19 15 12 20 23 21 Internet

25

26

In 000s

10 10 5 13

11

0 September October November December January February

Ex 1 Between which two months was there the greatest change in the number of Internet trades made?
Solution We calculate the change in the number of Internet trades between months (in 000s): Change = Number trades Month (n) – Number trades Month (n – 1) Between September and October October and November November and December December and January January and February Change 3 6 1 –9 1

From this we can see the greatest change in the number of Internet trades occurred between December and January. As the Question only referred to the change in the number of trades and not whether the change should be positive or negative, the change in number between December and January is the correct answer. Tip We should do these calculations mentally without resorting to a calculator. Once we have to resort to a calculator or pen and paper, we start losing time. Answer A September and October B October and November C November and December D December and January E January and February

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Ex 2 In September, approximately what proportion of the total number of trades was made up of Internet trades?
Solution We consider September data. We calculate what proportion of the total Proportion of trades = = = Answer A 25% B 31% C 34% D 37% E 43%

number of trades is made up of Internet trades (in 000s): Number of Internet Trades ÷ Total number of Trades 10 ÷ (10 + 22) 0.31 or 31%

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Test Questions
Photocopier service & running costs
(Assume 52 weeks per year, 13 weeks per quarter) Photocopier Expected average service & running cost per week ($) 120 125 130 160 140 175 Actual Spend ($) (Jan-March quarter) 1,872 1,975 1,924 2,400 2,716 3,605 Actual Spend ($) (April-June quarter) 1,968 1,425 2,340 2,112 2,772 2,730

PHTCPR01 PHTCPR02 PHTCPR03 PHTCPR04 PHTCPR05 PHTCPR06

1

By how much have the actual service and running costs for Photocopier PHTCPR06 exceeded those for Photocopier PHTCPR05 so far this year?
Solution We need to calculate the actual running costs for Photocopier PHTCPR06 and PHTCPR05. We then subtract the value calculated for Photocopier PHTCPR05 from the value calculated for Photocopier PHTCPR06: Value Exceeded = Actual Spend PHTCPR06 – Actual Spend PHTCPR05 = (3,605 + 2,730) – (2,716 + 2,772) = 3,605 + 2,730 – 2,716 – 2,772 (*) = $847 Answer A $847 B $876 C $898 D $913 E $925

2

On which photocopier is there the least amount of budget left to spend this year?
Solution We calculate the expected spend for the year per photocopier and subtract from this value the Actual Spend for that photocopier. The photocopier with the smallest difference will be the answer: Expected Spend Photocopier PHTCPR01 PHTCPR02 PHTCPR03 PHTCPR04 PHTCPR05 PHTCPR06 = (Average cost per week 52) – Actual Spend Difference 2400 3100 2496 3808 1792 2765

Expected spend 6240 6500 6760 8320 7280 9100

Actual Spend (to July) 3840 3400 4264 4512 5488 6335

From this table we can see photocopier PHTCPR05 or E is the correct answer.

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Tip Looking at the answers provided for Question 2 we can see PHTCPR06 is not a possible answer so we know we don't need to consider this photocopier in our calculations. An alternative way for calculating the answer is: For each of the remaining 5 photocopiers we compare the actual average weekly spend (to July) with the expected average weekly spend (for the year). The photocopier with the greatest proportionate change between expected and actual average weekly spend will be the photocopier for which the actual spend has so far been the greatest, which would then be the photocopier with the least amount of budget left. Example: For photocopier PHTCPR06, Expected average weekly cost = $140 Actual average weekly cost = $211.08 Proportionate change = 211.08/140 = 1.51 Answer A Photocopier A B Photocopier B C Photocopier C D Photocopier D E Photocopier E

3

By what proportion should the service and running costs for Photocopier PHTCPR05 be increased next year if the actual spend recorded between January and June is a true reflection of service and running costs?
Solution We calculate the percentage difference between the actual spend (to July) and the expected average spend (to July) for Photocopier PHTCPR05: Percentage difference = [((2,716 + 2,772) – (140 26)) ÷ (140 26)] 100% = 50.8% This is the increase we are looking for. Answer A 38.9% B 43.6% C 48.7% D 50.8% E 53.3%

4

By how much is the annual spend on Photocopier PHTCPR01 likely to exceed that of Photocopier PHTCPR02 if the current spend recorded between January and June is a true reflection of actual service and running costs?
Solution We calculate the difference in actual spend between the two photocopiers for the first six months. If the current spend is a true reflection, the spend in the next 6 months will be the same, so we double the difference calculated to get the answer: Difference in actual spend = [(1,872 + 1,968) – (1,975 + 1,425)] 2 = $880 Answer A $740 B $820 C $880 D $960 E $1,120

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Current Breakdown of Global Equity Fund (Total of Fund Value = $160 million)

US Japan Europe S.E. Asia

38% 18% 15% 12%

Emerging Markets 7% UK 10%

Note on Question 5 – Question 8 If we consider $100 million as 100% the Total of Fund Value will then be 160% or 1.6 times the percentage value. Thus the Total of Fund Value of the UK holding will be 1.6 10 = 16, which is $16 million. This ratio (1.6 per percentage point) can be used to speed up the calculations.

5

What is the value of the Japanese holding in the Fund?
Solution We calculate the value of the holding equivalent to 18% of the Total Fund Value (working in $ millions): Value Japanese holding = 18% 160 = 28.8 Tip See the note above. Answer A $24.0 million B $28.8 million C $32.8 Million D $44.8 million E $60.8 Million

6

Which of the following combinations of holdings has a value of $40 million?
Solution We calculate what percentage of Total Fund Value amounts to $40 million. We then see which holdings' share add up to this value: Percentage of Fund = (40/160) 100% = 25% UK holding and Europe holding add up to 25%. Tip See the note above Question 5. Answer A Emerging Markets & S. E. Asia B Emerging Markets & UK C Europe & Japan D Europe & S. E. Asia E Europe & UK

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7

By how much do the value of the holdings in the US and Japan exceed that of the value of all the other holdings combined?
Solution We need to calculate the vlaue of the difference in percentage between the US/Japan holdings and all the other holdings combined. We then calculate the value of $160 million equivalent to this percentage: Difference in percentage values = 38 + 18 – 15 – 12 – 7 – 10 = 12% Value equivalent to 12% (working in $ millions) = 160 12% = 19.2 Tip See the note above Question 5. Answer A $16.4 million B $18.8 million C $19.2 million D $22,8 million E $26.4 million

8

Last year, the value of the S. E. Asia holding was 10% less than what it is now. What was the value of the S.E. Asia holding last year?
Solution We calculate the value of the S.E. Asia holding and then calculate 90% of this to reach the answer (working in $ millions): Value of S.E. Asia holding = 12% 160 = 19.2 90% of Value = 90% 19.2 = 17.28 Tip See the note above Question 5. Answer A $17.28 million B $17.45 Million C $17.96 Million D $18.48 million E $19.05 million

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Actuarial Table: Frequency of Driver Accidents
Age Range Under 18 18-22 23-29 30-39 40-49 50-59 60-69 70 & over Accidents per Year Men 0.46 0.28 0.13 0.09 0.08 0.07 0.05 0.06 Women 0.27 0.16 0.09 0.07 0.06 0.06 0.05 0.04 Accidents per million kilometres Men 50 24 10 7 7 8 8 12 Women 52 27 15 14 12 14 13 12

9

In terms of kilometres driven, for which age range is the proportional difference between the number of accidents for men and women the greatest?
Solution We calculate the proportional difference between the number of accidents for men and women for each age range (in Accidents per million kilometres): Proportional Difference = (Accidents Women – Accidents Men) ÷ Accidents Men Age Range Under 18 18-22 23-29 30-39 40-49 50-59 60-69 70 & over Proportional Difference 0.040 0.125 0.500 1.000 0.714 0.750 0.625 0.000

From this we can see that the greatest proportional difference is for the age range 30-39. Tip From the answers provided we can ignore the proportionate differences for age ranges before 30-39. From the table provided for the question, a quick scan of the numbers involved should show that the range 30-39 is the answer. The easiest way to see this is to consider the equation: Proportionate change = Accidents Women ÷ Accidents Men For the numbers provided, these are all greater than 1. The proportionate difference is simply the Proportionate change minus 1. Answer A 30-39 B 40-49 C 50-59 D 60-69 E 70 & over

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10

How many more accidents in a year will a group of a thousand 18-22 year old men be likely to have compared with a group of a thousand 18-22 year old women?
Solution We consider the accidents per year for the age range 18-22. We first calculate the number of accidents for a group of thousand men and thousand women, and then calculate the difference: Accidents Men = 1000 0.28 = 280 Accidents Women = 1000 0.16 = 160 Difference = 120 Answer A 60 B 80 C 110 D 120 E 150

11

In terms of accidents per year, to what extent do women aged 40-49 have fewer accidents than men aged 23-29?
Solution We consider the accidents per year for women aged 40-49 and men aged 23-29. We calculate the percentage difference in accidents between men aged 23-29 and women aged 40-49, relative to the number of accidents for men aged 23-29: Percentage difference = [(0.06 – 0.13) ÷ 0.13] 100 = 53.8% The answer provided closest to this value is C: 54% less. Answer A 47% less B 49% less C 54% less D 58% less E 65% less

12

How many kilometres approximately does the average 18-22 year old woman drive per year?
Solution We consider woman in the age range 18-22 years. First we calculate the number of kilometres per accident. We then calculate what proportion of those kilometres will be driven in a year using the fact that only 0.16 accidents happen for women in the age group per year: Number of kilometres per accident = 1,000,000 ÷ 27 = 37,037 km Proportion of this driven in a year = 37,037 0.16 = 5,926 km Answer A 4,884 km B 4,929 km C 5,246 km D 5,624 km E 5,926 km

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Values of FTSE 100
6200
6100 6100 5950 5850 5750 5700

6000
5900 5850

5980

5800 Maximum

5600

5600 5550 5500

Median

5400

5400 5350 5250

5400

5350

Minimum
5200

5200

5000 Quarter 1 Quarter 2 Quarter 3 Quarter 4 Quarter 5 Quarter 6

13

Which quarter saw the greatest range in the value of the FTSE 100?
Solution We calculate the range for each of the quarters and from this establish the greatest range: Quarter 1 2 3 4 5 6 Range 600 550 700 350 350 400

From this we see the greatest range is in Quarter 3 Tip From the answers provided we see we can ignore Quarter 6. From the graph provided we can see that the range for Quarter 3 is the largest by considering the length of the line connecting minimum and maximum values. We can confirm this by checking the actual value difference between minimum and maximum for this Quarter and comparing it to some of the other Quarters if necessary. Answer A Quarter 1 B Quarter 2 C Quarter 3 D Quarter 4 E Quarter 5

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14

Between which two quarters was there the smallest proportional change in the median value of the FTSE 100?
Solution We calculate the proportional change in median value between each quarter respectively: Proportional change = (Median2 – Median1) ÷ Median1 Between Q1-Q2 Q2-Q3 Q3-Q4 Q4-Q5 Q5-Q6 Proportional change 0.037 0.068 –0.105 0.075 0.035

From this we see Quarter 5 to Quarter 6 has the smallest proportional change. Tip For the smallest proportional change we are looking for a small change and large median values. Both Q1-Q2 and Q5-Q6 fit this requirement and both have a difference in median value of 200. The median value of Q5 is larger than the median value of Q1 so Q5-Q6 would be our answer. Answer A Quarter 1 to Quarter 2 B Quarter 2 to Quarter 3 C Quarter 3 to Quarter 4 D Quarter 4 to Quarter 5 E Quarter 5 to Quarter 6

15

If in Quarter 7, the minimum value of the FTSE 100 increases by 15% but the maximum value increases by 20%, what will the difference be between the minimum and maximum value for that quarter?
Solution We calculate the associated increase for both the minimum and maximum values from Quarter 6 to Quarter 7. We then calculate the difference between the two values: For Q7: Increase in maximum value = Q6 maximum 120% = 6100 120% = 7320 Increase in minimum value = = = Q6 minimum 115% 5700 115% 6555

Difference in Q7 minimum and maximum value: Difference = 7320 – 6555 = 765 Answer A 745 B 755 C 760 D 765 E 775

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16

The median value of the FTSE 100 is expected to increase by 3% per quarter for the next 4 quarters, what will the median value be in Quarter 10?
Solution We need to calculate the compound increase of the median value over 4 quarters from Q6 to Q10: Increase in median value = Q6 median (1.03)^4 = 5950 1.1255 = 6697 Answer A 6,656 B 6,683 C 6,697 D 6,714 E 6,725

Average Value of US Dollar ($)
Currency Euro (Eur) Pound Sterling () Japanese Yen () Swiss Franc (CHF) Hong Kong Dollar (HK$) Year 1 0.78 0.53 112 1.21 7.75 Year 2 0.64 0.45 95 1.10 6.20

17

How much was a Japanese Yen worth in US dollars in Year 1?
Solution Considering Year 1 we know that 1 US Dollar is worth 112 Japanese Yen. We calculate the inverse to establish how much 1 Japanese Yen is worth in terms of US Dollars: US Dollars per Japanese Yen = 1/112 = 0.0089 Answer A $0.0009 B $0.0089 C $0.011 D $0.089 E $0.105

18

How much more Sterling could have been purchased with 2000 US Dollars in Year 1 compared to Year 2?
Solution Considering Pound Sterling, year 1 and Year 2. Difference = = = Answer A 145 B 150 C 155 D 160 E 165 we calculate the difference in how much Sterling, 2000 US Dollars you can purchase in 2000 2000 160 (Pound Sterling Year 1 – Pound Sterling Year 2) (0.53 – 0.45)

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19

Between Year 1 and Year 2, the value of the Swiss Franc in relation to the US dollar, moved:
Solution We calculate the change in value of the Swiss Franc in relation to the US Dollar from Year 1 to Year 2. We calculate this using: Percentage change = {[(1/Year 2 Value) – (1/Year 1 Value)] ÷ (1/Year 1 Value)} = {[(1/1.10) – (1/1.21)] ÷ (1/1.21)} 100% = 10% Answer E: Up 9.1% is the closest answer to this value.

100%

Tip A way to understand this calculation is to consider the following, simplified scenario: In Year 1 a shirt costs $1. In Year 2 the same shirt costs $2. Looking at this we can see that the value of the shirt has increased 100% in relation to the Dollar: Change of value of the shirt = [($2 – $1) ÷ $1] 100% = 100% Conversely, the value of the Dollar has decreased 50% in relation to the shirt: Change of value of Pound Sterling = {[(1/$2) – (1/$1)] ÷ (1/$1)} 100% = -50% We apply the same principle to calculate the change in value of the Swiss Franc in relation to the US Dollar by replacing the shirt with a US Dollar and the Dollar with Swiss Franc. Answer A Down 9.1% B Down 8.3% C Down 1.7% D Up 8.3% E Up 9.1%

20

In Year 1, US$ 200 was used to purchase a holding of Euros. What would the value of this holding be if exchanged for HK$ in Year 2?
Solution We have to calculate a series of conversions to get to the answer: 200 US Dollar -> Euro (Year 1) -> Dollar (Year 2) -> HK$ (Year 2) In terms of exchange values this series of conversions is: 200 -> 200 0.78 -> (200 0.78) ÷ 0.64 -> [(200 So, putting that together: 200 US Dollar (Year 1) in HK$ (Year 2)

0.78) ÷ 0.64]

6.2

= =

[(200 0.78) ÷ 0.64] HK$ 1,511

6.2

Answer A HK$ 1,313 B HK$ 1,412 C HK$ 1,511 D HK$ 1,610 E HK$ 1,709

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Business insurance claims per annum per 100,000 policies
Construction Hotels & Restaurants Transport & Distribution Manufacturing Agriculture & Fishing Other 0
5 9 9 12 18 15 23 20 21 17 36 32

USA

Europe

10

20

30

40

21

Assuming there were 1,250,000 insurance policies issued annually in Europe to cover the Manufacturing sector, how many claims would be expected per year?
Solution Considering cover for the Manufacturing sector, we know that 20 claims are made in Europe per 100,000 policies. We calculate the number of 100,000 policies in 1,250,000 policies and multiply by the number of claims per year: Number of expected claims = (1,250,000 ÷ 100,000) 20 = 250 Answer A 250 B 400 C 500 D 550 E 600

22

Comparing the USA with Europe, which business sector has the most similar number of claims on a proportional basis?
Solution We consider the proportional difference between claims per 100,000 for USA and Europe, for each industry: Proportional difference = (claims USA – claims Europe) ÷ (claims USA + claims Europe) Sector Construction Hotels & Restaurants Transport & Distribution Manufacturing Agriculture & Fishing Other Proportional difference 0.06 0.14 0.09 0.07 0.11 0.29

From this we can see Construction' has the most similar number of claims.

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Tip We could have derived the answer by looking at the chart provided. We are looking for a small difference in claims and a large number of total claims. As the differences for all sectors are either 3 or 4, we can see that Construction has by far the largest number of claims (considerably more than the others). This sector will thus give us the smallest proportional difference. Answer A Construction B Hotels & Restaurants C Transport & Distribution D Manufacturing E Agriculture & Fishing

23

There are on average 6 times as many claims made per annum in the USA compared to Europe in the Transport & Distribution sector. How many more policies are issued in the US compared to Europe in this sector?
Solution We consider the Transport & Distribution sector. If we assume only 15 claims have been made in Europe for this sector for the year, we can fix the number of policies sold in Europe to 100,000. On average 6 times as many claims are made per annum in the USA compared to Europe which amounts to 90 claims (6 15). From this we know that the number of policies sold in the USA is: Number of policies sold = (90 ÷ 18) 100,000 = 500,000 From this we know that 5.0 times more policies were sold in the USA compared to Europe. Answer A 3.3 times as many B 4 times as many C 4.3 times as many D 5.0 times as many E 5.5 times as many

24

Last year there were 630 claims made in the USA against policies written out in the Agriculture & Fishing sector. How many policies were issued that year for that sector?
Solution We consider the Agriculture & Fishing sector in the USA. We calculate the number of policies issued that year using the fact that 23 claims as made per 100,000 policies sold. This number of policies sold for 630 claims to be made is: Number of policies sold = (630 ÷ 21) 100,000 = 3,000,000 Answer A 2,670,000 B 3,000,000 C 4,330,000 D 5,000,000 E 6,300,000

Numerical Test Answer Booklet 2 is created by eFinancialCareers.com. eFinancialCareers.com 2007. No part of this booklet may be reproduced or transmitted in anyway without the written consent of eFinancialCareers.com

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