1. Place Value and Rounding
A district records 48,765 learners. Round this to the nearest thousand.
Show worked answer
The hundreds digit is 7, so increase the thousands digit by one. The estimate is 49,000 learners.
Original DAS BECE-style practice. Try each question before opening its worked solution. The complete NaCCA exemplar set is available in the Curriculum Reference and the source-page teaching lessons.
A district records 48,765 learners. Round this to the nearest thousand.
The hundreds digit is 7, so increase the thousands digit by one. The estimate is 49,000 learners.
Two bells ring every 12 and 18 minutes. They ring together at 8:00 a.m. When will they next ring together?
12 = 2² × 3 and 18 = 2 × 3². LCM = 2² × 3² = 36 minutes. They next ring together at 8:36 a.m.
Express 101101 base two in base ten.
Use powers of two: 1×32 + 0×16 + 1×8 + 1×4 + 0×2 + 1×1 = 45.
Of 40 learners, 25 like football and 18 like volleyball; 10 like both. How many like neither?
Number liking at least one = 25 + 18 − 10 = 33. Neither = 40 − 33 = 7 learners.
Which of −3, 0.75 and the square root of 2 is irrational?
−3 = −3/1 and 0.75 = 3/4 are rational. The square root of 2 cannot be expressed as a ratio of integers, so it is irrational.
Evaluate −6 + 3 × (8 − 5).
Calculate brackets first: 8 − 5 = 3. Multiply: 3 × 3 = 9. Add: −6 + 9 = 3.
Simplify 2⁵ × 2³ ÷ 2⁴.
For the same base, add indices on multiplication and subtract on division. 2^(5+3−4) = 2⁴ = 16.
Write 0.00072 in standard form.
Move the decimal four places right to obtain 7.2. Compensate by multiplying by 10 to the power −4. Answer: 7.2 × 10^(−4).
An operation is defined by a ★ b = 2a + b. Find 3 ★ 5.
Substitute a=3 and b=5 into the given rule: 2(3)+5 = 6+5 = 11.
A tank is 3/4 full. After 1/6 of the whole tank is used, what fraction remains?
Use denominator 12: 3/4 − 1/6 = 9/12 − 2/12 = 7/12.
A rope is 7.2 m long. How many pieces of length 0.45 m can be cut?
Number of pieces = 7.2 ÷ 0.45 = 720 ÷ 45 = 16 pieces.
A bag costs GH₵180 after a 10% discount. Find its original price.
The sale price is 90% of the original. Original = 180 ÷ 0.90 = GH₵200.
Find the simple interest on GH₵800 at 5% per annum for 3 years.
I = PRT/100 = 800 × 5 × 3 / 100 = GH₵120. The total amount is GH₵920.
Share GH₵360 between Ama and Kojo in the ratio 2:3.
Total parts = 2 + 3 = 5. One part = 360/5 = 72. Ama receives GH₵144; Kojo receives GH₵216.
Six equally productive workers finish a job in 10 days. How long will 15 workers take?
Work is constant: 6 × 10 = 60 worker-days. Time for 15 workers = 60/15 = 4 days.
A map scale is 1:50,000. Two villages are 6 cm apart on the map. Find the actual distance in kilometres.
Actual distance = 6 × 50,000 = 300,000 cm. Since 100,000 cm = 1 km, the distance is 3 km.
Find the next two terms of 5, 9, 13, 17, ...
The common difference is 4. Add 4 to 17 to obtain 21, and add 4 again to obtain 25.
Find the 20th term of 3, 7, 11, 15, ...
The first term is 3 and the difference is 4. T_n = 3 + 4(n−1) = 4n−1. T_20 = 80−1 = 79.
For y = 2x + 1, find y when x = −2, 0 and 3.
Substitute separately: y = −4+1 = −3; y = 0+1 = 1; y = 6+1 = 7. Plot (−2,−3), (0,1), (3,7).
Find the gradient of the line through (1,3) and (5,11).
Gradient = change in y / change in x = (11−3)/(5−1) = 8/4 = 2.
Simplify 3a + 2b − a + 5b, then evaluate when a=4 and b=−1.
Collect like terms: 2a + 7b. Substitute: 2(4) + 7(−1) = 8−7 = 1.
Factorise 6x² + 9x completely.
The highest common factor is 3x. Divide each term by 3x to obtain 2x and 3. Answer: 3x(2x+3).
Make h the subject of A = bh/2.
Multiply both sides by 2: 2A = bh. Divide by b: h = 2A/b, where b is not zero.
Solve 3(x−2) = 2x + 5.
Expand: 3x−6 = 2x+5. Subtract 2x: x−6 = 5. Add 6: x = 11. Check: both sides equal 27.
Two pens and one book cost GH₵13. One pen and one book cost GH₵9. Find each price.
Let p be a pen and b a book: 2p+b=13, p+b=9. Subtract to get p=4. Substitute: b=9−4=5. Pen GH₵4; book GH₵5.
Solve −2x + 3 < 11.
Subtract 3: −2x < 8. Divide by −2 and reverse the inequality: x > −4. Use an open circle at −4 and shade to the right.
Two angles on a straight line are 3x and 2x. Find the larger angle.
Their sum is 180°. Thus 5x = 180°, giving x = 36°. The larger angle is 3×36° = 108°.
Find each interior angle of a regular hexagon.
Sum of interior angles = (6−2)×180° = 720°. Each angle = 720°/6 = 120°.
Describe how to construct the perpendicular bisector of a line segment AB.
With the same compass radius greater than half AB, draw arcs above and below AB from A and B. Join the two arc intersections. The resulting line is perpendicular to AB and passes through its midpoint.
A triangular prism has 6 vertices and 9 edges. How many faces does it have?
Use Euler’s relation V−E+F=2. Therefore F=2−6+9=5. These are two triangular and three rectangular faces.
Find the area of a trapezium with parallel sides 8 cm and 14 cm, and height 5 cm.
A = (1/2)(a+b)h = (1/2)(8+14)×5 = 55 cm².
Find the area of a 90° sector of radius 14 cm. Use pi = 22/7.
Sector area = (90/360)×(22/7)×14² = (1/4)×616 = 154 cm².
Find the volume of a cylinder of radius 7 cm and height 10 cm. Use pi = 22/7.
V = pi × r² × h = (22/7)×49×10 = 1,540 cm³.
A 10 m ladder reaches 8 m up a wall. How far is its foot from the wall?
The ladder is the hypotenuse. x² + 8² = 10². Thus x²=36 and x=6 m. Use the positive root for a length.
A right triangle has opposite side 6 cm, adjacent side 8 cm and hypotenuse 10 cm relative to angle theta. Find sin theta and tan theta.
sin theta = opposite/hypotenuse = 6/10 = 3/5. tan theta = opposite/adjacent = 6/8 = 3/4.
Find the vector from A(−2,3) to B(4,−1).
Subtract coordinates of A from B: horizontal component = 4−(−2)=6; vertical component = −1−3=−4. Vector AB has components (6,−4).
Reflect P(3,−2) in the y-axis, then translate by vector (1,4).
Reflection in the y-axis changes the x sign: (−3,−2). Translation gives (−3+1,−2+4) = (−2,2).
Rotate (2,1) through 90° anticlockwise about the origin.
Use (x,y) → (−y,x). Therefore (2,1) maps to (−1,2). Distance from the origin remains the same.
The bearing of B from A is 065°. Find the bearing of A from B.
The reverse direction differs by 180°. 065°+180°=245°. The required bearing is 245°. Measure bearings clockwise from north.
A triangle of area 12 cm² is enlarged by scale factor 3. Find the new area.
Lengths are multiplied by 3, so areas are multiplied by 3² = 9. New area = 12×9 = 108 cm².
The scores are 2, 3, 2, 4, 3, 2, 5, 4. Make a frequency table.
Count each score: 2 occurs 3 times; 3 occurs 2 times; 4 occurs 2 times; 5 occurs once. Total frequency = 3+2+2+1 = 8.
In a class of 40, 15 learners choose science club. Find its pie-chart sector angle.
Fraction choosing science = 15/40. Sector angle = (15/40)×360° = 135°.
Find the mean, median, mode and range of 2, 4, 4, 5, 10.
Sum = 25, so mean = 25/5 = 5. Middle value is 4, so median = 4. Most frequent value is 4, so mode = 4. Range = 10−2 = 8.
A frequency table gives values 1, 2, 3 with frequencies 2, 5, 3. Find the mean.
Total frequency = 10. Sum of value × frequency = 1×2 + 2×5 + 3×3 = 21. Mean = 21/10 = 2.1.
Class intervals 0≤x<10 and 10≤x<20 have frequencies 3 and 7. Estimate the mean.
Midpoints are 5 and 15. Estimated total = 3×5 + 7×15 = 120. Total frequency = 10. Estimated mean = 120/10 = 12.
A fair die is thrown. Find the probability of obtaining a prime number.
Prime outcomes are 2, 3 and 5: 3 favourable outcomes out of 6. Probability = 3/6 = 1/2.
A bag contains 3 red and 7 blue balls. Find the probability of not choosing red.
Total balls = 10. P(red)=3/10. P(not red)=1−3/10=7/10, equal to the probability of blue.
Two fair coins are tossed. Find the probability of exactly one head.
The equally likely outcomes are HH, HT, TH, TT. Exactly one head occurs in HT and TH. Probability = 2/4 = 1/2.