Title: random choice
Course: Phil's test course

Q: This is a Multiple choice question.

Estimate the gradient of the curve shown in the plot at x = 2.

[Image: Curve]

Click on the answer closest to your estimate.
- 1 (This value is too low.)
- 2 (This value is too low.)
- 3 (This value is too low.)
- 4 (The gradient is indeed 4.)
- 5 (This value is too high.)
- 6 (This value is too high.)
A: 4
Feedback: [Image: Curve with tangent] Gradient of tangent = $$\frac{\Delta y}{\Delta x} = \frac{14 - 0}{4.0 - 0.5} = \frac{14}{3.5} = 4.0$$ The gradient of a curve at any point is the same as the gradient of a tangent drawn at that point, so the gradient of the curve is 4.0 at x = 2.
  Hint: Start by imagining a tangent drawn against the curve at x = 2. Your tangent should be exactly parallel to the curve at x = 2.
  Hint: Where does your tangent, drawn at x = 2, cross the x and y axes? Estimate these points then calculate the gradient as the difference in y values divided by the difference in x values.

Q: This is a Short answer question.

There are three different ways of increasing the rate of a chemical reaction. Two of them are

- to increase the temperature
- to increase the concentration of one or more of the reactants.

What is the third way?
A: add*catalyst* / use*catalyst*
Feedback ("add*catalyst*"): Your answer is correct.
Feedback ("use*catalyst*"): Your answer is correct.
Feedback ("*catalyst*"): Your answer appears to be on the right lines but perhaps you could phrase it differently.
Feedback (any other answer): Your answer is incorrect.
Feedback: The rate of a chemical reaction can be increased by any of - adding a catalyst - increasing the temperature - increasing the concentration of one or more of the reactants.
  Hint: Every chemical reaction has an energy barrier that must be overcome before reactant molecules can form products. One way to increase the rate is to lower this barrier. How might you achieve this?
  Hint: You will find more on lowering the energy barrier in Block 8 Section 10.5. Perhaps you would care to read that now before trying again.

Q: This is an OU multiple response question.

Which four of the following expressions are equivalent to $$4y$$?
- $$\frac{2y+6y}{2}$$
- $$4\frac{y^2}{y}$$
- $$\frac{1}{2}(2y+6y)$$
- $$2(y+y)$$
- $$2y+\frac{6y}{2}$$
- $$\frac{2y^2}{y}$$
- $$\sqrt{16}y^2$$
- $$6(y-2)$$
A: $$\frac{2y+6y}{2}$$ / $$4\frac{y^2}{y}$$ / $$\frac{1}{2}(2y+6y)$$ / $$2(y+y)$$
Feedback: The following four expressions may all be simplified to $$4y$$: $$\frac{2y+6y}{2}$$ = $$\frac{8y}{2}$$ = $$4y$$; $$4\frac{y^2}{y}$$ = $$4y$$ $$\frac{1}{2}(2y+6y)$$ = $$\frac{1}{2}8y$$ = $$4y$$; $$2(y+y)$$ = $$2(2y)$$ = $$4y$$; The other four expressions simplify to give the following: $$2y+\frac{6y}{2}$$ = $$2y+3y$$ = $$5y$$; $$\frac{2y^2}{y}$$ = $$2y$$ $$\sqrt{16}y^2$$ = $$4y^2$$; $$6(y-2)$$ = $$6y - 12$$
  Hint: Start by expanding all the expressions to remove the brackets.
  Hint: For your final attempt any incorrect choices that you have made will be removed.

Q: This is a Numerical question.

Solve $$x^2 + 3x - 28 = 0$$

Enter either of the possible answers.
A: -7 / 4
Feedback ("-7"): Your answer is correct.
Feedback ("4"): Your answer is correct.
Feedback (any other answer): Your answer is incorrect.
Feedback: The quadratic equation is $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ a = 1, b = 3 and c = -28. $$x = \frac{-3 \pm \sqrt{3^2 - 4\times 1\times(-28)}}{2\times a}$$ $$x = \frac{-3 \pm \sqrt{9 + 112}}{2}$$ $$x = \frac{-3 \pm \sqrt{121}}{2}$$ $$x = \frac{-3 \pm 11}{2}$$ $$x = \frac{-14}{2}$$ or $$\frac{8}{2}$$ and the solutions are x = -7 and x = 4. We can prove these are correct by substituting the values into the original equation. $$(-7)^2 + 3\times(-7) - 28 = 49 - 21 - 28 = 0$$ $$4^2 + 3\times4 - 28 = 16 + 12 -28 = 0$$
  Hint: You may need to be reminded of the quadratic formula $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
  Hint: You will need to substitute for a, b and c in the quadratic formula $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ a is the coefficient for $$x^2$$, b is coefficient for $$x$$ and c is the constant.

Q: This is a Drag and drop into text question.

Drag and drop the words from the list below to fill the blank spaces and correctly complete the sentence.

At 25°C all aqueous basic solutions have [...] ion concentrations less than [...]
mol litre^-1 and pH values [...] than [...].

Use the mouse to drag the boxed words into the empty boxes. Alternatively, use the tab key to select an empty box, then use the space key to cycle through the options.
Also offered: positive; hydroxide; negative; 107; 1; less
A: hydrogen; 10-7; greater; 7
Feedback: At 25 °C all aqueous basic solutions have hydrogen ion concentrations less than 10^−7 mol litre^−1 and pH values greater than 7. See Section 9 of S103 Discovering Science Block 8.
  Hint: You may wish to read Section 9 of Discovering Science Block 8.
  Hint: Any incorrect choices will be removed before your final try.

Q: This is a Matching question.

Listed below are seven quantities together with a list of units on the right. From the drop-down lists choose the appropriate units for each quantity.
- electric charge | coulomb
- electric current | amp
- energy | joule
- force | newton
- power | watt
- temperature | kelvin
- voltage difference | volt
Also offered: einstein; pascal
If right: Your answer is fully correct. Well done.
Feedback: Here are the quantities, in alphabetical order, together with their units. electric charge | coulomb electric current | amp energy | joule force | newton power | watt temperature | kelvin voltage difference | volt
  Hint: If in doubt do consult a reference book.
  Hint: Perhaps you would care to refer to Block 5 of Discovering Science. When you select Try again any incorrect selections that you have made will be cleared.

Q: This is a Drag and drop onto images question.

Identify the features in this cross-section by dragging the labels into the boxes.

Use the mouse to drag the boxed words into the empty boxes. Alternatively, use the tab key to select an empty box, then use the space key to cycle through the options.
- Zone 1 | mountainbelt
- Zone 2 | continentalshelf
- Zone 3 | oceantrench
- Zone 4 | abyssalplain
- Zone 5 | continentalslope
- Zone 6 | continentalrise
- Zone 7 | islandarc
- Zone 8 | mid-oceanridge
Feedback: The locations are now labelled on the diagram below. [Image: labelled ocean floor] More information about the major features of the Earth's surface can be found in Block 3, Section 6.2.
  Hint: Incorrect placements will be removed.
  Hint: - The abyssal plain is a flat almost featureless expanse of ocean floor 4km to 6km below sea-level. - The continental rise is the gently sloping part of the ocean floor beyond the continental slope. - The continental shelf is the gently sloping ocean floor just offshore from the land. - The continental slope is the relatively steep part of the ocean floor beyond the continental shelf. - A mid-ocean ridge is a broad submarine ridge several kilometres high. - A mountain belt is a long range of mountains. - An island arc is a chain of volcanic islands. - An oceanic trench is a deep trough in the ocean floor.

Q: This is a Pattern match question.

The photograph shows a layer of oil floating on top of a glass of water. Why does the oil float?

[Image: Oil on water]

Please give your answer as a short phrase or sentence.
- match_all ( match_mow (great*|high|higher|more|bigger|heavier|heavy dens*|[specific gravity]|sg) not ( match_mw (water|not|higher) ) ) (Your answer is incorrect. (Rule 10))
- match_all ( match_any ( match_mw (great*|high|higher|more|bigger|heavier|heavy dens*|[specific gravity]|sg than_water) match_mw (dens*|[specific gravity]|sg great*|high|higher|more|bigger|heavier|heavy than_water) match_mw (dens* water < dens* oil) ) not ( match_w (than_oil) ) ) (Your answer is incorrect. (Rule 11) The oil does not float because it is more dense than the water.)
- match_any ( match_mow (it|its|it's|oil|0il_great*|high|higher|more|bigger|heavier|heavy_dens*|[specific gravity]|sg*) match_mw (it|its|it's|oil|0il_dens*|[specific gravity]|sg_great*|high|higher|more|bigger|heavier|heavy) ) (Your answer is incorrect. (Rule 12) The oil does not float because it is more dense than the water.)
- match_all ( match_any ( match_mow (oil|it|its|it's great*|high|higher|more|bigger|heavier|heavy dens*|[specific gravity]|sg) match_mrxow (oil|it|its|it's_denser) match_mow (oil|it|its|it's not_less*|low* dens*|[specific gravity]|sg) ) not ( match_mw (water) ) ) (Your answer is incorrect. (Rule 13) The oil does not float because it is more dense than the water.)
- match_any ( match_mw (oil|it|its|it's great*|high|higher|more|bigger|heavier|heavy dens*|[specific gravity]|sg than water) match_mrxwp1 (oil|it|its|it's_denser than water) match_mwp0 (oil|it|its|it's not_less*|low*|light* dens*|[specific gravity]|sg than water) ) (Your answer is incorrect. (Rule 14) The oil does not float because it is more dense than the water.)
- match_any ( match_mw (oil|it|its|it's_great*|high|higher|more|bigger|heavier|heavy dens*|[specific gravity]|sg) match_mw (oil|it|its|it's_dens*|[specific gravity]|sg_great*|high|higher|more|bigger|heavier|heavy) ) (Your answer is incorrect. (Rule 15) The oil does not float because it is more dense than the water.)
- match_all ( match_mow (less*|smaller|low*|light* dens*|[specific gravity]|sg) not ( match_mw (water|not|higher) ) ) (Your answer is correct. (Rule 19))
- match_any ( match_mw (less*|smaller|low*|light* dens*|[specific gravity]|sg than water*) match_mw (dens*|[specific gravity]|sg less*|smaller|low*|light* than water) match_mw (isn't|not_dens*|[specific gravity]|sg as_water) match_mw (dens*|[specific gravity]|sg oil < dens*|[specific gravity]|sg water) ) (Your answer is correct. (Rule 20))
- match_all ( match_any ( match_mow (water great*|high|higher|more|bigger|heavier|heavy dens*|[specific gravity]|sg) match_mrxow (water denser) ) not ( match_mw (oil|oils|not) ) ) (Your answer is correct. (Rule 23))
- match_any ( match_mw (water great*|high|higher|more|bigger|heavier|heavy dens*|[specific gravity]|sg than oil) match_mw (dens*|[specific gravity]|sg water great*|high|higher|more|bigger|heavier|heavy than oil) match_mrxw (water denser than oil) match_mow (water_denser oil_floats) ) (Your answer is correct. (Rule 24))
- match_any ( match_mw (water|[water_is]_great*|high|higher|more|bigger|heavier|heavy_dens*|[specific gravity]|sg) match_mw (water_dens*|[specific gravity]|sg|[same volume]_great*|high|higher|more|bigger|heavier|heavy) match_mw (water same volume great*|high|higher|more|bigger|heavier|heavy) ) (Your answer is correct. (Rule 25))
- match_all ( match_any ( match_mw (oil 920* water 1000*|[1 000*]) match_mw (water 1000*|[1_000*] oil 920*) match_mw (oil 920* less) match_mw (water 1000*|[1_000*] more|greater) ) match_mwp4 (density|[mass_cubic|ml|m3|volume]) ) (Your answer is correct. (Rule 26))
- match_any ( match_mwp0 (less*|smaller|low*|light* dens*|[specific gravity]|sg than_oil) match_mwp0 (dens*|[specific gravity]|sg less*|smaller|low*|light* than_oil*) ) (Your answer is incorrect. (Rule 31))
- match_any ( match_mw (water_less*|smaller|low*|light*_dens*|[specific gravity]|sg) match_mw (water_dens*|[specific gravity]|sg_less*|smaller|low*|light*) ) (Your answer is incorrect. (Rule 32))
- match_any ( match_mw (mass oil|oils|it|its|per less*|smaller|low*|light* same|equiv*|per volume|size|amount|ml water) match_mw (mass oil|oils less*|smaller|low*|light* [water per|same metre|ml|litre|vol*]|[per|same metre|ml|litre|vol* water]) ) (Your answer is correct. (Rule 45))
- match_any ( match_mw (oil|oils|it|its mass [less*|smaller|low*|light* same|equiv*|per volume|size|amount|ml]|[same|equiv*|per volume|size|amount|ml|litre less*|smaller|low*|light*] water) match_mw (oil|oils|it|its mass less*|smaller|low*|light* same|equiv*|per|equal volume|size|amount|ml|quantity water) match_mw (oil|oils|it|its less*|smaller|low*|light* mass same|equiv*|per volume|size|amount|ml water) ) (Your answer is correct. (Rule 46))
- match_any ( match_mw ([oil|oils|it|its|it's_mass|weight]|[mass|weight_oil] less*|smaller|low*|light* water displace*|replac*) match_mw (oil|oils|it|its_less*|smaller|low*|light*_mass|weight water displace*) ) (Your answer is correct. (Rule 47))
- match_any ( match_mw (oil|oils|it|its less*|smaller|low*|light* water same volume|size) match_mw (oil|oils|it|its less*|smaller|low*|light* cubic metre water) match_mw (oil|oils|it|its volume less*|smaller|low*|light* mass) ) (Your answer is correct. (Rule 50))
- match_any ( match_mw (less*|smaller|low*|light* dens* floats) match_mw (great*|high|higher|more|bigger|heavier|heavy dens* sinks) match_mw (oil|oils|it|its less*|smaller|low*|light* cubic metre water) match_mw (mass cubic metre water greater) ) (Your answer is correct. (Rule 51))
- match_any ( match_mw (light*_same_volume_water) match_mw (lighter water replac*|displac*) match_mw (water replac*|displac* heav*|[greater_mass]) ) (Your answer is correct. (Rule 61))
- match_any ( match_mw (dens*|[specific gravity]|sg) ) (Your answer is incorrect. (Rule 71) You are right to say that the reason the oil floats has something to do with density, but you need to word your answer more carefully. The oil floats for the same reason that ice floats on water. See Section 4.2.1)
- match_all ( match_mw (mass|weight) not ( match_mw (density|[specific gravity]|sg) ) ) (Your answer is incorrect. (Rule 80) The reason that the oil floats has something to do with its mass, but you need to be more specific. The oil floats for the same reason that ice floats on water. See Section 4.2.1)
- match_all ( match_mw (light*) not ( match_mw (water) ) ) (Your answer is incorrect. (Rule 81) The reason that the oil floats has something to do with its mass, but you need to be more specific. The oil floats for the same reason that ice floats on water. See Section 4.2.1)
- match_mw (it|its|it's|oil_lighter) (Your answer is incorrect. (Rule 82) The reason that the oil floats has something to do with its mass, but you need to be more specific. The oil floats for the same reason that ice floats on water. See Section 4.2.1)
- match_any ( match_mw (it|its|it's|oil_float*) match_mw (light* than water) match_mw (water heavier|sink*) ) (Your answer is incorrect. (Rule 73) It is true that the water sinks to the bottom of the glass whilst the oil floats, but why is this? It is for the same reason that ice floats on water. See Section 4.2.1.)
- not ( match_mw (dens&|[specific gravity]|sg) ) (Your answer is incorrect. (Rule 89))
- match_any ( match_mow (mix|mixing|imiscible|hydrophobic|insoluble) match_mow (*n't|not_soluble|dissolve) ) (Your answer is incorrect. (Rule 74))
- * (Your answer is incorrect. (Rule 99))
A: match_all ( match_mow (less*|smaller|low*|light* dens*|[specific gravity]|sg) not ( match_mw (water|not|higher) ) ) / match_any ( match_mw (less*|smaller|low*|light* dens*|[specific gravity]|sg than water*) match_mw (dens*|[specific gravity]|sg less*|smaller|low*|light* than water) match_mw (isn't|not_dens*|[specific gravity]|sg as_water) match_mw (dens*|[specific gravity]|sg oil < dens*|[specific gravity]|sg water) ) / match_all ( match_any ( match_mow (water great*|high|higher|more|bigger|heavier|heavy dens*|[specific gravity]|sg) match_mrxow (water denser) ) not ( match_mw (oil|oils|not) ) ) / match_any ( match_mw (water great*|high|higher|more|bigger|heavier|heavy dens*|[specific gravity]|sg than oil) match_mw (dens*|[specific gravity]|sg water great*|high|higher|more|bigger|heavier|heavy than oil) match_mrxw (water denser than oil) match_mow (water_denser oil_floats) ) / match_any ( match_mw (water|[water_is]_great*|high|higher|more|bigger|heavier|heavy_dens*|[specific gravity]|sg) match_mw (water_dens*|[specific gravity]|sg|[same volume]_great*|high|higher|more|bigger|heavier|heavy) match_mw (water same volume great*|high|higher|more|bigger|heavier|heavy) ) / match_all ( match_any ( match_mw (oil 920* water 1000*|[1 000*]) match_mw (water 1000*|[1_000*] oil 920*) match_mw (oil 920* less) match_mw (water 1000*|[1_000*] more|greater) ) match_mwp4 (density|[mass_cubic|ml|m3|volume]) ) / match_any ( match_mw (mass oil|oils|it|its|per less*|smaller|low*|light* same|equiv*|per volume|size|amount|ml water) match_mw (mass oil|oils less*|smaller|low*|light* [water per|same metre|ml|litre|vol*]|[per|same metre|ml|litre|vol* water]) ) / match_any ( match_mw (oil|oils|it|its mass [less*|smaller|low*|light* same|equiv*|per volume|size|amount|ml]|[same|equiv*|per volume|size|amount|ml|litre less*|smaller|low*|light*] water) match_mw (oil|oils|it|its mass less*|smaller|low*|light* same|equiv*|per|equal volume|size|amount|ml|quantity water) match_mw (oil|oils|it|its less*|smaller|low*|light* mass same|equiv*|per volume|size|amount|ml water) ) / match_any ( match_mw ([oil|oils|it|its|it's_mass|weight]|[mass|weight_oil] less*|smaller|low*|light* water displace*|replac*) match_mw (oil|oils|it|its_less*|smaller|low*|light*_mass|weight water displace*) ) / match_any ( match_mw (oil|oils|it|its less*|smaller|low*|light* water same volume|size) match_mw (oil|oils|it|its less*|smaller|low*|light* cubic metre water) match_mw (oil|oils|it|its volume less*|smaller|low*|light* mass) ) / match_any ( match_mw (less*|smaller|low*|light* dens* floats) match_mw (great*|high|higher|more|bigger|heavier|heavy dens* sinks) match_mw (oil|oils|it|its less*|smaller|low*|light* cubic metre water) match_mw (mass cubic metre water greater) ) / match_any ( match_mw (light*_same_volume_water) match_mw (lighter water replac*|displac*) match_mw (water replac*|displac* heav*|[greater_mass]) )
Feedback: The oil floats because its density is less than that of the water, where $$density = \frac{mass}{volume}$$ See Section 4.2.1.
  Hint: The oil floats for the same reason that ice floats on water.
  Hint: See Section 4.2.1. Have another go, remembering to express your answer as a simple phrase or sentence.

Q: This is a Drag and drop markers question.

Please place the markers on the map of Milton Keynes to show the OU and all railway stations that are on the West Coast main line.
- Zone 1 | OU
- Zone 2 | Railway station
- Zone 3 | Railway station
- Zone 4 | Railway station
Feedback: The Open University is at the junction of Brickhill Street and Groveway. There are three railway stations on the West Coast main line, Wolverton, Milton Keynes Central and Bletchley. Bow Brickhill and Woburn Sands stations are on what remains of the old Oxford - Cambridge 'Universities line'.
  Hint: You should be looking to place four markers on the map.
  Hint: You should be looking to place markers on three railway stations.

Q: This is a Numerical question.

Solve $$x^2 + 3x - 28 = 0$$

Enter either of the possible answers.
A: -7 / 4
Feedback ("-7"): Your answer is correct.
Feedback ("4"): Your answer is correct.
Feedback: The quadratic equation is $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ a = 1, b = 3 and c = -28. $$x = \frac{-3 \pm \sqrt{3^2 - 4\times 1\times(-28)}}{2\times a}$$ $$x = \frac{-3 \pm \sqrt{9 + 112}}{2}$$ $$x = \frac{-3 \pm \sqrt{121}}{2}$$ $$x = \frac{-3 \pm 11}{2}$$ $$x = \frac{-14}{2}$$ or $$\frac{8}{2}$$ and the solutions are x = -7 and x = 4. We can prove these are correct by substituting the values into the original equation. $$(-7)^2 + 3\times(-7) - 28 = 49 - 21 - 28 = 0$$ $$4^2 + 3\times4 - 28 = 16 + 12 -28 = 0$$
  Hint: You may need to be reminded of the quadratic formula $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
  Hint: You will need to substitute for a, b and c in the quadratic formula $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ a is the coefficient for $$x^2$$, b is coefficient for $$x$$ and c is the constant.