1) H
Pancreatic juices contain amylases, lipases and proteases. Since cystic fibrosis prevents the secretion of pancreatic juices, the medication must contain amylases, lipases and proteases.


2) C
Q has electronic configuration of 2, 8, 2
Z has electronic configuration of 2, 7

Q needs to lose 2 electrons to have full outer shell and Z needs 1 electron to have full outer shell.

Q has a valency of 2 and Z has a valency of 1. (two Z atoms are needed for each Q atom)

So the Empirical formula would be QZ2

The bonding would be ionic because Q loses two electrons and each Z atom gains one electron.


3) E

1 – False. All light travel at the same speed in vacuum.

2 – True. Red light always has a larger wavelength than green light.

3 – False. All light travel faster in vacuum than in water.

4 – True. Frequency of blue light is always greater than frequency of red light.

So the answer is E (2 and 4).


4) B

48/40 is 6/5
m5 ÷ m2 = m3
p ÷ p3 = p-2

The simplification is 6/5 m3 p-2

So the answer is B


5) A

In Denitrification, nitrate ions are reduced to produce nitrogen gas (N2). This is only shown by process 1. So the answer is A.


6) C

1 – Wrong. Covalent bonds in position 1 are between one liquid and the other liquid. However, in position 3, only one of the liquids is obtained so the same covalent bonds are not formed.

2 – Correct. In position 3, the particles are in liquid form but in position 2, particles are in gas form. Particles are closer in liquid form than gas form.

3 – Wrong. In position 1, heat is being supplied to the mixture. So it is an endothermic process. Also, breaking bonds requires energy so it must be an endothermic change.

Since only statement 2 is correct, C is the answer.


7) E

Using c = f λ,

Speed of wave in material = 50 x 0.4 = 20 m/s

Time = Distance ÷ Speed = 100 ÷ 20 = 5 s


8) C

We need exactly 70p worth of coins. So that is one 50p coin and one 20p coin. Best way to do this is to draw a tree diagram.

We can only take the dashed paths shown above.

1st dashed path gives us = 2/7 x 5/6 = 10/42 = 5/21
2nd dashed path gives us = 5/7 x 2/6 = 10/42 = 5/21

Total = 5/21 + 5/21 = 10/21

So the answer is C.


9) B

We must first identify the type of cell division that occurred. 4 cells divided 5 times to form 128 cells. Mitosis produces 2 daughter cells after each division whilst meiosis produces 4 after each division.

If it was Mitosis,
4 – 8 – 16 – 32 – 64 – 128

If it was Meiosis,
4 – 16 – 64 – 256 – 1024 – 4096

So the type of cell division was mitosis.

1 – Correct. All cells would have the same number of chromosomes.

2 – Wrong. If the cells were gametes, they would have divided by meiosis instead of mitosis.

3 – Correct. Mitosis produces clones of the original cells.

4 – Wrong. It is mitosis, not meiosis.

So the answer is B (1 and 3).


10) C

In iron (III) oxide, iron has an oxidation state of +3. In iron (in products), iron has an oxidation state of 0. So iron is reduced, which means it has gained electrons. So A is wrong.

The oxide ions in iron (III) oxide each have a charge of -2. So B is wrong.

Iron (III) oxide oxidises carbon in CO. In CO, carbon has an oxidation state of +2 but in carbon monoxide, it has an oxidation state of +4. So Iron (III) oxide oxidises carbon from +2 to +4. So, Iron (III) oxide is an oxidising agent. C is the answer.

The value of ∆H is negative which indicates that this process is exothermic. D is wrong.

Electric current is not used in this reaction so this is not an example of electrolysis. So E is wrong.


11) E

Speed does not necessarily change when a vehicle changes direction, as speed is a scalar quantity. So A, B and C are wrong.

Velocity is a vector quantity so when a vehicle changes direction, its velocity changes.

Momentum is a vector quantity as its equation is p = mv, it depends on velocity. If a vehicle changes direction, its velocity changes so its momentum also changes.

Kinetic energy does not change, as it is a scalar quantity. Energy does not depend on direction and is always positive. So when a vehicle changes direction, it is not necessary that its kinetic energy will also change.

Only momentum and velocity change. So E is that answer.


12) F

It is important for you to realise that the diameter of the circle is the diagonal of the rectangle.

Let width of rectangle be W. So length is 3W

2(3W + W) = 24 cm
8W = 24 cm
W = 3 cm
Length = 3 x 3 = 9 cm

Diagonal = hypotenuse of one right-angled triangle.
Diagonal2 = 32 + 92
Diagonal2 = 9 + 81
Diagonal2 = 90
Diagonal = 3√10
Radius = (3√10 ÷ 2) cm

Area of circle = π (3√10 ÷ 2) 2 = 90/4 cm2
Area of rectangle = 9 x 3 = 27 cm2
Area of shaded area = (90/4) π – 27 cm2 

Obviously, this is not one of the options. However, F is the same as the answer. So F is correct.


13) H

All cells shown in the question form tissues. Epithelial cells together form epithelial tissue. Mature red blood cells form blood, which is a connective tissue. Muscle cells form muscle tissue.


14) B

The two spikes at 35 and 37 indicate two isotopes of element X.

The spike at 70 is due to two isotopes of mass 35 existing as a diatomic molecule. (35 + 35 = 70)

The spike at 72 is due to two isotopes of mass 35 and 37 existing as a diatomic molecule.
(35 + 37 = 72)

The spike at 74 is due to two isotopes of mass 37 existing as a diatomic molecule. (37 + 37 = 74)

So B is the answer.


15) E

PE at the top of the lift = 200 x 10 x 1.8 = 3600 J
All PE is converted to KE just before contact with the ground.
½ mv2 = 3600
(200) v2 = 7200
v2 = 36
v = 6 m/s


16) B

Let difference between r and r + 1 = M
Let difference between r + 1 and r + 2 = N
Let difference between r + 2 and r + 3 = O

The sum of the difference = 126
M + N + O = 126

By understanding the example given in the question, we can infer that:
N = M + 1
O = N + 1

Since N = M + 1,
O = M + 1 + 1 = M + 2
M + N + O = 126
M + M + 1 + M + 2 = 126
3M + 3 = 126
3M = 123
M = 41

M is the difference between rth and (r + 1)th triangular number.
From the example described in the question, we know that:
Difference = initial triangular number + 1.

E.g. Between 2nd triangular number and 3rd triangular number, difference = 3
Initial triangular number is the 2nd triangular number.

Therefore,
M = (rth triangular number) + 1
41 – 1 = rth triangular number
rth triangular number = 40

So the answer is B.


17) F

When diaphragm flattens, it contracts. So statement 2 is correct.

Volume of thorax increases. So statement 3 is correct.

Pressure decreases so statement 6 is correct.

F is the answer.


18) E

The double bond between 2nd and 3rd carbon atom becomes a single bond. The carbon atoms that were part of the double bond will each have a methyl group attached (CH3). This is only shown by option E.

In A, there are no methyl groups.

In B, there are methyl groups on every alternate carbon atom. So this is wrong.

In C, not all carbon atoms part of the polymer chain have a methyl group.

In D, not all carbon atoms part of the polymer chain have a methyl group.


19) B

Maximum current is when there is minimum resistance. The lowest value of resistance = 10 + 2 = 12 Ω

I = V ÷ R = 6 ÷ 12 = 0.5 A

Minimum current is when there is maximum resistance. The highest value of resistance = 10 + 20 = 30 Ω

I = V/R = 6 ÷ 30 = 0.2 A

Difference = 0.5 – 0.2 = 0.30 A


20) A

We can find the total mass of the first 16 sweets.

Total mass of 16 sweets = 16 x 9.5 = 152 g

Then 4 more sweets, each of mass ‘x’ were added.

Mass of the 4 sweets = 4x

Total mass of 20 sweets = 152 + 4x

Average mass = (152 + 4x) ÷ 20

Now substitute values given in the option into the equation.

Option A:
If x = 12,
Average mass = 10g
If x = 14.5,
Average mass = 10.5 g
So x must be greater than 12 but less than or equal to 14.5
So A is correct.

Option B is wrong because it says (x < 14.5). ‘x’ can be lesser than OR equal to 14.5, which is why option A is correct.

Option C is wrong because it says (12 ≤ x). ‘x’ cannot be equal to 12 as mean mass of a sweet MUST be greater than 10 g.

Options D, E and F are just completely wrong.


21) A

The chemical inhibits respiration, which means that glucose in the cell is not converted to energy. Therefore, there is glucose present in the cell.

Therefore, glucose does not move into the cell.

Water moves into the cell because there is a higher water concentration outside the cell than inside the cell.

So A is the answer.


22) C

Mass of oxygen = 105 – 57 = 48

To find the empirical formula, we divide the masses of the elements with their Ar.

For fluorine, 57/19 = 3
For oxygen, 48/16 = 3

Then, we divide all obtained values by the lowest value. In this case, we only receive one value, 3.

So the empirical formula is OF

Since it is relative molecular mass is double of empirical formula mass, the molecular formula would be O2F2.


23) B

Mass number only decreases by alpha decay. After each alpha decay, the mass number decreases by 4.

Decrease in mass number = 244 – 220 = 24
Number of alpha decays = 24 ÷ 4 = 6

After 6 alpha decays, the atomic number will decrease by 12.
So atomic number would decrease to 82 (94 – 12 = 82)

After each beta decay, the atomic number increases by 1. We know the atomic number of radon is 86. So there must be 4 beta decays to increase the atomic number from 82 to 86.

After 6 alpha decays and 4 beta decays, plutonium-244 becomes radon-220.

So the answer is B.


24) B

   x   1.     –           x2 + 3     1.5  
  x – 1             x2 + 2x – 3    

First we need to factorise x2 + 2x – 3 by splitting the middle term
x2 + 2x – 3
x2 + 3x – x – 3    
x (x + 3) – 1 (x + 3)
(x – 1) (x + 3)

 x   1.     –           x2 + 3     1.5
x – 1           (x – 1) (x + 3)

The common denominator is (x – 1)

We get,

     x   – (x2 + 3)
            (x + 3)   1.5
         x – 1

Solve the numerator first,

 x   – (x2 + 3) 
         (x + 3)                                                     

x (x + 3) – x2 – 3
        (x + 3)

x2 + 3x – x2 – 3
       (x + 3)                                

3x – 3
(x + 3)

3(x – 1)
(x + 3)

                                                  

The whole fraction is:

3(x – 1)   ÷ (x – 1)
(x + 3)

3(x – 1)   x        1       1.5
(x + 3)           (x – 1)

(x – 1)s cancel out

We are left with:

     3    5
( x + 3)

So B is the answer.


25) C

Let dominant allele be B
Let recessive allele be b

Genotype of male fly = Bb
Genotype of female fly = Bb

Simple monohybrid cross produces 3 genotypes: BB, Bb and bb (You should know how to do the cross).

Number of phenotypes is only 2. BB and Bb produce brown flies whilst bb produces black flies.

Max number of genotypes : Max number of phenotypes = 3 : 2

So C is the answer.


26) C

Xenon hexafluoride has 1 xenon atom and 6 fluorine atoms. (The prefix hexa- means six)

Fluorine is diatomic gas. In order to receive 6 fluorine atoms, we need to use three fluorine gas molecules (3F2)

Xenon is a noble gas so it is present as an atom only.

We can write down the equation too.
Xe + 3F2  ->   XeF6     ∆H = -330 kJ/mol

Number of F – F bonds broken = 3
Energy = 158 x 3 = 474 kJ/mol

There are 6 Xe – F bonds.
Let energy of 1 Xe – F bond = M (any letter)
474 + 330 = 6M
6M = 804
M = 134 kJ/mol


27) E

When the material is living the ratio is 1000 : 1015

When the material died, the ration became 100 : 1015

This means that the ratio decreased by a factor of 10.

1 half-life = 6000 years

Let number of half-lives be M

1000/100 = 2M

10 = 2M

We know 23 = 8 and 24 = 16

So number of half-lives is a decimal between 3 and 4. (Let us take it as 3.5)

Number of half-lives = 3.5
Number of years = 6000 x 3.5 = 21,000 years
This is close to 20,000 so E is the answer.
2018 Section 3