Chapter 6 exercise answers: Algebra Play

Open NCERT p. 140Checked by computer

1. In a 3-row pyramid, each upper cell is the sum of the two cells below it, so the top cell always works out to (first) + 2 × (middle) + (last).
2. For 4, 13, 8: top = 4 + 2×13 + 8 = 4 + 26 + 8 = 38.
3. For 7, 11, 3: top = 7 + 2×11 + 3 = 7 + 22 + 3 = 32.
4. For 10, 14, 25: top = 10 + 2×14 + 25 = 10 + 28 + 25 = 63.

Answer38, 32 and 63.

Watch this explained “Four cells, and a top found without a pyramid”, 7:11 into Number pyramids: what the top cell is made of

Open NCERT p. 140Checked by computer

1. Call the bottom row a, b, c, d.
2. The next row up is a+b, b+c, c+d.
3. The row above that is (a+b)+(b+c) = a+2b+c, and (b+c)+(c+d) = b+2c+d.
4. The top cell is their sum: (a+2b+c) + (b+2c+d) = a + 3b + 3c + d.

AnswerTop = a + 3b + 3c + d.

Watch this explained “Four cells, and a top found without a pyramid”, 7:11 into Number pyramids: what the top cell is made of

Open NCERT p. 140Checked by computer

1. For a 4-number bottom row a, b, c, d, the top of the pyramid is a + 3b + 3c + d (from Q2).
2. For 8, 19, 21, 13: top = 8 + 3×19 + 3×21 + 13 = 8 + 57 + 63 + 13 = 141.
3. For 7, 18, 19, 6: top = 7 + 3×18 + 3×19 + 6 = 7 + 54 + 57 + 6 = 124.
4. For 9, 7, 5, 11: top = 9 + 3×7 + 3×5 + 11 = 9 + 21 + 15 + 11 = 56.

Answer141, 124 and 56.

Watch this explained “Four cells, and a top found without a pyramid”, 7:11 into Number pyramids: what the top cell is made of

Open NCERT p. 140Checked by computer

1. The first three Virahāṅka-Fibonacci numbers are 1, 2 and 3. Write them in the bottom row.
2. Middle row: 1 + 2 = 3 and 2 + 3 = 5.
3. Top: 3 + 5 = 8.
4. So the six cells hold 1, 2, 3 (bottom row), 3, 5 (middle row) and 8 (top). The number 3 appears twice, so the different numbers are 1, 2, 3, 5 and 8.
5. The sequence runs 1, 2, 3, 5, 8, 13, …, so every one of them is a Virahāṅka-Fibonacci number. This is no accident: each cell above the bottom row is the sum of two neighbouring terms, and two neighbouring terms always add up to the next term.

AnswerThe grid holds 1, 2, 3, 3, 5 and 8; the number at the top is 8; and yes, all of them are Virahāṅka-Fibonacci numbers.

Watch this explained “Feeding a sequence in, and nothing leaving it”, 8:15 into Number pyramids: what the top cell is made of

Open NCERT p. 140One way to think about it

(i) The first four Virahāṅka-Fibonacci numbers are written in the bottom row

1. Bottom row: 1, 2, 3, 5 (the first four Virahāṅka-Fibonacci numbers).
2. Next row: 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8.
3. Next row: 3 + 5 = 8, 5 + 8 = 13.
4. Top: 8 + 13 = 21.
5. Every number that appears (1, 2, 3, 5, 8, 13, 21) is a Virahāṅka-Fibonacci number, and 21 is the 7th term of 1, 2, 3, 5, 8, 13, 21.

In shortEvery number in the pyramid is a Virahāṅka-Fibonacci number, and the number at the top is 21, the 7th term.

(ii) The first 29 Virahāṅka-Fibonacci numbers are written in the bottom row

1. Each cell above the bottom row is the sum of the two cells it rests on. When those two are neighbouring terms of the sequence, their sum is the next term, because that is how the sequence is made. So every cell is a Virahāṅka-Fibonacci number.
2. Look at what one row does. If a row holds term number k, term k + 1, term k + 2, …, then the row above holds term k + 2, term k + 3, …, since term k + term (k + 1) = term (k + 2). Each row is the row below moved 2 places along the sequence, and one cell shorter.
3. The bottom row starts at term 1. Climbing the 28 rows above it to the top of a 29-row pyramid moves 2 × 28 = 56 places, so the top cell is term 1 + 56 = 57.
4. This matches the small pyramids: 3 rows put term 5 (the number 8) on top and 4 rows put term 7 (the number 21) on top. (The 57th term is 591286729879, but the question only asks what we can say about it.)

In shortEvery number in the pyramid is a Virahāṅka-Fibonacci number, and the number at the top is the 57th Virahāṅka-Fibonacci number.

Watch this explained “Which term sits on top, and a 29-row pyramid”, 9:05 into Number pyramids: what the top cell is made of

Open NCERT p. 140One way to think about it

1. Each cell above the bottom row is the sum of the two cells it rests on, and in the Virahāṅka-Fibonacci sequence the sum of two neighbouring terms is the next term. So every number in the pyramid is a Virahāṅka-Fibonacci number.
2. If a row holds term number k, term k + 1, term k + 2, …, then the row above holds term k + 2, term k + 3, …, because term k + term (k + 1) = term (k + 2). So each row is the row below moved 2 places along the sequence, and one cell shorter.
3. The bottom row (row 1) holds terms 1 to n. Row 2 starts at term 3, row 3 at term 5, and in general row r starts at term 2r − 1.
4. The top is row n, a single cell, so it is term 2n − 1.
5. Check: n = 3 gives term 5, which is 8; n = 4 gives term 7, which is 21; n = 29 gives term 57.

In shortEvery number in the pyramid is a Virahāṅka-Fibonacci number, and the number at the top is the (2n − 1)th Virahāṅka-Fibonacci number.

Watch this explained “Which term sits on top, and a 29-row pyramid”, 9:05 into Number pyramids: what the top cell is made of

Open NCERT p. 144Checked by computer

1. The largest digit should be the one-digit multiplier, and the other two digits should form the two-digit number, with the larger of them in the tens place.
2. Largest digit among 1, 3, 7 is 7, so 7 is the multiplier.
3. The remaining digits 1 and 3, arranged in decreasing order, give the two-digit number 31.
4. Product = 31 × 7 = 217.

Answer31 × 7 = 217.

Watch this explained “Three digits, one frame, and a definite answer”, 0:00 into Arranging given digits to make the largest product

Open NCERT p. 144Checked by computer

1. Largest digit among 3, 5, 9 is 9, so 9 is the multiplier.
2. The remaining digits 3 and 5, arranged in decreasing order, give the two-digit number 53.
3. Product = 53 × 9 = 477.

Answer53 × 9 = 477.

Watch this explained “Tried on digits it has not seen”, 7:25 into Arranging given digits to make the largest product

Open NCERT p. 145Checked by computer

1. Let the number have tens digit a and ones digit b, with a ≠ b. The number is 10a + b and its reverse is 10b + a.
2. If a > b: (10a + b) − (10b + a) = 9a − 9b = 9(a − b), so dividing by 9 gives a − b.
3. If b > a: (10b + a) − (10a + b) = 9b − 9a = 9(b − a), so dividing by 9 gives b − a.
4. Either way, the quotient is the larger digit minus the smaller digit. Example: 74 − 47 = 27 and 27 ÷ 9 = 3 = 7 − 4.

AnswerThe quotient is the difference between the two digits, larger minus smaller, that is |a − b|. Yes: the quotient always equals this difference of the digits.

Watch this explained “Nine times what?”, 2:52 into Decoding a divisibility trick with algebra

Open NCERT p. 145Checked by computer

1. Let the two-digit number be 10a+b, so its reverse is 10b+a.
2. Their sum is (10a+b) + (10b+a) = 11a + 11b = 11(a+b).
3. Since 11(a+b) is 11 times a whole number, it is always divisible by 11.

AnswerSum = 11(a + b), which is always divisible by 11.

Watch this explained “Adding instead, and where eleven comes from”, 4:08 into Decoding a divisibility trick with algebra

Open NCERT p. 145Checked by computer

1. abc = 100a+10b+c, bca = 100b+10c+a, cab = 100c+10a+b.
2. Adding them: each of a, b, c appears once in the hundreds place, once in the tens place, and once in the units place, so the sum is 111a + 111b + 111c = 111(a+b+c).
3. 111 = 3 × 37, so the sum is always a multiple of both 37 and 3.

AnswerSum = 111(a + b + c). Since 111 = 3 × 37, the sum is always divisible by 3 as well as by 37.

Watch this explained “Three digits, cycled round”, 5:06 into Decoding a divisibility trick with algebra

Open NCERT p. 145Checked by computer

1. abcabc is the 3-digit number abc written twice, so it equals abc × 1000 + abc = abc × 1001.
2. 1001 = 7 × 11 × 13.
3. Dividing abcabc by 7, then 11, then 13 divides out the whole factor 1001, leaving just abc back again.

AnswerYou get back the original 3-digit number abc (that is, 100a + 10b + c), because abcabc = abc × 1001 = abc × 7 × 11 × 13.

Watch this explained “A block of three, written twice”, 6:13 into Decoding a divisibility trick with algebra

Open NCERT p. 145Checked by computerAnswers can differ: one example

1. Let x be the flowers he started with and y the number placed in each shrine.
2. Pond 1 doubles them to 2x; after shrine 1 he has 2x − y.
3. Pond 2 doubles that to 4x − 2y; after shrine 2 he has 4x − 3y.
4. Pond 3 doubles that to 8x − 6y, and all of it goes into shrine 3, so 8x − 6y = y, that is 8x = 7y.
5. The smallest whole numbers with 8x = 7y are x = 7 and y = 8. Any multiple also works (14 and 16, 21 and 24, and so on), but 7 and 8 is the smallest answer.
6. Check: 7 → 14, place 8, 6 left → 12, place 8, 4 left → 8, place all 8. Every shrine gets 8.

AnswerHe started with 7 flowers and placed 8 flowers in each shrine. (This is the smallest answer; 14 and 16, 21 and 24, … also work.)

Watch this explained “Naming what you do not know”, 8:03 into Decoding a divisibility trick with algebra

Open NCERT p. 146Checked by computer

1. With letter-numbers: let p be the number of horses and q the number of hens. Heads: p + q = 55. Legs: 4p + 2q = 150.
2. From the heads, q = 55 − p. Then 4p + 2(55 − p) = 150, so 2p + 110 = 150, 2p = 40 and p = 20. So q = 55 − 20 = 35.
3. Without letter-numbers: if all 55 animals were hens, there would be 55 × 2 = 110 legs.
4. That is 150 − 110 = 40 legs too few. Changing one hen into a horse adds 2 legs, so there are 40 ÷ 2 = 20 horses, and 55 − 20 = 35 hens.
5. Check: 20 + 35 = 55 heads, and 20 × 4 + 35 × 2 = 80 + 70 = 150 legs.

AnswerThere are 20 horses and 35 hens.

Watch this explained “Naming what you do not know”, 8:03 into Decoding a divisibility trick with algebra

Open NCERT p. 146Checked by computer

1. Let D = daughter's present age, so mother's present age = 5D.
2. In 6 years: mother's age = 5D+6, daughter's age = D+6.
3. The mother will be 3 times the daughter's age then: 5D+6 = 3(D+6).
4. 5D+6 = 3D+18, so 2D = 12, giving D = 6.

AnswerThe daughter is 6 years old now.

Watch this explained “Naming what you do not know”, 8:03 into Decoding a divisibility trick with algebra

Open NCERT p. 146Checked by computer

1. Let G = Gauri's cows and N = Naina's cows.
2. "You have twice as many cows as I do": N = 2G.
3. "If I gave you three of my cows, we would each have the same number": N − 3 = G + 3.
4. Substitute N = 2G: 2G − 3 = G + 3, so G = 6.
5. Then N = 2×6 = 12.

AnswerGauri has 6 cows and Naina has 12 cows.

Watch this explained “Naming what you do not know”, 8:03 into Decoding a divisibility trick with algebra

Open NCERT p. 146Checked by computer

(i) If I can sell 100 dosas a day, what should be …

1. Total cost for the day = rent + cost of 100 dosas = ₹5000 + 100×₹10 = ₹6000.
2. To make a profit of ₹2000, total revenue needed = ₹6000 + ₹2000 = ₹8000.
3. Selling price per dosa = ₹8000 ÷ 100 = ₹80.

Answer₹80 per dosa.

(ii) If my customers are willing to pay only ₹50 for a dosa

1. Let n = number of dosas sold. Revenue = 50n. Cost = 5000 + 10n.
2. Profit = revenue − cost = 50n − (5000+10n) = 40n − 5000.
3. Set profit = 2000: 40n − 5000 = 2000, so 40n = 7000, giving n = 175.

Answer175 dosas.

Watch this explained “Naming what you do not know”, 8:03 into Decoding a divisibility trick with algebra

Open NCERT p. 146Checked by computer

1. 1/3 is already 1/3.
2. (1 + 3)/(5 + 7) = 4/12 = 1/3.
3. (1 + 3 + 5)/(7 + 9 + 11) = 9/27 = 1/3.
4. Every fraction equals 1/3.
5. Why: in the nth fraction, the top is the sum of the first n odd numbers, which is n². The top and bottom together are the first 2n odd numbers, whose sum is (2n)² = 4n².
6. So the bottom is 4n² − n² = 3n², and the fraction is n²/3n² = 1/3 for every n.

AnswerAll three fractions equal 1/3, because the top is n² and the bottom is 4n² − n² = 3n².

Open NCERT p. 147Checked by computerReads two ways: both answers shown

(i) How many coins did Karim initially have?

1. Let Karim start with s coins. Each round the coins double, then he pays 8.
2. After round 1: 2s − 8. After round 2: 2(2s − 8) − 8 = 4s − 24. After the third doubling: 2(4s − 24) = 8s − 48.
3. At that moment he has only 8 coins, exactly the 8 he owes: 8s − 48 = 8, so 8s = 56 and s = 7.
4. Check: 7 → 14, pay 8 → 6 → 12, pay 8 → 4 → 8, pay 8 → nothing left.

AnswerKarim started with 7 coins.

(ii) For what cost per round should Karim agree to the deal

1. Say Karim has x coins and the cost is c coins per round. One round turns x into 2x − c.
2. He gains only if 2x − c > x, that is, c < x.
3. So the cost must be less than the coins he has; in whole coins, at most x − 1. His coins then grow every round, so the deal keeps helping him.

AnswerHe should agree only if the cost per round is less than the number of coins he has (x), that is, at most x − 1 coins.

(iii) How should the genie set the cost per round

1. The question does not say over how many rounds the genie wants to take all the coins, and the answer depends on it.
2. After 1 round Karim has 2x − c; after 2 rounds 4x − 3c; after 3 rounds 8x − 7c.
3. Over three rounds, as in the story. We lead with this because the story's trick is played over three rounds. 8x − 7c = 0, so c = 8x/7. With x = 7 this gives 8, exactly the story.
4. Over one round: 2x − c = 0, so c = 2x.
5. Over two rounds: 4x − 3c = 0, so c = 4x/3.
6. In general, over n rounds the genie should charge 2ⁿx ÷ (2ⁿ − 1) per round.

AnswerOver three rounds, as in the story: 8x/7 coins per round. Over one round: 2x. Over two rounds: 4x/3.

Watch this explained “Losing from the first circuit”, 9:10 into Decoding a divisibility trick with algebra