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v1 Algebra, STD2 A1 2010 HSC 24a

Margie tried to solve this equation and made a mistake in Line 2. 

\begin{array}{rl}
3(m+3)-2(m+4)=-5\ &\ \ \ \text{Line 1} \\
3m+9-2m+8=-5\ &\ \ \ \text{Line 2} \\
m+17=-5\ &\ \ \ \text{Line 3} \\
m=-12& \ \ \ \text{Line 4}
\end{array}

  1. Copy the equation in Line 1. Rewrite Line 2 correcting her mistake.   (1 mark)

  2. Continue your solution showing the correct working for Lines 3 and 4 to solve this equation for \(m\).   (1 mark)

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i.    \(3(m+3)-2(m+4)\) \(=-5\ \ \ \ \ \ \text{Line}\ 1\)
  \(3m+9-2m-8\) \(=-5\ \ \ \ \ \ \text{Line}\ 2\)
ii.    \(m+1\) \(=-5\)
  \(m\) \(=-6\)
Show Worked Solution
i.    \(3(m+3)-2(m+4)\) \(=-5\ \ \ \ \ \ \text{Line}\ 1\)
  \(3m+9-2m-8\) \(=-5\ \ \ \ \ \ \text{Line}\ 2\)

 

ii.    \(m+1\) \(=-5\ \ \ \ \ \ \text{Line}\ 3\)
  \(m\) \(=-6\ \ \ \ \ \ \text{Line}\ 4\)

v1 Algebra, STD2 A1 2013 HSC 29a

Jeremy tried to solve this equation and made a mistake in Line 2. 

\(\dfrac{M+3}{2}-\dfrac{2M-1}{5}\) \(=1\) \(\text{... Line 1}\)
\(5M+15-4M-2\) \(=10\) \(\text{... Line 2}\)
\(M+13\) \(=10\) \(\text{... Line 3}\)
\(M\) \(=-3\) \(\text{... Line 4}\)

  
Copy the equation in Line 1 and continue your solution to solve this equation for \(M\).

Show all lines of working.   (2 marks)

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\(\dfrac{M+3}{2}-\dfrac{2M-1}{5}\) \(=1\) `text(… Line 1)`
\(5M+15-4M+2\) \(=10\) `text(… Line 2)`
\(M+17\) \(=10\) `text(… Line 3)`
\(M\) \(=-7\) `text(… Line 4)`
Show Worked Solution
♦♦ Mean mark 27%
STRATEGY: The RHS of the equation increases from 1 to 10 (from Line 1 to Line 2), indicating both sides must have been multiplied by 10.
\(\dfrac{M+3}{2}-\dfrac{2M-1}{5}\) \(=1\) `text(… Line 1)`
\(5M+15-4M+2\) \(=10\) `text(… Line 2)`
\(M+17\) \(=10\) `text(… Line 3)`
\(M\) \(=-7\) `text(… Line 4)`

v1 Algebra, STD2 A1 2007 HSC 24b

The distance in kilometres (\(D\)) of an observer from the centre of a thunderstorm can be estimated by counting the number of seconds (\(t\)) between seeing the lightning and first hearing the thunder.

Use the formula  \(D=\dfrac{t}{3}\)  to estimate the number of seconds between seeing the lightning and hearing the thunder if the storm is 2.1 km away.   (1 mark)

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\(6.3\ \text{seconds}\)

Show Worked Solution

\(D=\dfrac{t}{3}\)

\(\text{When}\ \ D = 2.1,\)

\(\dfrac{t}{3}\) \(=2.1\)
\(t\) \(=6.3\ \text{seconds}\)

v1 Algebra, STD2 A1 2007 HSC 28b

This shape is made up of a right-angled triangle and a regular hexagon.
 

The area of a regular hexagon can be estimated using the formula  \(A=2.598H^2\)  where  \(H\)  is the side-length.

Calculate the total area of the shape using this formula.   (3 marks)

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\(51.264\ \text{cm}^2\)

Show Worked Solution

\(\text{Area}=2.598H^2\)

\(\text{Using Pythagoras}\)

\(H^2\) \(= 3^2+3^2\)
  \(=18\)
\(H\) \(=\sqrt{18}\)
\(\therefore\ \text{Area of hexagon}\) \(=2.598\times {\sqrt {18}}\ ^2\)
  \(=46.764\ \text{cm}^2\)
\(\text{Area of triangle}\) \(=\dfrac{1}{2}bh\)
  \(=\dfrac{1}{2}\times 3\times 3\)
  \(=4.5\ \text{cm}^2\)

 

\(\therefore\ \text{Total Area}\) \(=46.764+4.5\)
  \(=51.264\ \text{cm}^2\)

EXAMCOPY Algebra, STD2 A1 2004 HSC 11 MC

If  `d = 6t^2`, what is a possible value of  `t`  when  `d = 2400`?

  1. `0.05`
  2. `20`
  3. `120`
  4. `400`
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`B`

Show Worked Solution
`d` `= 6t^2`
`t^2` `= d/6`
`t` `= +- sqrt(d/6)`

 
`text(When)\ \ d = 2400:`

`t` `= +- sqrt(2400/6)`
  `= +- 20`

 
`=> B`

v1 Algebra, STD2 A1 2005 HSC 14 MC

Using the formula  \(d=6t^3-5\), Marcia tried to find the value of  \(t\)  when \(d=389\).

Here is her solution. She has made one mistake.
 

Which line does NOT follow correctly from the previous line?

  1. \(\text{Line}\ A\)
  2. \(\text{Line}\ B\)
  3. \(\text{Line}\ C\)
  4. \(\text{Line}\ D\)
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\(B\)

Show Worked Solution
\(d\) \(=6t^3-5\)  
\(389\) \(=6t^3-5\ \ \ \) \(\dots\text{ Line A}\)
\(394\) \(=6t^3\) \(\dots\text{ Line B}\)

  
\(\therefore\ \text{Line}\ B\ \text{does not follow on correctly.}\)

\(\Rightarrow B\)

v1 Algebra, STD2 A1 2015 HSC 24 MC

Consider the equation  \(\dfrac{5x}{2}-3=\dfrac{3x}{5}+1\).

Which of the following would be a correct step in solving this equation?

  1. \(\dfrac{5x}{2}-2=\dfrac{3x}{5}\)
  2. \(\dfrac{10x}{4}-4=\dfrac{6x}{5}\)
  3. \(\dfrac{5x}{2}=\dfrac{3x}{5}+4\)
  4. \(5x-3=\dfrac{6x}{5}+2\)
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\(C\)

Show Worked Solution
\(\dfrac{5x}{2}-3\) \(=\dfrac{3x}{5}+1\)
\(\dfrac{5x}{2}-3+3\) \(=\dfrac{3x}{5}+1+3\)
\(\dfrac{5x}{2}\) \(=\dfrac{3x}{5}+4\)

 
\(\Rightarrow C\)

v1 Algebra, STD2 A1 2015 HSC 28d

The formula  \(C=\dfrac{5}{9}(F-32)\)  is used to convert temperatures between degrees Fahrenheit \((F)\) and degrees Celsius \((C)\).

Convert 18°C to the equivalent temperature in Fahrenheit.  (2 marks)

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\(64.4\ \text{degrees}\ F\)

Show Worked Solution
\(C\) \(=\dfrac{5}{9}(F-32)\)
\(F-32\) \(=\dfrac{9}{5}C\)
\(F\)  \(=\dfrac{9}{5}C+32\)

 
\(\text{When}\ \ C = 18,\)

\(F\)  \(=\dfrac{9}{5}\times 18+32\)
  \(=64.4\ \text{degrees}\ F\)

EXAMCOPY Algebra, STD2 A1 SM-Bank 9

The volume of a sphere is given by  `V = 4/3 pi r^3`  where  `r`  is the radius of the sphere.

If the volume of a sphere is  `220\ text(cm)^3`, find the radius, to 1 decimal place.  (3 marks)

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`3.7\ \ text{cm  (to 1 d.p.)}`

Show Worked Solution
`V` `= 4/3 pi r^3`
`3V` `= 4 pi r^3`
`r^3` `= (3V)/(4 pi)`

 

`text(When)\ \ V = 220`

`r^3` `= (3 xx 220)/(4 pi)`
  `= 52.521…`
`:. r` `=root3 (52.521…)`
  `= 3.744…\ \ \ text{(by calc)}`
  `= 3.7\ \ text{cm   (to 1 d.p.)}`

v1 Algebra, STD2 A1 SM-Bank 7

If  \(S = V_0(1 - r)^n\), find \(S\) given  \(V_0 = $57\ 000\), \(r = 0.12\) and \(n=5\). (give your answer to the nearest cent).  (2 marks)

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\($30\ 080.72\ \text{(to nearest cent)}\)

Show Worked Solution
\(S\) \(=V_0(1 – r)^n\)
  \(=57\ 000 (1-0.12)^5\)
  \(=57\ 000 (0.88)^5\)
  \(=$30\ 080.719\dots\)
  \(=$30\ 080.72\ \text{(to nearest cent)}\)

v1 Algebra, STD2 A1 2018 HSC 28b

Solve the equation  \(\dfrac{3x}{4}+1=\dfrac{5x+1}{3}\), leaving your answer as a fraction.  (3 marks)

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\(\dfrac{8}{11}\)

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♦ Mean mark 35%.

\(\underbrace{\dfrac{3x}{4} + 1}_\text{multiply x 12}\) \(=\underbrace{\dfrac{5x+1}{3}}_\text{multiply x 12}\)
\(9x+12\) \(=20x+4\)
\(11x\) \(=8\)
\(x\) \(=\dfrac{8}{11}\)

v1 Algebra, STD2 A1 SM-Bank 10

For adults (18 years and older), the Body Mass Index is given by:
  

\(B=\dfrac{m}{h^2}\),    where  \(m=\) mass in kilograms and  \(h=\) height in metres.
  

The medically accepted healthy range for  \(B\)  is  \(21\leq B\leq 25\).

What is the minimum weight for a 172 cm adult female to be considered healthy, correct to 1 decimal place? (2 marks)

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\(62.1\ \text{kg}\)

Show Worked Solution

\(B=\dfrac{m}{h^2}\)

\(h=172\ \text{cm} =1.72\ \text{m}\)
 

\(\text{Given}\ \ 21\leq B\leq 25,\)

\(\rightarrow\ B = 21\ \text{for minimum healthy weight.}\)

\(21\) \(=\dfrac{m}{1.72^2}\)
\(\therefore\ m\) \(=21\times 1.72^2\)
  \(=62.1264\)
  \(=62.1\ \text{kg}\ \text{(1 d.p.)}\)

EXAMCOPY Algebra, STD1 A1 2019 HSC 34

Given the formula  `C = (A(y + 1))/24`, calculate the value of  `y`  when  `C = 120`  and  `A = 500`.  (3 marks)

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`4.76`

Show Worked Solution

`text(Make)\ \ y\ \ text(the subject:)`

`C` `= (A(y + 1))/24`
`24C` `= A(y + 1)`
`y + 1` `= (24C)/A`
`y` `= (24C)/A-1`
  `= (24 xx 120)/500-1`
  `= 4.76`

v1 Algebra, STD1 A1 2020 HSC 18

The distance, \(d\) metres, travelled by a car slowing down from \(u\) km/h to \(v\) km/h can be obtained using the formula

\(v^2=u^2-100 d\)

What distance does a car travel while slowing down from 100 km/h to 70 km/h?   (2 marks)

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\(51\ \text{metres}\)

Show Worked Solution

\(u=100 \ , \ v=70\)

\(v^2\) \(=u^2-100d\)
\(70^2\) \(=100^2-100d\)
\(100d\) \(=100^2-70^2\)
\(\therefore\ d\) \(=\dfrac{100^2-70^2}{100}\)
  \(=51\ \text{metres}\)

v1 Algebra, STD2 A1 2004 HSC 11 MC

If  \(m = 8n^2\), what is a possible value of  \(n\)  when  \(m=7200\)?

  1. \(0.0\dot{3}\)
  2. \(30\)
  3. \(240\)
  4. \(900\)
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\(B\)

Show Worked Solution
\(m\) \(=8n^2\)
\(n^2\) \(=\dfrac{m}{8}\)
\(n\) \(=\pm\sqrt{\dfrac{m}{8}}\)

 
\(\text{When}\ m=7200:\)

\(n\) \(=\pm\sqrt{\dfrac{7200}{8}}\)
  \(=\pm 30\)

 
\(\Rightarrow B\)

v1 Algebra, STD1 A1 2019 HSC 34

Given the formula  \(D=\dfrac{B(x+1)}{18}\), calculate the value of  \(x\)  when  \(D=90\)  and  \(B=400\).  (3 marks)

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\(3.05\)

Show Worked Solution

\(\text{Make}\ x\ \text{the subject:}\)

\(D\) \(=\dfrac{B(x+1)}{18}\)
\(18D\) \(=B(x+1)\)
\(x+1\) \(=\dfrac{18D}{B}\)
\(x\) \(=\dfrac{18D}{B}-1\)
\(\text{When }\) \(D=90, B=400\)
\(\therefore\ x\) \(=\dfrac{18\times 90}{400}-1=3.05\)

v1 Algebra, STD2 A1 SM-Bank 9

The volume of a sphere is given by  \(V=\dfrac{4}{3}\pi r^3\)  where  \(r\)  is the radius of the sphere.

If the volume of a sphere is  \(385\ \text{cm}^3\), find the radius, to 1 decimal place.  (3 marks)

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\(4.5\ \text{cm  (to 1 d.p.)}\)

Show Worked Solution
\(V\) \(=\dfrac{4}{3}\pi r^3\)
\(3V\) \(= 4\pi r^3\)
\(r^3\) \(=\dfrac{3V}{4\pi}\)

 

\(\text{When}\ \ V =385\)

\(r^3\) \(=\dfrac{3\times 385}{4\pi}\)
  \(=91.911\dots\)
\(\therefore\ r\) \(=\sqrt[3]{91.911\dots}\)
  \(=4.512\dots\ \ \text{(by calc)}\)
  \(=4.5\ \text{cm   (to 1 d.p.)}\)