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The price of beans later went up by 5% and that of maize remained constant. Utaati bought the same quantity of beans but spent the same total amount of money as before on the two items. State the new ratio of beans to maize.

Mathematics

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here's the solution to question 19.

a) Find P−1P^{-1}.
Given matrix P=[4758]P = \begin{bmatrix} 4 & 7 \\ 5 & 8 \end{bmatrix}.

  1. Calculate the determinant of PP.

    det⁡(P)=(4×8)−(7×5)=32−35=−3\det(P) = (4 \times 8) - (7 \times 5) = 32 - 35 = -3

  2. Calculate the inverse of PP.

    The formula for the inverse of a 2×22 \times 2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} is 1ad−bc[d−b−ca]\frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.

    P−1=1−3[8−7−54]P^{-1} = \frac{1}{-3} \begin{bmatrix} 8 & -7 \\ -5 & 4 \end{bmatrix} P−1=[−837353−43]P^{-1} = \begin{bmatrix} -\frac{8}{3} & \frac{7}{3} \\ \frac{5}{3} & -\frac{4}{3} \end{bmatrix}

    The inverse of P is [−837353−43]\boxed{\begin{bmatrix} -\frac{8}{3} & \frac{7}{3} \\ \frac{5}{3} & -\frac{4}{3} \end{bmatrix}}.

    b) (i) Form a matrix equation to represent above information.
    Let BB be the cost of a bag of beans and MM be the cost of a bag of maize.
    From Utaati school: 8B+14M=99,6008B + 14M = 99,600
    From Ngosini school: 10B+16M=120,00010B + 16M = 120,000
    The matrix equation is:

    [8141016][BM]=[99,600120,000]\boxed{\begin{bmatrix} 8 & 14 \\ 10 & 16 \end{bmatrix} \begin{bmatrix} B \\ M \end{bmatrix} = \begin{bmatrix} 99,600 \\ 120,000 \end{bmatrix}}

    b) (ii) Using P−1P^{-1} in (a) above, determine the cost of a bag of beans and a bag of maize.

  3. Relate the system matrix to PP.

    The system matrix from (b)(i) is A=[8141016]A = \begin{bmatrix} 8 & 14 \\ 10 & 16 \end{bmatrix}.
    Notice that A=2[4758]=2PA = 2 \begin{bmatrix} 4 & 7 \\ 5 & 8 \end{bmatrix} = 2P.

  4. Find the inverse of the system matrix AA.

    The inverse of AA is A−1=(2P)−1=12P−1A^{-1} = (2P)^{-1} = \frac{1}{2} P^{-1}.

    A−1=12[−837353−43]=[−437656−23]A^{-1} = \frac{1}{2} \begin{bmatrix} -\frac{8}{3} & \frac{7}{3} \\ \frac{5}{3} & -\frac{4}{3} \end{bmatrix} = \begin{bmatrix} -\frac{4}{3} & \frac{7}{6} \\ \frac{5}{6} & -\frac{2}{3} \end{bmatrix}

  5. Solve for BB and MM.

    [BM]=A−1[99,600120,000]\begin{bmatrix} B \\ M \end{bmatrix} = A^{-1} \begin{bmatrix} 99,600 \\ 120,000 \end{bmatrix} [BM]=[−437656−23][99,600120,000]\begin{bmatrix} B \\ M \end{bmatrix} = \begin{bmatrix} -\frac{4}{3} & \frac{7}{6} \\ \frac{5}{6} & -\frac{2}{3} \end{bmatrix} \begin{bmatrix} 99,600 \\ 120,000 \end{bmatrix} B=(−43)(99,600)+(76)(120,000)B = \left(-\frac{4}{3}\right)(99,600) + \left(\frac{7}{6}\right)(120,000) B=−4(33,200)+7(20,000)B = -4(33,200) + 7(20,000) B=−132,800+140,000B = -132,800 + 140,000 B=7,200B = 7,200 M=(56)(99,600)+(−23)(120,000)M = \left(\frac{5}{6}\right)(99,600) + \left(-\frac{2}{3}\right)(120,000) M=5(16,600)−2(40,000)M = 5(16,600) - 2(40,000) M=83,000−80,000M = 83,000 - 80,000 M=3,000M = 3,000

    The cost of a bag of beans is Ksh. 7,200\boxed{\text{Ksh. } 7,200} and the cost of a bag of maize is Ksh. 3,000\boxed{\text{Ksh. } 3,000}.

    c) The price of beans later went up by 5% and that of maize remained constant. Utaati bought the same quantity of beans but spent the same total amount of money as before on the two items. State the new ratio of beans to maize.

  6. Calculate the new price of beans.

    Original price of beans = Ksh. 7,200.
    New price of beans (B′B') = 7,200×(1+0.05)=7,200×1.05=7,5607,200 \times (1 + 0.05) = 7,200 \times 1.05 = 7,560 Ksh.
    The price of maize (M′M') remains constant at Ksh. 3,000.

  7. Calculate the money spent on beans in the new purchase.

    Utaati bought the same quantity of beans, which is 8 bags.
    Money spent on beans = 8×B′=8×7,560=60,4808 \times B' = 8 \times 7,560 = 60,480 Ksh.

  8. Calculate the money spent on maize in the new purchase.

    Utaati spent the same total amount of money as before, which is Ksh. 99,600.
    Money spent on maize = Total money - Money spent on beans
    Money spent on maize = 99,600−60,480=39,12099,600 - 60,480 = 39,120 Ksh.

  9. Calculate the new quantity of maize purchased.

    Quantity of maize = Money spent on maizePrice of maize=39,1203,000=13.04\frac{\text{Money spent on maize}}{\text{Price of maize}} = \frac{39,120}{3,000} = 13.04 bags.

  10. State the new ratio of beans to maize (quantities).

    Quantity of beans : Quantity of maize = 8:13.048 : 13.04
    To express this as a ratio of integers, multiply both sides by 100:
    800:1304800 : 1304
    Divide both sides by their greatest common divisor, which is 8:
    800÷8:1304÷8=100:163800 \div 8 : 1304 \div 8 = 100 : 163
    The new ratio of beans to maize is 100:163\boxed{100:163}.

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