# 2023 BECE Mathematics Questions with Solutions to Watch (Full Set)

In this post, we are sharing our first set of 10 BECE Mathematics Questions with a step-by-step solution to watch for 2023 BECE. Candidates and teachers should commit time and effort towards solving this and comparing the answers to the ones provided. Do well to revise the topics from which the questions were set as well.

## 10 BECE Mathematics Questions with a step-by-step solution to watch for 2023 BECE -Set 1

- Solve the following equation for x: 2x + 5 = 3(x – 1)
Solution: Distribute the 3 on the right side: 2x + 5 = 3x – 3

Subtract 2x from both sides: 5 = x – 3

Add 3 to both sides: 8 = x

**Therefore, x = 8 is the solution to the equation.** - Find the value of x in the equation: 3(2x + 1) = 4(x – 3) + 5
Solution: Distribute the 3 on the left side and the 4 on the right side: 6x + 3 = 4x – 12 + 5

Combine like terms: 6x + 3 = 4x – 7

Subtract 4x from both sides: 2x + 3 = -7

Subtract 3 from both sides: 2x = -10

Divide by 2: x = -5

## Therefore, x = -5 is the solution to the equation.

- Solve the following inequality and express the solution set in interval notation: 2x – 3 > 5
Solution: Add 3 to both sides: 2x > 8

Divide by 2: x > 4

The solution set in interval notation is (4, +∞), meaning x is greater than 4 but not including 4.

- Simplify the following expression: (3x² – 2x + 5) – (2x² + 4x – 1)
Solution: Remove the parentheses and combine like terms: 3x² – 2x + 5 – 2x² – 4x + 1

Combine like terms: (3x² – 2x²) + (-2x – 4x) + (5 + 1)

Simplify: x² – 6x + 6

#### Therefore, the simplified expression is x² – 6x + 6.

- Solve the system of equations: 2x + 3y = 10 4x – y = 8
Solution: Multiply the second equation by 3 to eliminate y: 12x – 3y = 24

Add the two equations together: (2x + 3y) + (12x – 3y) = 10 + 24 14x = 34

Divide by 14: x = 34/14

Simplify: x = 17/7

Substitute x back into one of the original equations: 2(17/7) + 3y = 10

Simplify: 34/7 + 3y = 10

Subtract 34/7 from both sides: 3y = 10 – 34/7

Find a common denominator: 3y = 70/7 – 34/7

Simplify: 3y = 36/7

Divide by 3: y = 36/21

Simplify: y = 12/7

### Therefore, the solution to the system of equations is x = 17/7 and y =

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## 10 BECE Mathematics Questions with a step-by-step solution to Watch for 2023 BECE SET 1

- A box contains 5 red balls, 3 blue balls, and 2 green balls. If a ball is randomly selected from the box, what is the probability of selecting a blue ball?
Solution: The total number of balls in the box is 5 + 3 + 2 = 10. The probability of selecting a blue ball is the number of favourable outcomes (blue balls) divided by the total number of outcomes. Therefore, the probability of selecting a blue ball is 3/10.

- A triangle has side lengths of 6 cm, 8 cm, and 10 cm. Determine whether the triangle is scalene, isosceles, or equilateral.
Solution: In a scalene triangle, all three sides have different lengths. In an isosceles triangle, two sides have the same length. In an equilateral triangle, all three sides have the same length. Comparing the side lengths, we find that 6 cm, 8 cm, and 10 cm are all different. Therefore, the triangle is a scalene triangle.

- Solve the following proportion for x: (2/3) = (x/12)
Solution: To solve the proportion, cross-multiply: 2 * 12 = 3 * x 24 = 3x

Divide both sides by 3: x = 8

## Therefore, the value of x that satisfies the proportion is 8.

- Find the area of a trapezium with a height of 5 cm, a base of 8 cm, and a top length of 4 cm.
Solution: The formula to calculate the area of a trapezium is given by: Area = (1/2) * (a + b) * h, where a and b are the lengths of the parallel sides and h is the height. Plugging in the given values: Area = (1/2) * (4 + 8) * 5 = (1/2) * 12 * 5 = 30 cm²

### Therefore, the area of the trapezium is 30 cm².

- Solve the following word problem: A shopkeeper bought a box of chocolates for GH₵ 120. He wants to sell them at a 25% markup. What should be the selling price of the chocolates?

Solution: To find the selling price, we need to add the markup percentage to the cost price. Markup amount = (25/100) * 120 = 0.25 * 120 = GH₵ 30 Selling price = Cost price + Markup amount = 120 + 30 = GH₵ 150

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## Therefore, the selling price of the chocolates should be GH₵ 150.

Thanks for taking the 2023 BECE 10 Mathematics Questions with solutions to watch seriously. Do well to revise the topics from which the questions were set as well.