⚡ Quick quiz

Test yourself in under a minute: Class 10 Math Practice Test

Expect the discriminant, similar-triangle area ratios, and standard trig angles here, the three ideas that decide most of the marks in the board's second half.

1. 1. A student says the square root of 4 is irrational because 4 is not a perfect cube. What is wrong with this reasoning?

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   A) 4 is a perfect cube too, so the student is rightB) The student's reasoning is completely correctC) Square roots of even numbers are always irrationalD) 4 is a perfect square, so √4 = 2 is rational

2. 2. A quadratic equation has two equal real roots. What does that tell you about where its graph meets the x-axis?

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   A) It touches the x-axis at exactly one pointB) It crosses the x-axis at two separate pointsC) It never touches the x-axis at allD) It touches the x-axis at every point

3. 3. In a finite AP, if you know only the first term, the last term, and how many terms there are, can you find the sum without knowing the common difference? Explain.

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4. 4. Two triangles have exactly the same three angles but different side lengths. Are they congruent, similar, both, or neither?

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   A) Both similar and congruentB) Congruent only, not similarC) Neither similar nor congruentD) Similar only, not congruent

5. 5. A tangent and a chord are both drawn from points on the same circle. Which one can pass through the circle's interior?

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6. 6. If sin(theta) = cos(theta) for an acute angle theta, what is theta, and how can you tell without a calculator?

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Real Numbers

Tests the Fundamental Theorem of Arithmetic for HCF/LCM, proving a number is irrational, and predicting whether a fraction's decimal expansion terminates.

1. 1. Find the HCF and LCM of 96 and 404 using the Fundamental Theorem of Arithmetic, and verify that HCF x LCM equals the product of the two numbers.

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   ▶Reveal answer

   HCF = 4, LCM = 9696🔊

   96 = 2⁵ x 3 and 404 = 2² x 101. The HCF takes the lowest power of each common prime factor (only 2 is common), giving 2² = 4. The LCM takes the highest power of every prime that appears in either number: 2⁵ x 3 x 101 = 9696. Checking the rule HCF x LCM = product of the numbers: 4 x 9696 = 38784, and 96 x 404 = 38784, so the two match.

2. 2. Show that 5 is an irrational number.

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   ▶Reveal answer

   5 is irrational (proof by contradiction)🔊

   Assume 5 is rational, so 5 = p/q where p and q are integers with no common factor and q is not 0. Squaring gives p² = 5q², so 5 divides p², which means 5 must divide p itself. Writing p = 5c and substituting back gives 25c² = 5q², so q² = 5c², meaning 5 divides q too. But that contradicts the assumption that p and q share no common factor, so 5 cannot be rational.

3. 3. Without actual division, state whether 129 / (2² x 5⁷ x 7⁵) has a terminating or a non-terminating repeating decimal expansion, and explain why.

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   Non-terminating, repeating🔊

   A fraction in lowest terms has a terminating decimal expansion only when its denominator's prime factorisation contains no primes other than 2 and 5. Here the denominator includes 7⁵, and 7 is neither 2 nor 5, so the decimal expansion does not terminate, it repeats instead.

4. 4. Find the least number divisible by both 15 and 24, using prime factorisation.

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   ▶Reveal answer

   120🔊

   15 = 3 x 5 and 24 = 2³ x 3. The least common multiple needs the highest power of every prime involved: 2³ x 3 x 5 = 120. A common mistake is multiplying 15 x 24 directly, which gives the product, not the smallest shared multiple.

Polynomials & Pair of Linear Equations in Two Variables

Covers finding zeroes of a quadratic polynomial and verifying the zero-coefficient relationship, plus solving and classifying pairs of linear equations.

1. 1. Find the zeroes of the polynomial p(x) = x² - 3x - 4, and verify the relationship between the zeroes and the coefficients.

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   Zeroes are 4 and -1🔊

   Factorising x² - 3x - 4 gives (x - 4)(x + 1), so the zeroes are 4 and -1. Checking the relationship: sum of zeroes = 4 + (-1) = 3, which equals -(coefficient of x)/(coefficient of x²) = -(-3)/1 = 3. Product of zeroes = 4 x (-1) = -4, which equals the constant term over the leading coefficient, -4/1 = -4. Both match.

2. 2. If one zero of the quadratic polynomial 2x² + px + 4 is 2, find the value of p.

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   ▶Reveal answer

   p = -6🔊

   Since 2 is a zero, substituting x = 2 into 2x² + px + 4 must give 0: 2(4) + 2p + 4 = 0, so 8 + 2p + 4 = 0, giving 2p = -12 and p = -6. Substituting a known zero directly into the polynomial is faster here than trying to factorise first.

3. 3. What is the maximum number of zeroes a quadratic polynomial can have, and what does each zero represent on its graph?

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   At most 2 zeroes; each zero is an x-intercept of the parabola🔊

   A quadratic polynomial's degree is 2, and a polynomial of degree n has at most n zeroes, so a quadratic has at most 2. Graphically, each real zero is a point where the parabola crosses or touches the x-axis, since that is exactly where the polynomial's value is 0.

4. 4. Write a quadratic polynomial whose sum and product of zeroes are 3 and -10 respectively.

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   x² - 3x - 10🔊

   A quadratic polynomial can be built directly from its sum and product of zeroes using the form x² - (sum)x + (product). Substituting sum = 3 and product = -10 gives x² - 3x + (-10), which simplifies to x² - 3x - 10.

5. 5. Solve the pair of linear equations by the elimination method: 2x + 3y = 11 and 2x - 4y = -24.

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   x = -2, y = 5🔊

   Subtracting the second equation from the first eliminates x: (2x + 3y) - (2x - 4y) = 11 - (-24), giving 7y = 35, so y = 5. Substituting back into 2x + 3(5) = 11 gives 2x = -4, so x = -2. Elimination works cleanly here because both equations already share the same coefficient of x.

6. 6. Without solving, determine whether the pair of equations 3x + 2y = 5 and 6x + 4y = 10 has a unique solution, no solution, or infinitely many solutions.

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   Infinitely many solutions🔊

   Comparing the ratios of coefficients: a1/a2 = 3/6 = 1/2, b1/b2 = 2/4 = 1/2, and c1/c2 = 5/10 = 1/2. All three ratios are equal, which means the two equations represent the same line, so every point on that line is a solution. If only the first two ratios matched but not the third, the lines would be parallel with no solution instead.

7. 7. The sum of a two-digit number and the number obtained by reversing its digits is 66. The digits differ by 2. Find the number.

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   ▶Reveal answer

   42 (or its reverse, 24)🔊

   Let the digits be x and y, so the number is 10x + y and its reverse is 10y + x. Their sum is 11x + 11y = 66, which simplifies to x + y = 6. Since the digits differ by 2, x - y = 2. Solving these together gives x = 4 and y = 2, so the number is 42 (the reverse, 24, also satisfies both conditions).

Quadratic Equations

Tests solving quadratic equations by factorisation and the quadratic formula, using the discriminant to determine the nature of roots.

1. 1. Solve by factorisation: x² - 7x + 12 = 0.

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   x = 3 or x = 4🔊

   x² - 7x + 12 factorises as (x - 3)(x - 4), since -3 and -4 multiply to 12 and add to -7. Setting each factor to 0 gives x = 3 and x = 4.

2. 2. Find the discriminant of 2x² - 4x + 3 = 0 and state the nature of its roots.

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   Discriminant = -8; no real roots🔊

   The discriminant is b² - 4ac = (-4)^2 - 4(2)(3) = 16 - 24 = -8. Since the discriminant is negative, the equation has no real roots, the parabola never touches the x-axis at all. A positive discriminant would mean two distinct real roots, and exactly 0 would mean one repeated real root.

3. 3. Solve using the quadratic formula: x² + 2x - 3 = 0.

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   x = 1 or x = -3🔊

   Using x = [-b ± √(b² - 4ac)] / 2a with a=1, b=2, c=-3: the discriminant is 4 - (-12) = 16, and √(16) = 4. So x = (-2 + 4)/2 = 1 or x = (-2 - 4)/2 = -3.

4. 4. The product of two consecutive positive integers is 240. Find the integers.

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   15 and 16🔊

   Let the smaller integer be n, so the equation is n(n+1) = 240, which expands to n² + n - 240 = 0. This factorises as (n - 15)(n + 16) = 0, giving n = 15 or n = -16. Since the problem asks for positive integers, n = 15 is the answer, making the integers 15 and 16.

5. 5. For what value of k does the quadratic equation kx² - 6x + 9 = 0 have equal roots?

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   ▶Reveal answer

   k = 1🔊

   Equal roots happen exactly when the discriminant is 0. Here, discriminant = (-6)^2 - 4(k)(9) = 36 - 36k. Setting this to 0 gives 36k = 36, so k = 1.

Arithmetic Progressions

Covers finding a specific term, the number of terms, and the sum of an AP, including real-world word problems.

1. 1. Find the 15th term of the AP: 3, 7, 11, 15, ...

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   59🔊

   Using a_n = a + (n-1)d with a = 3, d = 4, and n = 15: a_15 = 3 + 14(4) = 3 + 56 = 59.

2. 2. How many terms are there in the AP 7, 13, 19, ..., 205?

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   34 terms🔊

   The common difference is 6. Using a_n = a + (n-1)d: 205 = 7 + (n-1)(6), so (n-1)(6) = 198, giving n - 1 = 33 and n = 34.

3. 3. Find the sum of the first 20 terms of the AP whose first term is 5 and common difference is 3.

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   670🔊

   Using S_n = (n/2)[2a + (n-1)d] with n = 20, a = 5, d = 3: S_20 = 10[10 + 19(3)] = 10[10 + 57] = 10(67) = 670.

4. 4. A contractor pays a worker Rs. 200 on the first day, increasing the payment by Rs. 25 each following day. Find the total payment for the first 10 days.

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   Rs. 3125🔊

   The daily payments form an AP with a = 200 and d = 25. Using S_n = (n/2)[2a + (n-1)d] with n = 10: S_10 = 5[400 + 9(25)] = 5[400 + 225] = 5(625) = 3125.

Triangles: Similarity & Coordinate Geometry

Tests the Basic Proportionality Theorem, similarity criteria, area ratios of similar triangles, and the distance and section formulas.

1. 1. In triangle ABC, DE is parallel to BC, with AD = 2 cm, DB = 3 cm, and AE = 4 cm. Find EC using the Basic Proportionality Theorem.

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   EC = 6 cm🔊

   The Basic Proportionality Theorem says that when a line is parallel to one side of a triangle, it divides the other two sides in the same ratio: AD/DB = AE/EC. Substituting the known values, 2/3 = 4/EC, so EC = (4 x 3)/2 = 6 cm.

2. 2. In triangles ABC and DEF, AB/DE = BC/EF = CA/FD. Which similarity criterion applies?

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   SSS similarity🔊

   When all three pairs of corresponding sides of two triangles are in the same ratio, the triangles are similar by the SSS criterion, no angle information is needed for this one. This is different from SSS congruence, which requires the sides to be equal, not just proportional.

3. 3. Two similar triangles have corresponding sides in the ratio 3:5. If the area of the smaller triangle is 27 cm², find the area of the larger triangle.

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   75 cm²🔊

   For similar triangles, the ratio of areas equals the square of the ratio of corresponding sides: (3/5)^2 = 9/25. Setting 27/A = 9/25 and cross-multiplying gives 9A = 675, so A = 75 cm². A common mistake is using the side ratio directly instead of squaring it.

4. 4. Find the distance between the points (3, -2) and (-1, 1) using the distance formula.

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   5 units🔊

   Using the distance formula √[(x2-x1)^2 + (y2-y1)^2] with (3,-2) and (-1,1): √[(3-(-1))^2 + (-2-1)^2] = √[16 + 9] = √[25] = 5.

5. 5. Find the coordinates of the point which divides the line segment joining (2, -3) and (8, 5) internally in the ratio 1:3.

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   (3.5, -1)🔊

   The section formula for a point dividing (x1,y1) and (x2,y2) in ratio m:n is x = (m*x2 + n*x1)/(m+n), y = (m*y2 + n*y1)/(m+n). With m=1, n=3: x = (1(8) + 3(2))/4 = 14/4 = 3.5, and y = (1(5) + 3(-3))/4 = -4/4 = -1.

6. 6. Find the value of k so that the point (4, k) lies on the straight line joining (2, 3) and (6, -1).

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   k = 1🔊

   If three points lie on the same straight line, the slope between any two pairs of them must be equal. The slope from (2,3) to (6,-1) is (-1-3)/(6-2) = -1. Setting the slope from (2,3) to (4,k) equal to -1: (k-3)/(4-2) = -1, so k - 3 = -2, giving k = 1.

Trigonometry & Its Applications

Covers standard-angle trigonometric ratios, the Pythagorean identity, and heights-and-distances problems using angles of elevation and depression.

1. 1. Evaluate: sin 60° cos 30° + sin 30° cos 60°.

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   1🔊

   Substituting standard values: sin 60 = √3/2, cos 30 = √3/2, sin 30 = 1/2, cos 60 = 1/2. So the expression becomes (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1. This matches the identity sin(A+B), since sin 60 cos 30 + cos 60 sin 30 = sin(60+30) = sin 90 = 1.

2. 2. If sin(θ) = 3/5 for an acute angle θ, find cos(θ) and tan(θ).

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   cos(θ) = 4/5, tan(θ) = 3/4🔊

   If sin(θ) = 3/5, picture a right triangle with opposite side 3 and hypotenuse 5. Using the Pythagorean identity sin² + cos² = 1: cos²(θ) = 1 - 9/25 = 16/25, so cos(θ) = 4/5 (positive since θ is acute). Then tan(θ) = sin/cos = (3/5)/(4/5) = 3/4.

3. 3. Simplify: (1 - cos² A) x cosec² A.

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   1🔊

   1 - cos²(A) equals sin²(A) by the Pythagorean identity. Multiplying by cosec²(A), which is 1/sin²(A), gives sin²(A) x 1/sin²(A) = 1. Recognising 1 - cos²(A) as sin²(A) immediately, rather than expanding further, is what makes this simplify quickly.

4. 4. From a point on the ground 30 m away from the base of a tower, the angle of elevation to the top of the tower is 30°. Find the height of the tower.

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   10 √3 m (approximately 17.3 m)🔊

   Using tan(30) = height/distance: tan(30) = 1/√3 = h/30, so h = 30/√3 = 30 √3/3 = 10 √3 m. Rationalising the denominator (multiplying top and bottom by √3) is the step students most often skip, leaving an unsimplified answer.

5. 5. From the top of a 60 m high lighthouse, the angle of depression to a boat is 45°. Find the distance of the boat from the base of the lighthouse.

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   60 m🔊

   The angle of depression from the lighthouse top equals the angle of elevation from the boat, both 45 degrees, by alternate angles. Using tan(45) = height/distance: 1 = 60/d, so d = 60 m.

Circles & Areas Related to Circles

Tests the tangent-radius perpendicularity rule, equal tangent lengths from an external point, and the area and arc length of a sector.

1. 1. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center O at a point Q, such that OQ = 13 cm. Find the length of PQ.

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   12 cm🔊

   A tangent is always perpendicular to the radius at the point of contact, so triangle OPQ is right-angled at P. By the Pythagorean theorem: PQ² = OQ² - OP² = 13² - 5² = 169 - 25 = 144, so PQ = 12 cm.

2. 2. Two tangents are drawn from an external point to a circle, with lengths (x + 3) cm and 11 cm. Find x.

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   x = 8🔊

   The two tangents drawn from the same external point to a circle are always equal in length. Setting the two given lengths equal: x + 3 = 11, so x = 8.

3. 3. Find the area of a sector of a circle with radius 21 cm and a central angle of 60°. (Use π = 22/7.)

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   231 cm²🔊

   A sector's area is (central angle/360) x π x r². Substituting: (60/360) x (22/7) x 21² = (1/6) x (22/7) x 441 = (1/6) x 1386 = 231 cm².

4. 4. Find the length of an arc of a circle with radius 14 cm subtending a central angle of 90° at the center. (Use π = 22/7.)

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   22 cm🔊

   Arc length is (central angle/360) x 2 x π x r. Substituting: (90/360) x 2 x (22/7) x 14 = (1/4) x 88 = 22 cm.

Mensuration, Statistics & Probability

Covers the slant height of a cone in a combined solid, the mean of grouped data, identifying the median class, and simple probability.

1. 1. A solid is a cone mounted on a hemisphere of the same radius. If the radius is 7 cm and the height of the cone is 24 cm, find the slant height of the cone.

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   25 cm🔊

   The slant height of the cone is the hypotenuse of a right triangle formed by the radius and the height: l = √(r² + h²) = √(49 + 576) = √(625) = 25 cm. Since the cone sits on a hemisphere of the same radius, only the cone's own height and radius matter for its slant height, the hemisphere doesn't change this calculation.

2. 2. Find the mean of this grouped data using the direct method: class 0-10 (frequency 3), 10-20 (frequency 7), 20-30 (frequency 5), using class midpoints 5, 15, 25.

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   Mean is approximately 16.33🔊

   Using the direct method, mean = (sum of frequency x midpoint) / (sum of frequencies). Here that's [(3x5) + (7x15) + (5x25)] / (3+7+5) = [15 + 105 + 125] / 15 = 245/15, which is approximately 16.33.

3. 3. In a grouped frequency distribution with 40 total observations, what cumulative frequency value identifies the median class?

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   20🔊

   The median class is identified by finding n/2 and locating the first class whose cumulative frequency reaches or exceeds that value. With n = 40, n/2 = 20, so the median class is the one where the running cumulative frequency first hits 20 or more.

4. 4. A die is thrown once. Find the probability of getting a prime number.

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   1/2🔊

   A standard die has faces 1 through 6, and the prime numbers among them are 2, 3, and 5, giving 3 favourable outcomes out of 6 total. Probability = 3/6 = 1/2.

5. 5. A bag contains 5 red balls and 7 black balls. A ball is drawn at random. Find the probability that it is not red.

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   ▶Reveal answer

   7/12🔊

   There are 12 balls in total, 5 red and 7 black. The event 'not red' means drawing a black ball, so the probability is 7/12. This can also be found as 1 minus the probability of red (1 - 5/12 = 7/12).