Create New Account
Create New Account

Sign in to Innings2

or

New Account     Reset Password     

Real Numbers

    Exercise 1.1

Powered by Innings 2

Reset Progress

This will delete your progress and chat data for all chapters in this course, and cannot be undone!

Glossary

Select one of the keywords on the left…

10th class > Real Numbers > Exercise 1.1

Exercise 1.1

Express each number as a product of its prime factors:

(i)

(i)140

Solution

[Note: give the numbers in ascending order only]

Prime factorization of 140 = × 2 × 5 ×

= × 5 × 7

(ii)

(ii)156

Solution

[Note: give the numbers in ascending order only]

Prime factorization of 156 = 2 × 2 × ×

= × 3 × 13

(iii)

(iii)3825

Solution

[Note: give the numbers in ascending order only]

Prime factorization of 3825 = × 3 × 5 × ×

= × × 17

(iv)

(iv)5005

Solution

[Note: give the numbers in ascending order only]

Prime factorization of 5005 = × 7 × × 13

(v)

(v)7429

Solution

[Note: give the numbers in ascending order only]

Prime factorization of 7429 = × 19 ×

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i)

(i) 26 and 91

LCM × HCF = Product of the two numbers.

  • Step:1

To find the LCM and HCF of the given pairs of the integers, first, find the prime factors of the given pairs of numbers.

  • Step:2

Find the LCM

  • Step:3

Find the HCF

  • Step:4

Now, we have to verify LCM × HCF = product of the two numbers.

Solution

[Note: give the numbers in ascending order only]

Prime factors of 26 = 2 ×

Prime factors of 91 = × 13

HCF of 26 and 91 =

LCM of 26 and 91 = 2 × 7 × 13

= 14 × 13 =

Product of these two numbers = 26 × 91

=

LCM × HCF = 182 × 13 =

Thus, the product of two numbers = LCM × HCF = 2366.

(ii)

(ii) 510 and 92

Solution

[Note: give the numbers in ascending order only]

Prime factors of 510 = × 3 × × 17

Prime factors of 92 = × 2 ×

HCF of the two numbers =

LCM of the two numbers = 2 × 2 × 3 × 5 × 17 × 23 =

Product of these two numbers = 510 × 92 =

LCM x HCF = 2 × 23460 =

Thus, the product of two numbers = LCM × HCF = 46920

(iii)

(iii) 336 and 54

Solution

[Note: give the numbers in ascending order only]

Prime factors of 336 = × 2 × 2 × 2 × ×

Prime factors of 54 = × 3 × 3 ×

HCF of the two numbers =

LCM of the two numbers = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7

= 24 × 33 × 7 =

Product of these two numbers = 336 × 54 =

LCM x HCF = 3024 × 6 =

Thus, the product of two numbers = LCM × HCF = 18144

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

Solution

We know that LCM × HCF = product of two given integers.

Given, HCF (306, 657) =

We have to find, LCM (306, 657)

We have the given numbers 306 and 657.

Hence, we can find the product of 306 and 657.

The HCF of these two numbers is .

Substitute these values in the above formula and find the value of the unknown i.e. LCM.

LCM × 9 = 306 × 657

LCM = (306 × 657)9

LCM = 34 × 657

LCM =

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

Solution

If any number ends with the digit 0 that means it should be divisible by 5. That is, if 6n ends with the digit 0, then the prime factorisation of 6n would contain the prime number .

Prime factors of 6n = 2×3n = 2n × 3n

We can clearly observe, 5 is not present in the prime factors of 6n. That means 6n will not be divisible by .

Therefore, 6n cannot end with the digit 0 for any natural number n.

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

Solution

Now, simplify 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5.

On simplifying them, we find that both the numbers have more than factors.

So, if the number has more than two factors, it will be composite.

It can be observed that,

7 × 11 × 13 + 13 = 13 (7 × 11 + 1)

= 13(77 + 1)

= 13 × 78

= 13 × ×

= 13 × 13 × ×

The given number has 2, 3, 13, and 1 as its factors.

Therefore, it is a composite number.

Now, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1)

= 5 × (1008 + 1)

= 5 × 1009 × 1

1009 cannot be factorized further.

Therefore, the given expression has 5,1009 and 1 as its factors. Hence, it is a composite number.

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

Given:

  • Sonia takes 18 minutes to drive one round of the field.

  • Ravi takes 12 minutes for the same.

  • They both start at the same point and at the same time and go in the same direction.

Time taken by Sonia is more than Ravi to complete one round.

Now, we have to find after how many minutes will they meet again at the same point.

For this, there will be a number that is divisible by both 18 and 12, and that will be the time when both meet again at the starting point.

To find this we have to take LCM of both numbers.

Let's find LCM of 18 and 12 by prime factorization method.

[Note: give the numbers in ascending order only]

18 = × ×

12 = × ×

LCM of 12 and 18 = 2 × 2 × 3 × 3 =

Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.