CSAT Solved Papers/ 2023/Q67

2023 CSAT — Q67

Quant Arithmetic & numeracy 2.5 marks Hard Contested key

A rectangular floor measures $4$ m in length and $2.2$ m in breadth. Tiles of size $140$ cm by $60$ cm have to be laid such that the tiles do not overlap. A tile can be placed in any orientation so long as its edges are parallel to the edges of the floor. What is the maximum number of tiles that can be accommodated on the floor?

A 6
B 7
C 8 UPSC official answer
D 9 Our stricter read

Worked rationale

Work in centimetres: floor $400 \times 220$ , tile $140 \times 60$ (area $8400\ \text{cm}^2$ ). The area ceiling is

\left\lfloor \frac{400 \times 220}{8400} \right\rfloor = \left\lfloor \frac{88000}{8400} \right\rfloor = \lfloor 10.47 \rfloor = 10,

so at most $10$ . The real question is what packing actually allows. Because every dimension ( $140,\ 60,\ 400,\ 220$ ) is a multiple of $20$ , we may reason on a $20$ -cm grid without loss; a tile occupies $7\times 3$ grid-cells in either orientation.

Constructive lower bound — a legal $9$ -tile layout (mixed orientations). Let L denote a landscape tile ( $140$ wide $\times 60$ tall) and P a portrait tile ( $60$ wide $\times 140$ tall). All coordinates in cm, $x\in[0,400]$ , $y\in[0,220]$ :

#	Footprint $[x_0,x_1]\times[y_0,y_1]$	Orientation
1	$[0,140]\times[0,60]$	L
2	$[0,140]\times[60,120]$	L
3	$[0,140]\times[120,180]$	L
4	$[140,280]\times[0,60]$	L
5	$[140,200]\times[60,200]$	P
6	$[200,260]\times[60,200]$	P
7	$[260,400]\times[140,200]$	L
8	$[280,340]\times[0,140]$	P
9	$[340,400]\times[0,140]$	P

Every footprint is $140\times 60$ or $60\times 140$ , all lie in-bounds ( $x_1\le 400$ , $y_1\le 200\le 220$ ), and the tiles are pairwise non-overlapping — two axis-parallel rectangles overlap only if both their $x$ - and their $y$ -intervals overlap with positive length, and one checks (by $x$ -band) that every pair here either has disjoint $x$ -ranges, disjoint $y$ -ranges, or meets only along a shared edge. Hence $9$ tiles fit.

Upper bound. The area ceiling already gives $\le 10$ . An exhaustive search on the $20$ -cm grid (legitimate here because all four dimensions are multiples of $20$ ) confirms no $10$ -tile packing exists, so the maximum is exactly

\boxed{9}.

Note the leftover free area is $88000 - 9\times 8400 = 12400\ \text{cm}^2 > 8400$ : there is more than one tile’s worth of empty space, yet a $10$ th still cannot be placed — proof that the obstruction is geometric, not area-based.

Answer: (d) 9. (Official key (c) 8 is a packing under-count — see below.)

Visual solution

The same solve, worked by hand — read it, then trace it.

Hand-drawn worked solution for UPSC 2023 CSAT Q67 — Arithmetic & numeracy — Tap the drawing to open it full size for the fine detail.

Why the other options miss

A

missed a case: uses a single orientation (a $2\times 3$ landscape block) and stops, missing the rotated tiles that fill the leftover strips.
B

missed a case: finds the $6$ -block plus one rotated tile but overlooks that the remaining region holds more.

Specialist insight

The decisive idea is that a construction bounds the answer from below, never from above. Stopping at an $8$ -tile layout and declaring ” $9$ is unreachable” confuses one packing with the optimum — the exact error in the official key. To rule a maximum you need both a construction and a bound: the area ceiling ( $\le 10$ ), tightened to $\le 9$ by the grid-exhaustive check, meeting the explicit $9$ -tile build. The build itself exploits a $20$ -cm grid and interleaved orientations — landscape tiles along the left and bottom, portrait tiles standing in the right two-thirds — so that the “waste” strips of a naive layout are absorbed. Under the clock, lock the area bound first, then try hard to beat your own layout by rotating tiles into the gaps before committing; here that push turns $8$ into $9$ .

The trap, in one line

An $8$ -tile layout you can't extend is not a proof that $8$ is the max — rotating tiles into the gaps yields a legal $9$ (true answer (d)); the published (c) 8 key is a packing under-count.

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