(i) ₹ 6,400 were divided equally among x persons. Had this money been divided equally among (x + 14) persons, each would have got ₹ 28 less. Find the value of x.

(ii) ₹ 7,500 were divided equally among a certain number of children. Had there been 20 less children, each would have received ₹ 100 more. Find the original number of children.

(i) Given,

In first case :

₹ 6400 were divided equally among x persons

Money each person receives = $\dfrac{6400}{x}$

In second case:

If ₹ 6400 is divided among x + 14 persons, then each gets = $\dfrac{6400}{x + 14}$

In second case each person receives ₹ 28 less.

$\Rightarrow \dfrac{6400}{x} - \dfrac{6400}{x + 14} = 28 \\[1em] \Rightarrow \dfrac{6400(x + 14) - 6400x}{x(x + 14)} = 28 \\[1em] \Rightarrow \dfrac{6400x + 14 \times 6400 - 6400x}{x^2 + 14x} = 28 \\[1em] \Rightarrow 14 \times 6400 = 28(x^2 + 14x) \\[1em] \Rightarrow \dfrac{14 \times 6400}{28} = x^2 + 14x \\[1em] \Rightarrow 3200 = x^2 + 14x \\[1em] \Rightarrow x^2 + 14x - 3200 = 0 \\[1em] \Rightarrow x^2 + 64x - 50x - 3200 = 0 \\[1em] \Rightarrow x(x + 64) - 50(x + 64) = 0 \\[1em] \Rightarrow (x - 50)(x + 64) = 0 \\[1em] \Rightarrow (x - 50) = 0 \text{ or } (x + 64) = 0 \text{…..[Using zero-product rule]} \\[1em] \Rightarrow x = 50 \text{ or } x = -64.$

Since, the number of persons cannot be negative.

Thus, x = 50.

Hence, value of x = 50.

(ii) Given,

Let original number of children be x.

In first case:

₹ 7,500 is divided among x children.

Money each child receives = ₹ $\dfrac{7500}{x}$

In second case:

If ₹ 7500 is divided among (x - 20) children, then each gets = ₹ $\dfrac{7500}{x - 20}$

In second case, each student will receive ₹ 100 more.

$\Rightarrow \dfrac{7500}{x - 20} - \dfrac{7500}{x} = 100 \\[1em] \Rightarrow \dfrac{7500x - 7500(x - 20)}{x(x - 20)} = 100 \\[1em] \Rightarrow \dfrac{7500x - 7500x + 7500 \times 20}{x^2 - 20x} = 100 \\[1em] \Rightarrow 7500 \times 20 = 100(x^2 - 20x) \\[1em] \Rightarrow \dfrac{7500 \times 20}{100} = x^2 - 20x \\[1em] \Rightarrow 1500 = x^2 - 20x \\[1em] \Rightarrow x^2 - 20x - 1500 = 0 \\[1em] \Rightarrow x^2 + 30x - 50x - 1500 = 0 \\[1em] \Rightarrow x(x + 30) - 50(x + 30) = 0 \\[1em] \Rightarrow (x - 50)(x + 30) = 0 \\[1em] \Rightarrow (x - 50) = 0 \text{ or } (x + 30) = 0 \text{….[Using zero-product rule]} \\[1em] \Rightarrow x = 50 \text{ or } x = -30.$

Since, the number of children cannot be negative.

Thus, x = 50.

Hence, the original number of children = 50.