SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Financial Maths, STD2 EO-Bank 28

Leon opens a superannuation account to build up savings for retirement. At the end of each year he pays in $4000, and the account earns 5% per annum, compounded annually.

The spreadsheet below models the first 4 years of the account.

  
 

  1. Write down the formula used in cell C9, using appropriate grid references.   (1 mark)

  2. Determine the value that belongs in cell C9.   (1 mark)
  3. Starting from the end of year 4, Leon lifts his yearly payment from $4000 to $7000. Find the balance in the account at the end of year 7, and state how much larger this is than if he had stayed with $4000 payments.   (3 marks)
Show Answers Only

a.    \(\text{=E8*B3}\)

b.    \(\text{C9}=\$410.00\)

c.    \(\text{Balance at end of year 7}=\$42\,025.54\)

\(\text{Leon has}\ \$9457.50\ \text{more than at the standard contribution.}\)

Show Worked Solution

a.    \(\text{Formula: =E8*B3}\)
 

b.    \(\text{C9 (Year 3 interest)}=\text{balance at start}\times\text{rate}\)

\(\text{C9}=8200\times 0.05=\$410.00\)
  

c.    \(\text{Using}\ \ P+I+C\ \ \text{from end of year 4 balance}\ \$17\,240.50:\)

\(\text{Increased contributions of}\ \$7000\ \text{from year 5:}\)

\(\text{Year 5:}\ 17\,240.50+17\,240.50\times 0.05+7000=\$25\,102.53\)

\(\text{Year 6:}\ 25\,102.53+25\,102.53\times 0.05+7000=\$33\,357.66\)

\(\text{Year 7:}\ 33\,357.66+33\,357.66\times 0.05+7000=\$42\,025.54\)
  

\(\text{Standard contributions of}\ \$4000\ \text{from year 5:}\)

\(\text{Year 5:}\ 17\,240.50+17\,240.50\times 0.05+4000=\$22\,102.53\)

\(\text{Year 6:}\ 22\,102.53+22\,102.53\times 0.05+4000=\$27\,207.66\)

\(\text{Year 7:}\ 27\,207.66+27\,207.66\times 0.05+4000=\$32\,568.04\)
  

\(\text{Difference}=42\,025.54-32\,568.04=\$9457.50\)

\(\therefore\ \text{Leon has}\ \$9457.50\ \text{more by increasing his contributions.}\)

Financial Maths, STD2 EO-Bank 34

Farah wants to buy a commercial espresso machine for her cafe at a cost of $4200. She has decided to pay off whatever she owes in a single amount 75 days after the purchase, and is weighing up two ways to fund it.

  • A credit card that compounds interest daily at 17% per annum, but does not charge interest during the first 25 days after a purchase.
  • A personal loan on which simple interest is charged at 8.9% per annum.

Advise Farah on the cheaper option, backing up your recommendation with the relevant calculations.   (4 marks)

Show Answers Only

\(\text{Credit card interest}=\$98.93\)

\(\text{Personal loan interest}=\$76.81\)

\(\text{As }\$76.81\lt\$98.93,\text{ Farah should choose the personal loan.}\)

Show Worked Solution

\(\text{Days accruing interest}=75-25=50\)

\(\text{Credit card option (interest compounds daily):}\)

\(\text{Amount owing}\) \(=4200\left(1+\dfrac{0.17}{365}\right)^{50}\)
  \(=4298.931\ldots\)
  \(=\$4298.93\ \text{(nearest cent)}\)

 
\(\therefore\ \text{Interest}=4298.93-4200=\$98.93\)
  

\(\text{Personal loan option (simple interest applies):}\)

\(I=Prn\) \(=4200\times 0.089\times\dfrac{75}{365}\)
  \(=76.808\ldots\)
  \(=\$76.81\ \text{(nearest cent)}\)

  
\(\therefore\ \text{As }\$76.81\lt\$98.93,\text{ the personal loan}\)

\(\text{is cheaper, so Farah should choose it.}\)

Financial Maths, STD2 EO-Bank 26

Kiara wants to purchase a used motorbike selling for $2500 and will be in a position to settle the whole debt in a single payment 40 days after the purchase.

Two methods of financing the purchase are open to her.

  • Using a credit card, where interest of 20.5% per annum is compounded daily. There is no interest-free period, so interest is charged from the day after the purchase.
  • Taking out a 40-day personal loan, where simple interest is charged at 12% per annum.
  1. Determine the interest that would build up under each method across the 40 days.   (2 marks)
  2. A friend insists that the personal loan will leave Kiara better off by more than $25 compared with the credit card. Decide whether the friend is correct, justifying your answer with calculations.   (2 marks)
Show Answers Only

a.    \(\text{Credit card: }\$56.78\ \text{, Personal loan: }\$32.88\)

b.    \(\text{Difference}=\$23.90\)

\(\text{As }\$23.90\lt\$25,\text{ the friend’s claim is incorrect.}\)

Show Worked Solution

a.    \(\text{Credit card option (compounding daily):}\)

\(\text{Amount owing}\) \(=2500\left(1+\dfrac{0.205}{365}\right)^{40}\)
  \(=2556.784\ldots\)
  \(=\$2556.78\ \text{(nearest cent)}\)

 
\(\therefore\ \text{Interest}=2556.78-2500=\$56.78\)
 

\(\text{Personal loan option (simple interest):}\)

\(I=Prn\) \(=2500\times 0.12\times\dfrac{40}{365}\)
  \(=32.876\ldots\)
  \(=\$32.88\ \text{(nearest cent)}\)

 

b.    \(\text{Difference}=56.78-32.88=\$23.90\)

\(\text{As }\$23.90\lt\$25,\text{ the friend’s claim is incorrect.}\)

Financial Maths, STD2 EO-Bank 6 MC

Grace repays her credit card gradually over several months. She notices that, although the interest rate on her card stays the same, the amount of interest charged changes from month to month.

This occurs because a credit card is an example of a reducing balance loan.

Which statement best explains why a credit card is an example of a reducing balance loan?

  1. Interest is calculated on the outstanding balance, which reduces as repayments are made.
  2. The interest rate decreases each time a repayment is made.
  3. The same amount of interest is charged each month, regardless of the balance owing.
  4. A fixed portion of the balance is repaid each month, with no interest charged.
Show Answers Only

\(A\)

Show Worked Solution
  • A is correct: interest is charged on the outstanding balance, which reduces as repayments are made – the defining feature of a reducing balance loan.

Other options:

  • B is incorrect: the interest rate stays fixed; it is the balance that reduces, not the rate.
  • C is incorrect: the interest charged is not fixed – it depends on the balance owing.
  • D is incorrect: interest is charged on a credit card, and repayments are not a fixed portion of the principal.

\(\Rightarrow A\)

Financial Maths, STD2 EQ-Bank 29

Mei sets up a superannuation account to save for retirement. She contributes $5000 at the end of each year into the account which earns interest at 6% per annum, compounded annually.

The spreadsheet shown models the first 4 years of the account.
  

  1. Calculate the value in cell C9.   (1 mark)
  2. From the end of year 4, Mei increases her annual contribution from $5000 to $8000. Calculate the balance in her superannuation account at the end of year 7, and determine how much more this is than if she had continued contributing $5000.   (3 marks)
Show Answers Only

a.    \(\text{C9} = \$618.00\)

b.    \(\text{Balance at end of year 7} = \$51\,519.99\)

\(\text{Mei has}\ \$9550.80\ \text{more than at the standard contribution.}\)

Show Worked Solution

a.    \(\text{C9 (Year 3 interest)} = \text{balance at start} \times \text{rate}\)

\(\text{C9} = 10\,300 \times 0.06 = \$618.00\)
  

b.    \(\text{Using}\ \ P+I+C\ \ \text{from end of year 4 balance}\ \$21\,873.08:\)

\(\text{Increased contributions of}\ \$8000\ \text{from year 5:}\)

\(\text{Year 5:}\ 21\,873.08+21\,873.08 \times 0.06+8000=\$31\,185.46\)

\(\text{Year 6:}\ 31\,185.46+31\,185.46 \times 0.06+8000=\$41\,056.59\)

\(\text{Year 7:}\ 41\,056.59+41\,056.59 \times 0.06+8000=\$51\,519.99\)
  

\(\text{Standard contributions of}\ \$5000\ \text{from year 5:}\)

\(\text{Year 5:}\ 21\,873.08+21\,873.08 \times 0.06+5000=\$28\,185.46\)

\(\text{Year 6:}\ 28\,185.46+28\,185.46 \times 0.06+5000=\$34\,876.59\)

\(\text{Year 7:}\ 34\,876.59+34\,876.59 \times 0.06+5000=\$41\,969.19\)
  

\(\text{Difference}=51\,519.99-41\,969.19=\$9550.80\)

\(\therefore\ \text{Mei has}\ \$9550.80\ \text{more by increasing her contributions.}\)

Financial Maths, STD2 EO-Bank 29

Dao takes out a reducing balance loan of $5000. The loan has an interest rate of 12% per annum, compounded monthly, and Dao makes monthly repayments of $900.

The spreadsheet shown models the first 4 months of the loan.
  

  1. Calculate the value in cell C9.   (1 mark)
  2. By continuing the spreadsheet, determine the number of months it takes Dao to repay the loan in full, and calculate the value of the final repayment.   (3 marks)
Show Answers Only

a.    \(\text{C9} = \$41.50\)

b.    \(\text{The loan is repaid in}\ 6\ \text{months.}\)

\(\text{Final repayment} = \$670.78\)

Show Worked Solution

a.    \(\text{Monthly interest rate} = \dfrac{12\%}{12} = 1\%\)

\(\text{C9 (Month 2 interest)} = 4150 \times 0.01 = \$41.50\)
  

b.    \(\text{Continue the schedule using}\ \ P+I-R\ \ \text{each month:}\)

\(\text{Month 5:}\ 1548.65+1548.65 \times 0.01-900=\$664.14\)

\(\text{Month 6:}\ 664.14+664.14 \times 0.01=\$670.78\)
 

\(\text{The Month 6 balance owing}\ (\$670.78)\ \text{is less than the}\)

\(\text{usual}\ \$900\ \text{repayment, so this final repayment clears the loan.}\)

\(\therefore\ \text{The loan is repaid in}\ 6\ \text{months, with a final}\)

\(\text{repayment of}\ \$670.78.\)

Financial Maths, STD2 EO-Bank 3 MC

Leon borrows $2500 from a short-term loan company. The terms of the loan are:

  • Establishment fee: $120 charged when the loan starts
  • Monthly account-keeping fee: $45
  • Weekly repayments of $90 over 4 months (17 weeks)

What is the total amount Leon will repay over the term of the loan?

  1. \(\$1530\)
  2. \(\$1650\)
  3. \(\$1710\)
  4. \(\$1830\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Account-keeping fees} = 45 \times 4 = \$180\)

\(\text{Weekly repayments} = 90 \times 17 = \$1530\)

\(\text{Total repaid} = 120+180+1530 = \$1830\)
  

\(\Rightarrow D\)

Financial Maths, STD2 EO-Bank 28

Talia takes out a reducing balance loan of $12 000 to buy a boat. The loan has an interest rate of 7.2% per annum and Talia makes monthly repayments of $300.

The spreadsheet shown models the first 3 months of the loan.
  

  1. Calculate the value in cell C9.   (1 mark)
  2. From the end of month 3, Talia increases her monthly repayment from $300 to $500. Calculate the balance owing at the end of month 6, and determine how much less Talia owes compared to keeping repayments at $300.   (3 marks)
Show Answers Only

a.    \(\text{C9} = \$70.63\)

b.    \(\text{Balance at end of month 6} = \$10\,007.71\)

\(\text{Talia owes}\ \$603.61\ \text{less than at the standard repayment.}\)

Show Worked Solution

a.    \(\text{Monthly interest rate} = \dfrac{7.2\%}{12} = 0.6\%\)

\(\text{C9 (Month 2 interest)} = 11\,772 \times 0.006 = \$70.63\)
 

b.    \(\text{Using}\ \ P+I-R\ \ \text{from end of month 3 balance}\ \$11\,311.89:\)

\(\text{Increased repayments of}\ \$500\ \text{from month 4:}\)

\(\text{Month 4:}\ 11\,311.89+11\,311.89 \times 0.006-500=\$10\,879.76\)

\(\text{Month 5:}\ 10\,879.76+10\,879.76 \times 0.006-500=\$10\,445.04\)

\(\text{Month 6:}\ 10\,445.04+10\,445.04 \times 0.006-500=\$10\,007.71\)
 

\(\text{Standard repayments of}\ \$300\ \text{from month 4:}\)

\(\text{Month 4:}\ 11\,311.89+11\,311.89\times 0.006-300=\$11\,079.76\)

\(\text{Month 5:}\ 11\,079.76+11\,079.76\times 0.006-300=\$10\,846.24\)

\(\text{Month 6:}\ 10\,846.24+10\,846.24\times 0.006-300=\$10\,611.32\)
 

\(\text{Difference}=10\,611.32-10\,007.71=\$603.61\)

\(\therefore\ \text{Talia owes}\ \$603.61\ \text{less by increasing her repayments.}\)

Financial Maths, STD2 EO-Bank 20

Noah takes out a reducing balance loan of $10 000 to renovate his bathroom. The loan has an interest rate of 6% per annum and Noah makes monthly repayments of $400.

The spreadsheet shown models the first 4 months of the loan.
  

  1. Complete the missing values for cells C10, B11 and E11.   (3 marks)
  2. Calculate the total interest Noah pays over the first 4 months of the loan.   (1 mark)
Show Answers Only

a.    \(\text{C10} = \$46.49\)

\(\text{B11} = \$8944.74\)

\(\text{E11} = \$8589.46\)

b.    \(\$189.46\)

Show Worked Solution

a.    \(\text{Monthly interest rate} = \dfrac{6\%}{12} = 0.5\%\)

\(\text{C10 (Month 3 interest)} =9298.25 \times 0.005 = \$46.49\)

\(\text{B11 (Month 4 start)}=9298.25+46.49-400=\$8944.74\)

\(\text{E11 (Month 4 end)}=8944.74+44.72-400=\$8589.46\)
  

b.    \(\text{Total interest} = 50+48.25+46.49+44.72=\$189.46\)

Financial Maths, STD2 EO-Bank 24

Ravi takes out a short-term loan to buy a laptop with a cash price of $1800.

The loan has the following terms:

  • Establishment fee: $110 charged when the loan starts
  • Monthly account-keeping fee: $40
  • Weekly repayments of $75 over 6 months (26 weeks)
  1. Calculate the total amount Ravi will pay back over the term of the loan.   (2 marks)

  2. How much more than the cash price does Ravi pay for the laptop?   (1 mark)

Show Answers Only

a.    \(\$2300\)

b.    \(\$500\)

Show Worked Solution

a.    \(\text{Account-keeping fees} = 40 \times 6 = \$240\)

\(\text{Weekly repayments} = 75 \times 26 = \$1950\)

\(\text{Total paid} = 110+240+1950= \$2300\)
  

b.    \(\text{Extra paid} = 2300-1800 = \$500\)

Financial Maths, STD2 EQ-Bank 20_3

Mei uses a buy now, pay later payment option to make a purchase of $1600. Her repayments are split across 4 equal payments over 6 weeks. No interest is charged.

Mei misses her final payment and is charged a late fee of $68. Mei's payment schedule is shown, with her balance totalling $468.
 

  1. Find the total amount Mei pays for her purchase if repaying in full on 28 September 2026.   (1 mark)
  2. Mei's bank offers short-term loans where simple interest is charged at 18% per annum.
  3. Suppose Mei had borrowed $1600 from the bank to make this purchase on 3 August 2026 and repaid it in full 9 weeks later.
  4. How much would Mei have saved using this approach instead of the buy now, pay later option?   (2 marks)
Show Answers Only

a.    \($1668\)

b.    \($18.29\)

Show Worked Solution

a.    \(\text{If total owing paid on 28 September:}\)

\(\text{Total paid} = 400+400+400+468=$1668\)
 

b.    \(r=18\%=0.18,\ \ n=\dfrac{9 \times 7}{365} = \dfrac{63}{365}\)

\(I=Prn=1600 \times 0.18 \times \dfrac{63}{365} = 49.709… = $49.71 \)

\(\text{Amount saved} = 68-49.71=$18.29\)

Financial Maths, STD2 EQ-Bank 20_2

Tane uses a buy now, pay later payment option to make a purchase of $200. His repayments are split across 4 equal payments over 6 weeks. No interest is charged.

Tane misses his final payment and is charged a late fee of $22. Tane's payment schedule is shown, with his balance totalling $72.

  1. Find the total amount Tane pays for his purchase if repaying in full on 1 June 2026.   (1 mark)
  2. Tane's bank offers short-term loans where simple interest is charged at 14% per annum.
  3. Suppose Tane had borrowed $200 from the bank to make this purchase on 6 April 2026 and repaid it in full 7 weeks later.
  4. How much would Tane have saved using this approach instead of the buy now, pay later option?   (2 marks)
Show Answers Only

a.    \($222\)

b.    \($18.24\)

Show Worked Solution

a.    \(\text{If total owing paid on 1 June:}\)

\(\text{Total paid} = 50+50+50+72=$222\)
 

b.    \(r=14\%=0.14,\ \ n=\dfrac{7 \times 7}{365} = \dfrac{49}{365}\)

\(I=Prn=200 \times 0.14 \times \dfrac{49}{365} = 3.758… = $3.76 \)

\(\text{Amount saved} = 22-3.76=$18.24\)

Financial Maths, STD2 EQ-Bank 20_4

Idris uses a buy now, pay later payment option to make a purchase of $440. His repayments are split across 4 equal payments over 6 weeks. No interest is charged.

Idris misses his final payment and is charged a late fee of $26. Idris's payment schedule is shown, with his balance totalling $136.
 

  1. Find the total amount Idris pays for his purchase if repaying in full on 30 November 2026.  (1 mark)
  2. Idris's bank offers short-term loans where simple interest is charged at 12% per annum.
  3. Suppose Idris had borrowed $440 from the bank to make this purchase on 5 October 2026 and repaid it in full 8 weeks later.
  4. How much would Idris have saved using this approach instead of the buy now, pay later option?   (2 marks)
Show Answers Only

a.    \($466\)

b.    \($17.90\)

Show Worked Solution

a.    \(\text{If total owing paid on 30 November:}\)

\(\text{Total paid} = 110+110+110+136=$466\)
 

b.    \(r=12\%=0.12,\ \ n=\dfrac{8 \times 7}{365} = \dfrac{56}{365}\)

\(I=Prn=440 \times 0.12 \times \dfrac{56}{365} = 8.100… = $8.10 \)

\(\text{Amount saved} = 26-8.10=$17.90\)

Financial Maths, STD2 EQ-Bank 17

Jaxon buys a phone for $960 using a "buy now, pay later" service. The service charges no interest, and the cost is split into 8 equal fortnightly payments.

An account-keeping fee of $4 is added to every fortnightly payment. In addition, a late fee of $12 is charged for each payment that is made late.

  1. Calculate Jaxon's regular fortnightly payment, assuming it is paid on time.   (2 marks)
  2. Jaxon makes all 8 payments but 2 of them are late. Calculate the total amount Jaxon pays for the phone, and hence how much more this is than the original purchase price.   (2 marks)
Show Answers Only

a.    \(\$124.00\)

b.    \(\text{Total} = \$1016.00,\ \ \$56.00\ \text{more than the purchase price.}\)

Show Worked Solution

a.    \(\text{Base instalment} = \dfrac{960}{8} = \$120.00\)

\(\text{Fortnightly payment} = 120+4 = \$124.00\)
  

b.    \(\text{Account-keeping fees} = 8 \times \$4 = \$32.00\)

\(\text{Late fees} = 2 \times \$12 = \$24.00\)

\(\text{Total paid}= 960+32+24= \$1016.00\)

\(\text{Extra cost} = 1016-960 = \$56.00\)
  

\(\therefore\ \text{Jordan pays}\ \$56.00\ \text{more than the purchase price.}\)

Calculus, EXT1 EQ-Bank 19

The function  \(y=f(x)\)  has an inverse function  \(y=f^{-1}(x)\).

The tangent to  \(y=f(x)\) at the point \((2,3)\), \(\ell\), has a gradient of 1.

Show that the tangent to  \(y=f^{-1}(x)\)  at the point \((3,2)\) is parallel to \(\ell\).   (3 marks)

Show Answers Only

\(\text{See Worked Solution}\)

Show Worked Solution

\(\text{Since tangent to}\  f(x) \ \text {touches at}\ (2,3):\)

\(f(2)=3 \ \ \Rightarrow\ \ f^{-1}(3)=2\)
 

\(\text{For inverse functions,}\ \ \left(f^{-1}\right)^{\prime}(x)=\dfrac{1}{f^{\prime}\left(f^{-1}(x)\right)}\)

\(\text{Since tangent to} \ \ f(x) \ \ \text{at} \ \ x=2 \ \ \text{has gradient 1}\)

\(m_{\ell}=1 \ \Rightarrow \ f^{\prime}(2)=1\)
 

\(\text{Find gradient of}\ f^{-1}(x) \ \text{at} \ \ x=3:\)

\((f^{-1})^{\prime}(3)=\dfrac{1}{f^{\prime}\left(f^{-1}(3)\right)}=\dfrac{1}{f^{\prime}(2)}=1\)

\(\therefore \text{Gradient of tangent to} \ f^{-1}(x) \ \text {at } x=3\ \ \ \text {is parallel to} \ \ell\).

Calculus, EXT1 EQ-Bank 32

The polynomial  \(h(x)=x^3+2x+1\)  passes through the point \((1,4)\).

Find the gradient of the tangent to  \(f(x)=x h^{-1}(x)\)  at the point where \(x=4\).   (3 marks)

Show Answers Only

\(\dfrac{9}{5}\)

Show Worked Solution

\(h(1)=4 \ \ \Rightarrow\ \ h^{-1}(4)=1\)

\(f(x)=x h^{-1}(x)\)

\(\text{Using the product rule:}\)

\(f^{\prime}(x)=h^{-1}(x)+x\cdot \dfrac{d}{dx}\left(h^{-1}(x)\right)\)
 

\(\text{Find}\ \dfrac{d}{dx}\left(h^{-1}(x)\right):\)

\(\text{Let}\ \ y=h^{-1}(x)\ \ \Rightarrow\ \ x=h(y)\)

\(\dfrac{dx}{dy}=h^{\prime}(y)\ \ \Rightarrow\ \ \dfrac{dy}{dx}=\dfrac{1}{h^{\prime}(y)}\)

\(\dfrac{d}{dx}\left(h^{-1}(x)\right)=\dfrac{1}{h^{\prime}\left(h^{-1}(x)\right)}\)
 

\(f^{\prime}(x)=h^{-1}(x)+\dfrac{x}{h^{\prime}\left(h^{-1}(x)\right)}\)

\(f^{\prime}(4)=h^{-1}(4)+\dfrac{4}{h^{\prime}\left(h^{-1}(4)\right)}=1+\dfrac{4}{h^{\prime}(1)}\)
 

\(h(x)=x^3+2x+1\ \ \Rightarrow\ \ h^{\prime}(x)=3x^2+2\)

\(f^{\prime}(4)=1+\dfrac{4}{3(1)^2+2}=\dfrac{9}{5}\)

\(\therefore\ \text{Gradient of tangent}=\dfrac{9}{5}\)

Financial Maths, STD2 EQ-Bank 32

Yasmin is planning to purchase a Newcastle Knights corporate box for $5000. She intends to repay the entire amount 90 days after making the purchase.

She is comparing two credit options.

  • Credit card: compound interest charged daily at 18% per annum, with a 30-day interest-free period from the date of purchase.
  • Personal loan: simple interest charged at 9.5% per annum.

Determine which option Yasmin should choose, showing calculations to support your answer.   (4 marks)

Show Answers Only

\(\text{Credit card (daily compound interest):}\)

\(\text{Days accruing interest}\ =90-30=60\)

\(\text{Amount owing}\) \(= 5000\left(1+\dfrac{0.18}{365}\right)^{60}\)
  \(= 5150.118\ldots\)
  \(= \$5150.12\ \text{(nearest cent)}\)

  
\(\text{Interest} = 5150.12-5000 = \$150.12\)
  

\(\text{Personal loan (simple interest):}\)

\(I = Prn\) \(=5000 \times 0.095 \times \dfrac{90}{365}\)
  \(= 117.123\ldots\)
  \(= \$117.12\ \text{(nearest cent)}\)

 

\(\therefore\ \text{Since \$117.12 < \$150.12, Yasmin should choose the personal loan.}\)

Show Worked Solution

\(\text{Days accruing interest}\ =90-30=60\)

\(\text{Credit card (daily compound interest):}\)

\(\text{Amount owing}\) \(= 5000\left(1+\dfrac{0.18}{365}\right)^{60}\)
  \(= 5150.118\ldots\)
  \(= \$5150.12\ \text{(nearest cent)}\)

  
\(\text{Interest} = 5150.12-5000 = \$150.12\)
  

\(\text{Personal loan (simple interest):}\)

\(I = Prn\) \(=5000 \times 0.095 \times \dfrac{90}{365}\)
  \(= 117.123\ldots\)
  \(= \$117.12\ \text{(nearest cent)}\)

  

\(\therefore\ \text{Since \$117.12 < \$150.12, Yasmin should choose the personal loan.}\)

Financial Maths, STD2 EQ-Bank 26

A credit card has an interest-free period of 55 days from and including the date of purchase. Interest is charged on purchases, compounding daily at a rate of 16.8% per annum. Interest is charged from the day following the interest-free period.

Furniture was purchased for $1200 using this credit card. Full payment was made on the 78th day after the date of purchase. There were no other purchases on this credit card.

  1. For how many days is interest charged on the purchase? Give your answer to the nearest cent.   (1 mark)

  2. Calculate the total interest charged when the account was paid in full.   (2 marks)

Show Answers Only

a.    \(23 \text{ days}\)

b.    \(\$12.77\)

Show Worked Solution

a.    \(\text{Interest is charged after the 55-day interest-free period.}\)

\(\text{Days charged} = 78-55 = 23 \text{ days}\)
 

b.    \(\text{Daily interest rate} = \dfrac{0.168}{365}\)

\(\text{Amount owing} \) \(= 1200\left(1+\dfrac{0.168}{365}\right)^{23}\)
  \(= 1212.768\ldots\)
  \(= \$1212.77\ \text{(nearest cent)}\)

  

\(\therefore\ \text{Interest} = 1212.77-1200 = \$12.77\)

Financial Maths, STD2 EQ-Bank 5 MC

A credit card is an example of a reducing balance loan.

Which statement best explains why this is the case?

  1. Interest is charged at a fixed amount each month, regardless of the balance owing.
  2. Interest is charged on the outstanding balance, which decreases as repayments are made.
  3. The interest rate reduces each month as the balance is repaid.
  4. A fixed portion of the amount borrowed is repaid each month, with no interest charged.
Show Answers Only

\(B\)

Show Worked Solution

B is correct: Like any reducing balance loan, interest is charged on the outstanding balance, which decreases as repayments are made.

Other options:

  • A is incorrect: the interest is not a fixed amount; it depends on the balance.
  • C is incorrect: it is the balance that reduces, not the interest rate, which stays fixed.
  • D is incorrect: interest is charged on a credit card, and repayments are not a fixed portion of the principal.

\(\Rightarrow B\)

Financial Maths, STD1 EQ-Bank 4 MC

A credit card has the following fees:

  • annual fee of \(\$90\)
  • late payment fee of \(\$20\) each time a payment is late
  • cash advance fee of \(\$5\) each time a cash advance is made

During one year, the cardholder made 2 late payments and 3 cash advances.

What is the total amount of fees charged for the year?

  1. \(\$55\)
  2. \(\$90\)
  3. \(\$130\)
  4. \(\$145\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Annual fee} = \$90\)

\(\text{Late payment fees} = 2 \times \$20 = \$40\)

\(\text{Cash advance fees} = 3 \times \$5 = \$15\)

\(\text{Total fees} = 90+40+15 = \$145\)
  

\(\Rightarrow D\)

Networks, STD2 EQ-Bank 28

Lena is opening a new café. The project requires the completion of 7 activities, \(A\) to \(G\). The project is due to be completed in 15 days.

The directed network diagram shows these activities with their completion times in days.
  

   
  
  1. Use the information from the network diagram to complete the Gantt chart, including the critical path and all other activities required for the project.   (3 marks)
      

      
  2. List all activities, not on the critical path, which could be occurring at midday on day 7.   (1 mark)
Show Answers Only

a.

b.    \(B\text{, }D\text{ and }E\)

Show Worked Solution

a.

 
b.    \(\text{By inspection of the Gantt chart.}\)

\(\text{Activities which could be occurring at }\approx 6.5\ \text{on chart (midday day 7):}\)

\(B\text{, }D\text{ and }E\)

Networks, STD2 EQ-Bank 27

A project requires the completion of 8 activities, \(A\) to \(H\). The project is due to be completed in 16 days.

The directed network diagram shows these activities with their completion times in days.
  

   
  
  1. Use the information from the network diagram to complete the Gantt chart, including the critical path and all other activities required for the project.   (3 marks)
      


      

  2. List all activities, not on the critical path, which could be occurring at midday on day 9.   (1 mark)
Show Answers Only

a.

b.    \(C,  D\ \text{and}\ F\)

Show Worked Solution

a.

 
b.    \(\text{By inspection of the Gantt chart.}\)

\(\text{Activities which could be occurring at}\ \approx 8.5\ \text{on chart (midday day 9):}\)

\(C, D\ \text{and}\ F\)

Networks, STD2 EQ-Bank 2 MC

The Gantt chart below shows the activities involved in organising a school sports carnival. 
  

   
Which activity has the greatest float time?

  1. \(\text{B}\)
  2. \(\text{D}\)
  3. \(\text{E}\)
  4. \(\text{G}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{From the Gantt chart, the dashed extensions show:}\)

\(\text{B: float = 3 hours}\)

\(\text{D: float = 5 hours}\)

\(\text{E: float = 3 hours}\)

\(\text{G: float = 0 (critical path, no dashed extension)}\)

\(\text{Activity D has the greatest float time (5 hours).}\)

\(\Rightarrow B\)

Networks, STD2 EQ-Bank 6 MC

The Gantt chart below shows the activities involved in organising a community market day. Critical path activities are shown as solid bars and non-critical activities show available float as dashed extensions.
  


  

Activity \(D\) is delayed by 7 hours and activity \(B\) is delayed by 1 hour. What is the new minimum completion time for the project?

  1. 14 hours
  2. 15 hours
  3. 16 hours
  4. 22 hours
Show Answers Only

\(C\)

Show Worked Solution

\(\text{From the Gantt chart:}\)

\(\text{D has float of 5 hours. Delay of 7 exceeds float by 2.}\)

\(\text{B has float of 3 hours. Delay of 1 is within float.}\)

\(\text{Only the delay to D affects completion time.}\)

\(\text{New minimum} = 14+2 = 16 \text{ hours}\)

\(\Rightarrow C\)

Networks, STD2 EQ-Bank 25

The construction of a new reptile exhibit is a project involving nine activities, \(A\) to \(I\). The network diagram below shows the activities and their completion times in weeks. Some values are missing.

The Gantt chart below has been created for this project.
  


  
  1. Using the Gantt chart, identify the critical path.   (1 mark)
  2. Use the Gantt chart to determine the missing values in the network diagram.   (2 marks)
  3. Activity \(E\) is delayed by 8 weeks. Using the Gantt chart, explain whether this will affect the minimum completion time of the project.   (2 marks)
Show Answers Only

a.    \(ACDFGI\)

b.    \(\text{B} = 5 \text{ weeks, H} = 7 \text{ weeks}\)

c.    \(\text{From the Gantt chart, activity E has a float of 6 weeks (see the dashed}\)

\(\text{extension from week 10 to 16).}\)

\(\text{The delay of 8 weeks exceeds the float of 6 weeks.}\)

\(\text{The project will be delayed by } 8-6 = 2 \text{ weeks.}\)

\(\text{New minimum completion time} = 25+2 = 27 \text{ weeks.}\)

Show Worked Solution

a.    \(\text{The critical path is the continuous solid bar on row 1 of the Gantt chart.}\)

\(\text{Critical path:}\ ACDFGI\)
 

b.    \(\text{Activities B and H are not labelled in the network diagram.}\)

\(\text{From the Gantt chart:}\)

\(\text{B starts at week 0, ends at week 5} \to \text{duration} = 5 \text{ weeks}\)

\(\text{H starts at week 7, ends at week 14} \to \text{duration} = 7 \text{ weeks}\)
 

c.    \(\text{From the Gantt chart, activity E has a float of 6 weeks (see the dashed}\)

\(\text{extension from week 10 to 16).}\)

\(\text{The delay of 8 weeks exceeds the float of 6 weeks.}\)

\(\text{The project will be delayed by } 8-6 = 2 \text{ weeks.}\)

\(\text{New minimum completion time} = 25+2 = 27 \text{ weeks.}\)

Networks, STD2 EQ-Bank 18

A Gantt chart for a project with activities \(A, B, C, D, E, F, G\) and \(H\) has been created. The Gantt chart can be used to complete the missing information on the edges in the network diagram.
  

  
  1. Use the Gantt chart to determine the three missing values in the network diagram.   (3 marks)
  2. State the minimum completion time for the project.   (1 mark)
Show Answers Only

a.    \(\text{C} = 4 \text{ hours, D} = 5 \text{ hours, E} = 7 \text{ hours}\)

b.    \(26 \text{ hours}\)

Show Worked Solution

a.    \(\text{Using the Gantt chart}\)

\(\text{C (between A and F):} \)

\(\Rightarrow\ \text{Starts hour 7, ends hour 11 = 4 hours duration}\)

\(\text{D (between B and G):}\)

\(\Rightarrow\ \text{Starts hour 3, ends hour 8 = 5 hours duration}\)

\(\text{E (between B and H):}\)

\(\Rightarrow\ \text{Starts hour 3, ends hour 10 = 7 hours duration}\)
 

b.    \(\text{From the Gantt chart, the project ends at hour 26.}\)

\(\text{Minimum completion time} = 26 \text{ hours}\)

Statistics, EXT1 EQ-Bank 6 MC

The lifetime of a certain brand of batteries has a mean lifetime of 20 hours and a standard deviation of 2 hours. A random sample of 40 batteries is selected.

The probability that the mean lifetime of this sample of 40 batteries exceeds 19.5 hours is closest to

  1. 0.0571
  2. 0.5987
  3. 0.8944
  4. 0.9429
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{CLT applies}\)

\(\overline{X} \sim N\left( \mu, \dfrac{\sigma^2}{n}\right) \sim N\left( 20, \dfrac{4}{40}\right)\)

\(\sigma_{\bar{X}} = \dfrac{1}{\sqrt{10}} \)
 

\(\text{By the central limit theorem:}\)

\(Z=\dfrac{\overline{X}-20}{\frac{1}{\sqrt{10}}}\sim N(0,1)\)

\(Z=\dfrac{19.5-20}{\frac{1}{\sqrt{10}}}=-1.58 \ \text{(2 d.p.)}\)
 

\(\text{Using Normal Distribution Table of Values:}\)

\(\Pr(\overline{X}>19.5)=\Pr(Z>-1.58)\)

\(\phantom{\Pr(\overline{X}>19.5)}=0.9429\)

\(\Rightarrow D\)

Statistics, EXT1 EQ-Bank 29

The waiting time, \(T\) hours, to see a particular doctor at a clinic has a mean of 0.5 hours and a standard deviation of 0.3 hours.

A sample of 35 waiting times is chosen at random.

Use the standard normal distribution table (provided) to find the probability that the average waiting time of the sample of 35 patients is between 0.43 hours and 0.50 hours. Give your answer as a percentage correct to 1 decimal place.   (3 marks)

Show Answers Only

\(\Pr(0.43<\overline{T}<0.50) \approx 41.6\%\)

Show Worked Solution

\(\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{Central Limit Theorem applies}\)

\(\text{The sample mean,}\ \overline{T}, \text{for random samples of size 35 is}\)

\(\text{approximately normally distributed, where:}\)

\(\mu=0.5\ \ \text{and}\ \ \sigma=\dfrac{0.3}{\sqrt{n}}=\dfrac{0.3}{\sqrt{35}} \approx 0.0507\)
 

\(\text{By the central limit theorem:}\)

\(Z=\dfrac{\overline{T}-0.5}{\frac{0.3}{\sqrt{35}}} \sim N(0,1)\)

\(\Pr(0.43<\overline{T}<0.50)\) \(=\Pr\left(\dfrac{0.43-0.50}{0.0507}<Z<\dfrac{0.50-0.50}{0.0507}\right)\)
  \(=\Pr(-1.38<Z<0)\)
  \(=\Pr(0<Z<1.38)\)
  \(=0.9162-0.5000\)
  \(=0.4162 = 41.6\%\ \text{(1 d.p.)}\)

Statistics, EXT1 EQ-Bank 32

A company accountant has previously found that the mean amount owed on any individual unpaid invoice is \(\$800\) with a standard deviation of \(\$200\).

Using the normal distribution table (included), determine the probability that, in a random sample of 60 unpaid invoices, the total amount owed is more than \(\$47\,500\). Give your answer as a percentage correct to one decimal place.   (3 marks)

Show Answers Only

\(62.6\%\)

Show Worked Solution

\(\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{CLT applies}\)

\(\bar{X} \sim N\left(\mu, \dfrac{\sigma^2}{n} \right) \sim N\left(800, \dfrac{200^2}{60} \right)\)

\(P(\Sigma X) > 47\,500 = P\left(\bar{X}>\dfrac{47\,500}{60}\right)=P(\bar{X})>791.67 \)
 

\(\text{By the central limit theorem:}\)

\(Z=\dfrac{\bar{X}-800}{\frac{200}{\sqrt{60}}}\sim N(0,1)\)

\(Z=\dfrac{791.67-800}{\frac{200}{\sqrt{60}}}=-0.32 \ \text{(2 d.p.)}\)

 

\(\text{Using Normal Distribution Table of Values:}\)

\(P(\Sigma X > 47\,500)\) \(=P(Z>-0.32)\)
  \(=1-P(Z\leq-0.32)\)
  \(=1-0.3745\)
  \(=62.6\%\)

Statistics, EXT1 EQ-Bank 22

The gestation period of cats has a mean of 66 days and a variance of 9 days\(^2\).

A sample of 32 cats is chosen at random.

Using the Normal Distribution Table of Values, determine the probability that the sample has an average gestation period greater than 65 days.   (3 marks)

Show Answers Only

\(0.9706\)

Show Worked Solution

\(\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{CLT applies}\)

\(\overline{X} \sim N\left( \mu, \dfrac{\sigma^2}{n}\right) \sim N\left( 66, \dfrac{9}{32}\right)\)

\(\sigma_{\bar{X}} = \dfrac{3}{\sqrt{32}} \)
 

\(\text{By the central limit theorem:}\)

\(Z=\dfrac{\overline{X}-66}{\frac{3}{\sqrt{32}}}\sim N(0,1)\)

\(Z=\dfrac{65-66}{\frac{3}{\sqrt{32}}}=-1.89 \ \text{(2 d.p.)}\)
 

\(\text{Using Normal Distribution Table of Values:}\)

\(P(\overline{X}>65)\) \(=P(Z>-1.89)\)
  \(=1-P(Z\leq -1.89)\)
  \(=1-0.0294\)
  \(=0.9706\)

Statistics, EXT1 EQ-Bank 13

The weights of pumpkins on a particular farm are normally distributed with a mean of 13.5 kg and a standard deviation of 3 kg.

A random sample of 15 pumpkins is selected.

Let \(X_1, X_2, X_3, \ldots, X_{15}\) denote the weights of the pumpkins in the sample, and let

\(\overline{X}=\dfrac{X_1+X_2+X_3+\cdots+X_{15}}{15}\).

  1. Find the expected value of \(\overline{X}.\)  (1 mark)

  2. Find the standard deviation of \(\overline{X}.\)   (1 mark)
Show Answers Only

a.    \(13.5\)

b.    \(0.775\)

Show Worked Solution

a.    \(E(\overline{X})=\mu=13.5\)

b.    \(\sigma_{\overline{X}}=\dfrac{\sigma}{\sqrt{n}}=\dfrac{3}{\sqrt{15}} \approx 0.775\)

Statistics, EXT1 EQ-Bank 26

A research team is investigating the amount of sleep obtained by Year 12 students. Previous studies indicate that the sleep time of Year 12 students has a population mean of 7.4 hours and a population standard deviation of 2.42 hours.

A random sample of 100 Year 12 students is selected.

Using the normal distribution table (included), determine the probability that the mean sleep time of the sample is less than 7 hours. Give your answer correct to four decimal places.   (3 marks)

Show Answers Only

\(4.94 \%\)

Show Worked Solution

\(\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{Central Limit Theorem applies}\)

\(\text{The sample mean,}\ \overline{X}, \text{for random samples of size 100 is}\)

\(\text{approximately normally distributed, where:}\)

\(\mu=7.4\ \ \text{and}\ \ \sigma=\dfrac{2.42}{\sqrt{n}}=\dfrac{2.42}{\sqrt{100}}\)
 

\(\text{By the central limit theorem:}\)

\(Z=\dfrac{\overline{X}-7.4}{\frac{2.42}{\sqrt{100}}} \sim N(0,1)\)

\(Z=\dfrac{7-7.4}{\frac{2.42}{\sqrt{100}}}=-1.65 \ \text{(2 d.p.)}\)
 

\(\text{Using Normal Distribution Table of Values:}\)

\(P(\overline{X} < 7)=P(Z < -1.65)=0.0494=4.94 \%\)

Financial Maths, STD2 EQ-Bank 17

Mia wants to buy a tablet with a cash price of $960. She cannot pay for it upfront and is considering two options.
 

Option 1: Buy now, pay later

  • 4 equal payments of $240 over 8 weeks
  • No interest or fees if all payments are made on time

Option 2: Short-term loan

  • Establishment fee: $75
  • Monthly account-keeping fee: $30
  • Weekly repayments of $55 over 5 months (20 weeks)

Assume Mia will make all payments on time under either option.

  1. Calculate the total amount Mia will pay under each option.   (2 marks)

  2. Which option is cheaper, and by how much?   (1 mark)

Show Answers Only

a.    \(\text{Option 1: }\$960,\ \text{Option 2: }\$1325\)

b.    \(\text{Option 1 is cheaper by }\$365\)

Show Worked Solution

a.    \(\text{Option 1: } \)

\(\text{Repayments}=4 \times 240 = \$960\)
  

\(\text{Option 2:}\)

\(\text{Account-keeping} = 30 \times 5 = \$150\)

\(\text{Weekly repayments} = 55 \times 20 = \$1100\)

\(\text{Total} = 75+150+1100 = \$1325\)
  

b.    \(\text{Option 1 is cheaper.}\)

\(\text{Difference} = 1325-960 = \$365\)

Probability, STD2 EQ-Bank 31

A survey of 60 people found the following information about their exercise habits.

  • 35 enjoy hiking \((H)\)
  • 28 enjoy cycling \((C)\)
  • 8 enjoy neither hiking nor cycling
  1. Draw a Venn diagram to represent this information.   (2 marks)

  2. One person is selected at random. What is the probability that the person enjoys hiking or cycling?   (1 mark)

Show Answers Only

a.    

b.    \(\dfrac{13}{15}\)

Show Worked Solution

a.    \(\text{Number in either }H\text{ or }C = 60-8 = 52\)

\(\text{Number in both }H\text{ and }C = 35+28-52 = 11\)

\(H\text{ only} = 35-11 = 24\)

\(C\text{ only} = 28-11 = 17\)
  

b.    \(P(H \text{ or } C) = \dfrac{24+11+17}{60} = \dfrac{52}{60} = \dfrac{13}{15}\)

\(\text{OR, using the complement:}\)

\(P(H \text{ or } C) = 1-\dfrac{8}{60} = \dfrac{13}{15}\)

Probability, STD2 EQ-Bank 5 MC

The Venn diagram shows information about 40 people surveyed about whether they own a dog \((D)\) or a cat \((C)\).

Which two-way table correctly represents the information in the Venn diagram?

A.      \(\begin{array}{|l|c|c|c|} \hline & D & \text{Not }D & \text{Total} \\ \hline C & 12 & 8 & 20 \\ \hline \text{Not }C & 5 & 15 & 20 \\ \hline \text{Total} & 17 & 23 & 40 \\ \hline \end{array}\) B.    \(\begin{array}{|l|c|c|c|} \hline & D & \text{Not }D & \text{Total} \\ \hline C & 8 & 12 & 20 \\ \hline \text{Not }C & 15 & 5 & 20 \\ \hline \text{Total} & 23 & 17 & 40 \\ \hline \end{array}\)
C.    \(\begin{array}{|l|c|c|c|} \hline & D & \text{Not }D & \text{Total} \\ \hline C & 15 & 12 & 27 \\ \hline \text{Not }C & 8 & 5 & 13 \\ \hline \text{Total} & 23 & 17 & 40 \\ \hline \end{array}\) D.    \(\begin{array}{|l|c|c|c|} \hline & D & \text{Not }D & \text{Total} \\ \hline C & 5 & 12 & 17 \\ \hline \text{Not }C & 15 & 8 & 23 \\ \hline \text{Total} & 20 & 20 & 40 \\ \hline \end{array}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Reading the Venn diagram:}\)

\(\text{C and D (intersection)} = 8\)

\(\text{C only (not D)} = 12\)

\(\text{D only (not C)} = 15\)

\(\text{Neither} = 5\)
  

\(\text{In the two-way table:}\)

\(\text{Row }C\text{: }\ D\text{ column} = 8,\ \text{Not }D\text{ column} = 12,\ \text{Total} = 20\)

\(\text{Row Not }C\text{: }\ D\text{ column} = 15,\ \text{Not }D\text{ column} = 5,\ \text{Total} = 20\)
  

\(\Rightarrow B\)

Probability, STD2 EQ-Bank 2 MC

The Venn diagram shows information about 40 students and the subjects they study.
  

One student is selected at random.

What is the probability that the student studies Art but not Music?

  1. \(\dfrac{1}{8}\)
  2. \(\dfrac{1}{4}\)
  3. \(\dfrac{3}{8}\)
  4. \(\dfrac{1}{2}\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Number studying Art but not Music} = 15\)

\(\text{Total students} = 15+5+10+10=40\)

\(P(\text{Art but not Music}) = \dfrac{15}{40} = \dfrac{3}{8}\)
  

\(\Rightarrow C\)

Algebra, STD2 A4 EQ-Bank 27

SunPower Solutions is a business that installs solar panels. Fixed costs are $1200. Each panel costs $150.00 to install and generates revenue of $350.00.

The spreadsheet below models the business's costs and revenue for different numbers of panels installed.
  


  
  1. Calculate the spreadsheet values for the installation of 4 panels (cells B9, C9, D9) and 6 panels (cells B10, C10, D10).   (2 marks)
  2. Using the spreadsheet, identify the break-even point and explain what it means for the business.   (2 marks)
  3. SunPower Solutions has received a large order that will see them make a profit of $4200. Calculate the number of solar panels \((x)\) they will be installing.   (2 marks)
Show Answers Only

a.    \(\text{4 panels: TC (B9) = \$1800.00}\)

\(\text{Revenue (C9) = \$1400.00, Profit/Loss (D9) = }-\$400.00\)

\(\text{6 panels: TC (B10) = \$2100.00}\)

\(\text{Revenue (C10) = \$2100.00, Profit/Loss (D10) = \$0.00}\)

b.    \(\text{Break-even = 6 panels, See worked solution}\)

c.    \(27 \text{ panels}\)

Show Worked Solution

a.    \(\text{4 panels:}\)

\(\text{Total Cost (B9)} = \$1200+4\times\$150 = \$1800.00\)

\(\text{Revenue (C9)} = 4\times\$350 = \$1400.00\)

\(\text{Profit/Loss (D9)} = \$1400.00-\$1800.00 = -\$400.00\)

\(\text{6 panels:}\)

\(\text{Total Cost (B10)} = \$1200+6\times\$150 = \$2100.00\)

\(\text{Revenue (C10)} = 6\times\$350 = \$2100.00\)

\(\text{Profit/Loss (D10)} = \$2100.00-\$2100.00 = \$0.00\)
  

b.    \(\text{When 6 panels are installed:}\)

\(\text{Revenue = Total costs = \$2100.00  (breakeven)}\)

\(\text{This is the point at which the business covers all of its costs and}\)

\(\text{begins to make a profit.}\)

\(\text{OR}\)

\(\text{If fewer than 6 panels are installed the business will make a loss.}\)
  

c.    \(\text{Let } x = \text{the number of solar panels to be installed.}\)

\(\text{Profit}\) \( = \text{Revenue}-\text{Total costs}\)
\(4200\) \( = 350x-(1200+150x)\)
\(4200\) \(= 200x-1200\)
\(5400\) \(=200x\)
\(x\) \(=27\)

  
\(\text{SunPower Solutions will install 27 solar panels.}\)

Algebra, STD2 A4 EQ-Bank 16

Harmony Arts Festival is a cultural event with fixed costs of $510. Each ticket costs $8.00 to provide and sells for $25.00.

The spreadsheet below models the festival's costs and revenue for different numbers of tickets sold.
  


  
  1. Calculate the values for cells C12 and D12 in the spreadsheet.   (2 marks)
  2. Using the spreadsheet, identify the break-even point and explain what it means for the festival.   (2 marks)
Show Answers Only

a.    \(\text{C12} = \$1250.00 \quad \text{D12} = \$340.00\)

b.    \(\text{See worked solution}\)

Show Worked Solution

a.    \(\text{C12: Revenue} = 50 \times \$25.00 = \$1250.00\)

\(\text{D12: Profit/Loss} = \$1250.00-\$910.00 = \$340.00\)
 

b.    \(\text{When 30 tickets sold:}\)

\(\text{Revenue = Total costs = \$750.00  (Breakeven)}\)

\(\text{This is the point at which the festival covers all of its costs}\)

\(\text{and begins to make a profit.}\)

\(\text{OR}\)

\(\text{If less than 30 tickets are sold the festival will make a loss.}\)

Algebra, STD2 A4 EQ-Bank 3 MC

Beachside Brew is a cafe with fixed weekly costs of $200. Each cup of coffee costs $1.50 to make and sells for $4.00.

The spreadsheet below models the cafe's weekly costs and revenue for different numbers of cups sold.
  

How many cups of coffee must be sold each week to break even?

  1. 60
  2. 80
  3. 100
  4. 160
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Break-even occurs when Revenue = Total costs}\ \rightarrow\ \text{(i.e. Profit = \$0)}\)

\(\text{From the spreadsheet, Revenue = Total costs (\$320.00)}\)

\(\text{when 80 cups are sold.}\)

\(\Rightarrow B\)

Statistics, EXT1 EQ-Bank 25

In a large school, the average amount of money spent per student per day at the canteen is $8 with a standard deviation of 6.5 .

At the end of each day, 50 randomly chosen students are asked how much they spent at the canteen on that day.

Use the standard normal distribution table (included) to find the probability that the sample mean on a particular day is greater than $10.    (3 marks)

Show Answers Only

\(1.46 \%\)

Show Worked Solution

\(\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{Central Limit Theorem applies}\)

\(\text{The sample mean,}\ \overline{X}, \text{for random samples of size 50 is}\)

\(\text{approximately normally distributed, where:}\)

\(\mu=8\ \ \text{and}\ \ \sigma=\dfrac{6.5}{\sqrt{n}}=\dfrac{6.5}{\sqrt{50}}\)
 

\(\text{By the central limit theorem:}\)

\(Z=\dfrac{\overline{X}-8}{\frac{6.5}{\sqrt{50}}} \sim N(0,1)\)

\(Z=\dfrac{10-8}{\frac{6.5}{\sqrt{50}}}=2.18 \ \text{(2 d.p.)}\)
 

\(\text{Using Normal Distribution Table of Values:}\)

\(P(\overline{X} \geq 10)\) \(=1-P(Z \leq 2.18)\)
  \(=1-0.9854\)
  \(=1.46 \%\)

Vectors, EXT1 EQ-Bank 34

The position vector of a particle at time \(t\) is given by  \(\mathbf{r}(t)=n e^{-2 t}\,\mathbf{i}-t^2\,\mathbf{j}\), where \(n\) is a positive constant.

Determine \(n\) if the particle's acceleration is perpendicular to its velocity when  \(t=\dfrac{1}{2}\).   (3 marks)

Show Answers Only

\(n=\dfrac{e}{2}\)

Show Worked Solution
\(\underset{\sim}{r}(t)\) \(=n e^{-2 t} \underset{\sim}{i}-t^2 \underset{\sim}{j}\)
\(\underset{\sim}{v}(t)\) \(=-2n e^{-2t} \underset{\sim}{i}-2 t\underset{\sim}{i} \ \ \Rightarrow \ \ v\left(\frac{1}{2}\right)=-2 ne^{-1} \underset{\sim}{i}-\underset{\sim}{j}\)
\(\underset{\sim}{a}(t)\) \(=4 ne^{-2t} \underset{\sim}{i}-2 \underset{\sim}{j} \ \ \Rightarrow \ \ a\left(\frac{1}{2}\right)=4ne^{-1} \underset{\sim}{i}-2 \underset{\sim}{j}\)
 

\(\text{Velocity}\perp \text{acceleration at}\ \  t=\dfrac{1}{2}:\)

   \(\displaystyle \binom{-\tfrac{2 n}{e}}{-1}\binom{\tfrac{4 n}{e}}{-2}=0\)

\(-\dfrac{8 n^2}{e^2}+2=0 \ \ \Rightarrow \ \ n^2=\dfrac{e^2}{4} \ \ \Rightarrow \ \ n=\dfrac{e}{2}\ \ (n\gt 0)\)

Vectors, EXT1 EQ-Bank 26

The position vector of a particle that is moving along a curve at time `t` is given by 

`\mathbf{r}(t) = 3 cos (t) \mathbf{i} + 4 sin (t) \mathbf{j}, \ t >= 0`.

Determine the first time when the speed of the particle is a minimum.   (3 marks)

Show Answers Only

`t_1 = pi/2`

Show Worked Solution

`underset ~v(t)= -3 sin(t) underset ~i + 4 cos (t) underset ~j`

`text{Since speed =}\ |underset ~v(t)|:`

`|underset ~v(t)|` `= sqrt (9 sin^2(t) + 16 cos^2(t))`
  `= sqrt (9 sin^2(t) + 9 cos^2(t) + 7 cos^2(t))`
  `= sqrt (9 + 7 cos^2(t))`

 
`text(Minimised speed occurs when)\ cos(t) = 0:`

`:.  t_1 = pi/2`

COMMENT:
Undertilde notation in answers is allowed even when boldface vector notation appears in the question.

Vectors, EXT1 EQ-Bank 4 MC

The position of a body is given by  `\mathbf{r} = 3\mathbf{i} + \mathbf{j} `  metres at a particular time. The body moves with constant velocity and two seconds later its displacement is  `−\mathbf{i} + 5\mathbf{j} `  metres.

The velocity, in m s−1, of the body is

  1. `2\mathbf{i} + 6\mathbf{j}`
  2. `−2\mathbf{i} + 2\mathbf{j}`
  3. `−4\mathbf{i} + 4\mathbf{j}`
  4. `4\mathbf{i}-4\mathbf{j}`
Show Answers Only

`B`

Show Worked Solution

`Delta \mathbf{r}= (−1-3)\mathbf{i} + (5-1)\mathbf{j}= −4\mathbf{i} + 4\mathbf{j}`

`\mathbf{v} = (Delta \mathbf{r})/(Delta \mathbf{t})= (−4\mathbf{i} + 4\mathbf{j})/(2-0)=-2\mathbf{i}+\mathbf{j}`

`=> B`

Vectors, EXT1 EQ-Bank 5 MC

The acceleration vector of a particle that starts from rest is given by

`underset ~a(t) = −4 sin(2t) underset ~i + 20 cos (2t) underset ~j`, where `t >= 0`.

The velocity vector of the particle, `underset ~v(t)`, is given by

  1. `−8 cos(2t) underset ~i-40 sin(2t) underset ~j`
  2. `2 cos(2t) underset ~i + 10 sin(2t) underset ~j`
  3. `(8-8 cos(2t)) underset ~i-40 sin(2t) underset ~j`
  4. `(2 cos(2t)-2) underset ~i + 10 sin(2t) underset ~j`
Show Answers Only

`D`

Show Worked Solution

`underset ~v(t)= int underset ~a (t)\ dt= (2 cos (2t) + c_0) underset ~i + (10 sin (2t) + c_1) underset ~j`

`text(S)text(ince)\ \ v=0\ \ text(when)\ \ t=0:`

`0= (2 cos (0) + c_0) underset ~i + (10 sin (0) + c_1) underset ~j`

`0=(2 + c_0) underset ~i + c_1 underset ~j`

`=> c_0 = -2, \ \  c_1 = 0`
 

`:. underset ~v(t) = (2 cos (2t)-2) underset ~i + 10 sin(2t) underset ~j`

`=> D`

Vectors, EXT1 EQ-Bank 40

Determine the component of  \(\textbf{a} = 2\textbf{i}-\textbf{j} + 3\textbf{k}\)  that is perpendicular to  \(\textbf{b} = \textbf{i} + \textbf{j}-\textbf{k}.\)   (3 marks)

Show Answers Only

\(\textbf{a}-\operatorname{proj}_{\textbf{b}}\textbf{a}=\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right)+\dfrac{2}{3}\left(\begin{array}{c}1 \\ 1 \\ -1\end{array}\right)=\left(\begin{array}{c}2 \frac{2}{3} \\ -\frac{1}{3} \\ 2 \frac{1}{3}\end{array}\right)\)

Show Worked Solution

\(\textbf{a}=\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right), \ \textbf{b}=\left(\begin{array}{c}1 \\ 1 \\ -1\end{array}\right)\)

\(\textbf{a} \cdot \textbf{b}=2-1-3=-2\)

\(\abs{\textbf{b}}^2=1^2+1^2+(-1)^2=3\)
 

\(\text{Projection of} \ \textbf{a} \ \text{in the direction of} \  \textbf{b}\):

\(\operatorname{proj}_{\textbf{b}} \textbf{a}=\left(\dfrac{\textbf{a} \cdot \textbf{b}}{\abs{\textbf{b}}^2}\right) \textbf{b}=-\dfrac{2}{3}\left(\begin{array}{c}1 \\ 1 \\ -1\end{array}\right)\)
 

\(\text {Component of} \ \textbf{a} \ \text {that is perpendicular to} \ \textbf{b}\):

\(\textbf{a}-\operatorname{proj}_{\textbf{b}}\textbf{a}=\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right)+\dfrac{2}{3}\left(\begin{array}{c}1 \\ 1 \\ -1\end{array}\right)=\left(\begin{array}{c}2 \frac{2}{3} \\ -\frac{1}{3} \\ 2 \frac{1}{3}\end{array}\right)\)

Vectors, EXT1 EQ-Bank 8 MC

If  \(\underset{\sim}{u}=2 \underset{\sim}{i}-2 j+\underset{\sim}{k}\)  and  \(\underset{\sim}{v}=3 \underset{\sim}{i}-6 j+2 \underset{\sim}{k}\), the projection of \(\underset{\sim}{v}\) onto \(\underset{\sim}{u}\) is

  1. \(\dfrac{20}{49}(3 \underset{\sim}{i}-6 \underset{\sim}{j}+2 \underset{\sim}{k})\)
  2. \(\dfrac{20}{3}(2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k})\)
  3. \(\dfrac{20}{7}(3 \underset{\sim}{i}-6 \underset{\sim}{j}+2 \underset{\sim}{k})\)
  4. \(\dfrac{20}{9}(2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k})\)
Show Answers Only

\(D\)

Show Worked Solution

\(\underset{\sim}{u}=\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right), \ \ \underset{\sim}{v}=\left(\begin{array}{c}3 \\ -6 \\ 2\end{array}\right)\)

\(\underset{\sim}{u} \cdot \underset{\sim}{v}=6+12+2=20\)

\(\abs{\underset{\sim}{u}}^2=2^2+(-2)^2+1^2=9\)

\(\operatorname{proj}_{\underset{\sim}{u}} \underset{\sim}{v}=\left(\dfrac{\underset{\sim}{u} \cdot \underset{\sim}{v}}{\abs{\underset{\sim}{u}}^2}\right) \underset{\sim}{u}=\dfrac{20}{9}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)\)

\(\Rightarrow D\)

Vectors, EXT1 EQ-Bank 23

An object is travelling around a circular path of radius 5 m with position vector

\(\textbf{r} (t)=5 \cos \left(t^2\right)\textbf{i} +5 \sin \left(t^2\right)\textbf{j} \quad t \geq 0\)

where \(t\) is the time in seconds.

Find an expression for the speed of the object in terms of \(t\).   (2 marks)

Show Answers Only

\(\abs{\textbf{v} (t)}=10 t \ \text{ms}^{-1}\)

Show Worked Solution

\(\text {The velocity is given by}\)

\(\textbf{v}(t)=\dfrac{d}{d t} \textbf{r}(t)=-10 t\, \sin \left(t^2\right) \textbf{i} +10 t\, \cos \left(t^2\right) \textbf{j}\)

\(\text {The speed is the magnitude of the velocity:}\)

\(\abs{\textbf{v} (t)}\) \(=\sqrt{100 t^2\, \sin ^2\left(t^2\right)+100 t^2\, \cos ^2\left(t^2\right)}\)
  \(=10 t \sqrt{\sin ^2\left(t^2\right)+\cos ^2\left(t^2\right)}\)
  \(=10 t \ \text{m s}^{-1}\)

Vectors, EXT1 EQ-Bank 3 MC

Given that  \(\overrightarrow{OP}=\left(\begin{array}{c}-3 \\ 1 \\ -1\end{array}\right)\)  and  \(\overrightarrow{O Q}=\left(\begin{array}{c}2 \\ 5 \\ -3\end{array}\right)\), what is \(\overrightarrow{P Q}\) ?

  1. \(\left(\begin{array}{c}1 \\ -6 \\ 4\end{array}\right)\)
  2. \(\left(\begin{array}{c}-1 \\ 6 \\ -4\end{array}\right)\)
  3. \(\left(\begin{array}{c}5 \\ 4 \\ -2\end{array}\right)\)
  4. \(\left(\begin{array}{c}-5 \\ -4 \\ 2\end{array}\right)\)
Show Answers Only

\(C\)

Show Worked Solution

\(\overrightarrow{PQ}=\overrightarrow{O Q}-\overrightarrow{O P}=\left(\begin{array}{c}2 \\ 5 \\ -3\end{array}\right)-\left(\begin{array}{c}-3 \\ 1 \\ -1\end{array}\right)=\left(\begin{array}{c}5 \\ 4 \\ -2\end{array}\right)\)

\(\Rightarrow C\)

Vectors, EXT1 EQ-Bank 38

Let \(\underset{\sim}{a}=2 \underset{\sim}{i}-3 j+\underset{\sim}{k}\) and \(\underset{\sim}{b}=\underset{\sim}{i}+m j-\underset{\sim}{k}\), where \(m\) is an integer.

The vector resolute of \(\underset{\sim}{a}\) in the direction of \(\underset{\sim}{b}\) is \(-\dfrac{11}{18}(\underset{\sim}{i}+m\underset{\sim}{j}-\underset{\sim}{k})\).

  1. Find the value of  \(m\).   (3 marks)

  2. Find the component of \(\underset{\sim}{a}\) that is perpendicular to \(\underset{\sim}{b}\).   (1 mark)

Show Answers Only

a.    \(m=4\)

b.    \(\left(\begin{array}{c}2 \frac{11}{18} \\ -\frac{5}{9} \\ \frac{7}{18}\end{array}\right)\)

Show Worked Solution

a.    \(\underset{\sim}{a}=\left(\begin{array}{c}2 \\ -3 \\ 1\end{array}\right), \quad \underset{\sim}{b}=\left(\begin{array}{c}1 \\ m \\ -1\end{array}\right)\)

\(\underset{\sim}{b} \cdot \underset{\sim}{a}=2-3 m-1=1-3 m\)

\(\abs{\underset{\sim}{b}}=\sqrt{1^2+m^2+(-1)^2}=\sqrt{2+m^2}\)

\(\operatorname{proj}_{\underset{\sim}{b}}\underset{\sim}{a}=\left(\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{b}^2}\right) \underset{\sim}{b}=\dfrac{1-3 m}{2+m^2}\, \underset{\sim}{b}\)
 

\(\text{Equating projection vectors:}\)

\(\dfrac{1-3 m}{m^2+2}\) \(=-\dfrac{11}{18}\)  
\(18-54 m\) \(=-11 m^2-22\)  
\(11 m^2-54 m+40\) \(=0\)  
\((11 m-10)(m-4)\) \(=0\)  

 
\(\therefore m=4\ \left(m \neq \frac{10}{11}, m \in Z\right)\)
 

b.    \(\text{Component of \(\underset{\sim}{a}\) perpendicular to \(\underset{\sim}{b}\):}\)

\(\underset{\sim}{a}-\operatorname{proj}_{\underset{\sim}{b}} \underset{\sim}{a}=\left(\begin{array}{c}2 \\ -3 \\ 1\end{array}\right)+\dfrac{11}{18}\left(\begin{array}{c}1 \\ 4 \\ -1\end{array}\right)=\left(\begin{array}{c}2 \frac{11}{18} \\ -\frac{5}{9} \\ \frac{7}{18}\end{array}\right)\)

Vectors, EXT1 2024 SPEC1 4

Consider the vectors  \(\underset{\sim}{ a }=3 \underset{\sim}{ j }+3 \underset{\sim}{ k }\)  and  \(\underset{\sim}{ b }=2 \underset{\sim}{ i }-\underset{\sim}{ j }-2 \underset{\sim}{ k }\).

Find the angle between \(\underset{\sim}{ a }\) and \(\underset{\sim}{ b }\).   (2 marks)

Show Answers Only

\(\theta=\dfrac{3 \pi}{4}\left(\text{or} \ 135^{\circ}\right)\)

Show Worked Solution

\(\underset{\sim}{a}=\left(\begin{array}{l}0 \\ 3 \\ 3\end{array}\right) \Rightarrow \abs{\underset{\sim}{a}}=\sqrt{18}=3 \sqrt{2}\)

     \(\underset{\sim}{b}=\left(\begin{array}{c}2 \\ -1 \\ -2\end{array}\right) \Rightarrow\abs{\underset{\sim}{b}}=\sqrt{9}=3\)

     \(\cos \theta=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{\underset{\sim}{a}} \cdot \abs{\underset{\sim}{b}}}=\dfrac{-3-6}{3 \sqrt{2} \times 3}=-\dfrac{1}{\sqrt{2}}\)

    \(\therefore \theta=\cos ^{-1}\left(-\dfrac{1}{\sqrt{2}}\right)=\dfrac{3 \pi}{4}\left(\text{or} \ 135^{\circ}\right)\)

Vectors, EXT1 2014 SPEC1 1

Consider the vector  `underset ~a = sqrt 3 underset ~i-underset ~j-sqrt 2 underset ~k`, where `underset ~i, underset ~j` and `underset ~k` are unit vectors in the positive directions of the `x, y` and `z` axes respectively.

  1. Find the unit vector in the direction of  `underset ~a`.   (1 mark)

  2. Find the acute angle that `underset ~a` makes with the positive direction of the `x`-axis.   (2 marks)

  3. The vector  `underset ~b = 2 sqrt 3 underset ~i + m underset ~j-5 underset ~k`.
  4. Given that `underset ~b` is perpendicular to `underset ~a,` find the value of `m`.  (2 marks)

Show Answers Only

a.    `1/sqrt 6 (sqrt 3 underset ~i-underset ~j-sqrt 2 underset ~k)`

b.    `theta = 45^@`

c.    `m = 6 + 5 sqrt 2`

Show Worked Solution

a.    `|underset ~a|= sqrt((sqrt 3)^2 + (-1)^2 + (-sqrt 2)^2)= sqrt 6`

`hat underset ~a= underset ~a/|underset ~a|= 1/sqrt 6 (sqrt 3 underset ~i-underset ~j-sqrt 2 underset ~k)`
 

b.    `x text{-axis vectors include}\ (1,0,0).`

`underset ~a ⋅ underset ~i = ((\sqrt3),(-1),(-\sqrt2))((1),(0),(0))=\sqrt3`

  `underset ~a ⋅ underset ~i` `= |underset ~a||underset ~i| cos theta= sqrt 6 cos theta`
  `sqrt 3` `= sqrt 6 cos theta`
  `cos theta` `=1/sqrt 2`
  `:. theta` `= 45^@`

 
c.   `underset ~a ⋅ underset ~b = sqrt 3 (2 sqrt 3) + (-1)(m) + (-sqrt 2)(-5) = 0`

`6-m + 5 sqrt 2` `=0`  
`:. m` `=6 + 5 sqrt 2`  

Vectors, EXT1* V1 2025 HSC 11d

  1. Force \({\underset{\sim}{F}}_1\) has magnitude 12 newtons in the direction of vector  \(2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}\).   
  2. Show that  \({\underset{\sim}{F}}_1=8 \underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}\).   (1 mark)

  3. Force \({\underset{\sim}{F}}_1\) from part (i) and a second force,  \({\underset{\sim}{F}}_2=-6 \underset{\sim}{i}+12 \underset{\sim}{j}+4 \underset{\sim}{k}\), both act upon a particle.
  4. Show that the resultant force acting on the particle is given by:
  5.      \({\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}.\)   (1 mark)

  6. Calculate  \({\underset{\sim}{F}}_3 \cdot \underset{\sim}{d}\), where \({\underset{\sim}{F}}_3\) is the resultant force from part (ii) and  \(\underset{\sim}{d}=\underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k}\).   (1 mark)

Show Answers Only

i.    \(\text{Unit vector of the direction vector:}\)

\(\dfrac{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}{\abs{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}} = \dfrac{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}{\sqrt{2^2+(-2)^2 + 1^2}} = \dfrac{1}{3} \left( 2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k} \right)\)
 

\(\text{Since \({\underset{\sim}{F}}_1\) has magnitude 12:}\)

\({\underset{\sim}{F}}_1=12 \times \dfrac{1}{3}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)\)

\({\underset{\sim}{F}}_1=8\underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}\)
    

ii.    \({\underset{\sim}{F}}_3={\underset{\sim}{F}}_1+{\underset{\sim}{F}}_2=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)+\left(\begin{array}{c}-6 \\ 12 \\ 4\end{array}\right)=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\)

\({\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}\)
 

iii.  \({\underset{\sim}{F}}_3 \cdot d=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=2+4+16=22\)

Show Worked Solution

i.    \(\text{Unit vector of the direction vector:}\)

\(\dfrac{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}{\abs{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}} = \dfrac{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}{\sqrt{2^2+(-2)^2 + 1^2}} = \dfrac{1}{3} \left( 2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k} \right)\)
 

\(\text{Since \({\underset{\sim}{F}}_1\) has magnitude 12:}\)

\({\underset{\sim}{F}}_1=12 \times \dfrac{1}{3}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)\)

\({\underset{\sim}{F}}_1=8\underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}\)
 

ii.    \({\underset{\sim}{F}}_3={\underset{\sim}{F}}_1+{\underset{\sim}{F}}_2=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)+\left(\begin{array}{c}-6 \\ 12 \\ 4\end{array}\right)=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\)

\({\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}\)
 

iii.  \({\underset{\sim}{F}}_3 \cdot d=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=2+4+16=22\)

Vectors, EXT1* V1 2024 HSC 11c

Find the angle between the two vectors  \(\underset{\sim}{u}=\left(\begin{array}{c}1 \\ 2 \\ -2\end{array}\right)\) and  \(\underset{\sim}{v}=\left(\begin{array}{c}4 \\ -4 \\ 7\end{array}\right)\), giving your answer in radians, correct to 1 decimal place.   (2 marks)

Show Answers Only

\(\theta=2.3^c \ \ \text{(1 d.p.)}\)

Show Worked Solution

\(\underset{\sim}{u}=\left(\begin{array}{c}1 \\ 2 \\ -2\end{array}\right),\abs{\underset{\sim}{u}}=\sqrt{1+4+4}=3\)

\(\underset{\sim}{v}=\left(\begin{array}{c}4 \\ -4 \\ 7\end{array}\right),\abs{\underset{\sim}{v}}=\sqrt{16+16+49}=9\)

\(\cos \theta=\dfrac{\underset{\sim}{u} \cdot \underset{\sim}{v}}{|\underset{\sim}{u}||\underset{\sim}{v}|}=\dfrac{1 \times 4-2 \times 4-2 \times 7}{3 \times 9}=-\dfrac{2}{3}\)

\(\theta=\cos ^{-1}\left(-\dfrac{2}{3}\right)=2.30 \ldots=2.3^c \ \ \text{(1 d.p.)}\)

Vectors, EXT1* V1 2022 HSC 11d

A triangle is formed in three-dimensional space with vertices `A(1,-1,2)`, `B(0,2,-1)`  and `C(2,1,1)`.

Find the size of `/_ABC`, giving your answer to the nearest degree.   (3 marks)

Show Answers Only

`33°`

Show Worked Solution

`vec(BA)=((1),(-1),(2))-((0),(2),(-1))=((1),(-3),(3))`

`abs(vec(BA))=sqrt(1^2+3^2+3^2)=sqrt19`
 

`vec(BC)=((2),(1),(1))-((0),(2),(-1))=((2),(-1),(2))`

`abs(vec(BC))=sqrt(2^2+1^2+2^2)=sqrt9=3`
 

`vec(BA)*vec(BC)=1xx2+ -3xx-1+3xx2=11`

`cos/_ABC=(vec(BA)*vec(BC))/(abs{vec(BA)}abs{vec(BC)})=11/(3sqrt19)`

`:./_ABC=cos^(-1)(11/(3sqrt19))=32.733…=33°\ \ text{(nearest degree)}`

Vectors, EXT1* V1 2020 HSC 11d

Consider the two vectors  `underset~u = 2 underset~i-underset~j + 3 underset~k`  and  `underset~v = p underset~i +  underset~j + 2 underset~k`.
 
For what values of `p` are  `underset~u-underset~v`  and  `underset~u + underset~v`  perpendicular?   (3 marks)

Show Answers Only

`p= ± 3`

Show Worked Solution

`underset~u-underset~v = ((-2),(-1),(3))-((p),(1),(2)) = ((-2-p),(-2),(1))`
 

`underset~u + underset~v = ((-2),(-1),(3)) + ((p),(1),(2)) = ((p-2),(0),(5))`
 

`⊥ \ text{when} \ \ (underset~u-underset~v) · (underset~u + underset~v ) = 0 :`
 

`((-2-p),(-2),(1)) · ((p-2),(0),(5)) = 0`
 

`-(p + 2)(p-2) + 5` `= 0`
`-(p^2-4) + 5` `= 0`
`-p^2 + 9` `= 0`
`p^2` `= 9`
`p` `= ± 3`

Vectors, EXT1* V1 2024 HSC 12a

The vector \(\underset{\sim}{a}\) is \(\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)\) and the vector \(\underset{\sim}{b}\) is \(\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)\).

  1. Find \(\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}\).   (1 mark)

  2. Show that  \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}\)  is perpendicular to \(\underset{\sim}{b}\).   (2 marks)

Show Answers Only

i.     \(\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)\)

ii.    \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\)

\( \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0\)

\(\therefore\ \text {Vectors are perpendicular.}\)

Show Worked Solution

i.    \(\underset{\sim}{a}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right), \quad \underset{\sim}{b}=\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)\)
 

\(\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\dfrac{2+0-12}{4+0+16}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=-\dfrac{1}{2}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)\)

 
ii.    \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\)
 

\( \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\,\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0\)

 
\(\therefore\ \text{Vectors are perpendicular.}\)

Vectors, EXT1* V1 2023 HSC 11b

Find the angle between the vectors

\(\underset{\sim}{a}=\underset{\sim}{i}+2 \underset{\sim}{j}-3 \underset{\sim}{k}\)

\(\underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}+2 \underset{\sim}{k}\),

giving your answer to the nearest degree.   (3 marks)

Show Answers Only

\(87^{\circ} \)

Show Worked Solution

\[\underset{\sim}{a}=\left(\begin{array}{c} 1 \\ 2 \\ -3 \end{array}\right),\ \  \underset{\sim}{b}=\left(\begin{array}{c} -1 \\ 4 \\ 2 \end{array}\right) \]

\(\Big{|} \underset{\sim}{a} \Big{|} = \sqrt{1+4+9} = \sqrt{14} \)

\(\Big{|} \underset{\sim}{b} \Big{|} = \sqrt{1+16+4} = \sqrt{21} \)

\( \underset{\sim}{a} \cdot \underset{\sim}{b} = -1 + 8-6=1 \)

\(\cos\ \theta \) \(=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\Big{|}\underset{\sim}{a}\Big{|} \cdot \Big{|}\underset{\sim}{b}\Big{|}} \)  
  \(=\dfrac{1}{\sqrt{294}} \)  
\( \theta\) \(=\cos ^{-1} \Big{(}\dfrac{1}{\sqrt{294}}\Big{)} \)  
  \(=86.65…\)  
  \(=87^{\circ} \)  

Vectors, EXT1* V1 2021 HSC 11c

Find the angle between the vectors  `underset~a = ((2),(0),(4))`  and  `underset~b = ((-3),(1),(2))`, giving the angle in degrees correct to 1 decimal place.   (3 marks)

Show Answers Only

`83.1^@`

Show Worked Solution

`underset~a = ((2),(0),(4)) \ , \ |underset~a| \ = sqrt{2^2 + 4^2} = sqrt20`

`underset~b = ((-3),(1),(2)) \ , \ |underset~b| \ = sqrt{(-3)^2 + 1^2 + 2^2} = sqrt14`

`underset~a * underset~b` `= ((2),(0),(4)) ((-3),(1),(2)) = – 6 + 0 + 8 = 2`
`underset~a * underset~b` `= |underset~a| |underset~b| \ cos theta`
`2` `= sqrt20 sqrt14 \ cos theta`
`cos theta` `= 2/sqrt280`
`theta` `= cos^(-1) (1/sqrt70)`
  `= 83.1^@ \ text{(1 d.p,)}`

Vectors, EXT1 EQ-Bank 20

Two vectors are given by  `underset ~a = 4 underset ~i + m underset ~j - 3 underset ~k`  and  `underset ~b = −2 underset ~i + n underset ~j - underset ~k`, where `m`, `n in R^+`.

If  `|\ underset ~a\ | = 10`  and `underset ~a` is perpendicular to `underset ~b`, determine the exact values of `m` and `n`.   (3 marks)

Show Answers Only

`m=5sqrt3, \ n=\sqrt{3}/3`

Show Worked Solution

`text(Using)\ \ |\ underset ~a\ | = 10:`

`10` `= sqrt(4^2 + m^2 + (-3)^2)`
`100` `=m^2+75`
`m^2` `= 25`
`m` `=5sqrt3\ \ (m in R^+)`

 

`text(S)text(ince)\ \ underset ~a _|_ underset ~b\ \ =>\ \ underset ~a xx underset ~b=0`

`0` `=4 xx (−2) + mn + (−3) xx (−1)`
`0` `=n xx 5sqrt3-5`
`n` `=5/(5\sqrt{3})`
`n` `=1/\sqrt{3}=\sqrt{3}/3`