# Question 11:

## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\log_{10}S = \log_{10}(ab^t)$; $\log_{10}S = \log_{10}a + \log_{10}b^t$ | **M1** | Attempt to show reduction to linear form. Introduce logs on both sides and correctly split to sum of two terms |
| $\log_{10}S = t\log_{10}b + \log_{10}a$ | **A1** | Obtain correct equation. Condone no base. A0 for $\log_{10}bt$ unless previously seen as $t\log_{10}b$ |
| which is of the form $Y = mX + c$ | **A1** | Link to equation of straight line |

**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\log_{10}S = mt + c$; $S = 10^{mt+c}$ | **M1** | Attempt equation of straight line and attempt expression for $S$. Must use $\log_{10}S$ against $t$, base 10 |
| $S = 10^{mt} \times 10^c$ | **A1** | Correctly split into two terms |
| which is of the form $S = ab^t$ | **A1** | Link to exponential model |

## Question 11(b):

$m = \log_{10}b = 0.06$ so $b = 10^{0.06} = 1.15$ | **B1** | Link gradient of line of best fit to linear form and confirm $b \approx 1.15$; allow $m$ in range $[0.055, 0.065]$; or $\log_{10}1.15 = 0.06$ and compare to gradient

$c = \log_{10}a = 2.08$ so $a = 10^{2.08} = 120$ | **B1** | Link intercept of line of best fit to linear form and confirm $a \approx 120$; allow $c$ in range $[2.075, 2.085]$; or $\log_{10}120 = 2.08$ and compare to intercept

**[2]**

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## Question 11(c):

$S = 120 \times 1.15^7$ | **M1** | Substitute $t = 7$ into given model; soi

predicted sales are 319 items | **A1** | Conclude with integer value; accept 320 items

**[2]**

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## Question 11(d)(i):

GP with $a = 138$ and $r = 1.15$ | **B1** | State or imply sum of GP with $a$ as 120 or 138, and $r$ as 1.15; could be implied by attempt to use GP sum formula

$\dfrac{138(1-1.15^t)}{1-1.15} = 70000$ | **M1\*** | Attempt sum of GP with $a = 120$ or 138 and $r = 1.15$, related to 70000; must be correct sum formula; may have $n$ not $t$ throughout

$1.15^t = 77.087$ | **M1 dep\*** | Attempt to rearrange equation as far as $1.15^t =$; must now have $a = 138$; allow sign errors only; allow T&I as not DR

$t = 31.088\ldots$ hence 32 months | **A1** | Obtain 32 ('months not required'); if 32 given as **answer only** then allow full marks

**[4]**

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## Question 11(d)(ii):

Unlikely to be reliable as sales may not continue in same pattern as market could become saturated | **B1** | State or imply that the model is unlikely to be valid, with a sensible reason; e.g. decrease in demand, increase in competition, no values beyond $t=6$ so pattern unknown, sales likely to level off, other factors, seasonal variation

**[1]**

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