NCERT Solutions for Class 7th Maths Chapter 6 The black-and-white colouring argument — In-text Questions
Book page 159 – 160 Updated on2026-09-19
Q1.
What about this one?
Fig. 6.13, page 159.
Answer
It cannot be tiled — even though it has an even number of squares.
5 × 3 = 15 squares; remove 1 ⇒ 14 squares
14 is even, so 7 tiles would be needed
But no arrangement of 7 tiles works. Try it — one square is always stranded.
Counting alone cannot settle this, because the count passes the test. Something finer is needed: the black-and-white colouring of the next question.
Why it happens: when the middle square of the top row is removed, the two remaining top squares are cut off from each other — the only way to reach either of them is downwards. Follow the forced moves and you will always end up with a lone square somewhere. The colouring argument turns this feeling into a proof.
Tip: an even number of squares is only the first test. Passing it means "maybe"; failing it means "definitely not".
Q2.
Were you able to tile this? How can we be sure that this is not tileable? Can you find another unit square that, when removed from a 5 × 3 grid, makes it non-tileable?
Fig. 6.13, page 159 — the region to be tiled.
Answer
Colour the grid like a chessboard and count the two colours. If the counts differ, no tiling can exist.
Fig. 6.13 coloured like a chessboard. The counts are 8 and 6 — not equal — so no domino tiling can exist.
Colour so that neighbours always differ.
Every 2 × 1 tile covers two neighbouring squares
⇒ every tile covers exactly one white and one black square
⇒ a tiled region must have equal numbers of white and black squares
Fig. 6.13 has 8 white and 6 black
8 ≠ 6 ⇒ not tileable
Another square that spoils it. In a full 5 × 3 grid the four corners are the same colour, and that colour has 8 squares while the other has 7. Removing a square of the smaller colour class leaves 8 and 6 again. So the middle square of the bottom row also makes it non-tileable — and so does the centre square, or either end square of the middle row.
Square removed from the 5 × 3 grid
Colours left
Tileable?
A corner square
7 and 7
Yes
Middle of the top row (Fig. 6.13)
8 and 6
No
Middle of the bottom row
8 and 6
No
The centre square
8 and 6
No
Either end of the middle row
8 and 6
No
Why it happens: in an odd-by-odd grid the two colours are not balanced to begin with — one colour has one square more. Removing a square of the majority colour restores the balance; removing a square of the minority colour makes the gap two, and no set of dominoes can cover a region whose colours differ.
Q3.
If the plain grid is tileable, is the black-and-white-grid tileable?Fig. 6.14, page 159 — the same region shown as a plain grid and as a black-and-white grid, with the 2 × 1 tiles used in each case.
Answer
Yes. Take the tiling of the plain grid and just look at it in colour.
Suppose the plain region is tiled by 2 × 1 tiles.
Colour the region like a chessboard.
Each tile lies on two neighbouring squares, which have different colours.
So each tile already sits on one white square and one black square.
Give that tile the matching black-and-white colouring, and it fits.
⇒ the black-and-white grid is tileable.
Why it happens: nothing has to move. The colouring is drawn on top of an existing tiling, and a domino automatically straddles two neighbouring cells, which are always of opposite colours. So the colour rule is satisfied for free.
Q4.
If the black-and-white grid is tileable, is the plain grid tileable?Fig. 6.14, page 159 — the same region shown as a plain grid and as a black-and-white grid, with the 2 × 1 tiles used in each case.
Answer
Yes. Rub the colours out and the same arrangement is a tiling of the plain grid.
A tiling of the coloured grid is a set of 2 × 1 tiles
placed with no gaps and no overlaps.
Erasing the colours changes nothing about where the tiles are.
⇒ the same placement tiles the plain grid.
Why it happens: the colouring adds an extra rule, so the coloured problem is harder, not easier. Anything that solves the harder problem solves the easier one too. Together with the previous question this shows the two problems are equivalent — which is what lets us swap freely between them.
Tip: that equivalence is the whole point of the trick. We may study whichever version is easier — and the coloured one is easier, because it can be settled just by counting.
Q5.
Is the black-and-white region in Fig. 6.14 tileable? Any region tiled with black-and-white-tiles must have an equal number of black tiles and white tiles.
Fig. 6.14, page 159 — the region to be tiled, the same region coloured black and white, and the tiles.
Answer
No. The colours do not balance.
Fig. 6.14 has 8 white squares and 6 black squares
Total = 8 + 6 = 14 ✓ (the 5 × 3 grid with one square removed)
Each black-and-white tile covers exactly 1 white and 1 black square
7 tiles would cover 7 white and 7 black squares
But the region offers 8 white and 6 black
8 ≠ 7 and 6 ≠ 7
⇒ it can never be tiled
Why it happens: the mismatch is two squares — there is one extra white and one missing black. Even a single tile cannot be laid on two whites, so those two extra whites can never be paired off. This proves the plain Fig. 6.13 is untileable as well, because the two problems are equivalent.
Did you know? Making a problem more complicated — by adding colours — made it easier to settle. Mathematicians use this trick often: find an extra quantity that every legal move must preserve, then show the target does not preserve it.
Q6.
Use this idea to find another unit square that, when removed from a 5 × 3 grid, makes it non-tileable?
Answer
Remove any square of the minority colour — for example the middle square of the bottom row.
Colour the 5 × 3 grid like a chessboard, corners white.
Number of white squares = 8 · number of black squares = 7
(8 + 7 = 15 ✓)
Remove one black square ⇒ 8 white and 6 black remain
8 ≠ 6 ⇒ not tileable
Remove one white square ⇒ 7 white and 7 black remain
7 = 7 ⇒ the count test is passed (and a tiling does exist)
The 7 black squares are: the middle square of the top row (that is Fig. 6.13), the two end squares of the second row, the middle square of the third row, the two end squares of the fourth row, and the middle square of the bottom row. Removing any one of these makes the grid untileable.
Why it happens: in a 5 × 3 grid both sides are odd, so the two colours cannot be equal — one has 8 and the other 7. Taking away a square of the 8-colour evens things up; taking away a square of the 7-colour makes the difference 2, and a difference of even 1 is already fatal.
Check it yourself: colour a 5 × 3 grid in your notebook and count. You should get 8 of the corner colour and 7 of the other.