Chapter 3 exercise answers: The World of Numbers

Open NCERT p. 43Checked by computer

1. The rate is 15 ingots for every 2 bags, so one bag is worth 15 ÷ 2 = 7.5 ingots.
2. 12 bags is six lots of 2 bags each, since 12 ÷ 2 = 6.
3. Six lots at 15 ingots per lot gives 6 × 15 = 90 ingots.

Answer90 copper ingots.

Watch this explained “Fifteen for two”, 2:30 into Why India needed names for powers of ten · हिंदी में देखें

Open NCERT p. 43Checked by computer

1. Each of 11, 13, 17 and 19 has only two factors, 1 and itself, so all four are prime.
2. They are every prime between 10 and 20: 12, 14, 16 and 18 are even, and 15 = 3 × 5, so nothing else in that range is prime.
3. So the pattern is 'prime numbers', and the next three numbers are the next primes after 19.
4. Test the numbers after 19: 20 = 2 × 10, 21 = 3 × 7 and 22 = 2 × 11 are not prime; 23 is prime; 24, 25 = 5 × 5, 26, 27 = 3 × 9 and 28 are not; 29 is prime; 30 is not; 31 is prime.
5. Careful: copying the gaps (+2, +4, +2) would give 21 next, but 21 = 3 × 7 is not prime — follow the property, not the gaps.

AnswerThey are the prime numbers between 10 and 20; the next three are 23, 29 and 31.

Watch this explained “The trap at twenty-one”, 6:34 into One-to-one correspondence: counting without number words · हिंदी में देखें

Open NCERT p. 43Checked by computerAnswers can differ: one example

1. Closed under subtraction would mean: for any two natural numbers, subtracting one from the other always gives a natural number back.
2. Try 3 − 5: the result is −2, which is not a natural number.
3. Try 5 − 5: the result is 0, which is also not a natural number (natural numbers start at 1).

AnswerNo — natural numbers are not closed under subtraction; for example, 3 − 5 = −2 and 5 − 5 = 0, neither of which is a natural number.

Watch this explained “Where the pairing runs out”, 7:59 into One-to-one correspondence: counting without number words · हिंदी में देखें

Open NCERT p. 43Checked by computer

1. There are 4 fingers (not counting the thumb), each with 3 joints.
2. The thumb is used only to touch each joint in turn, so it is the counting tool, not one of the things being counted.
3. Total joints countable = 4 fingers × 3 joints = 12.

Answer12 joints on one hand — this is exactly why some ancient counting systems grouped numbers in twelves (base-12) instead of in tens.

Watch this explained “Twelve on one hand”, 3:22 into One-to-one correspondence: counting without number words · हिंदी में देखें

Open NCERT p. 46Checked by computer

1. Start at the noon temperature, 4 °C.
2. A drop of 15 °C means subtracting 15: 4 − 15.
3. 4 − 15 = −11.

Answer−11 °C

Watch this explained “A fall through nought”, 7:06 into Debts and fortunes: negative numbers close subtraction · हिंदी में देखें

Open NCERT p. 46Checked by computer

1. Write each event as a signed number: a loan (debt) of ₹850 is −850, a profit (fortune) of ₹1,200 is +1200, and a loss of ₹450 is −450.
2. Add them in order: −850 + 1200 − 450.
3. −850 + 1200 = 350, then 350 − 450 = −100.

Answer−850 + 1200 − 450 = −100, so the trader ends up ₹100 in debt.

Watch this explained “A week of trading”, 7:49 into Debts and fortunes: negative numbers close subtraction · हिंदी में देखें

Open NCERT p. 46Checked by computer

(i) (–12) × 5

1. A debt (−12) multiplied by a positive count (5) scales up the debt.
2. −12 × 5 = −60.

Answer−60

(ii) (–8) × (–7)

1. A debt multiplied by a debt becomes a fortune — multiplying two negatives gives a positive.
2. −8 × −7 = 56.

Answer56

(iii) 0 – (–14)

1. Taking away a debt is the same as adding the matching fortune: 0 − (−14) = 0 + 14.
2. 0 + 14 = 14.

Answer14

(iv) (–20) ÷ 4

1. Dividing a debt by a positive number keeps it a debt.
2. −20 ÷ 4 = −5.

Answer−5

Watch the lesson Why a debt times a debt is a fortune · हिंदी में देखें

Open NCERT p. 46One way to think about it

1. Suppose your net worth is ₹10, made up of ₹15 in cash and one debt slip of ₹5 (since 15 − 5 = 10).
2. If that ₹5 debt is cancelled (forgiven), no cash changes hands, but you no longer owe it, so your net worth becomes ₹15.
3. So cancelling a ₹5 debt raised your net worth from 10 to 15 — exactly as if ₹5 had been added.
4. That is why 10 − (−5) = 10 + 5 = 15: removing a debt has the same effect as adding the matching fortune.

In shortRemoving (cancelling) a debt raises your net worth by the size of the debt, the same effect as adding money — which is why subtracting a negative number equals adding the corresponding positive number.

Watch this explained “Tearing up the slip”, 8:45 into Debts and fortunes: negative numbers close subtraction · हिंदी में देखें

Open NCERT p. 49One way to think about it

(i) 2/3 and 4/6

1. Cross-multiply: 2 × 6 = 12 and 3 × 4 = 12.
2. The two products match, so 2/3 = 4/6.

In shortEqual — both cross-products are 12.

(ii) 5/4 and 10/8

1. Cross-multiply: 5 × 8 = 40 and 4 × 10 = 40.
2. The two products match, so 5/4 = 10/8.

In shortEqual — both cross-products are 40.

(iii) –3/5 and –6/10

1. Cross-multiply: (−3) × 10 = −30 and 5 × (−6) = −30.
2. The two products match, so −3/5 = −6/10.

In shortEqual — both cross-products are −30.

(iv) 9/3 and 3

1. Write 3 as the fraction 3/1.
2. Cross-multiply: 9 × 1 = 9 and 3 × 3 = 9.
3. The two products match, so 9/3 = 3.

In shortEqual — 9/3 simplifies to 3.

Watch this explained “The cross test”, 5:55 into What "rational" means, and why the denominator cannot be zero · हिंदी में देखें

Open NCERT p. 49Checked by computer

(i) 2/5 + 3/10

1. The common denominator of 5 and 10 is 10; rewrite 2/5 as 4/10.
2. 4/10 + 3/10 = 7/10.

Answer7/10

(ii) 7/12 + 5/8

1. The common denominator of 12 and 8 is 24; rewrite 7/12 as 14/24 and 5/8 as 15/24.
2. 14/24 + 15/24 = 29/24.

Answer29/24 (= 1 5/24)

(iii) –4/7 + 3/14

1. The common denominator of 7 and 14 is 14; rewrite −4/7 as −8/14.
2. −8/14 + 3/14 = −5/14.

Answer−5/14

Watch this explained “The four operations”, 6:57 into What "rational" means, and why the denominator cannot be zero · हिंदी में देखें

Open NCERT p. 49Checked by computer

(i) 5/6 – 1/4

1. The common denominator of 6 and 4 is 12; rewrite 5/6 as 10/12 and 1/4 as 3/12.
2. 10/12 − 3/12 = 7/12.

Answer7/12

(ii) 11/8 – 3/4

1. The common denominator of 8 and 4 is 8; rewrite 3/4 as 6/8.
2. 11/8 − 6/8 = 5/8.

Answer5/8

(iii) –7/9 – (–2/3)

1. Subtracting a negative means adding the positive: −7/9 − (−2/3) = −7/9 + 2/3.
2. The common denominator of 9 and 3 is 9; rewrite 2/3 as 6/9.
3. −7/9 + 6/9 = −1/9.

Answer−1/9

Watch this explained “The four operations”, 6:57 into What "rational" means, and why the denominator cannot be zero · हिंदी में देखें

Open NCERT p. 49Checked by computer

(i) 2/3 × 3/10

1. Multiply straight across: (2×3)/(3×10) = 6/30.
2. Reduce: 6/30 = 1/5.

Answer1/5

(ii) 7/11 × 5/8

1. Multiply straight across: (7×5)/(11×8) = 35/88.
2. 35 and 88 share no common factor, so this is already in lowest terms.

Answer35/88

(iii) –4/7 × 5/14

1. Multiply straight across, keeping the negative sign: (−4×5)/(7×14) = −20/98.
2. Reduce: −20/98 = −10/49.

Answer−10/49

Watch this explained “The four operations”, 6:57 into What "rational" means, and why the denominator cannot be zero · हिंदी में देखें

Open NCERT p. 49Checked by computer

(i) 2/3 ÷ 3/10

1. Turn the second fraction over and multiply: 2/3 ÷ 3/10 = 2/3 × 10/3.
2. 2/3 × 10/3 = 20/9.

Answer20/9 (= 2 2/9)

(ii) 7/11 ÷ 5/8

1. Turn the second fraction over and multiply: 7/11 ÷ 5/8 = 7/11 × 8/5.
2. 7/11 × 8/5 = 56/55.

Answer56/55 (= 1 1/55)

(iii) –4/7 ÷ 5/14

1. Turn the second fraction over and multiply: −4/7 ÷ 5/14 = −4/7 × 14/5.
2. −4/7 × 14/5 = −56/35, which reduces to −8/5.

Answer−8/5

Watch this explained “The four operations”, 6:57 into What "rational" means, and why the denominator cannot be zero · हिंदी में देखें

Open NCERT p. 50One way to think about it

1. Left side: 1/2 + 3/4 = 5/4 (common denominator 4), then 5/4 × 8/3 = 40/12 = 10/3.
2. Right side, first piece: 1/2 × 8/3 = 8/6 = 4/3.
3. Right side, second piece: 3/4 × 8/3 = 24/12 = 2.
4. Add the right-side pieces: 4/3 + 2 = 4/3 + 6/3 = 10/3.
5. Both sides equal 10/3, so the two expressions are equal.

In shortBoth sides equal 10/3, confirming that multiplication distributes over addition here.

Watch this explained “The four operations”, 6:57 into What "rational" means, and why the denominator cannot be zero · हिंदी में देखें

Open NCERT p. 50Checked by computer

1. Distributive property: a × (b − c) = a × b − a × c.
2. So 7/9 × (6/7 − 3/4) = 7/9 × 6/7 − 7/9 × 3/4.
3. First term: 7/9 × 6/7 = 42/63 = 2/3.
4. Second term: 7/9 × 3/4 = 21/36 = 7/12.
5. Subtract over the common denominator 12: 2/3 − 7/12 = 8/12 − 7/12 = 1/12.

Answer1/12

Watch this explained “The four operations”, 6:57 into What "rational" means, and why the denominator cannot be zero · हिंदी में देखें

Open NCERT p. 50Checked by computerAnswers can differ: one example

1. Expand the left side using the distributive property: 5/6 × x + 5/6 × 3/5 = 5/6 x + 5/6 × 3/5.
2. Work out 5/6 × 3/5 = 15/30 = 1/2, so the left side becomes 5/6 x + 1/2 — exactly the right side.
3. So the equation is true for every rational number x, not just one; x = 1 is one example that fits.

AnswerThe equation holds for every rational x — for example, x = 1 — because 5/6 × 3/5 already works out to 1/2.

Watch this explained “The four operations”, 6:57 into What "rational" means, and why the denominator cannot be zero · हिंदी में देखें

Open NCERT p. 52One way to think about it

1. Write all three numbers over one common denominator. 12 works for thirds, quarters and halves: 2/3 = 8/12, −5/4 = −15/12, and 1 1/2 = 3/2 = 18/12.
2. Cut every unit interval on the line into 12 equal parts.
3. Mark −5/4 by moving 15 twelfths to the left of 0; it lands between −2 and −1.
4. Mark 2/3 by moving 8 twelfths to the right of 0; it lands between 0 and 1.
5. Mark 1 1/2 by moving 18 twelfths to the right of 0; it lands halfway between 1 and 2.
6. Reading left to right: −5/4, then 0, then 2/3, then 1 1/2.

In shortOn one line cut into twelfths: −5/4 is 15 parts left of 0, 2/3 is 8 parts right of 0, and 1 1/2 is 18 parts right of 0.

Watch this explained “One cutting for several numbers”, 5:14 into Placing a rational number on the line, and distance as |a − b| · हिंदी में देखें

Open NCERT p. 52Checked by computerAnswers can differ: one example

1. Rewrite both ends over quarters: −1/2 = −2/4 and 1/4 = 1/4.
2. The integers strictly between the numerators −2 and 1 are −1 and 0, giving −1/4 and 0/4 = 0.
3. For a third number, cut finer into eighths: −1/2 = −4/8 and 1/4 = 2/8. The integers strictly between −4 and 2 are −3, −2, −1, 0 and 1, so 1/8 also lies strictly between them.
4. Many other answers are correct too; any three different rational numbers in this gap will do.

Answer−1/4, 0 and 1/8 (one possible answer).

Watch this explained “A second route”, 4:31 into Density: averaging always finds another rational in between · हिंदी में देखें

Open NCERT p. 53Checked by computer

1. The common denominator of 4 and 12 is 12; rewrite −1/4 as −3/12.
2. −3/12 + 5/12 = 2/12, which reduces to 1/6.

Answer1/6

Watch this explained “The four operations”, 6:57 into What "rational" means, and why the denominator cannot be zero · हिंदी में देखें

Open NCERT p. 53Checked by computer

1. Write both amounts as improper fractions: 15 3/4 = 63/4 metres of silk in total, and one kurta needs 2 1/4 = 9/4 metres.
2. Divide the total silk by the silk needed per kurta: (63/4) ÷ (9/4) = 63/4 × 4/9 = 63/9.
3. 63/9 = 7.

Answer7 kurtas.

Watch this explained “The four operations”, 6:57 into What "rational" means, and why the denominator cannot be zero · हिंदी में देखें

Open NCERT p. 53Checked by computerAnswers can differ: one example

1. 3.1415 and 3.1416 differ only in the fourth decimal place, so no number with four decimal places can sit strictly between them.
2. Go one decimal place finer: writing both to five places gives 3.14150 and 3.14160.
3. 3.14151, 3.14152 and 3.14153 all fall strictly between 3.14150 and 3.14160.

Answer3.14151, 3.14152 and 3.14153.

Watch this explained “Two close decimals”, 7:01 into Density: averaging always finds another rational in between · हिंदी में देखें

Open NCERT p. 53One way to think about it

1. There is more than one way. The book's way takes the average (a + b)/2; here is another.
2. Write both numbers over one common denominator. Any integer strictly between the two new numerators gives a rational number strictly between the original two.
3. Example, between 1/3 and 1/2: over sixths they are 2/6 and 3/6, and no integer lies strictly between 2 and 3.
4. So cut finer, into twelfths: 1/3 = 4/12 and 1/2 = 6/12. Since 5 lies between 4 and 6, 5/12 lies strictly between 1/3 and 1/2.
5. This always works: doubling the common denominator doubles both numerators, so they then differ by at least 2 and an integer fits between them.

In shortYes. Besides averaging, rewrite both numbers over a common denominator large enough that an integer sits between the numerators; for example, 5/12 lies between 1/3 and 1/2.

Watch this explained “A second route”, 4:31 into Density: averaging always finds another rational in between · हिंदी में देखें

Open NCERT p. 61Checked by computer

1. Write each fraction in lowest terms and factorise its denominator.
2. 20 = 2² × 5 and 250 = 2 × 5³ — both denominators use only the primes 2 and 5, so 7/20 and 13/250 stop (terminate).
3. 15 = 3 × 5 has a 3 in it, and no cancelling removes it (gcd(4,15)=1), so 4/15 never stops — it repeats.
4. Now divide to check: 7/20 = 0.35 (stops after 2 places).
5. 4/15 = 0.2666… — the 6 repeats forever.
6. 13/250 = 0.052 (stops after 3 places).

Answer7/20 = 0.35 (terminating); 4/15 = 0.26̅ = 0.2666… (repeating); 13/250 = 0.052 (terminating).

Watch this explained “The whole criterion”, 2:14 into Predicting the expansion from the denominator's prime factors · हिंदी में देखें

Open NCERT p. 62Checked by computer

1. Long division of 1 by 13 gives 1/13 = 0.076923076923…, so the repeating block is 076923 (6 digits).
2. Divide 2 by 13: 2/13 = 0.153846153846…, block 153846.
3. A block is a genuine rotation of 076923 only if it appears somewhere in 076923076923 (the block written twice). 153846 does not appear there, so 2/13 is NOT on the same cyclic ring as 1/13.
4. 3/13 = 0.230769230769… and 4/13 = 0.307692307692… — both of these ARE rotations of 076923.
5. So the twelve possible numerators split into two separate families of six: {1,3,4,9,10,12} share one ring of digits, and {2,5,6,7,8,11} share a different ring.

Answer1/13 = 0.0̅7̅6̅9̅2̅3̅ (repeating block 076923). 2/13 is not a rotation of that block, but 3/13 and 4/13 are.

Watch this explained “Trying it on thirteen”, 5:53 into Cyclic numbers: the hidden symmetry inside 1/7 · हिंदी में देखें

Open NCERT p. 62Checked by computer

(i) √81

1. √81 = 9, a whole number, so it is 9/1.

AnswerRational; 9/1.

(ii) √12

1. 12 is not a perfect square, so √12 = 2√3 cannot be written as p/q.

AnswerIrrational.

(iii) 0.33333 …

1. 0.33333… is a pure repeating decimal (block "3"), which always gives a fraction: x = 1/3.

AnswerRational; 1/3.

(iv) 0.123451234512345 …

1. The block "12345" (5 digits) repeats right after the point, so x = 12345/99999, which reduces to 4115/33333.

AnswerRational; 4115/33333.

(v) 1.01001000100001 … (Notice the pattern: Is it repeating a single block?)

1. The gaps of zeros keep growing (1, then 2, then 3, then 4 zeros, …), so there is no FIXED block that keeps repeating.
2. A decimal with no fixed repeating block cannot come from any fraction.

AnswerIrrational.

(vi) 23.560185612239874790120

1. The digits are given in full and then simply stop after the 21st decimal place — it is a terminating decimal.
2. Every terminating decimal is a fraction with a power of 10 as denominator: 23.560185612239874790120 = 589004640305996869753/25000000000000000000 (in lowest terms).

AnswerRational; 589004640305996869753/25000000000000000000.

Watch this explained “Six decimals, sorted”, 5:59 into Irrational decimals: an expansion with no stop and no repeating block · हिंदी में देखें

Open NCERT p. 62One way to think about it

1. Let x = 0.99999… (nines going on forever).
2. Multiply both sides by 10: 10x = 9.99999…
3. Subtract the first equation from the second: 10x − x = 9.99999… − 0.99999…
4. The infinite string of 9s after the point is identical on both numbers, so it cancels completely: 9x = 9.
5. Divide by 9: x = 1.

In shortx = 1, so 0.9̅ is not just close to 1 — it is exactly 1.

Watch this explained “Nine recurring”, 8:37 into Converting a terminating or repeating decimal back to p/q · हिंदी में देखें

Open NCERT p. 62Checked by computerAnswers can differ: one example

1. 1/7's block is cyclic because the repeating block is as long as it can possibly be: 6 digits, one less than the divisor 7. That uses up every possible remainder, so there is only one 'ring' and every numerator 1–6 shares it.
2. So look for other numbers n where the repeating block of 1/n has length exactly n − 1.
3. Testing small primes this way (by long division or by checking the block length): n = 17 gives a 16-digit cyclic block, n = 19 gives an 18-digit block, n = 23 gives a 22-digit block, and n = 29 gives a 28-digit block.
4. Each of these has only one ring of remainders, so — just like 1/7 — the multiples 1/n, 2/n, 3/n, … all turn out to be rotations of the very same block of digits.

Answern = 17, 19, 23, 29 (among others) — each gives a reciprocal whose repeating block is cyclic, because the block length equals n − 1.

Watch this explained “Tested both ways”, 7:27 into Cyclic numbers: the hidden symmetry inside 1/7 · हिंदी में देखें

Open NCERT p. 64Checked by computer

(i) 3/50

1. Divide 3 by 50: 3.000 ÷ 50 = 0.06 exactly, remainder becomes 0.

Answer0.06 (terminating).

(ii) 2/9

1. Divide 2 by 9: the remainder 2 keeps coming back at every step, so the digit 2 repeats forever.

Answer0.2̅ = 0.2222… (repeating).

Watch this explained “Watching the remainders”, 0:49 into Why long division must either stop or loop · हिंदी में देखें

Open NCERT p. 64One way to think about it

1. Suppose, for contradiction, that √5 IS rational. Then √5 = p/q for some integers p, q with no common factor (lowest terms) and q ≠ 0.
2. Squaring: 5 = p²/q², so p² = 5q². This means p² is a multiple of 5, and since 5 is prime, p itself must be a multiple of 5.
3. Write p = 5k for some integer k. Substitute back: (5k)² = 5q², so 25k² = 5q², giving q² = 5k².
4. This means q² is also a multiple of 5, so q itself must be a multiple of 5.
5. But now both p and q are multiples of 5 — they share the factor 5, which contradicts our assumption that p/q was in lowest terms with no common factor.
6. This contradiction shows the assumption was wrong, so √5 cannot be written as p/q — it is irrational.

In short√5 is irrational (proved by contradiction).

Watch this explained “Which numbers this works for”, 8:15 into Proof by contradiction: why √2 cannot be a ratio of integers · हिंदी में देखें

Open NCERT p. 64Checked by computer

(i) 12.6

1. 12.6 stops after one decimal place: 12.6 = 126/10 = 63/5.

Answer63/5

(ii) 0.0120

1. 0.0120 stops (the last 0 adds nothing): 0.0120 = 120/10000 = 3/250.

Answer3/250

(iii) 3.05̅2̅

1. Let x = 3.0525252…. One digit ('0') does not repeat, then '52' repeats (2 digits).
2. Multiply by 10 to move the non-repeating '0' past the point: 10x = 30.525252….
3. Multiply by 100 more to move one full block: 1000x = 3052.525252….
4. Subtract: 1000x − 10x = 3052.5252… − 30.5252… = 3022, so 990x = 3022 and x = 3022/990 = 1511/495.

Answer1511/495

(iv) 1.23̅5̅

1. Let x = 1.2353535…. One digit ('2') does not repeat, then '35' repeats.
2. 10x = 12.3535… and 1000x = 1235.3535….
3. 1000x − 10x = 1235 − 12 = 1223, so 990x = 1223 and x = 1223/990 (already in lowest terms).

Answer1223/990

(v) 0.2̅3̅

1. Let x = 0.232323…, a 2-digit block repeating right after the point.
2. 100x = 23.2323…, so 100x − x = 23, 99x = 23 and x = 23/99.

Answer23/99

(vi) 2.05̅

1. Let x = 2.05555…. One digit ('0') does not repeat, then '5' repeats.
2. 10x = 20.555… and 100x = 205.555….
3. 100x − 10x = 205 − 20 = 185, so 90x = 185 and x = 185/90 = 37/18.

Answer37/18

(vii) 2.125̅

1. Let x = 2.125555…. Two digits ('12') do not repeat, then '5' repeats.
2. 100x = 212.555… and 1000x = 2125.555….
3. 1000x − 100x = 2125 − 212 = 1913, so 900x = 1913 and x = 1913/900.

Answer1913/900

(viii) 3.125̅

1. Let x = 3.125555…, the same pattern as (vii).
2. 100x = 312.555… and 1000x = 3125.555….
3. 1000x − 100x = 3125 − 312 = 2813, so 900x = 2813 and x = 2813/900.

Answer2813/900

(ix) 2.1̅6̅2̅5̅

1. Let x = 2.16251625…, a 4-digit block '1625' repeating right after the point.
2. 10000x = 21625.1625…, so 10000x − x = 21625 − 2 = 21623, 9999x = 21623 and x = 21623/9999.

Answer21623/9999

Watch this explained “Two shifts, one subtraction”, 5:37 into Converting a terminating or repeating decimal back to p/q · हिंदी में देखें

Open NCERT p. 64One way to think about it

(i) 0.532

1. 0.532 = 532/1000 = 133/250, so it lies between 0 and 1.
2. Divide 0 to 1 into 10 equal parts: 0.532 lies between 0.5 and 0.6.
3. Divide 0.5 to 0.6 into 10 equal parts: 0.532 lies between 0.53 and 0.54.
4. Divide 0.53 to 0.54 into 10 equal parts: 0.532 is the 2nd mark after 0.53. (This is the same point as walking 133 of 250 equal parts from 0.)

In short0.532 is the 2nd of ten small marks between 0.53 and 0.54.

(ii) 1.15̅

1. 1.15̅ = 1.1555… (only the 5 repeats). Let x = 1.1555…; then 10x = 11.555… and 100x = 115.555….
2. 100x − 10x = 115 − 11 = 104, so 90x = 104 and x = 104/90 = 52/45 = 1 + 7/45.
3. So the point lies between 1 and 2: cut the unit from 1 to 2 into 45 equal parts and walk 7 of them to the right of 1.
4. Check by magnification: 1.1555… lies between 1.1 and 1.2, then between 1.15 and 1.16, then between 1.155 and 1.156.

In short1.15̅ = 52/45: the 7th of 45 equal parts from 1 towards 2, between 1.155 and 1.156.

Watch this explained “One cutting for several numbers”, 5:14 into Placing a rational number on the line, and distance as |a − b| · हिंदी में देखें

Open NCERT p. 65Checked by computerAnswers can differ: one example

1. Write 3 and 4 with a denominator fine enough to fit 6 numbers strictly between them: tenths work, since 3 = 30/10 and 4 = 40/10, leaving the whole numbers 31 to 39 available on top.
2. Pick any six: 31/10, 32/10, 33/10, 34/10, 35/10, 36/10 — that is 3.1, 3.2, 3.3, 3.4, 3.5, 3.6.

AnswerOne possible answer: 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 (i.e. 31/10, 32/10, 33/10, 34/10, 35/10, 36/10). Any six different rational numbers strictly between 3 and 4 are equally correct.

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Open NCERT p. 65Checked by computerAnswers can differ: one example

1. Over thirtieths, 2/5 = 12/30 and 3/5 = 18/30 — the whole numbers 13 to 17 sit strictly between 12 and 18, which is exactly 5 numbers.
2. So take 13/30, 14/30, 15/30, 16/30, 17/30.

AnswerOne possible answer: 13/30, 14/30, 15/30, 16/30, 17/30. Any five different rational numbers strictly between 2/5 and 3/5 are equally correct.

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Open NCERT p. 65Checked by computerAnswers can differ: one example

1. Over thirtieths, 1/6 = 5/30 and 2/5 = 12/30 — the whole numbers 6 to 11 sit strictly between 5 and 12 (six of them, plenty for 5).
2. Take any five: 6/30, 7/30, 8/30, 9/30, 10/30.

AnswerOne possible answer: 6/30 (=1/5), 7/30, 8/30 (=4/15), 9/30 (=3/10), 10/30 (=1/3). Any five different rational numbers strictly between 1/6 and 2/5 are equally correct.

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Open NCERT p. 65Checked by computer

1. Combine the left side over the common denominator 15: x/3 + x/5 = 5x/15 + 3x/15 = 8x/15.
2. So 8x/15 = 16/15, which means 8x = 16.
3. Divide both sides by 8: x = 2.

Answerx = 2.

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Open NCERT p. 65Checked by computer

1. From a + 1/b = 0, we get a = −1/b, i.e. a and 1/b are exact opposites (negatives of each other).
2. Multiply both sides of a = −1/b by b: ab = −1/b × b = −1.
3. So ab is always exactly −1, whatever nonzero values a and b actually take — it is negative.

Answerab is negative (in fact ab = −1 always).

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Open NCERT p. 65Checked by computer

1. Since the decimal has exactly 4 places and the last (4th) digit is non-zero, the whole number formed by pushing the decimal point 4 places is some integer p, giving the number = p/10⁴ = p/(2⁴·5⁴).
2. p's own last digit (units digit) equals that non-zero 4th decimal digit, so p is not a multiple of 10 — p cannot be divisible by both 2 and 5 at once.
3. So p is divisible by 2 alone (up to 2⁴), OR by 5 alone (up to 5⁴), OR by neither — never by both.
4. If p carries no factor of 5, the full 5⁴ survives in the reduced denominator. If p carries no factor of 2, the full 2⁴ survives. If p has neither, both survive.
5. In every case at least one of 2⁴ or 5⁴ survives completely in the lowest-form denominator — so yes, it must be divisible by 2⁴ or 5⁴ (though not necessarily by both).

AnswerYes — the lowest-form denominator must always be divisible by 2⁴ or by 5⁴ (an 'or', not necessarily both).

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Open NCERT p. 65Checked by computer

1. 125 = 5³, which uses only the prime 5 — so the decimal terminates.
2. The larger of the two exponents (twos: 0, fives: 3) is 3, so the decimal has exactly 3 places.
3. Check: multiply top and bottom by 2³=8 to reach a power of 10: 18×8/1000 = 144/1000 = 0.144.

AnswerTerminating, with exactly 3 decimal places (18/125 = 0.144).

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Open NCERT p. 65Checked by computer

1. The denominator is 2³ × 5¹ — only the primes 2 and 5, so the decimal terminates.
2. Topping up the smaller exponent (5¹) to match the larger (2³) needs multiplying by 5² to reach 2³×5³ = 10³.
3. So the number of decimal places equals the LARGER exponent, 3 — not the total count of prime factors (4).

Answer3 decimal places.

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Open NCERT p. 65Checked by computerAnswers can differ: one example

1. Over the LCM 12: a = 7/12 and b = 10/12. The numerators differ by only 3, but the question wants k₂ − k₁ > 6.
2. Multiply top and bottom of both by 3, so m = 36: a = 21/36 and b = 30/36. Now k₂ − k₁ = 30 − 21 = 9, which is more than 6. (m = 24 gives 14/24 and 20/24, a gap of exactly 6, which is not more than 6; larger multiples of 12 such as 48 also work.)
3. The whole numbers strictly between 21 and 30 are 22, 23, …, 29 — eight of them. Take any five: 22/36, 23/36, 24/36, 25/36, 26/36. Each lies between a and b because, over the same denominator 36, its numerator lies between 21 and 30.
4. Why the gap matters: the whole numbers strictly between k₁ and k₂ are k₁ + 1, k₁ + 2, …, k₂ − 1, which is k₂ − k₁ − 1 of them. To write n numbers this way we need k₂ − k₁ − 1 ≥ n, that is, k₂ − k₁ ≥ n + 1. If the gap is smaller, there are not enough whole-number numerators, and we must choose a larger m.
5. The condition k₂ − k₁ > n + 1 makes sure of this, with at least one to spare. (Strictly, a gap of exactly n + 1 is already enough: over 24ths, 14/24 and 20/24 leave exactly five numbers, 15/24 to 19/24.)

Answerk₁ = 21, k₂ = 30, m = 36 (a = 21/36, b = 30/36); five numbers: 22/36, 23/36, 24/36, 25/36, 26/36. Other valid choices exist (any multiple of 12 from 36 up). Between k₁ and k₂ there are k₂ − k₁ − 1 whole numbers, so n numbers need k₂ − k₁ ≥ n + 1; the condition k₂ − k₁ > n + 1 guarantees it.

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Open NCERT p. 65One way to think about it

1. Square the first equation: (x + y + z)² = 0² = 0.
2. Expand the left side: x² + y² + z² + 2(xy + yz + zx) = 0.
3. We are given xy + yz + zx = 0, so this becomes x² + y² + z² + 2(0) = 0, i.e. x² + y² + z² = 0.
4. Each of x², y², z² is a square of a rational number, so each is ≥ 0 (never negative).
5. A sum of three numbers that are each ≥ 0 can only equal 0 if every one of them is exactly 0.
6. So x² = 0, y² = 0, z² = 0, which forces x = 0, y = 0, z = 0.

In shortx = y = z = 0.

Open NCERT p. 66One way to think about it

1. Without loss of generality, take a < b (the two numbers are different — if a = b there is nothing strictly 'between' them).
2. Check it is above a: (a+b)/2 − a = (a + b − 2a)/2 = (b − a)/2. Since b > a, (b−a) is positive, so this difference is positive — meaning (a+b)/2 > a.
3. Check it is below b: b − (a+b)/2 = (2b − a − b)/2 = (b − a)/2, the SAME positive quantity — meaning (a+b)/2 < b.
4. So a < (a+b)/2 < b: the average always lies strictly between the two numbers.

In short(a+b)/2 always lies strictly between a and b (when a ≠ b).

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Open NCERT p. 66Checked by computer

1. The figure has 10 triangles fanned out from one shared corner. The very first (smallest) triangle has both legs equal to 1, so by Pythagoras its hypotenuse² = 1²+1² = 2, giving hypotenuse √2.
2. Every triangle after that reuses the PREVIOUS hypotenuse as one leg, and adds one fresh unit-length leg (marked '1' in the figure), so its hypotenuse² = (previous hypotenuse)² + 1².
3. Applying this ten times in a row: √2, √3, √4 (=2), √5, √6, √7, √8 (=2√2), √9 (=3), √10, √11.
4. So the fan of 10 triangles carries the hypotenuses √2 through √11, and it passes straight through the whole numbers 2 and 3 on the way.

Answer√2, √3, 2, √5, √6, √7, 2√2, 3, √10, √11 (in order, from the smallest triangle to the largest).

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