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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,10x = 6.666…
10x = 6 + x
9x = 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, 10x = 7.777…
10x = 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, 1000x = 1.001001…
1000x = 1 + x
999x = 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,
10x = 9.9999…. Eq. (b)
Eq.(b) – Eq.(a), we get
(10x = 9.9999)(x = 0.9999…)
9x = 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 nonterminating nonrecurring.
Solution:
We know that all irrational numbers are nonterminating nonrecurring. three numbers with decimal expansions that are nonterminating nonrecurring are:
Q.8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.
Solution:
Three different irrational numbers are:
Q.9. Classify the following numbers as rational or irrational according to their type:
(i)√23
Solution:
√23 = 4.79583152331…
Since the number is nonterminating nonrecurring 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 nonterminating but recurring, it is a rational number.
(v) 1.101001000100001…
Solution:
Since the number,1.101001000100001…, is nonterminating nonrepeating (nonrecurring), 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 nonterminating and nonrecurring.
Now, substituting the value of √5 in 2 –√5, we get,
2√5 = 22.2360679… = 0.2360679
Since the number, – 0.2360679…, is nonterminating nonrecurring, 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 nonterminating nonrecurring, 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 nonterminating nonrecurring, 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} = 93
= 6
(iii) (√5+√2)^{2}
Solution:
(√5+√2)^{2 }= √5^{2}+(2×√5×√2)+ √2^{2}
= 5+2×√10+2 = 7+2√10
(iv) (√5√2)(√5+√2)
Solution:
(√5√2)(√5+√2) = (√5^{2}√2^{2}) = 52 = 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 semicircle 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 semicircle), 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)(ab) = a^{2}b^{2}]
= (√7+√6)/(76)
= (√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)(ab) = a^{2}b^{2}]
= (√5√2)/(52)
= (√5√2)/3
(iv) 1/(√72)
Solution:
Multiply and divide 1/(√72) 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}
Solution:
64^{1/2} = (8×8)^{1/2}
= (8^{2})^{½}
= 8^{1} [⸪2×1/2 = 2/2 =1]
= 8
(ii) 32^{1/5}
Solution:
32^{1/5 }= (2^{5})^{1/5}
= (2^{5})^{⅕}
= 2^{1} [⸪5×1/5 = 1]
= 2
(iii) 125^{1/3}
Solution:
(125)^{1/3} = (5×5×5)^{1/3}
= (5^{3})^{⅓}
= 5^{1} (3×1/3 = 3/3 = 1)
= 5
Q.2. Find:
(i) 9^{3/2}
Solution:
9^{3/2} = (3×3)^{3/2}
= (3^{2})^{3/2}
= 3^{3} [⸪2×3/2 = 3]
=27
(ii) 32^{2/5}
Solution:
32^{2/5 }= (2×2×2×2×2)^{2/5}
= (2^{5})^{2⁄5}
= 2^{2} [⸪5×2/5= 2]
= 4
(iii)16^{3/4}
Solution:
16^{3/4 }= (2×2×2×2)^{3/4}
= (2^{4})^{3⁄4}
= 2^{3} [⸪4×3/4 = 3]
= 8
(iv) 125^{1/3}
Solution:
125^{1/3 }= (5×5×5)^{1/3}
= (5^{3})^{1⁄3}
= 5^{1} [⸪3×1/3 = 1]
= 1/5
Q.3. Simplify:
(i) 2^{2/3}×2^{1/5}
Solution:
2^{2/3}×2^{1/5 }= 2^{(2/3)+(1/5)} [⸪Since, a^{m}×a^{n}=a^{m+n}____ Laws of exponents]
= 2^{13/15} [⸪2/3 + 1/5 = (2×5+3×1)/(3×5) = 13/15]
(ii) (1/3^{3})^{7}
Solution:
(1/3^{3})^{7 }= (3^{3})^{7} [⸪Since,(a^{m})^{n }= a^{m x n}____ Laws of exponents]
= 3^{21}
(iii) 11^{1/2}/11^{1/4}
Solution:
11^{1/2}/11^{1/4 }= 11^{(1/2)(1/4)}
= 11^{1/4} [⸪(1/2) – (1/4) = (1×42×1)/(2×4) = 42)/8 = 2/8 = ¼ ]
(iv) 7^{1/2}×8^{1/2}
Solution:
7^{1/2}×8^{1/2} = (7×8)^{1/2} [⸪Since, (a^{m}×b^{m }= (a×b)^{m} ____ Laws of exponents]
= 56^{1/2}
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