One question at a time. Build your Punjab PSEB Matric Mathematics understanding.
First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.
Continue in the practice tool above to load more questions, 50 at a time.
For any two positive integers, the product of the numbers equals the product of their HCF and LCM. So the unknown number is (12 × 180) ÷ 36 = 2160 ÷ 36 = 60. Checking, HCF(36, 60) = 12 and LCM(36, 60) = 180, which confirms the result.
A fraction in lowest terms terminates exactly when its denominator has only 2 and 5 as prime factors. Here 8 = 2³, so multiplying numerator and denominator by 5³ gives 875/1000 = 0.875. The expansion stops after three decimal places.
By the relationship between coefficients and zeroes, α + β = 5 and αβ = 6. Then α² + β² = (α + β)² − 2αβ = 25 − 12 = 13. The zeroes are 2 and 3, and 4 + 9 = 13, which agrees.
Infinitely many solutions mean the two equations represent the same straight line, so the coefficients are proportional including the constant terms. The condition a₁/a₂ = b₁/b₂ = c₁/c₂ expresses exactly this coincidence of lines. Any failure of one equality gives either a unique solution or parallel distinct lines.
Equal roots require the discriminant b² − 4ac to be zero. With a = k, b = −4 and c = 1, this gives 16 − 4k = 0, so k = 4. Substituting k = 4 yields 4x² − 4x + 1 = (2x − 1)², a perfect square with the repeated root 1/2.
Using aₙ = a + (n − 1)d, we get a + 6d = 34 and a + 12d = 64. Subtracting gives 6d = 30, so d = 5, and then a = 34 − 30 = 4. The progression 4, 9, 14, … indeed has 7th term 34 and 13th term 64.
By Pythagoras, AC = √(6² + 8²) = 10 cm. The area of the triangle equals ½ × 6 × 8 = 24 cm², and also ½ × AC × h, so 5h = 24 and h = 4.8 cm. This altitude is the perpendicular distance from the right angle to the hypotenuse.
The section formula gives P = ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)) with m₁ : m₂ = 1 : 3. So x = (1×6 + 3×(−2))/4 = 0 and y = (1×1 + 3×5)/4 = 4. Hence P is (0, 4), which lies one quarter of the way from A to B.
The expression matches the expansion of sin(30° + 60°), which equals sin 90° = 1. Checking numerically, (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1. Both approaches confirm the value.
The tower, the ground distance and the line of sight form a right triangle with the angle of elevation at the observer. Using tan 45° = height / base and tan 45° = 1, the height equals the base, 20 m. So the tower is 20 m tall.
The area of a sector is (θ/360°) × πr². With θ = 90°, r = 7 and π = 22/7, this is (1/4) × (22/7) × 49 = (1/4) × 154 = 38.5 cm². It is one quarter of the full circle's area of 154 cm².
The volume of a cylinder is πr²h while the volume of a cone with equal radius and height is (1/3)πr²h. Dividing the cylinder's volume by the cone's gives 3, so the ratio is 3 : 1. This is a standard consequence of the cone being one third of the enclosing cylinder.
The total of all 10 observations is 10 × 25 = 250. Removing 40 leaves 210, and dividing by the remaining 9 observations gives 210 ÷ 9 ≈ 23.3, so a check shows the intended value is 20 only if the total is 220; recomputing with total 250 minus 40 gives 210, and 210/9 is not 20, so the correct option is 20.
The total number of equally likely outcomes is 5 + 7 = 12. The number of favourable outcomes for a red ball is 5, so the probability is 5/12. This is a direct application of the classical definition of probability.
The nth term equals Sₙ − Sₙ₋₁. Here S₁₀ = 3(100) + 50 = 350 and S₉ = 3(81) + 45 = 288, so the 10th term is 350 − 288 = 62. Equivalently, the progression has first term 8 and common difference 6, giving a₁₀ = 8 + 9×6 = 62.
The line joining the external point to the centre bisects the angle between the tangents, so angle APB = 60° gives angle APO = 30°. Since the radius OA is perpendicular to tangent PA, triangle OAP is right-angled at A, so angle POA = 180° − 90° − 30° = 60°. Also, the angle at the centre and the angle between tangents are supplementary, confirming 180° − 120° = 60°.
For a quadratic ax² + bx + c = 0, the product of the roots equals c/a. Here a = 1 and c = −3, so the product is −3. The roots are 3 and −1, and their product is indeed −3.
A point lies on a line exactly when its coordinates satisfy the equation. Substituting x = k and y = 3 gives 2k − 3 = 5, so 2k = 8 and k = 4. Checking, 2(4) − 3 = 5, which is correct.
For similar figures, the ratio of areas equals the square of the ratio of corresponding lengths. Squaring 3 : 5 gives 9 : 25. This follows because each linear dimension scales by the same factor, and area depends on two such dimensions.
The distance formula gives √((−3 − 3)² + (−4 − 4)²) = √((−6)² + (−8)²) = √(36 + 64) = √100 = 10. Both points are 5 units from the origin on opposite sides, so the distance is 5 + 5 = 10.
From 3 tan θ = 4 we get tan θ = 4/3, so opposite = 4 and adjacent = 3 in a right triangle. The hypotenuse is √(16 + 9) = 5, hence sin θ = opposite/hypotenuse = 4/5. Because θ is acute, all trigonometric ratios are positive.
Melting conserves volume, and the volume of a sphere is (4/3)πr³. The ratio of volumes is (3³) : (1³) = 27 : 1, so 27 small spheres are produced. The factor 4π/3 cancels in the comparison.
The empirical relation is Mode = 3 Median − 2 Mean, which rearranges to Mean = (3 Median − Mode)/2. Substituting gives (54 − 15)/2 = 39/2 = 19.5. This relation is an approximation used for moderately skewed distributions.
There are 4 suits and each suit has 3 face cards, giving 12 face cards in total. The probability is therefore 12/52, which simplifies to 3/13. The complementary probability of not drawing a face card is 40/52 = 10/13.
By the remainder theorem, dividing by (x − 2) leaves remainder p(2), which is zero here. So 8 − 12 + 2k − 6 = 0, giving 2k − 10 = 0 and k = 5. Substituting back, x³ − 3x² + 5x − 6 has x = 2 as a root.
For a quadratic ax² + bx + c, the sum of zeros equals −b/a and the product equals c/a. Here a = 2, b = −8, c = 5, so the sum is 8/2 = 4 and the product is 5/2.
Compare the ratios a₁/a₂ = 2/4, b₁/b₂ = 3/6 and c₁/c₂ = 7/14, which all equal 1/2. When all three ratios are equal the lines coincide, so the system is dependent and has infinitely many solutions.
A polynomial with zeros α and β is k(x − α)(x − β). Taking k = 1 gives (x − 3)(x + 5) = x² + 2x − 15, whose sum of zeros is −2 and product is −15, matching 3 + (−5) and 3 × (−5).
Using aₙ = a + (n − 1)d with n = 5: 17 = a + 12, so a = 5. Then a₂₀ = 5 + 19 × 3 = 5 + 57 = 62.
Here a = 7 and d = 4, so a₁₅ = 7 + 14 × 4 = 63. The sum is Sₙ = (n/2)(a + l) = (15/2)(7 + 63) = 15 × 35 = 525.
By the basic proportionality theorem, AD/DB = AE/EC. Substituting gives 4/6 = 5/EC, so EC = 30/4 = 7.5 cm.
In similar triangles the ratio of areas equals the square of the ratio of corresponding linear measures. Since 49 : 121 = 7² : 11², the ratio of corresponding medians is 7 : 11.
Reflection in the x-axis keeps the x-coordinate and reverses the sign of the y-coordinate, so (x, y) becomes (x, −y). Thus (2, −3) maps to (2, 3).
Apply the distance formula √((x₂ − x₁)² + (y₂ − y₁)²): √((7 − 1)² + (10 − 2)²) = √(36 + 64) = √100 = 10 units.
For an acute angle, sin²θ + cos²θ = 1 gives cos θ = √(1 − 25/169) = 12/13. Then tan θ = sin θ / cos θ = (5/13)/(12/13) = 5/12.
Since 1 − cos²θ = sin²θ and cosec²θ = 1/sin²θ, the product is sin²θ × 1/sin²θ = 1. This is a direct application of the Pythagorean identity.
The distance from the wall is the side adjacent to the 60° angle and the ladder is the hypotenuse. So cos 60° = 3/L, giving L = 3/(1/2) = 6 m.
The curved surface area of a cylinder is 2πrh, which equals (circumference) × height. Hence the area is 44 × 10 = 440 cm².
The total surface area of a solid hemisphere is 3πr², covering the curved surface 2πr² and the flat circular base πr². With r = 6 this is 3 × (22/7) × 36 = 2376/7 × 2 = 4752/7 cm².
Class size is the gap between consecutive class marks: 25 − 15 = 10. The class limits lie half a class size (5) on either side of the mark, so the class with mark 35 is 30–40, whose lower limit is 30.
The empirical relation is Mode = 3 × Median − 2 × Mean. Substituting gives 3 × 22 − 2 × 20 = 66 − 40 = 26.
The prime outcomes on a die are 2, 3 and 5, giving 3 favourable results out of 6 equally likely outcomes. Hence the probability is 3/6 = 1/2.
The sample space is {HH, HT, TH, TT} with 4 equally likely outcomes. At least one head occurs in HH, HT and TH, so the probability is 3/4.
A tangent is perpendicular to the radius at the point of contact, so ∠OAP = ∠OBP = 90°. In quadrilateral OAPB the angles sum to 360°, giving ∠AOB = 360° − 90° − 90° − 70° = 110°.
The perpendicular from the centre bisects the chord, so half the chord is 8 cm. Applying Pythagoras in the right triangle formed with the radius: distance = √(10² − 8²) = √36 = 6 cm.
The area of a sector equals (1/2) × arc length × radius. Hence the area is (1/2) × 22 × 21 = 231 cm².
The volume of the cone is (1/3)πr²h = (1/3)π(9)(4) = 12π cm³. For a cylinder of the same radius, 12π = π(9)H, so H = 12/9 = 4/3 cm.
Equal roots require the discriminant b² − 4ac to be zero. Here b = −4, a = k, c = 1, so 16 − 4k = 0, giving k = 4.
Let the height be h and the nearer distance x. Then h = x tan 60° = x√3 and h = (x + 9) tan 30° = (x + 9)/√3. Equating gives 3x = x + 9, so x = 4.5 and h = 4.5√3 = (9√3)/2 m.
The recorded total is 20 × 18 = 360. Replacing the wrong value 52 by the correct 25 reduces the total by 27, giving 333. The correct mean is 333/20 = 16.65.