Chapter 3:Pair of Linear Equations in Two Variables

Chapter 5: Arithmetic Progression

Chapter 8: Introduction to Trigonometry

Chapter 9:Some Applications of Trigonometry

Chapter 12:Area Related to Circles

Real Numbers

i. 135 and 225

ii. 196 and 38220

iii. 867 and 225

225 is greater than 135. Therefore, by Euclid’s division algorithm

225 = 135 × 1 + 90

remainder 90 ≠ 0, Again

135 = 90 × 1 + 45

Again, 45 ≠ 0,

90 = 45 × 2 + 0

The remainder is now zero in the last step, the divisor is 45.

Hence, the HCF of 225 and 135 is 45.

ii. 196 and 38220

38220>196,

therefore the by applying Euclid’s division algorithm and taking 38220 as divisor,

38220 = 196 × 195 + 0

remainder as 0 here So

Hence, the HCF of 196 and 38220 is 196.

iii. 867 and 225

867 is greater than 225. Now apply now Euclid’s division algorithm on 867,

867 = 225 × 3 + 102

Again Remainder 102 ≠ 0,

225 = 102 × 2 + 51

Again, 51 ≠ 0.

102 = 51 × 2 + 0

The remainder is now zero in the last step, the divisor is 51,

Hence, the HCF of 867 and 225 is 51.

by Euclid’s algorithm,

a = 6q + r, where q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6.

If r = 0, then a = 6q

Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively. here a = 6q, 6q+2, 6q+4,

a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.

Number of army band members = 32

The maximum number of columns in which they can march=HCF(616, 32).

By Using Euclid’s algorithm to find their HCF, we get,

Since, 616>32,

616 = 32 × 19 + 8

Since, 8 ≠ 0, therefore, taking 32 as new divisor

32 = 8 × 4 + 0

Now we have got remainder as 0 So,HCF (616, 32) = 8.

Hence, the maximum number of columns in which they can march is 8.

So, a= 3q + r for some integer q≥0 and

r = 0, 1, 2, as r ≥ 0 and r < 3.

Or a = 3q, 3q+1 and 3q+2

by squaring both the sides, we get,

a

=(9q

Let 3q

Therefore, a

a

a

a

Again,3q

a

from equation i,ii and iii,

we can say that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

q = 3q

or

3q + 1

or

3q + 2

Now, by taking the cube of all

Case (i): When r = 0, then,

a

Case (ii): When r = 1, then,

a

x

Putting = m, we get,

Putting (3q

a

Case (iii): When r = 2, then,

a

a

Putting (3q

a

it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.