CBSE Important Questions Class 10 Maths Chapter 1 – Real Numbers

Class 10 Maths Chapter 1 Real Numbers is the first chapter of the CBSE Class 10 Maths syllabus. It explains the Fundamental Theorem of Arithmetic, which says that every number can be written as a product of prime numbers in only one way. Students learn how to find the HCF and LCM by prime factorisation, how HCF and LCM are related, how to prove that numbers like √2 and √3 are irrational, and when a fraction gives a terminating decimal.

These CBSE important questions for Class 10 Maths Chapter 1 cover the maximum capacity of a container using HCF, the relation between HCF, LCM and the two numbers, conjugate surds, the condition for a terminating decimal, prime factorisation of the denominator, a word problem on frogs jumping together using LCM, and proving that √2 is irrational. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

CBSE Important Questions Class 10 Maths Chapter 1 – Real Numbers

Q1. A milkman has 75 litres of milk in one can and 45 litres in another can. Determine the maximum capacity of a container that can be used to measure the milk of either container an exact number of times. Justify your method.

Answer: The container must divide both 75 and 45 completely, and we want the largest such container. So we need the HCF of 75 and 45.
75 = 3 × 5 × 5
45 = 3 × 3 × 5
Common factors = 3 × 5 = 15
So the maximum capacity of the container is 15 litres. With it, the first can gives exactly 5 measures and the second can gives exactly 3 measures.

Q2. The last two digits of the number 3400 are
(a) 39
(b) 29
(c) 01
(d) 43

Answer: (c) 01
Look at the pattern of the last two digits of the powers of 3:
31 = 03, 32 = 09, 33 = 27, 34 = 81, 35 = 43, ... and 320 ends in 01.
So the last two digits repeat after every 20 powers. Since 400 is a multiple of 20, 3400 also ends in 01.

Q3. HCF and LCM of two numbers a and b (a > b) are 3 and 600 respectively and the difference of the two numbers is 51. Which of the following relations is correct?
(a) b² − 51b = 1800
(b) b² + 51b = 1800
(c) a² − 51a = 1800
(d) a + b = 1800

Answer: (b) b² + 51b = 1800
We know that HCF × LCM = product of the two numbers.
So, a × b = 3 × 600 = 1800
Also given a − b = 51, so a = b + 51.
Putting this in the first equation:
(b + 51) × b = 1800
b² + 51b = 1800

Q4. What is the conjugate number of 2 + √5?

Answer: The conjugate of a surd is obtained by changing the sign between the two terms.
So the conjugate of 2 + √5 is 2 − √5.
Conjugates are useful because their product is a rational number: (2 + √5)(2 − √5) = 4 − 5 = −1.

Q5. Write the condition for a rational number to have a terminating decimal expansion. Hence find whether 13/3125 has a terminating decimal or a non-terminating repeating decimal.

Answer: Condition: A rational number p/q, in its lowest form, has a terminating decimal expansion only if the denominator q can be written as 2m × 5n, where m and n are whole numbers.
For 13/3125:
3125 = 5 × 5 × 5 × 5 × 5 = 55, which is 20 × 55.
This matches the required form, so 13/3125 has a terminating decimal expansion.
In fact, 13/3125 = 0.00416.

Q6. Without actual division, state whether 17/8 is a terminating or a non-terminating rational number.

Answer: Look at the denominator and write it as a product of primes.
8 = 2 × 2 × 2 = 23, which can be written as 23 × 50.
Since the denominator is of the form 2m × 5n, the number is a terminating decimal.
(17/8 = 2.125)

Q7. Which of the following is a non-terminating repeating decimal?
(a) 3/8
(b) 7/20
(c) 46/400
(d) 5/6

Answer: (d) 5/6
Check the denominators:
8 = 2³, 20 = 2² × 5, 400 = 2⁴ × 5² — all of the form 2m × 5n, so these give terminating decimals.
6 = 2 × 3, which has the prime factor 3. Since 3 is neither 2 nor 5, 5/6 gives a non-terminating repeating decimal (5/6 = 0.8333...).

Q8. Determine the decimal expansion of 961/625 and explain why it has this form by analysing the prime factorisation of the denominator.

Answer: On dividing, 961/625 = 1.5376, which is a terminating decimal.
Reason: A rational number p/q has a terminating decimal expansion if the prime factorisation of q is of the form 2m × 5n.
Here the denominator is 625, and 625 = 5 × 5 × 5 × 5 = 54.
This can be written as 20 × 54, which matches the required form. That is why the decimal expansion terminates.

Q9. Five frogs start jumping together and their jump intervals are 4, 8, 12, 15 and 16 seconds respectively. Determine the total number of times they will jump together in 80 minutes, including their initial jump. Explain your method.

Answer: They jump together again after a time equal to the LCM of their intervals.
4 = 2², 8 = 2³, 12 = 2² × 3, 15 = 3 × 5, 16 = 2⁴
LCM = 2⁴ × 3 × 5 = 240 seconds = 4 minutes.
Total time = 80 minutes.
Number of times they jump together after the start = 80 ÷ 4 = 20.
Adding the first jump at the very beginning, the total = 20 + 1 = 21 times.

Q10. Prove that √2 is an irrational number.

Answer: We use the method of contradiction.
Suppose √2 is rational. Then it can be written as √2 = p/q, where p and q are integers with no common factor other than 1, and q ≠ 0.
Squaring both sides: 2 = p²/q², so p² = 2q².
This means p² is even, and therefore p must also be even. Let p = 2k.
Then (2k)² = 2q², so 4k² = 2q², which gives q² = 2k².
This means q² is even, so q is also even.
But now both p and q are even, so they have 2 as a common factor. This contradicts our assumption that they have no common factor.
So our assumption was wrong, and √2 is irrational.

Q11. Find the HCF and LCM of 96 and 404 by the prime factorisation method. Hence verify that HCF × LCM = product of the two numbers.

Answer: Prime factorisation:
96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3
404 = 2 × 2 × 101 = 22 × 101
HCF = product of the smallest powers of the common factors = 22 = 4
LCM = product of the greatest powers of all factors = 25 × 3 × 101 = 32 × 303 = 9696
Verification:
HCF × LCM = 4 × 9696 = 38784
96 × 404 = 38784
Both sides are equal, so the result is verified.

Q12. State the Fundamental Theorem of Arithmetic. Check whether 6n can end with the digit 0 for any natural number n.

Answer: Fundamental Theorem of Arithmetic: Every composite number can be written as a product of prime numbers, and this factorisation is unique except for the order in which the prime factors are written.
Can 6n end with 0?
If a number ends with the digit 0, it must be divisible by 10, that is by 2 × 5. So its prime factorisation must contain both 2 and 5.
Now, 6n = (2 × 3)n = 2n × 3n. Its only prime factors are 2 and 3.
By the uniqueness of prime factorisation, 5 can never appear in it.
So 6n can never end with the digit 0 for any natural number n.

Related Study Material on Extramarks

Q.1 Show that 5 3 is irrational.   

Marks:3
Ans

Let us assume, to the contrary, that 5 3 is rational.
That is, we can find co-prime a and b (b 0) such that 5 3 = a/b
Therefore, 5 – a/b = 3
Rearranging this equation, we get ?3 = 5 – (a/b) = (5b – a)/b
Since a and b are integers, we get 5 – (a/b) is rational, and so ?3 is rational.
But this contradicts the fact that 3 is irrational.
Therefore, our assumption that 5 3 is rational is incorrect.
So, we conclude that 5  3 is irrational.

Q.2 Show that the square of any positive integer is of the form 3m or 3m+1 for some integer m.

Marks:4
Ans

Let a be any positive integer. Then it is of the form 3q or 3q+1 or 3q+2.

Case I When a = 3q, we have
a2 = (3q)2 = 9q2 = 3(3q2) = 3m ; where m = 3q2

Case II When a = 3q+1, we have
a2 = (3q+1)2 = 9q2 +6q+1
or, a2 = 3q(3q+2)+1 = 3m+1; where m = 3q(3q+2)

Case III When a = 3q+2, we have
a2 = (3q+2)2 = 9q2+12q+4=9q2+12q+3+1
or, a2 = 3(3q2+4q+1)+1 = 3m+1 ; where m = (3q2+4q+1)

Hence, a is either of the form 3m or 3m+1 for some integer m.

Q.3 Use Euclid’s division algorithm to find the HCF of 210 and 55.

Marks:4
Ans

Clearly, 210 > 55.
On applying Euclid’s Division Lemma to 210 and 55
we get, 210 = 55×3 + 45
Since remainder 45 0, therefore on again applying Euclid’s Division Lemma we get,
55 = 45×1 + 10
Since remainder 10 0 therefore, on again applying Euclid’s Division Lemma we get,
45 = 10×4 + 5
Since remainder 5 0 therefore, on again applying Euclid’s Division Lemma we get,
10 = 5×2 + 0
Since remainder is 0 therefore, HCF = 5

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