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The following table shows part of a monthly income tax rates for a certain year. In a certain month an employee paid a net tax of Ksh. 2 200 after getting a tax relief of Ksh. 1 280. Calculate the emp

Accounting

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Answers

2 parts
a

(32,−2)\left(\frac{3}{2}, -2\right)

b

254π square units\frac{25}{4}\pi \text{ square units}

Step-by-step solution

5 steps

2. The following table shows part of a monthly income tax rates for a certain year.

Monthly taxable income (Ksh.)Tax rate (%)
0 to 11 18010
11 181 to 21 71415
21 715 to 32 24820

In a certain month an employee paid a net tax of Ksh. 2 200 after getting a tax relief of Ksh. 1 280. Calculate the employee's taxable income that month.

Step 1: Calculate the gross tax paid before tax relief.

Gross Tax=Net Tax+Tax Relief\text{Gross Tax} = \text{Net Tax} + \text{Tax Relief} Gross Tax=2200 Ksh+1280 Ksh\text{Gross Tax} = 2200 \text{ Ksh} + 1280 \text{ Ksh} Gross Tax=3480 Ksh\text{Gross Tax} = 3480 \text{ Ksh}

Step 2: Calculate the tax from the first income bracket.
The first bracket is from 0 to 11 180 Ksh at 10%.

Tax from 1st bracket=11180×10100=1118 Ksh\text{Tax from 1st bracket} = 11180 \times \frac{10}{100} = 1118 \text{ Ksh}

Step 3: Calculate the remaining tax to be paid after the first bracket.

Remaining Tax=Gross Tax−Tax from 1st bracket\text{Remaining Tax} = \text{Gross Tax} - \text{Tax from 1st bracket} Remaining Tax=3480 Ksh−1118 Ksh=2362 Ksh\text{Remaining Tax} = 3480 \text{ Ksh} - 1118 \text{ Ksh} = 2362 \text{ Ksh}

Step 4: Calculate the tax from the second income bracket.
The second bracket is from 11 181 to 21 714 Ksh at 15%.
The income range for this bracket is 21714−11180=1053421714 - 11180 = 10534 Ksh.

Tax from 2nd bracket=10534×15100=1580.10 Ksh\text{Tax from 2nd bracket} = 10534 \times \frac{15}{100} = 1580.10 \text{ Ksh}

Step 5: Calculate the remaining tax to be paid after the second bracket.

Remaining Tax=2362 Ksh−1580.10 Ksh=781.90 Ksh\text{Remaining Tax} = 2362 \text{ Ksh} - 1580.10 \text{ Ksh} = 781.90 \text{ Ksh}

Since there is still remaining tax, the employee's income extends into the third bracket.

Step 6: Calculate the income from the third bracket that generates the remaining tax.
The third bracket has a tax rate of 20%.
Let xx be the income in the third bracket.

x×20100=781.90x \times \frac{20}{100} = 781.90 x=781.90×10020x = \frac{781.90 \times 100}{20} x=7819020x = \frac{78190}{20} x=3909.50 Kshx = 3909.50 \text{ Ksh}

Step 7: Calculate the total taxable income.

Total Taxable Income=Income from 1st bracket+Income from 2nd bracket+Income from 3rd bracket\text{Total Taxable Income} = \text{Income from 1st bracket} + \text{Income from 2nd bracket} + \text{Income from 3rd bracket} Total Taxable Income=11180 Ksh+10534 Ksh+3909.50 Ksh\text{Total Taxable Income} = 11180 \text{ Ksh} + 10534 \text{ Ksh} + 3909.50 \text{ Ksh} Total Taxable Income=21714 Ksh+3909.50 Ksh\text{Total Taxable Income} = 21714 \text{ Ksh} + 3909.50 \text{ Ksh} Total Taxable Income=25623.50 Ksh\text{Total Taxable Income} = 25623.50 \text{ Ksh}

The employee's taxable income that month is:

25623.50 Ksh\boxed{25623.50 \text{ Ksh}}

13. The equation of a circle is given by x2+y2−3x+4y=0x^2 + y^2 - 3x + 4y = 0. Determine:

Part a· 3 steps

the coordinates of the centre of the circle;

  1. The general equation of a circle is x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0, where the centre is (−g,−f)(-g, -f).

  2. Compare the given equation x2+y2−3x+4y=0x^2 + y^2 - 3x + 4y = 0 with the general equation.

    2g=−3  ⟹  g=−322g = -3 \implies g = -\frac{3}{2} 2f=4  ⟹  f=22f = 4 \implies f = 2
  3. Determine the coordinates of the centre.

    Centre=(−g,−f)=(−(−32),−2)=(32,−2)\text{Centre} = (-g, -f) = \left(-\left(-\frac{3}{2}\right), -2\right) = \left(\frac{3}{2}, -2\right)

    The coordinates of the centre of the circle are:

    (32,−2)\boxed{\left(\frac{3}{2}, -2\right)}

Part b

254π square units\frac{25}{4}\pi \text{ square units}

the area of the circle in terms of π\pi.

  1. Calculate the radius rr of the circle using the formula r=g2+f2−cr = \sqrt{g^2 + f^2 - c}.

    From part (a), g=−32g = -\frac{3}{2}, f=2f = 2, and from the equation, c=0c = 0.

    r=(−32)2+(2)2−0r = \sqrt{\left(-\frac{3}{2}\right)^2 + (2)^2 - 0} r=94+4r = \sqrt{\frac{9}{4} + 4} r=94+164r = \sqrt{\frac{9}{4} + \frac{16}{4}} r=254r = \sqrt{\frac{25}{4}} r=52r = \frac{5}{2}
  2. Calculate the area of the circle using the formula A=πr2A = \pi r^2.

    A=π(52)2A = \pi \left(\frac{5}{2}\right)^2 A=π(254)A = \pi \left(\frac{25}{4}\right) A=254π square unitsA = \frac{25}{4}\pi \text{ square units}

    The area of the circle in terms of π\pi is:

    254π square units\boxed{\frac{25}{4}\pi \text{ square units}}
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