Question 6
Two online newspapers, X and Y, record their number of paying subscribers at the end of each month.
- Newspaper X had 18 000 subscribers at the end of month 1 and 20 225 subscribers at the end of month 3. Its number of subscribers grows geometrically.
- Newspaper Y had 30 000 subscribers at the end of month 1 and 33 480 subscribers at the end of month 5. Its number of subscribers grows arithmetically.
Newspaper X charges each subscriber $28 per month and Newspaper Y charges each subscriber $24 per month.
Determine the difference between the two newspapers’ monthly subscription income at the end of the first month in which Newspaper X has more subscribers than Newspaper Y.
Two ratios, four differences, then month by month
X starts lower and grows by a constant ratio while Y adds a constant difference, so the months are compared until X first has more subscribers.
Slide through the months
Move the slider to compare X and Y one month at a time. The three presets load three slips that cost marks.
This preset marks month 12. The income of X first passes the income of Y there ($956 732 against $949 680), while X still has fewer subscribers (34 169 against 39 570). The question compares subscribers, not incomes.
This preset uses $r = 20\,225 \div 18\,000 = 1.12361$, as if month 3 came straight after month 1. X then passes Y at month 7 (36 222 against 35 220), and the difference comes out at $168 936. The slider now starts at month 1.
This preset uses $d = 3480 \div 5 = 696$, but month 5 is only four differences on from month 1. X then passes Y at month 15 (40 696 against 39 744), and the difference comes out at $185 632.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Newspaper X grows geometrically, so use $t_n = t_1 r^{n-1}$. Month 3 is two ratios on from month 1, so solve for $r^2$ first.
$$t_3 = 18\,000\,r^{2} = 20\,225$$
$$r^2 = \frac{20\,225}{18\,000} = 1.12361$$
$$r = \sqrt{1.12361} = 1.060005 \approx 1.06$$
$$X_n = 18\,000(1.06)^{n-1}$$
$r = 1.060005$ is 6% growth per month to 4 decimal places, so use $r = 1.06$. This moves $X_{16}$ by only 3 subscribers and does not change the month in which X passes Y.
This mark may be implied by subsequent working.
Marker: correctly determines the common ratio for Newspaper X
Newspaper Y grows arithmetically, so use $t_n = t_1 + (n-1)d$. Month 5 is four differences on from month 1.
$$t_5 = 30\,000 + 4d = 33\,480$$
$$4d = 3480 \qquad d = 870$$
$$Y_n = 30\,000 + 870(n-1)$$
Marker: correctly determines the common difference for Newspaper Y
Substitute values of $n$ until $X_n$ first exceeds $Y_n$. Subscribers are whole people, so round $X_n$ to the nearest whole number.
| $n$ | $X_n$ | $Y_n$ |
|---|---|---|
| 10 | 30 411 | 37 830 |
| 14 | 38 393 | 41 310 |
| 15 | 40 696 | 42 180 |
| 16 | 43 138 | 43 050 |
X starts lower but grows by 6% each month. At $n = 10$, X is still 7419 behind and closes the gap by about 950 to 1450 a month, so jump to $n = 14$ and then step one month at a time.
Key 18000 × 1.06, then the power key ($x^y$ or ^) and the exponent $n - 1$. For example, 18000 × 1.06 ^ 13 gives $X_{14}$ for $n = 14$.
An alternative method also earns this mark, for example listing the terms of each sequence, a graphical method, or using $t_n = rt_{n-1}$ and $t_n = t_{n-1} + d$.
Marker: uses an appropriate method to compare the numbers of subscribers
At month 15, X is still behind ($40\,696 < 42\,180$). At month 16, X is ahead ($43\,138 > 43\,050$).
Newspaper X first has more subscribers than Newspaper Y at the end of month 16.
Follow-through applies from an error in earlier working.
Marker: determines the first month in which Newspaper X has more subscribers
Monthly income is the number of subscribers times the monthly price, both at month 16.
$$\text{Income X} = 43\,138 \times \text{\textdollar}28 = \text{\textdollar}1\,207\,864$$
$$\text{Income Y} = 43\,050 \times \text{\textdollar}24 = \text{\textdollar}1\,033\,200$$
$$\begin{aligned}\text{Difference} &= \text{\textdollar}1\,207\,864 - \text{\textdollar}1\,033\,200\\ &= \text{\textdollar}174\,664\end{aligned}$$
A value from $174 660 to $174 760, from rounding the subscribers or $r$, also earns this mark.
Marker: determines the difference in monthly subscription income
Set the response out in order: both models, the comparison, the month, then the incomes. End with a statement in context.
Newspaper X first has more subscribers than Newspaper Y at the end of month 16, with 43 138 subscribers against 43 050. At the end of month 16, Newspaper X’s monthly subscription income is $174 664 more than Newspaper Y’s.
Logical organisation includes a clear flow of the solution using appropriate mathematical terminology, symbols, conventions and representations.
Marker: shows logical organisation
If I had to guess where you might lose marks…
- Using $r = 20\,225 \div 18\,000 = 1.12361$ (no square root, as if month 3 came straight after month 1). X then passes Y at month 7 (36 222 against 35 220) and the difference comes out at $168 936. This loses Mark 1.
- Dividing 3480 by 5 instead of 4 gives $d = 696$. X then passes Y at month 15 (40 696 against 39 744) and the difference comes out at $185 632. This loses Mark 2.
- Using the exponent $n$ instead of $n - 1$, so $X_{15} = 18\,000(1.06)^{15} = 43\,138$. X then appears to pass Y at month 15 (against 42 180) and the difference comes out at $195 544. This loses Mark 4.
- Comparing incomes instead of subscribers. The income of X first passes the income of Y at month 12 ($956 732 against $949 680), but X still has fewer subscribers then (34 169 against 39 570). The difference comes out at $7052. This loses Mark 4.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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