ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE General & Further Mathematics => Topic started by: xtremeownage on July 05, 2011, 07:00:06 pm
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This thread will consist of question and answers, anyone who has trouble or would like to give suggested solutions on business maths reply to this thread.
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5. Calculate the amount of interest to be paid.
CD Player:
Price: $790
Deposit: $100
Interest rate: 12% p.a
Length of loan: 15months
Payment period: quarterely
Answer: $103.50
Can anyone supply solutions on how to get to this answer? thanks!
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Using the simple interest formula
I = P*r*t
I = Interest
P = Principal
r = Rate
t = time (years)
i.e.
I = x
P = 690 ($790 - $100 paid already)
r = 12%
t = 1.25years
x = (690)(0.12)(1.25)
x = $103.5
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thanks man, how stupid of me !!!
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No problem, the deposit can make the question look confusing :P
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Yep as usual further is full of tricks. Always have to be careful on these types of questions.
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Im pretty much following worked examples from book but they are giving me wrong answers ?
1. A machine originally costing $37000 is expected to produce 100 000 units. The output of the machine in each of the first three years was 5234, 6286 and 3987 units respectively. Its anticipated scrap value is $5000.
a) what is the unit for cost for this machine?
$0.32 (Good)
b) Find the total production over the first three years, and hence the book value at the end of three years.
3rd year: 33 316.6 - 3987 * 0.32
Book VALUE: = $32 037.76 (GOOD)
Total Production: 15 507 (good)
c)How many years it will be in use, if the average production during its life is 5169 units per year?
Further depreciation:
32 037.76 - 5000 = 27037.76
Remaining: 27037.76/5169 = 5.23 + 3 = 8.23 (WRONG - Answer = 19.3years ???)
4) A print machine costing $110 000 has a scrap value of $2400 after it has printer 4 million pages.
i) unit cost: $0.026875 (CORRECT)
ii)book value: $69 687.50 (CORRECT
iii) the annual deprecation charge of the machine if it prints 750 000 pages per year.
=???
Answer: $20 156.25
b) Find book value of the printing machine after five years if it prints on average 750 000 pg per year? = 9218.75 (CORRECT)
c) How many pages has the machine printed by the time the book value is $70 000, if it prints on average 750 000 pgs per year?
???????
Answer: 1 488 372 pages
7) A computer depreciates at a flat rate of 22.5% of the pruchase price per annum. Its purchase price is $5600.
a) What is book value of comp after 3 years
?????
Answer: $1820
b)After how long will the computer be written off if the value is nill?
????
Answer:4.44 years
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"Trihn has a part-time job at the local service station, where he is entitled to a staff discount of 10% on petrol and 15% on food and drink. On a particular day he buys 20 litres of petrol, which has a retail price of 153.4 cents per litre, and a can of coke and a packet of lollies, which cost $2.50 cents and $3.05, respectively. How much was his bill, after discount, for that day?"
The answer is $21.64. This is the only question in 20A (Essential 3rd edition) that I'm getting wrong.
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New price (petrol) = 90/100 x 20 x 97.8 c = 1760.40 c = $17.60
New price (coke) = 85/100 x 1.70 = $1.45
New price (lollies) = 85/100 x 3.05 = $2.59
Hence, bill after discount = 17.60 + 1.45 + 2.59 = $21.64
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Im pretty much following worked examples from book but they are giving me wrong answers ?
1. A machine originally costing $37000 is expected to produce 100 000 units. The output of the machine in each of the first three years was 5234, 6286 and 3987 units respectively. Its anticipated scrap value is $5000.
a) what is the unit for cost for this machine?
$0.32 (Good)
b) Find the total production over the first three years, and hence the book value at the end of three years.
3rd year: 33 316.6 - 3987 * 0.32
Book VALUE: = $32 037.76 (GOOD)
Total Production: 15 507 (good)
c)How many years it will be in use, if the average production during its life is 5169 units per year?
Further depreciation:
32 037.76 - 5000 = 27037.76
Remaining: 27037.76/5169 = 5.23 + 3 = 8.23 (WRONG - Answer = 19.3years ???)
100000/5169 = 19.34 years
4) A print machine costing $110 000 has a scrap value of $2400 after it has printer 4 million pages.
i) unit cost: $0.026875 (CORRECT)
ii)book value: $69 687.50 (CORRECT
iii) the annual deprecation charge of the machine if it prints 750 000 pages per year.
=???
Answer: $20 156.25
Total usage = 4,000,000 pages printed
Pages/year = 750,000
Useful life = 4,000,000/750,000 = 5.33333333 years
Annual dep'n = (Cost - Residual)/(Useful Life)
= (110000 - 2500)/(5.33333333)
= $20156.25p.a.
b) Find book value of the printing machine after five years if it prints on average 750 000 pg per year? = 9218.75 (CORRECT)
c) How many pages has the machine printed by the time the book value is $70 000, if it prints on average 750 000 pgs per year?
???????
Answer: 1 488 372 pages
Cost per page (unit cost) = $0.026875
Book value = $70,000
Cost = $110,000
Amt used = (110,000 - 70,000) = $40000 (Total cost)
Pages printed = Total cost / Unit cost
= 40,000/0.026875
= 1,488,372 pages printed
7) A computer depreciates at a flat rate of 22.5% of the pruchase price per annum. Its purchase price is $5600.
a) What is book value of comp after 3 years
?????
Answer: $1820
Depreciation expense per year = $5,600*22.5% = $1,260
Dep'n after 3 years = $1,260*3 = $3,780
Book value = Cost - Depreciation
= $5,600 - $3,780
= $1,820
b)After how long will the computer be written off if the value is nill?
????
Answer:4.44 years
[/color]
Useful life (years) = Cost/Depreciation(per year)
= $5,600/$1,260
= 4.44 years
Hope that makes sense
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New price (petrol) = 90/100 x 20 x 97.8 c = 1760.40 c = $17.60
New price (coke) = 85/100 x 1.70 = $1.45
New price (lollies) = 85/100 x 3.05 = $2.59
Hence, bill after discount = 17.60 + 1.45 + 2.59 = $21.64
How come for lollies you used the original retail price of 3.05 but for the coke you used 1.70? Also, how why/how did you use 97.8c for the petrol calculation? I understand that you're meant to find the discounts for all 3 then add together but out of the three separate product calculations, I only understand the lollies line.
I thought you were meant to go 0.85 X 2.50 for coke and 20 X 153.4 / 100 for petrol, then add.
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New price (petrol) = 90/100 x 20 x 97.8 c = 1760.40 c = $17.60
New price (coke) = 85/100 x 1.70 = $1.45
New price (lollies) = 85/100 x 3.05 = $2.59
Hence, bill after discount = 17.60 + 1.45 + 2.59 = $21.64
How come for lollies you used the original retail price of 3.05 but for the coke you used 1.70? Also, how why/how did you use 97.8c for the petrol calculation? I understand that you're meant to find the discounts for all 3 then add together but out of the three separate product calculations, I only understand the lollies line.
I thought you were meant to go 0.85 X 2.50 for coke and 20 X 153.4 / 100 for petrol, then add.
This is from my Essential Further Textbook:
"Trinh has a part-time job at the local service station, where he is entitled to a staff discount of 10% on petrol and 15% on food and drink. On a particular day he buys 20 litres of petrol, which has a retail price of 97.8 cents per litre, and a can of coke and a packet of lollies, which cost $1.70 cents and $3.05 respectively. How much was his bill, after discount, for that day?"
Perhaps our textbooks have different prices ??? (Although our textbooks have the same answer of $21.64 )
For the petrol calculation, since Trinh is buying 20 litres which costs 97.8 cents per litre and there is a 10% discount, we can say that the new price = 90/100 x 20 x 97.8 c = 1760.40 c = $17.60. Hope that helps! :)
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Wow that is weird. I checked my textbook again (essential further 3rd edition) and I have different prices yet the answer is the same! No wonder I wasn't getting the right answer.
Thanks for the clarification man.
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9) A stereo system purchased for $1200 incurs 12% per annum reducing balance depreciation.
a)find the book value after 7 years: correct ($490.41)
b)What is total depreciation after 7 years? (709.59) correct.
c)If the stereo has a scrap value of $215, in which year will this value be reached: Answer (14th year) ????
If any1 has suggested solutions for 21F Q9-28 for Further Maths Essentials 3rd edition please pm me.
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Can someone please help me with question 4a in Exercise 21B (Essential Further Mathematics Third Edition)?
The answer is $1.90. (However, I obtained an answer of $2.12).
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Can someone please help me with question 4a in Exercise 21B (Essential Further Mathematics Third Edition)?
The answer is $1.90. (However, I obtained an answer of $2.12).
BUMP please! :)
Edit: Refer to my previous post for the attachment!
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9) A stereo system purchased for $1200 incurs 12% per annum reducing balance depreciation.
a)find the book value after 7 years: correct ($490.41)
b)What is total depreciation after 7 years? (709.59) correct.
c)If the stereo has a scrap value of $215, in which year will this value be reached: Answer (14th year) ????
If any1 has suggested solutions for 21F Q9-28 for Further Maths Essentials 3rd edition please pm me.
14 years seems about right
use the calculator:
(http://i463.photobucket.com/albums/qq358/nightmareXD/maths.jpg)
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Can someone please help me with question 4a in Exercise 21B (Essential Further Mathematics Third Edition)?
The answer is $1.90. (However, I obtained an answer of $2.12).
i think book is wrong?
1 may - 6 may = 650.72 x (5/365) x (3/100) = 0.2674
6 may - 19 may = 445.82 x (13/365) x (3/100) = 0.4764
19 may - 1 june = 1241.37 x (13/365) x (3/100) = 1.2244
therefore, total interest in that month = 0.2674 + 0.4764 + 1.2244 = 1.9682 = $1.97
???
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The answer is $1.90. (However, I obtained an answer of $2.12).
I also think the book is wrong.
9) A stereo system purchased for $1200 incurs 12% per annum reducing balance depreciation.
a)find the book value after 7 years: correct ($490.41)
b)What is total depreciation after 7 years? (709.59) correct.
c)If the stereo has a scrap value of $215, in which year will this value be reached: Answer (14th year) ????
If any1 has suggested solutions for 21F Q9-28 for Further Maths Essentials 3rd edition please pm me.
14 years seems about right
use the calculator:
(http://i463.photobucket.com/albums/qq358/nightmareXD/maths.jpg)
I got the same answer, except I used the solve function.
solve(215=1200X (1-12/100)^x,x (find x)
x = 13.45, hence the answer is 14 years.