Firstly, I would like to caution students regarding the use of solutions when preparing for exams. They are certainly a double-edged sword and since you are planning to invest an enormous amount of time, effort and money into getting a degree, it would be a good idea to use solutions properly from now on. In all cases you should only consult a solution once you think your answer is correct or if you have spent a significant amount of time trying to reach an answer. Coming up with your own conclusion after several minutes of independent thought is far more worthwhile than learning directly from the solution itself. You should also make sure you are critical of all solutions presented to you. Not only will this help your understanding, but solutions may have errors. Therefore if you find that you can't reconcile a solution with yours then take the time to get in touch with the author to discuss the discrepancy. For example, please do contact me if you have spotted a mistake below or would like further guidance on any of the questions.
I will firstly set out a series of hints. Try to do the questions from these alone and if you are still having difficulty then scroll down to the full solutions.

Hints



1. Make an equation from the question and solve it.
2. Equate kinetic energy lost to work done.
3. Rearrange with the optimal number of steps.
4. Note that the upper resistor has the same potential difference across it before and after. Work with $P=\frac{V^2}{R}$.
5. Work out lengths and subtract the two triangles and trapezium from the square.
6. Work out how the gradient relates to the spring constant then use Hooke's law.
7. Write everything as a power of $3$.
8. Apply $Pressure=\frac{Force}{Area}$.
9. Using percentage multipliers, write the old and new salaries of the coach in terms of the old and new salaries of the star player.
10. Formulate a table with (say) $2N$ nuclei of $X$ and $N$ nuclei of $Y$ and work in lowest-common multiples of the half-lives.
11. Make a complete $Speed=\frac{Distance}{Time}$ calculation.
12. There can be no horizontal lines since the temperature must vary throughout the material. The better the conductor the higher the temperature will remain as you move through it.
13. Identify the relevant scale factors, make an equation from the question and solve it.
14. Draw a schematic diagram of the situation while taking care to interpret the voltages in the question correctly.
15. Favour square rooting both sides over expanding. Cancel factors of $10$ early.
16. Note that the particle in question only moves up and down, that the wavelength is a red herring and that one cycle consists of $4$ amplitudes.
17. Draw a rough diagram to determine the position of $Y$ then subtract $\overrightarrow{QY}$ from it
18. Define the unknown resistance as $Y$ and the potential difference of the cell to be $V$, make equations and solve for $Y$ then $V$ then $I$.
19. Simplify the expression given so that you only have one sign of $sinx$ in the power.
20. Using the idea that pressure, $p\alpha h$, where $h$ is the height of the liquid, work out the pressure at $\frac{d}{2}$ and apply $PV=constant$.
21. Solve the quadratic and substitute its roots in to the cubic.
22. Work out the wavelength in terms of the length of the pipe, $L$ in the two scenarios.
23. Subtract the sum of the first $10$ terms from both sides to make the initial equation easier to deal with.
24. Realise early that the horizontal component of velocity never changes so it has no bearing on the impulse.
25. Equate the common ratio to the first term divided by the second term and the second term divided by the third term and solve for $p$.
26. The force due to friction balances the component of the gravitational force down the slope.
27. Connect all points on the circumference to each other and apply angle in a semi-circle is a right angle twice.
28. Use $P=I^2R$ to determine the value of the power dissipated in $P$ and $S$.
29. Take $\sqrt{2}$ out as a factor and complete the square.
30. Take moments about the top of one of the planks.
31. Note that this is an even function, change the lower limit to $0$ and double your result.
32. Work out the acceleration of the whole apparatus then apply Newton's second law to $R$.
33. Differentiate twice, substitute in and solve the resulting simultaneous equations.
34. Note that $h(t)$ will be a parabola. Then apply conservation of energy and $GPE=mgh$.
35. Bring all the logarithms onto one side, simplify the resulting fraction then solve.
36. Use Ohm's law to work out the resistance of the conductor then apply $R=\frac{\rho L}{A}$.
37. Note that the powers of the two terms are each $4$. Save time by writing $^8C_4$ in terms of factorials rather than writing out Pascal's triangle.
38. The phase difference of $\pi$ means that the path difference must be half a wavelength in order to achieve constructive interference for the first time.
39. Note that the first equation is the unit circle. Now let $2x+3y=a$ touch the circle at a tangent and use Pythagoras' Theorem three times to solve for $a$.
40. Springs in series follow the mathematical pattern of resistances in parallel. Apply this then use $EPE=\frac{1}{2}kx^2$.
Full solutions

1. Make an equation from the question and solve it.
$4\pi R^2=10\times(2\pi r^2+2\pi r\times 4r)$
$R^2=25r^2$
2. Equate kinetic energy lost to work done.
$\frac{1}{2}\times 10000\times2^2=1000d$
3. Rearrange with the optimal number of steps.
$\frac{q-r}{s-x}=p-y$
$s-x=\frac{q-r}{p-y}$
4. Note that the upper resistor has the same potential difference across it before and after. Work with $P=\frac{V^2}{R}$.
From the hint the answer is $E$,$F$,$G$ or $H$. From $P=\frac{V^2}{R}$, the lower two resistors have half the voltage therefore one quarter of the power. $1+0.25+0.25=1.5$
5. Work out lengths and subtract the two triangles and trapezium from the square.
From the question, $WM=\frac{1}{3}$, $MX=\frac{2}{3}$, $XN=\frac{3}{4}$, $NY=\frac{1}{4}$, $YP=\frac{4}{5}$ and $PZ=\frac{1}{5}$.
Shaded$=1-\frac{1}{4}-\frac{1}{10}-\frac{1}{2}(\frac{1}{3}+\frac{1}{5})$
which simplifies to $\frac{45-6-16}{60}=\frac{23}{60}$.
6. Work out how the gradient relates to the spring constant then use Hooke's law.
Having a variable squared as the independent variable is unfamiliar but helpful in this question in that $E=\frac{1}{2}kx^2$ so the gradient is $\frac{k}{2}$.
$k=2\times\frac{0.015}{0.0025}$
$F=kx=\frac{2\times0.003\times0.05}{0.0005}=2\times0.3$
7. Write everything as a power of $3$.
$\frac{3^{6(x-2)}}{3^{4x-6}}=3^6$
$6x-12-4x+6=6$
8. Apply $Pressure=\frac{Force}{Area}$.
$\frac{mg}{12}=0.45$
$\frac{g}{12}\times12\times15\times\rho=0.45$
$\rho=\frac{0.03}{g}=0.003kgcm^{-3}$
9. Using percentage multipliers, write the old ($C$) and new ($C'$) salaries of the coach in terms of the old ($S$) and new ($S'$) salaries of the star player and spot cancellations.
$C=0.8S$
$C'=1.15C$
$S'=1.38S$
$\frac{C'}{S'}=\frac{1.15\times0.8}{1.38}=\frac{1.15\times0.4}{0.69}=\frac{0.5\times0.4}{0.3}=\frac{2}{3}$
10. Formulate a table with (say) $2N$ nuclei of $X$ and $N$ nuclei of $Y$ and work in lowest-common multiples of the half-lives.
$X$ starts off at $2N$ and after $6$ days is at $\frac{N}{4}$. $Y$ starts off at $N$ and after $6$ days is at $\frac{N}{4}$.
11. Make a complete $Speed=\frac{Distance}{Time}$ calculation. Work in kilometres.
Speed=$\frac{0.4}{\frac{0.1}{6}+\frac{0.2}{10}+\frac{0.1}{20}}=\frac{24}{1+1.2+0.3}=\frac{24}{2.5}=\frac{48}{5}$
12. There can be no horizontal lines since the temperature must vary throughout the material. The better the conductor the higher the temperature will remain as you move through it.
Therefore the gradient should be shallower between $0$ and $a$.
13. Identify the relevant scale factors, make an equation from the question and solve it.
The scale factor for length is $\sqrt2$ therefore the scale factor for volume is $2\sqrt2$.
$Y=7\sqrt2+X=2\sqrt2X$
$X=\frac{7\sqrt2}{2\sqrt2-1}$
$=\frac{7\sqrt2(2\sqrt2+1)}{7}$
14. Draw a schematic diagram of the situation while taking care to interpret the voltages in the question correctly.
Defining turns ratio to be $N_p:N_s$, the turns ratio for the first transformer is $1:5$. Taking into account the ideal nature of the second transformer, the ratio of the currents must be the reverse of the turns ratio, so the turns ratio is $10:3$. Take a moment to check that this is a step-down transformer. the voltage for distribution in the town is $33000V$ so the initial voltage is:
$\frac{33000\times 10}{5\times 3}$
15. Favour square rooting both sides over expanding. Cancel factors of $10$ early.
$\frac{10000a+2000a}{0.3}=\pm10000\sqrt{80}$
with hindsight I could have divided both sides by $10^8$ before this first step.
$\frac{10a+2a}{0.3}=\pm10\sqrt{80}$
$120a=\pm30\sqrt{80}$
$a=\pm\frac{\sqrt{80}}{4}$
therefore the difference is $\frac{\sqrt{80}}{2}$.
16. Note that the particle in question only moves up and down, that the wavelength is a red herring and that one cycle consists of $4$ amplitudes. Work in centimetres.
$24$ oscillations.
$\frac{24\times12}{2}$
17. Draw a rough diagram to determine the position of $Y$ then subtract $\overrightarrow{QY}$ from it
The diagram does not need to be accurate to the situation. Draw a generic vector from $X$ to $P$ then one double its length from $P$ to $Y$. From the information, the coordinates of $Y$ are $(0,-1)$
18. Define the unknown resistance as $Y$ and the potential difference of the cell to be $V$, make equations and solve for $Y$ then $V$ then $I$.
$\frac{V^2}{10+Y}=6$
$\frac{VY}{10+Y}=4$
square the second equation
$\frac{V^2Y^2}{(10+Y)^2}=16$
Divide this by the first equation
$\frac{Y^2}{10+Y}=\frac{16}{6}$
$160+6Y=6Y^2$
$3Y^2-8Y-80=0$
$(Y+4)(3Y-20)=0$
Discard the negative resistance and substitute $Y=\frac{20}{3}$ into the second equation.
$\frac{20V}{10+\frac{20}{3}}=12$
$20V=120+80$
$V=10$
Applying $P=I^2R$
$6=I^2(10+\frac{20}{3})$
$I^2=\frac{6}{10+\frac{20}{3}}=\frac{18}{50}=0.36$

19. Simplify the expression given so that you only have one sign of $sinx$ in the power.
$(\frac{3}{2})^{-sinx}$
Noting that $\frac{3}{2}>1$, this is maximised when the power is as high as possible, so equal to $1$ in this scenario which is possible in this range for $x$
20. Using the idea that pressure, $p\alpha h$, where $h$ is the height of the liquid, work out the pressure at $\frac{d}{2}$ and apply $PV=constant$.
$P_{top}=atm$
$P_d=8\times atm$
$p_{\frac{d}{2}}=4\frac{1}{2}\times atm$
$P_{\frac{d}{2}}\times V_{\frac{d}{2}}=P_{top}\times V_{top}$
$V_{\frac{d}{2}}=\frac{V_{top}}{4.5}=\frac{720\times2}{9}$
21. Solve the quadratic and substitute its roots in to the cubic.
$(x-3)(x+2)$
$27+9a+6+b=0$
$-8+4a-4+b=0$
$5a=-45$
$a=-9$
$b=48$
22. Work out the wavelength in terms of the length of the pipe, $L$ in the two scenarios.
$\lambda=4L$
$f=\frac{v}{4L}=4000$
Opening the other end of the pipe halves the wavelength, thereby doubling the frequency.
23. Subtract the sum of the first $10$ terms from both sides to make the initial equation easier to deal with.
$\frac{9}{2}(2a+8d)=a+10d$
$9a+36d=a+10d$
$8a=-26d$
24. Realise early that the horizontal component of velocity never changes so it has no bearing on the impulse.
$v^2=u^2+2as=900$
$v=30$, $mv=30\times 2.6$
25. Equate the common ratio to the first term divided by the second term and the second term divided by the third term and solve for $p$.
$\frac{2p}{p-3}=\frac{p-3}{p-7}$
$2p^2-14p=p^2-6p+9$
$(p-9)(p+1)=0$
$p=9$ doesn't converge
$\frac{1}{1-\frac{1}{2}}$
26. The force due to friction balances the component of the gravitational force down the slope.
$60gsin30\times 15$
27. Connect all points on the circumference to each other and apply angle in a semi-circle is a right angle twice.
Labelling the points from the topmost one clockwise as $A$, $B$, $C$, $D$ and $E$;
$AEC$ is a right angle so $ECA=35$.
$BED$ is a right angle so $EDB=50$.
$ECB=50$ (angles in the same segment).
$ACB=ECB-ECA=15$
$DCB$ is a right angle, so $DCA=90-15$.
28. Use $P=I^2R$ to determine the value of the power dissipated in $P$ and $S$.
Using the above approach, the power dissipated in $R$ is $4$ and the power dissipated in each of $P$ and $S$ is $16$.
29. Take $\sqrt{2}$ out as a factor and complete the square.
$\sqrt2(x^2-3\sqrt2+2\sqrt2)$
$\sqrt2((x-\frac{3\sqrt2}{2})^2-\frac{9}{2}+2\sqrt2)$
$\sqrt2(x-\frac{3\sqrt2}{2})^2-\frac{9\sqrt2}{2}+4$
30. Take moments about the top of one of the planks.
$mg\frac{l}{2}cos60=lFcos30$
$F=\frac{mg}{2\sqrt3}$
31. Note that this is an even function, change the lower limit to $0$ and double your result.
$\frac{2}{3(3-\sqrt5)}[x^3]^{3-\sqrt5}_0$
$\frac{2(3-\sqrt5)^2}{3}$
$\frac{2(14-6\sqrt5)}{3}$
32. Work out the acceleration of the whole apparatus then apply Newton's second law to $R$.
$a=\frac{20}{4+2+2}=2$
Let $R_Q$ be the force exerted by $R$ on $Q$ and $Q_R$ be the force exerted by $Q$ on $R$.
$R_Q=Q_R$ from Newton's third law.
From the acceleration of $R$:
$Q_R-1=m_Ra=2\times2=4$
33. Differentiate twice, substitute in and solve the resulting simultaneous equations.
$f'(x)=-\frac{a}{x^2}-\frac{2b}{x^3}$
$f''(x)=\frac{2a}{x^3}+\frac{6b}{x^4}$
$-a-2b=2$
$-2a+6b=-2$
$b=-\frac{3}{5}$, $a=-\frac{4}{5}$
34. Note that $h(t)$ will be a parabola. Then apply conservation of energy and $GPE=mgh$.
Statement 1 is false. In statement 2, $KE\alpha h$ and the graph starts high and ends low as expected. In statement 3, $GPE$ starts high then decreases linearly with $s$.
35. Bring all the logarithms onto one side, simplify the resulting fraction then solve.
$\frac{x^2+3x+2}{x^2+2x}=9$
$\frac{(x+1)(x+2)}{x(x+2)}=9$
$x+1=9x$
36. Use Ohm's law to work out the resistance of the conductor then apply $R=\frac{\rho L}{A}$.
$\frac{20}{20+R}=\frac{1}{5}$
$100=20+R$
$R=80$
The length is increased by a factor of $4$ which reduces the cross-sectional area by a factor of $4$, both of which multiply the resistance by $4$.
37. Note that the powers of the two terms are each $4$. Save time by writing $^8C_4$ in terms of factorials rather than writing out Pascal's triangle.
$^8C_4\times\frac{1}{16}$
$\frac{8\times7\times6\times5}{4\times3\times2}\times\frac{1}{16}$
$\frac{70}{16}$
38. The phase difference of $\pi$ means that the path difference must be half a wavelength in order to achieve constructive interference for the first time.
$\frac{\lambda}{2}=d(\frac{1}{cos30}-1)$
39. Note that the first equation is the unit circle. Now let $2x+3y=a$ touch the circle at a tangent and use Pythagoras' Theorem three times to solve for $a$.
Draw a unit circle centred on the origin. Now draw a straight line in the first quadrant (i.e. it has positive $x$ and $y$ axis intercepts) which touches the circle at a tangent. Label this point $T$, the $x$ intercept $X$ and the $y$ intercept $Y$. The value of the $x$ intercept is $\frac{a}{2}$ and the value of the $y$ intercept is $\frac{a}{3}$.
$YT^2=\frac{a^2}{9}-1$
$TX^2=\frac{a^2}{4}-1$
$YX^2=\frac{a^2}{9}+\frac{a^2}{4}$
Since $YX=YT+TX$;
$\sqrt{\frac{a^2}{9}+\frac{a^2}{4}}=\sqrt{\frac{a^2}{9}-1}+\sqrt{\frac{a^2}{4}-1}$
$\frac{a^2}{9}+\frac{a^2}{4}=\frac{a^2}{9}-1+\frac{a^2}{4}-1+2\sqrt{\frac{a^2}{9}-1}\sqrt{\frac{a^2}{4}-1}$
Rearrange to leave the square root term alone then square again
$1=(\frac{a^2}{9}-1)(\frac{a^2}{4}-1)$
$36=(a^2-9)(a^2-4)=a^4-13a^2+36$
$a^2=\pm 13$ or $0$
40. Springs in series follow the mathematical pattern of resistances in parallel. Apply this then use $EPE=\frac{1}{2}kx^2$.
$k_{Total}=\frac{1}{\frac{1}{200}+\frac{1}{600}}=\frac{600}{3+1}=150$
$75\times(\frac{4}{5})^2$