A filling machine and an insufficient tail · source reference

Printed question

(i) A certain machine A pumps liquid into an empty container of total volume $850\mathrm{m}^3$. The first pumping action fills it with $4\mathrm{m}^3$. Each subsequent action fills it with $0.5\mathrm{m}^3$ more than the previous action. Pumping continues until it is completely filled. Find the overflow volume at the final pumping action. [3]

(ii) Suppose after A’s 50th action, a different machine B fills the remaining volume. B first fills $5\mathrm{m}^3$, then each subsequent action fills $\frac56$ the amount filled in the previous action.

(a) Find the total volume filled after B’s 10th action. [3]

(b) Explain whether the container would be completely filled eventually. [2]

Printed working

(i) Total volume $=850\mathrm{m}^3$.

$$S_n=\frac n2[2(4)+(n-1)0.5]\ge850\Rightarrow n^2+15n-3400\ge0.$$

Roots −66.29 and 51.29 give $n\le-66.29$ (reject, $n>0$) or $n\ge51.29$. Thus action 52 completely fills the container with some overflow.

$$S_{52}=\frac{52}2[2(4)+51(0.5)]=871,$$

overflow $=(871-850)\mathrm{m}^3=21\mathrm{m}^3$.

(ii)(a) For A, $S_{50}=\frac{50}2[2(4)+49(0.5)]=812.5$.

For B,

$$S_{10}=\frac{5[1-(5/6)^{10}]}{1-5/6}=\frac{156941}{6239}\approx25.1548\mathrm{m}^3\quad\text{as printed}.$$

Total $812.5+25.1548=837.6548\mathrm{m}^3\approx838\mathrm{m}^3$ (3 s.f.).

(ii)(b) Ratio $5/6<1$, so sum to infinity exists:

$$S_{\infty}=\frac5{1-5/6}=30.$$

Volume remaining for B $=850-812.5=37.5\mathrm{m}^3$. The theoretical maximum B adds is only $30\mathrm{m}^3<37.5\mathrm{m}^3$, so it never fills completely.

Teacher notes

AP $S_n=\frac n2[2a+(n-1)d]$, here $a=4,d=0.5$, or $\frac n2(a+u_n)$ with $u_n=4+(n-1)0.5$.

GC roots: apps → 9:PlySmlt2 → 1:Polynomial Root Finder → Order 2 → graph (Next). Key coefficients; graph (solve).

GP sum $a(1-r^n)/(1-r)$; B has $a=5,r=5/6$. Its theoretical maximum is the GP infinite sum $a/(1-r)$.

The source writes an equality to the displayed fraction $156941/6239$, a close calculator rational approximation rather than the exact finite sum: the exact ten-term sum is $\frac{30(6^{10}-5^{10})}{6^{10}}$, approximately 25.1548. The decimal and final answer are consistent with the correct formula.

Source tables and calculator displays

Pump A	Amount ($\mathrm m^3$)
1st	4
2nd	$4+0.5$
3rd	$4+2(0.5)$
…	…
$n$th	$4+(n-1)(0.5)$

Pump B	Amount ($\mathrm m^3$)
1st	5
2nd	$5(5/6)$
…	…
$n$th	$5(5/6)^{n-1}$

Source diagrams redrawn

Printed answer key

(i) $21\mathrm{m}^3$; (ii)(a) $838\mathrm{m}^3$ (3 s.f.); (b) No

Mathematical review and qualifications

(i) $S_A(n)=n[8+0.5(n-1)]/2=n(n+15)/4$. At 51 actions the volume is 841.5; at 52 it is 871. Overflow is $871-850=21\,\mathrm{m}^3$.

(ii)(a) After 50 A actions, volume is 812.5. Ten B actions add $5[1-(5/6)^{10}]/(1-5/6)$, giving total about $837.65\,\mathrm{m}^3$, or 838 (3 s.f.). (b) B adds at most $5/(1-5/6)=30$, so total remains below 842.5, itself below 850. It never fills completely.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2013/CJC/Prelim/II/1] Question PDF pages 3; solution PDF pages 15, 16. The original files remain private QA records outside this native content.