0041-BASIC-MATHEMATICS-2024-NECTA

Objectives: 0041-BASIC-MATHEMATICS-2024-NECTA-SOLVING

Basic Mathematics — Form Four (2024)

Questions ..

Answer all questions in this section.

    1. Equal squares which are as large as possible are drawn on a rectangular board measuring 54 cm by 78 cm. Find the size of each square.
    2. In a class of 40 students, 17 students are boys and the rest are girls. Determine:
      1. the percentage of girls and boys.
      2. the number of boys in decimal.
    1. Express (5 + √7) / (3 + √7) in the form a + b√c, where a, b, and c are integers.
    2. A line is defined by a logarithmic equation 3 · log3(x + 4) = log3(24). Without using a mathematical table or calculator, find the value of x.
    1. In a school of 100 students, 45 students prefer Music subject, 40 students prefer Theatre Arts subject and 5 students prefer both subjects. Find the number of students who prefer none of the two subjects by using formula.
    2. The following table shows the number of tables in twenty offices from a certain company:
      Number of Tables Number of Offices
      14
      25
      36
      41
      52
      62
      If one office is picked randomly, find the probability that the office has:
      1. two tables,
      2. at least five tables.
    1. A quadrilateral has the vertices, A (6, −4), B (8, 4), P (6, 1) and Q (7, 5). Determine whether or not \u2192AB is parallel to \u2192PQ.
    2. 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.
    1. 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 by rounding off the answer to the nearest:
      1. tens,
      2. tenths.
    2. A firm reserved two triangular plots for the construction of more offices as shown in the diagram.
      Figure: Triangles AEB and CDE with a transversal; AE = 24 m, BE = 36 m, DE = 32 m.
      1. Prove that the triangles AEB and CDE are similar.
      2. Using similarity properties, find the length of CD in metres.
Fully Worked Solutions – Clear, Step-by-Step (HTML + a little JS)

Fully Worked, Well-Explained Answers

Every part includes the formulas used, why those formulas fit, and a short symbol guide. Where a second valid method exists, it is shown too.

1. Arithmetic and Percentages

1(a) A rectangle is 54 cm × 78 cm. Equal squares as large as possible are drawn on it. Find the size of each square.

Method 1 — Greatest Common Divisor (Euclidean Algorithm)

The side of the largest square that tiles the rectangle exactly equals gcd(54, 78).

  1. Compute: 78 ÷ 54 = 1 remainder 24.
  2. Then 54 ÷ 24 = 2 remainder 6.
  3. Then 24 ÷ 6 = 4 remainder 0. Stop. The last non-zero remainder is 6.

Answer: side of each square = 6 cm.

Method 2 — Prime Factorisation

54 = 2 × 33 and 78 = 2 × 3 × 13. Common prime factors are 2 × 3 so gcd = 6.

Extra: Number of such squares = area of rectangle ÷ area of square = (54×78) ÷ 62 = 4212 ÷ 36 = 117.

Formulas and Symbols

gcd(a,b)
greatest common divisor of a and b (largest number dividing both without remainder)
÷
division
×
multiplication
remainder
what is left after division that does not fit evenly
1(b) In a class of 40, boys = 17, girls are the rest. Find (i) percentages of girls and boys, (ii) boys in decimal.
  1. Percentages
    Boys: (17/40) × 100% = 42.5%
    Girls: ((40−17)/40) × 100% = (23/40) × 100% = 57.5%
  2. “Number of boys in decimal”
    Interpreting as the proportion of boys written as a decimal: 17/40 = 0.425.
    (If the intention was literally the count as a decimal numeral, it is 17.0.)

Formula and Symbols

%
“percent” = per hundred
fraction → decimal
divide numerator by denominator

2. Surds and Logarithms

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

Method — Rationalise the Denominator

Multiply top and bottom by the conjugate (3 − √7) to use the difference of squares.

Numerator: (5 + √7)(3 − √7) = 15 − 5√7 + 3√7 − 7 = 8 − 2√7
Denominator: (3 + √7)(3 − √7) = 9 − 7 = 2
Value: (8 − 2√7)/2 = 4 − √7

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

Formulas and Symbols

√x
square root of x
conjugate
replace + with − (or vice-versa): p+q√r ↔ p−q√r
(u+v)(u−v)=u2−v2
difference of squares identity
2(b) Solve 3·log3(x) + 4 = log3(24) without tables or a calculator.

Method 1 — Convert all constants into base-3 logs

Move 4 to the right using 4 = log3(34) = log3(81): 3·log3(x) = log3(24) − log3(81) = log3(24/81) = log3(8/27).
Divide both sides by 3 using k·logb(t)=logb(tk):
log3(x) = (1/3)·log3(8/27) = log3((8/27)1/3) = log3(2/3).
Therefore x = 2/3.

Method 2 — Exponent form

Let y = log3(x). Then 3y + 4 = log3(24) means 3y = log3(24) − 4 = log3(24) − log3(81) = log3(8/27). So y = (1/3)log3(8/27) = log3(2/3), hence x = 3y = 2/3.

Answer: x = 2/3.

Formulas and Symbols

logb(t)
the power on base b that produces t
log laws
logb(M)−logb(N)=logb(M/N), k·logb(t)=logb(tk)

3. Sets and Probability

3(a) In a school of 100, Music =45, Theatre =40, both =5. How many prefer none?

By inclusion–exclusion: |M ∪ T| = |M| + |T| − |M ∩ T| = 45 + 40 − 5 = 80. Therefore “none” = 100 − 80 = 20.

Answer: 20 students.

Symbols

|S|
number of elements in set S
∪
union (either preference)
∩
intersection (both preferences)
3(b) Distribution of tables in 20 offices:
Number of TablesNumber of Offices
14
25
36
41
52
62
  1. Two tables: favourable offices = 5. Total = 20.
    P = 5/20 = 1/4 = 0.25.
  2. At least five tables (5 or 6): favourable = 2 + 2 = 4.
    P = 4/20 = 1/5 = 0.20.

Formula and Symbols

P(E)
probability of event E
P(E) = favourable / total
classical probability for equally likely outcomes
“at least”
greater than or equal to

4. Coordinate Geometry and Vectors

4(a) A (6, −4), B (8, 4), P (6, 1), Q (7, 5). Decide if AB is parallel to PQ.

Method 1 — Compare slopes

slope(AB) = (4 − (−4)) / (8 − 6) = 8/2 = 4
slope(PQ) = (5 − 1) / (7 − 6) = 4/1 = 4
Equal slopes ⇒ lines are parallel.

Method 2 — Vector multiples

\u2192AB = (8−6, 4−(−4)) = (2, 8), \u2192PQ = (7−6, 5−1) = (1, 4).
Since \u2192AB = 2·\u2192PQ, they are parallel.

Conclusion: AB ∥ PQ (parallel).

Symbols

∥
is parallel to
(x,y)
Cartesian coordinates
\u2192UV
vector from point U to point V
4(b) Vectors: a = 2i − 7j, and 2a + 3b = 13i − 2j. Find b.

Compute 2a = 4i − 14j.
Then 3b = (13i − 2j) − (4i − 14j) = 9i + 12j.
Hence b = 3i + 4j.

Answer: b = 3i + 4j.

Symbols

i, j
unit vectors along x- and y-axes
component form
write a vector as xi + yj

5. Regular Hexagon and Similar Triangles

5(a) A regular 6-sided plot (hexagon) is inscribed in a circle of radius r. Its area is 720 m². Find r to the nearest tens and tenths.

Method 1 — Standard hexagon area

For a regular hexagon with side s, area A = (3√3/2)·s2. When it is inscribed in a circle, s = r. So: 720 = (3√3/2)·r2 ⇒ r2 = 720×2/(3√3) = 480/√3 ⇒ r = √(480/√3) ≈ 16.647 m.

Method 2 — Split into 6 equilateral triangles

The hexagon equals 6 congruent equilateral triangles of side r. Area of one is (√3/4)·r2; multiply by 6: 6·(√3/4)·r2 = (3√3/2)·r2 = 720 — same result for r.

Answer: r ≈ 16.647 m.
To the nearest tens: 20 m.
To the nearest tenths: 16.6 m.

Symbols

√3
square root of 3
A
area
s, r
side length and circle radius
5(b) Two triangular plots are shown. Given AE = 24 m, BE = 36 m, DE = 32 m.
Triangles AEB and CDE with a transversal through E; right angles at A and D.
  1. Prove triangles AEB and CDE are similar
    Angle ∠AEB equals ∠CED (they are vertically opposite).
    Also ∠A and ∠D are right angles (each 90°).
    Two angles equal ⇒ triangles are similar by the AA criterion.
  2. Find CD using two different correct methods

    Method 1 — Scale factor via matching legs

    Correspondence (by the geometry around E): AE ↔ DE and BE ↔ CE.
    Scale factor from △AEB to △CDE is k = DE / AE = 32 / 24 = 4/3.
    In △AEB, the right angle is at A, so BE is the hypotenuse and AB is the other leg. Compute AB by Pythagoras: AB = √(BE2 − AE2) = √(362 − 242) = √(1296 − 576) = √720 = 12√5.
    The corresponding side in △CDE is CD, so CD = k·AB = (4/3)·12√5 = 16√5 ≈ 35.78 m.

    Method 2 — Use hypotenuse ratio then Pythagoras in △CDE

    From similarity, hypotenuse ratio equals any side ratio: CE / BE = DE / AE = 32/24 = 4/3 ⇒ CE = (4/3)·BE = (4/3)·36 = 48 m.
    Now apply Pythagoras in △CDE (right angle at D): CD = √(CE2 − DE2) = √(482 − 322) = √(2304 − 1024) = √1280 = 16√5 ≈ 35.78 m.

    Answer: CD = 16√5 m ≈ 35.78 m.

Formulas and Symbols

AA similarity
two equal angles ⇒ triangles similar
Pythagoras
for a right triangle with legs a,b and hypotenuse c: a2 + b2 = c2
k
scale factor between similar triangles
∠
angle

All computations are exact where shown (for example 16√5) and approximated where decimals are requested.

Reference Book: N/A

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