CTVET 2026 MATHEMATICS HEAT QUESTIONS ✅ SOLVE Q1. A student scored 45 marks in a test which is 60%. a) Find the total marks b) How many more marks are needed to reach 80%? Q2. A sum of GH₵360 is shared among Kofi, Ama, and Yaw in the ratio 2 : 3 : 4. Find: a) Kofi’s share b) Ama’s share c) Yaw’s share Q3. A rectangular garden is 20 m long and 15 m wide. a) Find the perimeter b) Find the area c) Find the cost of fencing it at GH₵5 per metre Q4. Two angles on a straight line Find(x+10)°and (3x-20) ° a) Find x b) If one of the angles is part of a triangle, find the third angle if the other two are 50° and that angle. Q5. A shirt was sold for GH₵180 after a 10% discount. What was the original price? b) A shopkeeper bought goods for GH₵450 and sold them for GH₵540. Find the profit percentage. Q6. The marks of 12 students are: 42, 55, 50, 48, 55, 62, 38, 50, 55, 60, 45, 48 i. Find the mode ii. Find the range b) A bag contains 4 red, 3 blue and 3 green balls. One ball is drawn at random. Find the probability that it is: i. Blue ii. Not green Q7. a) A taxi company charges a fixed fee of GH₵10 plus GH₵1.20 per kilometer. How much will a passenger pay for a trip of 15 km? b) A rectangular field is 85 m long and 60 m wide. A farmer wants to put a fence around the field and also divide the field into 5 equal smaller plots by lines parallel to the shorter side. i. What is the perimeter of the field? ii. How many metres of fencing will be needed altogether? JOIN OUR GHANA 🇬🇭CTVET PAGE FOR TECHNICAL STUDENTS🚧👨‍🔧🛠️ https://t.me/ctvetGH https://t.me/ctvet2026 FOLLOW THE MAIN 🇬🇭CTVET |NABPTEX 🛠️👨🏽‍🔧🚧 WHATSAPP https://whatsapp.com/channel/0029VaSrl7I7tkjCsPAoRW11 SHARE WITH YOUR FRIENDS TO INFORM THEM_

Here are the solutions to your questions:

Q1.

a) Find the total marks

Step 1: Let the total marks be $T$. We are given that 45 marks is 60% of the total marks.

$$0.60 \times T = 45$$

Step 2: Divide by 0.60 to find $T$.

$$T = \frac{45}{0.60}$$ $$T = 75$$

The total marks are 75.

$$\boxed{\text{75 marks}}$$

b) How many more marks are needed to reach 80%?

Step 1: Calculate 80% of the total marks.

$$80\% \text{ of } 75 = 0.80 \times 75$$ $$0.80 \times 75 = 60$$

Step 2: Find the difference between 80% of the total marks and the student's current marks.

$$\text{Marks needed} = 60 - 45$$ $$\text{Marks needed} = 15$$

The student needs 15 more marks to reach 80%.

$$\boxed{\text{15 marks}}$$

Q2.

a) Kofi’s share

b) Ama’s share

c) Yaw’s share

Step 1: Find the total number of ratio parts.

$$\text{Total ratio parts} = 2 + 3 + 4 = 9$$

Step 2: Find the value of one ratio part.

$$\text{Value per part} = \frac{\text{GH₵}360}{9} = \text{GH₵}40$$

Step 3: Calculate each person's share.

a) Kofi's share:

$$\text{Kofi's share} = 2 \times \text{GH₵}40 = \text{GH₵}80$$ $$\boxed{\text{GH₵80}}$$

b) Ama's share:

$$\text{Ama's share} = 3 \times \text{GH₵}40 = \text{GH₵}120$$ $$\boxed{\text{GH₵120}}$$

c) Yaw's share:

$$\text{Yaw's share} = 4 \times \text{GH₵}40 = \text{GH₵}160$$ $$\boxed{\text{GH₵160}}$$

Q3.

a) Find the perimeter

Step 1: The formula for the perimeter of a rectangle is $P = 2(L+W)$.

$$P = 2(20 \text{ m} + 15 \text{ m})$$ $$P = 2(35 \text{ m})$$ $$P = 70 \text{ m}$$

The perimeter is 70 m.

$$\boxed{\text{70 m}}$$

b) Find the area

Step 1: The formula for the area of a rectangle is $A = L \times W$.

$$A = 20 \text{ m} \times 15 \text{ m}$$ $$A = 300 \text{ m}^2$$

The area is 300 m$^2$.

$$\boxed{\text{300 m}^2}$$

c) Find the cost of fencing it at GH₵5 per metre

Step 1: Multiply the perimeter by the cost per metre.

$$\text{Cost} = 70 \text{ m} \times \text{GH₵}5/\text{m}$$ $$\text{Cost} = \text{GH₵}350$$

The cost of fencing is GH₵350.

$$\boxed{\text{GH₵350}}$$

Q4.

a) Find x

Step 1: Angles on a straight line sum to 180°.

$$(x+10)^\circ + (3x-20)^\circ = 180^\circ$$

Step 2: Combine like terms.

$$4x - 10 = 180$$

Step 3: Add 10 to both sides.

$$4x = 190$$

Step 4: Divide by 4.

$$x = \frac{190}{4}$$ $$x = 47.5$$

The value of $x$ is 47.5.

$$\boxed{x = 47.5}$$

b) If one of the angles is part of a triangle, find the third angle if the other two are 50° and that angle.

Step 1: Calculate the values of the two angles on the straight line.
First angle: $x+10 = 47.5+10 = 57.5^\circ$
Second angle: $3x-20 = 3(47.5)-20 = 142.5-20 = 122.5^\circ$
Let's use the first angle, $57.5^\circ$, as "that angle" for the triangle.

Step 2: The sum of angles in a triangle is 180°.

$$\text{Third angle} = 180^\circ - 50^\circ - 57.5^\circ$$ $$\text{Third angle} = 130^\circ - 57.5^\circ$$ $$\text{Third angle} = 72.5^\circ$$

The third angle is 72.5°.

$$\boxed{\text{72.5}^\circ}$$

Q5.

a) What was the original price?

Step 1: A 10% discount means the shirt was sold for $100\% - 10\% = 90\%$ of its original price.
Let the original price be $P$.

$$0.90 \times P = \text{GH₵}180$$

Step 2: Divide by 0.90 to find $P$.

$$P = \frac{\text{GH₵}180}{0.90}$$ $$P = \text{GH₵}200$$

The original price was GH₵200.

$$\boxed{\text{GH₵200}}$$

b) A shopkeeper bought goods for GH₵450 and sold them for GH₵540. Find the profit percentage.

Step 1: Calculate the profit.

$$\text{Profit} = \text{Selling Price} - \text{Cost Price}$$ $$\text{Profit} = \text{GH₵}540 - \text{GH₵}450$$ $$\text{Profit} = \text{GH₵}90$$

Step 2: Calculate the profit percentage.

$$\text{Profit Percentage} = \frac{\text{Profit}}{\text{Cost Price}} \times 100\%$$ $$\text{Profit Percentage} = \frac{\text{GH₵}90}{\text{GH₵}450} \times 100\%$$ $$\text{Profit Percentage} = 0.2 \times 100\%$$ $$\text{Profit Percentage} = 20\%$$

The profit percentage is 20%.

$$\boxed{20\%}$$

Q6.
i. Find the mode

Step 1: List the marks and count their frequencies:
42 (1), 55 (3), 50 (2), 48 (2), 62 (1), 38 (1), 60 (1), 45 (1)

Step 2: The mode is the mark that appears most frequently.
The mark 55 appears 3 times, which is more than any other mark.
The mode is 55.

$$\boxed{55}$$

ii. Find the range

Step 1: Identify the highest and lowest marks.
Highest mark = 62
Lowest mark = 38

Step 2: Calculate the range.

$$\text{Range} = \text{Highest mark} - \text{Lowest mark}$$ $$\text{Range} = 62 - 38$$ $$\text{Range} = 24$$

The range is 24.

$$\boxed{24}$$

b) A bag contains 4 red, 3 blue and 3 green balls. One ball is drawn at random.

Step 1: Find the total number of balls.

$$\text{Total balls} = 4 (\text{red}) + 3 (\text{blue}) + 3 (\text{green}) = 10$$

i. Find the probability that it is: Blue

Step 2: The probability of drawing a blue ball is the number of blue balls divided by the total number of balls.

$$P(\text{Blue}) = \frac{\text{Number of blue balls}}{\text{Total balls}}$$ $$P(\text{Blue}) = \frac{3}{10}$$

The probability of drawing a blue ball is $\frac{3}{10}$.

$$\boxed{\frac{3}{10}}$$

ii. Not green

Step 2: The number of balls that are not green is the sum of red and blue balls.

$$\text{Not green balls} = 4 (\text{red}) + 3 (\text{blue}) = 7$$

Step 3: The probability of drawing a ball that is not green is the number of non-green balls divided by the total number of balls.

$$P(\text{Not green}) = \frac{\text{Number of not green balls}}{\text{Total balls}}$$ $$P(\text{Not green}) = \frac{7}{10}$$

The probability of drawing a ball that is not green is $\frac{7}{10}$.

$$\boxed{\frac{7}{10}}$$

Q7.

a) A taxi company charges a fixed fee of GH₵10 plus GH₵1.20 per kilometer. How much will a passenger pay for a trip of 15 km?

Step 1: Calculate the variable cost for 15 km.

$$\text{Variable cost} = 15 \text{ km} \times \text{GH₵}1.20/\text{km}$$ $$\text{Variable cost} = \text{GH₵}18.00$$

Step 2: Add the fixed fee to the variable cost.

$$\text{Total cost} = \text{Fixed fee} + \text{Variable cost}$$ $$\text{Total cost} = \text{GH₵}10.00 + \text{GH₵}18.00$$ $$\text{Total cost} = \text{GH₵}28.00$$

A passenger will pay GH₵28.00 for a 15 km trip.

$$\boxed{\text{GH₵28.00}}$$

b) A rectangular field is 85 m long and 60 m wide. A farmer wants to put a fence around the field and also divide the field into 5 equal smaller plots by lines parallel to the shorter side.
i. What is the perimeter of the field?

Step 1: The formula for the perimeter of a rectangle is $P = 2(L+W)$.

$$P = 2(85 \text{ m} + 60 \text{ m})$$ $$P = 2(145 \text{ m})$$ $$P = 290 \text{ m}$$

The perimeter of the field is 290 m.

$$\boxed{\text{290 m}}$$

ii. How many metres of fencing will be needed altogether?

Step 1: The fencing around the field is its perimeter, which is 290 m.

Step 2: To divide the field into 5 equal smaller plots by lines parallel to the shorter side (60 m), 4 dividing lines are needed. Each dividing line will be 60 m long.

$$\text{Length of dividing lines} = 4 \times 60 \text{ m} = 240 \text{ m}$$

Step 3: Add the perimeter and the length of the dividing lines to find the total fencing needed.

$$$$