A pen costs Ksh 4 more than a pencil. Kering purchased (2x-1) pens for Ksh 180 and Naraya purchased (3x+1) pencils for Ksh 200. Find the cost of a pencil.

Multiply the entire equation by $(2x-1)(3x+1)$ to eliminate denominators.

$$180(3x+1) = 200(2x-1) + 4(2x-1)(3x+1)$$ $$540x + 180 = 400x - 200 + 4(6x^2 + 2x - 3x - 1)$$ $$540x + 180 = 400x - 200 + 4(6x^2 - x - 1)$$ $$540x + 180 = 400x - 200 + 24x^2 - 4x - 4$$

Rearrange into a standard quadratic form $ax^2 + bx + c = 0$:

$$0 = 24x^2 + (400x - 4x - 540x) + (-200 - 4 - 180)$$ $$0 = 24x^2 - 144x - 384$$

Divide the entire equation by 24:

$$\frac{24x^2}{24} - \frac{144x}{24} - \frac{384}{24} = 0$$ $$x^2 - 6x - 16 = 0$$

We need two numbers that multiply to -16 and add to -6. These numbers are -8 and 2.

$$(x-8)(x+2) = 0$$

This gives two possible solutions for $x$:
$x-8 = 0 \implies x = 8$
$x+2 = 0 \implies x = -2$

Since $x$ represents the price of a pencil, it must be a positive value. Also, the number of items purchased $(2x-1)$ and $(3x+1)$ must be positive. If $x=-2$, the price would be negative, and the number of pens would be $2(-2)-1 = -5$, which is not possible.
Therefore, the valid value is $x = 8$.

The value of $x$ is $\boxed{8}$.

Question 30(c):

From part (b), $x=8$.
Original price of one pencil = $x = 8$ Ksh.
Original price of one pen = $x+4 = 8+4 = 12$ Ksh.

The price of a pen went up by 25%.
New price of one pen = $12 \times (1 + 0.25) = 12 \times 1.25 = 15$ Ksh.
The price of a pencil remained unchanged.
New price of one pencil = 8 Ksh.

Let $N_P$ be the number of pens bought by the school.
Let $N_{Pc}$ be the number of pencils bought by the school.

Condition 1: The school spent the same amount of money on pens as on pencils.

$$N_P \times (\text{New price of pen}) = N_{Pc} \times (\text{New price of pencil})$$ $$15N_P = 8N_{Pc} \quad \text{(Equation 1)}$$

Condition 2: The total number of both pens and pencils bought was 46.

$$N_P + N_{Pc} = 46 \quad \text{(Equation 2)}$$

From Equation 2, express $N_{Pc}$ in terms of $N_P$:

$$N_{Pc} = 46 - N_P$$

Substitute this into Equation 1:

$$15N_P = 8(46 - N_P)$$ $$15N_P = 368 - 8N_P$$ $$15N_P + 8N_P = 368$$ $$23N_P = 368$$ $$N_P = \frac{368}{23}$$ $$N_P = 16$$

The number of pens bought by the school is $\boxed{16}$.