Category : 10th Class
Real Numbers
Real-Numbers
For any two given positive integer a and b there exist unique integer q and r satisfying
\[a\text{ }=\text{ }bq\text{ }+\text{ }r\], where \[0\,\,\le \,\,r\,\,<\text{ }b\]
Here a is known as dividend, b as divisor, q as quotient and r as remainder.
To find the H.C.F. and L.C.M. of numbers using the method of Fundamental Theorem of Arithmetic (Prime Factorisation Method):
Express each one of the given numbers as a product of prime factors. Then,
H.C.F. = Product of the smallest powers of each common prime factor in the numbers
L.C.M. = Product of the greatest powers of each prime factor, involved in the numbers
A rational number, in the simplest form \[\frac{p}{q}\], where p and q are integers and \[q\,\,\ne \,\,0\] is:
A terminating decimal if prime factorisation of q is of the form \[\left( {{2}^{m}}\text{ }\times \text{ }{{5}^{n}} \right)\], where m and n are non-negative integers.
(ii) A non-terminating repeating decimal if prime factorisation of q is not of the form \[\left( {{2}^{m}}\text{ }\times \text{ }{{5}^{n}} \right)\], where m and n are non-negative integers.
Snap Test
(a) 1990 (b) 1980
(c) 1955 (d) 1985
(e) None of these
Ans. (c)
Explanation: \[9775\,\,=\,\,52\,\,\times ~\,\,17~\,\,\times ~\,\,23\]
\[11730\text{ }=\text{ }2~\,\,\times \,\,3\,\,\times \,\,~5\,\,~\times \,\,17\,\,~\times \,\,23\]
\[\therefore \] H.C.F. \[\left( 9775,\text{ }11730 \right)\text{ }=\text{ }5~\,\,\times \,\,17\text{ }\times \text{ }23\text{ }=\text{ }1955.\]
(a) Terminating
(b) Non-terminating repeating
(c) Non-terminating non-repeating
(d) Can’t be find
(e) None of these
Ans. (b)
Explanation: The given rational number is in the form \[\frac{p}{q}\], where \[p\text{ }=\text{ }41\] and \[q\text{ }=\text{ }37500.\]
After prime factorisation of 37500 we get:
\[37500\text{ }=\text{ }({{2}^{2}}~\,\,\times \,\,3\,\,~\times \,\,{{5}^{2}})\] which is not of the form \[({{2}^{m}}~\times \,\,{{5}^{n}})\]
Hence, \[\frac{41}{37500}\] is a non-terminating repeating decimal.
(a) \[\frac{785}{330}\] (b) \[\frac{785}{333}\]
(c) \[\frac{785}{335}\] (d) \[\frac{780}{335}\]
(e) None of these
Ans. (b)
Explanation:
Let \[x~~~~=~~~2.357357357\]….. ... (i)
Then, \[1000x~~=~~~2357.357357357\]…. ... (ii)
Subtracting (i) from (ii), we get:
\[999x=2355~~\] \[\Rightarrow \,\,\,\,\,x\,\,\,=\,\,\frac{2355}{999}=\frac{785}{333}\]
(a) \[\frac{26}{45}\] (b) \[\frac{26}{40}\]
(c) \[\frac{45}{26}\] (d) \[\frac{45}{22}\]
(e) None of these
Ans. (a)
Explanation:
\[Let\text{ }\,\,x~~=~~0.5\overline{7}~\,\,=\text{ }0.57777\] ... (i)
Then, \[10x\text{ }=\text{ }5.7777...\] ... (ii)
\[100x\text{ }=\text{ }57.7777...\] ... (iii)
Subtracting (ii) from (iii), we get:
\[90x\,\,=\,\,52\,\,\Rightarrow \,x\,\,=\,\,\frac{52}{90}\,\,=\,\,\frac{26}{45}\]
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