(A) 2 (B) – 2 (C) ¼ (D) ½ (A) 2 In case ½ is a foundation of the situation \[x2\text{ }+\text{ }kx\text{ }\text{ }5/4\text{ }=\text{ }0\] then, at that point, subbing the worth of ½ instead of x...
Solution: Real numbers are simply the combination of rational and irrational numbers, in the number system. In general, all the arithmetic operations can be performed on these numbers and they...
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![\[\mathbf{380}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cb340828e6c6101fbde249e54a9cd1cb_l3.png "Rendered by QuickLaTeX.com")
Solution: On dividing \[380\]by \[16\]we get,\[23\]as the quotient and\[12\]as the remainder. Now, since the LCM is not exactly divisible by the HCF its can be said that two numbers cannot have...
![\[\mathbf{84},\text{ }\mathbf{90}\text{ }\mathbf{and}\text{ }\mathbf{120}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ccfecca6ebaa7e44c50bda4ca8c75939_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{24},\text{ }\mathbf{15}\text{ }\mathbf{and}\text{ }\mathbf{36}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-94f859fd8a71046e251dab20e2d08dcb_l3.png "Rendered by QuickLaTeX.com")
Solution: First, Find the prime factors of the given integers: \[84,\text{ }90\text{ }and\text{ }120\] For, \[\begin{array}{*{35}{l}} ~84\text{...
![\[\mathbf{8},\text{ }\mathbf{9}\text{ }\mathbf{and}\text{ }\mathbf{25}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-78297b7f285a41777161d421fedfd300_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{40},\text{ }\mathbf{36}\text{ }\mathbf{and}\text{ }\mathbf{126}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-17883cf51f0a9421203bd7e5576dc901_l3.png "Rendered by QuickLaTeX.com")
Solution: First, find the prime factors of the given integers: 8, 9 and 25 For, \[8\text{ }=\text{ }2\text{ }\times...
(i) 12, 15 and 21(ii) \[\mathbf{17},\text{ }\mathbf{23}\text{ }\mathbf{and}\text{ }\mathbf{29}\] Solution: First, find the prime factors of the given integers: \[12,\text{ }15\text{ }and\text{...
(i) \[\mathbf{26}\text{ and }\mathbf{91}\](ii) \[\mathbf{510}\text{ and }\mathbf{92}\] Solution: Given integers are\[\mathbf{26}\text{ and...
![\[\surd 5\text{ }+\text{ }\surd 3~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0efb8e4ff13d751cc9c194f8bc9b1b82_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[\surd 5\text{ }+\text{ }\surd 3\] is a rational number. Then, there exist co prime positive integers a and b such that\[\surd 5\text{ }+\text{ }\surd...
![\[\mathbf{2}\text{ }-\text{ }\mathbf{3}\surd \mathbf{5}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-492fc7220996be5495f8ab5970d52502_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[2\text{ }\text{ }3\surd 5\]is a rational number. Then, there exist co prime positive integers a and b such that \[2\text{ }\text{ }3\surd 5\text{...
![\[\mathbf{5}\text{ }-\text{ }\mathbf{2}\surd \mathbf{3}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4fe99f6574a18dae472d801d66331d39_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[5\text{ }\text{ }2\surd 3\] is a rational number. Then, there exist co prime positive integers a and b such that \[5\text{ }\text{ }2\surd 3\text{...
![\[\mathbf{4}\text{ }-\text{ }\mathbf{5}\surd \mathbf{2}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-728fc3aae21f5413024bcf441fffe566_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[4\text{ }\text{ }5\surd 2\] is a rational number. Then, there exist co prime positive integers a and b such that \[4\text{ }\text{ }5\surd 2\text{...
![\[\mathbf{3}\text{ }+\text{ }\surd \mathbf{2}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fc1c82b2f160b7a1dc310519dbde254a_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[3\text{ }+\text{ }\surd 2\] is a rational number. Then, there exist co prime positive integers a and b such that \[3\text{ }+\text{ }\surd 2=\text{...
![\[\mathbf{2}\text{ }-\text{ }\surd \mathbf{3}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d0a39d7c5e26f9ab889ba6084fa2705f_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[2\text{ }\text{ }\surd 3\] is a rational number. Then, there exist co prime positive integers a and b such that\[2\text{ }\text{ }\surd 3=\text{ }a/b\]...
![\[\mathbf{4}\text{ }+\text{ }\surd \mathbf{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e72e18bc9159e65a933a5182ce232b1e_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{5}\surd \mathbf{2}~~~~~~~~~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f66067211a2d0fa3d76358415a69dc68_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[4\text{ }+\text{ }\surd 2\] is a rational number. Then, there exist co prime positive integers a and b such that \[4\text{ }+\text{ }\surd 2\text{...
![\[\mathbf{6}\text{ }+\text{ }\surd \mathbf{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3751ac6d34393366c041894ce911cbaa_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[6+\surd 2\] is a rational number. Then, there exist co prime positive integers a and b such that \[6\text{ }+\text{ }\surd 2\text{ }=\text{ }a/b\]...
Arrangements: (I) 13/3125 13/3125 = 0.00416 (ii) 17/8 17/8 = 2.125 (iii) 64/455 has a Non ending decimal development (iv)15/1600 15/1600 = 0.009375 (v) 29/343 has a Non ending decimal development...
Solutions: Note: If the denominator has just factors of 2 and 5 or as 2m ×5n then it has ending decimal extension. On the off chance that the denominator has factors other than 2 and 5, it has a...
Solution: Since, Both Sonia and Ravi move a similar way and simultaneously, the technique to make when they will meet again at the beginning stage is LCM of 18 and 12. Consequently, LCM(18,12) =...
Solution: By the meaning of composite number, we know, assuming a number is composite, it implies it has factors other than 1 and itself. In this way, for the given articulation; 7 × 11 × 13 + 13...
Solution: If the number 6n closures with the digit zero (0), then, at that point it ought to be distinct by 5, as we probably are aware any number with unit place as 0 or 5 is detachable by 5. Prime...
Solutiont: As we realize that, HCF×LCM=Product of the two given numbers Accordingly, 9 × LCM = 306 × 657 LCM = (306×657)/9 = 22338 Consequently, LCM(306,657) = 22338
(I) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25 Solution: (I) 12, 15 and 21 Composing the result of prime components for every one of the three numbers, we get, 12=2×2×3 15=5×3 21=7×3...
(I) 26 and 91 (ii) 510 and 92 (iii) 336 and 54 Solution: (I) 26 and 91 Communicating 26 and 91 as result of its excellent elements, we get, 26 = 2 × 13 × 1 91 = 7 × 13 × 1 In this manner, LCM (26,...
(I) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429 Solution (I) 140 By Taking the LCM of 140, we will get the result of its great factor. Subsequently, 140 = 2 × 2 × 5 × 7 × 1 = 22×5×7 (ii) 156 By...