x̄ - > KCSE Maths 2007 Paper 1 — Section II (Worked Solutions)

KCSE Mathematics — Paper 1 (2007) — Section II

Worked solutions: stepwise reasoning, clear conclusions, and exam-style presentation.

Author: Zacharia Maganga · Format: Responsive Blogger post · Rendered with MathJax

Contents

1. Question 11 — Compound Interest
2. Question 12 — Circle Theorem & Chords
3. Question 13 — Simultaneous Equations
4. Question 14 — Trigonometry: Bearings
5. Question 15 — Statistics: Grouped Data
6. Question 16 — Mensuration: Frustum

Q11Compound Interest and Growth (10 marks)

A sum of money, \(K\), is deposited in a bank earning 10% per annum compound interest. After 3 years, the amount grows to Ksh 14641.

Find (1) the principal \(K\); (2) the amount after 5 years.

Show worked solution

1. Recall compound interest formula: \(A=P(1+\frac{r}{100})^n\).
2. Given \(A=14641,\ r=10,\ n=3\) so \(14641=P(1.1)^3=P(1.331)\).
3. Thus \(P=\dfrac{14641}{1.331}=11000\). (Principal = Ksh 11,000)
4. Amount after 5 years: \(A_5=11000(1.1)^5=11000(1.61051)\approx 17715.61\Rightarrow\) Ksh 17,716 (approx).

Final: Principal = Ksh 11,000. Amount after 5 years ≈ Ksh 17,716.

Q12Geometry — Circle Theorem & Chords (10 marks)

A circle centre \(O\) has chords \(AB\) and \(CD\) intersecting at \(X\) inside the circle. \(AX=4\,cm,\ BX=6\,cm,\ CX=3\,cm\). Find \(XD\), and radius \(r\) given that \(OD=r\) and the line \(OD\) is perpendicular to \(CD\).

Show worked solution

1. Intersecting chords theorem: \(AX\times BX=CX\times DX\).
2. So \(4\times6=3\times DX\Rightarrow 24=3DX\Rightarrow DX=8\,cm.\)
3. Since \(OD\perp CD\), \(OD\) bisects chord \(CD\). Thus \(CD=3+8=11\,cm\) and half-chord \(CM=5.5\,cm\).
4. Radius: \(r^2=OM^2+CM^2=OM^2+(5.5)^2\). If more data (e.g. \(OX\) or \(OM\)) is provided, substitute to find numeric \(r\). Otherwise leave in terms of \(OM\).

Final: \(XD=8\,cm.\) Radius \(r=\sqrt{OM^2+5.5^2}\) (requires an additional distance to evaluate numerically).

Q13Simultaneous Equations — Quadratic & Linear (12 marks)

Solve \(y=2x+3\) and \(x^2+y^2=65\).

Show worked solution

1. Substitute \(y=2x+3\) into \(x^2+y^2=65\): \(x^2+(2x+3)^2=65\).
2. Expand: \(x^2+4x^2+12x+9=65\Rightarrow5x^2+12x-56=0\).
3. Quadratic formula: \(x=\dfrac{-12\pm\sqrt{144+1120}}{10}=\dfrac{-12\pm\sqrt{1264}}{10}\). \(\sqrt{1264}\approx35.556\).
4. Solutions: \(x\approx2.3556\Rightarrow y\approx7.7112;\quad x\approx-4.7556\Rightarrow y\approx-6.5112\).

Final: \((x,y)\approx(2.36,7.71)\) and \((-4.76,-6.51)\) (round as needed for marking).

Q14Trigonometry — Bearings & Distances (12 marks)

Ship A sails due east from a port; Ship B sails N 30° E from the same port. After both sail 100 km, find distance between the ships and bearing of B from A.

Show worked solution

1. Angle between paths = 60° (east vs. N30°E). Use cosine rule: \(AB^2=100^2+100^2-2(100)(100)\cos60^\circ\).
2. Compute: \(AB^2=20000-20000(0.5)=10000\Rightarrow AB=100\,km.\)
3. Bearing from A: by symmetry the triangle is isosceles and the bearing of B from A is approximately N 30° E.

Final: Distance = 100 km. Bearing of B from A = N 30° E.

Q15Statistics — Frequency Distribution (12 marks)

Grouped data for 40 students given in classes 0–9, 10–19, ..., 50–59 with frequencies 2,4,8,12,10,4 respectively. Find mean, median, histogram and estimate mode.

Show worked solution

1. Midpoints: 4.5, 14.5, 24.5, 34.5, 44.5, 54.5. Compute \(fx\) and sum: \(\Sigma f=40,\ \Sigma fx=1340\).
2. Mean = \(\dfrac{1340}{40}=33.5\).
3. Median: locate class containing the 20th value: cumulative freqs 2,6,14,26 so median class = 30–39. \(\text{Median}=L+\left(\dfrac{N/2-c.f_{before}}{f_m}\right)h\ =30+\dfrac{20-14}{12}\times10=35.\)
4. Mode (grouped): \(L+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times h = 30+\dfrac{12-8}{24-8-10}\times10\approx36.7.\)

Final: Mean = 33.5; Median = 35; Mode ≈ 36.7.

Q16Mensuration — Frustum of a Cone (12 marks)

Frustum height = 12 cm, bottom radius = 7 cm, top radius = 3.5 cm. Take \(\pi=3.142\). Find volume and total surface area (TSA).

Show worked solution

1. Volume: \(V=\dfrac{1}{3}\pi h(R^2+Rr+r^2)\). Substitute: \(=\dfrac{1}{3}\pi(12)(49+24.5+12.25)=343\pi\approx1078.7\,cm^3.\)
2. Slant height \(l=\sqrt{h^2+(R-r)^2}=\sqrt{144+12.25}=12.5\,cm.\)
3. Curved surface area \(=\pi(R+r)l=\pi(10.5)(12.5)=131.25\pi\approx412.3\,cm^2.\) Base and top areas \(=\pi(R^2+r^2)=61.25\pi\approx192.4\,cm^2.\)
4. Total \(\approx412.3+192.4=604.7\,cm^2.\)

Final: Volume ≈ 1078.7 cm³; TSA ≈ 604.7 cm².

Prepared as a revision aid — faithful to the KCSE Paper structure. Edit freely for Blogger publishing.