CBSE Class 10th chapter Real number Solved questions

X Real Numbers : Topics:

1. Euclid's Division Lemma/Algorithm
2. Fundamental Theorem of Arithmetic
3. Irrational Numbers
4. Decimal expression of Rational Number

Q.1. Based on Euclid’s algorithm: a = bq + r where 0 ≤ rÐb

Solved example:Using Euclid’s algorithm: Find the HCF of 825 and 175.

Explanation:

Step 1. Since 825>175. Divide 825 by 175.

We get, quotient = 4 and remainder = 125. This can be written as 825 = 175 x 4 + 125

Step II. Now divide 175 by the remainder 125. We get quotient = 1 and remainder = 50.

So we write 175 = 125 x 1 + 50.

Step III. Repeating the above step we now divide 125 by 50 and get quotient = 2 and remainder = 25.

so 125 = 50 x 2 + 25

Step IV.

Now divide 50 by 25 to get quotient = 2 and remainder 0. Since remainder has become zero we stop here.

Since divisor at this stage is 25, so the HCF of 825 and 175 is 25.

Solution: This is how a student should write answer in his answer sheet:

Since 825>175, we apply division lemma to 825 and 175 to get

825 = 175 x 4 + 125.

Since r ≠ 0, we apply division lemma to 175 and 125 to get

175 = 125 x 1 + 50

Again applying division lemma to 125 and 50 we get,

125 = 50 x 2 + 25.

Once again applying division lemma to 50 and 25 we get.

50 = 25 x 2 + 0.

Since remainder has now become 0, this implies that HCF of 825 and 125 is 25.

Problems for practice;

1. Find HCF of the following pairs using Euclid’s division Lemma

2. 34782 and 1892

3. 588 and 240

4. 80784 and 628

Q.2. Based on Showing that every positive integer is either of the given forms:

Prove that every odd positive integer is either of the form 4q + 1 or 4q + 3 for some integer q.

Explanation:   Euclid’s division lemma a= bq + r. 

Let b = 4 then r = 0,1,2,3

odd positive integer  =  (4q + 1)  and (4q + 1)

Even positive integers will be 4q or 4q + 2

Practice questions:

   

i. Write the possible remainders when a number is divided by
a. 5            b. 3              c. 7                    d. 2.

ii. Prove that every even positive integer is either of the form 6q , 6q + 2 or 6q + 4.    

iii. Prove that every positive integer is either of the form 3q, 3q + 1 or 3q + 2 for some integer q.

Q. 3. Based on LCM and HCF:

Formula: LCM x HCF = product of numbers Or product of numbers = LCM x HCF

a. Find HCF (26,91) if LCM(26,91) is 182

Sol: We know that LCM x HCF = Product of numbers.

or 182 x HCF = 26 x 91

or HCF = (26 x 91)/182 = 13

b. LCM and HCF of two numbers are 3024 and 6 respectively. If one of the number is 336 find the other number.

Sol: We know that      :   Product of numbers = LCM x HCF

Or other Number = (LCM x HCF)/ginen number = Or number = (3024 x 6)/336 = 54

Solve similar questions from your text book.

Q. 4. Based on irrational numbers. Prove that √5 is irrational.

Solution:
Let  √5 is rational. => √5 = p/q where p and q are integers ,q≠0 and p and q are co-primes 

 Squaring both side , 5 = a^2/ /b^2.

5b2 = a2.          -----------------------(i)

This means a^2 is multiple of 5 => a will be multiple of 5

Now, Let a = 5x  we can write a^2 = 5x^2 for some integer x

Putting this value in expression (i) ,we get

5b2 = (5x)2

Or 5b2 = 25x2

Or b2 = 5x2.          This means b^2 is multiple of 5 => b will be multiple of 5

So, a and b have at least 5 as a common factor. 

But this contradicts the fact that a and b are co-primes. (6th jingle)

This contradiction has arisen because of our incorrect assumption that √5 is rational. (7th jingle)

Hence it follows that √5 is irrational. (8th jingle)

Q. show that there is no positive integer n so that root (n -­1) + root (n+1) is rational
Solution:

10th Real Numbers CBSE Test Papers     Download File