Class 10 Maths · Original EZT practice
Real Numbers (Class 10): original worked practice
A long division, a prime factorisation and a proof each need a different kind of check. Work a question on paper before opening the steps, then try a new question without the answer in view.
8 original worked examples, 8 fresh self-check questions and 8 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 17 October 2026:
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Euclid: follow the remainder
Choose the method: For positive integers a and b, write a = bq + r with 0 ≤ r < b. Repeat with divisor and remainder. The last non-zero divisor is the HCF. This is extra revision here.
EZT practice question
Find the HCF of 1008 and 396.
Show the worked solution
- 1008 = 396 × 2 + 216.
- 396 = 216 × 1 + 180.
- 216 = 180 × 1 + 36; 180 = 36 × 5 + 0.
Answer: 36
Check: 1008 ÷ 36 = 28 and 396 ÷ 36 = 11. Since 28 and 11 are coprime, no larger common factor remains.
Watch for this: Stop at the last non-zero divisor, not at zero.
Now try it yourself
Find the HCF of 714 and 306.
Check your answer
102
714 = 306 × 2 + 102, then 306 = 102 × 3.
When division ends immediately
Choose the method: If the smaller positive number divides the larger, it is their HCF.
EZT practice question
Find the HCF of 176 and 4928.
Show the worked solution
- 4928 = 176 × 28 + 0.
- The smaller number is already a divisor of both.
Answer: 176
Check: Any common factor must divide 176, so none can exceed 176.
Watch for this: An exact division needs no extra Euclid step.
Now try it yourself
Find the HCF of 145 and 3770.
Check your answer
145
3770 = 145 × 26.
Prime factors, HCF and LCM
Choose the method: Prime factorisation is unique apart from factor order. Take minimum shared exponents for HCF and maximum exponents for LCM.
EZT practice question
Find the HCF and LCM of 84 and 150.
Show the worked solution
- 84 = 2² × 3 × 7; 150 = 2 × 3 × 5².
- HCF = 2 × 3. LCM = 2² × 3 × 5² × 7.
Answer: HCF 6; LCM 2100
Check: 6 × 2100 = 12600 = 84 × 150. Also 2100 is divisible by both numbers.
Watch for this: The product identity is for two positive integers, not a general three-number rule.
Now try it yourself
Find the HCF and LCM of 45 and 72.
Check your answer
HCF 9; LCM 360
45 = 3² × 5 and 72 = 2³ × 3²; 9 × 360 = 45 × 72.
Recover a missing number
Choose the method: For two positive integers, first number × second number = HCF × LCM. Then verify both HCF and LCM.
EZT practice question
Two positive integers have HCF 14 and LCM 1260. One is 140. Find the other.
Show the worked solution
- Product = 14 × 1260 = 17640.
- Other number = 17640 ÷ 140 = 126.
Answer: 126
Check: 140 = 14 × 10 and 126 = 14 × 9. Since 10 and 9 are coprime, HCF 14 and LCM 14 × 10 × 9 = 1260.
Watch for this: The product alone is not enough if a question gives inconsistent HCF/LCM data.
Now try it yourself
HCF 9, LCM 630, one number 90. Find the other.
Check your answer
63
9 × 630 ÷ 90 = 63; 90 = 9 × 10 and 63 = 9 × 7.
A contradiction proof
Choose the method: If a prime divides the square of an integer, it divides that integer. Use this fact with a fraction in lowest terms.
EZT practice question
Prove that √7 is irrational.
Show the worked solution
- Suppose √7 = p/q for coprime integers p and q, q ≠ 0.
- Squaring gives p² = 7q², so 7 divides p. Write p = 7k.
- Then 49k² = 7q², so q² = 7k². Thus 7 divides q as well.
- Both p and q have factor 7, contradicting their being coprime.
Answer: √7 is irrational.
Check: The contradiction concerns the assumed lowest-terms fraction, not the arithmetic of square roots.
Watch for this: Do not claim this prime-divisibility step for an arbitrary composite number.
Now try it yourself
Prove that √11 is irrational.
Check your answer
It is irrational.
Assume √11 = p/q in lowest terms; p² = 11q² forces 11 to divide both p and q, a contradiction.
Can the decimal end?
Choose the method: Reduce the positive-denominator fraction first. A rational number terminates exactly when its reduced denominator has only prime factors 2 and 5. This section is extra revision.
EZT practice question
Does 19/1250 terminate? Find its decimal.
Show the worked solution
- 1250 = 2 × 5⁴, so the reduced denominator uses only 2 and 5.
- Multiply numerator and denominator by 8: 19/1250 = 152/10000.
Answer: 0.0152, terminating
Check: 0.0152 × 1250 = 19.
Watch for this: A denominator condition applies after simplification.
Now try it yourself
Does 23/400 terminate? Find its decimal.
Check your answer
Yes: 0.0575
400 = 2⁴ × 5²; 23/400 = 575/10000.
A remaining prime changes the answer
Choose the method: A reduced denominator containing a prime other than 2 or 5 gives a non-terminating recurring decimal.
EZT practice question
Does 17/189 terminate?
Show the worked solution
- 17 and 189 share no factor.
- 189 = 3³ × 7, so the reduced denominator contains primes other than 2 and 5.
Answer: No. It is non-terminating recurring.
Check: It is still rational: recurring decimals are not irrational.
Watch for this: Never use "non-terminating" alone as a test for irrationality.
Now try it yourself
Does 35/245 terminate?
Check your answer
No.
35/245 = 1/7; factor 7 remains.
Can a power end in zero?
Choose the method: An integer ending in zero must have factors 2 and 5. Check the prime factors available.
EZT practice question
Can 14ⁿ end in zero for a non-negative integer n?
Show the worked solution
- For n ≥ 1, 14ⁿ = 2ⁿ × 7ⁿ, with no factor 5.
- For n = 0, 14⁰ = 1.
Answer: No.
Check: The positive powers end in 4 or 6, never 0; the prime-factor argument covers every n.
Watch for this: State the domain of the exponent; a negative power is not generally an integer.
Now try it yourself
Can 35ⁿ end in zero for non-negative integer n?
Check your answer
No.
For n ≥ 1 it has factors 5 and 7 but no 2; at n = 0 it equals 1.
Try a mixed set
Choose a method or explanation before opening each answer.
1. HCF of 432 and 180?
Show answer and reason
36
432 = 180 × 2 + 72; 180 = 72 × 2 + 36.
2. LCM of 28 and 45?
Show answer and reason
1260
They are coprime, so LCM is their product.
3. HCF 8, LCM 480, one number 96: find the other.
Show answer and reason
40
8 × 480 ÷ 96 = 40; gcd(96, 40) = 8.
4. Is √49 irrational?
Show answer and reason
No: √49 = 7.
A perfect-square root is an integer.
5. Does 21/150 terminate?
Show answer and reason
Yes: 0.14.
Reduce to 7/50, whose denominator is 2 × 5².
6. Does 18/132 terminate?
Show answer and reason
No.
Reduce to 3/22; factor 11 remains.
7. Can 21ⁿ end in zero for non-negative integer n?
Show answer and reason
No.
No factor 2 or 5 occurs; at n = 0 the result is 1.
8. Is HCF × LCM = product valid for 6, 10, 15?
Show answer and reason
No.
HCF 1 and LCM 30 give 30, but 6 × 10 × 15 = 900.
Know the rule but lose track of the steps?
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Questions about this practice page
- Is every topic here in the current syllabus?
- No. The linked NCERT rationalised-content list drops the Euclid and rational-decimal sections. They remain useful extra revision, but use your current school topic list for assessed coverage.
- How should I check an LCM?
- For two positive integers, check that each divides your LCM and that HCF × LCM equals their product. The product identity does not extend unchanged to three numbers.
- Why must a fraction be in lowest terms?
- A denominator factor may cancel. For example, 14/35 reduces to 2/5 and terminates even though 35 contains factor 7.
- What does an irrationality proof need?
- State the rational assumption and lowest-terms condition, show why a shared factor follows, then state the contradiction. A decimal approximation is not a proof.
- Are these NCERT exercise answers?
- No. These are original practice questions with worked explanations, not an exercise-number answer key. The official sources are linked for topic and edition context.