KCSE 2025 Maths P1 Q20 — Matrices: Transport, Cost and Profit

KCSE 2025 Form 4 Algebra

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The Question

“A transporter was contracted to move 133 tonnes of sand from site A to site B using two lorries, a 7-tonne lorry and a 14-tonne lorry. The 7-tonne lorry costs Ksh 3,000 per trip and the 14-tonne lorry costs Ksh 4,000 per trip, and the total cost was Ksh 47,000. The 7-tonne lorry made x trips and the 14-tonne lorry made y trips. (a) Write two equations to represent the information. (b) Use the matrix method to solve the equations. (c) The transporter was paid Ksh 500 per tonne delivered; calculate the profit made.”

1

Form the two equations

Each trip of the 7-tonne lorry carries 7 tonnes, so x trips carry 7x tonnes; likewise the 14-tonne lorry carries 14y tonnes. Together they move all 133 tonnes, which gives the tonnage equation. The cost works the same way: 3,000 shillings per trip of the small lorry and 4,000 per trip of the big one add up to 47,000. Dividing each equation by a common factor keeps the numbers small and easy to handle.

7x+14y=133⇒x+2y=19(equation 1)7x + 14y = 133 \Rightarrow x + 2y = 19 \quad \text{(equation 1)}
3000x+4000y=47000⇒3x+4y=47(equation 2)3000x + 4000y = 47000 \Rightarrow 3x + 4y = 47 \quad \text{(equation 2)}
2

Write the equations as a matrix equation

Collect the coefficients of x and y into a 2 by 2 matrix, the unknowns into a column, and the constants into another column. This lets us solve the pair using a single inverse matrix instead of elimination or substitution.

(1234)(xy)=(1947)\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 19 \\ 47 \end{pmatrix}
3

Find the determinant

For a 2 by 2 coefficient matrix, the determinant is the product of the main diagonal minus the product of the other diagonal. It must be non-zero for the inverse to exist, and here it is minus two.

det⁡=(1)(4)−(2)(3)=4−6=−2\det = (1)(4) - (2)(3) = 4 - 6 = -2
4

Multiply both sides by the inverse matrix

The inverse of a 2 by 2 matrix swaps the main-diagonal entries, negates the other two, and divides everything by the determinant. Multiplying the constants column by this inverse gives the values of x and y directly.

(xy)=1−2(4−2−31)(1947)\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} \begin{pmatrix} 19 \\ 47 \end{pmatrix}
=1−2((4)(19)+(−2)(47)(−3)(19)+(1)(47))=1−2(−18−10)= \frac{1}{-2} \begin{pmatrix} (4)(19) + (-2)(47) \\ (-3)(19) + (1)(47) \end{pmatrix} = \frac{1}{-2} \begin{pmatrix} -18 \\ -10 \end{pmatrix}
x=9,y=5x = 9, \quad y = 5
5

Calculate the profit

The transporter is paid 500 shillings for every tonne delivered, and 133 tonnes were delivered, so the income is the payment rate times the tonnage. Profit is what remains after subtracting the transport cost of 47,000 shillings that we were given.

Income=500×133=66500\text{Income} = 500 \times 133 = 66500
Profit=66500−47000=19500\text{Profit} = 66500 - 47000 = 19500

Final Result

The 7-tonne lorry made 9 trips and the 14-tonne lorry made 5 trips (x = 9, y = 5), and the transporter made a profit of Ksh 19,500.

Why this method works

The matrix method works because a system of two linear equations can be written as a single equation A times the unknown vector equals the constant vector. As long as the coefficient matrix A has a non-zero determinant, it has an inverse, and multiplying both sides by that inverse isolates the unknown vector exactly the way dividing isolates a variable in ordinary algebra. Because the determinant here is minus two, the inverse exists and delivers one unique pair of values for x and y.

✓

Substitute back: 7(9) + 14(5) = 63 + 70 = 133 tonnes, and 3000(9) + 4000(5) = 27000 + 20000 = 47000 shillings, both correct.