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Ebimo deposited an amount of 20a6150,000.00 at a bank with an annual simple interest rate of 5% for 5 years. What is the interest value.

Mathematics

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1. Ebimo deposited an amount of N150,000.00\text{N}150,000.00 at a bank with an annual simple interest rate of 5%5\% for 5 years. What is the interest value?

  1. Identify the given values.

    Principal (PP) =N150,000= \text{N}150,000
    Rate (RR) =5%=5100=0.05= 5\% = \frac{5}{100} = 0.05
    Time (TT) =5= 5 years

  2. Use the simple interest formula I=P×R×TI = P \times R \times T.

    I=150,000×0.05×5I = 150,000 \times 0.05 \times 5
  3. Calculate the interest.

    I=150,000×0.25I = 150,000 \times 0.25 I=37,500I = 37,500

    The interest value is N37,500\text{N}37,500.

    Answer: N37,500\boxed{\text{N}37,500}

    2. The area of a rectangular shop is 2423m224\frac{2}{3} \text{m}^2. If the width of the shop is 313m3\frac{1}{3} \text{m}, what is the length of the shop?

  4. Convert the mixed numbers to improper fractions.

    Area (AA) =2423=(24×3)+23=72+23=743m2= 24\frac{2}{3} = \frac{(24 \times 3) + 2}{3} = \frac{72 + 2}{3} = \frac{74}{3} \text{m}^2
    Width (WW) =313=(3×3)+13=9+13=103m= 3\frac{1}{3} = \frac{(3 \times 3) + 1}{3} = \frac{9 + 1}{3} = \frac{10}{3} \text{m}

  5. Use the formula for the area of a rectangle, A=L×WA = L \times W, and rearrange to find the length (LL).

    L=AWL = \frac{A}{W}
  6. Substitute the values and calculate the length.

    L=743103L = \frac{\frac{74}{3}}{\frac{10}{3}} L=743×310L = \frac{74}{3} \times \frac{3}{10} L=7410L = \frac{74}{10} L=375L = \frac{37}{5} L=725 mL = 7\frac{2}{5} \text{ m}

    The length of the shop is 725 m7\frac{2}{5} \text{ m}.

    Answer: 725 m\boxed{7\frac{2}{5} \text{ m}}

    3. If the number 200 is 125%125\% of 160, then what percent of 200 is 160?

  7. Verify the given statement.

    125%125\% of 160=125100×160=1.25×160=200160 = \frac{125}{100} \times 160 = 1.25 \times 160 = 200. The statement is true.

  8. Set up an equation to find what percent of 200 is 160.

    Let the unknown percentage be xx.

    x% of 200=160x\% \text{ of } 200 = 160 x100×200=160\frac{x}{100} \times 200 = 160
  9. Solve for xx.

    2x=1602x = 160 x=1602x = \frac{160}{2} x=80x = 80

    So, 80%80\% of 200 is 160.

    Answer: 80%\boxed{80\%}

    4. A student thinks of a number. He multiplies this number by 4, subtracts 20 from the result, and the answer is 36. What is the number?

  10. Let the unknown number be xx.

  11. Translate the problem into an algebraic equation.

    Multiplying the number by 4 gives 4x4x.
    Subtracting 20 from the result gives 4x−204x - 20.
    The answer is 36, so:

    4x−20=364x - 20 = 36
  12. Solve the equation for xx.

    Add 20 to both sides of the equation:

    4x=36+204x = 36 + 20 4x=564x = 56
  13. Divide both sides by 4

    x=564x = \frac{56}{4} x=14x = 14

    The number is 14.

    Answer: 14\boxed{14}

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