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Mr. Manohar gave 20% part of his income to his elder son and 30% part to his younger son. He gave 10% of the balance as donation to a school. He still had Rs. 1,80,000 for himself. What was Mr. Manoha

Mathematics

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1) Mr. Manohar gave 20% part of his income to his elder son and 30% part to his younger son. He gave 10% of the balance as donation to a school. He still had Rs. 1,80,000 for himself. What was Mr. Manohar's income?

  1. Let Mr.

    Manohar's total income be II.
    Amount given to elder son = 20%20\% of I=0.20II = 0.20I
    Amount given to younger son = 30%30\% of I=0.30II = 0.30I

  2. Calculate the total percentage of income given to his sons.

    Total percentage given to sons = 20%+30%=50%20\% + 30\% = 50\%
    Total amount given to sons = 0.20I+0.30I=0.50I0.20I + 0.30I = 0.50I

  3. Calculate the balance remaining after giving to his sons.

    Balance = Total Income - Amount given to sons
    Balance = I−0.50I=0.50II - 0.50I = 0.50I

  4. Calculate the amount donated to the school.

    Donation = 10%10\% of the balance
    Donation = 0.10×(0.50I)=0.05I0.10 \times (0.50I) = 0.05I

  5. Calculate the amount remaining for himself.

    Amount remaining = Balance - Donation
    Amount remaining = 0.50I−0.05I=0.45I0.50I - 0.05I = 0.45I

  6. Set up the equation with the given remaining amount and solve for II.

    We are given that he still had Rs. 1,80,000 for himself.

    0.45I=1,80,0000.45I = 1,80,000 I=1,80,0000.45I = \frac{1,80,000}{0.45}

    To simplify the division, multiply the numerator and denominator by 100:

    I=1,80,000×1000.45×100I = \frac{1,80,000 \times 100}{0.45 \times 100} I=18,000,00045I = \frac{18,000,000}{45} I=400,000I = 400,000

    Mr. Manohar's total income was Rs. 4,00,000.

    Mr. Manohar’s income was Rs. 4,00,000.\boxed{\text{Mr. Manohar's income was Rs. 4,00,000.}}

    2) If (x+3):(x+11)=(x−2):(x+1)(x+3) : (x+11) = (x-2) : (x+1), then find the value of xx.

  7. Write the proportion as a fraction.

    x+3x+11=x−2x+1\frac{x+3}{x+11} = \frac{x-2}{x+1}
  8. Cross-multiply the terms.

    (x+3)(x+1)=(x−2)(x+11)(x+3)(x+1) = (x-2)(x+11)
  9. Expand both sides of the equation.

    Left side:

    x(x+1)+3(x+1)=x2+x+3x+3=x2+4x+3x(x+1) + 3(x+1) = x^2 + x + 3x + 3 = x^2 + 4x + 3

    Right side:

    x(x+11)−2(x+11)=x2+11x−2x−22=x2+9x−22x(x+11) - 2(x+11) = x^2 + 11x - 2x - 22 = x^2 + 9x - 22
  10. Set the expanded expressions equal to each other.

    x2+4x+3=x2+9x−22x^2 + 4x + 3 = x^2 + 9x - 22
  11. Solve for xx.

    Subtract x2x^2 from both sides:

    4x+3=9x−224x + 3 = 9x - 22

    Subtract 4x4x from both sides:

    3=5x−223 = 5x - 22

    Add 22 to both sides:

    3+22=5x3 + 22 = 5x 25=5x25 = 5x

    Divide by 5:

    x=255x = \frac{25}{5} x=5x = 5 x=5\boxed{x = 5}
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