HSC 2016 General Maths Marathon (1 Viewer)

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InteGrand

Well-Known Member
what a random question lol
the 'trick' is that super isnt really a deduction?
You deduct everything.

I realised davidgoes4wce didn't write his deduction of super, but the final answer he got includes this deduction (so he probably just forgot to type the super amount, i.e. typo).

davidgoes4wce

Well-Known Member
You deduct everything.

I realised davidgoes4wce didn't write his deduction of super, but the final answer he got includes this deduction (so he probably just forgot to type the super amount, i.e. typo).
I fixed it up. I do have a background in accounting but working out what is tax-deductible and non-deductible items has never been my strong point.

davidgoes4wce

Well-Known Member
$\bg_white New Century Ex 1-01 Q 7$

$\bg_white Natalie borrowed \14,500 over 3 years at 13.5 \% p.a. flat interest to help towards her wedding. If she adds the \37.50 stamp duty and \115 per year loan insurance to the amount borrowed, how much does she repay per month?$

$\bg_white Ans \ 580.84$

anyone can explain the step by step in order to get to this answer

BLIT2014

The pessimistic optimist.
Moderator
$\bg_white New Century Ex 1-01 Q 7$

$\bg_white Natalie borrowed \14,500 over 3 years at 13.5 \% p.a. flat interest to help towards her wedding. If she adds the \37.50 stamp duty and \115 per year loan insurance to the amount borrowed, how much does she repay per month?$

$\bg_white Ans \ 580.84$

anyone can explain the step by step in order to get to this answer
Interest borrowed:
I=P*r*n (Simple interest formula)
=14500*13.5%*3
=5872.5

Total amount borrowed
=Principle+interest
=14500+5872.50
=$20372.5 Loan Insurance 115*3=$345

Months in 3 years : 36

Monthly Payment =($20372.5+$437.50+$345) divided by 36 =$576.527777778

I did this but it doesn't get the answer in the book.

davidgoes4wce

Well-Known Member
This was the example under the New Century text for that section

$\bg_white My thinking is the answer should be \571.85.$

$\bg_white Total amount borrowed= 14500+37.50+115=14652.50$

$\bg_white Interest Paid = P r n = 14652.50 \times 0.135 \times 3= \ \ 5,934.26$

$\bg_white Total Amount to Repay= 5,934.26+14,652.50=\20,586.76$

$\bg_white Monthly repayment= \20,586.76 \div 36 =\ 571.85$

davidgoes4wce

Well-Known Member
$\bg_white Fcuk me I finally got it! Had to multiply the insurance by 3 years duration$

$\bg_white Total Amount Borrowed =14,500+37.50 +115 \times 3=14,882.50$

$\bg_white Interest Paid = P r n = 14882.50 \times 0.135 \times 3= \ \ 6,027.41$

$\bg_white Total Amount to Repay= 6,027.41+14,882.50=\20,909.91$

$\bg_white Monthly repayment= \20,909.91 \div 36 =\ 580.84$

davidgoes4wce

Well-Known Member
$\bg_white New Century Ex 1.02 Q 2$

$\bg_white Scott purchased an engagement ring for \7560, paying \100 deposit and 7.2 \% p.a. interest for 18 months. What was the total amount paid for the ring? Select A,B,C or D$

$\bg_white A) \ \8365.68 B) \ \8265.68 \ C) \7460 \ D) \8056.80$

davidgoes4wce

Well-Known Member
$\bg_white I got the answer$

$\bg_white My thinking with this question was to subtract the difference between the deposit and the purchased item= 7560-100=7460$

$\bg_white I=PrN=7460 \times 0.072 \times 1.5 =805.68$

$\bg_white The total amount Paid= 7460+805.68+100=\8365.68$

Choice A should be the right answer

davidgoes4wce

Well-Known Member

I didn't understand this question from the Year 11 Oxford Book, (Q11)

I know that the median is half of the total frequency of scores, so I looked for the 55th score. My choice would have been then to work out the centre score of the 12-3pm interval, as this is where the 55th score lies. So technically, I would have had 1.30pm as my answer.

The books answer is: A

davidgoes4wce

Well-Known Member

Thought the answer was D but the student got a cross against it.

davidgoes4wce

Well-Known Member
$\bg_white I'm thinking with this question \angle ROS =95^\circ$

$\bg_white Using the Area of a triangle for ROS = \frac{1}{2} ab sin C$

$\bg_white A= \frac{1}{2} \times 4 \times 6 \times sin 95^\circ$

$\bg_white =11.954m^2$

$\bg_white The question wanted the nearest metre, so I then rounded is to 12 m^2$

My Pick would have been D but that was marked wrong.

davidgoes4wce

Well-Known Member
$\bg_white Lebron Curry measured a piece of material for her HSC major work at 200 mm correct to the nearest millimetre. What is the percentage error in her measurement ?$

$\bg_white A) \pm 0.0025\%$

$\bg_white B) \pm 0.005\%$

$\bg_white C) \pm 0.25\%$

$\bg_white D) \pm 0.5\%$

Now I have a table from my notes last year:

PERCENTAGE ERROR

$\bg_white 193 \ cm \implies Error=\pm \ 0.50$

$\bg_white 139.4 m \implies Error=\pm \ 0.05$

$\bg_white 138.32 m \implies Error=\pm \ 0.005$

$\bg_white 138.654 m \implies Error=\pm \ 0.0005$

Now I'm not sure if those notes if that is referring to percentages or not.

What would be your guys response to that Multiple Choice Question?

InteGrand

Well-Known Member
$\bg_white Lebron Curry measured a piece of material for her HSC major work at 200 mm correct to the nearest millimetre. What is the percentage error in her measurement ?$

$\bg_white A) \pm 0.0025\%$

$\bg_white B) \pm 0.005\%$

$\bg_white C) \pm 0.25\%$

$\bg_white D) \pm 0.5\%$

Now I have a table from my notes last year:

PERCENTAGE ERROR

$\bg_white 193 \ cm \implies Error=\pm \ 0.50$

$\bg_white 139.4 m \implies Error=\pm \ 0.05$

$\bg_white 138.32 m \implies Error=\pm \ 0.005$

$\bg_white 138.654 m \implies Error=\pm \ 0.0005$

Now I'm not sure if those notes if that is referring to percentages or not.

What would be your guys response to that Multiple Choice Question?
Those notes aren't referring to percentage error, they're essentially referring to implied absolute error (half the limit of reading). So to get the percentage error, we would divide these by the measured value (200 mm for the question at hand).

By the way, for your notes on Error, they should say like ±0.5 m, i.e. include the units of the measurement (so if the measurement is in mm or kg, the absolute error is also in mm or kg). This is because these are absolute errors, e.g. how many metres we would be off by. This'll also help you remember that these errors are indeed absolute errors. Percentage errors on the other hand have no units (since they are just percentages like 1% error etc.) since the units cancel out when we divide.

davidgoes4wce

Well-Known Member
Those notes aren't referring to percentage error, they're essentially referring to implied absolute error (half the limit of reading). So to get the percentage error, we would divide these by the measured value (200 mm for the question at hand).

By the way, for your notes on Error, they should say like ±0.5 m, i.e. include the units of the measurement (so if the measurement is in mm or kg, the absolute error is also in mm or kg). This is because these are absolute errors, e.g. how many metres we would be off by. This'll also help you remember that these errors are indeed absolute errors. Percentage errors on the other hand have no units (since they are just percentages like 1% error etc.) since the units cancel out when we divide.
The answer is D but im not 100% sure why it is.

So I guess after what you say there

$\bg_white 200 mm \pm 0.50 mm is the error there$

I still don't get how they get to $\bg_white \pm 0.50 \%$ as their answer.

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davidgoes4wce

Well-Known Member
Also the question where I got this , they had the question prior wrong for the radial survey, that's why I am having an open mind with these questions because I'm not sure if these markers marked it correctly or not.

davidgoes4wce

Well-Known Member
This is an example I found online :

How To Find Percent Error: To correctly put together the percent error equation, take the difference from accepted value which is also your result minus the accepted value, divide by accepted value and multiply this result by 100. This formula is always expressed as %.
Example How To Calculate Percent Error: You estimated your monthly car payment to be $315. The actual car payment turned out to be$300. Calculate the percent error in these payments: First, take 315 and subtract 300 = 15. Next, take 15 and divide by the correct monthly car payment. 15/300=0.05. Finally, multiply 0.05 by 100=5%. The final percent error in your car payment estimate equals 5%.

davidgoes4wce

Well-Known Member
Those notes aren't referring to percentage error, they're essentially referring to implied absolute error (half the limit of reading). So to get the percentage error, we would divide these by the measured value (200 mm for the question at hand).

By the way, for your notes on Error, they should say like ±0.5 m, i.e. include the units of the measurement (so if the measurement is in mm or kg, the absolute error is also in mm or kg). This is because these are absolute errors, e.g. how many metres we would be off by. This'll also help you remember that these errors are indeed absolute errors. Percentage errors on the other hand have no units (since they are just percentages like 1% error etc.) since the units cancel out when we divide.

$\bg_white Based on what you said there =\frac{0.50}{200} \times 100 \% = 0.25 \% would that be your answer ?$

InteGrand

Well-Known Member
The answer is D but im not 100% sure why it is.

So I guess after what you say there

$\bg_white 200 mm \pm 0.50 mm is the error there$

I still don't get how they get to $\bg_white \pm 0.50 \%$ as their answer.
Since it's +/- 0.5 mm, the total uncertainty in the reading is 1 mm, so the answer is whatever 1 mm is as a percentage of 200 mm, which is 0.5%.

InteGrand

Well-Known Member
Actually realised it said +/- 0.5%. In the case of +/-, I reckon the +/- 0.25% is more correct. This would be because measuring to the nearest millimetre means that the measurement is 200 mm +/- 0.5 mm. So the error should be +/- 0.25%. Maybe General Maths has some formula for these Q's that dictates otherwise.

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