BECE 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?

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Q1.

a)

  1. Let the total marks be TT.

    We are given that 45 marks represent 60% of the total marks.

    0.60×T=450.60 \times T = 45

  2. Divide both sides by 0.60 to find TT.

    T=450.60T = \frac{45}{0.60} T=75T = 75

    The total marks are 75.

    b)

  3. Calculate 80% of the total marks.

    80% of 75=0.80×75=6080\% \text{ of } 75 = 0.80 \times 75 = 60

  4. Find the difference between 80% of the total marks and the student's current score.

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

    The student needs 15 more marks to reach 80%.

    Q2.

  5. Find the total number of ratio parts.

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

  6. Find the value of one ratio part.

    Value per part=GH₵3609=GH₵40\text{Value per part} = \frac{\text{GH₵}360}{9} = \text{GH₵}40

    a)

  7. Calculate Kofi's share.

    Kofi’s share=2×GH₵40=GH₵80\text{Kofi's share} = 2 \times \text{GH₵}40 = \text{GH₵}80

    Kofi's share is GH₵80.

    b)

  8. Calculate Ama's share.

    Ama’s share=3×GH₵40=GH₵120\text{Ama's share} = 3 \times \text{GH₵}40 = \text{GH₵}120

    Ama's share is GH₵120.

    c)

  9. Calculate Yaw's share.

    Yaw’s share=4×GH₵40=GH₵160\text{Yaw's share} = 4 \times \text{GH₵}40 = \text{GH₵}160

    Yaw's share is GH₵160.

    Q3.
    Given: Length (LL) = 20 m, Width (WW) = 15 m.

    a)

  10. Use the formula for the perimeter of a rectangle.

    P=2(L+W)P = 2(L + W) P=2(20 m+15 m)P = 2(20 \text{ m} + 15 \text{ m}) P=2(35 m)P = 2(35 \text{ m}) P=70 mP = 70 \text{ m}

    The perimeter is 70 m.

    b)

  11. Use the formula for the area of a rectangle.

    A=L×WA = L \times W A=20 m×15 mA = 20 \text{ m} \times 15 \text{ m} A=300 m2A = 300 \text{ m}^2

    The area is 300 m2^2.

    c)

  12. Calculate the cost of fencing using the perimeter and the cost per metre.

    Cost=Perimeter×Cost per metre\text{Cost} = \text{Perimeter} \times \text{Cost per metre} Cost=70 m×GH₵5/m\text{Cost} = 70 \text{ m} \times \text{GH₵}5/\text{m} Cost=GH₵350\text{Cost} = \text{GH₵}350

    The cost of fencing is GH₵350.

    Q4.
    Given: Two angles on a straight line are (x+10)∘(x+10)^\circ and (3x−20)∘(3x-20)^\circ.

    a)

  13. Angles on a straight line sum to 180∘180^\circ.

    (x+10)+(3x−20)=180(x+10) + (3x-20) = 180

  14. Combine like terms.

    4x−10=1804x - 10 = 180

  15. Add 10 to both sides.

    4x=1904x = 190

  16. Divide by 4.

    x=1904x = \frac{190}{4} x=47.5x = 47.5

    The value of xx is 47.5.

    b)

  17. Calculate the values of the two angles using x=47.5x=47.5.

    First angle=(x+10)∘=(47.5+10)∘=57.5∘\text{First angle} = (x+10)^\circ = (47.5+10)^\circ = 57.5^\circ Second angle=(3x−20)∘=(3×47.5−20)∘=(142.5−20)∘=122.5∘\text{Second angle} = (3x-20)^\circ = (3 \times 47.5 - 20)^\circ = (142.5 - 20)^\circ = 122.5^\circ

  18. Assume "that angle" refers to the first angle, 57.5∘57.5^\circ.

    The sum of angles in a triangle is 180∘180^\circ.

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

    The third angle is 72.5∘^\circ.

    Q5.

    a)

  19. Let the original price be PP.

    A 10% discount means the selling price is 90% of the original price.

    0.90×P=GH₵1800.90 \times P = \text{GH₵}180

  20. Divide by 0.90 to find PP.

    P=GH₵1800.90P = \frac{\text{GH₵}180}{0.90} P=GH₵200P = \text{GH₵}200

    The original price was GH₵200.

    b)

  21. Calculate the profit.

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

  22. Calculate the profit percentage.

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

    The profit percentage is 20%.

    Q6.
    Given marks: 42, 55, 50, 48, 55, 62, 38, 50, 55, 60, 45, 48.
    i.

  23. Identify the frequency of each mark.

    • 38: 1
    • 42: 1
    • 45: 1
    • 48: 2
    • 50: 2
    • 55: 3
    • 60: 1
    • 62: 1 The mark that appears most frequently is 55. The mode is 55.

    ii.

  24. Identify the highest and lowest marks.

    Highest mark = 62
    Lowest mark = 38

  25. Calculate the range.

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

    The range is 24.

    b)
    Given: 4 red, 3 blue, 3 green balls.

  26. Calculate the total number of balls.

    Total balls=4+3+3=10\text{Total balls} = 4 + 3 + 3 = 10

    i.

  27. Calculate the probability of drawing a blue ball.

    P(Blue)=Number of blue ballsTotal number of ballsP(\text{Blue}) = \frac{\text{Number of blue balls}}{\text{Total number of balls}} P(Blue)=310P(\text{Blue}) = \frac{3}{10}

    The probability of drawing a blue ball is 310\boxed{\frac{3}{10}}.

    ii.

  28. Calculate the probability of drawing a ball that is not green.

    Number of non-green balls=Number of red balls+Number of blue balls\text{Number of non-green balls} = \text{Number of red balls} + \text{Number of blue balls} Number of non-green balls=4+3=7\text{Number of non-green balls} = 4 + 3 = 7 P(Not green)=Number of non-green ballsTotal number of ballsP(\text{Not green}) = \frac{\text{Number of non-green balls}}{\text{Total number of balls}} P(Not green)=710P(\text{Not green}) = \frac{7}{10}

    The probability of drawing a ball that is not green is 710\boxed{\frac{7}{10}}.

    Q7.

    a)

  29. Identify the fixed fee and the cost per kilometer.

    Fixed fee = GH₵10
    Cost per km = GH₵1.20
    Distance = 15 km

  30. Calculate the total cost.

    Total cost=Fixed fee+(Cost per km×Distance)\text{Total cost} = \text{Fixed fee} + (\text{Cost per km} \times \text{Distance}) Total cost=GH₵10+(GH₵1.20/km×15 km)\text{Total cost} = \text{GH₵}10 + (\text{GH₵}1.20/\text{km} \times 15 \text{ km}) Total cost=GH₵10+GH₵18\text{Total cost} = \text{GH₵}10 + \text{GH₵}18 Total cost=GH₵28\text{Total cost} = \text{GH₵}28

    A passenger will pay GH₵28 for a trip of 15 km.

    b)
    Given: Length (LL) = 85 m, Width (WW) = 60 m.
    i.

  31. Use the formula for the perimeter of a rectangle.

    P=2(L+W)P = 2(L + W) P=2(85 m+60 m)P = 2(85 \text{ m} + 60 \text{ m}) P=2(145 m)P = 2(145 \text{ m}) P=290 mP = 290 \text{ m}

    The perimeter of the field is 290 m.

    ii.

  32. The fencing around the field is its perimeter.

    Perimeter fencing=290 m\text{Perimeter fencing} = 290 \text{ m}

  33. To divide the field into 5 equal smaller plots by lines parallel to the shorter side (60 m), 4 internal fences are needed.

    Number of internal fences=5−1=4\text{Number of internal fences} = 5 - 1 = 4

  34. Each internal fence will have the length of the shorter side.

    Length of each internal fence=60 m\text{Length of each internal fence} = 60 \text{ m}

  35. Calculate the total length of internal fencing.

    Total internal fencing=4×60 m=240 m\text{Total internal fencing} = 4 \times 60 \text{ m} = 240 \text{ m}

  36. Calculate the total fencing needed.

    Total fencing=Perimeter fencing+Total internal fencing\text{Total fencing} = \text{Perimeter fencing} + \text{Total internal fencing} Total fencing=290 m+240 m=530 m\text{Total fencing} = 290 \text{ m} + 240 \text{ m} = 530 \text{ m}

    A total of 530 m of

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