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 Mathematics Class Ten Class Ten Subject List
Rational Numbers
Polynomials
Pair of Linear Equations
Exercise 1
Exercise 2
Exercise 3
Exercise 4
Exercise 1
Exercise 2
Exercise 3
Arithmetic Progressions
Exercise 1
Exercise 2
Exercise 3
Triangles
Exercise 1
Exercise 2
Exercise 3
Exercise 4
Coordinate Geometry
Introduction to Trigonometry
Exercise 1
Exercise 2
Exercise 3
Exercise 4
Applications of Trigonometry
Circles
Areas of Circle
Exercise 1
Exercise 2
Exercise 3
Surface Area & Volumes
Exercise 1
Exercise 2
Exercise 3
Statistics
Probability

# Rational Numbers

1. Use Euclid’s division algorithm to find the HCF of:

(i) 135 and 225

Solution:

As there is no reminder after third step

So, HCF = 45

(ii) 196 and 38220

Solution:

So, HCF = 196

(iii) 867 and 255

Solution:

So, HCF = 51

2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Solution:

Let us assume q = 1

Then, 6q+1 = 7

6q+3 = 9

6q+5 = 11

It is clear that 7, 9 and 11 are consecutive negative integers.

3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution:

To get the solution we need to find the HCF of 616 and 32

So, HCF = 8

So, the maximum number of column is 8

4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Solution:

4 = 3m+1

Next square number is 9

9 = 3m

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution:

Next number is 27

27 = 9m

Next number is 64

64 = 9m + 1

Next number is 125

125 = 9m + 8

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

(i) 140

Solution:

(ii) 156

Solution:

(iii) 3825

Solution:

(iv) 5005

Solution:

(v) 7429

Solution:

7. 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

Solution: Factorization of numbers gives following:

(ii) 510 and 92

Solution:

(iii) 336 and 54

Solution:

8. Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) 12, 15 and 21

Solution:

(ii) 17, 23 and 29

Solution:

Since all are prime numbers so HCF = 1

(iii) 8, 9 and 25

Solution:

HCF = 1

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

Solution: As you know

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

Solution: As you know for getting 0 at unit’s place of a number there should be at least a 5 and a 2 in the factor.

Now, 6 is having 2 as one of its factors so we need a 5 to fulfill the condition. If n = 5 or product of 5 then there will be 0 at unit’s place.

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

Solution: As per fundamental theorem of arithmetic every composite number is a product of primes.

The first example will be an even number because two odd numbers equal an even number when added. So, it is a composite number.

The second example will be a product of 5, hence a composite number.

As they are not prime numbers so it is obvious that they are composite numbers.

12. 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: We need to calculate the LCM to find the answer.

After 36 minutes they will be together at the starting point.

13. Prove that is irrational.

Solution: Let us prove it by the method of contradiction. Let us assume that is rational. Let us assume that = , where a and b are co-prime that is they have not common factors other than 1.

Or, b = a

Or, b²5= a²

So, a² is divisible by 5

And a is also divisible by 5.

So a= 5c for some integer c.

Substituting the value of a from above equation we get

b²5 = 25c²

Or, b²= 5c²

This means that b² and b is divisible by 5.

So, a and b have common factors other than 1.

This contradicts our earlier assumption of a and b being co-prime.

The contradiction came because is an irrational number.

14. Prove that 3 +2 is irrational.

Solution: The addition of an irrational number and a rational number will always give an irrational number as answer. Other steps will be same as the previous question.

15. Prove that the following are irrationals :

(i)

(ii) 7

(iii) 6 +

Solution: When an irrational number is added to, deleted from, multiplied by or divided by a rational number, the result is always an irrational number.