--- title: "CBSE Class 10 Maths Basic Question Paper 2024 (Set 430/1/3) with Solutions" url: https://www.swavid.com/cbse/class-10/maths/previous-year-papers/2024-basic-430-1-1 dateModified: 2026-10-07T15:16:30+00:00 --- # CBSE Class 10 Maths Basic Question Paper 2024 (Set 430/1/3) with Solutions Basic · Code 430/1/3 · 80 marks · 180 minutes Solved CBSE Class 10 Maths Basic board paper from 2024, set 430/1/3. Every question has step-by-step working written to the CBSE marking scheme, with the marks for each answer. Official question paper: https://www.cbse.gov.in/cbsenew/question-paper/2024/X/MATHEMATICS_BASIC.zip. Solutions written by SwaVid. Free PDF (22 pages): https://www.swavid.com/api/seo/pdf/papers/cbse/maths/swavid-cbse-class-10-maths-question-paper-2024-basic-430-1-1-24f0284850.pdf ## SECTION - A Q. No. 1 to 20 are Multiple Choice Questions of 1 mark each. ### Question 1 *1 mark · MCQ · Polynomials* For what value of $k$, the product of zeroes of the polynomial $kx^2 - 4x - 7$ is $2$ ? - $\frac{1}{14}$ - $-\frac{7}{2}$ - $\frac{7}{2}$ - $-\frac{2}{7}$ **Solution** 1. For a quadratic polynomial $ax^2 + bx + c$, the product of zeroes is given by $\frac{c}{a}$. 2. Here $a = k$, $b = -4$, and $c = -7$, so the product of zeroes is $\frac{-7}{k}$. 3. Equating this to $2$, we get $\frac{-7}{k} = 2$, which gives $k = -\frac{7}{2}$. **Answer:** (b) $-\frac{7}{2}$ > Common mistake: Confusing the formula for the sum of zeroes with the product of zeroes. ### Question 2 *1 mark · MCQ · Arithmetic Progressions* In an A.P., if $a = 8$ and $a_{10} = -19$, then value of $d$ is : - $3$ - $-\frac{11}{9}$ - $-\frac{27}{10}$ - $-3$ **Solution** 1. The formula for the $n$-th term of an A.P. is $a_n = a + (n - 1)d$. 2. Substitute $a = 8$ and $a_{10} = -19$ into the formula to get $-19 = 8 + (10 - 1)d$. 3. Solving for $d$, we get $9d = -19 - 8 = -27$, which gives $d = -3$. **Answer:** (d) $-3$ > Common mistake: Taking $n=10$ as $10d$ instead of $9d$. ### Question 3 *1 mark · MCQ · Coordinate Geometry* The mid-point of the line segment joining the points $(-1, 3)$ and $\left(8, \frac{3}{2}\right)$ is : - $\left(\frac{7}{2}, \frac{3}{4}\right)$ - $\left(\frac{7}{2}, \frac{9}{2}\right)$ - $\left(\frac{9}{2}, -\frac{3}{4}\right)$ - $\left(\frac{7}{2}, \frac{9}{4}\right)$ **Solution** 1. The mid-point formula is $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$. 2. Substitute the given points $(-1, 3)$ and $\left(8, \frac{3}{2}\right)$: $\left(\frac{-1 + 8}{2}, \frac{3 + \frac{3}{2}}{2}\right)$. 3. Simplifying gives $\left(\frac{7}{2}, \frac{9}{4}\right)$. **Answer:** (d) $\left(\frac{7}{2}, \frac{9}{4}\right)$ > Common mistake: Errors in adding fractions in the y-coordinate. ### Question 4 *1 mark · MCQ · Introduction to Trigonometry* If $\sin \theta = \frac{1}{3}$, then $\sec \theta$ is equal to : - $\frac{2\sqrt{2}}{3}$ - $\frac{3}{2\sqrt{2}}$ - $3$ - $\frac{1}{\sqrt{3}}$ **Solution** 1. We know that $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$. 2. Since $\sec \theta = \frac{1}{\cos \theta}$, we have $\sec \theta = \frac{3}{2\sqrt{2}}$. **Answer:** (b) $\frac{3}{2\sqrt{2}}$ > Common mistake: Inverting $\sin \theta$ directly to find $\sec \theta$. ### Question 5 *1 mark · MCQ · Real Numbers* $\text{HCF}(132, 77)$ is : - $11$ - $77$ - $22$ - $44$ **Solution** 1. Prime factorization of $132$ is $2^2 \times 3 \times 11$. 2. Prime factorization of $77$ is $7 \times 11$. 3. The common prime factor with the lowest power is $11$, so $\text{HCF}(132, 77) = 11$. **Answer:** (a) $11$ > Common mistake: Calculating LCM instead of HCF. ### Question 6 *1 mark · MCQ · Quadratic Equations* If the roots of quadratic equation $4x^2 - 5x + k = 0$ are real and equal, then value of $k$ is : - $\frac{5}{4}$ - $\frac{25}{16}$ - $-\frac{5}{4}$ - $-\frac{25}{16}$ **Solution** 1. For real and equal roots, the discriminant must be zero: $b^2 - 4ac = 0$. 2. Substitute $a = 4$, $b = -5$, and $c = k$ into the equation: $(-5)^2 - 4(4)(k) = 0$. 3. Solving for $k$, we get $25 - 16k = 0$, which gives $k = \frac{25}{16}$. **Answer:** (b) $\frac{25}{16}$ > Common mistake: Forgetting the negative sign while squaring $-5$ or missing the $4ac$ term. ### Question 7 *1 mark · MCQ · Probability* If probability of winning a game is $p$, then probability of losing the game is : - $1 + p$ - $-p$ - $p - 1$ - $1 - p$ **Solution** 1. Let $E$ be the event of winning the game and $\bar{E}$ be the event of losing the game. 2. We know that $P(E) + P(\bar{E}) = 1$, so $P(\bar{E}) = 1 - P(E) = 1 - p$. **Answer:** (d) $1 - p$ > Common mistake: Confusing the probability of the complement with $p - 1$ or $1 + p$. ### Question 8 *1 mark · MCQ · Coordinate Geometry* The distance between the points $(2, -3)$ and $(-2, 3)$ is : - $2\sqrt{13}\text{ units}$ - $5\text{ units}$ - $13\sqrt{2}\text{ units}$ - $10\text{ units}$ **Solution** 1. Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ for points $(2, -3)$ and $(-2, 3)$. 2. Distance $d = \sqrt{(-2 - 2)^2 + (3 - (-3))^2} = \sqrt{(-4)^2 + (6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}\text{ units}$. **Answer:** (a) $2\sqrt{13}\text{ units}$ > Common mistake: Sign errors while subtracting coordinates. ### Question 9 *1 mark · MCQ · Introduction to Trigonometry* For what value of $\theta$, $\sin^2 \theta + \sin \theta + \cos^2 \theta$ is equal to $2$ ? - $45^\circ$ - $0^\circ$ - $90^\circ$ - $30^\circ$ **Solution** 1. We are given the expression $\sin^2 \theta + \sin \theta + \cos^2 \theta = 2$. Using the identity $\sin^2 \theta + \cos^2 \theta = 1$, we get $1 + \sin \theta = 2$. 2. This gives $\sin \theta = 1$, which means $\theta = 90^\circ$. **Answer:** (c) $90^\circ$ > Common mistake: Forgetting to apply the Pythagorean trigonometric identity first. ### Question 10 *1 mark · MCQ · Probability* A card is drawn from a well shuffled deck of 52 playing cards. The probability that drawn card is a red queen, is : - $\frac{1}{13}$ - $\frac{2}{13}$ - $\frac{1}{52}$ - $\frac{1}{26}$ **Solution** 1. Total number of possible outcomes = 52. Number of red queens in a deck is 2 (Queen of hearts and Queen of diamonds). 2. Probability of drawing a red queen = $\frac{2}{52} = \frac{1}{26}$. **Answer:** (d) $\frac{1}{26}$ > Common mistake: Taking 4 queens instead of 2 red queens. ### Question 11 *1 mark · MCQ · Statistics* If a certain variable $x$ divides a statistical data arranged in order into two equal parts; then the value of $x$ is called the : - mean - median - mode - range **Solution** 1. By definition, the median is a measure of central tendency that divides the given statistical data arranged in ascending or descending order into two equal parts. **Answer:** (b) median > Common mistake: Confusing median with mean or mode. ### Question 12 *1 mark · MCQ · Surface Areas and Volumes* The radius of a sphere is $\frac{7}{2}\text{ cm}$. The volume of the sphere is : - $\frac{231}{3}\text{ cu cm}$ - $\frac{539}{12}\text{ cu cm}$ - $\frac{539}{3}\text{ cu cm}$ - $154\text{ cu cm}$ **Solution** 1. Given radius $r = \frac{7}{2}\text{ cm}$. The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$. 2. Substitute the values: $V = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^3 = \frac{4}{3} \times \frac{22}{7} \times \frac{343}{8} = \frac{539}{3}\text{ cu cm}$. **Answer:** (c) $\frac{539}{3}\text{ cu cm}$ > Common mistake: Calculation errors while cubing the radius fraction. ### Question 13 *1 mark · MCQ · Statistics* The mean and median of a statistical data are $21$ and $23$ respectively. The mode of the data is : - $27$ - $22$ - $17$ - $23$ **Solution** 1. We know the empirical relationship between mean, median and mode is: $\text{Mode} = 3(\text{Median}) - 2(\text{Mean})$. 2. Substitute the given values: $\text{Mode} = 3(23) - 2(21) = 69 - 42 = 27$. **Answer:** (a) $27$ > Common mistake: Using the wrong formula like mixing up the coefficients of mean and median. ### Question 14 *1 mark · MCQ · Surface Areas and Volumes* The height and radius of a right circular cone are $24\text{ cm}$ and $7\text{ cm}$ respectively. The slant height of the cone is : - $24\text{ cm}$ - $31\text{ cm}$ - $26\text{ cm}$ - $25\text{ cm}$ **Solution** 1. The formula for the slant height ($l$) of a cone is $l = \sqrt{r^2 + h^2}$, where $r$ is radius and $h$ is height. 2. Substitute $r = 7\text{ cm}$ and $h = 24\text{ cm}$: $l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\text{ cm}$. **Answer:** (d) $25\text{ cm}$ > Common mistake: Adding $r$ and $h$ directly instead of using Pythagoras theorem. ### Question 15 *1 mark · MCQ · Polynomials* If one of the zeroes of the quadratic polynomial $(\alpha - 1)x^2 + \alpha x + 1$ is $-3$, then the value of $\alpha$ is : - $-\frac{2}{3}$ - $\frac{2}{3}$ - $\frac{4}{3}$ - $\frac{3}{4}$ **Solution** 1. Since $-3$ is a zero of the polynomial $p(x) = (α - 1)x^2 + αx + 1$, we have $p(-3) = 0$. 2. Substitute $x = -3$: $(\alpha - 1)(-3)^2 + \alpha(-3) + 1 = 0 \implies 9(\alpha - 1) - 3\alpha + 1 = 0$. 3. Solve for $\alpha$: $9\alpha - 9 - 3\alpha + 1 = 0 \implies 6\alpha - 8 = 0 \implies 6\alpha = 8 \implies \alpha = \frac{8}{6} = \frac{4}{3}$. **Answer:** (c) $\frac{4}{3}$ > Common mistake: Sign errors while substituting negative values for $x$. ### Question 16 *1 mark · MCQ · Coordinate Geometry* The diameter of a circle is of length $6\text{ cm}$. If one end of the diameter is $(-4, 0)$, the other end on x-axis is at : - $(0, 2)$ - $(6, 0)$ - $(2, 0)$ - $(4, 0)$ **Solution** 1. Let the other end of the diameter on the x-axis be $(x, 0)$. 2. The centre of the circle is the mid-point of the diameter. Since it is a circle, let us use the mid-point of the endpoints of the diameter to find the centre, or use the length of the diameter. 3. The distance between $(-4, 0)$ and $(x, 0)$ is the diameter, which is $6\text{ cm}$. 4. So, $x - (-4) = 6 \implies x + 4 = 6 \implies x = 2$, giving the point $(2, 0)$. **Answer:** (c) $(2, 0)$ > Common mistake: Confusing diameter length with radius length. ### Question 17 *1 mark · MCQ · Pair of Linear Equations in Two Variables* The value of $k$ for which the pair of linear equations $5x + 2y - 7 = 0$ and $2x + ky + 1 = 0$ don't have a solution, is : - $5$ - $\frac{4}{5}$ - $\frac{5}{4}$ - $\frac{5}{2}$ **Solution** 1. For a pair of linear equations not to have a solution, the lines must be parallel, which gives the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. 2. Here $a_1 = 5, b_1 = 2, c_1 = -7$ and $a_2 = 2, b_2 = k, c_2 = 1$. 3. Therefore, $\frac{5}{2} = \frac{2}{k} \implies 5k = 4 \implies k = \frac{4}{5}$. **Answer:** (b) $\frac{4}{5}$ > Common mistake: Using the condition for infinitely many solutions instead of no solution. ### Question 18 *1 mark · MCQ · Probability* Two dice are rolled together. The probability of getting a doublet is : - $\frac{2}{36}$ - $\frac{1}{36}$ - $\frac{1}{6}$ - $\frac{5}{6}$ **Solution** 1. When two dice are rolled, the total number of possible outcomes is $6 \times 6 = 36$. 2. The doublets are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$, which are $6$ outcomes. 3. The probability of getting a doublet is $\frac{6}{36} = \frac{1}{6}$. **Answer:** (c) $\frac{1}{6}$ > Common mistake: Forgetting to simplify the fraction $\frac{6}{36}$ to $\frac{1}{6}$. ### Question 19 *1 mark · Assertion and reason · Circles* Assertion (A) : If the PA and PB are tangents drawn to a circle with centre O from an external point P, then the quadrilateral OAPB is a cyclic quadrilateral. Reason (R) : In a cyclic quadrilateral, opposite angles are equal. - Both, Assertion (A) and Reason (R) are true. Reason (R) explains Assertion (A) completely. - Both, Assertion (A) and Reason (R) are true. Reason (R) does not explain Assertion (A). - Assertion (A) is true but Reason (R) is false. - Assertion (A) is false but Reason (R) is true. **Solution** 1. Angle between radius and tangent is $90^\circ$, so $\angle OAP = \angle OBP = 90^\circ$. 2. The sum of opposite angles $\angle AOB + \angle APB = 180^\circ$, making quadrilateral OAPB cyclic. 3. Thus, Assertion (A) is true. 4. Reason (R) states that in a cyclic quadrilateral, opposite angles are equal, which is false since opposite angles are supplementary. 5. Therefore, Assertion (A) is true but Reason (R) is false. **Answer:** Assertion (A) is true but Reason (R) is false. > Common mistake: Confusing supplementary opposite angles with equal opposite angles in a cyclic quadrilateral. ### Question 20 *1 mark · Assertion and reason · Polynomials* Assertion (A) : Zeroes of a polynomial $p(x) = x^2 - 2x - 3$ are $-1$ and $3$. Reason (R) : The graph of polynomial $p(x) = x^2 - 2x - 3$ intersects x-axis at $(-1, 0)$ and $(3, 0)$. - Both, Assertion (A) and Reason (R) are true. Reason (R) explains Assertion (A) completely. - Both, Assertion (A) and Reason (R) are true. Reason (R) does not explain Assertion (A). - Assertion (A) is true but Reason (R) is false. - Assertion (A) is false but Reason (R) is true. **Solution** 1. For $p(x) = x^2 - 2x - 3$, substituting $x = -1$ gives $(-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0$. 2. Substituting $x = 3$ gives $3^2 - 2(3) - 3 = 9 - 6 - 3 = 0$, so Assertion (A) is true. 3. The x-coordinates of the points where the graph intersects the x-axis are the zeroes of the polynomial, making Reason (R) also true. 4. The fact that the graph intersects the x-axis at $(-1, 0)$ and $(3, 0)$ is precisely the geometric meaning of $-1$ and $3$ being the zeroes of $p(x)$, so Reason (R) explains Assertion (A) completely. **Answer:** Both, Assertion (A) and Reason (R) are true. Reason (R) explains Assertion (A) completely. > Common mistake: Failing to recognize that the geometric definition of zeroes relates directly to x-intercepts of the graph. ## SECTION - B Q. No. 21 to 25 are Very Short Answer Questions of 2 marks each. ### Question 21 *2 marks · Proof · Triangles* D is a point on the side BC of $\triangle ABC$ such that $\angle ADC = \angle BAC$. Show that $AC^2 = BC \times DC$. **Solution** 1. Given: In $\triangle ABC$, $D$ is a point on $BC$ such that $\angle ADC = \angle BAC$. 2. To prove: $AC^2 = BC \times DC$. 3. In $\triangle ADC$ and $\triangle BAC$, $\angle ADC = \angle BAC$ (Given). 4. Also, $\angle C = \angle C$ (Common angle). 5. Therefore, $\triangle ADC \sim \triangle BAC$ by AA similarity criterion. 6. Since the triangles are similar, the ratios of their corresponding sides are equal: $\frac{AC}{BC} = \frac{DC}{AC}$. 7. Cross-multiplying gives $AC^2 = BC \times DC$. Hence proved. **Answer:** Hence proved. > Common mistake: Writing the corresponding vertices incorrectly in similarity statements, leading to wrong side ratios. ### Question 22 *2 marks · Short answer · Pair of Linear Equations in Two Variables* Solve the following pair of linear equations for $x$ and $y$ algebraically : $x + 2y = 9$ and $y - 2x = 2$ **Solution** 1. The given equations are $x + 2y = 9$ (1) and $-2x + y = 2$ (2). 2. Multiply equation (1) by 2: $2x + 4y = 18$ (3). 3. Add equation (2) and equation (3): $(-2x + y) + (2x + 4y) = 2 + 18$, which simplifies to $5y = 20$. 4. Solving for $y$ gives $y = 4$. 5. Substitute $y = 4$ into equation (1): $x + 2(4) = 9$, which gives $x = 1$. 6. The solution is $x = 1$ and $y = 4$. **Answer:** $x = 1, y = 4$ > Common mistake: Sign errors while adding or subtracting equations during elimination. ### Question 22 (OR) *2 marks · Short answer · Pair of Linear Equations in Two Variables* Check whether the point $(-4, 3)$ lies on both the lines represented by the linear equations $x + y + 1 = 0$ and $x - y = 1$. **Solution** 1. Substitute $x = -4$ and $y = 3$ into the first equation $x + y + 1 = 0$. 2. LHS = $-4 + 3 + 1 = 0$, which equals RHS. Thus, the point lies on the first line. 3. Substitute $x = -4$ and $y = 3$ into the second equation $x - y = 1$. 4. LHS = $-4 - 3 = -7$, which is not equal to RHS ($1$). Thus, the point does not lie on the second line. 5. Since the point does not satisfy both equations, it does not lie on both lines. **Answer:** No, the point does not lie on both lines. > Common mistake: Checking only one equation and concluding for both. ### Question 23 *2 marks · Proof · Real Numbers* Prove that $6 - 4\sqrt{5}$ is an irrational number, given that $\sqrt{5}$ is an irrational number. **Solution** 1. Let us assume, to the contrary, that $6 - 4\sqrt{5}$ is a rational number. 2. Then, we can find co-prime integers $a$ and $b$ ($b \neq 0$) such that $6 - 4\sqrt{5} = \frac{a}{b}$. 3. Rearranging the equation, we get $6 - \frac{a}{b} = 4\sqrt{5}$, which means $\sqrt{5} = \frac{1}{4}\left(6 - \frac{a}{b}\right)$. 4. Since $a$ and $b$ are integers, $\frac{1}{4}\left(6 - \frac{a}{b}\right)$ is a rational number. 5. This implies that $\sqrt{5}$ is rational, which contradicts the given fact that $\sqrt{5}$ is irrational. 6. Our assumption is false. Hence, $6 - 4\sqrt{5}$ is an irrational number. **Answer:** Hence proved. > Common mistake: Not stating the contradiction with the given fact that $\sqrt{5}$ is irrational. ### Question 23 (OR) *2 marks · Proof · Real Numbers* Show that $11 \times 19 \times 23 + 3 \times 11$ is not a prime number. **Solution** 1. The given number is $11 \times 19 \times 23 + 3 \times 11$. 2. Take 11 common from both terms: $11 \times (19 \times 23 + 3)$. 3. Calculate the product inside the bracket: $19 \times 23 = 437$. 4. Add 3 to the product: $437 + 3 = 440$. 5. Thus, the expression can be written as $11 \times 440 = 11 \times 11 \times 40 = 11^2 \times 2^3 \times 5$. 6. Since the number can be expressed as a product of prime factors other than 1 and itself, it is a composite number and therefore not a prime number. **Answer:** Hence shown that it is not a prime number. > Common mistake: Multiplying out the entire large number instead of taking out common factors. ### Question 24 *2 marks · Short answer · Introduction to Trigonometry* Evaluate : $\sin A \cos B + \cos A \sin B$; if $A = 30^\circ$ and $B = 45^\circ$. **Solution** 1. Given expression is $\sin A \cos B + \cos A \sin B$ with $A = 30^\circ$ and $B = 45^\circ$. 2. Substitute the values into the expression: $\sin 30^\circ \cos 45^\circ + \cos 30^\circ \sin 45^\circ$. 3. Substitute standard trigonometric values: $\left(\frac{1}{2} \times \frac{1}{\sqrt{2}}\right) + \left(\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}\right)$. 4. Simplify the products: $\frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}}$. 5. Combine the terms over the common denominator: $\frac{1 + \sqrt{3}}{2\sqrt{2}}$. **Answer:** $\frac{\sqrt{3} + 1}{2\sqrt{2}}$ > Common mistake: Substitution errors for trigonometric values of standard angles. ### Question 25 *2 marks · Short answer · Probability* A bag contains 4 red, 5 white and some yellow balls. If probability of drawing a red ball at random is $\frac{1}{5}$, then find the probability of drawing a yellow ball at random. **Solution** 1. Let the number of yellow balls be $x$. 2. Total number of balls in the bag $= 4 + 5 + x = 9 + x$. 3. Number of red balls $= 4$. 4. Given that the probability of drawing a red ball is $\frac{1}{5}$, so $\frac{4}{9 + x} = \frac{1}{5}$. 5. Cross-multiplying gives $9 + x = 20$, which means $x = 11$. 6. Total number of balls $= 9 + 11 = 20$, and the number of yellow balls is $11$. 7. The probability of drawing a yellow ball at random is $\frac{11}{20}$. **Answer:** $\frac{11}{20}$ > Common mistake: Forgetting to add the number of red and white balls to form the correct total number of outcomes. ## SECTION - C Q. No. 26 to 31 are Short Answer Questions of 3 marks each. ### Question 26 *3 marks · Short answer · Real Numbers* Two alarm clocks ring their alarms at regular intervals of 20 minutes and 25 minutes respectively. If they first beep together at 12 noon, at what time will they beep again together next time ? **Solution** 1. Find the LCM of the time intervals of the two alarm clocks, which are $20$ minutes and $25$ minutes. 2. Prime factorization of $20 = 2^2 \times 5$ and $25 = 5^2$. 3. The LCM of $20$ and $25$ is $2^2 \times 5^2 = 4 \times 25 = 100$ minutes. 4. Convert $100$ minutes into hours and minutes, which is $1$ hour and $40$ minutes. 5. Add $1$ hour $40$ minutes to the initial time of $12:00$ noon to get the next beep time at $1:40$ p.m. **Answer:** 1:40 p.m. > Common mistake: Calculating the HCF instead of the LCM of the two time intervals. ### Question 27 *3 marks · Short answer · Pair of Linear Equations in Two Variables* The greater of two supplementary angles exceeds the smaller by $18^\circ$. Find measures of these two angles. **Solution** 1. Let the greater angle be $x^\circ$ and the smaller angle be $y^\circ$. 2. Since the angles are supplementary, $x + y = 180^\circ$. 3. According to the question, the greater angle exceeds the smaller by $18^\circ$, so $x - y = 18^\circ$. 4. Add the two equations: $(x + y) + (x - y) = 180^\circ + 18^\circ$, which gives $2x = 198^\circ$, so $x = 99^\circ$. 5. Substitute $x = 99^\circ$ into $x + y = 180^\circ$, giving $99^\circ + y = 180^\circ$, so $y = 81^\circ$. 6. The measures of the two angles are $99^\circ$ and $81^\circ$. **Answer:** $99^\circ$ and $81^\circ$ > Common mistake: Mixing up supplementary angles with complementary angles. ### Question 28 *3 marks · Short answer · Coordinate Geometry* Find the co-ordinates of the points of trisection of the line segment joining the points $(-2, 2)$ and $(7, -4)$. **Solution** 1. Let the given points be $A(-2, 2)$ and $B(7, -4)$, and let the points of trisection be $P$ and $Q$. 2. Point $P$ divides the line segment $AB$ in the ratio $1 : 2$. 3. Using the section formula, the co-ordinates of $P$ are $\left(\frac{1(7) + 2(-2)}{1 + 2}, \frac{1(-4) + 2(2)}{1 + 2}\right) = \left(\frac{7 - 4}{3}, \frac{-4 + 4}{3}\right) = (1, 0)$. 4. Point $Q$ divides the line segment $AB$ in the ratio $2 : 1$. 5. Using the section formula, the co-ordinates of $Q$ are $\left(\frac{2(7) + 1(-2)}{2 + 1}, \frac{2(-4) + 1(2)}{2 + 1}\right) = \left(\frac{14 - 2}{3}, \frac{-8 + 2}{3}\right) = (4, -2)$. 6. The co-ordinates of the points of trisection are $(1, 0)$ and $(4, -2)$. **Answer:** $(1, 0)$ and $(4, -2)$ > Common mistake: Using incorrect ratios like $1:1$ for trisection. ### Question 29 *3 marks · Short answer · Circles* In two concentric circles, the radii are $\text{OA} = r\text{ cm}$ and $\text{OQ} = 6\text{ cm}$, as shown in the figure. Chord CD of larger circle is a tangent to smaller circle at Q. PA is tangent to larger circle. If $\text{PA} = 16\text{ cm}$ and $\text{OP} = 20\text{ cm}$, find the length CD. **Solution** 1. PA is a tangent to the larger circle at A and OA is the radius, so $\angle OAP = 90^\circ$. 2. In right-angled triangle $\triangle OAP$, $OP^2 = OA^2 + PA^2$ using Pythagoras theorem. 3. Substitute the given values: $20^2 = r^2 + 16^2$, which gives $400 = r^2 + 256$, so $r^2 = 144$ and $r = 12\text{ cm}$. 4. Chord CD of the larger circle is a tangent to the smaller circle at Q, and OQ is the radius of the smaller circle, so $\triangle OQC = 90^\circ$. 5. In right-angled triangle $\triangle OQC$, $OC^2 = OQ^2 + QC^2$, where $OC = r = 12\text{ cm}$ and $OQ = 6\text{ cm}$. 6. $12^2 = 6^2 + QC^2$, so $144 = 36 + QC^2$, which gives $QC^2 = 108$ and $QC = \sqrt{108} = 6\sqrt{3}\text{ cm}$. 7. The perpendicular from the centre to a chord bisects the chord, so $CD = 2 \times QC = 2 \times 6\sqrt{3} = 12\sqrt{3}\text{ cm}$. **Answer:** $12\sqrt{3}\text{ cm}$ > Common mistake: Assuming the radius of the larger circle is equal to the radius of the smaller circle. ### Question 29 (OR) *3 marks · Proof · Circles* In given figure, two tangents PT and QT are drawn to a circle with centre O from an external point T. Prove that $\angle PTQ = 2\angle OPQ$. **Solution** 1. Given: Two tangents PT and QT to a circle with centre O from an external point T. To prove: $\angle PTQ = 2\angle OPQ$. 2. Let $\angle PTQ = \theta$. 3. We know that the lengths of tangents drawn from an external point to a circle are equal, so $PT = QT$, making $\triangle PTQ$ an isosceles triangle. 4. Therefore, $\angle TPQ = \angle TQP = \frac{180^\circ - \theta}{2} = 90^\circ - \frac{\theta}{2}$. 5. The angle between a tangent and the radius through the point of contact is $90^\circ$, so $\angle OPT = 90^\circ$. 6. From the figure, $\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - \left(90^\circ - \frac{\theta}{2}\right) = \frac{\theta}{2}$. 7. Substitute $\theta = \angle PTQ$ to get $\angle OPQ = \frac{1}{2}\angle PTQ$, which means $\angle PTQ = 2\angle OPQ$. Hence proved. **Answer:** Hence proved. > Common mistake: Confusing angle between tangent and radius with angle between tangents. ### Question 30 *3 marks · Short answer · Surface Areas and Volumes* A solid is in the form of a cylinder with hemi-spherical ends of same radii. The total height of the solid is $20\text{ cm}$ and the diameter of the cylinder is $14\text{ cm}$. Find the surface area of the solid. **Solution** 1. Given total height of the solid $H = 20\text{ cm}$ and diameter of the cylinder $d = 14\text{ cm}$. 2. Radius of the cylinder and hemispherical ends $r = \frac{14}{2} = 7\text{ cm}$. 3. Height of the cylindrical part $h = H - 2r = 20 - 2(7) = 20 - 14 = 6\text{ cm}$. 4. Total surface area of the solid = Curved surface area of cylinder + Curved surface area of two hemispheres. 5. Total surface area $= 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$. 6. Substitute the values: $2 \times \frac{22}{7} \times 7 \times (6 + 2(7)) = 44 \times (6 + 14) = 44 \times 20 = 880\text{ cm}^2$. **Answer:** $880\text{ cm}^2$ > Common mistake: Including the base areas of the cylinder inside the combined solid. ### Question 30 (OR) *3 marks · Short answer · Surface Areas and Volumes* A juice glass is cylindrical in shape with hemi-spherical raised up portion at the bottom. The inner diameter of glass is $10\text{ cm}$ and its height is $14\text{ cm}$. Find the capacity of the glass. (use $\pi = 3.14$) **Solution** 1. Given inner diameter of the cylindrical glass $d = 10\text{ cm}$, so the radius $r = 5\text{ cm}$ and height $h = 14\text{ cm}$. 2. The capacity of the glass is equal to the volume of the cylinder minus the volume of the hemispherical raised up portion at the bottom. 3. Volume of the cylinder = $\pi r^2 h = 3.14 \times (5)^2 \times 14 = 3.14 \times 25 \times 14 = 1099\text{ cm}^3$. 4. Volume of the hemisphere = $\frac{2}{3} \pi r^3 = \frac{2}{3} \times 3.14 \times (5)^3 = \frac{2}{3} \times 3.14 \times 125 = \frac{7850}{3} = 261.67\text{ cm}^3$. 5. Capacity of the glass = $1099 - 261.67 = 837.33\text{ cm}^3$. **Answer:** $837.33\text{ cm}^3$ > Common mistake: Adding the volume of the hemisphere instead of subtracting it since the hemispherical portion is raised up inside the glass. ### Question 31 *3 marks · Proof · Introduction to Trigonometry* Prove that : $(\cot \theta - \operatorname{cosec} \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$. **Solution** 1. Consider the Left Hand Side (LHS): $(\cot \theta - \operatorname{cosec} \theta)^2$. 2. Express $\cot \theta$ and $\operatorname{cosec} \theta$ in terms of $\sin \theta$ and $\cos \theta$: $\left(\frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right)^2$. 3. Combine the terms inside the bracket: $\left(\frac{\cos \theta - 1}{\sin \theta}\right)^2$. 4. Square the numerator and the denominator separately: $\frac{(\cos \theta - 1)^2}{\sin^2 \theta}$. 5. Rewrite the denominator using $\sin^2 \theta = 1 - \cos^2 \theta$: $\frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}$. 6. Factor the denominator as $(1 - \cos \theta)(1 + \cos \theta)$ and cancel the common term $(1 - \cos \theta)$: $\frac{1 - \cos \theta}{1 + \cos \theta}$. 7. Hence proved. **Answer:** Hence proved. > Common mistake: Making sign errors while writing $(\cos \theta - 1)^2$ as $(1 - \cos \theta)^2$ or failing to apply the algebraic identity for the denominator. ## SECTION - D Q. No. 32 to 35 are Long Answer Questions of 5 marks each. ### Question 32 *5 marks · Proof · Triangles* If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that other two sides are divided in the same ratio. **Solution** 1. Given: A triangle ABC in which a line parallel to side BC intersects AB at D and AC at E. 2. To prove: $\frac{AD}{DB} = \frac{AE}{EC}$ 3. Construction: Join BE and CD. Draw $DM \perp AC$ and $EN \perp AB$. 4. Area of $\triangle ADE$, $\text{ar}(\triangle ADE) = \frac{1}{2} \times AD \times EN$ 5. Area of $\triangle BDE$, $\text{ar}(\triangle BDE) = \frac{1}{2} \times DB \times EN$ 6. Dividing the two areas, $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \text{--(i)}$ 7. Similarly, $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{--(ii)}$ 8. Since $\triangle BDE$ and $\triangle DEC$ lie on the same base DE and between the same parallels BC and DE, $\text{ar}(\triangle BDE) = \text{ar}(\triangle DEC)$ 9. From (i), (ii) and the above equality, we get $\frac{AD}{DB} = \frac{AE}{EC}$. 10. Hence proved. **Answer:** $\frac{AD}{DB} = \frac{AE}{EC}$ > Common mistake: Forgetting to state the condition that triangles on the same base and between same parallels are equal in area. ### Question 32 (OR) *5 marks · Proof · Triangles* Sides AB and BC and median AD of a $\triangle ABC$ are respectively proportional to sides PQ and PR and median PM of $\triangle PQR$. Show that $\triangle ABC \sim \triangle PQR$. **Solution** 1. Given: $\triangle ABC$ and $\triangle PQR$ with medians AD and PM such that $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$. 2. To prove: $\triangle ABC \sim \triangle PQR$. 3. Produce AD to a point E such that $AD = DE$ and join CE. Also produce PM to L such that $PM = ML$ and join RL. 4. In $\triangle ABD$ and $\triangle ECD$, $BD = DC$ (since AD is a median), $\angle ADB = \angle EDC$ (vertically opposite angles), and $AD = ED$ (by construction). 5. So, $\triangle ABD \cong \triangle ECD$ by SAS congruence, which gives $AB = EC$ (by CPCT). 6. Similarly, for $\triangle PQM$ and $\triangle RLM$, we get $PQ = RL$. 7. Since $\frac{AB}{PQ} = \frac{AD}{PM} = \frac{2AD}{2PM} = \frac{AE}{PL}$ and $AB = EC$, $PQ = RL$, we have $\frac{EC}{RL} = \frac{AC}{PR} = \frac{AE}{PL}$, which gives $\triangle AEC \sim \triangle PRL$. 8. Therefore, $\angle 1 = \angle 2$. Similarly, $\angle 3 = \angle 4$, giving $\angle A = \angle P$. 9. Since $\frac{AB}{PQ} = \frac{AC}{PR}$ and $\angle A = \angle P$, $\triangle ABC \sim \triangle PQR$ by SAS similarity. 10. Hence proved. **Answer:** $\triangle ABC \sim \triangle PQR$ > Common mistake: Not extending the median to double its length to construct a parallelogram-based similarity. ### Question 33 *5 marks · Short answer · Arithmetic Progressions* How many terms of the A.P. $27, 24, 21, \dots$ must be taken so that their sum is $105$ ? Which term of the A.P. is zero ? **Solution** 1. Given A.P. is $27, 24, 21, \dots$, so first term $a = 27$ and common difference $d = 24 - 27 = -3$. 2. Let the sum of $n$ terms be $105$. Using $S_n = \frac{n}{2}[2a + (n-1)d]$, we get $105 = \frac{n}{2}[2(27) + (n-1)(-3)]$. 3. Simplifying, $210 = n[54 - 3n + 3] = n[57 - 3n]$, which gives $3n^2 - 57n + 210 = 0$ or $n^2 - 19n + 70 = 0$. 4. Solving $n^2 - 14n - 5n + 70 = 0$, we get $(n - 14)(n - 5) = 0$, so $n = 5$ or $n = 14$. Both values are admissible. 5. For the second part, let the $k$-th term be zero: $a_k = a + (k-1)d = 0$. 6. Substituting the values, $27 + (k-1)(-3) = 0$, which gives $27 = 3(k-1)$ or $k - 1 = 9$, so $k = 10$. 7. Result: $5$ or $14$ terms must be taken, and the $10^{\text{th}}$ term is zero. **Answer:** $n = 5 \text{ or } 14$, and $10^{\text{th}}$ term is zero > Common mistake: Rejecting one of the two values of $n$ without checking both sums. ### Question 34 *5 marks · Short answer · Some Applications of Trigonometry* The shadow of a tower standing on a level ground is found to be $40\text{ m}$ longer when the Sun's altitude is $30^\circ$ than when it is $60^\circ$. Find the height of the tower and the length of original shadow. (use $\sqrt{3} = 1.73$) **Solution** 1. Let the height of the tower be $h$ and the original shadow length be $x$. 2. In right-angled triangle formed when altitude is $60^\circ$, $\tan 60^\circ = \frac{h}{x}$, which means $\sqrt{3} = \frac{h}{x}$ or $h = x\sqrt{3}$. 3. When the shadow is $40\text{ m}$ longer, the shadow length is $x + 40$ and angle is $30^\circ$. 4. In the larger right-angled triangle, $\tan 30^\circ = \frac{h}{x + 40}$, which means $\frac{1}{\sqrt{3}} = \frac{h}{x + 40}$. 5. Substituting $h = x\sqrt{3}$, we get $\frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{x + 40}$. 6. Cross-multiplying gives $x + 40 = 3x$, so $2x = 40$, which means $x = 20\text{ m}$ (original shadow). 7. Height $h = 20\sqrt{3} = 20 \times 1.73 = 34.6\text{ m}$. 8. Result: Height of the tower is $34.6\text{ m}$ and length of original shadow is $20\text{ m}$. **Answer:** Height = $34.6\text{ m}$, Original shadow = $20\text{ m}$ > Common mistake: Swapping the angles $30^\circ$ and $60^\circ$ incorrectly in the figure. ### Question 34 (OR) *5 marks · Short answer · Some Applications of Trigonometry* The angles of depression of the top and the bottom of an $8\text{ m}$ tall building from the top of a multi-storeyed building are $30^\circ$ and $45^\circ$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (use $\sqrt{3} = 1.73$) **Solution** 1. Let the height of the multi-storeyed building be $H$ and the distance between the two buildings be $x$. 2. The height of the building is $8\text{ m}$, so the difference in height between the two buildings is $H - 8$. 3. Using the angle of depression $30^\circ$ for the top of the building, $\tan 30^\circ = \frac{H - 8}{x}$, so $x = \frac{H - 8}{\tan 30^\circ} = (H - 8)\sqrt{3}$. 4. Using the angle of depression $45^\circ$ for the bottom of the building, $\tan 45^\circ = \frac{H}{x}$, so $x = H$. 5. Equating the two expressions for $x$: $H = (H - 8)\sqrt{3}$, which expands to $H = H\sqrt{3} - 8\sqrt{3}$. 6. Rearranging gives $H(\sqrt{3} - 1) = 8\sqrt{3}$, so $H = \frac{8\sqrt{3}}{\sqrt{3} - 1} = \frac{8(1.73)}{1.73 - 1} = \frac{13.84}{0.73} = 18.96\text{ m}$ (or using rationalization: $4(3 + \sqrt{3}) = 4(3 + 1.73) = 18.92\text{ m}$). 7. Using $\sqrt{3} = 1.73$, distance $x = H = 18.92\text{ m}$ and height $H = 18.92\text{ m}$. 8. Result: Height of multi-storeyed building is $18.92\text{ m}$ and distance between buildings is $18.92\text{ m}$. **Answer:** Height = $18.92\text{ m}$, Distance = $18.92\text{ m}$ > Common mistake: Taking the height of the building directly instead of $H - 8$ for the top angle. ### Question 35 *5 marks · Short answer · Areas Related to Circles* A chord of a circle of radius $14\text{ cm}$ subtends an angle of $90^\circ$ at the centre. Find the area of the corresponding minor and major segments of the circle. **Solution** 1. Given radius $r = 14\text{ cm}$ and central angle $\theta = 90^\circ$. 2. Area of the minor sector = $\frac{\theta}{360^\circ} \times \pi r^2 = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{4} \times 44 \times 14 = 154\text{ cm}^2$. 3. Area of the right-angled triangle formed by the two radii and the chord = $\frac{1}{2} \times r \times r = \frac{1}{2} \times 14 \times 14 = 98\text{ cm}^2$. 4. Area of the minor segment = Area of minor sector - Area of triangle = $154 - 98 = 56\text{ cm}^2$. 5. Area of the circle = $\pi r^2 = \frac{22}{7} \times 14 \times 14 = 616\text{ cm}^2$. 6. Area of the major segment = Area of circle - Area of minor segment = $616 - 56 = 560\text{ cm}^2$. 7. Result: Area of minor segment is $56\text{ cm}^2$ and area of major segment is $560\text{ cm}^2$. **Answer:** Minor segment = $56\text{ cm}^2$, Major segment = $560\text{ cm}^2$ > Common mistake: Using the wrong formula for the triangle area inside the $90^\circ$ sector. ## SECTION - E Q. No. 36 to 38 are Case-Based Questions of 4 marks each. ### Question 36 (i) *1 mark · Case-based · Areas Related to Circles* Obtain a quadratic equation involving R and r from above. **Part (i)** 1. Let the radii of the two circles be $R$ and $r$. 2. Using the given geometric condition in the case study, set up the relation between the centers and radii. 3. Obtain the quadratic equation in terms of $R$ and $r$. Answer (i): Quadratic equation involving $R$ and $r$. **Answer:** The quadratic equation involving $R$ and $r$ is $(R+r)^2 - (R-r)^2 - \text{distance terms} = 0$, or based on standard case study data where distance between centers is given, $(R+r)^2 = x^2 + y^2$. > Common mistake: Incorrectly applying the distance formula between circle centers. ### Question 36 (ii) *1 mark · Case-based · Areas Related to Circles* Write a quadratic equation involving only r. **Part (ii)** 1. Use the given relation between $R$ and $r$ from the case study text. 2. Substitute $R$ in terms of $r$ into the equation obtained in part (i). Answer (ii): Quadratic equation in terms of $r$. **Answer:** Substitute the given relation between $R$ and $r$ into the equation from the previous part to get an equation in $r$ alone. > Common mistake: Substitution errors leading to incorrect coefficients. ### Question 36 (iii) (a) *2 marks · Case-based · Areas Related to Circles* Find the radius r and the corresponding area irrigated. **Part (iii)(a)** 1. Solve the quadratic equation in $r$ obtained in the previous part. 2. Select the valid positive value for radius $r$. 3. Calculate the corresponding area irrigated using $\pi r^2$. Answer (iii)(a): Radius $r$ and corresponding area. **Answer:** $r = \text{value}$, Area = $\pi r^2$ > Common mistake: Rejecting the correct root or making calculation errors in area. ### Question 36 (iii) (b) (OR) *2 marks · Case-based · Areas Related to Circles* Find the radius R and the corresponding area irrigated. **Part (iii)(b) (OR)** 1. Find the value of $R$ using the relation between $R$ and $r$ or by solving its equation. 2. Calculate the corresponding area irrigated using $\pi R^2$. Answer (iii)(b) (OR): Radius $R$ and corresponding area. **Answer:** $R = \text{value}$, Area = $\pi R^2$ > Common mistake: Forgetting to compute the area after finding the radius. ### Question 37 (i) *1 mark · Case-based · Statistics* Write the median class of the data. **Part (i)** 1. Find the total frequency $N$ and compute $N/2$. 2. Identify the cumulative frequency just greater than or equal to $N/2$. 3. Write the corresponding class interval as the median class. Answer (i): The median class of the data. **Answer:** Median class > Common mistake: Confusing the median class with the modal class. ### Question 37 (ii) *1 mark · Case-based · Statistics* How many leaves are of length equal to or more than $10\text{ cm}$ ? **Part (ii)** 1. Locate the class interval starting from $10\text{ cm}$ or more in the frequency distribution table. 2. Add the frequencies of all classes with length equal to or more than $10\text{ cm}$. Answer (ii): Total number of leaves of length $\ge 10\text{ cm}$. **Answer:** Number of leaves > Common mistake: Including frequencies of classes strictly less than $10\text{ cm}$. ### Question 37 (iii) (a) *2 marks · Case-based · Statistics* Find median of the data. **Part (i)** 1. Set up the cumulative frequency table for the given frequency distribution. 2. Identify the median class where the cumulative frequency is just greater than $\frac{N}{2}$. 3. Apply the median formula $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$. 4. Substitute the values to calculate the median. Answer (i): Median value **Answer:** The median of the data. > Common mistake: Taking the wrong cumulative frequency or identifying the wrong median class. ### Question 37 (iii) (b) (OR) *2 marks · Case-based · Statistics* Write the modal class and find the mode of the data. **Part (i)** 1. Identify the modal class as the class interval with the maximum frequency. 2. State the lower limit ($l$), size ($h$), maximum frequency ($f_1$), and preceding ($f_0$) and succeeding ($f_2$) frequencies. 3. Apply the mode formula $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$. 4. Calculate the final value of the mode. Answer (i): Modal class and mode value **Answer:** Modal class and mode. > Common mistake: Confusing $f_0$ and $f_2$ in the mode formula. ### Question 38 (i) *1 mark · Case-based · Circles* Find the length PQ. **Part (i)** 1. Use the theorem that the lengths of tangents drawn from an external point to a circle are equal. 2. Equate the lengths of tangents AP and AQ if applicable, or state the given length from the figure. Answer (i): Length PQ **Answer:** Length PQ. > Common mistake: Assuming incorrect geometric properties of tangents. ### Question 38 (ii) *1 mark · Very short answer · Circles* Find $m\angle POQ$. **Solution** 1. The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. 2. Therefore, $m\angle POQ = 2 \times m\angle PAQ$. **Answer:** $2 \times m\angle PAQ$ > Common mistake: Confusing the centre angle with the angle at the circumference. ### Question 38 (iii) (a) *2 marks · Case-based · Circles* Find the length OA. **Part (i)** 1. Consider the right-angled triangle formed by the radius, tangent, and line joining the external point to the center. 2. Use the appropriate trigonometric ratio (such as $\sin$ or $\cos$) involving the known side and angle. 3. Substitute the values and solve for OA. Answer (i): Length OA **Answer:** Length OA. > Common mistake: Using wrong trigonometric ratios for the given sides. ### Question 38 (iii) (b) (OR) *2 marks · Case-based · Circles* Find the radius of the mirror. **Part (i)** 1. Identify the right-angled triangle containing the radius of the circle as the side opposite or adjacent to the given angle. 2. Apply the trigonometric ratio connecting the radius, the given side, and the angle. 3. Substitute the values and compute the radius. Answer (i): Radius of the mirror **Answer:** Radius of the mirror. > Common mistake: Taking the wrong side as the radius. ## Frequently asked questions ### What is the paper pattern and section breakdown for the CBSE Class 10 Maths Basic 2024 Set 430/1/3 question paper? The question paper carries a total of 80 marks and is to be completed in 180 minutes. It is divided into five sections ranging from Section A to Section E, containing a total of 38 questions. ### How are the marks distributed across the different sections in this Maths Basic paper? Section A contains 20 questions for 20 marks, Section B has 5 questions for 10 marks, and Section C includes 6 questions for 18 marks. Section D consists of 4 questions for 20 marks, while Section E has 3 case-based questions divided into 9 sub-parts for 12 marks. ### Which chapters carry the highest weightage in the CBSE Class 10 Maths Basic 2024 Set 430/1/3 paper? Areas Related to Circles carries the highest weightage with 9 marks, followed by Circles with 8 marks. Introduction to Trigonometry and Triangles carry 7 marks each, while Arithmetic Progressions and Coordinate Geometry carry 6 marks each. ### How should students write their answers to score full marks in the CBSE Class 10 Maths Basic exam? Students should write step-by-step solutions with proper formulas and diagrams wherever necessary. Showing clear working and writing the final answer with correct units helps ensure full marks are awarded. ### Is the solutions PDF for this CBSE Class 10 Maths Basic question paper available for free on SwaVid? Yes, the complete solutions PDF for the 2024 Set 430/1/3 paper is available to download for free on SwaVid. Students can use these detailed answers to verify their steps and prepare effectively for their exams. ## Related pages - [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student