ERC BECE MATHEMATICS 2025

Question 1:
Write 3025 in Roman numerals.

A. MMMXXV
B. MMMCCV
C. MMMXXV
D. CCXXV

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Answer: A (or C)

Explanation:
Break the number into thousands, hundreds, tens, and units:
$3000 = \text{MMM}$
$0 = \text{ (no symbol)}$
$20 = \text{XX}$
$5 = \text{V}$
Combining them gives **MMMXXV**.

Question 2:
What is the value of the digit 8 in the figure 6.085?

A. hundred
B. tens
C. hundredth
D. tenth

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Answer: C (hundredth)

Explanation:
In decimals, the positions to the right of the point are:
6 (Units) . 0 (Tenths) 8 (Hundredths) 5 (Thousandths).
Therefore, 8 represents $8 \times \frac{1}{100}$.

Question 3:
Write 0.07465 in its standard form.

A. $7.45 \times 10^2$
B. $7.465 \times 10^3$
C. $7.465 \times 10^{-2}$
D. $7.465 \times 10^{-3}$

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Answer: C (7.465 x 10^-2)

Explanation:
To write in standard form, move the decimal point until there is one non-zero digit to its left. Moving the point 2 places to the right gives $7.465$. Since we moved to the right, the exponent is negative: $7.465 \times 10^{-2}$.

Question 4:
Express 504 as a product of its prime factors in index form.

A. $2^2 \times 3^3 \times 7^1$
B. $2^4 \times 3^3 \times 7^1$
C. $2^3 \times 3^2 \times 7^1$
D. $2^5 \times 3^4 \times 7^1$

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Answer: C (2^3 x 3^2 x 7^1)

Explanation:
Perform prime factorization:
$504 \div 2 = 252$
$252 \div 2 = 126$
$126 \div 2 = 63$
$63 \div 3 = 21$
$21 \div 3 = 7$
$7 \div 7 = 1$
So, $504 = 2 \times 2 \times 2 \times 3 \times 3 \times 7 = 2^3 \times 3^2 \times 7^1$.

Question 5:
Find the HCF of 21, 35 and 56?

A. 2
B. 5
C. 3
D. 7

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Answer: D (7)

Explanation:
List the factors for each number:
Factors of 21: 1, 3, **7**, 21
Factors of 35: 1, 5, **7**, 35
Factors of 56: 1, 2, 4, **7**, 8, 14, 28, 56
The Highest Common Factor shared by all three is **7**.

Question 6:
The number of students in a Secondary school in Bwari area Council is 5400. Express this figure as a product of its prime factor in the index form.

A. $2^3 \times 3^3 \times 5^2$
B. $2^2 \times 3^3 \times 5^2$
C. $2^2 \times 3^2 \times 5^2
D. $2^3 \times 3^3 \times 5$

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Answer: A ($2^3 \times 3^3 \times 5^2$)

Explanation:
Break 5400 down into prime factors:
$5400 = 54 \times 100$
$54 = 2 \times 3 \times 3 \times 3 = 2^1 \times 3^3$
$100 = 2 \times 2 \times 5 \times 5 = 2^2 \times 5^2$
Combine them: $2^{(1+2)} \times 3^3 \times 5^2 = 2^3 \times 3^3 \times 5^2$.

Question 7:
Which of the following numbers is completely divisible by 9?

A. 18089
B. 18099
C. 18980
D. 19898

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Answer: B (18099)

Explanation:
A number is divisible by 9 if the sum of its digits is divisible by 9:
A: $1+8+0+8+9 = 26$ (No)
B: $1+8+0+9+9 = 27$ (Yes, $27 \div 9 = 3$)
C: $1+8+9+8+0 = 26$ (No)
D: $1+9+8+9+8 = 35$ (No)

Question 8:
Determine the numerator, when $18 \frac{7}{8}$ is converted to an improper fraction.

A. 127
B. 147
C. 151
D. 161

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Answer: C (151)

Explanation:
To convert a mixed number to an improper fraction, multiply the whole number by the denominator and add the numerator:
$\text{Numerator} = (18 \times 8) + 7$
$18 \times 8 = 144$
$144 + 7 = 151$
The improper fraction is $\frac{151}{8}$.

Question 9:
Simplify the expression $x^3 \times x^5$.

A. $x^{-8}$
B. $x^{-1}$
C. $x^{-2}$
D. $x^8

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Answer: D ($x^8$)

Explanation:
According to the laws of indices, when multiplying terms with the same base, you add the exponents:
$x^a \times x^b = x^{a+b}$
$x^3 \times x^5 = x^{3+5} = x^8$.

Question 10:
Ade scored 16 out of 30 sums correct. Express his score in percentage.

A. 85.3%
B. 73.3%
C. 63.3%
D. 53.3%

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Answer: D (53.3%)

Explanation:
To find the percentage, divide the score by the total and multiply by 100:
$\frac{16}{30} \times 100 = \frac{160}{3} = 53.333...\%$
To one decimal place, this is **53.3%**.

Question 11:
Evaluate $(111_2)^2$.

A. $110001_2$
B. $110010_2$
C. $11001_2$
D. $1010_2$

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Answer: A ($110001_2$)

Explanation:
First, convert $111_2$ to base 10:
$(1 \times 2^2) + (1 \times 2^1) + (1 \times 2^0) = 4 + 2 + 1 = 7_{10}$.
Square the result: $7^2 = 49_{10}$.
Convert 49 back to binary:
$49 \div 2 = 24$ R 1
$24 \div 2 = 12$ R 0
$12 \div 2 = 6$ R 0
$6 \div 2 = 3$ R 0
$3 \div 2 = 1$ R 1
$1 \div 2 = 0$ R 1
Reading remainders upwards: **$110001_2$**.

Question 12:
Simplify $(2\frac{1}{2} + 1\frac{2}{3}) \div 1\frac{3}{4}$.

A. $3\frac{1}{2}$
B. $4\frac{1}{6}$
C. $2\frac{1}{2}$
D. $2\frac{8}{21}$ (Correct calculation result)

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Answer: D ($2\frac{8}{21}$)

Explanation:
1. Solve the bracket first: $\frac{5}{2} + \frac{5}{3} = \frac{15 + 10}{6} = \frac{25}{6}$.
2. Convert $1\frac{3}{4}$ to $\frac{7}{4}$.
3. Divide: $\frac{25}{6} \div \frac{7}{4} = \frac{25}{6} \times \frac{4}{7}$.
4. $\frac{100}{42} = \frac{50}{21} = 2\frac{8}{21}$.

Question 13:
Change 0.95 to fraction in its lowest term.

A. 95/100
B. 19/20
C. 15/20
D. 29/50

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Answer: B (19/20)

Explanation:
0.95 is written as $\frac{95}{100}$.
Divide both numerator and denominator by their HCF (5):
$95 \div 5 = 19$
$100 \div 5 = 20$
Result: **19/20**.

Question 14:
Find the sum of $10110_2$ and $10101_2$.

A. $100101_2$
B. $100111_2$
C. $101010_2$
D. $101011_2$

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Answer: B ($100111_2$)

Explanation:
Perform binary addition:
  1 0 1 1 0
+ 1 0 1 0 1
-----------
1 0 0 1 1 1
(Note: $1+1 = 10$ in binary, so you write 0 and carry 1).

Question 15:
Find the range of values of $y$ which satisfy the inequality: $3y - 2 \le 5y + 4$

A. $y \ge -3$
B. $y \ge 3$
C. $y \le 2$
D. $y \le 3$

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Answer: A ($y \ge -3$)

Explanation:
1. Collect like terms: $3y - 5y \le 4 + 2$
2. Simplify: $-2y \le 6$
3. Divide by -2 (and flip the inequality sign): $y \ge \frac{6}{-2}$
4. Result: $y \ge -3$

Question 16:
What inequality does the number line below represent?

A. $x \le 2$
B. $x \le -2$
C. $x \ge -2$
D. $x \ge 2$

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Answer: C ($x \ge -2$)

Explanation:
A solid (shaded) circle indicates "equal to" ($\le$ or $\ge$). Since the arrow points to the right (towards larger numbers), it represents "greater than or equal to -2".

Question 17:
Find the $\sqrt{441}$

A. 3
B. 9
C. 17
D. 21

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Answer: D (21)

Explanation:
You can check by squaring the options:
$20 \times 20 = 400$
$21 \times 21 = 441$
Therefore, $\sqrt{441} = 21$.

Question 18:
Write 37500 in standard form.

A. $3.75 \times 10^4$
B. $3.75 \times 10^3$
C. $3.75 \times 10^2$
D. $3.75 \times 10^0$

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Answer: A ($3.75 \times 10^4$)

Explanation:
Move the decimal point from the end of the number to the left until it is after the first digit (3).
$37500 \rightarrow 3.7500$ (moved 4 places).
Since we moved left, the exponent is positive 4.

Question 19:
Evaluate $3^4$

A. 81
B. 64
C. 12
D. 7

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Answer: A (81)

Explanation:
$3^4$ means multiplying 3 by itself four times:
$3 \times 3 \times 3 \times 3$
$9 \times 9 = 81$.

Question 20:
Simplify the expression $x^3 \times x^5$.

A. $x^{-8}$
B. $x^{-1}$
C. $x^{-2}$
D. $x^8$

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Answer: D ($x^8$)

Explanation:
According to the First Law of Indices (Product Rule), when multiplying terms with the same base, you add their exponents:
$x^a \times x^b = x^{a+b}$
Applying this to the problem:
$x^3 \times x^5 = x^{3+5} = x^8$

Question 21:
The population of a state is 3,652,473. Write this figure to the nearest hundreds.

A. 3,700,000
B. 3,650,000
C. 3,652,500
D. 3,600,000

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Answer: C (3,652,500)

Explanation:
To round to the nearest hundred, look at the digit in the tens place (7). Since 7 is 5 or greater, round up the digit in the hundreds place (4 becomes 5) and replace the following digits with zeros.

Question 22:
Evaluate $(p^2 + 3q)/2r$ given $p = 4, q = 3$ and $r = -5$.

A. -5
B. -5/2
C. 5/7
D. -11/5

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Answer: B (-5/2)

Explanation:
Substitute the values: $\frac{4^2 + 3(3)}{2(-5)} = \frac{16 + 9}{-10} = \frac{25}{-10} = -\frac{5}{2}$.

Question 23:
Express 0.00853 to 1 significant figure.

A. 0.009
B. 0.0081
C. 0.0085
D. 0.008

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Answer: A (0.009)

Explanation:
The first non-zero digit is 8. The following digit is 5, so we round the 8 up to 9. The result is 0.009.

Question 24:
Express 7/19 as a percentage, correct to 1 decimal place.

A. 2.7%
B. 3.7%
C. 27.1%
D. 36.8%

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Answer: D (36.8%)

Explanation:
Multiply the fraction by 100: $(7 \div 19) \times 100 \approx 36.842\%$. Rounded to one decimal place, this is 36.8%.

Question 25:
Simplify: 3(a - 2b + c) - 2(a + 3b - 2c)

A. a - 3
B. a - 12b + 7c
C. a - 12b + 3c
D. a - 12b - 3c

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Answer: B (a - 12b + 7c)

Explanation:
1. Expand the first bracket: $3(a) + 3(-2b) + 3(c) = 3a - 6b + 3c$.
2. Expand the second bracket (be careful with the -2): $-2(a) - 2(3b) - 2(-2c) = -2a - 6b + 4c$.
3. Combine the results: $3a - 6b + 3c - 2a - 6b + 4c$.
4. Group like terms: $(3a - 2a) + (-6b - 6b) + (3c + 4c)$.
5. Final result: $a - 12b + 7c$.

Question 26:
Find the next 2 terms in the sequence 4, 8, 12, 16 —

A. 20, 24
B. 28, 32
C. 36, 40
D. 20, 22

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Answer: A (20, 24)

Explanation:
The sequence follows a pattern of adding 4 to each preceding term (Arithmetic Progression).
$16 + 4 = 20$
$20 + 4 = 24$

Question 27:
Find the simple interest of ₦500 for 3 years at 4%.

A. ₦45
B. ₦60
C. ₦70
D. ₦72

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Answer: B (₦60)

Explanation:
Use the formula: $I = \frac{P \times R \times T}{100}$
$I = \frac{500 \times 4 \times 3}{100} = 5 \times 4 \times 3 = 60$.

Question 28:
Mr. Obi sold a car for ₦84,000 at a loss of 4%; for how much did Mr. Obi buy the car?

A. ₦80,500
B. ₦80,640
C. ₦87,360
D. ₦87,500

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Answer: D (₦87,500)

Explanation:
The selling price (₦84,000) represents 96% of the cost price ($100\% - 4\% = 96\%$).
$Cost Price = \frac{Selling Price}{96} \times 100 = \frac{84,000}{96} \times 100 = 87,500$.

Question 29:
In a boutique, two (2) pairs of trousers and a packet shirt cost ₦17,000. The difference between three (3) pairs of trousers and a packet shirt is ₦8,000. Find the cost of a pair of trousers and two (2) packet shirts.

A. ₦19,000
B. ₦12,000
C. ₦9,000
D. ₦7,000

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Answer: A (₦19,000)

Explanation:
Let $x$ be trousers and $y$ be shirts:
1) $2x + y = 17,000$
2) $3x - y = 8,000$
Add both equations: $5x = 25,000 \Rightarrow x = 5,000$.
Substitute $x$ into (1): $2(5,000) + y = 17,000 \Rightarrow y = 7,000$.
Cost of 1 trouser ($5,000$) and 2 shirts ($14,000$) = ₦19,000.

Question 30:
Adams runs a distance of 800m in 2½ minutes, calculate his speed.

A. 300m/minutes
B. 320m/minutes
C. 400m/minutes
D. 600m/minutes

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Answer: B (320m/minutes)

Explanation:
$Speed = \frac{Distance}{Time}$.
$Time = 2.5$ minutes.
$Speed = \frac{800}{2.5} = 320$ m/min.

Question 31:
It takes five (5) girls eating the same ration six (6) days to finish a supply of food. How long will the supply last three (3) girls if they maintain the same ration?

A. 4 days
B. 5 days
C. 8 days
D. 10 days

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Answer: D (10 days)

Explanation:
This is inverse proportion. Fewer girls will take more time to finish the food.
Total food "man-days" = $5 \text{ girls} \times 6 \text{ days} = 30 \text{ units of food}$.
Days for 3 girls = $30 \div 3 = 10 \text{ days}$.

Question 32:
What number should be in the box to make $13 + \square = 22$?

A. 9
B. 11
C. 13
D. 15

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Answer: A (9)

Explanation:
To find the missing number, subtract 13 from 22:
$22 - 13 = 9$.

Question 33:
Find the sum of angles in an octagon.

A. 800°
B. 900°
C. 1080°
D. 1260°

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Answer: C (1080°)

Explanation:
An octagon has $n = 8$ sides. Use the formula for the sum of interior angles: $(n - 2) \times 180°$.
$(8 - 2) \times 180° = 6 \times 180° = 1080°$.

Question 34:
The ratio of birds to livestock on Mr. Oni’s farm is 3:2. If there are 150 birds on the farm, how many livestock are there?

A. 600
B. 500
C. 250
D. 100

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Answer: D (100)

Explanation:
The ratio is $\frac{\text{Birds}}{\text{Livestock}} = \frac{3}{2}$.
$\frac{150}{x} = \frac{3}{2}$
Cross multiply: $3x = 300 \Rightarrow x = 100$.

Question 35:
Find the value of $x$ in the figure below.

A. 22°
B. 56°
C. 34°
D. 112°

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Answer: D (112°)

Explanation:
The angles $x$ and 68° lie on a straight line. The sum of angles on a straight line is 180°.
$x + 68° = 180°$
$x = 180° - 68°$
$x = 112°$.

Question 36:
The adjacent sides of a right angle triangle are 5cm and 12cm respectively. What is the length of the hypotenuse?

A. 13cm
B. 17cm
C. 45cm
D. 60cm

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Answer: A (13cm)

Explanation:
Using the Pythagorean theorem: $a^2 + b^2 = c^2$, where $c$ is the hypotenuse.
$5^2 + 12^2 = c^2$
$25 + 144 = c^2$
$169 = c^2$
$c = \sqrt{169} = 13$ cm.

Question 37:
Find the size of the missing angle x° in the quadrilateral, in the figure below.

A. 104°
B. 360°
C. 256°
D. 420°

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Answer: A (104°)

Explanation:
The sum of interior angles in any quadrilateral is 360°.
$97° + 74° + 85° + x = 360°$
$256° + x = 360°$
$x = 360° - 256°$
$x = 104°$.

Question 38:
Calculate the area of a parallelogram of base 12cm and 7cm high.

A. 19cm²
B. 38cm²
C. 42cm²
D. 84cm²

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Answer: D (84cm²)

Explanation:
The area of a parallelogram is calculated using the formula: $Area = \text{base} \times \text{height}$.
$Area = 12\text{cm} \times 7\text{cm} = 84\text{cm}^2$.

Question 39:
What is the perimeter of the figure below?

A. 54cm
B. 44cm
C. 49cm
D. 32cm

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Answer: A (54cm)

Explanation:
The perimeter is the total distance around the outside of the shape. To find it, we need the lengths of all sides:
1. Bottom: 15cm
2. Left: 12cm
3. Right: 7cm
4. Middle horizontal: 10cm
5. Top horizontal: Total width (15cm) - Middle width (10cm) = 5cm
6. Middle vertical: Total height (12cm) - Right height (7cm) = 5cm

Summing all sides: $15 + 12 + 5 + 5 + 10 + 7 = 54$cm.

Question 40:
Suppose that $\theta$ is an acute angle for which $\tan \theta = 3/5$, find $\sin \theta$.

A. $3/\sqrt{34}$
B. $5/\sqrt{34}$
C. $3/4$
D. $5/3$

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Answer: A ($3/\sqrt{34}$)

Explanation:
1. Recall that $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{5}$.
2. Find the hypotenuse ($h$) using Pythagoras' theorem: $h^2 = 3^2 + 5^2$.
3. $h^2 = 9 + 25 = 34$, so $h = \sqrt{34}$.
4. Since $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$, then $\sin \theta = \frac{3}{\sqrt{34}}$.

Question 41:
In the figure below, two equal angles are shown. These angles are:

A. alternate angles
B. corresponding angles
C. acute angles
D. vertically opposite angles

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Answer: A (alternate angles)

Explanation:
The angles are located on opposite sides of the transversal line and between the two parallel lines. These are specifically called alternate interior angles, and they are equal when the lines are parallel.

Question 42:
What is the value of x in the figure below?

A. 180°
B. 85°
C. 90°
D. 60°

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Answer: D (60°)

Explanation:
The figure shows a triangle with three equal sides (indicated by the tick marks). This is an equilateral triangle. In an equilateral triangle, all three interior angles are equal. Since the sum of angles in a triangle is 180°:
$x + y + z = 180°$
$3x = 180°$
$x = 180° / 3 = 60°$.

Question 43:
What is the diameter of a circle whose area is 616cm²?

A. 30cm
B. 25cm
C. 28cm
D. 14cm

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Answer: C (28cm)

Explanation:
1. Area of a circle $A = \pi r^2$. Given $A = 616$ and $\pi = 22/7$:
$616 = \frac{22}{7} \times r^2$
2. $r^2 = \frac{616 \times 7}{22} = 28 \times 7 = 196$.
3. $r = \sqrt{196} = 14$cm.
4. Diameter $d = 2r = 2 \times 14 = 28$cm.

Question 44:
The radius of a cone is 6cm. Given that the height of the cone is 14cm and $\pi$ is 22/7. Calculate the volume of the cone.

A. 250cm³
B. 350cm³
C. 500cm³
D. 528cm³

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Answer: D (528cm³)

Explanation:
Use the formula: $V = \frac{1}{3}\pi r^2h$
$V = \frac{1}{3} \times \frac{22}{7} \times 6^2 \times 14$
$V = \frac{1}{3} \times \frac{22}{7} \times 36 \times 14$
$V = 22 \times 12 \times 2 = 528$cm³.

Question 45:
The adjacent sides of a triangle are 12cm and 5cm respectively, calculate its hypotenuse.

A. 60cm
B. 30cm
C. 17cm
D. 13cm

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Answer: D (13cm)

Explanation:
Using Pythagoras' theorem ($a^2 + b^2 = c^2$):
$12^2 + 5^2 = c^2$
$144 + 25 = 169$
$c = \sqrt{169} = 13$cm.

Question 46:
Find the value of $x$ in the figure below.

A. 180°
B. 21°
C. 138°
D. 18°

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Answer: C (138°)

Explanation:
1. First, find the interior angle adjacent to 159°. Since they are on a straight line: $180° - 159° = 21°$.
2. Now you have two interior angles of the triangle: $21°$ and $21°$.
3. Find the third interior angle (let's call it $y$): $180° - (21° + 21°) = 180° - 42° = 138°$.
4. Since $x$ is vertically opposite to the third interior angle $y$, $x = 138°$.

Question 47:
Find the sum of interior angles of a regular pentagon.

A. 180°
B. 720°
C. 540°
D. 900°

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Answer: C (540°)

Explanation:
Use the formula for the sum of interior angles: $(n - 2) \times 180°$.
For a pentagon, $n = 5$:
$(5 - 2) \times 180° = 3 \times 180° = 540°$.

Question 48:
Given that $v = u + at$, make $t$ the subject.

A. $t = (v - u)/a$
B. $t = (v + u)/a$
C. $t = (u - v)/a$
D. $t = u - av$

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Answer: A (t = (v - u)/a)

Explanation:
1. Start with $v = u + at$.
2. Subtract $u$ from both sides: $v - u = at$.
3. Divide both sides by $a$: $t = \frac{v - u}{a}$.

Question 49:
Solve the simultaneous equations: $3x - y = 4$ and $x + 2y = 6$.

A. $x = 2, y = 2$
B. $x = 2, y = 3$
C. $x = 3, y = 2$
D. $x = 1, y = 2$

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Answer: A (x = 2, y = 2)

Explanation:
Multiply the first equation by 2: $6x - 2y = 8$.
Add the second equation ($x + 2y = 6$) to it: $7x = 14 \Rightarrow x = 2$.
Substitute $x = 2$ into the first equation: $3(2) - y = 4 \Rightarrow 6 - y = 4 \Rightarrow y = 2$.

Question 50:
The bearing of a school farm from the library is 075°. Obtain the bearing of the library from the farm.

A. 015°
B. 075°
C. 105°
D. 255°

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Answer: D (255°)

Explanation:
To find the back bearing, add 180° if the given bearing is less than 180°.
$75° + 180° = 255°$.

Question 51:
If the probability of scoring in a match is 0.8. What is the probability that they will not score in that particular match?

A. 1
B. 0.4
C. 0.2
D. 0

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Answer: C (0.2)

Explanation:
The sum of the probability of an event happening and not happening is always 1.
$P(\text{not scoring}) = 1 - P(\text{scoring})$
$1 - 0.8 = 0.2$.

Question 52:
The sum of two numbers is 12. Their difference is 2, find the numbers.

A. 7, 5
B. 7, 8
C. 9, 3
D. 9, 4

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Answer: A (7, 5)

Explanation:
Let the numbers be $x$ and $y$.
$x + y = 12$
$x - y = 2$
Adding both equations: $2x = 14 \Rightarrow x = 7$.
If $x = 7$, then $7 + y = 12 \Rightarrow y = 5$.

Question 53:
If 6 students bought a pen and a novel for ₦50 and ₦300 respectively. Find the total cost of the items.

A. ₦2100
B. ₦1100
C. ₦1000
D. ₦900

👉 Click to view answer & explanation

Answer: A (₦2100)

Explanation:
Cost for one student = $₦50 (\text{pen}) + ₦300 (\text{novel}) = ₦350$.
Total cost for 6 students = $350 \times 6 = ₦2100$.

Question 54:
If 4 is the mean of the following set of numbers 1, 2, x, 7, 6. What is the value of x?

A. 7
B. 8
C. 4
D. -4

👉 Click to view answer & explanation

Answer: C (4)

Explanation:
$\text{Mean} = \frac{\text{Sum of numbers}}{\text{Count of numbers}}$
$4 = \frac{1 + 2 + x + 7 + 6}{5}$
$4 = \frac{16 + x}{5}$
Multiply both sides by 5: $20 = 16 + x \Rightarrow x = 4$.

Question 55:
What is the range of the following set of numbers -2, -1, -5, 0, 2?

A. 7
B. 4
C. -4
D. 6

👉 Click to view answer & explanation

Answer: A (7)

Explanation:
$\text{Range} = \text{Highest value} - \text{Lowest value}$
Highest = 2, Lowest = -5
$\text{Range} = 2 - (-5) = 2 + 5 = 7$.

Question 56:
If $b$ is directly proportional to $a$ and $b = 10$ when $a = 2$, find $a$ if $b = 35$.

A. 7
B. 10
C. 14
D. 5

👉 Click to view answer & explanation

Answer: A (7)

Explanation:
1. Relationship: $b = ka$
2. Find constant $k$: $10 = k(2) \Rightarrow k = 5$.
3. Find $a$ when $b = 35$: $35 = 5a \Rightarrow a = 7$.

The pie chart below shows the record of animals in a ranch, use the chart to answer questions 57 to 60.

Question 57:
If there are 36 animals in the ranch, how many rabbits are there?

A. 120
B. 36
C. 24
D. 12

👉 Click to view answer & explanation

Answer: D (12)

Explanation:
The angle for Rabbit is 120°. Total angle in a circle is 360°.
Number of rabbits = $\frac{120}{360} \times 36$
$= \frac{1}{3} \times 36 = 12$.

Question 58:
What is the size of the angle represented by sheep?

A. 140°
B. 100°
C. 80°
D. 60°

👉 Click to view answer & explanation

Answer: C (80°)

Explanation:
The sum of angles in a circle is 360°. The "Cow" section has a right-angle symbol (90°).
$\text{Sheep} = 360° - (\text{Cow} + \text{Goat} + \text{Rabbit})$
$= 360° - (90° + 70° + 120°) = 360° - 280° = 80°$.

Question 59:
If an animal is picked at random from the ranch, what is the probability that it is a cow?

A. 1/2
B. 1/3
C. 2/5
D. 1/4

👉 Click to view answer & explanation

Answer: D (1/4)

Explanation:
Probability = $\frac{\text{Angle of Cow}}{\text{Total Angle}}$
$= \frac{90}{360} = \frac{1}{4}$.

Question 60:
How many goats are in the ranch?

A. 6
B. 7
C. 8
D. 10

👉 Click to view answer & explanation

Answer: B (7)

Explanation:
Number of goats = $\frac{\text{Angle of Goat}}{360} \times \text{Total Animals}$
$= \frac{70}{360} \times 36$
$= \frac{70}{10} = 7$.

Section B: Essay - Question 1

Question 1:
The scores obtained by some students in a mathematics test are given below:
4, 5, 6, 9, 7, 5, 8, 6, 5, 7, 6, 7, 5, 6, 10, 5, 6, 7, 6, 8, 4, 6, 10, 9, 8, 7, 9, 8, 6, 7

a. Prepare a frequency distribution table.
b. Find: i. Range of the scores, ii. Mode of the scores, iii. Mean of the scores, iv. Probability of selecting a student who scored 7.

👉 Click to view Frequency Table & Solutions

a. Frequency Distribution Table

Score ($x$)	Tally	Frequency ($f$)	$f \times x$
4	||	2	8
5	|||| |	5	25
6	|||| |||	8	48
7	|||| |	6	42
8	||||	4	32
9	|||	3	27
10	||	2	20
Total		$\sum f = 30$	$\sum fx = 202$

b. Solutions:

i. Range: Highest (10) - Lowest (4) = 6.

ii. Mode: The score with the highest frequency (8) is 6.

iii. Mean: $\bar{x} = \frac{\sum fx}{\sum f} = \frac{202}{30} = \mathbf{6.73}$

iv. Probability: $P(\text{score of 7}) = \frac{6}{30} = \mathbf{1/5}$ (or 0.2).

Section B: Essay - Question 2

Question 2:
a. If the sum of two numbers is 60 and the positive difference between the numbers is 20. Determine the numbers.
b. Make $f$ the subject of the formula: $\frac{1}{u} = \frac{1}{f} - \frac{1}{v}$

👉 Click to view solutions

a. Determining the Numbers:

Let the two numbers be $x$ and $y$.
Equation 1: $x + y = 60$
Equation 2: $x - y = 20$

Adding Equation 1 and Equation 2:
$2x = 80$
$x = \frac{80}{2} = 40$

Substitute $x = 40$ into Equation 1:
$40 + y = 60$
$y = 60 - 40 = 20$

The numbers are 40 and 20.

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b. Changing the Subject of the Formula:

Given: $\frac{1}{u} = \frac{1}{f} - \frac{1}{v}$

1. Isolate $\frac{1}{f}$ by adding $\frac{1}{v}$ to both sides:
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

2. Find a common denominator for the right side ($uv$):
$\frac{1}{f} = \frac{v + u}{uv}$

3. Take the reciprocal of both sides to solve for $f$:
$f = \frac{uv}{v + u}$ (or $f = \frac{uv}{u + v}$)

Section B: Essay - Question 3

Question 3:
a. A ladder 5m long leans against an electric pole. If the distance between the foot of the ladder and that of the electric pole is 4m, how far up the electric pole does the ladder reach?
b. Calculate the simple interest on ₦450.00 for 1½ years at 8% per annum.
c. Convert $110111_{two}$ to base ten.

👉 Click to view solutions

a. Pythagorean Theorem Problem:

This forms a right-angled triangle where the ladder is the hypotenuse ($c = 5$m) and the distance on the ground is the base ($b = 4$m). We need to find the height ($a$).
$a^2 + b^2 = c^2$
$a^2 + 4^2 = 5^2$
$a^2 + 16 = 25$
$a^2 = 25 - 16 = 9$
$a = \sqrt{9} = 3$m.
The ladder reaches 3m up the pole.

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b. Simple Interest Calculation:

Principal ($P$) = ₦450, Rate ($R$) = 8%, Time ($T$) = 1.5 years (1½)
$I = \frac{P \times R \times T}{100}$
$I = \frac{450 \times 8 \times 1.5}{100}$
$I = \frac{450 \times 12}{100}$
$I = \frac{5400}{100} = 54$
The simple interest is ₦54.00.

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c. Number Base Conversion ($110111_2$ to Base 10):

Multiply each digit by 2 raised to its position power:
$(1 \times 2^5) + (1 \times 2^4) + (0 \times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (1 \times 2^0)$
$= 32 + 16 + 0 + 4 + 2 + 1$
$= 55$
$110111_{two} = 55_{ten}$.

Section B: Essay - Question 4

Question 4:
a. How many liters can be held by a cylindrical can 7cm in diameter and 30cm high? (Take $\pi = 22/7$)
b. Calculate the number of sides in a regular polygon whose one interior angle is 160°.

👉 Click to view solutions

a. Volume of a Cylinder (in Liters):

1. Given: Diameter = 7cm, so Radius ($r$) = $7/2$cm = 3.5cm. Height ($h$) = 30cm.
2. Formula: $V = \pi r^2 h$
$V = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 30$. (check denominator that can divide the numerator and cancel out to reduce the numbers)
$V = 11 \times 7 \times 15 = 1155$ cm³.
3. Convert to Liters: Since $1000 \text{cm}^3 = 1 \text{Liter}$:
$V = \frac{1155}{1000} = 1.155$
The can can hold 1.155 liters.

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b. Number of Sides of a Regular Polygon:

1. Find the exterior angle first (Interior + Exterior = 180°):
$\text{Exterior angle} = 180° - 160° = 20°$.
2. Use the formula: $n = \frac{360°}{\text{Exterior angle}}$:
$n = \frac{360}{20} = 18$
The polygon has 18 sides.