FBISE Class 10 Maths Practice: Worked Problems and Common Mistakes

Mathematics revision works best when you can explain your method, carry it out accurately and check whether the answer fits the original problem. Looking at a completed solution can make a topic feel familiar, but independent practice reveals which steps you can actually produce.

This guide offers original problems involving equations, quadratics, functions, coordinate geometry, simultaneous equations and units. It is a focused practice article rather than a complete Class 10 syllabus or an official model paper. Use only the topics relevant to your own course.

Before planning examination revision, confirm the applicable mathematics framework through the FBISE assessment and model-paper index. The official site contains resources with different curriculum and scheme labels. Match those details with your school’s guidance instead of assuming that every Class 10 paper uses the same structure.

Prepare an answer page that shows your reasoning

For each question, write the given information, the method, the calculation and a check. You do not need a separate heading for every short step, but the reasoning should be easy to follow. Use one equation per line when several operations are involved.

Keep an error notebook beside your practice book. Record whether a mistake came from choosing the wrong method, a sign error, arithmetic, a missing condition or a unit. A useful correction names the cause and includes a new problem that tests the same skill.

Try each example before reading its solution. If you get stuck, identify the exact step that is unclear. “I cannot factor this expression” is a more useful starting point for help than “I am bad at maths.”

Problem one: solve a linear equation

Solve 3(2x − 1) = 15. Begin by expanding the bracket: 6x − 3 = 15. Add three to both sides to obtain 6x = 18. Divide both sides by six, giving x = 3.

Check the result in the original equation. Substituting three gives 3(6 − 1) = 3 × 5 = 15. The left side equals the right side, so the value satisfies the equation.

A common error is expanding 3(2x − 1) as 6x − 1. The multiplier applies to both terms inside the bracket. For further practice, solve 2(3x + 4) = 26. The correct value is x = 3, which gives 2(9 + 4) = 26.

Problem two: factor a quadratic

Solve x² − 5x + 6 = 0. Look for two numbers whose product is six and whose sum is negative five. Those numbers are negative two and negative three, so the expression factors as (x − 2)(x − 3).

A product is zero when at least one of its factors is zero. Therefore x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3. Check both values in the original expression: four minus ten plus six is zero, and nine minus fifteen plus six is also zero.

Do not stop after finding one root. Unless a condition excludes one value, both belong in the solution set. Also remember that the zero-product step applies after the equation has been written with zero on one side.

Problem three: distinguish an expression from an equation

The expression x² − 5x + 6 can be factored, but it has no stated equality to solve. The equation x² − 5x + 6 = 0 asks for values that make an equality true. Reading this distinction prevents you from answering a different task.

If the instruction says “factor,” a suitable result is (x − 2)(x − 3). If the instruction says “solve the equation,” provide the values of x. Where the question requests a check, include substitution or another valid verification.

For practice, factor y² + 7y + 12. The result is (y + 3)(y + 4). If the equation is y² + 7y + 12 = 0, its roots are negative three and negative four. The algebra is connected, but the requested final response differs.

Problem four: use function notation

Let f(x) = 2x + 1 and g(x) = x². Find f(g(3)). First evaluate the inner function: g(3) = 9. Then substitute that output into f: f(9) = 19. Therefore f(g(3)) = 19.

Now reverse the order. For g(f(3)), first calculate f(3) = 7. Then g(7) = 49. This gives g(f(3)) = 49. The different results show why the order of composition matters.

Read the notation from the inside outward. Avoid multiplying f(3) by g(3), because composition is not the same operation as multiplication. As a quick check, f(g(2)) = f(4) = 9, while g(f(2)) = g(5) = 25.

Problem five: find a gradient

Find the gradient of the line through A(2, 3) and B(6, 11). Gradient is the change in y divided by the change in x. Using the same point order in numerator and denominator gives (11 − 3)/(6 − 2) = 8/4 = 2.

You could instead use (3 − 11)/(2 − 6), which also equals two. The important point is consistency. Reversing the order in only one part changes the sign and produces an incorrect result.

A positive gradient means y increases as x increases along this line. If two points have the same x-coordinate, the denominator is zero and the line is vertical; its gradient is undefined. Check for that condition before dividing.

Problem six: solve simultaneous equations

Solve x + y = 7 and x − y = 1. Add the equations to eliminate y. The result is 2x = 8, so x = 4. Substitute into the first equation: 4 + y = 7, giving y = 3.

Check the pair in both original equations. Four plus three equals seven, and four minus three equals one. A pair that satisfies only one equation does not solve the simultaneous system.

Elimination is convenient here because the y-coefficients already cancel. In another system, substitution may be simpler. Choose the method that fits the equations, and preserve the equality when multiplying, adding or subtracting entire equations.

Problem seven: interpret a word problem

A rectangle has length three centimetres more than its width and perimeter thirty centimetres. Find its dimensions. Let the width be w centimetres, so the length is w + 3 centimetres. The perimeter formula gives 2w + 2(w + 3) = 30.

Simplifying gives 4w + 6 = 30, then 4w = 24 and w = 6. The length is nine centimetres. Check the perimeter: 2(6 + 9) = 30 centimetres.

The wording “three more” creates an addition relationship, while “perimeter” requires the distance around the shape. Using area instead would solve a different problem. Include units in the final answer: width six centimetres and length nine centimetres.

Problem eight: calculate a mean

Find the arithmetic mean of 6, 8, 10 and 12. Add the four values to obtain thirty-six, then divide by four. The mean is nine. The divisor is the number of observations, not the largest value or the number of gaps between the observations.

Check whether nine is reasonable. It lies between the smallest and largest values and matches the symmetry of this simple set. A mean of thirty-six would signal that the division step had been missed.

For a frequency table, each value must be multiplied by its frequency before summing. For example, two scores of four and one score of ten have total eighteen across three observations, so the mean is six. They are not just two equally weighted values.

Try this short independent set

One: solve 4x + 7 = 23. Two: factor x² − 7x + 12. Three: for f(x) = 3x − 2, find f(5). Four: find the gradient between (1, 2) and (4, 8).

Five: solve a + b = 9 and a − b = 3. Six: a rectangle has length eight centimetres and width five centimetres; find its area and perimeter. Seven: find the mean of 3, 5, 7 and 9.

Write complete workings before checking the answers. If you use this as a timed exercise, record the time honestly, but do not mistake this small set for an official examination section. Its purpose is to identify the next skill that needs practice.

Check the independent answers

One: x = 4, because 4x = 16. Two: (x − 3)(x − 4), because negative three and negative four have product twelve and sum negative seven. Three: f(5) = 15 − 2 = 13.

Four: gradient = (8 − 2)/(4 − 1) = 6/3 = 2. Five: adding gives 2a = 12, so a = 6 and b = 3. Six: area = 8 × 5 = 40 square centimetres; perimeter = 2(8 + 5) = 26 centimetres.

Seven: the total is twenty-four and there are four values, so the mean is six. If your answer differs, locate the first incorrect step rather than replacing the whole solution without understanding the cause.

Improve your checking routine

Use substitution for equations, estimation for numerical calculations and a sketch when geometry relationships are difficult to track. Check units separately: area uses square units, perimeter uses length units, and a numerical ratio may have different unit behaviour depending on the quantities involved.

For fractions, make sure the denominator is not zero. For word problems, reject a result that violates the situation, such as a negative physical length. For graphs, label axes and use a consistent scale that allows the relevant points to be read.

A check supports a solution but should not hide the method. Keep enough working for another reader to understand how the answer was obtained. This also makes it easier for you to find an error when the result fails the check.

Explain a solution to someone else

Choose one completed problem and talk through it without reading every line. Explain why the first step is valid, what each later operation achieves and how the final check connects to the original question. If you cannot explain a step, return to that part rather than repeating the whole solution automatically.

For example, in a simultaneous system, adding two equations is useful when the coefficients allow a variable to cancel. Saying “add because that is the method” does not explain the choice. Identifying the opposite coefficients shows that you understand why the operation simplifies the system.

Next, change one number in the question and decide whether the same method remains convenient. This checks whether your understanding extends beyond one remembered example. You can do this with a classmate, a teacher or a short written explanation to yourself. Keep any changed problem clearly separate from the original so that the two sets of numbers do not become mixed in your working.

Plan your next week of maths

Choose one weak skill from today’s work. Spend the next session reviewing a clear explanation, then attempt a few fresh problems without the worked example beside you. Return to the original mistake later to see whether you can now handle it independently.

Include mixed practice once individual skills are more secure. Mixed questions require you to recognise which method applies. This is different from completing a page where every problem uses the same procedure.

When you use a model paper, match it to your current course and review its instructions directly. The official FBISE model-paper page provides the starting point. Clear methods, regular checks and honest correction give each practice session a useful purpose.