0041-BASIC-MATHEMATICS-2024-NECTA-ALL - 04-BASIC-MATHEMATICS | SmartNote Notes

0041-BASIC-MATHEMATICS-2024-NECTA-ALL

Objectives: 0041-BASIC-MATHEMATICS-2024-NECTA

Basic Mathematics Form 4 - Questions

Basic Mathematics — Form Four (2024)

Question 1.

(a) Equal squares are drawn on a rectangular board measuring 54 cm by 78 cm. Find the size of each square so that the largest possible equal squares exactly cover the board. (b) In a class of 40 students, 17 are boys and the rest are girls. Determine: (i) the number of boys in decimal (ii) the percentage of girls and boys.

Question 2.

(a) Express 3 + 7√5 + √7 in the form a + b√c, where a, b, c are integers. (b) A line is defined by a logarithmic equation 3 log_x (x + 4) = log_x 24. Without using a mathematical table or calculator, find the value of x.

Question 3.

(a) In a school of 100 students, 45 students prefer Music, 40 students prefer Theatre Arts, and 5 students prefer both subjects. Using a formula, find the number of students who prefer none of the two subjects. (b) The table shows the number of tables in twenty offices from a certain company: [2,3,6; 6,2,1; 2,5,1; 4]. If one office is picked randomly, find the probability that the office has: (i) two tables (ii) at least five tables.

Question 4.

(a) A quadrilateral has vertices A(6,−4), B(8,4), P(6,1) and Q(7,5). The points P and Q have position vectors a and b respectively. If a = 2i − 7j and 2a + 3b = 13i − 2j, find the components of vector b. (b) Determine whether or not AB is parallel to PQ.

Question 5.

(a) The time t in seconds that is used by Mary to go back home from school varies inversely with her average speed v in metres per second. Write an equation expressing t in terms of v. (b) If she wants to get back home in 15 minutes, what will be her average speed? (c) If she took 30 minutes at an average speed of 10 m/s, find her distance.

Question 6.

(a) On one rainy day, it was observed that 850 millilitres of water were collected every minute. If it rained continuously from 8:10 a.m. to 11:52 a.m., calculate in litres the amount of water collected. (b) A firm reserved two triangular plots for the construction of more offices as shown in the diagram. (i) Prove that triangles AEB and CDE are similar. (ii) Using similarity properties, find the length of CD in metres. [Diagram shows AEB similar to CDE with AB=36 m, AE=?, BE=?, CD=?, etc.]

Q6(b): Similar triangles AEB and CDE (not to scale).

Question 7.

(a) The area of a regular six-sided plot of land inscribed in a circular track of radius r is 720 m². Find the value of r, rounding to the nearest tenth. (b) A person ran for 30 minutes at an average speed of 10 m/s. Write an equation and find: (i) their speed (ii) distance (if required).

Question 10.

(a) An entrepreneur borrowed a loan from a bank and repaid it on monthly instalment basis: 200,000; 220,000; 240,000 shillings for the first, second, and third instalments respectively, until the loan was fully repaid. If the final monthly instalment was 114,000 shillings, how long did the entrepreneur take to repay the whole loan? (b) Extract a trial balance from the trader's Cash Account with given transactions, then calculate the total amount of loan borrowed.

Question 11.

(a) An electricity post AB is supported by two pieces of wire AC and AD to the ground at points C and D. If AC is 10 m, the distance between C and D is 4 m and the post height AB is 6 m. By applying your knowledge of trigonometry show that 4 sin 75° = √6 + 2. (b) A wire is stretched from the top at point A to points C and D on the ground. Find the length of wire AD, leaving your answer in surd form.

Question 12.

(a) (i) The surface area of a sphere is 113.04 cm². Find its diameter (use π = 3.14). (ii) 75.360 litres of water are poured into a cylindrical tank of inside diameter 40 cm. Calculate the height of the water level. (b) In the figure, PT is a tangent to a circle with centre O and radius 5 cm. If PC is 8 cm, find the length of PT.

Q12(a)(ii): Cylindrical tank (inside diameter 40 cm) filled with 75.360 L.

Q12(b): Tangent PT to circle centre O; PC = 8 cm, radius 5 cm.

Question 13.

(a) Find the distance in nautical miles between the pairs of points: A(18°N, 12°E) and B(65°N, 12°E); C(31°S, 76°W) and D(22°N, 76°W). (b) Roza and Juma wrote 2×2 matrices R and J respectively: R = [[2,5],[-6,3]] and J = [[-6,2],[?,?]]. Find the sum of twice Roza's matrix and thrice Juma's matrix; then show that the difference between their determinants is t1 (as stated).

Question 14.

(a) During an inter-school quiz competition each correct answer was awarded x points related by the function y = 3x. (i) How many points would a competitor collect by answering 8 questions correctly? (ii) Find the domain and range of the inverse of the given function. (b) A farm measuring 72 m by 88 m is enlarged by a scale factor of 4. What will be the area of the enlarged farm? (c) A cook wishes to mix two types of foods, I and II, such that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents per kilogram are given in the table. (i) Formulate the linear programming model to minimize the cost. (ii) What is the minimum cost of the mixture?

Q14(b): Rectangle 72 m × 88 m enlarged by scale factor 4 (area comparison).

Basic Mathematics — Worked Solutions (Simple HTML)

BASIC MATHEMATICS — Worked Solutions (Simple)

1(a) Equal squares as large as possible on a 54 cm by 78 cm board — size of each square?

We need the greatest common divisor (GCD) of 54 and 78 — that gives the side of the largest equal square.

54 = 2 × 3³ = 2 × 27
78 = 2 × 3 × 13 = 2 × 39
GCD = 2 × 3 = 6

Answer: each square has side 6 cm.

Explanation of symbols: × means multiply. GCD = greatest common divisor (largest integer dividing both numbers).

1(b) In a class of 40 students, 17 are boys and rest girls.

Girls = 40 − 17 = 23

Percentages: percentage = (part / total) × 100.

% boys = (17 / 40) × 100 = 42.5%
% girls = (23 / 40) × 100 = 57.5%

Answer: Boys 42.5%, Girls 57.5%. Number of boys (decimal) = 17.0.

2(a) Express (5 + √7) / (3 + √7) in form a + b√c (integers a,b,c)

Rationalize denominator by multiplying numerator and denominator by the conjugate (3 − √7).

(5+√7)/(3+√7) × (3−√7)/(3−√7) = (5+√7)(3−√7) / (3² − (√7)²)
Numerator = 15 − 5√7 + 3√7 − 7 = 8 − 2√7
Denominator = 9 − 7 = 2 → (8 − 2√7)/2 = 4 − √7

Answer: 4 − √7 (so a = 4, b = −1, c = 7).

Explanation: conjugate of (p + q) is (p − q); multiplying removes the square root from denominator because (p+q)(p−q)=p²−q².

2(b) Solve 3 log₃ x + 4 = log₃ 24

Use log rules: a·log₃ x = log₃(xᵃ), and subtract logs: log₃ A − log₃ B = log₃(A/B).

3 log₃ x + 4 = log₃ 24
4 = log₃ 3⁴ = log₃ 81, so 3 log₃ x = log₃ 24 − log₃ 81 = log₃(24/81) = log₃(8/27)
⇒ log₃(x³) = log₃(8/27) ⇒ x³ = 8/27 ⇒ x = ∛(8/27) = 2/3

Answer: x = 2/3.

Explanation of symbols: log₃ means logarithm base 3. 3 log₃ x = log₃(x³).

3(a) From 100 students: 45 prefer Music (M), 40 prefer Theatre (T), 5 prefer both. How many prefer none?

Use formula for union of two sets: n(M ∪ T) = n(M) + n(T) − n(M ∩ T).

n(M ∪ T) = 45 + 40 − 5 = 80 → none = 100 − 80 = 20

Answer: 20 students prefer neither subject.

3(b) Table of 20 offices: (data given) — (i) probability 2 tables; (ii) probability at least 5 tables

Given frequencies: offices with 2 tables = 5, offices with at least 5 tables (counts for 5 and 6 etc) = 3

(i) P(2 tables) = 5 / 20 = 1/4
(ii) P(at least 5 tables) = 3 / 20

Answer: (i) 1/4, (ii) 3/20.

4(a) Vertices A(6,−4), B(8,4), P(6,1), Q(7,5). Is AB ∥ PQ?

Compute slope m = (y₂ − y₁)/(x₂ − x₁). Equal slopes → parallel.

slope(AB) = (4 − (−4)) / (8 − 6) = 8 / 2 = 4
slope(PQ) = (5 − 1) / (7 − 6) = 4 / 1 = 4

Answer: Yes, AB is parallel to PQ (both slopes = 4).

4(b) Position vectors: a = 2i − 7j and 2a + 3b = 13i − 2j. Find b.

2a = 4i − 14j → 2a + 3b = 13i − 2j → 4i − 14j + 3b = 13i − 2j
3b = (13i − 2j) − (4i − 14j) = 9i + 12j → b = 3i + 4j

Answer: b = 3i + 4j.

5(a) Area of regular hexagon inscribed in circle of radius r is 720 m². Find r (nearest tens and tenths).

Formula: Area of regular hexagon = (3√3 / 2) r² when hexagon is regular and r is circumradius.

(3√3 / 2) r² = 720 → r² = 720×2 / (3√3) = 1440 / (3√3) = 480 / √3
Using √3 ≈ 1.732 → r² ≈ 277.136 → r ≈ √277.136 ≈ 16.647

Answers: nearest tens = 20 m; nearest tenths = 16.6 m.

Symbols: √ means square root. Approximation used for √3.

5(b) (Diagram problem) Prove triangles AEB and CDE similar and find CD (sketch/logic omitted here; final result shown).

Using similarity properties (angles equal as indicated by diagram) you can set up proportion and solve. (If you want the full diagram-based steps, they can be added.)

Answer: (follows from similarity) — length CD found using the given ratios (not enough numeric detail in the text to compute here).

6(a) 850 ml of water per minute from 8:10 a.m. to 11:52 a.m. How many litres collected?

Duration: 8:10 → 11:52 = 3 hours 42 minutes = 222 minutes.

Volume = 850 ml/min × 222 min = 188700 ml = 188.7 litres

Answer: 188.7 L.

6(b) Time t (s) varies inversely with speed v (m/s). She gets home in 30 min at 10 m/s.

Inverse variation: t = k / v. Find k from known values: 30 min = 1800 s at v = 10 m/s → k = 1800×10 = 18000.

t = 18000 / v

If t = 15 min = 900 s, then v = 18000 / 900 = 20 m/s.

Answer: (i) t = 18000 / v. (ii) v = 20 m/s for 15 min.

7(a) If (A + B) : (A − B) = 5 : 4, find A : B.

(A + B)/(A − B) = 5/4 → 4(A + B) = 5(A − B) → 4A + 4B = 5A − 5B → A = 9B

Answer: A : B = 9 : 1.

7(b) Trial balance from cash account (start capital 1,000,000 TSh on 1 Jul 2021)

Summary trial balance (values from the provided workings):

Account	Dr (TSh)	Cr (TSh)
Cash / Balance c/d	310000
Capital		1000000
Purchases	1335000
Rent	110000
Wages	55000
Sales		810000
Totals	1,810,000	1,810,000

This matches the provided totals.

8(a) Monthly installments form an AP: 20000, 22000, 24000, ... final installment 114000. How many months?

AP: a₁ = 20000, common difference d = 2000. Formula: a_n = a₁ + (n−1)d.

114000 = 20000 + (n−1)×2000 → 94000 = (n−1)×2000 → n−1 = 47 → n = 48

Answer: 48 months.

8(b) Total amount of loan (sum of AP)

Sum formula: S_n = (n/2)(a₁ + a_n).

S_48 = (48/2)(20000 + 114000) = 24 × 134000 = 3,216,000

Answer: 3,216,000 shillings.

9(a) Show that 4 sin 75° = √6 + √2.

Use angle addition: sin(45°+30°) = sin45·cos30 + cos45·sin30.

sin75 = sin(45+30) = (√2/2)(√3/2) + (√2/2)(1/2) = (√6/4) + (√2/4) = (√6+√2)/4
4 sin75 = √6 + √2

Answer: Verified: 4 sin75° = √6 + √2.

9(b) Electricity post AB height 6 m supported by AC = 10 m, distance between C and D = 4 m. Find AD (leave surd form).

Right triangle ABC has AB = 6 (vertical), AC = 10 (hypotenuse) so BC = √(AC² − AB²) = √(100 − 36) = 8.

Point D is 4 m from C horizontally, so BD = BC + CD = 8 + 4 = 12.

In right triangle ABD (right at B): AD = √(AB² + BD²) = √(6² + 12²) = √(36 + 144) = √180 = 6√5.

Answer: AD = 6√5 m.

10(a) Candidate has minimum average 53 in two tests. Scored 54 in first. Least possible second mark?

Average ≥ 53: (54 + x)/2 ≥ 53 → 54 + x ≥ 106 → x ≥ 52.

Answer: least mark = 52.

10(b) Rectangular fence perimeter 56 m, area 171 m² — dimensions?

Let l and w be sides. l + w = 28 and lw = 171. Solve quadratic:

l(28 − l) = 171 → l² − 28l + 171 = 0 → Discriminant = 784 − 684 = 100 → √100 = 10
l = (28 ± 10)/2 → l = 19 or 9 → corresponding w = 9 or 19

Answer: dimensions 19 m by 9 m.

11(a)(i) From cumulative frequency table find mean (assumed mean 77), class width 5.

Cumulative frequencies given; convert to class frequencies f. Midpoints x used; compute d = (x − 77)/5 and fd.

Classes & midpoints: 65–69(67, f=10), 70–74(72, f=12), 75–79(77, f=21), 80–84(82, f=6), 85–89(87, f=9), 90–94(92, f=4), 95–99(97, f=4).
Σf = 66, Σ(fd) = 20 → Mean = 77 + (Σfd / Σf) × class width = 77 + (20/66)×5 ≈ 77 + 1.515 = 78.515 ≈ 78.52

Answer: mean ≈ 78.52 (to 2 decimal places).

11(a)(ii) Draw cumulative frequency curve (ogive)

Key points (upper class boundaries with cumulative frequencies): (65,0), (70,10), (75,22), (80,43), (85,49), (90,58), (95,62), (100,66). Plot smoothly to form ogive.

(Graphical — follow points on graph paper or software.)

11(a)(iii) Mode (to 3 decimal places)

Modal class = 75–79 with f_m = 21, previous class f₁ = 12, next class f₂ = 6, class width h = 5.

Mode ≈ lower boundary + [(f_m − f₁) / ((f_m − f₁) + (f_m − f₂))] × h
= 75 + [ (21−12) / ((21−12) + (21−6)) ] × 5 = 75 + (9/24)×5 = 75 + 1.875 = 76.875

Answer: Mode ≈ 76.875.

11(b) PT tangent to circle centre O radius 5 cm; PC = 8 cm. Find PT.

Right triangle OPT with OT = radius to point of contact C perpendicular to tangent PT. Use Pythagoras in triangle POC where OC = 5, PC = 8.

PT² + 5² = 8² → PT² = 64 − 25 = 39 → PT = √39

Answer: PT = √39 cm.

13(a)(i) Surface area of sphere is 113.04 cm². Find diameter (π = 3.14)

Surface area formula: 4πr² = given → solve for r then diameter = 2r.

4×3.14×r² = 113.04 → r² = 113.04 / (12.56) ≈ 9 → r ≈ 3 → diameter = 6 cm

Answer: 6 cm.

13(a)(ii) 75,360 litres into cylindrical tank of inside diameter 40 cm — find height of water

Convert litres → cm³: 1 L = 1000 cm³ so 75,360 L = 75,360,000 cm³. Cylinder volume V = π r² h, r = 20 cm.

75,360,000 = 3.14 × 20² × h = 3.14 × 400 × h = 1256 × h → h = 75,360,000 / 1256 ≈ 60,000 cm = 600 m

Answer: height ≈ 600 m. (Check units: extremely large tank — numbers from source.)

13(b) Distances in nautical miles between pairs of latitudes (same longitude)

Distance between two points with same longitude = difference in latitude degrees × 60 nautical miles (1° latitude = 60 NM).

(i) A(18°N) to B(65°N): Δ = 65 − 18 = 47° → 47×60 = 2820 NM
(ii) C(31°S) to D(22°N): Δ = 31 + 22 = 53° → 53×60 = 3180 NM

Answers: (i) 2820 NM, (ii) 3180 NM.

14(a)(i) Roza's 2×2 matrix R = [[2,3],[5,−6]]; Juma's J = [[−6,2],[4,3]]. Find 2R + 3J.

2R = [[4,6],[10,−12]]; 3J = [[−18,6],[12,9]]
2R + 3J = [[4−18,6+6],[10+12,−12+9]] = [[−14,12],[22,−3]]

Answer: [[−14,12],[22,−3]].

14(a)(ii) Show difference between determinants is ±1

Compute determinants: det(R) = 2(−6) − 3·5 = −12 − 15 = −27. det(J) = (−6)·3 − 2·4 = −18 − 8 = −26.

Difference: −27 − (−26) = −1 (or reverse = +1). So difference is ±1.

14(b) Farm 72 m by 88 m enlarged by scale factor 4 — new area?

Original area = 72×88 = 6,336 m². Area scales by factor k² when linear scale is k.

New area = 6,336 × 4² = 6,336 × 16 = 101,376 m²

Answer: 101,376 m².

15(a)(i) y = 3x². How many points for x = 8?

y(8) = 3×8² = 3×64 = 192

Answer: 192 points.

15(a)(ii) Domain & range of inverse function of y = 3x² (consider principal inverse)

If y = 3x² with x ≥ 0 (for inverse to be a function), inverse x = √(y/3). Domain of inverse (input y) must be y ≥ 0; range (x) is x ≥ 0.

Answer: Domain of inverse: [0, ∞). Range of inverse: [0, ∞).

15(b) Linear programming mixture problem (minimize cost)

Let x = kg of Food I, y = kg of Food II. Minimize cost C = 6000x + 10000y subject to nutrient constraints:

x + 2y ≥ 10 (vitamin A)
2x + 2y ≥ 12 (vitamin B)
x + y ≥ 6 (vitamin C alternative written in text)
3x + y ≥ 8 (as additional constraint in source)
x ≥ 0, y ≥ 0

Feasible corner points (from source): (2,4) and (1,5) etc. Compute cost at vertices:

At (2,4): C = 6000·2 + 10000·4 = 12,000 + 40,000 = 52,000
At (1,5): C = 6000·1 + 10000·5 = 6,000 + 50,000 = 56,000

Minimum cost: 52,000 TSh at (x,y) = (2,4).

Reference Book: N/A

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