- Real Numbers Important Definition
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- Euclid's Division Lemma
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- Proof of Euclid's Division Lemma
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- HCF (Highest common factor)
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- What is Prime Numbers
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- What is Composite Numbers
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- Fundamental Theorem of Arithmetic
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- HCF and LCM by prime factorization method
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- Irrational Numbers
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- How to prove the irrational numbers or Rational numbers

In this page we have *NCERT solutions for Class 10 Maths Chapter 1:Real Numbers* for EXERCISE 1.1.

**Formula Used**

__Euclid’s Division Lemma__

For a and b any two positive integer, we can always find unique integer q and r such that

$a=bq + r \; ,\; \; 0 \leq r < b$

If r =0, then b is divisor of a.

__HCF (Highest common factor)__

HCF of two positive integers can be find using the Euclid’s Division Lemma algorithm

We know that for any two integers a,b. we can write following expression

$a=bq + r \; \; \; 0 \leq r < b$

If r=0 ,then

HCF( a,b) =b

If r≠0 , then

$HCF ( a,b) = HCF ( b,r)$

Again expressing the integer b,r in Euclid’s Division Lemma, we get

$ b=pr + r_1$

$HCF ( b,r)=HCF ( r,r_1)$

Similarly successive Euclid ‘s division can be written until we get the remainder zero, the divisor at that point is called the HCF of the a and b

This exercise contains Questions on finding HCF using Euclid Division Lemma

Hope you like them and do not forget to like , social_share and comment at the end of the page.

For a and b any two positive integer, we can always find unique integer q and r such that

$a=bq + r \; ,\; \; 0 \leq r < b$

If r =0, then b is divisor of a.

HCF of two positive integers can be find using the Euclid’s Division Lemma algorithm

We know that for any two integers a,b. we can write following expression

$a=bq + r \; \; \; 0 \leq r < b$

If r=0 ,then

HCF( a,b) =b

If r≠0 , then

$HCF ( a,b) = HCF ( b,r)$

Again expressing the integer b,r in Euclid’s Division Lemma, we get

$ b=pr + r_1$

$HCF ( b,r)=HCF ( r,r_1)$

Similarly successive Euclid ‘s division can be written until we get the remainder zero, the divisor at that point is called the HCF of the a and b

This exercise contains Questions on finding HCF using Euclid Division Lemma

Hope you like them and do not forget to like , social_share and comment at the end of the page.

(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

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i)We should always start with Larger number

So in this case the larger number is 225

Now we can write 135 and 225 in Euclid division algorithm

$225=135 \times 1 +90$

Now $HCF (225,135) = HCF ( 135,90)$

Again writing 135,90 is Euclid division formula

$135=90 \times 1 +45$

Now $HCF (135,90)= HCF (90,45)$

Again writing 90,45 is Euclid division formula

$90=45 \times 2 +0$

Now r =0, 45 is HCF (90,45)

45 is HCF ( 225,135)

ii) We should always start with Larger number

So in this case the larger number is 38220

Now we can write 38220 and 196 in Euclid division algorithm

$38220=196 \times 195 +0$

Now r =0, 196 is HCF (196,38220)

iii) We should always start with Larger number

So in this case the larger number is 867

Now we can write 867 and 255 in Euclid division algorithm

$867=255 \times 3 +102$

Now $HCF (867,255) = HCF ( 255,102)$

Again writing 255,102 is Euclid division formula

$255=102 \times 2 +51$

Now $HCF (255,102)= HCF (102,51)$

Again writing 102,51 is Euclid division formula

$102=51 \times 2 +0$

Now r =0, 51 is HCF (102,51)

51 is HCF ( 867,255)

For a and b any two positive integer, we can always find unique integer q and r such that

$ a=bq + r \; ,\; \; 0 \leq r < b$

Let us take a as any positive integers and b=6 ,Then using Euclid’s algorithm we get,

$a = 6q + r$ as $0 \leq r < 6$,possible reminders are r = 0, 1, 2, 3, 4, 5

So total possible forms will $6q + 0 , 6q + 1 , 6q + 2,6q + 3, 6q + 4, 6q + 5$

$a=6q+ r, 0 \leq r < 6$

$a=6q $, This is a even number

$a=6q+1$ , This is a odd number

$a=6q+2$ ,This is an even number as 6 and 2 are divisible by 2

$a=6q+3$, it is not divisible by 2

$a=6q+4$, it is divisible by 2

$a=6q+5$ , it is not divisible by 2

So $6q,6q+2,6q +4$ are even number

$6q+1,6q+3,6q+5$ are odd numbers

According to the questions, we need to find the maximum number of column

maximum number of column is the HCF of number 32 and 616

So question has reduced to finding the HCF of 32 and 616 using Euclid division algorithm

We should always start with Larger number

So in this case the larger number is 616

Now we can write 616 and 32 in Euclid division algorithm

$616=32 \times 19 +8$

Now HCF of numbers (616,32) = HCF ( 32,8)

Again writing 32,8 is Euclid division formula

$32=8 \times 4+0$

As r=0, 8 is the HCF of 616 and 32

For a and b any two positive integer, we can always find unique integer q and r such that

$ a=bq + r \; ,\; 0 \leq r < b$

Let us take a as any positive integers and b=3 ,Then using Euclid’s algorithm we get,

$a = 3q + r$ as $0 \leq r < 3$,possible reminders are r = 0, 1, 2

So total possible forms will $3q + 0 , 3q + 1 , 3q + 2$

$a=3q \;,\; a^2=9q^2 =3m \; \\ where \; m=3q^2$

$a=3q+1\;,\; a^2=9q^2 +6q+1 \\=3(3q^2+2q) +1 =3m+1 \; where \; m=3q^2+2q$

$a=3q+2\;,\; a^2=9q^2 +12q+4 \\=3(3q^2+4q+1) +1 =3m+1 \; where \; m=3q^2+4q+1$

For a and b any two positive integer, we can always find unique integer q and r such that

$ a=bq + r \; ,\; 0 \leq r < b$

Let us take a as any positive integers and b=3 ,Then using Euclid’s algorithm we get,

$a = 3q + r$ as $0 \leq r < 3$,possible reminders are r = 0, 1, 2

So total possible forms will $3q + 0 , 3q + 1 , 3q + 2$

$a=3q\;,\; a^3=27q^3 =9m \; \\ where \; m=3q^3$

$a=3q+1\;,\; a^3=27q^3 +27q^2 +9q +1 \\=9(3q^3+3q+q) +1 =9m+1 \; \\where \; m=3q^3+3q+q$

$a=3q+2\;,\; a^3=27q^3 +54q^2 +36q +8 \\=9(3q^3+6q+4q) +8 =9m+8 \; \\ where \; m=3q^3+6q+4q$

Download Real Numbers Exercise 1.1 as pdf

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.