1
In 17 = 5q + r the pair allowed by the lemma is —
💡 Hint:
The remainder must sit from 0 up to 4.
Explanation:
17 = 5×3 + 2, and 0 ≤ 2 < 5. r = 7 is larger than the divisor.
Class 10 · Maths · Chapter 1 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
Real Numbers
Real Numbers
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- 1 Euclid’s division lemma
- 2 HCF by Euclid — part of Exercise 1.1
- 3 Forms of odds, squares and cubes — rest of Exercise 1.1
- 4 The fundamental theorem — start of Exercise 1.2
- 5 HCF and LCM — rest of Exercise 1.2
- 6 Irrational numbers — Exercise 1.3
- 7 Decimal expansions — Exercise 1.4
- Chapter winner — every lesson at mastery ★
🎬 Shorts
This chapter has no Short on the channel yet, so ten empty cards are not shown. The notes, diagrams and checks are complete.
📄 PDF download — notes + answer key
Euclid’s division lemma
यूक्लिड की विभाजन प्रमेयिका · NCERT 1.2 · the lemma · the egg puzzle
The remainder stays smaller than the divisorNotes
For positive integers a and b, Euclid’s division lemma says there are whole numbers q and r such that
a = bq + r, 0 ≤ r < b
q is the quotient and r is the remainder. r may be zero, but it cannot be equal to b or larger. Divide 17 by 5: 17 = 5×3 + 2. Here q = 3 and r = 2, and 2 is smaller than 5.
The book’s puzzle — eggs in a basketActivity
A trader’s basket holds at most 150 eggs. Count by twos and 1 is left, by threes 2, by fours 3, by fives 4, by sixes 5, and by sevens nothing is left.
Write that in the language of the lemma: a = 2u+1 = 3t+2 = 4s+3 = 5w+4 = 6q+5 = 7p. Each time the remainder is 1 less than the divisor, and the remainder on 7 is 0.
Worked exampleExample
Question: Find the number a ≤ 150 that meets the conditions above.
Formula: a = 7p, and a = 6q+5, a = 5w+4, a = 4s+3, a = 3t+2, a = 2u+1.
Check: 119 = 7×17 + 0. 119 = 6×19 + 5. 119 = 5×23 + 4. 119 = 4×29 + 3. 119 = 3×39 + 2. 119 = 2×59 + 1.
119 is smaller than 150. The next such number would be 119+420, which is larger than 150. So there are 119 eggs.
10-second revision
- a = bq + r and 0 ≤ r < b
- r = 0 means b divides a exactly
- The egg puzzle answers 119
Board tip · BSEBBoard tip
In the answer write both q and r, and also show the line r < b.
Board tip · CBSEBoard tip
Writing a remainder equal to the divisor breaks the lemma. 17 = 5×2 + 7 is the wrong form.
Check your understandingall correct = mastery ★
2
In a = bq + r the remainder r may equal b.
Explanation:
False — the condition is 0 ≤ r < b. r does not reach b.
3
For positive integers a and b, in a = bq + r the remainder condition is 0 ≤ r < ______.
💡 Hint:
The remainder is smaller than the divisor.
Explanation:
0 ≤ r < b.
4
In the puzzle with remainder 0 on sevens, and a ≤ 150, the number of eggs is —
💡 Hint:
119 = 7×17.
Explanation:
119 meets every remainder condition and is smaller than 150.
5
Write 38 divided by 7 in the form of Euclid’s lemma. Show q, r and the condition.
Model answer (board format)
38 = 7×5 + 3. Here q = 5, r = 3, and 0 ≤ 3 < 7.
Check your answer:
HCF by Euclid — part of Exercise 1.1
यूक्लिड से HCF — अभ्यास 1.1 के भाग · Exercise 1.1 · questions 1 and 3
Repeat the same division on the remainderNotes
If a > b, write a = bq + r from the lemma. If r = 0 then HCF = b. If r ≠ 0 then HCF(a, b) = HCF(b, r). Repeat until the remainder is 0. The last divisor is the HCF.
This is different from the lemma. The lemma writes one division. The algorithm carries that division on until the HCF.
Exercise 1.1 — questions 1 and 3Activity
In question 1, find the HCF by Euclid’s algorithm: (i) 135 and 225, (ii) 196 and 38220, (iii) 867 and 255.
In question 3, a contingent of 616 and a band of 32 march in the same number of columns. The greatest number of columns is HCF(616, 32).
Worked exampleExample
Question: HCF of 135 and 225.
Steps: 225 = 135×1 + 90.
135 = 90×1 + 45.
90 = 45×2 + 0.
The remainder is 0, so the HCF is 45. Check: 135 = 45×3 and 225 = 45×5.
| Pair | Last divisor | HCF |
|---|---|---|
| 867, 255 | 51 | 51 |
| 616, 32 | 8 | 8 |
| 196, 38220 | 196 | 196 |
10-second revision
- HCF(a, b) = HCF(b, r)
- Stop when the remainder is 0
- Parade columns = HCF(616, 32) = 8
Board tip · BSEBBoard tip
In every line write dividend, divisor, quotient and remainder. Marks are cut for only the final number.
Board tip · CBSEBoard tip
For 135 and 225 make the larger number the first dividend. Starting the other way adds steps.
Check your understandingall correct = mastery ★
1
The HCF of 135 and 225 is —
💡 Hint:
In the third line the remainder is 0 and the divisor is 45.
Explanation:
225 = 135×1 + 90, 135 = 90×1 + 45, 90 = 45×2 + 0. HCF = 45.
2
The HCF of 867 and 255 is 51.
Explanation:
True — 867 = 255×3 + 102, 255 = 102×2 + 51, 102 = 51×2 + 0.
3
The greatest number of columns for 616 and 32 members is ______.
💡 Hint:
616 = 32×19 + 8.
Explanation:
616 = 32×19 + 8 and 32 = 8×4 + 0, so 8.
4
Find the HCF of 867 and 255 by Euclid’s steps.
Model answer (board format)
867 = 255×3 + 102. 255 = 102×2 + 51. 102 = 51×2 + 0. HCF = 51.
Check your answer:
5
196 divides 38220 exactly. The HCF is —
💡 Hint:
The remainder is already 0 at the first step.
Explanation:
38220 = 196×195 + 0, so the HCF is 196.
6
While finding an HCF, if the first remainder is 0 then the HCF is the divisor.
Explanation:
True — the algorithm stops there. That is the case of 196 and 38220.
Forms of odds, squares and cubes — rest of Exercise 1.1
विषम, वर्ग और घन के रूप — अभ्यास 1.1 शेष · Exercise 1.1 · questions 2, 4, 5
Write the form by dividing every integer by bNotes
Any integer leaves remainder 0, 1, 2, 3, 4 or 5 on division by 6. The even forms are 6q, 6q+2 and 6q+4. The odd forms are 6q+1, 6q+3 and 6q+5.
For a square, take divisor 3. For a cube, take divisor 3 as well, then expand the cube. A proof that skips the list of remainders stays incomplete.
Exercise 1.1 — questions 2, 4 and 5Activity
Question 2: every positive odd integer is of the form 6q+1, 6q+3 or 6q+5.
Question 4: the square of a positive integer is of the form 3m or 3m+1. Hint: the number is 3q, 3q+1 or 3q+2. Square all three.
Question 5: the cube is of the form 9m, 9m+1 or 9m+8.
Worked exampleExample
Question: Why is the square of a positive integer only of the form 3m or 3m+1?
Forms: the number is 3q, 3q+1 or 3q+2.
(3q)² = 9q² = 3(3q²), which is 3m.
(3q+1)² = 9q² + 6q + 1 = 3(3q²+2q) + 1, which is 3m+1.
(3q+2)² = 9q² + 12q + 4 = 3(3q²+4q+1) + 1, which is also 3m+1.
In all three cases the square is 3m or 3m+1. It never becomes 3m+2.
10-second revision
- Odds: 6q+1, 6q+3, 6q+5
- Squares: 3m or 3m+1
- Cubes: 9m, 9m+1 or 9m+8
Board tip · BSEBBoard tip
Expand the square or the cube of all three forms. Stopping after one form is an incomplete proof.
Board tip · CBSEBoard tip
Calling 3m+2 a square is a common slip. The three expansions show that form does not arise.
Check your understandingall correct = mastery ★
1
A positive odd integer has the form —
💡 Hint:
6q+2 and 6q+4 are even.
Explanation:
After the even forms are removed, the odd forms 6q+1, 6q+3 and 6q+5 remain.
2
Show that the cube of a positive integer is of the form 9m, 9m+1 or 9m+8.
Model answer (board format)
The number is 3q, 3q+1 or 3q+2. (3q)³ = 27q³ = 9(3q³). (3q+1)³ = 27q³+27q²+9q+1 = 9(3q³+3q²+q)+1. (3q+2)³ = 27q³+54q²+36q+8 = 9(3q³+6q²+4q)+8. So the form is 9m, 9m+1 or 9m+8.
Check your answer:
3
The square of a positive integer can be of the form 3m+2.
Explanation:
False — the squares of the three forms give only 3m or 3m+1.
4
A number of the form 6q+4 is ______.
💡 Hint:
4 is even and 6q is even.
Explanation:
Even. The odd forms are 6q+1, 6q+3 and 6q+5.
The fundamental theorem — start of Exercise 1.2
अंकगणित की आधारभूत प्रमेय — अभ्यास 1.2 की शुरुआत · Theorem 1.2 · Exercise 1.2 questions 1, 5, 6
Prime factors are the same apart from orderNotes
Every composite number can be written as a product of primes, and this writing is unique apart from order. 12 = 2×2×3 = 3×2×2. Both are the same factors.
This is why 4ⁿ cannot end in 0. In 4ⁿ = 2²ⁿ the only prime is 2. A final 0 needs both 2 and 5. 6ⁿ = 2ⁿ×3ⁿ has no 5 either.
Exercise 1.2 — questions 1, 5 and 6Activity
Question 1: write 140, 156, 3825, 5005 and 7429 as products of prime factors.
Question 5: can 6ⁿ end with the digit 0?
Question 6: explain why 7×11×13+13 and 7×6×5×4×3×2×1+5 are composite. Take the common factor out.
Worked exampleExample
Question: Why is 7×11×13 + 13 composite?
Factor: 13 is in both terms. 13(7×11 + 1) = 13(77+1) = 13×78.
Both 13 and 78 are greater than 1, so the number is composite.
The other number: 7! + 5 = 5(7×6×4×3×2×1 + 1) = 5×1009. Both 5 and 1009 are greater than 1, so it is composite too.
10-second revision
- The prime factorisation of a composite number is unique
- 6ⁿ does not end in 0, because 5 is not a factor
- A common factor taken out shows a composite
Board tip · BSEBBoard tip
Write prime powers: 140 = 2²×5×7. Just 2×2×5×7 is also correct, but powers stay clear.
Board tip · CBSEBoard tip
A final 0 means divisibility by 10, so both 2 and 5. Being even is not enough.
Check your understandingall correct = mastery ★
1
The prime factorisation of 140 is —
💡 Hint:
4 and 35 are not prime.
Explanation:
140 = 2×70 = 2×2×35 = 2²×5×7.
2
For some natural number n, 6ⁿ can end with the digit 0.
Explanation:
False — 6ⁿ = 2ⁿ×3ⁿ has no 5, so it is not divisible by 10.
3
5005 = 5 × 7 × 11 × ______.
💡 Hint:
5×7×11 = 385, and 5005÷385 = 13.
Explanation:
13. Also 7429 = 17×19×23 and 3825 = 3²×5²×17.
4
7×11×13+13 is composite because —
💡 Hint:
13 comes out of the bracket.
Explanation:
13(77+1) = 13×78, two factors greater than 1.
5
Write the prime factors of 156 and 3825. Then say why 4ⁿ does not end in 0.
Model answer (board format)
156 = 2²×3×13. 3825 = 3²×5²×17. 4ⁿ = 2²ⁿ has only the prime 2. A final 0 also needs the factor 5, which is absent.
Check your answer:
HCF and LCM — rest of Exercise 1.2
HCF और LCM — अभ्यास 1.2 शेष · Exercise 1.2 · questions 2, 3, 4, 7
The product of two numbersNotes
For two positive integers a and b,
HCF(a, b) × LCM(a, b) = a × b
So LCM = a×b / HCF. For three numbers that product sentence does not hold. 6×72×120 is not equal to the product of their HCF and LCM. The book writes this caution for the reader.
Do not use the product rule on three numbersCaution
The HCF of 12, 15 and 21 is 3 and the LCM is 420. 12×15×21 = 3780, while 3×420 = 1260. They are not equal. Find the LCM of three numbers from the highest power of each prime.
Exercise 1.2 — questions 2, 3, 4 and 7Activity
Question 2: find HCF and LCM of the pairs 26 and 91, 510 and 92, 336 and 54, and check HCF×LCM = product.
Question 3: 12, 15, 21; then 17, 23, 29; then 8, 9, 25. Question 4: HCF(306, 657) = 9 is given, find the LCM.
Question 7: Sonia takes 18 minutes for one round and Ravi takes 12. The time until they meet again at the start is LCM(18, 12). The NCERT reprint marks this question as not for the examination.
Worked exampleExample
Question: If HCF(306, 657) = 9, find the LCM.
Formula: LCM = (first × second) / HCF.
Substitution: LCM = (306 × 657) / 9. 306/9 = 34, so 34 × 657.
657×30 = 19710 and 657×4 = 2628. The sum is 22338.
10-second revision
- For two: LCM = product / HCF
- For three, that formula does not hold
- Sonia and Ravi meet after 36 minutes
Board tip · BSEBBoard tip
Write the check line: 13×182 = 26×91. Question 2 is incomplete without the check.
Board tip · CBSEBoard tip
In the LCM take the highest power of each prime. In the HCF take the lowest.
Check your understandingall correct = mastery ★
1
The HCF and LCM of 26 and 91 are —
💡 Hint:
Both contain 13.
Explanation:
26 = 2×13, 91 = 7×13. HCF = 13, LCM = 2×7×13 = 182.
2
For three numbers too, HCF × LCM is always equal to the product of the three.
Explanation:
False — the book gives this caution for 6, 72 and 120. 12×15×21 ≠ 3×420.
3
If HCF(306, 657) = 9, then LCM = ______.
💡 Hint:
Divide 306×657 by 9.
Explanation:
22338.
4
Sonia and Ravi will meet again at the starting point after —
💡 Hint:
The LCM of 18 and 12.
Explanation:
18 = 2×3², 12 = 2²×3, LCM = 2²×3² = 36 minutes.
5
Find the HCF and LCM of 336 and 54 by prime factors and check with the product.
Model answer (board format)
336 = 2⁴×3×7, 54 = 2×3³. HCF = 2×3 = 6. LCM = 2⁴×3³×7 = 3024. Check: 6×3024 = 18144 and 336×54 = 18144.
Check your answer:
6
The HCF of 17, 23 and 29 is —
💡 Hint:
All three are prime and distinct.
Explanation:
There is no common prime, so HCF = 1 and LCM = 17×23×29 = 11339.
Irrational numbers — Exercise 1.3
अपरिमेय संख्याएँ — अभ्यास 1.3 · Theorem 1.3 · all three questions of Exercise 1.3
A prime that divides a square also divides the numberNotes
If a prime p divides a², then p also divides a. This follows from the fundamental theorem. It is how √2, √3 and √5 are proved irrational.
A rational can be written as p/q with q ≠ 0, and in a proof p and q are taken coprime. An irrational cannot be written in that form. √2, √3, π and 0.101101110… are irrational examples.
Exercise 1.3 — all three questionsActivity
Question 1: prove that √5 is irrational.
Question 2: prove that 3 + 2√5 is irrational.
Question 3: prove that (i) 1/√2, (ii) 7√5, (iii) 6 + √2 are irrational.
Worked exampleExample
Question: √5 is irrational.
Suppose √5 = a/b, where a and b are coprime integers and b ≠ 0.
Then a² = 5b². So 5 divides a², and by the theorem 5 divides a. Put a = 5k.
25k² = 5b², so b² = 5k². Then 5 also divides b.
Both a and b became divisible by 5, which clashes with the coprime assumption. So √5 is irrational.
10-second revision
- If p is prime and p | a², then p | a
- The clash with the coprime assumption is the proof
- Rational ± irrational, and irrational times a non-zero rational, stay irrational
Board tip · BSEBBoard tip
First prove the square root irrational, then lean on it for a sum such as 3+2√5.
Board tip · CBSEBoard tip
“It is irrational because the decimal is long” is not a proof. Write the contradiction step.
Check your understandingall correct = mastery ★
1
If √5 = a/b is assumed with a and b coprime, the next correct step is —
💡 Hint:
The theorem carries the prime from the square onto the number.
Explanation:
From a² = 5b², 5 divides a² and then a.
2
If 3 + 2√5 were rational, then √5 would be rational too.
Explanation:
True — √5 = (that rational − 3)/2. That is the clash, because √5 is irrational.
3
Show that 7√5 is irrational.
Model answer (board format)
Suppose 7√5 = a/b is rational, b ≠ 0. Then √5 = a/(7b), which would be rational. That clashes with √5 being irrational. So 7√5 is irrational.
Check your answer:
4
To show that 1/√2 is irrational, write it as ______ / 2.
💡 Hint:
Multiply numerator and denominator by √2.
Explanation:
√2 / 2. If this were rational, √2 would be rational.
Decimal expansions — Exercise 1.4
दशमलव प्रसार — अभ्यास 1.4 · Theorems 1.5 to 1.7 · Exercise 1.4
The decimal stops when the denominator is only 2 and 5Notes
In lowest terms p/q, if q = 2ⁿ5ᵐ with n and m non-negative integers, the decimal terminates. If q has any other prime, the decimal is non-terminating and repeating.
Every rational decimal either terminates or repeats. A decimal that neither stops nor repeats is irrational. In the Bihar book this is Exercise 1.4. The NCERT reprint of 2026-27 does not print this exercise.
Exercise 1.4 — all three questionsActivity
Question 1: without long division, say whether 13/3125, 17/8, 64/455, 15/1600, 29/343, 23/(2³5²), 129/(2²5⁷7), 6/15, 35/50 and 77/210 terminate or repeat.
Question 2: write the decimals of those that terminate. Question 3: 43.123456789 is a terminating rational; 0.120120012000… neither stops nor repeats, so it is irrational; if 43.123456789 carries a repeat bar, it is a repeating rational and the denominator has a prime other than 2 and 5.
Worked exampleExample
Question: Does 15/1600 terminate? Write the decimal.
Formula: if the lowest denominator is 2ⁿ5ᵐ, it terminates.
Substitution: 15/1600 = 3/320, because a 5 cancels. 320 = 2⁶×5. Only 2 and 5 appear, so it terminates.
To make the denominator 10⁶, multiply numerator and denominator by 5⁵: 3×3125 / 10⁶ = 9375/1000000 = 0.009375.
10-second revision
- Terminates ⇔ lowest denominator = 2ⁿ5ᵐ
- Another prime means non-terminating repeating
- Neither terminating nor repeating means irrational
Board tip · BSEBBoard tip
Simplify the denominator first. 6/15 = 2/5 terminates. Calling it repeating because you saw a 3 before cancelling is a mistake.
Board tip · CBSEBoard tip
77/210 = 11/30. 30 contains a 3, so it repeats. 210 had both 7 and 3.
Check your understandingall correct = mastery ★
1
Without dividing, the decimal of 17/8 is —
💡 Hint:
8 = 2³.
Explanation:
The denominator is 2³, so it terminates. 17/8 = 2.125.
2
The decimal of 64/455 terminates.
Explanation:
False — 455 = 5×7×13. 7 and 13 remain, so it is non-terminating repeating.
3
The lowest denominator of a terminating decimal has only the primes ______ and 5.
💡 Hint:
10 = 2×5.
Explanation:
2 and 5.
4
0.120120012000120000… is —
💡 Hint:
The run of zeros keeps growing.
Explanation:
It neither stops nor repeats one block, so it is irrational.
5
Which of 13/3125, 29/343 and 35/50 terminate? Write the type or the decimal with a reason.
Model answer (board format)
3125 = 5⁵, so 13/3125 terminates and equals 0.00416. 343 = 7³, so 29/343 is non-terminating repeating. 35/50 = 7/10, denominator 2×5, so it terminates and the decimal is 0.7.
Check your answer:
❓ Full question bank — with answers and explanations — 65 questions
Pick a type. The 35 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
BSEB modelpractice · not annual60 tried · 0 correctMultiple choiceMCQ120 tried · 0 correctTrue / falseTrue / false80 tried · 0 correctFill in the blanksFill80 tried · 0 correctMatchMatch20 tried · 0 correctAssertion–reasonAssertion–reason50 tried · 0 correctBalancingBalancing40 tried · 0 correctIdentifyIdentify20 tried · 0 correctShort and longVSA, SA, LA120 tried · 0 correctCBSE-stylenot a PYQ60 tried · 0 correct
No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.
Multiple choice
0/12Pick one option.
1
In a = bq + r the correct condition on r is —
💡 Hint:
The remainder is smaller than the divisor and not negative.
Explanation:
0 ≤ r < b.
2
The HCF of 225 and 135 is —
💡 Hint:
The third line of Euclid.
Explanation:
HCF = 45.
3
The HCF of 616 and 32 is —
💡 Hint:
616 = 32×19 + 8.
Explanation:
8 columns.
4
The square of a positive integer cannot be —
💡 Hint:
Look at the squares of the three forms.
Explanation:
3m+2 does not arise.
5
140 = —
💡 Hint:
Every factor should be prime.
Explanation:
2²×5×7.
6
6ⁿ does not end in 0 because —
💡 Hint:
A final 0 means 2×5.
Explanation:
6ⁿ = 2ⁿ×3ⁿ has no 5.
7
If HCF(306, 657) = 9, the LCM is —
💡 Hint:
Divide the product by the HCF.
Explanation:
22338.
8
For 26 and 91, HCF × LCM equals —
💡 Hint:
The rule for two numbers.
Explanation:
13×182 = 26×91.
9
The LCM of 12, 15 and 21 is —
💡 Hint:
The product of the highest powers.
Explanation:
2²×3×5×7 = 420.
10
The contradiction in proving √2 irrational is that —
💡 Hint:
The coprime assumption breaks.
Explanation:
Both become divisible by 2.
11
The decimal of 17/8 —
💡 Hint:
8 = 2³.
Explanation:
It terminates, 2.125.
12
After simplifying 77/210, the denominator still has —
💡 Hint:
77/210 = 11/30.
Explanation:
30 = 2×3×5 contains 3, so it repeats.
True or false
0/81
In Euclid’s lemma the remainder may be negative.
Explanation:
False — 0 ≤ r < b.
2
The HCF of 38220 and 196 is 196.
Explanation:
True — 196 divides 38220 exactly.
3
4ⁿ can end with the digit 0 for some natural n.
Explanation:
False — 5 is not a factor.
4
The HCF of 510 and 92 is 2.
Explanation:
True — 510 = 2×3×5×17 and 92 = 2²×23.
5
3 + 2√5 is rational.
Explanation:
False — that assumption would make √5 rational.
6
The decimal of 13/3125 terminates.
Explanation:
True — 3125 = 5⁵.
7
0.120120012000… is a repeating rational.
Explanation:
False — the number of zeros grows, so no block repeats.
8
The LCM of 18 and 12 is 36.
Explanation:
True — 2²×3² = 36.
Fill in the blanks
0/81
Euclid: a = ______ + r.
💡 Hint:
a = bq + r.
Explanation:
q is the quotient.
2
The HCF of 135 and 225 is ______.
💡 Hint:
45.
Explanation:
The last divisor.
3
156 = 2² × 3 × ______.
💡 Hint:
13.
Explanation:
156÷12 = 13.
4
For two numbers, LCM = product / ______.
💡 Hint:
HCF.
Explanation:
Only for two numbers.
5
If a prime p divides a², then p divides ______.
💡 Hint:
a.
Explanation:
From the square onto the number.
6
The denominator of a terminating decimal has the form 2ⁿ ______.
💡 Hint:
5ᵐ.
Explanation:
The other prime is 5.
7
The terminating decimal of 35/50 is ______.
💡 Hint:
0.7.
Explanation:
7/10.
8
7429 = 17 × 19 × ______.
💡 Hint:
23.
Explanation:
17×19 = 323.
Match
0/21
Match the statement with its meaning.
Column B: A. Prime factors are unique · B. a = bq + r · C. Denominator = 2ⁿ5ᵐ · D. HCF until remainder 0
1. The lemma
2. The algorithm
3. The fundamental theorem
4. A terminating decimal
Explanation:
The lemma is one division, the algorithm is the HCF, the theorem is unique factors, and a terminating denominator is 2ⁿ5ᵐ.
2
Match the number with its decimal type.
Column B: A. Terminating · B. Non-terminating repeating · C. Irrational · D. Repeating rational
1. 17/8
2. 1/3
3. √2
4. 22/7
Explanation:
17/8 terminates, 1/3 repeats, √2 is irrational, and 22/7 is a repeating rational.
Assertion–reason
0/5Check both statements first.
1
Assertion (A): The HCF of 135 and 225 is 45.
Reason (R): 225 = 135×1 + 90, 135 = 90×1 + 45 and 90 = 45×2 + 0.
Explanation:
Both are true and R is the calculation that gives A.
2
Assertion (A): 6ⁿ does not end in 0.
Reason (R): For two numbers, HCF × LCM equals their product.
Explanation:
Both are true, but R does not explain A. The reason for A is that 5 is missing from the factors.
3
Assertion (A): The decimal of 13/3125 terminates.
Reason (R): 7 is a factor of 3125.
Explanation:
A is true because 3125 = 5⁵. R is false.
4
Assertion (A): For three numbers, HCF × LCM always equals the product of the three.
Reason (R): For two numbers, HCF × LCM equals the product of those two.
Explanation:
A is false. R is true.
5
Assertion (A): √5 is irrational.
Reason (R): Assuming √5 = a/b makes both a and b divisible by 5.
Explanation:
Both are true and R is the proof of A.
Coefficient practice
0/4This is coefficient practice. It is not a result of this maths chapter. These are small equations of water and carbon dioxide forming.
1
This is coefficient practice and not a result of this maths chapter. Balance the equation of water forming. Write coefficients only in the coefficient places.
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
2
This is coefficient practice and not a result of this maths chapter. Balance the equation of carbon dioxide forming from carbon.
C + O2 → CO2
Explanation: C + O2 → CO2
3
This is coefficient practice and not a result of this maths chapter. Balance the equation of carbon dioxide forming from carbon monoxide.
CO + O2 → CO2
Explanation: 2CO + O2 → 2CO2
4
This is coefficient practice and not a result of this maths chapter. Burning methane forms carbon dioxide and water. Balance the equation.
CH4 + O2 → CO2 + H2O
Explanation: CH4 + 2O2 → CO2 + 2H2O
Classify
0/21
Place each number as terminating, repeating, or irrational.
1/2
1/3
√2
7/8
Explanation:
1/2 and 7/8 terminate. 1/3 repeats. √2 is irrational.
2
Place each statement with the lemma or the algorithm.
a = bq + r
0 ≤ r < b
Repeat until remainder 0
HCF(a, b) = HCF(b, r)
Explanation:
The first two lines are the lemma. The last two are the algorithm.
Very short answer
0/51
Write Euclid’s division lemma in one line.
Model answer (board format)
For positive integers a and b, a = bq + r, where 0 ≤ r < b.
Check your answer:
2
What is the HCF of 196 and 38220? Write one reason.
Model answer (board format)
196, because 38220 = 196×195 + 0.
Check your answer:
3
Write the relation between HCF and LCM for two numbers.
Model answer (board format)
HCF × LCM = the product of the two numbers.
Check your answer:
4
Write the form of the denominator of a terminating decimal.
Model answer (board format)
In lowest terms the denominator is 2ⁿ5ᵐ.
Check your answer:
5
Why is 7×6×5×4×3×2×1 + 5 composite?
Model answer (board format)
5(7×6×4×3×2×1 + 1) = 5×1009. Both factors are greater than 1.
Check your answer:
Short answer
0/41
Find the HCF and LCM of 510 and 92 and write the check.
Model answer (board format)
510 = 2×3×5×17, 92 = 2²×23. HCF = 2, LCM = 2²×3×5×17×23 = 23460. Check: 2×23460 = 510×92 = 46920.
Check your answer:
2
Show that 6 + √2 is irrational.
Model answer (board format)
Suppose 6 + √2 is rational. Then √2 = that rational − 6, which would be rational. This clashes with √2 being irrational.
Check your answer:
3
Write the HCF and LCM of 8, 9 and 25 from prime factors. Why is the product rule not used here?
Model answer (board format)
8 = 2³, 9 = 3², 25 = 5². HCF = 1, LCM = 2³×3²×5² = 1800. For three numbers, HCF × LCM is not equal to the product of the three.
Check your answer:
4
Write the decimal type of 6/15 and of 129/(2²×5⁷×7), with a reason.
Model answer (board format)
6/15 = 2/5, denominator 5, so it terminates and the decimal is 0.4. The other denominator contains 7, so it is non-terminating repeating.
Check your answer:
Long answer
0/31
Find the HCF of 867 and 255 by Euclid’s algorithm. Then say what this kind of number does in the parade question. Also find the HCF of 616 and 32.
Model answer (board format)
867 = 255×3 + 102, 255 = 102×2 + 51, 102 = 51×2 + 0, so the HCF is 51. In the parade the greatest number of columns is an HCF. 616 = 32×19 + 8, 32 = 8×4 + 0, so 8 columns.
Check your answer:
2
Prove that √5 is irrational. Using that, prove that 3 + 2√5 is irrational.
Model answer (board format)
Suppose √5 = a/b with a and b coprime. From a² = 5b², 5 divides a. Put a = 5k and 5 also divides b. Contradiction. If 3 + 2√5 were rational, √5 = (rational − 3)/2 would be rational, which clashes with the first result.
Check your answer:
3
Write the rule for terminating and non-terminating repeating decimals. Apply it to 15/1600 and 64/455. Find the decimal of 15/1600.
Model answer (board format)
If the lowest denominator is 2ⁿ5ᵐ the decimal terminates; otherwise it repeats. 15/1600 = 3/320 and 320 = 2⁶×5, so it terminates. 3×5⁵/10⁶ = 0.009375. 455 = 5×7×13 contains 7 and 13, so 64/455 is non-terminating repeating.
Check your answer:
BSEB model paper · practice
0/6This model set is for practice. It is not a question from any year’s annual examination.
1
The greatest number of columns for 616 and 32 in the parade is —
💡 Hint:
Take the HCF.
Explanation:
8.
2
One prime factor of 3825 is —
💡 Hint:
3825 = 3²×5²×17.
Explanation:
17 is prime.
3
LCM(18, 12) = ______ minutes.
💡 Hint:
36.
Explanation:
Sonia and Ravi.
4
Write the three forms of a positive odd integer.
Model answer (board format)
6q+1, 6q+3 and 6q+5, where q is an integer.
Check your answer:
5
This is coefficient practice and not a result of this maths chapter. Balance the equation of water forming.
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
6
Write the HCF of 135 and 225 by Euclid. Then find the LCM from HCF(306, 657) = 9. Add the three-number caution in one line.
Model answer (board format)
225 = 135×1 + 90, 135 = 90×1 + 45, 90 = 45×2 + 0, HCF = 45. LCM = 306×657/9 = 22338. For three numbers, HCF × LCM is not equal to their product.
Check your answer:
CBSE-style questions
0/6These are competency-based practice questions. They are not copies of a past paper.
1
A student wrote 17 = 5×2 + 7. The mistake is —
💡 Hint:
r < b is required.
Explanation:
7 > 5, so this is not the lemma’s form. The correct writing is 17 = 5×3 + 2.
2
A shopkeeper wants to stack 6ⁿ boxes so that the last row is a full ten. This —
💡 Hint:
A full ten means a final 0.
Explanation:
6ⁿ has no 5, so it is not divisible by 10.
3
Assertion (A): The decimal of 15/1600 terminates.
Reason (R): If a 3 is visible in the denominator before cancelling, the decimal is always non-terminating.
Explanation:
A is true, 15/1600 = 3/320 = 0.009375. R is false, because the 3 cancels and the denominator remains 2⁶×5.
4
Assertion (A): Two buses leave the same stop every 18 and 12 minutes. They leave together again after 36 minutes.
Reason (R): The next shared departure is the LCM of the two intervals.
Explanation:
Both are true and R explains A.
5
A student says the LCM of 12, 15 and 21 equals 12×15×21 / HCF. Put the numbers in and show where this breaks, and write the correct LCM.
Model answer (board format)
HCF = 3. 12×15×21 / 3 = 1260, but 1260 is not their LCM. From prime powers the LCM is 2²×3×5×7 = 420. The two-number formula does not hold for three numbers.
Check your answer:
6
A cashier must split bundles of 867 and 255 rupees into identical piles of the greatest size. Find the pile size and the number of piles.
Model answer (board format)
The pile size is the HCF, 51 rupees. 867/51 = 17 piles and 255/51 = 5 piles. That is 22 piles in all.
Check your answer:
🏛️ Board exam corner — Bihar Board (BSEB)— CBSE
Switch board with BSEB | CBSE above. The lessons follow the same NCERT chapter.
BSEB model · practiceBoard tip
This page has no verified annual-exam question, because no source page has been added. The model set below is practice in the board pattern.
🏛️ Model questions on one page →
The verified label will be used only when a source page for the question is available.
CBSE-style · competency-based
These are case and assertion-reason practice items. Do not treat them as past CBSE questions.
🔁 Spaced review — today’s questions
Wrong questions return soon; correct ones return after a few days.
🧠 What you learned + equation sheet
What you learned
- Euclid: a = bq + r, with 0 ≤ r < b.
- The HCF is the last divisor that leaves remainder zero.
- Every composite number has a prime factorisation that is unique apart from order.
- For two numbers, HCF × LCM equals the product of the numbers.
- If a prime p divides a², then p divides a. This is why √2 and √5 are irrational.
- A terminating decimal has denominator of the form 2ⁿ5ᵐ. Otherwise a rational decimal repeats.
| What | Keep this |
|---|---|
| Lemma | a = bq + r, 0 ≤ r < b |
| HCF steps | HCF(a, b) = HCF(b, r) |
| Two numbers | HCF × LCM = product |
| Prime and square | p | a² ⇒ p | a |
| Terminating | q = 2ⁿ5ᵐ |
| Repeating | q has another prime |
The notes are original writing. The textbook was used only for activity order and numbers. “Verified” will be used only on a question that has a source page.