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### 2015 A Level H1 Chemistry (8872) Paper 1 Suggested Solutions

All solutions here are SUGGESTED. Mr. Lee will hold no liability for any errors. Comments are entirely personal opinions.

Many thanks to the student who sent us the question paper. 🙂

You can post any questions regarding the answers but please do not ask me how many marks to secure an A for the examination. Nobody will know so there is no point in discussing this.

### 2015 A Level H2 Chemistry (9647) Paper 1 Suggested Solutions

All solutions here are SUGGESTED. Mr. Lee will hold no liability for any errors. Comments are entirely personal opinions.

Question 1:   B

The element must be from Group IV so the electronic configuration must be either 1s22s22p2 or 1s22s22p63s23p2. The four electrons of highest energy refer to the electrons at shells with higher principal quantum number. The best answer is B.

Question 2:   B

Sigma bonds can be formed via head-on overlap between two s orbitals, two p orbitals and between one s and one p orbitals. In hydrocarbons, the carbon atoms have hybrid orbitals consisting of s and p orbitals. Hence, sigma bonds can be formed by either s or p orbitals while the pi bonds can only be formed by side-way overlap between two p orbitals.

Question 3:   C

Oxygen atom has 8 electrons. Hydrogen atom has 1 electron. Since hydroxide anion has an extra electron gained, the total number of electrons is 10. Oxygen atom has 8 neutrons while hydrogen atom has no neutrons. Hence, the answer is C.

Question 4:   C

A higher charge on Ca2+ would mean that the electrostatic forces of attraction between the cation and the sea of mobile electrons are stronger as compared to Na+. Option D is wrong because the electrons in the calcium ion are not mobile and even if calcium ion has more electrons than sodium ion, it does not explain the stronger metallic bonding. The correct explanation should be calcium metal has more mobile electrons than sodium metal, instead of ion.

Question 5:   B

In zwitterion, –NH2 (trigonal pyramidal, 107°) group becomes –NH3+ (tetrahedral, 109.5°).

Question 6:   D

Lattice energy is the enthalpy changed when 1 mole of solid ionic compound is formed from its constituent gaseous ions.

Question 7:   B

Number of moles of NaOH = 12.5 / 1000 x 0.0500 = 6.25 x 10-4 mol (limiting reagent)

Number of moles of HCl = 25.0 / 1000 x 0.100 = 2.50 x 10-3 mol (in excess)

Number of moles of HCl remaining = 1.875 x 10-3 mol

Concentration of HCl remaining = 1.875 x 10-3 / (37.5 / 1000) = 0.0500 mol dm-3

Question 8:   C

CH4 + 2O2 → CO2 + 2H2O

CH4 + 3/2 O2 → CO + 2H2O

9CH4 + 18O2 → 9CO2 + 18H2O

CH4 + 3/2 O2 → CO + 2H2O

Therefore, 10 moles of CH4 react with 19.5 moles of oxygen to produce 9 moles of CO2 and 1 mole of CO (9 : 1 ratio)

Hence, 1 dm3 of CH4 reacts with 1.95 dm3 of oxygen.

Question 9:   A

Formation of hydrogen bonds indicates that the reaction is exothermic. ΔH is negative.

The product (1 mole) is more ordered as compared to the reactants (1 mole of CO2 and n moles of H2O). Hence, ΔS is negative.

Question 10: B

Standard conditions means that all concentrations must be 1 mol dm-3.

Question 11: D

Equilibrium concentration for CH3CO2H = C(1 – α)

Equilibrium concentration for CH3CO2 = Cα

Equilibrium concentration for H+ = Cα

Kc = (Cα)2 / C(1 – α) = α2C / (1 – α)

Question 12: C

Increasing the pressure will shift the equilibrium to the right because the forward reaction produces fewer number of moles of gaseous substances. Decreasing the temperature for the exothermic reaction will shift the equilibrium to the right because the forward reaction will release heat energy to counteract the decrease in temperature.

Question 13: D

Total number of moles = 2.0 mol

Number of moles of H2 = 33.3 / 100 x 2.0 = 0.666

Number of moles of CO2 = 33.3 / 100 x 2.0 = 0.666

Percentage of H2O = (100% – 33.3% – 33.3%) / 2 = 16.7%

Number of moles of H2O = 16.7 / 100 x 2.0 = 0.334

Number of moles of CO = 16.7 / 100 x 2.0 = 0.334

Kc = (0.666)2 / (0.334)2 = 3.98 ≈ 4.0

Question 14: B

The reaction is second order with respect to NO. Hence, when concentration decreases by 2, the initial rate will decrease by 4. Hence, x = 6.0 / 4 = 1.5

The reaction is first order with respect to H2. Hence when concentration increases by 2, the initial rate will increase by 2. Hence, y = 1.5 x 2 = 3.0

The reaction is second order with respect to NO. Hence, when the initial rate decreases by (3.0 / 0.75) = 4, the concentration of NO will decrease by 2. Hence, z = 0.5.

Question 15: D

Thermal stability of nitrates increases down Group II due to lower charge density of Group II metal cations as the size of the cations increases.

Question 16: D

Phosphorus exists as P4. Option A is wrong because sulfur can also form two acidic oxides, SO2 and SO3. Option B is wrong because argon should have the highest ionisation energy. Option C is wrong because chlories of silicon (SiCl4) and sulfur (S2Cl2) can react with water to form HCl solution which is acidic.

Question 17: A

Only Option A is correct. Option B is wrong because it is not a precipitate. Option C is wrong because it is a blue precipitate instead of a pale blue solution. Option D is wrong because Cu+ complex should not have any colour because it has completely filled 3d orbitals so no d-d transition is possible.

Question 18: C

Atomic radius should decrease across the period from Mg to P while for melting point, Si has the highest, followed by Al, Mg and P.

Question 19: B

Only dilute sulfuric acid is used to acidify aqueous potassium dichromate(VI). Ethanol can be oxidised to form ethanoic acid.

Question 20: A

PCl5 will react with hydroxyl groups and carboxylic acid groups. The Br atom does not react with PCl5

Question 21: B

The 3 structural isomers are:

Question 22: D

Question 23: B

NaBH4 will reduce aldehyde to alcohol but it does not reduce alkene to alkane.

Question 24: A

Molecular formula of citric acid = C6H8O7

Condensation reaction will lose one molecule of water.

Hence, C6H8O7 + C3H6O2 → C9H12O8 + H2O

Hence, option A is the best answer as the structural formula of C2H5CO2H fits the molecular formula of C3H6O2

Question 25: D

Ethanoyl chloride is an acid chloride. It will react with water to produce ethanoic acid and Cl  and the Clion will form a white precipitate (AgCl) with Ag+

Question 26: D

pH 11 is alkaline so the functional groups will be deprotonated.

Question 27: D

Addition of molecular mass of 14 suggests that one oxygen atom is added but two hydrogen atoms are removed. This indicates that primary alcohol (–CH2OH) is converted to carboxylic acid group (–COOH).

Question 28: C

Cold NaOH(aq) is able to react with the acidic phenol group but not with the alcohol group.

Question 29: B

The phenol group is an activating group and is 2,4,6-directing.

Question 30: A

The carbon-carbon double bond will attack the Br–Br first to form a carbocation containing one Br atom.

Question 31: D

Mn2+: [Ar] 3d5 (no paired 3d electrons)

Fe2+: [Ar] 3d6 (one paired 3d electrons)

Co3+: [Ar] 3d6 (one paired 3d electrons)

Question 32: A

Simple covalent compounds have low boiling point. AlBr3 has a simple molecular structure because it has a low boiling point even though it has a metal and non-metal. If AlBr3 is ionic, the boiling point should be 4 digit. Hence, all 3 compounds are covalent.

Question 33: C

The molecules of the gas have mass, so option 1 is wrong.

Question 34: A

Zinc will be oxidised while NH4+ will be reduced. So the overall potential will be +1.50V (energetically feasible). Option 2 and 3 are correct also.

Question 35: B

Rate constants are affected by activation energy and temperature based on Arrhenius equation k = Ae–Ea / RT. Introducing a catalyst will increase both kf and kb equally while increasing the temperature for an exothermic reaction will increase BOTH kb and kf but will increase kb more than kf so equilibrium shifts to the left. Rate constant does not depend on concentration of the reactants.

Question 36: A

Methanal is trigonal planar with respect to carbon and the carbon has an oxidation number of 0 (hydrogen has an oxidation number of +1 while oxygen has an oxidation number of –2). The equation for complete combustion of methanal is:

CH2O + O2 → CO2 + H2O

Hence, 1 mole of oxygen is required for 1 mole of methanal.

Question 37: C

Radicals do not have a lone pair of electrons. They have unpaired electrons and can be formed by homolytic fission of a covalent bond.

Question 38: C

Geometric isomerism is possible if there are two different groups attached to the sp2 carbon in the carbon-carbon double bond. Hence, Y cannot be H and can be Br so that the carbon-carbon double bond with Y can exist as geometric isomers. The carbon-carbon double bonds with Z and X cannot exist as geometric isomers since the groups attached to the sp2 carbon in the carbon-carbon double bond are the same.

Question 39: A

Thymol has a phenol group so the acidic phenol group can react with the alkali present in alkaline KMnO4 via neutralisation. The phenol group can also react with ethanoyl chloride to form an ester via condensation and it can also react with sodium metal to form phenoxide ion.

Question 40: B

Option 1 and 2 have no line of symmetry while option 3 has a line of symmetry. Hence, option 3 is not optically active.

Please do let me know of any mistakes or typing errors that I made while rushing this. Much appreciated and thanks!

### 2015 A Level H2 Chemistry (9647) Paper 3 Suggested Solutions

All solutions here are SUGGESTED. Mr. Lee will hold no liability for any errors. Comments are entirely personal opinions.

[Please do not ask me how many marks for A. The bell curve is something that is out of your control so there is no point in estimating your grade based on the number of marks you have lost. No one can tell you any accurate information on this. I hope you can learn from your mistakes here and do not make the same mistakes in Paper 1.]

Question 1

1          (a)       (i)        Amount of menthol = 1.32 x 10-2 x 2 = 0.0264 mol

Mass of menthol = 0.0264 mol x (10 x 12 + 20 + 16)

=          4.1184 g

Percentage mass       =          4.1184 / 10.0   x 100%

=          41.2% (3 sf)

(ii)       3 chiral centres, 23 = 8 optical isomers

(b)          (i)

(ii)

(iii)    The catalyst is in a solid state and it functions as a heterogeneous catalyst as it is in a different phase than menthone (liquid state) and hydrogen (gaseous state). Menthone and hydrogen will undergo adsorption at the active sites of the surface of the catalyst, forming weak bonds between the reactant molecules and the catalyst and this causes the bonds in menthone and hydrogen to be weakened. This provides an alternative pathway which has a lower activation energy. The reactant molecules are now in closer proximity with one another and hence, the frequency of effective collisions between menthone and hydrogen gas increases and this increases the speed of reaction.

(iv)    Iron might be suitable as a catalyst. Iron is a transition metal with incompletely filled 3d orbitals.

(c)     All three isomers undergo electrophilic addition with bromine water. Hence, they contain carbon-carbon double bond.

All three isomers contain carbonyl group as they undergo condensation reaction with 2,4-dinitrophenylhydrazine.

All three isomers do not contain aliphatic aldehyde group.

All three isomers undergo reduction with hydrogen gas to form menthol with molecular formula C10H20O. Hence, all three isomers contain two functional groups that can be reduced by hydrogen gas as 4 hydrogen atoms are added.

Menthol contains a secondary alcohol, hence, the 3 isomers contain a ketone group. The other functional group that is reduced must be carbon-carbon double bond.

When reacted with hot concentrated KMnO4, strong oxidation or oxidative cleavage occurs.

A gives D, which is a ketone, and E which contains two ketone groups (one ketone group is originally present in A).

B gives F, which contains two ketone groups (one ketone group is originally present in B). B contains a terminal carbon-carbon double bond.

C contains a carbon-carbon double bond in a ring. G contains a carboxylic acid group and two ketone groups (one ketone group is originally present in C).

D, F and G contains CH3C=O group as they undergo positive iodoform test (oxidation), forming yellow precipitate of CHI3.

Question 2

(a)     The volatilities of the halogens decrease from chlorine to iodine (i.e. melting and boiling point increase from chlorine to iodine).

From chlorine to iodine, the number of electrons increases. Hence the size of the electron cloud increases which increases the strength of van der Waals’ forces of attraction between the halogen molecules. Hence, the volatilities decrease from chlorine to iodine.

(b)    3Cl2(g) + 6OH(aq) → 5Cl(aq) + ClO3(aq) + 3H2O(l)

The oxidation number of chlorine increases from 0 in Cl2 to +5 in ClO3

The oxidation number of chlorine decreases from 0 in Cl2 to -1 in Cl

(c)     (i)     The bond energy for H–F is 562 kJ mol-1 while the bond energy for H–Cl to H–I decreases from 431 to 366 to 299 kJ mol-1. H–F bond has the highest bond energy so the bond strength is the strongest which require the most energy to break. Hence, HF does not dissociate completely in water to produce hydrogen ions.

(ii)          pH of HCl = – lg (0.50) = 0.301

Ka = [H+]2 / [HF]

[H+]2 = 5.6 x 10-4 x 0.50 = 2.8 x 10-4

pH = – lg [(2.8 x 10-4)1/2] = 1.78

(d)    When aqueous silver nitrate is added to chloride ions, a white precipitate of silver chloride (AgCl) is formed.

Ag+(aq) + Cl(aq) → AgCl(s)

When aqueous ammonia is added to silver chloride, the white precipitate dissolves to form a colourless solution of [Ag(NH3)2]+

AgCl(s) ⇌ Ag+(aq) + Cl(aq)

Ag+(aq) + 2NH3(aq) → [Ag(NH3)2]+

When aqueous silver nitrate is added to iodide ions, a yellow precipitate of silver iodide (AgI) is formed.

Ag+(aq) + l(aq) → Agl(s)

When aqueous ammonia is added to silver iodide, the yellow precipitate is insoluble in aqueous ammonia.

(e)       (i)        The value of pV remains constant.

(ii)       pV = nRT

12.0 x 105 x V = 0.40 x 8.31 x 300

V = 8.31 x 10-4 m3

(iii)     9.26 x 105

8.88 x 105

pV = 9.11 x 105 Pa dm3

V = 0.759 dm3

(iv)    There is presence of permanent dipole-permanent dipole interactions between hydrogen chloride molecules due to the net dipole moment present as the chlorine atom is electronegative. These intermolecular forces of attraction are significant and will deviate from the ideal gas properties. Hence, the molecules are closer together and occupy a smaller volume as compared to an ideal gas.

(f)        (i)        ΔG = ΔH – TΔS

0 = +16.8 – 188ΔS

ΔS = +0.089362 kJ K-1 mol-1 ≈ +0.0894 kJ K-1 mol-1 (3 sf)

The entropy change is positive because there is an increase in disorderness when the liquid state of hydrogen chloride is changed to the gaseous state as the molecules have an increased number of ways of arranging themselves in the gaseous state.

(ii)       ΔG = +16.8 – (298)(0.089362) = –9.83 kJ mol-1 (3 sf)

A negative sign for ΔG means that the reaction is feasible and can take place spontaneously.

Question 3

(a)     (i)      Magnesium burns with a brilliant white flame when it reacts with oxygen to form magnesium oxide.

2Mg(s) + O2(g) → 2MgO(s)

Calcium burns with a brick-red flame when it reacts with oxygen to form calcium oxide.

2Ca(s) + O2(g) → 2CaO(s)

(ii)     Magnesium oxide is sparingly soluble in water, forming magnesium hydroxide while calcium oxide is partially soluble in water, forming aqueous calcium hydroxide which is alkaline in water.

MgO(s) + H2O(l) → Mg(OH)2(aq)

CaO(s) + H2O(l) → Ca(OH)2(aq)

(b)       (i)        2.0 x 10-1 mol dm-3

(ii)       Ca(OH)2 ⇌ Ca2+(aq) + 2OH(aq)

Concentration of OH = 2.5 x 10-2 x 2 = 5.0 x 10-2 mol dm-3

pOH = – lg (5.0 x 10-2) = 1.30

pH = 14 – 1.30 = 12.7

(iii)      Ksp = [Mg2+][OH]2

Ksp = (s) (2s)2 = (1.6 x 10-4)(3.2 x 10-4)2 = 1.64 x 10-11 mol3 dm-9

(iv)    White precipitate of magnesium hydroxide would be observed. Barium hydroxide is more soluble in water than magnesium hydroxide. Hence, the concentration of hydroxide ions increases. Common ion effect occurs and hence, the solubility of magnesium hydroxide decreases so less magnesium hydroxide is able to dissolve in water.

(c)       (i)        Aspartate and glutamate

Ionic bonding

(ii)     The alpha helix is held in place due to hydrogen bonding  formed between the N–H group of each amino acid and the fourth C=O group following it along the chain.

(iii)      H2SO4(aq), heat under reflux / NaOH(aq), heat under reflux

(iv)      gly-asp-gly-tyr-ile-ser

(d)       (i)

(ii)       Step 4: Reagent: excess concentrated ethanolic NH3

Condition: heat in sealed tube

Step 5: Reagent: H2SO4(aq)

Condition: heat under reflux

Question 4

(a)       Step 1: hydrolysis

Step 2: dehydration

Step 3: reduction

(b)       C6H8O(l) + 15/2 O2(g) → 6CO2(g) + 4H2O(l)

Energy taken in for bond breaking

= 8 (C – H) + 3 (C – C) + 2 (C = C) + 2 (C – O) + 15/2 (O = O)

= 8(410) + 3(350) + 2(610) + 2(360) + 15/2 (496)

= 9990 kJ mol-1

Energy released for bond formation

= 12 (C = O) + 8 (O – H)

= 12(805) + 8(460)

= 13340 kJ mol-1

Enthalpy change of combustion = 9990 – 13340 =  –3350 kJ mol-1

(c)       Q = mcΔT = (200)(4.18)(32) = 26752 J

100% heat = 26752 / 80  x 100 = 33440 J

Amount of DMF = 1.00 / 96 = 0.01042 mol

Experimental enthalpy change of combustion

= 33.440 / 0.01042 = –3209 kJ mol-13210 kJ mol-1

The enthalpy change of combustion obtained in (b) uses bond energies from the data booklet which are average values of bond energies obtained from a range of molecules containing that bond. In addition, some of the heat energy released from the burning of DMF is lost through the heating of the container itself or through the surroundings. Hence, the experimental value is less exothermic. Hence, there is a slight difference in both values.

(d)       (i)        R–CH2OH + H+ → R–CH2OH2+          (fast)

R–CH2OH2+ → R–CH2+ + H2O           (slow)

R–CH2+ + Cl → R–CH2Cl                  (fast)

The second step (R–CH2OH2+ → R–CH2+ + H2O) is the rate determining step.

(ii)       Reagents: K2Cr2O7(aq), H2SO4(aq)

Condition: heat under reflux

[KMnO4(aq), H2SO4(aq) is not accepted due to oxidative cleavage]

(e)       (i)        Ester

(ii)       HO–CH2–CH2–OH

(iii)      Reagents: concentrated sulfuric acid

Condition: heat under reflux

(f)        Name of mechanism: Free radical substitution

Initiation

Cl–Cl → 2 Cl·

Propagation

Cl· + R–CH3 → R–CH2· + HCl

R–CH2· + Cl2 → R–CH2Cl + Cl·

Termination

Cl· + Cl· → Cl2

R–CH2· + R–CH2· → R–CH2CH2–R

R–CH2· + Cl· → R–CH2Cl

Question 5

(a)       (i)        Proton number is the number of protons in the nucleus of the atom.

Nucleon number is the number of protons and neutrons in the nucleus of the atom.

(ii)       Let the relative abundance of 7Li be k

Relative abundance of 6Li = 1 – k

k x 7.016  +  (1 – k)(6.015) = 6.942

7.016k + 6.015 – 6.015k = 6.942

1.001k + 6.015 = 6.942

1.001k = 0.927

k = 0.9261

Therefore, relative percentage abundance of 7Li = 92.61%

Relative percentage abundance of 6Li = 7.39%

(iii)      X = 3He           Y = 7Li

(b)       (i)        Ionic bonding. The lithium atom is able to transfer its valence electron to the carbon atom in graphite which has one unpaired electron, forming positive Li+ cation and anionic graphite. The bonding between the positive Li+ cation and anionic graphite is ionic.

(ii)       Before discharge: +4

After the cell is totally discharged: +3

(iii)    The shape is tetrahedral with respect to boron. There are 4 bond pairs of electrons and 0 lone pair of electrons for boron. Hence, the shape is tetrahedral.

(iv)    Cold, alkaline KMnO4(aq)

(v)     Condensation

(c)       (i)        2Li2O2 + 2CO2 → 2Li2CO3 + O2

(ii)     Lithium has similar chemical properties as magnesium because the atomic radius and electronegativity are similar. Lithium is smallest in size for Group I metal and hence, lithium ion has the largest charge density for Group I metal ions. Therefore, it is able to polarise and distort the electron cloud of the carbonate anion to a larger extent and the distorted electron cloud of the carbonate anion is more readily decomposed by heat energy. Hence, the thermal stability of lithium carbonate is low and can be easily decomposed.

(d)       (i)        CH3CH2CH2CHO        + CH3CH2Br

OR

CH3CH2CHO + CH3CH2CH2Br

(ii)       (CH3CH2)2CO + CH3Br

OR

CH3CH2COCH3 + CH3CH2Br

(e)     Only P and Q will turn orange acidified potassium dichromate(VI) to green. R, being a tertiary alcohol will not be oxidised and will not turn orange K2Cr2O7 to green.

After oxidation, 2,4-dinitrophenylhydrazine can be added. For Q, after oxidation, the product is a ketone and will form an orange precipitate with 2,4-dinitrophenylhydrazine. For P, after oxidation, the product is a carboxylic acid and will not form an orange precipitate.

Please do let me know of any mistakes or typing errors that I made while rushing this. Much appreciated and thanks!

### A little on bell curve

Since many students were asking, I thought I’ll roughly explain the bell curve and its distribution properties.

Before that, I thought I should mention that I’m not really sure myself if Private Candidates and School Candidates share a different bell curve. I doubt so, since all take the same subject code, there should be no differentiation as there will be bias. And I do not really see a particular need for them if all the students take the same paper. There is no differentiation between taking the paper again and first time too, it doesn’t mean that you take it twice, you have a higher chance/ lower chance. They should be competing in all fairness still.

A bit of intuition of how this bell curve thing works. We can’t have too many A’s. If ALL the JC students were to score zero mark and no student deviate (cheat), then all of you will get A. Similarly, if God answered all your prayers and ALL JC student were to score full marks, then all will get A. The curve attempts to distribute proportionally the results of everybody based on the marks, and place you in the correct percentile (probability).

Normal Distribution Generator

Logistics wise, you can click the link above which gives us a normal distribution. Here, you can input the Mean and Variance so you can create your own bell curve. I can’t really advise the mean and variance score since I have no data. Its really why we only see letter grade.

Next we look at the bell curve

First of all, this is a standardised normal, the $Z \sim \mathrm{N}(0,1)$ curve that you all know. To simplify things, just consider the cumulative percent at the bottom. 80% means you 80th percentile, and that in general warrants an A I’ll say. So consider we have 20 or 21 JC (I don’t know), each having approximately 700 students taking, we can find the population size. From here we can then take a percentage and figure out how many A’s are available.

Next, what does it mean when the bell curve shifts?

What we see above is a Normal curve centred about 0 (mark), and we know that Normal curve is centred about $\mu$ (mean/ modal mark). So if you can figure out the mean mark, by taking a sufficient sample size of students and how they did (we lack the information here since we only receive letter grades), we can that find the unbiased estimate of the population mean. And then the unbiased estimate of the population variance, which is crucial too.

Now since we all squeeze into the same bell curve and due to some friends scoring full marks or near full marks, yes there are definitely students that have full marks in a few particular JCs, I know a handful of my students above 90 actually… This results in the mean mark shifting to the right. And in general, the curve should not shift left, at least not in our society. We are Asians.

Here is why, it is really impossible to say what’s the range for A unless you can confidently tell me the population mean and variance. But I do consider the case of 25% of students should get an A.

I hope this clarifies a bit. And yes, this is what you guys actually learnt in A-levels, if you understood your content of course as what I’ve explained in class before. To spice things up, we can also perform hypothesis testing (simple case can be just with $\alpha$, while we should also look at $\beta$ )to see if the $\mu$ we found is the best unbiased estimate. And something out of syllabus will be to test if $\sigma^2$ is also best unbiased estimate. The word Best adds one more criterion for our estimate actually.

Now you can go at play with the above link and fit in what you believe the mean and standard deviation is, and then you can calculate the percentile (probability) you are in by putting the score you think you have.

Take note the standard deviation should be not go too big, as we can’t have people scoring more than 100. 100 marks should be $3.1 \sigma$ away from $\mu$.

Again, please don’t ask me whats the $\mu$ and $\sigma$. I don’t want to advise such things. You can open your sampling notes, perform stratified sampling then collect unbiased estimates of population mean and variance yourself, and put what you learnt to real (finally) use.

If you’re lazy, then just check what’s your school percentage of A then consider the population of your school, that roughly how well you will do. 🙂 I usually say this in class, that if your school delivers 50%, then you need to find a friend to not do well at least.

My take for the bell curve is that the bell curve will always help the best students. As for it helping the weaker students, it is highly dependent on the cohort on the whole and the difficulty of the paper. Consider me scoring 90/100 marks and everybody else scores above 95 marks, I’ll be the weakest and will end up failing. I hope these does not discourage or crush the confidence. Remember that there is a paper 2 still, with equal weightage.

So I took this image of google, it does not represent the A-levels. Please take note. But this is what happens when too many students do well. If you’re keen, you can read more on skewed normal distribution independently. I might do more on Stat 101 here when I’m free.

### 2015 A’level Suggested Solutions

Congratulations on the completion of A’levels for the 2015 batch!

As for those who still, are taking A’levels 2016, we hope you find this site helpful. Read the comments (ignore the trolls who have yet grown up) and learn from our mistakes.

It has been a pleasure for all us to interact with you guys. Do check back after your release of A’levels, as we will love to hear from you! In the meantime, you can always read through some of the posts regarding undergraduate & postgraduate courses in Mathematics (KS), Economics (KS), Physics (Casey) & Chemistry (Eric) as we share some of our past experiences, along with some of our works today.

You’ve come a long way, now party hard. We wish you all the best in your future endeavours.