10th Class math’s solution Chapter 1.2

 

10th Class math’s solution Chapter 1

 

 

 NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers Ex1.2

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 1.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 1.1

         

           

Access Answers of Maths NCERT Class 10 Chapter 1 – Real Number Exercise 1.2

1.   Express each number as a product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

 

Solutions :

(i)             140 = 2 × 2 × 5 × 7 × 1

  = 2²×5×7

(ii)            156 = 2 × 2 × 13 × 3 × 1

  = 2²× 13 × 3

(iii)          3825 = 3 × 3 × 5 × 5 × 17 × 1

         = 3²×5²×17

(iv)          5005 = 5 × 7 × 11 × 13 × 1

         = 5 × 7 × 11 × 13

(v)           7429 = 17 × 19 × 23 × 1

         = 17 × 19 × 23.

 

2. Find the LCM and HCF of the following pairs of integers and verify that

 LCM × HCF = product of the two numbers.

       (i) 26 and 91

      (ii) 510 and 92

      (iii) 336 and 54

 

Solutions:

      (i) By prime factorisation, we get

             26 = 2 × 13 × 1

             91 = 7 × 13 × 1

             Therefore, LCM (26, 91) = 2 × 7 × 13 × 1 = 182

             And HCF (26, 91) = 13.

        Verification:

   Now, product of 26 and 91 = 26 × 91 = 2366

  And Product of LCM and HCF = 182 × 13 = 2366

  Hence, LCM × HCF = product of the 26 and 91.

 

          (ii) By prime factorisation, we get

               510 = 2 × 3 × 17 × 5 × 1

                 92 = 2 × 2 × 23 × 1

            Therefore, LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460

            And HCF (510, 92) = 2

         Verification

            Now, product of 510 and 92 = 510 × 92 = 46920

            And Product of LCM and HCF = 23460 × 2 = 46920

            Hence, LCM × HCF = product of the 510 and 92.

        (iii) By prime factorisation, we get

            336 = 2 × 2 × 2 × 2 × 7 × 3 × 1

              54 = 2 × 3 × 3 × 3 × 1

         Therefore, LCM(336, 54) = = 3024

         And HCF(336, 54) = 2×3 = 6

      Verification

        Now, product of 336 and 54 = 336 × 54 = 18,144

        And Product of LCM and HCF = 3024 × 6 = 18,144

        Hence, LCM × HCF = product of the 336 and 54.

3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

      (i) 12, 15 and 21

     (ii) 17, 23 and 29

     (iii) 8, 9 and 25

Solutions:

(i) By prime factorisation, we get

     12=2×2×3

     15=5×3

     21=7×3

  Therefore,

    HCF(12,15,21) = 3

    LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420

 

(ii) By prime factorisation, we get

   17=17×1

   23=23×1

   29=29×1

Therefore,

   HCF(17,23,29) = 1

   LCM(17,23,29) = 17 × 23 × 29 = 11339

(iii) By prime factorisation, we get

   8=2×2×2×1

  9=3×3×1

 25=5×5×1

Therefore,

  HCF(8,9,25)=1

  LCM(8,9,25) = 2×2×2×3×3×5×5 = 1800

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

Solution: As we know that,

HCF×LCM=Product of the two given numbers

Therefore,

9 × LCM = 306 × 657

LCM = (306×657)/9 = 22338

Hence, LCM(306,657) = 22338

5. Check whether 6ⁿ can end with the digit 0 for any natural number n.

Solution: 

Prime factorization of 6ⁿ = (2×3)ⁿ

Therefore, the prime factorization of 6ⁿ doesn’t contain prime number 5.

Hence, it is clear that for any natural number n, 6ⁿ is not divisible by 5 and

 thus it proves that 6ⁿ cannot end with the digit 0 for any natural number n.

6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Solution:  for the given expression;

7 × 11 × 13 + 13

Taking 13 as common factor, we get,

=13(7×11×1+1) = 13(77+1) = 13×78 = 13×3×2×13

Hence, 7 × 11 × 13 + 13 is a composite number.

Now let’s take the other number,

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Taking 5 as a common factor, we get,

=5(7×6×4×3×2×1+1) = 5(1008+1) = 5×1009

Hence, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution: The minimum time when Sonia and Ravi meet again at starting point will be the :

 LCM(18,12) = 2×3×3×2×1=36

Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.

 

 

 

Class 10 Maths Real Numbers

Rational numbers and irrational numbers are taken together form the set of real numbers. The set of real numbers is denoted by R. Thus every real number is either a rational number or an irrational number. In either case, it has a non–terminating decimal representation. In the case of rational numbers, the decimal representation is repeating (including repeating zeroes) and if the decimal representation is non–repeating, it is an irrational number. For every real number, there corresponds a unique point on the number line ‘l’ or we may say that every point on the line ‘l’ corresponds to a real number (rational or irrational).

From the above discussion we may conclude that:
To every real number there corresponds a unique point on the number line and conversely, to every point on the number line there corresponds a real number. Thus we see that there is one–to–one correspondence between the real numbers and points on the number line ‘l’, that is why the number line is called the ‘real number line’.