RS Aggarwal Solutions Class 9 Ex 1A

RS Aggarwal Class 9 Ex 1A Chapter 1

Question 1: What are irrational numbers? How do they differ from rational numbers? Give examples.

Solution:

A number that can neither be expressed as a terminating decimal nor be expressed as a repeating decimal is called an irrational number. A rational number, on the other hand is always a terminating decimal, and if not, it is a repeating decimal.

Examples of irrational numbers:

0.101001000..

0.232332333..

Question 2: Classify the following numbers as rational or irrational. Give reasons to support your SOLUTION. 


(i) 4

(ii) 196

(iii) 21 = 3 x 7

(iv) 43

(v) 3 + 3

(vi) 7 – 2

(vii) 23 6

(viii) .66666666

(ix) 1.232332333…

(x) 3.040040004…

(xi) 3.2576

(xii) 2.356565656…

(xiii) π

(xiv) 227

Solution:

(i) 4 = 2

It is a rational number.

(ii) 196 = 14

It is a rational number.

(iii) 21 = 3 x 7 = 4.58257…

It is an irrational number.

(iv) 43

If a is a positive integer, which is not a perfect square, then a is an irrational number.

Here, 43 is not a perfect square,  so it is irrational.

(v) 3 + 3

The sum of a rational number and an irrational number is an irrational number.

So, it is an irrational number.

(vi) 7 – 2

The difference of an irrational number and a rational number is an irrational number, so it is an irrational number.

(vii) 23 6

The product of a rational number and an irrational number is an irrational number, so it is an irrational number.

(viii) .66666666

It is a rational number because it is a repeating decimal.

(ix) 1.232332333…

It is an irrational number because it is a non-terminating, non- repeating decimal.

(x) 3.040040004…

It is an irrational number because it is a non- terminating, non- repeating decimal.

(xi) 3.2576

It is a rational number because it is a terminating decimal.

(xii) 2.356565656…

It is a rational number because it is repeating.

(xiii) π = 3.14285…

It is an irrational number because it is a non – terminating, non – repeating decimal.

(xiv) 227 is a rational number because it can be expressed in the pq form.

Question 3: Represent 2 , 3 and 5 on the real line .

Solution:

https://lh6.googleusercontent.com/5gPsUeefYaWOJJBHIDpvH5e2kSc2mWSpxpu24rvwuA0bo6GPMk1ZGtUgrQrghQrjuH88FPoBApE_qIWOJCJZSuATS7tWfkciGYMnreuHA1OJ7Ir8ChR2CaVoceBq1EiredS2loMFUwo0e9ngGA

Let X’ OX be a horizontal line taken as the x – axis and O be the origin representing 0.

Take OA = 1 unit and AB OA such that AB = 1 unit.

Join OB

Now,

OB = OA2+AB2=12+12=2units

Taking O as the center and OB as the radius, draw an arc, meeting OX at P

We have:

OP = OB = 2 units

Thus, point P represents 2 on the number line.

Now, draw BC OB such that BC = 1 unit.

Join OC

We have:

OC = OB2+BC2=(2)2+12 = 3 units.

Taking O as the centre and OC as the radius, draw an arc, meeting OX at Q.

We have:

OQ = OC = 3 units.

Thus, point Q represents 3 on the number line.

Now, draw CD OC such that CD = 1 unit.

Join OD

We have:

OD = OC2+OD2=(3)2+12 = 4 = 2 units.

Now, draw DE OD such that DE = 1 unit.

Join OE.

We have:

OD2+DE2=22+12 = 5 units.

Taking O as the center and OE as the radius, draw an arc, meeting OX at R.

We have:

OR = OE = 5 units

Thus, point R represents 5 on the number line.

Question 4: Represent 6 , 7 and 5 on the real line .

Solution:

https://lh5.googleusercontent.com/mYQpv6_9ODKHOjubTnotoH2-qAJZdcrDa7f0Ryozr1HFM9vcAyNqOCBlKRLE-QSJS9TU6j0Oy8mKZ6Cf8mF_QkmM21DM4hS6AexwYNmM4x2FWDjHr5-_R4YpnKMPf3FucB7JqXnmKq0CKJEfbA

Let X’OX be a horizontal line taken as the x – axis and O be the origin representing 0.

Take OA = 2 units and AB OA such that AB = 1 unit.

Now join OB.

We have: OB = OA2+AB2=22+12 = 5 Units

Taking O as the center and OB as the radius, draw an arc, meeting OX at P .

Thus, we have:

OP = OB = 5 units

Here, point P represents 5 on the number line.

Now, draw BC OB such that BC = 1 unit.

Join OC

We have: OC = OB2+BC2=(5)2+12=6 units.

Taking O as the centre and OC as the radius, draw an arc, meeting OX at Q.

Thus, we have:

OQ = OC = 6 units

Here, point Q represents 6 on the number line.

Now, draw CD OC such that CD = 1 unit.

Join OD.

We have: OD = OC2+OD2=(6)2+12=7 units.

Taking O as the centre and OE as the radius, draw an arc, meeting OX at R.

Now,

OR = OD  = 7

Thus, point R represents 7 on the number line.

Question 5: Giving reason in each case, show that each of the following members is irrational.

  1. 4 + 5
  2. – 3 + 6
  3. 57
  4. 38
  5. 25
  6. 43

Solution:

(i)  4 + 5 because the sum of a rational number and an irrational number is an irrational number.

(ii) – 3 + 6 because the difference of a rational number and an irrational number is an irrational number.

(iii) 57 because the product of a rational number and an irrational number is an irrational number.

(iv) 38 because the product of a rational number and an irrational number is an irrational number.

(v) 25 because the quotient of a rational number and an irrational number is an irrational number.

(vi) 43 because the quotient of a rational number and an irrational number is an irrational number.

 

Question 6: State in each case, whether the given statement is true or false.

  1. The sum of two rational number is rational.
  2. The sum of two irrational number is irrational.
  3. The product of two rational number is rational.
  4. The product of two irrational numbers is irrational.
  5. The sum of a rational number and irrational number is irrational.
  6. The product of rational number and irrational number is a rational number.
  7. Every real number is rational.
  8. Every real number is rational or irrational.
  9. π is irrational and 227 is rational.

Solution:

(i) True

(ii) False

Example: ( 2 + 3) + (2 – 3 ) = 4

Here, 4 is a rational number.

(iii) True

(iv) False

Example: 3 x 3 = 3

Here, 3 is a rational number.

(v) True

(vi) False

Example: (4) x 5 = 45

Here, 45 is an irrational number.

(vii) False

Real numbers can be divided into rational and irrational numbers.

(viii) True

(ix) True

Question 7: ADD;

(i) 23 – 52 and 3 + 22

(ii) 22 + 53 – 75 and 332 + 5

(iii) 237122 +611  and 137 + 32211

Solution:

(i) 23 – 52 + 3 + 22

= (23 + 3) + (22 – 52)

= 33 – 32

(ii) 22 + 53 – 75 + 332 + 5

= 222 + 53 + 33 + 5 – 75

= 2 + 83 – 65

(iii) 237122 +611  + 137 + 32211

= 237 + 137  – 11 + 611 + 322122

= 7 + 511 + 2

Question 8: MULTIPLY;

  1. 35 by 2 5
  2. 6 15 by 4 3
  3. 2 6 by 3 3
  4. 3 8 by 3 2
  5. 10 by 40
  6. 3 28 by 2 7

Solution:

(i) 35 x 2 5 = 3 x 2 x 5 x 5 = 6 x 5 = 30

(ii) 6 15 x 4 3 = 6 x 4 x 5 x 3 x 3 = 24 x 3 x 5 = 72 5

(iii) 2 6 x 3 3 = 2 x 3 x 2 x 3 x 3 = 6 x 3 x 2 = 182

(iv) 3 8 x 3 2 = 3 x 3 x 2 x 2 x 2 x 2 = 9 x 4 = 36

(v) 10 x 40 = 2 x 5 x 2 x 2 x 2 x 5 = 2 x 2 x 2 x 2 x 5 x 5 = 2 x 2 x 5 = 20

(vi) 3 28 x 2 7 = 6 7×4 x 7 = 6 x 7 x 4 = 42 x 2 = 84

Question 9: DIVIDE ;

  1. 166,by42
  2. 166,by42
  3. 166,by42

Solution:

(i) 16642 = 162342 = 4 3

(ii) 16642 = 125×343 = 3 5

(iii) 16642 = 187367 = 3 3


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