Solution: $\begin{array}{l} \mathrm{A}=\{1,2,3,4\} \text { and } \mathrm{R}=\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,3) \\ (3,2)\} \text { (Given) } \end{array}$ $\mathrm{R}$ is reflexive if $\mathrm{a}...
Solution: $A=\{1,2,3\}$ and $\bar{R}=\{(1,1),(2,2),(3,3),(1,2),(2,3)\}$ (Given) $\mathrm{R}$ is reflexive if $\mathrm{a} \in \mathrm{A}$ and $(\mathrm{a}, \mathrm{a}) \in \mathrm{R}$ Here,...
Solution: $R=\{(a, b): a>b\}$ on $N$ (given) Non-Reflexivity: Assume $a$ be an arbitrary element of $\mathrm{N}$ a cannot be greater than $a$ $\Rightarrow(\mathrm{a}, \mathrm{a}) \notin...
for all 
Show that R satisfies none of reflexivity, symmetry and transitivity.Solution: $\mathrm{R}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}=\mathrm{b}_{2}\right\}$ for all $\mathrm{a}, \mathrm{b} \in \mathrm{N}$ (As given) Non-Reflexivity: Assume $a$ be an arbitrary...
be the set of all real numbers. Show that the relation 
is symmetric but neither reflexive nor transitive.Solution: $\mathrm{R}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}_{2}+\mathrm{b}_{2}=1\right\}$ and $\mathrm{a}, \mathrm{b} \in \mathrm{S}$ (As given) Non-Reflexivity: Assume $a$ be an arbitrary...
Solution: $\mathrm{R}=\{(\mathrm{A}, \mathrm{B}): \mathrm{d}(\mathrm{A}, \mathrm{B})<2$ units $\}$, where $\mathrm{d}(\mathrm{A}, \mathrm{B})$ is the distance between the points $\mathrm{A}$ and...
be the set of all real numbers and let 
and 
Show that 
is an equivalence relation on .Solution: $\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{S}$ and $\mathrm{a}=\pm \mathrm{b}\} .$ (Given) If $\mathrm{R}$ is Reflexive, Symmetric and Transitive, then...
on 
, defined by 
is an equivalent relation.Solution: If $R$ is Reflexive, Symmetric and Transitive, then $R$ is an equivalence relation. Reflexivity: Suppose $a$ and $\mathrm{b}$ be an arbitrary element of $\mathrm{N} \times \mathrm{N}$...
Solution: $\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{A}$ and $|\mathrm{a}-\mathrm{b}|$ is even $\}$ where $\mathrm{A}=$ (As given) If $\mathrm{R}$ is Reflexive,...
be the set of all triangles in a plane. Show that the relation 
is an equivalence relation on 
.Solution: Suppose $\mathrm{R}=\left\{\left(\Delta 1, \Delta_{2}\right): \Delta_{1} \sim \Delta_{2}\right\}$ be a relation defined on A. (As given) If $\mathrm{R}$ is Reflexive, Symmetric and...
Answer: Using factorization, 660 = 2 × 2 × 3 × 5 × 11
Answer: Six bells toll together at intervals of 2,4, 6, 8, 10 and 12 minutes. Using prime factorization, 2 = 2 4 = 2 × 2 6 = 2 × 3 8 = 2 × 2 × 2 10 = 2 × 5 12 = 2 × 2 × 3 LCM (2, 4, 6, 8, 10, 12) =...
Answer: Beep duration of first device = 60 seconds Beep duration of second device = 62 seconds The interval of beeping together = LCM (60, 62) Using prime factorization, 60 = 22 × 3 × 5 62 = 2 × 31...
Answer: Length of the three measuring rods are 64cm, 80cm and 96cm. The length of cloth that can be measured an exact number of times = LCM (64, 80, 96) Using prime factorization, 64 = 26 80 = 24 ×...
Answer: We know that, Length of a tile = 15m 17m = 1517cm [∵ 1m = 100cm] Breadth of a tile = 9m 2m = 902cm ∴ Side of each square tile = HCF (1517, 902) Using prime factorization, 1517 = 37 × 41 902...
Answer: Given lengths are 7m (700cm), 3m 85cm (385cm) and 12m 95m (1295cm). The required length = HCF (700, 385, 1295) Using prime factorization, 700 = 2 × 2 × 5 × 5 × 7 = 22 × 52 × 7 385 = 5 × 7 ×...
Answer: Total number of English books = 336 Total number of mathematics books = 240 Total number of science books = 96 ∴ Number of books stored in each stack = HCF (336, 240, 96) Using prime...
Answer: Minimum number of rooms required = ???????????????????? ???????????????????????? ???????? ???????????????????????????????????????????????? / ???????????? (60,84,108) Prime factorization of...
Answer: Using prime factorization: 15 = 3 × 5 24 = 23 × 3 36 = 22 × 32 LCM = 23 × 32× 5 LCM = 360 The greatest four-digit number is 9999. Divide 9999 by 360, 279 - remainder 9999 – 279 = 9720 is...
Answer: The smallest number which when increased by 17 is exactly divisible by both 468 and 520 is obtained by subtracting 17 from the LCM of 468 and 520. Prime factorization of 468 and 520: 468 =...
Answer: Consider x as the required number, Apply Euclid’s lemma, x = 28p + 8 and x = 32q + 12 [p and q - quotients] 28p + 8 = 32q + 12 28p = 32q + 4 7p = 8q + 1….. (1) p = 8n – 1 and q = 7n –...
Answer: Least number which can be divided by 35, 56 and 91 is LCM of 35, 56 and 91. Prime factorization of 35, 56 and 91: 35 = 5 × 7 56 = 23 × 7 91 = 7 × 13 LCM = 23 × 5 × 7 × 13 LCM = 3640 Least...
Answer: Given, The number divides 315 (320 – 5) and 450 (457 – 7). ∴ Required number = HCF (315, 450) Apply Euclid’s algorithm, 315 ) 450 ( 1 - 315 ____________ 135 ) 315 ( 2 - 270...
Answer: Largest number which divides 438 and 606, leaving remainder 6 438 – 6 = 432 606 – 6 = 600 [remainder – 0] HCF of 432 and 600 gives the largest number. Prime factors of 432 and 600: 432 = 24...
![\[(i)\frac{1095}{1168}(ii)\frac{368}{496}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-32f0a63581456d210bcc76f5723fbff1_l3.png "Rendered by QuickLaTeX.com")
Answers: (i) Prime factorization of 1095 and 1168 is: 1095 = 3 × 5 × 73 1168 = 24 × 73 $\eqalign{ & \frac{{1095}}{{1168}} = \frac{{3 \times 5 \times 73}}{{{2^4} \times 73}} \cr & = >...
Answer: No, it is not possible to have two numbers whose HCF is 18 and LCM is 760. HCF must be a factor of LCM, but 18 is not factor of 760.
Answer: HCF of two numbers = 18 Product of two numbers = 12960 Consider, LCM - x Product of two numbers = HCF × LCM 12960 = 18 × x $$\eqalign{ & x = \frac{{12960}}{{18}} \cr & \therefore x =...
Answer: HCF of two numbers = 145 LCM of two numbers = 2175 Consider the one number as 725 and other be x. Product of two numbers = HCF × LCM 725 × x = 145 × 2175 $$\matrix {x = \frac{{145...
Answer: Consider a and b as the two numbers, a - 161 HCF = 23 LCM = 1449 a × b = HCF × LCM 161 × b = 23 × 1449 $$\matrix {b = \frac{{23 \times 1449}}{{161}}} \hfill \\ {b =...
Answers: (i) 30 = 2 × 3 × 5 72 = 2 × 2 × 2 × 3 × 3 => 23 × 32 432 = 2 × 2 × 2 × 2 × 3 × 3 × 3 => 24 × 33 HCF = 2 × 3 HCF = 6 LCM = 24 × 33 × 5 LCM = 2160 (ii) 21 = 3 × 7 28 = 2 × 2 × 7 =>...
Answers: (i) 17 = 17 23 = 23 29 = 29 HCF = 1 LCM = 17 × 23 × 29 LCM = 11339 (ii) 24 = 2 × 2 × 2 × 3 => 23 × 3 36 = 2 × 2 × 3 × 3 => 22 × 32 40 = 2 × 2 × 2 × 5 => 23 × 5 HCF = 22 HCF = 4 LCM...
Answers: (i) Using prime factorization, 396 = 22 × 32 × 11 1080 = 23 × 33 × 5 HCF = 22 × 32 HCF = 36 LCM = 23 × 33 × 5 × 11 LCM = 11880 (ii) Using prime factorization, 1152 = 27 × 32 1664 = 27 × 13...
Answers: (i) 8 = 2 × 2 × 2 => 23 9 = 3 × 3 => 32 25 = 5 × 5 => 52 HCF = 1 LCM = 23 × 32 × 52 LCM = 1800 (ii) 12 = 2 × 2 × 3 => 22 × 3 15 = 3 × 5 21 = 3 × 7 HCF = 3 LCM = 22 × 3 × 5 × 7...
Answers: (i) Using prime factorization, 96 = 25 × 3 404 = 22 × 101 HCF = 22 HCF = 4 LCM = 25 × 3 × 101 LCM = 9696 (ii) Using prime factorization, 144 = 24 × 32 198 = 2 × 32 × 11 HCF = 2 × 32 HCF =...
Answers: (i) Using prime factorization, 36 = 22 × 3 84 = 22 × 3 × 7 HCF = 22 × 3 HCF = 12 LCM = 22 × 32 × 7 LCM = 252 (ii) Using prime factorization, 23 = 23 31 = 31 HCF = 1 LCM = 23 × 31 =...
![\[(i)\frac{69}{92}(ii)\frac{473}{645}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f4c72db715036c18e34216306f8263e8_l3.png "Rendered by QuickLaTeX.com")
Answers: (i) Prime factorization of 69 and 92: 69 = 3 × 23 92 = 22 × 23 $\eqalign{ & \frac{{69}}{{92}} = \frac{{3 \times 23}}{{{2^2} \times 23}} \cr & = > \frac{3}{{{2^2}}} \cr & ...