12 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 145, Q1
In the reverse-and-subtract trick, what is the quotient when you divide by 9? Is there a relationship between the two digits and the quotient?
Hint. You have already shown what the difference equals — read the quotient straight off it.
Read it off the algebra. The difference has been shown to be
9 × |a − b|
so dividing by 9 gives a quotient of
|a − b| — the difference of the two digits.
Check with the chapter's example. 74 − 47 = 27, and 27 ÷ 9 = 3. The digits are 7 and 4, whose difference is 3 ✓
More checks.
| Number | Reverse | Difference | ÷ 9 | Digit difference |
|---|
| 15 | 51 | 36 | 4 | 5 − 1 = 4 ✓ |
| 29 | 92 | 63 | 7 | 9 − 2 = 7 ✓ |
| 83 | 38 | 45 | 5 | 8 − 3 = 5 ✓ |
| 62 | 26 | 36 | 4 | 6 − 2 = 4 ✓ |
The relationship, stated plainly. The quotient is the gap between the two digits. So a performer told only the quotient learns immediately how far apart the digits are — although not which digits they were, since 15, 26, 37, 48 and 59 all give a quotient of 4.
A stronger version of the trick. Ask for the quotient and one digit, and the whole number is determined.
✦ The quotient is |a − b|, the difference of the two digits — for 74 − 47 = 27 the quotient 3 is exactly 7 − 4.
23 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 145, Q2
Instead of the difference, take the sum of the two-digit number and its reverse. Examples: 31 + 13 = 44, 28 + 82 = 110, 12 + 21 = 33. All are divisible by 11. Is this always true? Justify with algebra.
Hint. Add the expanded forms instead of subtracting them.
Add the expanded forms.
(10a + b) + (10b + a)
= 10a + a + b + 10b
= 11a + 11b
= 11(a + b)
Since a + b is a whole number, the sum is 11 times a whole number, so it is always divisible by 11 — for every two-digit number, with no exceptions.
Check the three examples.
| Number | Reverse | Sum | 11 × (a + b) | Quotient a + b |
|---|
| 31 | 13 | 44 | 11 × 4 | 3 + 1 = 4 ✓ |
| 28 | 82 | 110 | 11 × 10 | 2 + 8 = 10 ✓ |
| 12 | 21 | 33 | 11 × 3 | 1 + 2 = 3 ✓ |
The bonus fact. Just as the difference gave a quotient of |a − b|, the sum gives a quotient of a + b — the sum of the digits. So dividing the sum by 11 hands you the digit sum directly.
Why the two tricks are twins. Subtracting cancelled the units digits and left 9 lots of the difference; adding reinforces them and leaves 11 lots of the sum. It is the same expanded-form calculation with one sign changed:
- difference → (10 − 1) = 9 times (b − a)
- sum → (10 + 1) = 11 times (a + b)
Note this works even when a + b reaches 10 or more, as in 28 + 82 = 110 = 11 × 10 — carrying does not disturb the algebra.
✦ Yes, always. The sum is (10a + b) + (10b + a) = 11(a + b), a multiple of 11, and the quotient on dividing by 11 is the sum of the digits.
33 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 145, Q3
Take any 3-digit number abc, cycle the digits to make bca and cab, and add all three. Justify using algebra that the sum is always divisible by 37. Will it always be divisible by 3 as well?
Hint. Write each of the three numbers in expanded form and add the columns.
Write all three in expanded form.
- abc = 100a + 10b + c
- bca = 100b + 10c + a
- cab = 100c + 10a + b
Add them, collecting each letter.
Coefficient of a: 100 + 1 + 10 = 111
Coefficient of b: 10 + 100 + 1 = 111
Coefficient of c: 1 + 10 + 100 = 111
Sum = 111(a + b + c)
Every digit ends up in the hundreds place once, the tens place once and the units place once, which is why each coefficient is 111.
Factorise 111.
111 = 3 × 37
So the sum = 3 × 37 × (a + b + c), which is divisible by 37 and also by 3 — both, always, for every three-digit number.
Check with 152.
152 + 521 + 215 = 888
888 ÷ 37 = 24, and 888 ÷ 3 = 296. Also 111 × (1 + 5 + 2) = 111 × 8 = 888 ✓
Check with 407.
407 + 074 + 740 = 1221
1221 = 111 × 11, and 4 + 0 + 7 = 11 ✓; 1221 ÷ 37 = 33 and 1221 ÷ 3 = 407.
A further consequence. The sum is also divisible by 111 itself, and hence by 3 and 37 together. And if the digit sum a + b + c happens to be a multiple of 3, the total is divisible by 9 as well.
✦ The sum is 111(a + b + c) = 3 × 37 × (a + b + c), so it is always divisible by 37 and always divisible by 3. Each digit occupies the hundreds, tens and units place exactly once across the three numbers, which is what produces the coefficient 111.
43 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 145, Q4
Take a 3-digit number abc and repeat the digits to make the 6-digit number abcabc. Divide it by 7, then by 11, then by 13. What do you get, and why does it work?
Hint. Write abcabc in terms of abc, and factorise the multiplier.
Express the six-digit number in terms of the three-digit one. Let N = abc = 100a + 10b + c. Then abcabc has N in the top three places and N again in the bottom three:
abcabc = N × 1000 + N = N × 1001
Factorise 1001.
1001 = 7 × 143 = 7 × 11 × 13
Check: 7 × 11 = 77, and 77 × 13 = 1001 ✓ — one of the prettiest factorisations in arithmetic, since 7, 11 and 13 are three consecutive primes.
So the divisions unwind the product.
abcabc = N × 7 × 11 × 13
- ÷ 7 → N × 11 × 13 = N × 143
- ÷ 11 → N × 13
- ÷ 13 → N
You get back the original three-digit number, with no remainder at any stage.
Check with 486.
486486 ÷ 7 = 69498
69498 ÷ 11 = 6318
6318 ÷ 13 = 486 ✓
Check with 205.
205205 ÷ 7 = 29315; ÷ 11 = 2665; ÷ 13 = 205 ✓
Why it feels magical. The three divisions look arbitrary, and there is no obvious reason for 7, 11 and 13 to divide anything. The single fact 1001 = 7 × 11 × 13 explains all of it — and it explains why the order of the three divisions does not matter either.
✦ abcabc = abc × 1001 = abc × 7 × 11 × 13, so dividing by 7, then 11, then 13 returns the original three-digit number exactly. For example 486486 ÷ 7 ÷ 11 ÷ 13 = 486.
54 marksGanita Prakash Cl-8 Part 2, Figure it Out, pages 145-146, Q5
There are 3 shrines, each with a magical pond in front that doubles any flowers dipped in it. A person dips all his flowers in pond 1 and leaves some at shrine 1, dips the rest in pond 2 and leaves some at shrine 2, dips the rest in pond 3 and leaves them all at shrine 3. He left an equal number at each shrine. How many flowers did he start with, and how many did he leave at each?
Hint. Call the starting number x and the number left at each shrine k, and track what remains after each shrine.
Set up two letters. Let the starting number of flowers be x and the number left at each shrine be k.
| Stage | Flowers in hand |
|---|
| Start | x |
| After pond 1 | 2x |
| After shrine 1 | 2x − k |
| After pond 2 | 2(2x − k) = 4x − 2k |
| After shrine 2 | 4x − 3k |
| After pond 3 | 2(4x − 3k) = 8x − 6k |
| After shrine 3 | 8x − 6k − k = 8x − 7k |
Use the last condition. At the third shrine he places all the remaining flowers, and that amount must also be k. So the hand ends empty:
8x − 7k = 0, that is 8x = 7k
Solve in whole numbers. 8x = 7k means 8 divides 7k, and since 7 and 8 share no factor, 8 must divide k. The smallest possibility is
k = 8, giving x = 7
Check.
| Stage | Flowers |
|---|
| Start | 7 |
| Double | 14 → leave 8 → 6 left |
| Double | 12 → leave 8 → 4 left |
| Double | 8 → leave 8 → 0 left |
Eight flowers at each of the three shrines ✓ and nothing left over ✓
Other solutions. Any multiple works: k = 16 with x = 14, k = 24 with x = 21, and so on — the answer is really the ratio x : k = 7 : 8. The smallest and intended answer is 7 flowers to begin with.
The idea worth keeping. The doubling makes the earliest offerings the most expensive: the k flowers left at shrine 1 would have doubled twice more, so they cost 4k of the final total, while those at shrine 2 cost 2k and those at shrine 3 only k. Adding, 4k + 2k + k = 7k must equal 8x — the same equation, seen from the other side.
✦ He started with 7 flowers and left 8 at each shrine. The equation is 8x = 7k, so x : k = 7 : 8, and any multiple (14 and 16, 21 and 24, …) also works.
63 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 146, Q6
A farm has horses and hens. The total number of heads is 55 and the total number of legs is 150. How many of each? Solve it with algebra and also without letter-numbers.
Hint. Every animal has one head; horses have four legs and hens two.
With algebra. Let there be h horses and n hens.
- Heads: h + n = 55
- Legs: 4h + 2n = 150
From the first equation n = 55 − h. Substituting into the second:
4h + 2(55 − h) = 150
4h + 110 − 2h = 150
2h = 40
h = 20, so n = 55 − 20 = 35
Check. 20 + 35 = 55 heads ✓ and 4(20) + 2(35) = 80 + 70 = 150 legs ✓
Without letter-numbers. Suppose for a moment that all 55 animals were hens. Then the legs would number
55 × 2 = 110
But there are actually 150 legs, so there are
150 − 110 = 40 extra legs
Each horse has 4 legs instead of a hen's 2, so turning one hen into a horse adds 2 legs. To account for 40 extra legs we need
40 ÷ 2 = 20 horses
and therefore 55 − 20 = 35 hens — the same answer.
Why the two methods are really the same. The 'all hens' assumption is exactly the substitution n = 55 − h in disguise: assuming all hens sets h = 0, and each unit increase in h adds 2 to the leg count, which is precisely the 2h left in the equation after the substitution. The arithmetic method is often quicker in the head; the algebraic one generalises to problems with more animals or more constraints.
✦ There are 20 horses and 35 hens. Without algebra: if all 55 were hens there would be 110 legs, the 40 extra legs each come in pairs from replacing a hen by a horse, so there are 40 ÷ 2 = 20 horses.
73 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 146, Q7
A mother is 5 times her daughter's age. In 6 years' time, the mother will be 3 times her daughter's age. How old is the daughter now?
Hint. One letter is enough — write both ages now and both ages in 6 years in terms of it.
Set up with one letter. Let the daughter's present age be d years. Then the mother's present age is 5d.
In 6 years' time both ages increase by 6:
- daughter: d + 6
- mother: 5d + 6
Write the second condition. The mother will then be 3 times the daughter:
5d + 6 = 3(d + 6)
Solve.
5d + 6 = 3d + 18
5d − 3d = 18 − 6
2d = 12
d = 6
So the daughter is 6 now and the mother is 5 × 6 = 30.
Check. In 6 years the daughter will be 12 and the mother 36, and 36 = 3 × 12 ✓
A remark on why the multiple falls. The gap between the two ages never changes — it is 30 − 6 = 24 years, now and always. What changes is the ratio: as both ages grow, the fixed gap of 24 becomes a smaller fraction of the daughter's age. That is why the multiple slides from 5 down to 3, and it will keep falling: when the daughter is 12 the ratio is 3, when she is 24 it will be 2, and it approaches 1 but never reaches it.
The constant gap gives a second route to the answer: if the mother is 3 times the daughter and the gap is 24, then 2 × (daughter's age then) = 24, so she is 12 then and 6 now ✓
✦ The daughter is 6 years old now and the mother is 30.
83 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 146, Q8
Gauri says to Naina, 'You have twice as many cows as I do.' Naina replies, 'True, but if I gave you three of my cows, we would each have the same number.' How many cows does each have?
Hint. Use one letter for Gauri's cows and write Naina's in terms of it.
Set up. Let Gauri have g cows. Naina has twice as many, so Naina has 2g.
After the transfer of three cows:
- Gauri would have g + 3
- Naina would have 2g − 3
These are to be equal:
2g − 3 = g + 3
Solve.
2g − g = 3 + 3
g = 6
So Gauri has 6 cows and Naina has 2 × 6 = 12.
Check. Naina has twice as many ✓ And after the transfer: Gauri 6 + 3 = 9, Naina 12 − 3 = 9 — equal ✓
Seeing it without algebra. Handing over 3 cows moves Gauri up by 3 and Naina down by 3, so it closes a gap of 6. The gap between them must therefore have been 6 to start with. Since Naina has twice as many, the gap equals Gauri's own herd, so Gauri has 6 and Naina has 12.
A caution the problem rewards. It is tempting to write 'Naina gives 3, so subtract 3 from each' — but the 3 cows do not vanish, they arrive at Gauri's. Getting the direction of a transfer right is where most of the marks are lost in problems of this shape.
✦ Gauri has 6 cows and Naina has 12.
94 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 146, Q9
A dosa cart costs ₹5000 per day in rent, and each dosa costs ₹10 to make. (i) If 100 dosas are sold a day, what should the selling price be to make a profit of ₹2000? (ii) If customers will pay only ₹50 per dosa, how many dosas must be sold a day to make a profit of ₹2000?
Hint. Profit = money in − money out, and the money out has a fixed part and a per-dosa part.
The structure of the costs. There are two kinds:
- a fixed cost of ₹5000 a day (the rent), which does not depend on how many dosas are sold
- a variable cost of ₹10 per dosa (ingredients and fuel)
So for n dosas sold at ₹p each:
Profit = np − (5000 + 10n)
(i) 100 dosas, profit ₹2000. Find p.
Total cost = 5000 + 10 × 100 = 5000 + 1000 = ₹6000
Required revenue = cost + profit = 6000 + 2000 = ₹8000
100p = 8000
p = ₹80
Check: 100 × 80 = ₹8000 revenue; costs ₹6000; profit ₹2000 ✓
(ii) Price fixed at ₹50, profit ₹2000. Find n.
50n − (5000 + 10n) = 2000
50n − 10n = 2000 + 5000
40n = 7000
n = 175 dosas
Check: revenue 175 × 50 = ₹8750; costs 5000 + 175 × 10 = 5000 + 1750 = ₹6750; profit = 8750 − 6750 = ₹2000 ✓
The idea behind part (ii). Each dosa sold at ₹50 costs ₹10 to make, so it contributes ₹40 towards the rent and profit. The rent of ₹5000 plus the target profit of ₹2000 comes to ₹7000, and
7000 ÷ 40 = 175 dosas
That ₹40 is called the contribution per unit, and it is how every small business works out how much it must sell. It also gives the break-even point at once: to merely cover the rent, 5000 ÷ 40 = 125 dosas must be sold, and every dosa after the 125th is pure profit.
✦ (i) The selling price must be ₹80 per dosa. (ii) At ₹50 a dosa, 175 dosas must be sold — each contributes ₹40 towards rent and profit, and ₹7000 ÷ ₹40 = 175.
104 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 146, Q10
Evaluate 1/3, (1 + 3)/(5 + 7), (1 + 3 + 5)/(7 + 9 + 11). What do you observe? Can you explain why it happens?
Hint. Recall what the sum of the first n odd numbers comes to.
Evaluate the three fractions.
- 1/3 = 1/3
- (1 + 3)/(5 + 7) = 4/12 = 1/3
- (1 + 3 + 5)/(7 + 9 + 11) = 9/27 = 1/3
The observation: every one of them is 1/3.
Look at the structure. In the nth fraction:
- the numerator is the sum of the first n odd numbers
- the denominator is the sum of the next n odd numbers
| n | Numerator | Denominator | Value |
|---|
| 1 | 1 | 3 | 1/3 |
| 2 | 1 + 3 = 4 | 5 + 7 = 12 | 1/3 |
| 3 | 1 + 3 + 5 = 9 | 7 + 9 + 11 = 27 | 1/3 |
| 4 | 16 | 9 + 11 + 13 + 15 = 48 | 1/3 |
The explanation. The sum of the first n odd numbers is n² — a fact from earlier work, and one the hint points to.
- Numerator = sum of the first n odd numbers = n²
- Denominator = (sum of the first 2n odd numbers) − (sum of the first n odd numbers) = (2n)² − n² = 4n² − n² = 3n²
So every fraction is
n²/(3n²) = 1/3
exactly, for every n — the n² cancels completely.
Check the pattern once more with n = 5. Numerator 1 + 3 + 5 + 7 + 9 = 25 = 5². Denominator 11 + 13 + 15 + 17 + 19 = 75 = 3 × 25 ✓ and 25/75 = 1/3 ✓
Why n² is the sum of the first n odd numbers. Picture a square of dots grown one L-shaped border at a time: a 1 × 1 square needs 1 dot, growing to 2 × 2 needs 3 more, to 3 × 3 needs 5 more, and so on. Each new border is the next odd number, so after n borders there are n² dots.
✦ All three fractions equal 1/3, and so does every later one. The numerator is the sum of the first n odd numbers = n², and the denominator is the sum of the next n = (2n)² − n² = 3n², so the ratio is n²/3n² = 1/3 exactly.
115 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 147, Q11
Karim's coins double each time he goes round a banyan tree, but he must pay the genie 8 coins per round. After the third doubling he was left with exactly 8 coins — precisely what he owed. (i) How many coins did Karim start with? (ii) What cost per round should he agree to if he wants his money to grow? (iii) How should the genie set the cost so that it gets all of Karim's coins?
Hint. Track the coins through one round as a formula, then apply it three times.
One round as a formula. If Karim starts a round with x coins, he ends it with
2x − 8
(i) How many coins to begin with? Let the starting number be x.
| Stage | Coins |
|---|
| Start | x |
| After round 1 | 2x − 8 |
| After round 2 | 2(2x − 8) − 8 = 4x − 24 |
| Third doubling | 2(4x − 24) = 8x − 48 |
The story says that after the third doubling he had exactly 8 coins:
8x − 48 = 8
8x = 56
x = 7
Check the whole story.
| Stage | Coins |
|---|
| Start | 7 |
| Double | 14, pay 8 → 6 |
| Double | 12, pay 8 → 4 |
| Double | 8 — exactly what he owes → 0 |
Karim ends with nothing ✓
(ii) What cost would let his money grow? With a cost of c per round, one round takes x to 2x − c. His money grows when
2x − c > x, that is c < x
So the cost must be less than the number of coins he currently holds. With 7 coins to start, any charge of 6 or fewer would have worked — and once it grows, it keeps growing, because x rises while c stays fixed, so the inequality only becomes easier to satisfy.
The three cases are worth stating together:
- c < x → his coins grow, and grow faster every round
- c = x → he stays at exactly x forever, since 2x − x = x
- c > x → his coins shrink every round, and he is eventually ruined
The break-even point c = x is what makes the genie's offer look harmless: 8 coins sounds trifling until you notice Karim only had 7.
(iii) How should the genie set the cost to take everything? To leave Karim with nothing after k rounds, the genie needs
2ᵏx − c(2ᵏ⁻¹ + … + 2 + 1) = 0
2ᵏx − c(2ᵏ − 1) = 0
c = 2ᵏx / (2ᵏ − 1)
- k = 1: c = 2x — double the coins Karim has; he goes round once and hands over everything.
- k = 2: c = 4x/3
- k = 3: c = 8x/7 — with x = 7 this gives c = 8, exactly the genie's charge ✓
So the genie, knowing x, picks the number of rounds it wants the game to last and sets the cost by that formula. Choosing k = 3 and c = 8x/7 is the cruellest choice available with whole numbers: Karim watches his money double three times and still ends with nothing.
✦ (i) Karim started with 7 coins. (ii) He gains only if the cost per round is less than the coins he holds (c < x); at c = x he stands still and at c > x he loses. (iii) To strip him in exactly k rounds the genie must set c = 2ᵏx/(2ᵏ − 1) — with k = 3 and x = 7 that is exactly the 8 coins it charged.
Solutions written by the tuition.in editorial team and checked against NCERT Ganita Prakash Grade 8 Part 2 (hegp206.pdf), Chapter 6 'Algebra Play', pages 135-147. HAND-WRITTEN throughout. Part 2 books carry NO printed answer key, so every answer was derived from first principles and independently recomputed in Python. FIGURES READ OFF HIGH-DPI RENDERS AND RE-SOLVED: the three p.138 pyramids (bottom rows 6,2 / 3,4,3 / 5,4,5,0 giving tops 8, 14 and 32); the p.138 top-down pyramid 10 / 4,6 / 1,3,3; the p.139 letter-number pyramid 60 / 32,28 / 12,20,8; the three p.139 four-row pyramids, whose bottom rows solve to 4,9,6,1 (top 50), 5,14,7,2 (top 70) and 3,5,5,2 (top 35); the p.140 Figure it Out bottom rows 4,13,8 / 7,11,3 / 10,14,25 (tops 38, 32, 63) and 8,19,21,13 / 7,18,19,6 / 9,7,5,11 (tops 141, 124, 56); and the two p.142 algebra grids, which solve to square 11 and circle 5 with a third-row total of 21, and circle 4 and diamond 7 with a third-row total of 15 (the fourth row of that grid is printed blank for the student to design). The Virahanka-Fibonacci result was proved and then checked by computation for n = 2 to 6: every entry of the pyramid is a Virahanka-Fibonacci number, row k from the bottom is the consecutive run V(2k-1) to V(n+k-1), and the top is V(2n-1) - so a 29-row pyramid tops out at V57 = 591286729879. All six arrangements were enumerated for each Largest Product question: 32 x 5 = 160, 31 x 7 = 217 and 53 x 9 = 477, each beating its runner-up by exactly the margin the algebra predicts (4, 4 and 12). The date trick decodes as 100M + 165 + D, giving 4 November, 29 February and 31 January for 1269, 394 and 296. The puzzle answers were each verified by substitution: 7 flowers with 8 per shrine, 20 horses and 35 hens, a daughter of 6 and mother of 30, Gauri 6 cows and Naina 12, a dosa price of Rs 80 or a target of 175 dosas, the fraction sequence equal to 1/3 because n^2/(4n^2 - n^2) = 1/3, and Karim starting with 7 coins with the genie's general ruinous charge being c = 2^k x / (2^k - 1).. Questions are referenced from the NCERT textbook for identification.
