We read your photo as:

Form two equations to represent the given information and find the cost of 1 bag of each item using the matrix method.

Mathematics

ScanSolve AI Answer

Step-by-step solution

8 steps

a) i) Form two equations to represent the above information.

Let xx be the cost of a bag of maize.
Let yy be the cost of a bag of beans.

From Musamali's purchase:
Musamali bought 100 bags of maize and 80 bags of beans for a total of Ksh 694,000.

100x+80y=694000(1)100x + 80y = 694000 \quad (1)

From Cheruiyot's purchase:
Cheruiyot bought 100+20=120100 + 20 = 120 bags of maize.
Cheruiyot bought 50 bags of beans.
Cheruiyot paid Ksh 116,500 less than Musamali.
Amount paid by Cheruiyot = 694000−116500=577500694000 - 116500 = 577500 Ksh.

120x+50y=577500(2)120x + 50y = 577500 \quad (2)

The two equations are:

100x+80y=694000\boxed{100x + 80y = 694000} 120x+50y=577500\boxed{120x + 50y = 577500}

a) ii) Find the cost of 1 bag of each item using the matrix method.

  1. Write the system of equations in matrix form.

    [1008012050][xy]=[694000577500]\begin{bmatrix} 100 & 80 \\ 120 & 50 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 694000 \\ 577500 \end{bmatrix}

  2. Find the determinant of the coefficient matrix A=[1008012050]A = \begin{bmatrix} 100 & 80 \\ 120 & 50 \end{bmatrix}.

    det(A)=(100)(50)−(80)(120)=5000−9600=−4600\text{det}(A) = (100)(50) - (80)(120) = 5000 - 9600 = -4600

  3. Find the inverse of the coefficient matrix A−1A^{-1}.

    A−1=1det(A)[50−80−120100]=1−4600[50−80−120100]A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} 50 & -80 \\ -120 & 100 \end{bmatrix} = \frac{1}{-4600} \begin{bmatrix} 50 & -80 \\ -120 & 100 \end{bmatrix}

  4. Multiply A−1A^{-1} by the constant matrix B=[694000577500]\mathbf{B} = \begin{bmatrix} 694000 \\ 577500 \end{bmatrix} to find [xy]\begin{bmatrix} x \\ y \end{bmatrix}.

    [xy]=1−4600[50−80−120100][694000577500]\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-4600} \begin{bmatrix} 50 & -80 \\ -120 & 100 \end{bmatrix} \begin{bmatrix} 694000 \\ 577500 \end{bmatrix} [xy]=1−4600[(50)(694000)+(−80)(577500)(−120)(694000)+(100)(577500)]\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-4600} \begin{bmatrix} (50)(694000) + (-80)(577500) \\ (-120)(694000) + (100)(577500) \end{bmatrix} [xy]=1−4600[34700000−46200000−83280000+57750000]\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-4600} \begin{bmatrix} 34700000 - 46200000 \\ -83280000 + 57750000 \end{bmatrix} [xy]=1−4600[−11500000−25530000]\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-4600} \begin{bmatrix} -11500000 \\ -25530000 \end{bmatrix}

  5. Calculate the values of xx and yy.

    x=−11500000−4600=2500x = \frac{-11500000}{-4600} = 2500 y=−25530000−4600=5550y = \frac{-25530000}{-4600} = 5550

    The cost of 1 bag of maize is Ksh 2500.
    The cost of 1 bag of beans is Ksh 5550.

    Cost of maize (x) = Ksh 2500\boxed{\text{Cost of maize (x) = Ksh 2500}} Cost of beans (y) = Ksh 5550\boxed{\text{Cost of beans (y) = Ksh 5550}}

    b) In the subsequent month, the price of maize increased by 20% while that of beans reduced by 10%. Find the total amount he spent if he bought 40 bags of maize and 25 bags of beans.

  6. Calculate the new price of maize.

    New price of maize = 2500×(1+0.20)=2500×1.20=30002500 \times (1 + 0.20) = 2500 \times 1.20 = 3000 Ksh.

  7. Calculate the new price of beans.

    New price of beans = 5550×(1−0.10)=5550×0.90=49955550 \times (1 - 0.10) = 5550 \times 0.90 = 4995 Ksh.

  8. Calculate the total amount spent for the new purchase.

    Total amount spent = (40 bags of maize ×\times new maize price) + (25 bags of beans ×\times new beans price)
    Total amount spent = (40×3000)+(25×4995)(40 \times 3000) + (25 \times 4995)
    Total amount spent = 120000+124875120000 + 124875
    Total amount spent = 244875 Ksh.

    The total amount spent is Ksh 244875\boxed{\text{Ksh 244875}}.

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QA school bought 28 bags of maize and 15 bags of beans for a total of sh.263400. In the following…QThe cost of 2 bags of maize and 3 bags of beans is Ksh 12,500. The cost of 3 bags of maize and 5…QForm two simplified simultaneous equations from the given information about production costs in two…QForm a matrix equation to represent the information and use matrix methods to find the price of a…QRepresent the number of items bought by Juma, Tom, and Mary using a 3×2 matrix. Represent the prices…
Browse by topic:Linear Algebra
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.