Declining sales prices and holding costs · source reference

Printed question

A hamster farm has 500 hamsters to sell. The farmer sells $k$ at the end of each week, where $k$ is a constant factor of 500 (e.g. 5,100,etc.). Price is 10 dollars in week 1 and drops by 5% each subsequent week. There is neither birth nor death.

(i) State the total number of weeks to sell all in terms of $k$. [1]

(ii) Show the total proceeds are $200k(1-0.95^{500/k})$ dollars. [3]

(iii) Rearing costs 0.50 dollars per hamster per week. Find the total rearing cost when all are sold. [3]

(iv) If other costs are negligible, find the least $k$ for profit. [2]

Printed working

(i) Total weeks $=500/k$.

(ii) Let $T_n$ denote each week’s earnings in dollars:

$$T_1=10k,\quad T_2=10k(0.95),\quad T_3=10k(0.95)^2,\quad\ldots,\quad T_n=10k(0.95)^{n-1}.$$

Total proceeds

$$S=\frac{a(1-r^n)}{1-r}=\frac{10k[1-0.95^{500/k}]}{1-0.95}=200k[1-0.95^{500/k}].$$

(iii) Let $C_n$ denote cost in week $n$:

$$\begin{aligned}C_1&=0.50(500)=250,\\C_2&=0.50(500-k)=250+(-0.5k),\\C_3&=0.50(500-2k)=250+2(-0.5k),\\C_n&=0.50[500-(n-1)k]=250+(n-1)(-0.5k),\\C_{500/k}&=0.50[500-(500/k-1)k]=250+(500/k-1)(-0.5k).\end{aligned}$$ $$\begin{aligned}\text{total cost}&=\frac{500/k}2[2(250)+(500/k-1)(-0.5k)]\\&=\frac{250}k[500-250+0.5k]=\frac{250}k[\frac{500}2+\frac12k]=\frac{125}k(500+k).\end{aligned}$$

(iv) For profit,

$$200k[1-0.95^{500/k}]>\frac{125}k(500+k),$$ $$200k[1-0.95^{500/k}]-\frac{125}k(500+k)>0.$$

GC table suggests $k\ge22$. Since $k>0$ and divides 500, least allowed $k=25$. Must sell at least 25 each week.

Teacher notes

GC: enter expression in Y=; 2nd window → TblStart:1, ΔTbl:1 → 2nd graph. Scroll down to the rows where values change from negative to positive.

Source tables and calculator displays

$k$	Profit expression
21	−139.6 <0
22	62.669 >0
23	249.31 >0

Printed answer key

(i) $500/k$; (iii) $125(500+k)/k$; (iv) 25

Mathematical review and qualifications

(i) $N=500/k$ weeks.

(ii) Proceeds are $10k\sum_{j=0}^{N-1}0.95^j=200k(1-0.95^N)$.

(iii) Weekly holding counts are $500,500-k,\ldots,k$. Cost is $0.5\cdot N(500+k)/2=\frac{125}{k}(500+k)$ dollars.

(iv) Evaluate profit $200k(1-0.95^{500/k})-125(500+k)/k$ on the allowed divisors. The smaller divisors $1,2,4,5,10,20$ all give losses; at $k=20$ profit is about −359.56 and at $k=25$ about 582.57. Thus the least permissible $k$ is 25.

Round a numerical threshold only after imposing k divides 500; an arbitrary integer such as 22 is not allowed.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2012/PJC/Prelim/I/9] Question PDF pages 5; solution PDF pages 21, 22, 23. The original files remain private QA records outside this native content.