(iii)
Ans:
= 4.125 (Terminating)
(iv) 3/13
Ans:
(v) 2/11
Ans:
(vi) 329/400
Ans:
= 0.8225 (Terminating)
Q2. You know that 1/7 = . 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.]
Ans:
Q3. Express the following in the form p/q, where p and q are integers and q ≠ 0.
(i)
Ans:
Assume that x = 0.666…
Then,10x = 6.666…
10x = 6 + x
9x = 6
x = 2/3
(ii)
Ans:
= (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
(iii)
Ans:
Assume that x = 0.001001…
Then, 1000x = 1.001001…
1000x = 1 + x
999x = 1
x = 1/999
Q4. 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.
Ans: 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.
Q5. 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.
Ans: 1/17
Dividing 1 by 17:
There are 16 digits in the repeating block of the decimal expansion of 1/17.
Q6. 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?
Ans: 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.
Q7. Write three numbers whose decimal expansions are nonterminating nonrecurring.
Ans: We know that all irrational numbers are nonterminating nonrecurring. three numbers with decimal expansions that are nonterminating nonrecurring are:
Q8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.
Ans:
Three different irrational numbers are:
Q9. Classify the following numbers as rational or irrational according to their type:
(i)√23
Ans: √23 = 4.79583152331…
Since the number is nonterminating nonrecurring therefore, it is an irrational number.
(ii)√225
Ans: √225 = 15 = 15/1
Since the number can be represented in p/q form, it is a rational number.
(iii) 0.3796
Ans: Since the number, 0.3796, is terminating, it is a rational number.
(iv) 7.478478
Ans: The number, 7.478478, is nonterminating but recurring, it is a rational number.
(v) 1.101001000100001…
Ans: Since the number, 1.101001000100001…, is nonterminating nonrepeating (nonrecurring), it is an irrational number.
Q1. Classify the following numbers as rational or irrational:
(i) 2 –√5
Ans: 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
Ans: (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
Ans: 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
Ans: 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
Ans: 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.
Q2. Simplify each of the following expressions:
(i) (3 + √3) (2 + √2)
Ans: (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)
Ans: (3 + √3) (3  √3 ) = 3^{2 } (√3)^{2} = 9  3
= 6
(iii) (√5 + √2)^{2}
Ans: (√5 + √2)^{2 }= √5^{2 }+ (2 × √5 × √2) + √2^{2}
= 5 + 2 × √10 + 2 = 7 + 2√10
(iv) (√5  √2)(√5 + √2)
Ans: (√5  √2)(√5 + √2) = (√5^{2 } √2^{2}) = 5  2 = 3
Q3. 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?
Ans: 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…
Q4. Represent (√9.3) on the number line.
Ans:
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.
Q5. Rationalize the denominators of the following:
(i) 1/√7
Ans: Multiply and divide 1/√7 by √7
(1×√7)/(√7×√7) = √7/7
(ii) 1/(√7√6)
Ans: 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)
Ans: 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/(√72)
Ans: 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
Ans: 64^{1/2} = (8 × 8)^{1/2}
= (8^{2})^{½}
= 8^{1} [⸪2 × 1/2 = 2/2 = 1]
= 8
(ii) 32^{1/5}
Ans: 32^{1/5 }= (2^{5})^{1/5}
= (2^{5})^{⅕}
= 2^{1} [⸪ 5 × 1/5 = 1]
= 2
(iii) 125^{1/3}
Ans: (125)^{1/3} = (5 × 5 × 5)^{1/3}
= (5^{3})^{⅓}
= 5^{1} (3 × 1/3 = 3/3 = 1)
= 5
Q2. Find:
(i) 9^{3/2}
Ans: 9^{3/2} = (3 × 3)^{3/2}
= (3^{2})^{3/2}
= 3^{3} [⸪ 2 × 3/2 = 3]
=27
(ii) 32^{2/5}
Ans: 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}
Ans: 16^{3/4 }= (2 × 2 × 2 × 2)^{3/4}
= (2^{4})^{3⁄4}
= 2^{3} [⸪ 4 × 3/4 = 3]
= 8
(iv) 125^{1/3}
Ans: 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}
Ans: 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}
Ans: (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}
Ans: 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}
Ans: 7^{1/2}× 8^{1/2} = (7 × 8)^{1/2} [⸪ Since, (a^{m }× b^{m }= (a × b)^{m} ____ Laws of exponents]
= 56^{1/2}
1. How can I solve Exercise 1.3 in the NCERT Number System book? 
2. What is the significance of Exercise 1.4 in the NCERT Number System book? 
3. What are the key topics covered in Exercise 1.5 of the NCERT Number System book? 
4. How can Exercise 1.6 in the NCERT Number System book help in understanding decimal representation? 
5. How can the NCERT Number System book help in preparing for competitive exams? 
62 videos438 docs102 tests

62 videos438 docs102 tests
