CBSE Important Questions for Class 10 Maths (2025-26) Chapter 1: Real Numbers with Answers, Download PDF

Particulars

Details

Board 

Central Board of Secondary Examination (CBSE)

Exam Mode

Offline

Subject

Mathematics

Paper Name

Maths Standard and Maths Basic

Medium/ Language

English and Hindi

Exam Duration

3 Hours

Total Marks

100

Theory Paper

80 Marks

Internal Assessment

20 Marks

Question No.

Answer

1.

c) √5

2.

a) 2^2×3×7

3.

b) 72

4.

b) (9, 28)

5.

(7)

6.

60 =2^2×3×7

7.

a=bq+r, where 0≤r<b

8.

(r) is the remainder

9.

Rational (0 = 0/1).

10.

(|a|) (the absolute value of (a)

Question No

Answer

1.

HCF of 1078 and 462 using Euclid's Algorithm

We apply the division algorithm repeatedly: a=bq+r, where 0≤r<b.

1. 1078=462×2+154

2. 462=154×3+0

Since the remainder is 0, the divisor at this stage is the HCF.

HCF(1078,462)=154

2.

Show that √7 is irrational

We use proof by contradiction.

• Assume √7 is rational. This means we can write √7=ba, where a and b are coprime integers (b=0).

• Square both sides: 7=b2/a2, which gives a2=7b2.

• This means a2 is divisible by 7. By the theorem (if a prime number p divides a2, then p divides a), a is divisible by 7.

• Since a is divisible by 7, we can write a=7c for some integer c.

• Substitute this back into a2=7b2: (7c)2=7b2, 49c2=7b2, 7c2=b2

• This means b2 is divisible by 7, and thus b is also divisible by 7.

• Since both a and b are divisible by 7, they have a common factor of 7. This contradicts our initial assumption that a and b are coprime.

• Therefore, our initial assumption is false.
  √7 is an irrational number.

3.

4.

Express 2310 as a product of its prime factors

2310=2×3×5×7×11

5.

6.

If √p is prime, show √p is irrational

This is the general form of question 2. We use proof by contradiction.

• Assume √p is rational: p=ba, where a and b are coprime integers (b=0).

• Squaring both sides gives p=b2/a2, so a2=pb2.

• This means a2 is divisible by p. Since p is a prime number, a must be divisible by p.

• Let a=pk for some integer k.

• Substitute a=pk into a2=pb2: (pk)2=pb2, p2k2=pb2, pk2=b2

• This means b2 is divisible by p, and thus b must also be divisible by p.

• Since both a and b are divisible by p, they share a common factor of p. This contradicts the assumption that a and b are coprime.

• Therefore, the assumption is false.

√p is an irrational number.

7.

Using Euclid's Division Lemma, find HCF of 1234 and 56

We apply the division algorithm:

1. 1234=56×22+2

2. 56=2×28+0

Since the remainder is 0, the divisor at this stage is the HCF.

HCF(1234,56)=2

8.

Prove that there are infinitely many primes

We use Euclid's proof by contradiction (circa 300 BC).

1. Assume there is only a finite number of prime numbers. Let this complete list of all primes be P1,P2,P3,…,Pn.

2. Consider the number N formed by multiplying all these primes together and adding 1:
   N=(P1×P2×P3×⋯×Pn)+1

3. When N is divided by any prime Pi from our assumed finite list, the remainder will always be 1. This means N is not divisible by any prime in the list.

4. By the Fundamental Theorem of Arithmetic, every integer greater than 1 is either prime or can be written as a unique product of primes.

5. Therefore, N must be either:

• A prime number itself (but it's not in our list, which contradicts the assumption that the list was complete).

• Divisible by a new prime number (a prime factor not in the list, which again contradicts the assumption that the list was complete).

6. Since the assumption leads to a contradiction, the assumption is false.
   There are infinitely many primes.

9.

10.

Question No.

Answer

1.

2.

Deriving HCF(a,b)=HCF(b,r) using Euclid’s Division Lemma

The Euclid's Division Lemma states: a=bq+r, where 0≤r<b.

1. Show HCF(a,b) divides r: Let d be a common divisor of a and b (so d divides both a and b). Since r=a−bq, and d divides both terms (a and bq), d must divide their difference, r. Conclusion: Any common divisor of (a,b) is also a common divisor of (b,r).

2. Show HCF(b,r) divides a: Let d′ be a common divisor of b and r (so d′ divides both b and r). Since a=bq+r, and d′ divides both terms (bq and r), d′ must divide their sum, a. Conclusion: Any common divisor of (b,r) is also a common divisor of (a,b).

3. Final Derivation: Since the set of common divisors for the pair (a,b) is exactly the same as the set of common divisors for the pair (b,r), their greatest common divisor must be equal.
   Therefore, HCF(a,b)=HCF(b,r).

3.

We use the Prime Factorization Method.

1. Prime Factorization:

• 450=2×225=2×3^2×5^2

• 600=6×100=(2×3)×(2^2×5^2)=2^3×3^1×5^2

• 675=3×225=3×3^2×5^2=3^3×5^2

2. HCF (Highest Common Factor): Take the lowest power of the common prime factors (3 and 5):
   HCF=2min(1,3,0)×3min(2,1,3)×5min(2,2,2)
   HCF=2^0×3^1×5^2=1×3×25=75

3. LCM (Least Common Multiple): Take the highest power of all prime factors (2,3,5):
   LCM=2max(1,3,0)×3max(2,1,3)×5max(2,2,2)
   LCM=2^3×3^3×5^2=8×27×25
   LCM=200×27=5400

4.

The relationship between HCF, LCM, and two numbers (a and b) is:

HCF(a,b)×LCM(a,b)=a×b

Given: HCF=14, LCM=1680, a=84. Find b.

14×1680=84×b

b=14×1680/84

Since 84=6×14, we simplify:

b=1680/6

b=280

5.

6.

Show that 2m=3n has No Non-zero Integer Solution

1. Assumption: Assume there exist positive integers m and n such that 2m=3n.

2. Prime Factorization: The expression 2^m is a number whose only prime factor is 2. The expression 3^n is a number whose only prime factor is 3.

3. Contradiction: The Fundamental Theorem of Arithmetic states that every integer greater than 1 has a unique prime factorization. If 2^m were equal to 3^n, it would mean this single number has two different unique prime factorizations (one containing only 2s and one containing only 3s).
   This violates the Fundamental Theorem of Arithmetic, proving the equality has no solution.

7.

Given that a divides b, we can write b=k⋅a for some positive integer k.

HCF(a,b)

• Since a divides both a and b, a is a common divisor.

• Since the divisors of a cannot exceed a, a must be the greatest common divisor.
  HCF(a,b)=a

LCM(a,b)

• The multiples of b are b,2b,3b,…

• Since b=k⋅a, b is a multiple of a.

• Since b is the smallest positive multiple of itself, and b is also a multiple of a, b must be the least common multiple.
  LCM(a,b)=b

8.

Prove the Uniqueness of Prime Factorization (Fundamental Theorem of Arithmetic)

The theorem states that every integer N>1 can be uniquely (except for the order) written as a product of primes.

Proof of Uniqueness (by Contradiction):

1. Assume there exists a smallest integer N with two different prime factorizations:
   N=p1p2…pm=q1q2…qn
   where all pi and qj are primes, and the factorizations are distinct.

2. Since p1 divides N, p1 must divide the product q1q2…qn.

3. A key property of primes is: if a prime divides a product, it must divide at least one factor. Thus, p1 must divide some qj. Since both p1 and qj are primes, they must be equal: p1=qj (after reordering the q's, let's say p1=q1).

4. Divide both sides by p1(=q1):
   N′=p2p3…pm=q2q3…qn

5. N′ is an integer smaller than N that also has two distinct prime factorizations. This contradicts the initial assumption that N was the smallest integer with non-unique factorization.
   Therefore, the prime factorization must be unique.

9.

10.

Show that among any n+1 positive integers, there exist two whose difference is divisible by n.

This is proven using the Pigeonhole Principle.

1. Pigeonholes (Remainders): When any positive integer is divided by n, the set of possible remainders is {0,1,2,…,n−1}. There are exactly n possible remainders (the pigeonholes).

2. Pigeons (Integers): We are given n+1 positive integers (the pigeons).

3. Application of the Principle: Since we have n+1 integers (pigeons) and only n possible remainders (pigeonholes), by the Pigeonhole Principle, at least two of the integers must have the same remainder when divided by n.

4. Conclusion: Let a and b be the two integers that share the same remainder, r.

• a=nk1+r

• b=nk2+r The difference between the two integers is:
  ∣a−b∣=∣(nk1+r)−(nk2+r)∣=∣nk1−nk2∣=n∣k1−k2∣
  Since the difference ∣a−b∣ is a multiple of n, it is divisible by n.

Therefore, among any n+1 integers, there exist two whose difference is divisible by n.

CBSE Class 10 Maths Chapter 1 Real Numbers: Download PDF of Important Questions and Answers