If the number of male unsubscribed viewers in Town-D is \(\frac{200}{3}\%\)…

Concept

When a table gives each category's percentage share of a whole, the whole is recovered from any single category whose absolute count is known: Total = known count ÷ (that category's share written as a decimal). Every other category's count is then its own percentage of that same total. Within any row, non-subscribers = total viewers − subscribers. And when one part exceeds another by \(p\%\), the two parts stand in the ratio \((100+p) : 100\), so a known sum divides in exactly that proportion.

Application

1. Town-E accounts for 20% of all viewers and has 1200 viewers, so the total number of viewers across the five towns = 1200 ÷ 0.20 = 6000.

2. Town-A: viewers = 12% of 6000 = 720; subscribers = 440; non-subscribers = 720 − 440 = 280.

3. Town-C: viewers = 28% of 6000 = 1680; subscribers = 880; non-subscribers = 1680 − 880 = 800.

4. Town-D: viewers = 25% of 6000 = 1500; subscribers = 700; non-subscribers = 1500 − 700 = 800.

5. In Town-D the male non-subscribers are \(\frac{200}{3}\%\) more than the female non-subscribers, so male : female = \(\left(100+\frac{200}{3}\right) : 100 = \frac{500}{3} : 100 = 5 : 3\).

6. Splitting the 800 Town-D non-subscribers in the ratio 5 : 3 gives males = 800 × 5/8 = 500 and females = 800 × 3/8 = 300.

7. Non-subscribers in Town-A and Town-C together = 280 + 800 = 1080.

8. Required ratio = 500 : 1080; dividing both terms by 20 gives 25 : 54.

Town-wise figures used

Cross-check

• 500 is indeed \(\frac{200}{3}\%\) more than 300, because \(\frac{200}{3}\%\) of 300 = 200 and 300 + 200 = 500.

• Males + females = 500 + 300 = 800, which restores the Town-D non-subscriber count found in step 4.

• The listed shares add to 12 + 15 + 28 + 25 + 20 = 100%, confirming 6000 as the whole, and 25 : 54 is in lowest terms because 25 and 54 share no common factor other than 1.