Kaliyuga Ekalavya

NCERT Class X Chapter 7: Coordinate Geometry Example 5 Question: Find a point on the y-axis which is equidistant from the points A(6, 5) and B(–4, 3). Given: Point \(A(6, 5)\) Point \(B(-4, 3)\) To Find: A point on the y-axis which is equidistant from \(A\) and \(B\). Formula: Distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by: $$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Solution: Step 1: Let the required point on the y-axis be \(P(0, y)\). Step 2: Since \(P\) is equidistant from \(A\) and \(B\), we have: $$ PA = PB $$ $$ \sqrt{(6 - 0)^2 + (5 - y)^2} = \sqrt{(-4 - 0)^2 + (3 - y)^2} $$ Step 3: Squaring both sides to remove the square roots: $$ (6)^2 + (5 - y)^2 = (-4)^2 + (3 - y)^2 $$ $$ 36 + (5 - y)^2 = 16 + (3 - y)^2 $$ Step 4: Expand both sides: $$ 36 + [25 - 10y + y^2] = 16 + [9 - 6y + y^2] $$ $$ (36 + 25 - 10y + y^2) = (16 + 9 - 6y + y^2) $$ $$ (61 - 10y + y^2...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Example 5 Question: Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Represent this situation algebraically and graphically by the method of substitution. Given: Let the present age of Aftab be \( x \) years. Let the present age of his daughter be \( y \) years. Seven years ago, Aftab's age was \( x - 7 \) years, daughter's age was \( y - 7 \) years. Seven years ago, Aftab was seven times as old as his daughter: \( x - 7 = 7(y - 7) \). Three years from now, Aftab's age will be \( x + 3 \) years, daughter's age will be \( y + 3 \) years. Three years from now, Aftab will be three times as old as his daughter: \( x + 3 = 3(y + 3) \). To Find: The present ages of Aftab (\( x \)) and his daughter (\( y \)). Algebraic representation of the situation. Grap...

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower. Given: Difference in shadow lengths = 40 m Sun's altitude (angle of elevation) = 30° and 60° To Find: Height of the tower Formula: \[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \] Solution: Step 1: Let the height of the tower be \( h \) meters. Let the length of the shadow when the Sun's altitude is 60° be \( x \) meters. Step 2: When the Sun's altitude is 60°, using the definition of tangent: \[ \tan 60^\circ = \frac{h}{x} \] \[ \sqrt{3} = \frac{h}{x} \] \[ x = \frac{h}{\sqrt{3}} \] Step 3: When the Sun's altitude is 30°, the length of the shadow is \( x + 40 \) meters. \[ \tan 30^\circ = \frac{h}{x + 40} \] \[ \frac{1}{\sqrt{3}} = \frac{h}{x + 40} \] \[ x + 40 = h \sqrt{3} \] St...

NCERT Class X Chapter 1: Real Numbers Example 5 Question: Prove that \( \sqrt{3} \) is irrational. Given: We need to prove that \( \sqrt{3} \) is an irrational number. To Find: Show that \( \sqrt{3} \) cannot be expressed as a ratio of two integers. Formula: No specific formula is required. We use proof by contradiction and properties of rational and irrational numbers. Solution: Step 1: Assume, for contradiction, that \( \sqrt{3} \) is rational. That is, $$ \sqrt{3} = \frac{a}{b} $$ where \( a \) and \( b \) are coprime integers and \( b \neq 0 \). Step 2: Square both sides to remove the square root: $$ (\sqrt{3})^2 = \left( \frac{a}{b} \right)^2 $$ So, $$ 3 = \frac{a^2}{b^2} $$ Step 3: Multiply both sides by \( b^2 \) to clear the denominator: $$ 3b^2 = a^2 $$ Step 4: Since 3 divides \( a^2 \), 3 must also divide \( a \) (because 3 is prime). Let \( a = 3k \) for some integer \( k \). Step 5: ...

NCERT Class X Chapter 14: Probability Example 5 Question: Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match? Given: Probability of Sangeeta winning the match = 0.62 To Find: Probability of Reshma winning the match Formula: Since there are only two players,  the sum of probabilities of Sangeeta winning and Reshma winning must be equal to 1 (or 100%).  Therefore, P(Sangeeta) + P(Reshma) = 1 Solution: Let P(S) be the probability of Sangeeta winning and  P(R) be the probability of Reshma winning. Given P(S) = 0.62 We know that P(S) + P(R) = 1 ⇒ 0.62 + P(R) = 1 ⇒ P(R) = 1 - 0.62 ⇒ P(R) = 0.38 Result: The probability of Reshma winning the match is 0.38 Next question solution: NCERT Class X Chapter 14: Probability Example 6 Explore more in Probability: Click this link to explore more N...

NCERT Class X Chapter 5: Arithmetic Progression Example 6 Question: Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . Given: The list of numbers is 5, 11, 17, 23, . . . To Find: Whether 301 is a term of the given list of numbers. Formula: The nth term of an arithmetic progression is given by a n = a + (n-1)d,  where  a is the first term and  d is the common difference. Solution: The given list of numbers forms an arithmetic progression. First term, a = 5 Common difference, d = 11 - 5 = 6 Let 301 be the nth term of the arithmetic progression.  Then, 301 = 5 + (n - 1)6 ⇒ 301 - 5 = (n - 1)6 ⇒ 296 = (n - 1)6 ⇒ n - 1 = 296 6 ⇒ n - 1 = 49.333... ⇒ n = 50.333.... Since n must be an integer, 301 is not a term of the given arithmetic progression. Result: 301 is not a term of the list of numbers 5, 11, 17, 23, . . . Next question solution: NCERT Class X Chapter 5: Arithmetic Progression...

NCERT Class X Chapter 5: Arithmetic Progression Example 5 Question: Determine the AP whose 3rd term is 5 and the 7th term is 9. Given: 3rd term (a 3 ) = 5 7th term (a 7 ) = 9 To Find: The arithmetic progression (AP). Formula: a n = a 1 + (n-1)d where, a n is the nth term,  a 1 is the first term, and  d is the common difference. Solution: We have  a 3 = a 1 + 2d = 5 and  a 7 = a 1 + 6d = 9. Subtracting the first equation from the second equation, we get: (a 1 + 6d) - (a 1 + 2d) = 9 - 5  a 1 + 6d - a 1 - 2d = 9 - 5  ⇒ 6d - 2d = 4  ⇒ 4d = 4 ⇒ d = 1 Substituting d = 1 into a 1 + 2d = 5,  we get: a 1 + 2(1) = 5  ⇒ a 1 = 3 Therefore, the AP is 3, 4, 5, 6, 7, 8, 9, ... Result: The arithmetic progression is 3, 4, 5, 6, 7, 8, 9, ... Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Example 6. Explore more in Arithmetic Progressions chapter: Click t...

NCERT Class X Chapter 4: Quadratic Equation Example 5 Question: Find the roots of the quadratic equation:Find the roots of the quadratic equation 3x 2 - 2√6x - 2 = 0 . Given: The quadratic equation : 3x 2 - 2√6x - 2 = 0 To Find: The roots of the equation. Formula: WKT, 2√ax = √ax + √ax Solution: The given equation is 3x 2 - 2√6x - 2 = 0. Split the middle term: using the formula 2√ax = √ax+ √ax ⇒ 3x 2 - 2√6x - 2 = 3x 2 - (√6x + √6x) - 2 ⇒ 3x 2 - 2√6x - 2 = 3x 2 - √6x - √6x - 2 Grouping common factors in 3x 2 - √6x - √6x - 2 we have ⇒ (3x 2 - √6x) + (- √6x - 2) = √3x(√3x-√2) -√2(√3x-√2) ⇒ 3x 2 - 2√6x - 2 = (√3x-√2) (√3x-√2) Since 3x 2 - 2√6x - 2 = 0, (√3x-√2) (√3x-√2) = 0 Case 1: ⇒ (√3x-√2) = 0 ⇒ √3x = √2 ⇒ x = √2 √3 Case 2: ⇒ (√3x-√2) = 0 ⇒ √3x = √2 ⇒ x = √2 √3 Therefor the two roots are: x = √2 √3...