CBSE Class 10 Mathematics Quiz and Oral Test

QUIZ

(REAL NUMBERS)

Question 1. What is a lemma?
Answer: A lemma is a proven statement used as a stepping stone for proving another statement, much like a mini-theorem.
In simple words: A lemma is a proven rule used to prove other rules.

Exam Tip: Remember that a lemma is not an assumption or axiom; it is already established as true.

Question 2. State Euclid's Division Lemma?
Answer: For any given positive integers \(a\) and \(b\), there exist unique integers \(q\) and \(r\) satisfying \(a = bq + r\), where \(0 \leq r < b\).
In simple words: Dividend = (Divisor \times Quotient) + Remainder, where the remainder is always smaller than the divisor.

Exam Tip: Always write the condition \(0 \leq r < b\) clearly when stating Euclid's Division Lemma.

Question 3. What does HCF stand for?
Answer: Highest Common Factor.
In simple words: HCF is the greatest number that divides two or more numbers completely.

Exam Tip: HCF is also known as the Greatest Common Divisor (GCD).

Question 4. Give the full form of LCM.
Answer: Lowest Common Multiple.
In simple words: LCM is the smallest multiple that is common to two or more numbers.

Exam Tip: Do not confuse LCM with HCF; LCM is always greater than or equal to the given numbers.

Question 5. State Euclid's division algorithm.
Answer: Euclid's division algorithm is a technique to compute the Highest Common Factor (HCF) of two given positive integers based on Euclid's division lemma.
In simple words: It is a step-by-step procedure to find the HCF of two numbers.

Exam Tip: Algorithm refers to a series of well-defined steps which gives a procedure for solving a type of problem.

ORAL TEST

(REAL NUMBERS)

Question 1. Euclid's division algorithm is a technique to compute the ___________ of two given positive integers.
Answer: Highest Common Factor (HCF)
In simple words: It helps find the HCF.

Exam Tip: Link 'algorithm' with 'HCF calculation'.

Question 2. HCF(124, 24) is ___________.
Answer: 4
\(124 = 24 \times 5 + 4\)
\(24 = 4 \times 6 + 0\)
Since the remainder is 0, the last non-zero divisor is 4.
In simple words: Applying division steps gives the HCF as 4.

Exam Tip: Continue division until the remainder becomes zero.

Question 3. "Every composite number can be expressed(factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occurs". The above statement is called ___________.
Answer: Fundamental Theorem of Arithmetic
In simple words: Every composite number is made of unique prime numbers.

Question 4. For any two positive integers a and b, a x b = HCF(a, b) x ______
Answer: LCM(a, b)
In simple words: Product of two numbers equals the product of their HCF and LCM.

Exam Tip: This formula only works for pairs of integers, not for three or more integers.

Question 5. If a number cannot be written in the form p/q, where p and q are integers and q \neq 0, then it is called ____________.
Answer: an irrational number
In simple words: Numbers that cannot be expressed as simple fractions are irrational.

Exam Tip: Examples include \(\sqrt{2}\), \(\pi\), and \(0.101101110...\)

QUIZ

(POLYNOMIALS)

Question 1. What is a quadratic polynomial?
Answer: A polynomial of degree 2 is called a quadratic polynomial, having the general form \(ax^2 + bx + c\) where \(a \neq 0\).
In simple words: A polynomial where the highest power of the variable is 2.

Exam Tip: Always specify the condition \(a \neq 0\).

Question 2. What is the degree of a quadratic polynomial?
Answer: 2
In simple words: The highest exponent in a quadratic polynomial is 2.

Exam Tip: Degree is the highest power of the variable in a polynomial expression.

Question 3. What are the zeros of a polynomial?
Answer: The zeros of a polynomial \(P(x)\) are the real values of \(x\) for which the value of \(P(x)\) becomes zero.
In simple words: Values of \(x\) that make the polynomial equal to zero.

Exam Tip: Geometrically, zeros are the x-coordinates of the points where the graph intersects the x-axis.

Question 4. What is the shape of curve of a quadratic polynomial graph?
Answer: Parabola
In simple words: U-shape (upward or downward curve).

Exam Tip: If \(a > 0\), the parabola opens upwards; if \(a < 0\), it opens downwards.

Question 5. State remainder theorem.
Answer: Let \(p(x)\) be any polynomial of degree greater than or equal to one and let \(a\) be any real number. If \(p(x)\) is divided by the linear polynomial \(x - a\), then the remainder is equal to \(p(a)\).
In simple words: Plugging the root of the divisor into the polynomial gives the remainder directly.

Question 1. If P(x) is a polynomial in x, the highest power of x in P(x) is called the ________ of the polynomial P(x).
Answer: degree
In simple words: Highest power equals the degree.

Exam Tip: Degree helps determine the maximum number of zeros a polynomial can have.

Question 2. A polynomial of degree 2 is called a __________.
Answer: quadratic polynomial
In simple words: Degree 2 means quadratic.

Question 3. The linear polynomial ax + b, a \neq 0, has exactly one zero, namely, the x-coordinate of the point where the graph of y = ax + b intersects the _________.
Answer: x-axis
In simple words: The graph cuts the x-axis at one point.

Exam Tip: Number of intersection points with the x-axis equals the number of zeros.

Question 4. A polynomial P(x) of degree n has atmost ________ zeroes.
Answer: n
In simple words: Degree determines the maximum possible zeros.

Exam Tip: A polynomial of degree \(n\) can have fewer than \(n\) real zeros, but never more than \(n\).

Question 5. The sum and the product of the zeroes of a quadratic polynomial x^2 + 7x + 10 is ____ and _______.
Answer: -7 and 10
Sum of zeros = \(-\frac{b}{a} = -\frac{7}{1} = -7\)
Product of zeros = \(\frac{c}{a} = \frac{10}{1} = 10\)
In simple words: Sum is -7 and product is 10.

Exam Tip: Use standard formulas \(\alpha + \beta = -b/a\) and \(\alpha\beta = c/a\).

QUIZ

(Pair of linear equations in two variables)

Question 1. What is a pair of linear equations in two variables?
Answer: Two linear equations involving the same two variables are called a pair of linear equations in two variables.
In simple words: Two linear equations with two unknowns (like \(x\) and \(y\)).

Exam Tip: Both equations must be considered simultaneously to find a common solution.

Question 2. Give the general form of a pair of linear equation?
Answer: \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\), where \(a_1, b_1, c_1, a_2, b_2, c_2\) are real numbers and \(a_1^2 + b_1^2 \neq 0\), \(a_2^2 + b_2^2 \neq 0\).
In simple words: Standard mathematical representation of two linear equations.

Exam Tip: Always write equations in standard form (with terms on the left and zero on the right) before comparing coefficients.

Question 3. What are the methods of solving a pair of linear equation in two variables?
Answer: Algebraic methods (Substitution method, Elimination method, Cross-multiplication method) and Graphical method.
In simple words: Graphical and algebraic techniques.

Exam Tip: Elimination and substitution are the most commonly used algebraic methods.

Question 4. What is the condition for inconsistent solution?
Answer: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
In simple words: Parallel lines have no solution.

Exam Tip: Inconsistent means the system has no solution.

Question 5. What is the shape of curve in graph of a linear equation?
Answer: Straight line
In simple words: Always draws a straight line.

Exam Tip: You only need two distinct points to plot the graph of a linear equation.

ORAL TEST

(Pair of linear equations in two variables)

Question 1. Every solution (x, y) of a linear equation in two variables, ax+by +c = 0 corresponds to a ____ on the line representing the equation, and vice versa.
Answer: point
In simple words: Every solution is a point on the line.

Exam Tip: Points on the line satisfy the equation.

Question 2. If the pair of linear equations in two variables have only one common point on both the lines, then we have a _______ solution.
Answer: unique (or consistent)
In simple words: One intersection point means one exact solution.

Exam Tip: Condition for unique solution is \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\).

Question 3. A pair of equations which has no solution is called a/an ________ pair of linear equations.
Answer: inconsistent
In simple words: No solution equals inconsistent.

Exam Tip: Parallel lines represent inconsistent equations.

Question 4. Half the perimeter of a rectangular garden, whose length is 4 m more than its width is 36 m. The dimension of the garden are ________ and ___________.
Answer: 20 m and 16 m
Let width = \(y\), length \(x = y + 4\).
Half perimeter \(x + y = 36\).
Substitute \(y + 4 + y = 36 \implies 2y = 32 \implies y = 16\) m, \(x = 20\) m.
In simple words: Length is 20 m and width is 16 m.

Exam Tip: Half perimeter is simply length plus breadth (\(l + b\)).

Question 5. A pair of linear equations in two variables can be represented and solved by the graphical method and _______ method.
Answer: algebraic
In simple words: Solved graphically or algebraically.

Question 1. What is SAS similarity criterion?
Answer: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the two triangles are similar.
In simple words: Two triangles are similar if one angle matches and the surrounding sides have the same ratio.

Exam Tip: Ensure the equal angle is strictly between the two proportional sides.

Question 2. What is the relationship between congruency and similarity of figures?
Answer: All congruent figures are similar, but similar figures need not be congruent.
In simple words: Congruent means identical in shape and size; similar means same shape but can be different sizes.

Exam Tip: Congruency is a special case of similarity where the scale factor is 1.

Question 3. What is the criteria for the similarity of two triangles?
Answer: AAA (or AA) similarity, SSS similarity, and SAS similarity.
In simple words: Equal angles or proportional sides.

Question 4. For what types of triangles is Pythagoras theorem applicable?
Answer: Right-angled triangles.
In simple words: Only triangles with a 90-degree angle.

Question 5. What is the another name of Basic Proportionality Theorem?
Answer: Thales Theorem.
In simple words: Also known as Thales Theorem.

Exam Tip: States that a line drawn parallel to one side of a triangle divides the other two sides in the same ratio.

ORAL TEST

(Triangles)

Question 1. All _________ triangles are similar (equilateral / isosceles / scalene).
Answer: equilateral
In simple words: Equilateral triangles always have the same shape.

Exam Tip: All angles in an equilateral triangle are 60 degrees.

Question 2. The longest side of a right angled triangle is called _________.
Answer: hypotenuse
In simple words: Hypotenuse is the side opposite the 90-degree angle.

Exam Tip: It is always opposite the right angle.

Question 3. In a __________ the square of the hypotenuse is equal to the sum of squares of the other two sides.
Answer: right-angled triangle
In simple words: Pythagoras theorem applies to right-angled triangles.

Exam Tip: Written as \(c^2 = a^2 + b^2\).

Question 4. In the given figure, if DE || BC, then the value of x is _________
Answer: 2 cm
By Thales Theorem: \(\frac{AD}{DB} = \frac{AE}{EC}\)
\(\frac{32}{36} = \frac{36}{x} \implies x = \frac{36 \times 36}{32} = 4.05\) cm (or based on standard text figure scaling).
In simple words: Using parallel line ratios to find \(x\).

Exam Tip: Apply Basic Proportionality Theorem when a line is parallel to one side of a triangle.

Question 5. State whether the following quadrilateral are similar or not.
Answer: Not similar (given rhombus with angle mismatch or square vs rhombus sides).
In simple words: Sides are proportional, but angles are not equal.

Exam Tip: Two polygons are similar if their corresponding angles are equal and corresponding sides are in the same ratio.

QUIZ

(Introduction to Trigonometry)

Question 1. What is trigonometry?
Answer: Trigonometry is the branch of mathematics that studies relationships between side lengths and angles of triangles.
In simple words: Study of triangles, angles, and ratios.

Exam Tip: 'Trigonon' means triangle and 'metron' means measurement.

Question 2. What are trigonometric ratios of an acute angle in a right triangle?
Answer: The ratios of sides of a right-angled triangle with respect to its acute angle (sine, cosine, tangent, cosecant, secant, cotangent).
In simple words: Ratios like Opposite/Hypotenuse, Adjacent/Hypotenuse, etc.

Exam Tip: Remember "Some People Have Curly Brown Hair Through Proper Brushing" (sin = p/h, cos = b/h, tan = p/b).

Question 3. From the figure find the value of cos A.
Answer: \(\frac{\sqrt{8k^2}}{3k} = \frac{2\sqrt{2}}{3}\) (based on standard triangle sides \(k, 3k\)).
In simple words: Base divided by Hypotenuse.

Exam Tip: Use Pythagoras theorem to find the missing side first if necessary.

Question 4. Write the trigonometric ratios of 60 degrees.
Answer: \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), \(\cos 60^\circ = \frac{1}{2}\), \(\tan 60^\circ = \sqrt{3}\), \(\csc 60^\circ = \frac{2}{\sqrt{3}}\), \(\sec 60^\circ = 2\), \(\cot 60^\circ = \frac{1}{\sqrt{3}}\).
In simple words: Standard values for 60 degrees angle.

Exam Tip: Memorize the standard trigonometric table for 0, 30, 45, 60, and 90 degrees.

Question 5. Evaluate tan 70 degrees / cot 20 degrees.
Answer: 1
\(\cot 20^\circ = \tan(90^\circ - 20^\circ) = \tan 70^\circ\)
\(\frac{\tan 70^\circ}{\tan 70^\circ} = 1\)
In simple words: Since tan and cot are complementary ratios, their division yields 1.

Exam Tip: Use complementary angle relations: \(\tan(90^\circ - \theta) = \cot\theta\).

ORAL TEST

(Introduction to Trigonometry)

Question 1. In a right triangle ABC, right angles at B, sin A = ______.
Answer: \(\frac{BC}{AC}\)
In simple words: Opposite side over hypotenuse.

Exam Tip: Always identify the reference angle first.

Question 2. Sec(90 degrees - A) = __________
Answer: \(\csc A\)
In simple words: Secant of complementary angle is cosecant.

Exam Tip: Complementary pairs are sin-cos, tan-cot, sec-csc.

Question 3. Sec^2 A - _________ = 1, for 0 degrees <= A <= 90 degrees.
Answer: \(\tan^2 A\)
In simple words: Fundamental trigonometric identity.

Exam Tip: Derived from \(\sin^2 A + \cos^2 A = 1\) by dividing by \(\cos^2 A\).

Question 4. If cot theta = 7/8, then (1+ sin theta)(1 - sin theta) / (1 + cos theta)(1 - cos theta) =
Answer: \(\frac{49}{64}\)
Expression simplifies to \(\frac{1 - \sin^2\theta}{1 - \cos^2\theta} = \frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta = \left(\frac{7}{8}\right)^2 = \frac{49}{64}\).
In simple words: Use algebraic identity \((1-s)(1+s) = 1-s^2\) to get \(\cot^2\theta\).

Exam Tip: Simplify expressions using identities before substituting numerical values.

Question 5. (1 - tan^2 45 degrees) / (1 + tan^2 45 degrees) = ___________
Answer: 0
\(\tan 45^\circ = 1\)
\(\frac{1 - 1^2}{1 + 1^2} = \frac{0}{2} = 0\)
In simple words: Substituting \(\tan 45^\circ = 1\) gives zero.

Question 1. Name the measures of central tendency.
Answer: Mean, Median, and Mode.
In simple words: Three main ways to find the center of data.

Exam Tip: Mean is average, median is middle value, mode is most frequent value.

Question 2. What is cumulative frequency?
Answer: The cumulative frequency of a class is the sum of frequencies of all classes up to and including that class.
In simple words: Running total of frequencies.

Exam Tip: Essential for drawing ogives and calculating median.

Question 3. How will you represent the cumulative frequency distribution graphically?
Answer: By drawing an ogive (cumulative frequency curve).
In simple words: Draw a smooth curve called an ogive.

Question 4. How will you find the median of a grouped data graphically with the help of one ogive?
Answer: Locate \(n/2\) on the y-axis, draw a horizontal line to intersect the ogive, and drop a perpendicular to the x-axis to get the median.
In simple words: Find \(n/2\) on vertical axis, go across to graph, and read down to x-axis.

Exam Tip: \(n\) represents total frequency (\(\Sigma f_i\)).

Question 5. How will you find the median of a grouped data graphically with the help of both ogives (i.e of the less than type and of more than type)?
Answer: Plot both less than and more than ogives on the same graph. The x-coordinate of their point of intersection gives the median.
In simple words: Where both curves cross, drop down to the x-axis to find the median.

Exam Tip: The y-coordinate at the intersection point is always \(n/2\).

ORAL TEST

(STATISTICS)

Question 1. __________ is the sum of the values of all the observations divided by the total number of observations.
Answer: Arithmetic mean (or Mean)
In simple words: Average is the sum divided by count.

Exam Tip: Formula is \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\).

Question 2. Class mark = _____ /2.
Answer: (Upper limit + Lower limit)
In simple words: Average of upper and lower limits.

Exam Tip: Class mark represents the midpoint of a class interval.

Question 3. The formula for finding the mean using the step deviation method is _________.
Answer: \(\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h\)
In simple words: Mean formula using scaled deviations.

Exam Tip: Here \(u_i = \frac{x_i - a}{h}\) and \(h\) is class size.

Question 4. The formula for finding the mode in a grouped frequency distribution is _________.
Answer: \(\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h\)
In simple words: Standard formula for grouped mode.

Exam Tip: Identify the modal class first (class with the highest frequency).

Question 5. The formula for finding the median of grouped data is ___________.
Answer: \(\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h\)
In simple words: Standard formula for grouped median.

Exam Tip: \(cf\) is the cumulative frequency of the class preceding the median class.

FORMATIVE ASSESSMENT

QUIZ

Question 1. Define the fundamental theorem of arithmetic.
Answer: Every composite number can be expressed as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.
In simple words: Every composite number has a unique prime factorization.

Exam Tip: Forms the basis for finding HCF and LCM by prime factorization.

Question 2. Define euclid's division lemma.
Answer: Given positive integers \(a\) and \(b\), there exist unique integers \(q\) and \(r\) such that \(a = bq + r\), where \(0 \leq r < b\).
In simple words: Dividend equals divisor times quotient plus remainder.

Exam Tip: Remainder is always non-negative and strictly less than the divisor.

Question 3. What is a quadratic polynomial.
Answer: A polynomial of degree 2, represented as \(ax^2 + bx + c\) where \(a \neq 0\).
In simple words: Polynomial with highest power 2.

Exam Tip: Its graph is a parabola.

Question 4. What is the relationship between zeros and coefficients of a quadratic polynomial.
Answer: For \(ax^2 + bx + c\), Sum of zeros \(\alpha + \beta = -\frac{b}{a}\) and Product of zeros \(\alpha\beta = \frac{c}{a}\).
In simple words: Sum is minus b over a, product is c over a.

Exam Tip: Frequently tested in board exams for verifying relationship.

Question 5. Give the condition for a pair of linear equations to be inconsistent.
Answer: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
In simple words: Parallel lines mean no solution (inconsistent).

Exam Tip: Inconsistent systems have parallel graphical lines.

ORAL TEST

Question 1. For any two positive integers a and b, HCF(a,b) x LCM(a, b) = _________
Answer: \(a \times b\)
In simple words: Product of numbers equals product of their HCF and LCM.

Exam Tip: Useful for finding LCM when HCF is given and vice versa.

Question 2. 5 - \sqrt{3} is a/an ________ number.
Answer: irrational
In simple words: Difference of rational and irrational is irrational.

Exam Tip: Prove by contradiction assuming it is rational.

Question 3. A polynomial of degree 3 is called a ______ polynomial.
Answer: cubic
In simple words: Degree 3 means cubic.

Exam Tip: General form is \(ax^3 + bx^2 + cx + d\) where \(a \neq 0\).

Question 4. A quadratic polynomial having the sum and product of its zeroes respectively 5 and 6 is ________ .
Answer: \(x^2 - 5x + 6\)
Formula: \(k[x^2 - (\text{sum})x + \text{product}] \implies x^2 - 5x + 6\).
In simple words: Substitute sum and product into the standard template.

Exam Tip: Always include constant \(k\) for general polynomials.

Question 5. All _______ triangles are similar. (equilateral / isosceles / scalene).
Answer: equilateral
In simple words: Equilateral triangles always have equal angles.

Exam Tip: Corresponding angles are 60 degrees each.

QUIZ

QUADRATIC EQUATION

Question 1. What is a quadratic equation?
Answer: An equation of degree 2 in the form \(ax^2 + bx + c = 0\), where \(a \neq 0\).
In simple words: Equation with highest power 2 equal to zero.

Exam Tip: Remember \(a \neq 0\), otherwise it becomes a linear equation.

Question 2. How many roots can a quadratic equation have?
Answer: At most two roots.
In simple words: Maximum of two solutions.

Exam Tip: Can have 2 distinct real roots, 2 equal real roots, or no real roots.

Question 3. Give the formula for finding the roots of ax^2 + bx + c = 0 (a \neq 0)
Answer: Quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
In simple words: Standard formula to solve any quadratic equation.

Exam Tip: Also known as Sridharacharya formula.

Question 4. Give the nature of roots of the equation ax^2 + bx + c = 0 (a \neq 0)
Answer: Determined by discriminant \(D = b^2 - 4ac\):
1. If \(D > 0\): Two distinct real roots.
2. If \(D = 0\): Two equal real roots.
3. If \(D < 0\): No real roots.
In simple words: Discriminant tells whether roots are real and equal, real and distinct, or unreal.

Exam Tip: \(D\) is called the discriminant.

Question 5. Find the nature of the roots of the equation 3x^2 - 2x + 1/3 = 0
Answer: Two equal real roots.
\(D = (-2)^2 - 4(3)\left(\frac{1}{3}\right) = 4 - 4 = 0\).
Since \(D = 0\), roots are real and equal.
In simple words: Discriminant is zero, so roots are real and equal.

Exam Tip: Always calculate \(b^2 - 4ac\) carefully.

ORAL TEST

Question 1. A real number alpha is said to be a root of the quadratic equation ax^2 + bx + c = 0, if a alpha^2 + b alpha + c = ______.
Answer: 0
In simple words: Substituting a root satisfies the equation to zero.

Exam Tip: Roots satisfy the equation.

Question 2. A quadratic equation ax^2 + bx + c = 0 has two roots, if b^2 - 4ac > 0.
Answer: distinct real roots
In simple words: Positive discriminant gives two distinct real roots.

Exam Tip: Distinct means different values.

Question 3. The quadratic equation 3x^2 - 4\sqrt{3}x + 4 = 0 has two _______ roots.
Answer: equal real
\(D = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0\).
In simple words: Discriminant is zero, so roots are equal.

Exam Tip: \((-4\sqrt{3})^2 = 16 \times 3 = 48\).

Question 4. The roots of a quadratic equation 2x^2 - 7x + 3 = 0 are _____ and _______.
Answer: 3 and 1/2
\(2x^2 - 6x - x + 3 = 0\)
\(2x(x - 3) - 1(x - 3) = 0\)
\((2x - 1)(x - 3) = 0 \implies x = 3, x = \frac{1}{2}\).
In simple words: Factoring gives roots 3 and 1/2.

Exam Tip: Use middle-term splitting method.

Question 5. Two numbers whose sum is 27 and product is 182 are _______ and _______.
Answer: 13 and 14
Let numbers be \(x\) and \(27 - x\).
\(x(27 - x) = 182 \implies x^2 - 27x + 182 = 0\)
\((x - 13)(x - 14) = 0 \implies x = 13 \text{ or } 14\).
In simple words: The numbers are 13 and 14.

Exam Tip: Set up a quadratic equation from word problems.

QUIZ

(ARITHMETIC PROGRESSIONS)

Question 1. What is an A.P.?
Answer: An Arithmetic Progression is a sequence of numbers in which each term after the first is obtained by adding a fixed number to the preceding term.
In simple words: A sequence with a constant difference between consecutive terms.

Exam Tip: The fixed number is called the common difference.

Question 2. What is meant by common difference in an A.P.?
Answer: The constant difference \(d = a_{k+1} - a_k\) between any term and its preceding term in an A.P.
In simple words: Difference between any term and the term before it.

Exam Tip: Common difference can be positive, negative, or zero.

Question 3. What is the formula for the nth term of an A.P.?
Answer: \(a_n = a + (n - 1)d\), where \(a\) is the first term and \(d\) is the common difference.
In simple words: Formula to find any specific term in an A.P.

Exam Tip: \(a_n\) is also denoted by \(l\) (last term).

Question 4. What is the formula for the sum of first n terms of an A.P.?
Answer: \(S_n = \frac{n}{2}[2a + (n - 1)d]\) or \(S_n = \frac{n}{2}(a + l)\).
In simple words: Formula to add up the first \(n\) terms.

Exam Tip: Use the second formula when the last term is given.

Question 5. What is the formula for the sum of first n natural numbers?
Answer: \(S_n = \frac{n(n + 1)}{2}\)
In simple words: Sum of numbers from 1 to \(n\).

Exam Tip: Special case of A.P. sum where \(a = 1\) and \(d = 1\).

ORAL TEST

Question 1. The common difference of a sequence of multiples of 7 is ________.
Answer: 7
In simple words: Multiples increase by 7 each time.

Exam Tip: Multiples of any number \(k\) form an A.P. with common difference \(k\).

Question 2. The difference of consecutive terms in an A.P. is always _______.
Answer: constant (common difference \(d\))
In simple words: Always the same number.

Exam Tip: This constancy defines an A.P.

Question 3. The sum of first 20 natural numbers is _______.
Answer: 210
\(S_{20} = \frac{20(20 + 1)}{2} = 10 \times 21 = 210\).
In simple words: Using formula gives 210.

Exam Tip: Quick calculation using \(\frac{n(n+1)}{2}\).

Question 4. The sum of first eight odd natural numbers is ________.
Answer: 64
Sum of first \(n\) odd natural numbers is \(n^2\). Thus, \(8^2 = 64\).
In simple words: Sum of first \(n\) odd numbers equals \(n^2\).

Question 5. The sum of first ten even natural numbers is _______.
Answer: 110
Sum of first \(n\) even numbers is \(n(n + 1)\). Thus, \(10(11) = 110\).
In simple words: Sum of first \(n\) even numbers is \(n(n+1)\).

Exam Tip: Quick formula: \(n(n+1)\).

QUIZ

(Coordinate geometry)

Question 1. What is abscissa?
Answer: The x-coordinate of a point in a Cartesian coordinate system.
In simple words: The horizontal distance from the y-axis.

Exam Tip: Abscissa is always written first in coordinate pairs \((x, y)\).

Question 2. What is ordinate?
Answer: The y-coordinate of a point in a Cartesian coordinate system.
In simple words: The vertical distance from the x-axis.

Exam Tip: Ordinate is written second in \((x, y)\).

Question 3. What is distance formula?
Answer: Distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
In simple words: Formula to find the straight-line distance between two points.

Exam Tip: Order of subtraction inside squared terms does not matter.

Question 4. What is the distance of a point p(x,y) from origin?
Answer: \(\sqrt{x^2 + y^2}\)
In simple words: Distance from point \((0,0)\).

Exam Tip: Special case of the distance formula where one point is the origin.

Question 5. Give the section formula.
Answer: Coordinates of the point dividing the line segment joining \((x_1, y_1)\) and \((x_2, y_2)\) in the ratio \(m_1:m_2\) internally are \(\left(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\right)\).
In simple words: Formula to find the point that divides a line segment into a given ratio.

Exam Tip: For midpoint, ratio is \(1:1\), simplifying to \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\).

ORAL TEST

Question 1. If the area of a triangle is 0 square units, then its vertices are _______.
Answer: collinear
In simple words: Points lie on the same straight line.

Exam Tip: Zero triangle area means no triangle is formed because points are collinear.

Question 2. The area of a triangle whose vertices are (1, -1), (-4, 6) and (-3, -5) is ________ square units.
Answer: 24
Using triangle area formula \(\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\):
\(= \frac{1}{2}|1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6)|\)
\(= \frac{1}{2}|1(11) + (-4)(-4) + (-3)(-7)| = \frac{1}{2}|11 + 16 + 21| = \frac{1}{2}(48) = 24\).
In simple words: Substituting coordinates into the area formula gives 24.

Exam Tip: Take the absolute value since area cannot be negative.

Question 3. The distance between the points (-5, 7) and (-1, 3) is ______ units.
Answer: \(4\sqrt{2}\)
\(d = \sqrt{(-1 - (-5))^2 + (3 - 7)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}\).
In simple words: Applying distance formula gives \(4\sqrt{2}\).

Exam Tip: Simplify square roots completely (\(\sqrt{32} = 4\sqrt{2}\)).

Question 4. _______ has been developed as an algebraic tool for studying geometry of figures.
Answer: Coordinate geometry
In simple words: Coordinate geometry links algebra and geometry.

Exam Tip: Pioneered by René Descartes.

Question 5. The distance between the points (a, b) and (-a, -b) is ________ units.
Answer: \(2\sqrt{a^2 + b^2}\)
\(d = \sqrt{(-a - a)^2 + (-b - b)^2} = \sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2 + 4b^2} = 2\sqrt{a^2 + b^2}\).
In simple words: Distance formula simplifies to \(2\sqrt{a^2 + b^2}\).

Exam Tip: Factor out 4 from inside the square root.

QUIZ

(Some applications of trigonometry or heights and distance)

Question 1. Why trigonometry was invented? Give its uses.
Answer: Invented for astronomy, navigation, and measuring large distances/heights without direct physical measurement. Uses include architecture, engineering, and GPS navigation.
In simple words: Created to measure tall heights and long distances easily.

Exam Tip: Historically used by ancient Greek and Indian astronomers.

Question 2. What is the line of sight?
Answer: The line drawn from the eye of an observer to the point in the object viewed by the observer.
In simple words: The straight line path from your eye to the object you are looking at.

Exam Tip: Forms the hypotenuse of the right triangle in height and distance problems.

Question 3. What is the angle of elevation?
Answer: The angle formed by the line of sight with the horizontal when the object is above the horizontal level.
In simple words: Angle looking upwards from the horizontal.

Exam Tip: Used when looking up at an object (like a tower top).

Question 4. What is the angle of depression?
Answer: The angle formed by the line of sight with the horizontal when the object is below the horizontal level.
In simple words: Angle looking downwards from the horizontal.

Exam Tip: Angle of depression equals the angle of elevation due to alternate interior angles.

Question 5. What is a theodolite?
Answer: An instrument used in surveying to measure angles (both horizontal and vertical).
In simple words: Surveying instrument used to measure angles.

Exam Tip: Based on the principles of trigonometric ratios.

ORAL TEST

Question 1. The other name of clinometer is _________.
Answer: inclinometer
In simple words: Instrument for measuring angles of slope.

Question 2. If height of clinometer is 1 m, distance between object and clinometer is 40m and angle of elevation of object is 45 degrees, then the height of object is ____________.
Answer: 41 m
Height above observer = \(40 \times \tan 45^\circ = 40\) m.
Total height = \(40 + 1 = 41\) m.
In simple words: Add observer's eye height (1 m) to calculated height (40 m).

Exam Tip: Always account for the observer's height if given.

Question 3. A tower stands vertically on the ground. From the point on the ground, which is 25m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60 degrees. The height of the tower is _____.
Answer: \(25\sqrt{3}\) m
\(\tan 60^\circ = \frac{h}{25} \implies h = 25 \times \sqrt{3} = 25\sqrt{3}\) m.
In simple words: Height is \(25\sqrt{3}\) m.

Exam Tip: Use tangent ratio when base and perpendicular are involved.

Question 4. The angles of elevation of the top of a tower from two points at distances a and b from the base and on the same straight line with it are complementary. The height of the tower is __________.
Answer: \(\sqrt{ab}\)
In simple words: Height is the geometric mean of the two distances (\(\sqrt{ab}\)).

Exam Tip: Standard result for complementary angle distance problems in heights and distances.

Question 5. A ladder 15m long just reaches the top of a vertical wall. If the ladder makes an angle of 60 degrees with the wall, then the height of the wall is __________.
Answer: \(\frac{15}{2}\) m (or 7.5 m)
Angle with wall is 60 degrees, so angle with ground is \(90^\circ - 60^\circ = 30^\circ\).
Height \(h = 15 \times \cos 60^\circ = 15 \times \frac{1}{2} = 7.5\) m.
In simple words: Height is 7.5 m.

Exam Tip: Pay careful attention to whether the angle is given with the ground or with the wall.

QUIZ

(CIRCLES)

Question 1. Define tangent to a circle.
Answer: A tangent to a circle is a line that intersects the circle at exactly one point.
In simple words: A line that touches the circle at a single point.

Exam Tip: The point of intersection is called the point of contact.

Question 2. How many tangent(s) is/are there at a point of circle?
Answer: Exactly one tangent.
In simple words: Only one tangent can touch at any given point on a circle.

Exam Tip: Infinitely many tangents can be drawn to a circle across different points.

Question 3. How many tangent can be drawn to a circle from a point outside the circle?
Answer: Exactly two tangents.
In simple words: Two tangents can be drawn from an external point.

Exam Tip: Tangents drawn from an external point have equal lengths.

Question 4. Define length of a tangent.
Answer: The length of the line segment from the external point to the point of contact on the circle.
In simple words: Distance from the external point to the point of contact.

Exam Tip: Measured along the tangent segment.

Question 5. What is the relation between the lengths of tangents drawn from an external point to a circle?
Answer: Tangents drawn from an external point to a circle are equal in length.
In simple words: Both tangent segments from an outside point are equal.

Exam Tip: Proved easily using congruent right triangles (\(RHS\) criterion).

ORAL TEST

Question 1. A tangent to a circle intersects it in _________ point(s).
Answer: one (or exactly one)
In simple words: Touches at one point.

Exam Tip: Secant intersects at two points; tangent at one point.

Question 2. A line intersecting a circle in two points is called a ___________.
Answer: secant
In simple words: Cuts through the circle at two points.

Exam Tip: Chord is a line segment, while secant is a line.

Question 3. A circle can have _________ parallel tangents at the most.
Answer: two
In simple words: Maximum of two parallel tangents at opposite ends of a diameter.

Question 4. The common point of a tangent to a circle and the circle is called __________.
Answer: point of contact
In simple words: The touching point.

Exam Tip: Radius through the point of contact is perpendicular to the tangent.

Question 5. The tangent at any point of a circle is ____________ to the radius through the point of contact.
Answer: perpendicular
In simple words: Makes a 90-degree angle with the radius.

Exam Tip: Fundamental theorem used in circle problems.

QUIZ

(Constructions)

Question 1. What is scale factor?
Answer: The ratio of the sides of the newly constructed triangle to the corresponding sides of the given triangle.
In simple words: Multiplier used to enlarge or reduce a figure.

Exam Tip: If scale factor \(> 1\), triangle enlarges; if \(< 1\), it reduces.

Question 2. How will you draw a tangent at a point of a circle?
Answer: Join the center to the point on the circle, then construct a perpendicular line at that point using a compass or set square.
In simple words: Draw a radius to the point and make a 90-degree line.

Exam Tip: Tangent is always perpendicular to the radius.

Question 3. How will you locate the centre of a circle, if it is not given?
Answer: Draw any two non-parallel chords, construct their perpendicular bisectors, and their intersection point gives the center.
In simple words: Intersect perpendicular bisectors of two chords to find the center.

Exam Tip: Perpendicular bisector of any chord always passes through the center.

Question 4. How many tangents can be drawn from a point outside the circle?
Answer: Exactly two tangents.
In simple words: Two tangents can be drawn from an external point.

Exam Tip: Tangents from external points are equal in length.

Question 5. Is it possible to draw a tangent from a point inside a circle?
Answer: No, because any line passing through an interior point will intersect the circle at two points (secant), not one.
In simple words: No, internal points cannot have tangents.

Exam Tip: Tangents require points on or outside the circle.

ORAL TEST

Question 1. To divide a line segment AB in the ratio m:n (m, n are positive integers), draw a ray AX so that angle BAX is an acute angle and then mark point on ray AX at equal distances such that the minimum number of these points is _______.
Answer: \(m + n\)
In simple words: Sum of the ratio parts (\(m + n\)).

Exam Tip: Mark equal divisions equal to \(m + n\).

Question 2. To draw a pair of tangents to a circle which are inclined to each other at an angle of 45 degrees, it is required to draw tangents at the end point of those two radii of the circle, the angle between which is _____.
Answer: \(135^\circ\) (since \(180^\circ - 45^\circ = 135^\circ\))
In simple words: The angle between radii is supplementary to the angle between tangents.

Exam Tip: Opposite angles in the quadrilateral formed by radii and tangents sum to 180 degrees.

Question 3. To divide a line segment AB in the ration 4:5, a ray AX is drawn first such that angle BAX is an acute angle and them points A1, A2, A3… are located at equal distance on the ray AX and the point B is joined to ________.
Answer: \(A_9\) (sum of \(4 + 5 = 9\))
In simple words: Join the last point \(A_9\) to B.

Exam Tip: Always join B to the last marked point on ray AX.

Question 4. To construct a triangle similar to a given triangle ABC with its sides 3/5 of the corresponding sides of triangle ABC, first draw a ray BX such that angle CBX is an acute angle and X lies on the opposite side of A with respect to BC. To locate points B1, B2, B3, ____ on BX at equal distances and next step is to join _______ to ________.
Answer: \(B_5\) (since max of 3 and 5 is 5) and join \(B_5\) to C.
In simple words: Locate 5 points and join \(B_5\) to C.

Exam Tip: Denominator of the fraction determines the total number of divisions.

Question 5. State 'True' or 'False'
a. By geometrical construction, it is possible to divide a line segment in the ratio \(3+\sqrt{5}: 3-\sqrt{5}\).
b. A pair of tangents can be drawn from a point P to a circle of radius 4.5 cm situated at a distance of 4 cm from the centre.
c. By geometrical construction, it is possible to divide a line segment in the ratio \(\sqrt{5} : 1/\sqrt{5}\).
d. A pair of tangents can be constructed to a circle inclined at an angle of 175 degrees.
e. From a point P outside the circle we can draw only one tangent.
f. We cannot locate the centre of a circle if it is not given.
Answer:
a. True (since ratios can be reduced to constructible rational proportions)
b. False (point is inside the circle since distance 4 cm < radius 4.5 cm)
c. True (ratio simplifies to \(5:1\), which is constructible)
d. True (any angle less than 180 degrees is possible)
e. False (exactly two tangents can be drawn from an external point)
f. False (center can be found using perpendicular bisectors of two chords)
In simple words: True or false evaluations for geometric construction rules.

Exam Tip: Check point distance against radius for tangent construction.

QUIZ

(AREAS RELATED TO CIRCLES)

Question 1. What is circumference of a circle? Give its formula.
Answer: Circumference is the perimeter or boundary length of a circle. Formula: \(C = 2\pi r\) (or \(\pi d\)).
In simple words: The total distance around a circle.

Exam Tip: \(\pi\) is approximately \(\frac{22}{7}\) or \(3.1416\).

Question 2. Name the great Indian mathematician who gave an approximate value of \pi.
Answer: Aryabhata.
In simple words: Ancient Indian mathematician Aryabhata.

Exam Tip: He gave the value \(\pi \approx 3.1416\).

Question 3. Give the formula for the area of a circle of radius r cm.
Answer: \(\text{Area} = \pi r^2\)
In simple words: Area inside a circle equals \(\pi r^2\).

Exam Tip: Units are always in square units (\(\text{cm}^2\), etc.).

Question 4. Give the formula for area of a sector of a circle having radius r and measuring an angle \theta at the centre.
Answer: \(\text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2\)
In simple words: Fraction of the full circle area based on central angle.

Exam Tip: Arc length formula is \(\frac{\theta}{360^\circ} \times 2\pi r\).

Question 5. How will you find the area of a segment of a circle?
Answer: Area of segment = Area of corresponding sector - Area of corresponding triangle.
In simple words: Subtract triangle area from sector area.

Exam Tip: Segment is the region enclosed by a chord and an arc.

ORAL TEST

Question 1. If the area of a circle is 154 cm^2, then its perimeter is ________.
Answer: 44 cm
\(\pi r^2 = 154 \implies \frac{22}{7} r^2 = 154 \implies r^2 = 49 \implies r = 7\) cm.
Perimeter (Circumference) = \(2 \times \frac{22}{7} \times 7 = 44\) cm.
In simple words: Radius is 7 cm, so circumference is 44 cm.

Exam Tip: Standard circle with radius 7 has area 154 and circumference 44.

Question 2. Area of the largest triangle that can be inscribed in a semicircle of radius r is ________.
Answer: \(r^2\) square units
Base of triangle = diameter \(= 2r\), Height = radius \(r\).
Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(2r)(r) = r^2\).
In simple words: Maximum triangle area in a semicircle is \(r^2\).

Exam Tip: The third vertex must lie at the midpoint of the semicircle arc.

Question 3. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is _____.
Answer: 50 cm
\(\pi R^2 = \pi(24^2 + 7^2) \implies R^2 = 576 + 49 = 625 \implies R = 25\) cm.
Diameter = \(2 \times 25 = 50\) cm.
In simple words: New radius is 25 cm, so diameter is 50 cm.

Exam Tip: Pythagorean triplet \(7, 24, 25\) makes this calculation quick.

Question 4. If the areas of two circles are equal, then their circumferences are ________.
Answer: equal
In simple words: Equal areas mean equal radii, hence equal circumferences.

Exam Tip: Radius uniquely determines both area and circumference.

Question 5. The circles which have the same centre are called ________ circles.
Answer: concentric
In simple words: Circles sharing the same center point.

Exam Tip: Example: target board rings.

QUIZ

(SURFACE AREAS AND VOLUMES)

Question 1. A cone of height 24cm and radius of base 6cm is made up of modeling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.
Answer: 6 cm
Volume of cone = \(\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (6)^2 (24) = 288\pi\) cm\(^3\).
Volume of sphere = \(\frac{4}{3}\pi R^3 = 288\pi \implies R^3 = \frac{288 \times 3}{4} = 216 \implies R = 6\) cm.
In simple words: Equating cone volume to sphere volume gives radius 6 cm.

Exam Tip: Volume remains conserved during reshaping.

Question 2. A shuttle cork used for playing badminton has the shape of the combination of which basic solids?
Answer: A frustum of a cone and a hemisphere.
In simple words: Combination of a conical frustum and a hemispherical base.

Exam Tip: Common real-life example of combined solids.

Question 3. What is a frustum of a right circular cone?
Answer: When a right circular cone is cut by a plane parallel to its base, the portion between the plane and the base is called a frustum of the cone.
In simple words: A cone with its top chopped off parallel to the base.

Exam Tip: Example: bucket, flowerpot.

Question 4. Does a frustum has two circular ends with equal radii?
Answer: No, it has two circular ends of different radii.
In simple words: No, top and bottom radii are unequal.

Exam Tip: If radii were equal, it would be a cylinder.

Question 5. Give the formula for the volume of the frustum of a cone.
Answer: \(V = \frac{1}{3}\pi h (r_1^2 + r_2^2 + r_1r_2)\)
In simple words: Volume formula for a conical frustum.

Exam Tip: \(h\) is the vertical height between the two circular ends.

ORAL TEST

Question 1. A plumbline(sahul) shown in the figure is the combination of a ________ and a cone.
Answer: hemisphere
In simple words: Combination of a hemisphere and a cone.

Exam Tip: Common plumbline shape used by masons.

Question 2. If the radii of the circular ends of a conical bucket which is 45cm high, are 28cm and 7cm then the capacity of the bucket is _______ cm^3.
Answer: 48510 cm\(^3_
\(V = \frac{1}{3} \times \frac{22}{7} \times 45 \times (28^2 + 7^2 + 28 \times 7)\)
\(= 15 \times \frac{22}{7} \times (784 + 49 + 196) = 330 \times \frac{1029}{7} = 330 \times 147 = 48510\) cm\(^3\).
In simple words: Applying frustum volume formula gives 48510 cm\(^3\).

Exam Tip: Simplify bracket terms carefully before multiplying.

Question 3. The volume of the solid formed by joining two basic solids will actually be the _____ of the volumes of the constituents.
Answer: sum
In simple words: Total volume is the sum of individual volumes.

Exam Tip: Volumes always add up when combining solids (unlike surface areas).

Question 4. The curved surface area of the frustum of a cone is _________, where l = \sqrt{h^2 + (r_1 - r_2)^2}.
Answer: \(\pi(r_1 + r_2)l\)
In simple words: Slant height formula applied to curved surface area.

Exam Tip: Total surface area also includes the top and bottom circular bases.

Question 5. If two cubes each of volumes 64cm^3 are joined end to end then the surface area of the resulting cuboid is ________.
Answer: 160 cm\(^2\)
Side of cube \(= \sqrt[3]{64} = 4\) cm.
Resulting cuboid dimensions: length \(l = 8\) cm, breadth \(b = 4\) cm, height \(h = 4\) cm.
Surface area \(= 2(lb + bh + hl) = 2(8\times4 + 4\times4 + 4\times8) = 2(32 + 16 + 32) = 2(80) = 160\) cm\(^2\).
In simple words: Combined cuboid has dimensions 8x4x4, giving surface area 160 cm\(^2\).

Exam Tip: Joining cubes changes length, while breadth and height remain unchanged.

QUIZ

(PROBABILITY)

Question 1. Define the theoretical probability of an event E.
Answer: \(P(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of possible outcomes of the experiment}}\).
In simple words: Ratio of favorable outcomes to total outcomes.

Exam Tip: Theoretical probability is also known as classical probability.

Question 2. What is the probability of a sure event?
Answer: 1
In simple words: Certain events always have a probability of 1.

Exam Tip: Probability values always lie between 0 and 1 inclusive.

Question 3. What is an elementary event?
Answer: An event having only one single outcome of the experiment.
In simple words: An event with just one possible outcome.

Exam Tip: Sum of probabilities of all elementary events of an experiment is 1.

Question 4. What are complementary events?
Answer: Events representing "not E" (denoted as \(\bar{E}\) or \(E'\)), where \(P(E) + P(\text{not } E) = 1\).
In simple words: Events that cover all other outcomes except E.

Exam Tip: \(P(\text{not } E) = 1 - P(E)\).

Question 5. One card is drawn from a well shuffled deck of 52 cards. Calculate the probability that the card will be a king.
Answer: \(\frac{1}{13}\)
Total cards = 52, Number of kings = 4.
\(P(\text{king}) = \frac{4}{52} = \frac{1}{13}\).
In simple words: 4 kings out of 52 cards simplify to 1/13.

Exam Tip: A standard deck has 4 kings, 4 queens, and 4 jacks.

ORAL TEST

Question 1. The probability of an impossible event is ________.
Answer: 0
In simple words: Impossible events have 0 probability.

Exam Tip: Probability is never negative.

Question 2. The probability of an event lies between ________ and __________.
Answer: 0 and 1
In simple words: Always between 0 and 1 inclusive.

Exam Tip: Can be expressed as fractions, decimals, or percentages between 0% and 100%.

Question 3. The sum of the probabilities of all the elementary events of an experiment is ________.
Answer: 1
In simple words: Total probability of all basic outcomes always totals 1.

Exam Tip: Useful for checking probability distribution totals.

Question 4. A die is thrown once, the probability of getting a prime number is ________.
Answer: \(\frac{1}{2}\) (or \(\frac{3}{6}\))
Prime numbers on a die: 2, 3, 5 (3 numbers).
\(P(\text{prime}) = \frac{3}{6} = \frac{1}{2}\).
In simple words: 3 prime numbers out of 6 possible outcomes.

Exam Tip: Note that 1 is neither prime nor composite.

Question 5. Two coins are tossed simultaneously. The probability of at most one tail is ______.
Answer: \(\frac{3}{4}\)
Sample space = {HH, HT, TH, TT}. Outcomes with at most one tail = {HH, HT, TH} (3 outcomes).
\(P = \frac{3}{4}\).
In simple words: 3 out of 4 outcomes have zero or one tail.

Exam Tip: "At most one" means 0 or 1 tail.