Access Answers of Maths NCERT class 9 Chapter 1 – Number Systems Exercise 1.3
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:
(iii)
Solution:
= 4.125 (Terminating)
(iv) 3/13
Solution:
(non terminate Recurring )
(v) 2/11
Solution:
(vi) 329/400
Solution:
= 0.8225 (Terminating)
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:
Hence
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
(iii)
Solution:
= (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)
Solution:
Assume that x = 0.001001…
Then, 1000x = 1.001001…
1000x = 1 + x
999x = 1
x = 1/999
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 (1)
Multiplying both sides by 10,
10x = 9.9999…. Eq. (2)
Eq.(2) – Eq.(1), 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.
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.
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 = 21
7/8 = 0. 875, denominator q =23
4/5 = 0. 8, denominator q = 51
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.
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
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…
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.
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