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**Q.1. Write the following in decimal form and say what kind of decimal expansion each has:** **(i) 36/100****Solution:****= 0.36 (Terminating) **

**(ii)1/11**

**Solution:****Solution:**

= 4.125 (Terminating)

**(iv) 3/13 Solution:**

**(v) 2/11 ****Solution:**

**(vi) 329/400 ****Solution:**

= 0.8225 (Terminating)**Q.2. You know that 1/7 = 0.142857. Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, 6/7 are, without actually doing the long division? If so, how?****[Hint: Study the remainders while finding the value of 1/7 carefully.]****Solution:****Q.3. Express the following in the form p/q, where p and q are integers and q 0.****(i) ****Solution:**

Assume that *x* = 0.666…

Then,10*x* = 6.666…

10*x* = 6 + *x*

9*x* = 6*x* = 2/3

**(ii) [latex]0.4\overline{7}[/latex]**

Solution:

[latex]0.4\overline{7} = 0.4777..[/latex]

= (4/10)+(0.777/10)

Assume that *x* = 0.777…

Then, 10*x* = 7.777…

10*x* = 7 + *x**x* = 7/9

(4/10) + (0.777../10) = (4/10) + (7/90) ( x = 7/9 and x = 0.777…0.777…/10 = 7/(9 × 10) = 7/90)

= (36/90) + (7/90) = 43/90

**Solution:**

Assume that *x* = 0.001001…

Then, 1000*x* = 1.001001…

1000*x* = 1 + *x*

999*x* = 1*x* = 1/999**Q.4. Express 0.99999…. in the form p/q . Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.**

Solution:

Assume that *x* = 0.9999…..Eq (a)

Multiplying both sides by 10,

10*x* = 9.9999…. Eq. (b)

Eq.(b) – Eq.(a), we get

(10*x* = 9.9999)-(*x* = 0.9999…)

9*x* = 9*x* = 1

The difference between 1 and 0.999999 is 0.000001 which is negligible.

Hence, we can conclude that, 0.999 is too much near 1, therefore, 1 as the answer can be justified.**Q.5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17 ? Perform the division to check your answer.Solution: **

1/17

Dividing 1 by 17:

There are 16 digits in the repeating block of the decimal expansion of 1/17.**Q.6. Look at several examples of rational numbers in the form p/q (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?****Solution:**

We observe that when q is 2, 4, 5, 8, 10… Then the decimal expansion is terminating. For example:

1/2 = 0. 5, denominator q = 2^{1}

7/8 = 0. 875, denominator q =2^{3}

4/5 = 0. 8, denominator q = 5^{1}

We can observe that the terminating decimal may be obtained in the situation where prime factorization of the denominator of the given fractions has the power of only 2 or only 5 or both.**Q.7. Write three numbers whose decimal expansions are non-terminating non-recurring.Solution:**

We know that all irrational numbers are non-terminating non-recurring. three numbers with decimal expansions that are non-terminating non-recurring are:

- √3 = 1.732050807568
- √26 =5.099019513592
- √101 = 10.04987562112

**Q.****8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.****Solution:**

Three different irrational numbers are:

- 0.73073007300073000073…
- 0.75075007300075000075…
- 0.76076007600076000076…

**Q.****9. Classify the following numbers as rational or irrational according to their type:****(i)√23****Solution:**

√23 = 4.79583152331…

Since the number is non-terminating non-recurring therefore, it is an irrational number.**(ii)√225**

**Solution:**

√225 = 15 = 15/1

Since the number can be represented in p/q form, it is a rational number.**(iii) 0.3796****Solution:**

Since the number,0.3796, is terminating, it is a rational number.**(iv) 7.478478****Solution:**

The number,7.478478, is non-terminating but recurring, it is a rational number.

**(v) 1.101001000100001…****Solution:**

Since the number,1.101001000100001…, is non-terminating non-repeating (non-recurring), it is an irrational number.

**Q.1. Visualise 3.765 on the number line, using successive magnification.****Solution:****Q.2. Visualise on the number line, up to 4 decimal places. ****Solution: **

**Q.1. Classify the following numbers as rational or irrational:****(i) 2 –√5****Solution:**

We know that, √5 = 2.2360679…

Here, 2.2360679…is non-terminating and non-recurring.

Now, substituting the value of √5 in 2 –√5, we get,

2-√5 = 2-2.2360679… = -0.2360679

Since the number, – 0.2360679…, is non-terminating non-recurring, 2 –√5 is an irrational number.

**(ii) (3 +√23)- √23****Solution:**

(3 +**√**23) –√23 = 3+**√**23–√23

= 3

= 3/1

Since the number 3/1 is in p/q form, (**3 +√23)- √23** is rational.

**(iii) 2√7/7√7****Solution:**

2√7/7√7 = ( 2/7)× (√7/√7)

We know that (√7/√7) = 1

Hence, ( 2/7)× (√7/√7) = (2/7)×1 = 2/7

Since the number, 2/7 is in p/q form, 2√7/7√7 is rational.

**(iv) 1/√2****Solution:**

Multiplying and dividing numerator and denominator by √2 we get,

(1/√2) ×(√2/√2)= √2/2 ( since √2×√2 = 2)

We know that, √2 = 1.4142…

Then, √2/2 = 1.4142/2 = 0.7071..

Since the number , 0.7071..is non-terminating non-recurring, 1/√2 is an irrational number.

**(v) 2****Solution:**

We know that, the value of = 3.1415

Hence, 2 = 2×3.1415.. = 6.2830…

Since the number, 6.2830…, is non-terminating non-recurring, 2 is an irrational number.**Q.2. Simplify each of the following expressions:****(i) (3+√3)(2+√2)****Solution:**

(3+√3)(2+√2 )

Opening the brackets, we get, (3×2)+(3×√2)+(√3×2)+(√3×√2)

= 6+3√2+2√3+√6

**(ii) (3+√3)(3-√3 )****Solution:**

(3+√3)(3-√3 ) = 3^{2}-(√3)^{2} = 9-3

= 6

**(iii) (√5+√2) ^{2}**

(√5+√2)

= 5+2×√10+2 = 7+2√10

**(iv) (√5-√2)(√5+√2)****Solution:**

(√5-√2)(√5+√2) = (√5^{2}-√2^{2}) = 5-2 = 3**Q.3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter, (say d). That is, π = c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?****Solution:**

There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize whether c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…**Q.4. Represent (√9.3) on the number line.****Solution:**

Step 1: Draw a 9.3 units long line segment, AB. Extend AB to C such that BC = 1 unit.

Step 2: Now, AC = 10.3 units. Let the centre of AC be O.

Step 3: Draw a semi-circle of radius OC with centre O.

Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD.

Step 5: OBD, obtained, is a right angled triangle.

Here, OD 10.3/2 (radius of semi-circle), OC = 10.3/2 , BC = 1

OB = OC – BC

⟹ (10.3/2)-1 = 8.3/2

Using Pythagoras theorem,

We get,

OD^{2 }= BD^{2 }+ OB^{2}

⟹ (10.3/2)^{2} = BD^{2}+(8.3/2)^{2}

⟹ BD^{2 = }(10.3/2)^{2}-(8.3/2)^{2}

⟹ (BD)^{2 }= (10.3/2)-(8.3/2)(10.3/2)+(8.3/2)

⟹ BD^{2} = 9.3

⟹ BD = √9.3

Thus, the length of BD is √9.3.

Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O as shown in the figure.

**Q.5. Rationalize the denominators of the following:****(i) 1/√7****Solution:**

Multiply and divide 1/√7 by √7

(1×√7)/(√7×√7) = √7/7

**(ii) 1/(√7-√6)****Solution:**

Multiply and divide 1/(√7-√6) by (√7+√6)

[1/(√7-√6)]×(√7+√6)/(√7+√6) = (√7+√6)/(√7-√6)(√7+√6)

= (√7+√6)/√7^{2}-√6^{2 }[denominator is obtained by the property, (a+b)(a-b) = a^{2}-b^{2}]

= (√7+√6)/(7-6)

= (√7+√6)/1

= √7+√6

**(iii) 1/(√5+√2)****Solution:**

Multiply and divide 1/(√5+√2) by (√5-√2)

[1/(√5+√2)]×(√5-√2)/(√5-√2) = (√5-√2)/(√5+√2)(√5-√2)

= (√5-√2)/(√5^{2}-√2^{2}) [denominator is obtained by the property, (a+b)(a-b) = a^{2}-b^{2}]

= (√5-√2)/(5-2)

= (√5-√2)/3

**(iv) 1/(√7-2)****Solution:**

Multiply and divide 1/(√7-2) by (√7+2)

1/(√7 - 2)×(√7 + 2)/(√7 + 2) = (√7 + 2)/(√7 - 2)(√7 + 2)

= (√7 + 2)/(√7^{2 }- 2^{2}) [denominator is obtained by the property, (a + b)(a - b) = a^{2}-b^{2}]

= (√7 + 2)/(7 - 4)

= (√7 + 2)/3

**Q.1. Find:**

**(i) 64 ^{1/2}**

64

= (8

= 8

= 8

**(ii) 32 ^{1/5}**

32

= (2

= 2

= 2

**(iii) 125 ^{1/3}**

(125)

= (5

= 5

= 5

9

= (3

= 3

=27

**(ii) 32 ^{2/5}**

32

= (2

= 2

= 4

**(iii)16 ^{3/4}**

16

= (2

= 2

= 8

**(iv) 125 ^{-1/3}**

125

= (5

= 5

= 1/5

2

= 2

**(ii) (1/3 ^{3})^{7}**

(1/3

= 3

**(iii) 11 ^{1/2}/11^{1/4}**

11

= 11

**(iv) 7 ^{1/2}×8^{1/2}**

7

= 56

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