10 Hardest ESAT Questions with Worked Solutions

1. The three regions outside $PQR$ are trapezium $APRD$, triangle $PBQ$, and triangle $QCR$. Their areas are

   $$2(x+2),\qquad \frac{x(4-x)}2,\qquad 4-x.$$

2. Subtract these from the square’s area:

   $$\text{Area}(PQR)=8-3x+\frac{x^2}{2}.$$

   Set this equal to the given area:

   $$2x^2-12x+17=0.$$

3. The quadratic formula gives

   $$x=3\pm\frac{\sqrt2}{2}.$$

   Both values lie between 0 and 4. Their positive difference is $\boxed{\sqrt2}$.

1. Let $K$ be the number of red counters transferred. There are ten equally likely unordered pairs in X. One pair is red–red, six are mixed, and three are blue–blue.

   Possible transfers and their consequences

   $K$	Transfer probability	Y after transfer	Probability of a same-colour pair from Y
   0	$3/10$	1 red, 4 blue	$6/10$
   1	$6/10$	2 red, 3 blue	$4/10$
   2	$1/10$	3 red, 2 blue	$4/10$

2. So the probability of the observed same-colour pair is

   $$\frac3{10}\frac6{10}+\frac6{10}\frac4{10}+\frac1{10}\frac4{10}=\frac{46}{100}.$$

3. The probability of transferring two reds and then observing a same-colour pair is $4/100$. Restricting to the observed outcomes gives

   $$\frac{4/100}{46/100}=\boxed{\frac2{23}}.$$

1. The two given sums are

   $$\frac{a}{1-r}=12,\qquad \frac{ar}{1-r^2}=-4.$$

   Divide the second equation by the first:

   $$\frac{r}{1+r}=-\frac13.$$

2. Hence $3r=-1-r$, so $r=-\frac14$ and

   $$a=12(1-r)=15.$$

3. The squared terms have first term $a^2$ and ratio $r^2$:

   $$\frac{a^2}{1-r^2}=\frac{225}{1-1/16}=\boxed{240}.$$

1. Write $f(x)=x^3-3x^2+2x$. Since $f'(a)=3a^2-6a+2$, the tangent is

   $$y=(3a^2-6a+2)x-2a^3+3a^2.$$

2. Subtracting this line from the curve and factorising gives

   $$f(x)-y_{\rm tangent}=(x-a)^2(x+2a-3).$$

   The second intersection is therefore $b=3-2a$. As $a>1$, we have $a>b$, so

   $$a-b=3a-3=3.$$

   Thus $a=2$, $b=-1$, and the tangent is $y=2x-4$.

3. Between these intersections, the curve lies above the tangent because

   $$f(x)-(2x-4)=(x-2)^2(x+1)\geq0.$$

4. The required area is

   $$\int_{-1}^{2}(x^3-3x^2+4)\,\mathrm dx$$
   $$=\left[\frac{x^4}{4}-x^3+4x\right]_{-1}^{2}=\boxed{\frac{27}{4}}.$$

1. Let the battery voltage be $V$. With S open, the parallel branches each have resistance $2R$. Their combined resistance is $R$, giving total resistance $2R$.

   Therefore A has voltage $V/2$, so

   $$P_{A,\rm open}=\frac{V^2}{4R}=25.$$

   Hence $V^2/R=100\text{ W}$.

2. Closing S bypasses D. The parallel branches are now $2R$ and $R$, with combined resistance $2R/3$. The total resistance is $5R/3$.

   The voltage across B is the voltage across this parallel combination:

   $$V_B=V\frac{2R/3}{5R/3}=\frac{2V}{5}.$$

3. Using $P=V^2/R$ for resistor B gives

   $$P_B=\frac{(2V/5)^2}{2R}=\frac{2}{25}\frac{V^2}{R}=\boxed{8\text{ W}}.$$

You can obtain a parallel resistance from the current rules: at voltage $U$, branches $R$ and $2R$ draw total current $U/R+U/(2R)$, equivalent to resistance $2R/3$.

1. Momentum conservation gives the speed just after the collision:

   $$1(6)=(1+3)v,\qquad v=1.5\text{ m s}^{-1}.$$

   The kinetic energy available to compress the springs is

   $$\frac12(4)(1.5)^2=4.5\text{ J}.$$

2. Let the softer spring’s compression be $x$. The force is the same through both massless springs, so the stiffer spring compresses by $x/3$.

3. At maximum compression, all the joined carts’ kinetic energy is stored in the springs:

   $$\frac12(300)x^2+\frac12(900)\left(\frac{x}{3}\right)^2=4.5.$$
   $$200x^2=4.5,\qquad x^2=0.0225.$$

   Taking the positive root gives $\boxed{x=0.15\text{ m}}$.

1. Initial HCl is $0.0400\text{ mol}$. The remaining HCl is $0.0200\times0.500=0.0100\text{ mol}$, because HCl and NaOH react 1:1.

   So the salts consumed $0.0300\text{ mol}$ of HCl.

2. Let $x$ and $y$ be the moles of carbonate and hydrogencarbonate respectively. Each gives one mole of carbon dioxide per mole of salt. Therefore

   $$x+y=\frac{0.480}{24.0}=0.0200.$$

   Carbonate consumes two moles of HCl per mole, while hydrogencarbonate consumes one:

   $$2x+y=0.0300.$$

3. Subtracting gives $x=0.0100$, hence $y=0.0100$. The mass of the two salts is

   $$(0.0100)(106)+(0.0100)(84)=1.90\text{ g}.$$

4. The impurity has mass $2.50-1.90=0.60\text{ g}$. Its percentage is

   $$\frac{0.60}{2.50}\times100=\boxed{24\%}.$$

1. For one mole of alcohol, the carbon products are $n/3$ moles of $\mathrm{CO_2}$ and $2n/3$ moles of $\mathrm{CO}$. The hydrogen produces $n+1$ moles of water.

2. The products therefore contain this many moles of oxygen atoms:

   $$2\left(\frac n3\right)+\frac{2n}3+(n+1)=\frac{7n}{3}+1.$$

   One mole of oxygen atoms comes from the alcohol itself. The oxygen gas required per mole of alcohol is consequently

   $$\frac12\left(\frac{7n}{3}+1-1\right)=\frac{7n}{6}.$$

3. The alcohol’s molar mass is $14n+18\text{ g mol}^{-1}$. Use the measured oxygen consumption:

   $$\frac{3}{14n+18}\times\frac{7n}{6}=\frac{4.20}{24.0}=\frac7{40}.$$

   This simplifies to $20n=14n+18$, giving $n=3$.

   The molecular formula is $\boxed{\mathrm{C_3H_8O}}$.

1. The original cross is $Aa\times Aa$. Among its purple offspring, the proportions are

   $$P(AA)=\frac13,\qquad P(Aa)=\frac23.$$

2. If the selected parent is $AA$, both first offspring must be purple. If it is $Aa$, the probability that both are purple is $(3/4)^2=9/16$.

   The two routes to the observed result therefore have weights

   $$AA:\ \frac13=\frac8{24},\qquad Aa:\ \frac23\frac9{16}=\frac9{24}.$$

3. Restricting to those routes, the probability that the selected parent is $Aa$ is $9/(8+9)=9/17$.

   Only an $Aa$ parent can produce a white offspring by self-pollination; its probability of doing so is $1/4$. Hence

   $$\frac9{17}\times\frac14=\boxed{\frac9{68}}.$$

1. Water uptake is the volume swept by the bubble: $\pi d^2L/4$. Therefore the ratio of uptake rates per unit leaf area is

   $$\frac{\text{Y}}{\text{X}}=\left(\frac{1.20}{0.80}\right)^2\frac{32}{24}\frac48\frac{60}{90}.$$
   $$=\frac94\times\frac43\times\frac12\times\frac23=1.$$

2. Only 80% of Y’s uptake and 90% of X’s uptake are transpired. The required ratio is therefore

   $$1\times\frac{0.80}{0.90}=\boxed{\frac89}.$$

A potometer measures water uptake. This question supplies the retained fractions so that actual water loss can be calculated.