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NCERT Solutions for Class 7th Maths Chapter 6 Tiling grids with 2 × 1 tiles — In-text Questions

Book page 158 Updated on2026-09-19

Q1.
Complete the justification. (that a 5 × 7 grid cannot be tiled by 2 × 1 tiles — the grid has 35 unit squares and each tile covers exactly 2 unit squares)
Answer

Here is the completed argument.

Suppose the 5 × 7 grid could be tiled with n tiles.
Each tile covers exactly 2 unit squares, and tiles never overlap.
So the total number of squares covered = 2 × n, which is an even number.

The grid has 5 × 7 = 35 unit squares, and 35 is odd.
An even number can never equal an odd number.

So no such n exists ⇒ the 5 × 7 grid cannot be tiled with 2 × 1 tiles.
Why it happens: the argument is a contradiction. We assume a tiling exists, follow the consequences, and reach something impossible (even = odd). The only way out is that our assumption was wrong. Notice how strong this is — it rules out every possible arrangement at once.
Q2.
Is an m × n grid tileable with 2 × 1 tiles, if both m and n are even? If yes, come up with a general strategy to tile it.
Answer

Yes, always. Fill each column with vertical tiles.

m = number of rows (even), n = number of columns (even)
A column has m squares. Since m is even, m ÷ 2 is a whole number.
So each column takes exactly m ÷ 2 vertical tiles.

Total tiles = (m ÷ 2) × n = mn ÷ 2
Check: total squares = mn, and mn ÷ 2 tiles × 2 = mn ✓

Example. A 6 × 8 grid: each column of 6 takes 3 vertical tiles; 3 × 8 = 24 tiles; 24 × 2 = 48 = 6 × 8 ✓

Why it happens: when the number of rows is even, a column is an exact stack of tiles — nothing is left over at the bottom. Repeating that for every column covers the whole rectangle. The strategy does not depend on how large m and n are, which is why it counts as a general strategy.
Tip: you could equally fill each row with horizontal tiles, since n is even too. With both sides even you have both options.
Q3.
Is an m × n grid tileable with 2 × 1 tiles, if one of m and n is even and the other is odd? If yes, come up with a general strategy to tile it.
Answer

Yes. Tile along the even direction.

Say m (rows) is even and n (columns) is odd.
Each column has m squares, and m is even
⇒ each column takes m ÷ 2 vertical tiles, exactly
Total tiles = (m ÷ 2) × n = mn ÷ 2

If instead n is even and m odd, use horizontal tiles along each row:
each row takes n ÷ 2 tiles ⇒ total = m × (n ÷ 2) = mn ÷ 2

Example. 4 × 7: rows = 4 (even) ⇒ 2 vertical tiles per column × 7 columns = 14 tiles. 14 × 2 = 28 = 4 × 7 ✓

Why it happens: only one even side is needed. The even side tells you which way to lay the tiles; the odd side just tells you how many lines of tiles there are. Note that mn is even whenever one of m, n is even, so the count of squares is even too — the necessary condition and the construction agree.
Q4.
Is an m × n grid tileable with 2 × 1 tiles, if both m and n are odd? Give reasons.
Answer

No — never.

m is odd and n is odd
odd × odd = odd
So the number of unit squares mn is odd

Each tile covers 2 squares, so any number of tiles covers an even number of squares
even ≠ odd
⇒ tiling is impossible

Examples. 3 × 3 = 9, 5 × 7 = 35, 7 × 9 = 63 — all odd, all impossible.

Why it happens: the product of two odd numbers is always odd. (5 × 7 = 35; 3 × 3 = 9.) And a heap of dominoes always covers an even number of cells. So the two counts can never match. This is a complete answer for the whole family of grids at once — we never have to look at a single arrangement.
Summary of all three cases: an m × n grid is tileable by 2 × 1 tiles exactly when mn is even, that is, when at least one of m and n is even.
Q5.
Here is a 5 × 3 grid, with a unit square removed. Now, it has an even number of unit squares. Is it tileable with 2 × 1 tiles?
Answer

Yes. With the top-right corner square removed, 14 squares are left and 7 tiles cover them.

5 × 3 = 15 squares; remove 1 ⇒ 14 squares
14 ÷ 2 = 7 tiles

A tiling. The top row now has only 2 squares — cover them with one horizontal tile. What is left is a full 4 × 3 block (rows 2 to 5). Each of its 3 columns is 4 squares tall, so it takes 2 vertical tiles. That is 6 more tiles.

7 tiles cover all 14 squares
One horizontal tile in the short top row, then two vertical tiles in each of the three columns below.
1 horizontal tile + (2 tiles × 3 columns) = 1 + 6 = 7 tiles ✓
7 × 2 = 14 squares ✓
Why it happens: removing a corner square left the top row exactly 2 squares long — a perfect fit for one tile — and the rest a clean 4 × 3 rectangle with an even number of rows. An even count of squares does not guarantee a tiling, but here an actual arrangement exists, so the answer is yes.
Q6.
Is the following region tileable with 2 × 1 tiles? (the full 5 × 3 grid)
Answer

No. This is the complete 5 × 3 grid, with nothing removed.

Number of unit squares = 5 × 3 = 15
Both 5 and 3 are odd, so their product is odd
Any number of 2 × 1 tiles covers an even number of squares
15 is odd ⇒ not tileable
Why it happens: even if you fill 14 squares neatly with 7 tiles, one square is always left uncovered, and no tile can sit on a single square. The parity of the count decides the matter before you place a single tile.
Tip: notice the pattern. An even number of squares is necessary for a tiling but, as the next page shows, it is not sufficient.
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