KCSE 2024 Mathematics Paper 2 Marking Scheme

Section I (50 Marks)

Answer all the questions in the spaces provided.

1. An arithmetic progression (AP) is given as \(600, 650, 700, 750, \dots\)

Determine:

(a) The \(30^{\text{th}}\) term of the AP

The formula for the \(n\)-th term of an arithmetic progression:

\(T_n = a + (n-1)d\)

Given:

\(a = 600 \\ n = 30, \\ d = 700 - 650 \\ = 50\)

Substituting the values:

\(T_{30} = 600 + 50(30 - 1)\)

\(T_{30} = 600 + 50 \times 29\)

\(T_{30} = 2050\)

Answer: \(T_{30} = 2050\)

(b) The sum of the first 30 terms of the AP

Using the formula:

\(S_n = \frac{n}{2} [a + l]\)

where \(l\) is the last term (\(T_{30}\)):

\(S_{30} = \frac{30}{2} [600 + 2050]\)

\(S_{30} = 15 \times 2650\)

\(S_{30} = 39750\)

Alternatively, using the other formula:

\(S_n = \frac{n}{2} [2a + (n-1)d]\)

Substituting:

\(S_{30} = \frac{30}{2} \left[ (600 \times 2) + 50(30 - 1) \right]\)

\(S_{30} = 15 [1200 + 1450]\)

\(S_{30} = 15 \times 2650\)

\(S_{30} = 39750\)

Answer: \(S_{30} = 39750\)

2. The quadratic equation \(5x^2 + kx + 20 = 0\) has only one root.

Determine the possible values of \(k\). (2 marks)

For a repeated root, the discriminant is given by:

\(b^2 - 4ac = 0\)

Thus, \(b^2 = 4ac\).

Substituting the values:

\(a = 5, \, b = k, \, c = 20\)

\(k^2 = 4 \times 5 \times 20\)

\(k^2 = 400\)

Taking the square root of both sides:

\(k = \pm \sqrt{400} = \pm 20\)

Therefore, \(k = 20 \, \text{or} \, -20\).

3. Without using mathematical tables or a calculator, evaluate:

\(\frac{\log 125 + \log 64}{\log \sqrt{5} + \log \sqrt[3]{2}}\) (3 marks)

Solution:

First, express the given numbers in terms of powers of prime numbers:

\(125 = 5^3, \, 64 = 2^6, \, \sqrt{5} = 5^{\frac{1}{2}}, \, \sqrt[3]{2} = 2^{\frac{1}{3}}\)

Rewrite the expression:

\(\frac{\log (5^3 \cdot 2^6)}{\log (5^{\frac{1}{2}} \cdot 2^{\frac{1}{3}})}\)

Using the logarithmic property \(\log(ab) = \log a + \log b\):

\(\frac{\log 5^3 + \log 2^6}{\log 5^{\frac{1}{2}} + \log 2^{\frac{1}{3}}}\)

Using the logarithmic property \(\log a^n = n \log a\):

\(\frac{3\log 5 + 6\log 2}{\frac{1}{2}\log 5 + \frac{1}{3}\log 2}\)

Simplify the numerator and denominator:

\(\frac{3\log 5 + 6\log 2}{\frac{1}{6}(3\log 5 + 6\log 2)}\)

Cancel common terms and simplify further:

\(\frac{3}{\frac{1}{6}} = 3 \times 6 = 18\)

Alternatively:

Rewrite the original expression using properties of logarithms:

\(\frac{\log 5^3 + \log 2^6}{\log 5^{\frac{1}{2}} + \log 2^{\frac{1}{3}}}\)

Evaluate each logarithmic term:

\(\frac{3\log 5 + 6\log 2}{\frac{1}{2}\log 5 + \frac{1}{3}\log 2}\)

Factorize and simplify as above to get:

\(18\)

Final Answer:

\(18\)

4. Make \(x\) the subject of the formula \(y = \frac{a}{b^x}\) (3 marks)

Start with the equation:

\(y b^x = a \implies b^x = \frac{a}{y}\)

Taking logarithms on both sides:

\(x \log b = \log \left( \frac{a}{y} \right)\)

Using the logarithmic property \(\log \left( \frac{a}{b} \right) = \log a - \log b\):

\(x \log b = \log a - \log y\)

Divide through by \(\log b\):

\(x = \frac{\log a - \log y}{\log b}\)

Alternatively:

Start with the equation:

\(y b^x = a \implies b^x = \frac{a}{y}\)

Taking logarithms on both sides:

\(x \log b = \log \frac{a}{y}\)

Again, use the property \(\log \frac{a}{b} = \log a - \log b\):

\(x \log b = \log a - \log y\)

Divide by \(\log b\):

\(x = \frac{\log a - \log y}{\log b}\)

Final Answer:

\(x = \frac{\log a - \log y}{\log b}\)

5. The unshaded region on the Cartesian plane satisfies the inequalities

\(x + 3y \leq 21\), \(-3x + y < 3\), \(-3x + y \geq -13\) and \(x + 3y \geq 11\).

6. An aircraft flew due west from point A (39.64°N, 50°E) to B (39.64°N, 20°W). Calculate the distance covered by the aircraft correct to the nearest km. (Take \( \pi=\frac{22}{7}\) and \(R=6370~km\))

Longitude difference, \( \theta=50^{\circ}+20^{\circ}=70^{\circ}\) (A and B are on either side of the prime meridian)

\(\beta\)=latitude angle, \(R\)=Radius of the earth=6370 km

Distance along a parallel of latitude

\(=\frac{\theta}{360^{\circ}}\times2\pi R~cos~\beta\)

\(=\frac{70^{\circ}}{360^{\circ}}\times2\times\frac{22}{7}\times6370\times cos~39.64^{\circ},\)

\(=\frac{70070}{9}\times0.7701\)

=5995.6563km

≈ 5996 km

7. In the following figure, A, B, C, D, and E are points on the circumference of the circle. Line AB is parallel to line DC, and line FAG is tangent to the circle at A. Angle GAB = 25° and ∠ABC = 100°.

a) \(\angle BAC\) = 180 - (100 + 25) = 55°(sum of angles in a triangle is \(180^{o}\))

(b) \(\angle DCA = 55°\) (alternate angles are equal)

\(\angle AED\) = 180 - 55 (opposite angles of a cyclic quadrilateral)

= 125°

8. The triangle ABC with vertices A(1,0), B(3,0), and C(1,2) is transformed by the matrix

\( T = \begin{pmatrix} 3k & 1.6 \\ 3k & -0.9 \end{pmatrix} \)

onto triangle A'B'C'. Given that the area of triangle A'B'C' is 6 square units, determine the value of k.(3 marks)

Area of ΔABC = \( \frac{1}{2} \times 2 \times 2 \)

= 2 sq units

\( \det T = \begin{vmatrix} 3k & 1.6 \\ 3k & -0.9 \end{vmatrix} \)

\( = 3k(-0.9) - 3k(1.6) \)

\( = -2.7k - 4.8k \)

\( = -7.5k \)

ASF = Area of ΔA'B'C' / Area of ΔABC

\( = \frac{6}{2} \)

\( = 3 \)

Det T = ASF, hence -7.5k = 3

\( k = \frac{3}{-7.5} \)

\( = -\frac{1}{4} \text{ or } -0.25 \)

9. Solve the equation \(6 \cos^2 x + \sin x = 4\) for \(0^\circ \leq x \leq 180^\circ\), giving the answer correct to 2 decimal places.(3 marks)

\( \sin^2 x + \cos^2 x = 1\\ \implies \cos^2 x = 1 - \sin^2 x \)

\( 6(1 - \sin^2 x) + \sin x = 4 \)

\( 6 - 6 \sin^2 x + \sin x = 4 \)

\( 6 \sin^2 x - \sin x - 2 = 0 \)

\( 3 \sin x - 2 = 0 \) or \( 2 \sin x + 1 = 0 \)

\( 3 \sin x = 2 \) or \( 2 \sin x = -1 \)

\( \sin x = 0.6667 \) or \( \sin x = -0.5 \)

Since sine is positive in the range \( 0^\circ \leq x \leq 180^\circ \),

\( x = \sin^{-1} 0.6667 \)

\( = 41.81^\circ, 138.19^\circ \)

10. The figure below shows a circle. Lines CP and CQ are tangents to the circle at points P and Q respectively.

The circle is to be inscribed in a triangle ABC. Point B lies on CQ produced and \(\angle CBA=90°\). Use a ruler and a pair of compasses only to:

(a) locate point O, the centre of the circle;

(b) complete triangle ABC.

11. The deviations of the masses of 10 students from an assumed mean are: -10, -5, -2, 1, 4, 5, 7, 8, 9, 13

The mass of the heaviest student was 58 kg. Calculate the mean mass of the students.(3 marks)

Let the assumed mean be A : \(58 - A = 13 \\ \Rightarrow A = 58 - 13 \\ = 45\)

\(\overline{x}=A+\frac{\sum fd}{\sum f} \\=45+\frac{30}{10}=48\)

12. The following table shows part of a monthly income tax rates for a certain year.

In a certain month an employee paid a net tax of Ksh. 2 200 after getting a tax relief of Ksh. 1 280. Calculate the employee’s taxable income that month. (3 marks)

Gross tax: \( \left(\frac{10}{100}\times11180\right)+ \\ \left(\frac{15}{100}\times10534\right)+ \\ \left(\frac{20}{100}\times y\right) \\ =2200+1280 \)

\(1118 + 1580.10 + 0.2y \\ = 3480 \)

\(0.2y = 781.9 \)

\(y = \text{Ksh. } 3909.50 \)

Taxable income = Ksh. (21714 + 3909.50) = Ksh. 25 623.50

13. The equation of a circle is given by \(x^2 + y^2 - 3x + 4y = 0\). Determine:

(a) the coordinates of the centre of the circle;

Compare \(x^2 + y^2 - 3x + 4y = 0\) with

\(x^2 + y^2 - 2ax + 2by + a^2 + b^2 - r^2 = 0\)

By comparing:

\(-2a = -3 \Rightarrow a = \frac{-3}{-2} = 1.5\)

\(-2b = 4 \Rightarrow b = \frac{4}{-2} = -2\)

Centre (1.5, -2)

Alternatively

Rearranging: \(x^2 - 3x + y^2 + 4y = 0\)

\(x^2 - 3x + (\frac{-3}{2})^2 + y^2 + 4y + (\frac{4}{2})^2 = (\frac{-3}{2})^2 + (\frac{4}{2})^2\)

\((x - 1.5)^2 + (y + 2)^2 = 6.25\)

\((x - a)^2 + (y - b)^2 = r^2\)

\(-a = -1.5 \Rightarrow a = 1.5\)

\(-b = 2 \Rightarrow b = -2\)

Centre (1.5, -2)

(b) the area of the circle in terms of π.

\(r^2 = 6.25\) hence area of circle = πr² = 6.25π sq.units

or using a² + b² - r² = c, r² = a² + b² - c

r² = 1.5² + (-2)² - 0

= 6.25

Area of circle = πr² = 6.25π sq.units

14. The position vectors of points A, B and C are such that \(OA = 3i - 5j - 4k\), \(OB = j + 8k\) and \(OC = -2i + 5j + 16k\). Show that the points A, B and C are collinear.

Let AC and AB be 2 vectors such that \(AC = kAB\).

\[AC = OC - OA = (-2i + 5j + 16k) - (3i - 5j - 4k)\]

\[= -5i + 10j + 20k\]

\[AB = OB - OA = (j + 8k) - (3i - 5j - 4k)\]

\[= -3i + 6j + 12k\]

\[-5i + 10j + 20k = k(-3i + 6j + 12k)\]

\[k = \frac{-5}{-3} = \frac{10}{6} = \frac{20}{12} = \frac{5}{3}\]

\(AC = \frac{5}{3}AB\) hence \(AC || AB\) and A is a common point therefore A, B and C are collinear.

Alternatively

\[\begin{bmatrix} -2 \\ 5 \\ 16 \end{bmatrix} - \begin{bmatrix} 3 \\ -5 \\ -4 \end{bmatrix} = \begin{bmatrix} -5 \\ 10 \\ 20 \end{bmatrix}\]

\[\begin{bmatrix} 0 \\ 1 \\ 8 \end{bmatrix} - \begin{bmatrix} 3 \\ -5 \\ -4 \end{bmatrix} = \begin{bmatrix} -3 \\ 6 \\ 12 \end{bmatrix}\]

\[\frac{-5}{-3} = \frac{10}{6} = \frac{20}{12} = \frac{5}{3}\]

\(AC = \frac{5}{3}AB\) hence \(AC || AB\) and A is a common point therefore A, B and C are collinear.

17. A poultry dealer has two types of chicken feeds: type A and type B. He sells 1 kg of type A at Ksh. 45 and 1 kg type B at Ksh. 30. He makes a profit of 20% per kg of type A feed sold and 25% per kg of type B feed sold. He also sells mixtures of type A and type B feeds.

(a) Determine the amount of profit made by the dealer for selling 1 kg of:

(i) type A feed.

Profit = \(\frac{20}{120} \times 45\) Ksh. = 7.50 Ksh.

OR

Profit = 45 - \(\frac{100 \times 45}{120}\) = Ksh. 7.50

(ii) type B feed.

Profit = \(\frac{25}{125} \times 30\) = Ksh. 6

OR

Profit = 30 - \(\frac{100 \times 30}{125}\) = Ksh. 6

(b) Type A and type B feeds were mixed in the ratio 3:7. Calculate:

(b) Type A and type B feeds were mixed in the ratio 3:7. Calculate:

(i) the selling price of 1 kg of the mixture;

Selling price = \(\frac{(3 \times 45) + (7 \times 30)}{3 + 7}\) = \(\frac{345}{10}\) = Ksh. 34.50

(ii) the profit made by the dealer in selling 50 kg of the mixture.

Profit in selling 1 kg of mixture = \(\frac{(3 \times 7.50) + (7 \times 6)}{10}\) = Ksh. 6.45

Profit in selling 50 kg of mixture = 50 x 6.45 = Ksh. 322.50

(c) The dealer made a profit of Ksh. 1 387.50 for the sale of 200 kg of a different mixture of type A and type B feeds. Determine the ratio of type A feed to that of type B feed in the mixture.

Let the ratio of A:B in the mixture be x:y.

\[200\left(\frac{7.50x+6y}{x+y}\right)=1387.50\]

Dividing both sides by 200...

\[\frac{7.50x+6y}{x+y}=\frac{111}{16}\]

\[16(7.50x+6y)=111(x+y)\]

\[120x+96y=111x+111y\]

\[9x=15y\]

Dividing both sides by 9y...

\[\frac{x}{y}=\frac{5}{3}\]

A:B = 5:3

18.(a) A quantity P is partly constant and partly varies as the square root of a quantity Q. Given that P = 20 when Q = 4 and that P = 60 when Q = 100, find Q when P = 22.

\[P=C+k\sqrt{Q}\]

\[20=C+k\sqrt{4} \Rightarrow 20=C+2k\] ...(i)

\[60=C+k\sqrt{100} \Rightarrow 60=C+10k\] (ii)

(ii) - (i): \(8k=40\)

\[k=5\]

Using (i), \(C=20-(2\times5)=10\)

\[P=10+5\sqrt{Q}\]

When P = 22;

\[5\sqrt{Q}=22-10\]

\[5\sqrt{Q}=12\]

\[\sqrt{Q}=2.4\]

\[Q=2.4^{2}\]

\[=5.76\]

(b) Three quantities, T, U and V are such that T varies directly as the square of (10-U) and inversely as the cube root of V. When T = 12, U = 4 and V = 8.

(i) Determine the equation connecting T, U and V.(3 marks)

\[T \propto \frac{(10-U)^2}{\sqrt[3]{V}}\]

\[T = \frac{k(10-U)^2}{\sqrt[3]{V}}\]

Given T = 12 when U = 4 and V = 8;

\[12 = \frac{k(10-4)^2}{\sqrt[3]{8}}\]

\[12 = \frac{k \times 6^2}{2}\]

\[36k = 24\]

\[k = \frac{2}{3}\]

Therefore, the equation connecting T, U and V is:

\[T = \frac{2(10-U)^2}{3\sqrt[3]{V}}\]

(ii) Find U when \(T = 5\frac{2}{5}\) and \(V = 15\frac{5}{8}\).

\[\frac{27}{5} = \frac{2(10-U)^2}{3 \times \sqrt[3]{15.625}}\]

\[27 \times 3 \times \sqrt[3]{15.625} = 5 \times 2(10-U)^2\]

\[27 \times 3 \times 2.5 = 10(10-U)^2\]

\[\frac{27 \times 3 \times 2.5}{10} = (10-U)^2 \times \frac{1}{10}\]

\[\sqrt{20.25} = \sqrt{(10-U)^2}\]

\[\pm4.5 = 10-U\]

\[U = 10 - 4.5 = 5.5\] Or \[U = 10 + 4.5 = 14.5\]

21.(a) Fadhili deposited Ksh. 400,000 in an account that paid compound interest on deposits at a rate of 7% p.a. At the end of 3 years, he withdrew all the money from the account.

(i) Calculate the amount that Fadhili withdrew.

Accumulated amount = \(400,000\left(1 + \frac{7}{100}\right)^3\)

= \(400,000 \times 1.225043\)

= Ksh. 490,017.20

∴ Amount he withdrew is Ksh. 490,017.20

(ii) Fadhili invested the withdrawn amount in shares. The value of the shares depreciated at a rate of 1.5% every 6 months. Determine the value of the shares at the end of 2 years correct to 2 decimal places.

\[A=P\left(1-\frac{r}{100}\right)^{n}\]

\[n=2\times2=4\]

Amount = \(490,017.20\left(1-\frac{1.5}{100}\right)^{4}\)

= \(490,017.20\times0.94133655\)

= Ksh. 461,217.10

(iii) Determine the gain or loss from Fadhili’s investments in the 5 years.

Negative if loss, positive if gain.

Gain/Loss = Amount at the end of 5 years - Initial amount

= Ksh. (461,217.10 - 400,000)

= Ksh. 61,217.10 (gain)

(b) Nyambuto invested Ksh. 400,000 in a financial institution that paid compound interest at the rate of 6% per annum. After n years, the amount had accumulated to Ksh. 500,000. Calculate the value of n, correct to the nearest whole number.

From \(A=P\left(1+\frac{r}{100}\right)^{n} \Rightarrow \left(1+\frac{r}{100}\right)^{n}=\frac{A}{P}\)

Taking logs on both sides;

\[n\log\left(1+\frac{r}{100}\right)=\log A - \log P\]

\[n=\frac{\log A - \log P}{\log\left(1+\frac{r}{100}\right)}\]

\[n=\frac{\log 500000-\log 400000}{\log\left(1+\frac{6}{100}\right)}\]

\[=\frac{5.69897-5.60206}{0.0253059}\]

\[=3.8295\]

\[≈4 years\]

22. The probabilities of obtaining scores 1, 2, 3, 4 and 5 using a biased pentagonal spinner were recorded as shown in the following table.

(a) Determine:

(i) the value of \(k\).

If all possible values of a random variable are included in the probability distribution then the sum of the probabilities is 1

\[k + 0.1 + 0.25 + 2k + 0.2 = 1\]

\[3k + 0.55 = 1\]

\[3k = 1 - 0.55\]

\[3k = 0.45\]

\[k = \frac{0.45}{3} = 0.15\]

(ii) the probability of obtaining a score of 4.

\(P(score = 4) = 2k\)

\(= 2 \times 0.15 = 0.3\)

(b) The spinner was spun twice.

(i) Work out the probability of obtaining an even number in the first spin and an odd number in the second spin.

Probability space = 25

Favourable outcomes = 6

\(P(\text{even and odd}) = \frac{6}{25} = 0.24\)