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It takes y years for N 87,300.00 to amount to N 99,522.00 at 7% per annum simple interest. Find the value of y.

Mathematics

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Answer

C

Step-by-step solution

9 steps

Here are the solutions for questions 3, 4, and 5.

Part 3· 3 steps

It takes yy years for N 87,300.00 to amount to N 99,522.00 at 7% per annum simple interest. Find the value of yy.

  1. Calculate the simple interest (I).

    The simple interest is the difference between the amount and the principal.

    I=Amount−PrincipalI = \text{Amount} - \text{Principal} I=N 99,522.00−N 87,300.00=N 12,222.00I = \text{N } 99,522.00 - \text{N } 87,300.00 = \text{N } 12,222.00
  2. Use the simple interest formula to find yy.

    The formula for simple interest is I=P×R×TI = P \times R \times T, where PP is the principal, RR is the annual interest rate (as a decimal), and TT is the time in years.
    Given P=N 87,300P = \text{N } 87,300, R=7%=0.07R = 7\% = 0.07, and I=N 12,222I = \text{N } 12,222.

    12,222=87,300×0.07×y12,222 = 87,300 \times 0.07 \times y
  3. Solve for yy.

    12,222=6111×y12,222 = 6111 \times y y=12,2226111y = \frac{12,222}{6111} y=2y = 2

    The value of yy is 2 years.

    The correct option is C.

    C. 2\boxed{\text{C. } 2}

Part 4· 3 steps

Given that 3a−2b=123a - 2b = 12 and a+5b=−13a + 5b = -13, find the value of (4a−b)(4a - b).

  1. Solve the system of linear equations for aa and bb.

    We have two equations:

    1. 3a−2b=123a - 2b = 12
    2. a+5b=−13a + 5b = -13

    From equation (2), express aa in terms of bb:

    a=−13−5ba = -13 - 5b

    Substitute this expression for aa into equation (1):

    3(−13−5b)−2b=123(-13 - 5b) - 2b = 12 −39−15b−2b=12-39 - 15b - 2b = 12 −39−17b=12-39 - 17b = 12 −17b=12+39-17b = 12 + 39 −17b=51-17b = 51 b=51−17b = \frac{51}{-17} b=−3b = -3
  2. Substitute the value of bb back into the expression for aa.

    a=−13−5(−3)a = -13 - 5(-3) a=−13+15a = -13 + 15 a=2a = 2
  3. Calculate the value of (4a−b)(4a - b).

    4a−b=4(2)−(−3)4a - b = 4(2) - (-3) 4a−b=8+34a - b = 8 + 3 4a−b=114a - b = 11

    The correct option is D.

    D. 11\boxed{\text{D. } 11}

Part 5· 3 steps

A line in the x−yx-y plane passes through the point (−9,10)(-9, 10). If the gradient of the line is 13\frac{1}{3}, find the equation of the line.

  1. Use the point-slope form of a linear equation.

    The point-slope form is y−y1=m(x−x1)y - y_1 = m(x - x_1), where (x1,y1)(x_1, y_1) is a point on the line and mm is the gradient (slope).
    Given point (x1,y1)=(−9,10)(x_1, y_1) = (-9, 10) and gradient m=13m = \frac{1}{3}.

  2. Substitute the given values into the formula.

    y−10=13(x−(−9))y - 10 = \frac{1}{3}(x - (-9)) y−10=13(x+9)y - 10 = \frac{1}{3}(x + 9)
  3. Simplify the equation to the slope-intercept form (y=mx+cy = mx + c).

    y−10=13x+13×9y - 10 = \frac{1}{3}x + \frac{1}{3} \times 9 y−10=13x+3y - 10 = \frac{1}{3}x + 3

    Add 10 to both sides of the equation:

    y=13x+3+10y = \frac{1}{3}x + 3 + 10 y=13x+13y = \frac{1}{3}x + 13

    This can also be written as y=x3+13y = \frac{x}{3} + 13.

    The correct option is C.

    C. y=x3+13\boxed{\text{C. } y = \frac{x}{3} + 13}
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