Distinguish full capacity from capacity ceilings · source reference

Printed question

A small garment factory manufactures dresses, blouses and T-shirts. The production process includes cutting, sewing and packaging. The table below shows the amounts of time in minutes required to produce each piece of garment.

Garment	Cutting (min)	Sewing (min)	Packaging (min)
Dress	45	70	8
Blouse	50	60	6
T-shirt	20	20	3

The total maximum amounts of time spent on cutting, sewing and packaging are 3350, 4300 and 510 minutes per week respectively.

(i) How many pieces of each garment are produced each week for the factory to operate at full capacity? [4]

(ii) Suppose the packaging process is no longer needed and at least 20 pieces of dresses are produced each week. What is the minimum and maximum possible number of T-shirts that can be produced? Assume that the total maximum amounts of time spent on cutting and sewing remains unchanged. [3]

[modified 2016/JJC/Promo/1]

Printed working

(i) Let $x,y,z$ be the number of dresses, blouses and T-shirts produced respectively. Then

$45x+50y+20z=3350$.

$70x+60y+20z=4300$.

$8x+6y+3z=510$.

From GC, $x=30,y=20,z=50$.

Thus the factory produces 30 dresses, 20 blouses and 50 T-shirts at full capacity.

(ii) If packaging is no longer needed, we only consider

$45x+50y+20z=3350$.

$70x+60y+20z=4300$.

From GC, $x=17.5+0.25z$, $y=51.25-0.625z$.

Since $x\ge20$ and $y\ge0$, we have

$17.5+0.25z\ge20$ and $51.25-0.625z\ge0$,

i.e. $z\ge10$ and $z\le82$.

Thus, minimum possible number of T-shirts is 10, and maximum possible is 82.

Note: When $z=10,x=20,y=45$ (whole number). So $z=10$ is a possible number. When $z=82,x=38,y=0$. So $z=82$ is a possible number.

Printed answer key

(i) 30 dresses, 20 blouses, 50 T-shirts.

(ii) minimum possible is 10, maximum possible is 82.

Editorial qualification

The printed part (ii) uses full cutting and sewing capacity, hence equalities and the printed extrema 10 and 82. If the stated maximum amounts are ceilings and unused capacity is allowed, take 20 dresses and no blouses. Cutting requires $900+20z\le3350$, giving integer $z\le122$; sewing requires $1400+20z\le4300$, giving $z\le145$. Both constraints apply, so cutting is the binding limit: 0–122 T-shirts are feasible. At $z=122$, cutting uses 3340 minutes and sewing uses 3840 minutes. The earlier editorial upper bound145 checked only sewing and was incorrect. Under full-capacity equalities, whole-number production also requires $z\equiv2\pmod8$; the printed minimum 10 and maximum 82 satisfy this. The source wording leaves the equality-versus-ceiling distinction implicit.

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: EQUATIONS. Question sheet printed pages 15–19; solution sheet printed pages 1–12. Original exam attribution is retained above. Currency dollar signs are spelled out outside maths to avoid confusing currency with TeX delimiters.