A theatre had sold adult and child tickets in the ratio 9 : 4. On the day of the show it sold 20 more adult tickets and no more child tickets, and the ratio became 8 : 3. How many of each type had been sold before, and how many adult tickets were sold in total?

Step-by-step solution

1. 1

   Turn the first ratio into expressions with a single unknown. A ratio fixes only the proportion, so introduce a multiplier $k$:

   $\text{adult}=9k,\qquad \text{child}=4k.$

   Using two separate unknowns would leave the system underdetermined; the multiplier is what makes one equation enough.

2. 2

   Write the situation after the extra sales. Only the adult count changes:

   $\text{adult}=9k+20,\qquad \text{child}=4k.$

3. 3

   Set the new ratio equal to 8 : 3.

   $\frac{9k+20}{4k}=\frac{8}{3}.$

4. 4

   Cross-multiply and solve for k.

   $3(9k+20)=8(4k)\ \Longrightarrow\ 27k+60=32k\ \Longrightarrow\ 5k=60\ \Longrightarrow\ k=12.$

5. 5

   Convert back to actual ticket counts.

   $\text{adult before}=9(12)=108,\qquad \text{child}=4(12)=48,$
   $\text{adult after}=108+20=128.$

6. 6

   Check both ratios. Before: $108:48$, and dividing both by $12$ gives $9:4$ ✓. After: $128:48$, and dividing both by $16$ gives $8:3$ ✓. The adult share rose from $\tfrac{9}{13}$ to $\tfrac{8}{11}$ of all tickets, consistent with adding adults only.