Class 10 Math Chapter 1 Real Numbers Solutions

Here, Class 10 Math Chapter 1 Real Numbers Solutions are provided. Subject matter experts have created these solutions to help students in their board exam. They have created solutions in such a way that students can revise the entire solutions in very little time and can score good marks in the board exam. Every solution is provided in detail and in very simple language.

Class 10 Math Chapter 1 Real Numbers Solutions are very important for board exam. Solving these real numbers solutions will boost the confidence and students can score better in exam. Solutions are provided according to latest NCERT syllabus. Experts created these step-by-step solutions, which will definitely be helpful for students.

Real Numbers Solutions

Exercise 1.1

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Exercise 1.2

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Chapter 2: Polynomials Solutions

We have given solutions of all execrise of class 10 math chapter 1. These solutions will be helpful for students understanding core concept of Real Numbers chapters.

To understand the chapter thoroughly, it is essential to practice all the questions provided within it, as well as any additional questions. Solving previous years’ question papers also proves to be very helpful in exam preparation. You can also find other study material resources on our website.

Important Questions for NCERT Chapter 1: Real Numbers

Question :Given that HCF (306, 657) = 9, find LCM (306, 657).

Answer :

HCF (306,657)=9\text{HCF } (306, 657) = 9
We know that, LCM×HCF=Product of two numbers\text{We know that, } \text{LCM} \times \text{HCF} = \text{Product of two numbers}
∴LCM×HCF=306×657\therefore \text{LCM} \times \text{HCF} = 306 \times 657
LCM=306×657HCF=306×6579\text{LCM} = \frac{306 \times 657}{\text{HCF}} = \frac{306 \times 657}{9}
LCM=22338\text{LCM} = 22338

Question : Express each number as product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Answer :

(i) 140=2×2×5×7=22×5×7\text{(i) } 140 = 2 \times 2 \times 5 \times 7 = 2^2 \times 5 \times 7
(ii) 156=2×2×3×13=22×3×13\text{(ii) } 156 = 2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13
(iii) 3825=3×3×5×5×17=32×52×17\text{(iii) } 3825 = 3 \times 3 \times 5 \times 5 \times 17 = 3^2 \times 5^2 \times 17
(iv) 5005=5×7×11×13\text{(iv) } 5005 = 5 \times 7 \times 11 \times 13
(v) 7429=17×19×23\text{(v) } 7429 = 17 \times 19 \times 23

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