NCERT Solutions for Class 8 Maths Chapter 13 Algebra Play (PDF) – 2027

Yes — the answer is 2 every single time. Try three very different starting numbers:

Trying numbers shows that it works. To see why, do the same five steps to a letter-number x:

Change only step 3 — the number you add. Add 6 to finish at 3, and add 10 to finish at 5.

Yes. The recipe is simple: write the expression next to every step, and make sure the last step removes every x that is left.

A longer one, with a detour that changes nothing — it still ends at 10:

He undid the steps with algebra. Let the month be M and the day be D, and follow the instructions on the letter-numbers:

Mukta’s answer was 291, so

She started with 25 December (25/12). Subtract 165 and split off the last two digits as the day.

Subtract 165 from each answer, then read the last two digits as the day and the rest as the month.

Yes. The numbers you multiply the month by must have product 100; everything added along the way piles up into one constant, and that is the number to subtract at the end.

Here is a shorter trick. Multiply the month by 5, add 3, multiply by 20, then add the day:

Design it backwards: decide what expression you want at each stage, then write the instruction that produces it.

A trick that always ends at 7:

A trick that hands the number back:

Each box is the sum of the two boxes directly below it, so work upwards, adding each neighbouring pair.

(i) 6 + 2 = 8.

(ii) 3 + 4 = 7 and 4 + 3 = 7, then 7 + 7 = 14.

(iii) 5 + 4 = 9, 4 + 5 = 9, 5 + 0 = 5; then 9 + 9 = 18 and 9 + 5 = 14; finally 18 + 14 = 32.

Here the bottom row is not given, so read the rule backwards. “A box is the sum of the two below it” also says “a missing box below is the box above minus its known neighbour”.

So the pyramid is: bottom 1, 3, 3; middle 4, 6; top 10.

No box has both the numbers it needs, so name the empty boxes a, b, c and write equations.

Call the missing bottom entries x and y, write what the given boxes say, and solve.

(i) Bottom row 4, x, 6, y.

(ii) Bottom row 5, x, 7, y.

(iii) Bottom row 3, 5, x, y.

The top is always a fixed combination of the bottom row. Each bottom number is counted as many times as there are paths from it up to the top, so the end numbers count once and the middle ones more often.

With bottom row a, b, c the middle row is a + b and b + c, so the top is their sum:

Use the three-row formula top = a + 2b + c, doubling only the middle entry.

With bottom row a, b, c, d:

Use top = a + 3b + 3c + d.

The first three Virahāṅka-Fibonacci numbers are 1, 2, 3.

The numbers appearing are 1, 2, 3, 3, 5, 8. The number at the top is 8. Yes — every one of them is a Virahāṅka-Fibonacci number, and 8 is the 5th one in the sequence 1, 2, 3, 5, 8.

(i) The bottom row is 1, 2, 3, 5.

All the numbers are Virahāṅka-Fibonacci numbers; each row is a run of consecutive ones starting two places later than the row below. The top is 21, the 7th Virahāṅka-Fibonacci number.

(ii) Again every number is a Virahāṅka-Fibonacci number, each row starting two places later. After 28 steps up, the top is the (1 + 2 × 28) = 57th Virahāṅka-Fibonacci number:

Every number in the pyramid is a Virahāṅka-Fibonacci number: each row is a run of consecutive ones, one number shorter than the row below and starting two places later. Going up n – 1 rows moves the start from position 1 to 1 + 2(n – 1), so the number at the top is the (2n – 1)th Virahāṅka-Fibonacci number.

Yes — the sum decides all four numbers. Let a be the top-left date of the 2 × 2 square.

The example on the page checks out: for 6, 7, 13, 14 the sum is 40, and (40 – 16) ÷ 4 = 6 — the top-left date.

Yes. Solve 4a + 16 = 36.

Sample answer — a 3 × 3 square trick. Write every cell in terms of one letter-number. Let c be the middle date of any 3 × 3 square on the calendar:

Try it on 11, 12, 13 / 18, 19, 20 / 25, 26, 27: the sum is 171, and 171 ÷ 9 = 19, the middle date. ✓

Read each row as an equation and start with the row that has only one kind of shape.

So the blue square is 9 and the red circle is 5.

Grid 1: ■ = 11, ● = 5; the empty square is 21.

Grid 2: ● = 4, ◆ = 7; row 3 sum is 15.

The largest is 32 × 5 = 160. The six possible arrangements are:

So the largest digit, 5, is the multiplier and the other two digits form 32, in decreasing order.

Compare in pairs, grouping by the multiplier.

Yes, always. Let the digits be p < q < r. Group the six products by multiplier:

In each pair the multiplicand with the larger tens digit wins, leaving rq × p, rp × q and qp × r.

So the largest product is qp × r = (10q + p) × r: the largest digit is the multiplier and the other two are in decreasing order.

Use the largest digit, 7, as the multiplier and arrange the other two in decreasing order to form 31. Here p = 1, q = 3, r = 7.

Use the largest digit, 9, as the multiplier and arrange the other two in decreasing order to form 53. Here p = 3, q = 5, r = 9.

Yes — there is never a remainder, for any starting two-digit number.

Since 9(b – a) is 9 times a whole number, dividing it by 9 always leaves remainder 0.

Exactly the same thing — you just subtract the other way round.

So in both cases the difference is 9 × (the difference of the two digits), and dividing by 9 leaves no remainder.

The quotient is the difference of the two digits (larger digit – smaller digit). For 47: 74 – 47 = 27 and 27 ÷ 9 = 3 = 7 – 4.

Yes, it is always true, and the algebra is one line.

So the sum is always a multiple of 11, and the quotient is a + b, the sum of the digits.

Multiples of 37 are 37, 74, 111, … so

So the sum is always divisible by 37, and yes, always divisible by 3 too (in fact by 111).

You get back the original three-digit number abc, every time — and none of the three divisions leaves a remainder.

Dividing by 7, then 11, then 13 peels off exactly those three factors and leaves abc.

He started with 7 flowers and placed 8 flowers in each shrine (the smallest answer). Let x be the flowers at the start and k the number placed at each shrine.

Since 8x = 7k with whole numbers, x must be a multiple of 7. The smallest is x = 7, giving k = 8. (Any pair in the ratio 7 : 8, such as 14 and 16, also works.)

20 horses and 35 hens.

With letter-numbers. Let there be h horses (4 legs) and n hens (2 legs).

Without letter-numbers — yes (the hint’s method):

The daughter is 6 years old now, and the mother is 30.

Gauri has 6 cows and Naina has 12.

(i) ₹80 per dosa. (ii) 175 dosas a day.

(i) Let the selling price be ₹p per dosa.

(ii) Let the number of dosas sold be n, each at ₹50.

Every fraction equals 1/3.

Why: the sum of the first n odd numbers is n².

(i) 7 coins. Let x be his coins at the start.

(ii) The cost must be less than the coins he has. With x coins and cost c, one round leaves 2x – c, and 2x – c > x means c < x. With 7 coins, any cost of 6 or less works.

(iii) The genie makes the coins after the last round equal the cost. For three rounds, as in the story: 2(2(2x – c) – c) – c = 8x – 7c = 0, so c = 8x ÷ 7. With x = 7 this is 8 — exactly the genie’s price. (To take everything in one round, it could simply charge c = 2x.)