Kaliyuga Ekalavya

NCERT Class X Chapter 7: Coordinate Geometry Example 3 Question: Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively. Do you think they are seated in a line? Give reasons for your answer. Given: Coordinates of Ashima (A): (3, 1) Coordinates of Bharti (B): (6, 4) Coordinates of Camella (C): (8, 6) To Find: Whether the points A, B, and C are collinear. Formula: The slope of a line passing through points \((x_1, y_1)\) and \((x_2, y_2)\) is given by: $$ m = \frac{y_2 - y_1}{x_2 - x_1} $$ If the slopes between all pairs of points are equal, the points are collinear. Solution: Step 1: Find the slope of AB. $$ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 1}{6 - 3} = \frac{3}{3} = 1 $$ Step 2: Find the slope of BC. $$ m_{BC} = \frac{6 - 4}{8 - 6} = \frac{2}{2} = 1 $$ Step 3: Compare the slopes. $$ m_{AB} = m_{BC} = 1 $$ Since the slopes are equal, the points A, B, and C are collinear....

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Example 3 Question: Champa went to a ‘Sale’ to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased”. Help her friends to find how many pants and skirts Champa bought. Given: Let the number of pants Champa bought be \( x \). Let the number of skirts Champa bought be \( y \). The number of skirts is two less than twice the number of pants: \( y = 2x - 2 \). The number of skirts is four less than four times the number of pants: \( y = 4x - 4 \). To Find: The number of pants (\( x \)) and skirts (\( y \)) that Champa bought. Formula: Form equations based on the statements: \( y = 2x - 2 \) \( y = 4x - 4 \) Solve the system of linear equations to find \( x \) and...

NCERT Class X Chapter 12: Surface Areas And Volumes Example 3 Question: A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in Fig. 12.8. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Use \( \pi = 3.14 \)) Given: Height of rocket, \( H = 26 \) cm Height of conical part, \( h_c = 6 \) cm Diameter of conical base, \( d_c = 5 \) cm \( \Rightarrow r_c = 2.5 \) cm Diameter of cylindrical base, \( d_{cy} = 3 \) cm \( \Rightarrow r_{cy} = 1.5 \) cm \( \pi = 3.14 \) To Find: Area to be painted orange (curved surface area of conical part) Area to be painted yellow (curved surface area of cylindrical part) Formula: Curved surface area of con...

NCERT Class X Chapter 9: Some Applications of Trigonometry Example 3 Question: An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney? Given: Height of observer = \(1.5\,\text{m}\) Distance from observer to chimney = \(28.5\,\text{m}\) Angle of elevation = \(45^\circ\) To Find: Height of the chimney Formula: \[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \] Solution: Step 1: Let the height of the chimney above the observer's eyes be \( h \) meters. Step 2: Using the right triangle formed, apply the tangent formula: $$ \tan 45^\circ = \frac{h}{28.5} $$ Step 3: Since \( \tan 45^\circ = 1 \), substitute and solve for \( h \): $$ 1 = \frac{h}{28.5} \implies h = 28.5 $$ Step 4: Total height of the chimney is the sum of \( h \) and the observer's height. $$ \text{Total height} = h + ...

NCERT Class X Chapter 1: Real Numbers Example 3 Question: Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM. Given: Numbers: 96 and 404 To Find: HCF and LCM of 96 and 404 using the prime factorisation method. Formula: $$\text{HCF} \times \text{LCM} = \text{Product of the two numbers}$$ Solution: Step 1: Find the prime factorisation of 96. $$ 96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3 $$ Step 2: Find the prime factorisation of 404. $$ 404 = 2 \times 2 \times 101 = 2^2 \times 101 $$ Step 3: Find the HCF by taking the lowest power of common prime factors. Common prime factor: 2 Lowest power of 2 in both numbers: \(2^2\) $$ \text{HCF}(96,\,404) = 2^2 = 4 $$ Step 4: Use the formula to find the LCM. $$ \text{HCF} \times \text{LCM} = 96 \times 404 $$ $$ 4 \times \text{LCM} = 38784 $$ Step 5: Calculate the LCM. $$ \text{LCM} = \frac{38784}{4} = 9696 $$ Result: \(\text{HCF}(96,\...

NCERT Class X Chapter 14: Probability Example 3 Question: Suppose we throw a die once. (i) What is the probability of getting a number greater than 4? (ii) What is the probability of getting a number less than or equal to 4? Given: A die is thrown once. The possible outcomes are {1, 2, 3, 4, 5, 6}. To Find: (i) Probability of getting a number greater than 4. (ii) Probability of getting a number less than or equal to 4. Formula: Probability = Number of favorable outcomes Total number of possible outcomes Solution: (i) Numbers greater than 4 are 5 and 6.  Number of favorable outcomes = 2.  Total number of possible outcomes = 6. ⇒ Probability = 2 6 = 1 3 (ii) Numbers less than or equal to 4 are 1, 2, 3, and 4.  Number of favorable outcomes = 4.  Total number of possible outcomes = 6. ⇒ Probability = 4 6 = 2 3 Result: (i) The probability of getting a number greater than 4 is 1 3 . (ii) The probability of getting a number le...

NCERT Class X Chapter 5: Arithmetic Progression Example 3 Question: Find the 10th term of the AP : 2, 7, 12, . . . Given: Arithmetic Progression (AP): 2, 7, 12, . . . To Find: The 10th term of the given AP. Formula: The nth term of an AP is given by: a n = a + (n - 1)d, where, a is the first term,  n is the term number, and  d is the common difference. Solution: Here,  the first term a = 2 and  the common difference d = 7 - 2 = 5. We need to find the 10th term, so n = 10. Using the formula a n = a + (n - 1)d  ⇒ a 10 = 2 + (10 - 1)5  ⇒ a 10 = 2 + 9 × 5  ⇒ a 10 = 2 + 45  ⇒ a 10 = 47 Result: The 10th term of the AP is 47. Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Example 4 Explore more in Arithmetic Progressions chapter: Click this link to explore more NCERT Class X Chapter 5 Arithmetic Progressions solutions Explore more: Click this link to explo...

NCERT Class X Chapter 4: Quadratic Equation Example 3 Question: Find the roots of the equation 2x 2 – 5x + 3 = 0, by factorisation. Given: The quadratic equation: 2x 2 – 5x + 3 = 0 To Find: The roots of the equation by factorisation. Formula: For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs . Solution: The given equation is 2x 2 – 5x + 3 = 0. Here, a = 2, b = -5, c = 3. Product (a × c) = 2 × 3 = 6. Sum (b) = -5. We need to...