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Paramhansh Trivedi

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  1. Asked: October 29, 2020In: Class 10

    A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.

    Best Answer
    Paramhansh Trivedi
    Added an answer on February 28, 2023 at 7:16 am

    Total number of balls = 12 Total number of black balls = x P (getting a black ball) = x/12 If 6 more black balls are put in the box, then Total number of balls = 12 + 6 = 18 Total number of black balls = x + 6 P (getting a black ball now) = (x + 6)/18 According to the condition given in the questionRead more

    Total number of balls = 12
    Total number of black balls = x
    P (getting a black ball) = x/12
    If 6 more black balls are put in the box, then
    Total number of balls = 12 + 6 = 18
    Total number of black balls = x + 6
    P (getting a black ball now) = (x + 6)/18
    According to the condition given in the question,
    2(x/12) = (x + 6)/18
    ⇒ 3x = x + 6
    ⇒ 2x = 6
    ⇒ x = 3

    See here for video explanation🙌

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  2. Asked: October 29, 2020In: Class 10

    A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

    Best Answer
    Paramhansh Trivedi
    Added an answer on February 28, 2023 at 7:16 am

    Let the number of blue balls be x. Number of red balls = 5 Total number of balls = x + 5 P (getting a red all) = 5/(5 + x) P (getting a blue ball) = x/(5 + x) Given that, 2(5/(5 + x)) = (x/(5 + x)) ⇒ 10(x + 5) = x² + 5x - 50 = 0 ⇒ x² - 10x + 5x - 50 = 0 ⇒ x(x - 10) + 5(x - 10) = 0 ⇒ (x - 10)(x + 5)Read more

    Let the number of blue balls be x.
    Number of red balls = 5
    Total number of balls = x + 5
    P (getting a red all) = 5/(5 + x)
    P (getting a blue ball) = x/(5 + x)
    Given that,
    2(5/(5 + x)) = (x/(5 + x)) ⇒ 10(x + 5) = x² + 5x – 50 = 0
    ⇒ x² – 10x + 5x – 50 = 0 ⇒ x(x – 10) + 5(x – 10) = 0
    ⇒ (x – 10)(x + 5) = 0
    ⇒ Either x – 10 = 0 or x + 5 = 0
    ⇒ x – 10 or x = -5
    However, the number of balls cannot be negative.
    Hence, number of blue balls = 10

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  3. Asked: February 21, 2023In: Class 10 Maths

    Find the area of the shaded region in Figure.

    Best Answer
    Paramhansh Trivedi
    Added an answer on February 21, 2023 at 11:43 am

    QR is diameter of circle. Therefore, ∠RPQ = 90° [Angle in semicircle is right angle] In APQR, by Pythagoras theorem, RP² + PQ² = RQ² (7)² + (24)² = RQ² ⇒ RQ² = 576 + 49 = 625 ⇒ RQ = √625 = 25 Therefore, the radius of circle = RQ/2 = 25/2 cm Area of shaded region = Area of semicircle - Area of APQR =Read more

    QR is diameter of circle.
    Therefore, ∠RPQ = 90° [Angle in semicircle is right angle]
    In APQR, by Pythagoras theorem,
    RP² + PQ² = RQ²
    (7)² + (24)² = RQ²
    ⇒ RQ² = 576 + 49 = 625
    ⇒ RQ = √625 = 25
    Therefore, the radius of circle = RQ/2 = 25/2 cm
    Area of shaded region = Area of semicircle – Area of APQR
    = 1/2 × πr² – 1/2 PR × PQ = 1/2 × π(25/2)² – 1/2 × 7 × 24 = 1/2 × 22/7 × 25/2 × 25/2 – 7 × 12
    = 6875/28 – 84 = (6875 – 2352)/28 = 4523/28 cm²

    Here is the video explanation ✌😇

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  4. Asked: December 21, 2020In: Class 10 Maths

    Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. -1/4,1/4

    Best Answer
    Paramhansh Trivedi
    Added an answer on January 5, 2023 at 6:51 am

    Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = (-1)/4 = (-b)/a αβ = 1/4 = c/a On comparing, a = 4, b = 1 and c = 1 Hence, the required quadratic polynomial is 4x² + x + 1. For Video Solution See Here 👀

    Let α and β are the zeroes of the polynomial ax² + bx + c, then we have
    α + β = (-1)/4 = (-b)/a
    αβ = 1/4 = c/a
    On comparing,
    a = 4, b = 1 and c = 1
    Hence, the required quadratic polynomial is 4x² + x + 1.

    For Video Solution See Here 👀

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  5. Asked: December 22, 2020In: Class 10 Maths

    Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. 4,1

    Best Answer
    Paramhansh Trivedi
    Added an answer on January 5, 2023 at 6:51 am

    Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = 4 = 4/1 = (-b)/a αβ = 1 = 1/1 = c/a On comparing, a = 1, b= - 4 and c = 1 Hence, the required quadratic polynomial is x² - 4x + 1. See Here 😃✌

    Let α and β are the zeroes of the polynomial ax² + bx + c, then we have
    α + β = 4 = 4/1 = (-b)/a
    αβ = 1 = 1/1 = c/a
    On comparing,
    a = 1, b= – 4 and c = 1
    Hence, the required quadratic polynomial is x² – 4x + 1.

    See Here 😃✌

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