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A cuboid of length 25 cm and breadth 8 cm holds exactly 1 litre of water. Find the height of the cuboid. 21. The roots of a quadratic equation are -2/3 and -1/4. Find the equation.

Mathematics

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20.

  1. Convert the volume from litres to cubic centimetres.

    We know that 1 litre=1000 cm31 \text{ litre} = 1000 \text{ cm}^3.
    So, the volume of the cuboid is V=1000 cm3V = 1000 \text{ cm}^3.

  2. Use the formula for the volume of a cuboid.

    The volume of a cuboid is given by V=length×breadth×heightV = \text{length} \times \text{breadth} \times \text{height}.
    Given length L=25 cmL = 25 \text{ cm} and breadth B=8 cmB = 8 \text{ cm}. Let the height be HH.

    1000 cm3=25 cm×8 cm×H1000 \text{ cm}^3 = 25 \text{ cm} \times 8 \text{ cm} \times H

  3. Calculate the product of length and breadth.

    1000 cm3=200 cm2×H1000 \text{ cm}^3 = 200 \text{ cm}^2 \times H

  4. Solve for the height HH.

    H=1000 cm3200 cm2H = \frac{1000 \text{ cm}^3}{200 \text{ cm}^2} H=5 cmH = 5 \text{ cm}

    The height of the cuboid is 5 cm.

    The correct option is A.

    A. 5 cm\boxed{\text{A. } 5 \text{ cm}}

    21.

  5. Identify the given roots.

    Let the roots be α=−23\alpha = -\frac{2}{3} and β=−14\beta = -\frac{1}{4}.

  6. Calculate the sum of the roots.

    α+β=−23+(−14)\alpha + \beta = -\frac{2}{3} + \left(-\frac{1}{4}\right) α+β=−23−14\alpha + \beta = -\frac{2}{3} - \frac{1}{4}

    To add these fractions, find a common denominator, which is 12.

    α+β=−2×43×4−1×34×3\alpha + \beta = -\frac{2 \times 4}{3 \times 4} - \frac{1 \times 3}{4 \times 3} α+β=−812−312\alpha + \beta = -\frac{8}{12} - \frac{3}{12} α+β=−1112\alpha + \beta = -\frac{11}{12}

  7. Calculate the product of the roots.

    αβ=(−23)×(−14)\alpha\beta = \left(-\frac{2}{3}\right) \times \left(-\frac{1}{4}\right) αβ=2×13×4\alpha\beta = \frac{2 \times 1}{3 \times 4} αβ=212\alpha\beta = \frac{2}{12} αβ=16\alpha\beta = \frac{1}{6}

  8. Form the quadratic equation using the general form x2−(α+β)x+αβ=0x^2 - (\alpha + \beta)x + \alpha\beta = 0.

    Substitute the sum and product of the roots:

    x2−(−1112)x+16=0x^2 - \left(-\frac{11}{12}\right)x + \frac{1}{6} = 0 x2+1112x+16=0x^2 + \frac{11}{12}x + \frac{1}{6} = 0

  9. Clear the denominators by multiplying the entire equation by the least common multiple (LCM) of 12 and 6, which is 12.

    12(x2+1112x+16)=12×012 \left(x^2 + \frac{11}{12}x + \frac{1}{6}\right) = 12 \times 0 12x2+12(1112)x+12(16)=012x^2 + 12 \left(\frac{11}{12}\right)x + 12 \left(\frac{1}{6}\right) = 0 12x2+11x+2=012x^2 + 11x + 2 = 0

    The quadratic equation is 12x² + 11x + 2 = 0.

    The correct option is A.

    A. 12x2+11x+2=0\boxed{\text{A. } 12x^2 + 11x + 2 = 0}
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