{"id":170472,"date":"2024-05-24T12:49:22","date_gmt":"2024-05-24T07:19:22","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=170472"},"modified":"2024-05-24T14:16:52","modified_gmt":"2024-05-24T08:46:52","slug":"ml-aggarwal-class-11-maths-solutions-section-a-chapter-1-mcqs","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-11-maths-solutions-section-a-chapter-1-mcqs\/","title":{"rendered":"ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets MCQs"},"content":{"rendered":"

Interactive ISC Mathematics Class 11 Solutions<\/a> Chapter 1 Sets MCQs engage students in active learning and exploration.<\/p>\n

ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets MCQs<\/h2>\n

Choose the correct answer from the given four options in questions (1 to 29) :<\/p>\n

Question 1.
\nWhich of the following collection of objects is not a set?
\n(a) The collection of all even integers.
\n(b) The collection of all months of a year beginning with letter J.
\n(c) The collection of most talented writers of India.
\n(d) The collection of all prime numbers less than 20.
\nAnswer:
\n(c) The collection of most talented writers of India.<\/p>\n

The collection of most talented writers of India is not a well defined collection. Since according to some one, one witer is talented and according to some other the same writer is non-talented so the word talented is vague.
\n\u2234 Thus collection does not represent a set.<\/p>\n

Question 2.
\nIf A = {1, 2, 3, 4, 5}, then which of the following is not true?
\n(a) 0 \u2260 A
\n(b) 3 \u2208 A
\n(c) {3} \u2208 A
\n(d) {3} \u2282 A
\nAnswer:
\n(c) {3} \u2208 A<\/p>\n

Clearly 0, 3 are members of A so both are belongs to set A and {3} is a subset of A but {3} \u2209 A.<\/p>\n

\"ML<\/p>\n

Question 3.
\nWhich of the following is a null set?
\n(a) {x : x \u2208 N, 2x – 1 = 3}
\n(b) {x : x \u2208 N, x2<\/sup> < 20}
\n(c) (x : x \u2208 N, x is a factor of 128}
\n(d) {x : x \u2208 I, x \u2264 7}
\nAnswer:
\n(c) (x : x \u2208 N, x is a factor of 128}<\/p>\n

For option (a) ;
\n2x – 1 = 3
\n\u21d2 2x = 4
\n\u21d2 x = 2
\n\u2234 {x : x \u2208 N, 2x – 1 = 3} = {2) which is not an empty set.<\/p>\n

For option (b) ;
\nx2<\/sup> < 20 and x \u2208 N
\n\u2234 x = 1 ;
\n\u2235 12<\/sup> < 20 ;
\nx = 3
\n\u2235 32<\/sup> < 20 ;
\nx = 2
\n\u2235 22<\/sup> < 20 ;
\nx = 4
\n\u2235 42<\/sup> < 20
\n\u2234 {x : x \u2208 N ; x2<\/sup> < 20} = {1, 2, 3, 4}<\/p>\n

For option (d) :
\n3x + 7 = 1
\n\u21d2 x = – 2 {x : x \u2208 N ; 3x + 7 = 1}
\n= {- 2} which is not a null set<\/p>\n

For option (c) ;
\nsince there is only one even prime number, which is 2 and their is no even prime number which is > 2.
\n\u2234 given set is a null set.<\/p>\n

Question 4.
\nWhich of the following is a finite set?
\n(a) {x : x = 2n, n\u00b0 \u2208 N}
\n(b) {x : x is a prime number}
\n(c) {x : x \u2208 N, x is a factor of 128}
\n(d) {x : x \u2208 I, x \u2264 7}
\nAnswer:
\n(c) {x : x \u2208 N, x is a factor of 128}<\/p>\n

xisafactorof 128 and x \u2208 N
\n\u2234 x = 1, 2, 4, 8, 16, 32, 64, 128
\nThus {x : x \u2208 N, x is a factor of 128} = {1, 2, 4, 8, 16, 32, 64, 128}
\nwhich contains 8 elements and hence it is a finite set.<\/p>\n

Question 5.
\nGiven set A = {1, 3, 5, 7, 9}, B = {0, 2, 4, 6} and C = {7, 8, 9}. Which of the following may be taken as universal set for all the three sets A, B and C?
\n(a) {0, 1, 2, 3, 4, 5, 6, 7, 8}
\n(b) {1, 2, 3, 4, 5, 6, 7, 8, 9}
\n(c) {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
\n(d) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
\nAnswer:
\n(d) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}<\/p>\n

Now, universal set contains all elements of A three sets A, B and C
\n\u2234 U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
\nSo universal set contain some more elements other than elements of A, B and C.<\/p>\n

\"ML<\/p>\n

Question 6.
\nNumber of proper subsets of a set containing 4 elements is
\n(a) 42<\/sup>
\n(b) 42<\/sup> – 1
\n(c) 24<\/sup>
\n(d) 24<\/sup> – 1
\nAnswer:
\n(d) 24<\/sup> – 1<\/p>\n

We know that, the number of proper subsets of set containing n elements = 2n<\/sup> – 1
\n\u2234 No. of subsets of a set containing 4 elements = 24<\/sup> – 1 = 15.<\/p>\n

Question 7.
\nWhich of the following is not correct?
\n(a) N \u2282 R
\n(b) N \u2282 Q
\n(c) Q \u2282 R
\n(d) N \u2282 T
\nAnswer:
\n(d) N \u2282 T<\/p>\n

Since every natural number is a real number
\nas well as rational number
\n\u2234 N \u2282 R and N \u2282 Q both are true.
\nAlso rational and irrational numbers combine to form real number
\n\u2234 Q \u2282 R is correct<\/p>\n

Question 8.
\nOn real axis if A = [1, 5] and B = [3, 9], then A – B is
\n(a) (5, 9)
\n(b) (1, 3)
\n(c) [5, 9)
\n(d) [1, 3)
\nAnswer:
\n(d) [1, 3)<\/p>\n

A = [1, 5]<\/p>\n

\"ML<\/p>\n

and B = [3, 9]<\/p>\n

\"ML<\/p>\n

\u2234 A – B = A \u2229 Bc<\/sup>
\n= A – A \u2229 B
\n= [1, 3)<\/p>\n

\"ML<\/p>\n

Question 9.
\nIf n (A – B) = 10, n (B – A) = 23, n (A \u222a B) = 50, then n (A \u2229 B) is
\n(a) 7
\n(b) 17
\n(c) 27
\n(d) 33
\nAnswer:
\n(b) 17<\/p>\n

\u2234 n (A \u222a B) = n (A – B) + n (A \u2229 B) + n (B – A)<\/p>\n

\"ML<\/p>\n

\u21d2 50 = 10 + n (A \u2229 B) + 23
\n\u21d2 n (A \u2229 B) = 50 – 33 = 17<\/p>\n

Question 10.
\nTwo finite sets A and B are such that A \u2282 B. then which of the following is not correct ?
\n(a) A \u222a B = B
\n(b) A \u2229 B = A
\n(c) A – B = \u03a6
\n(d) B – A = \u03a6
\nAnswer:
\n(d) B – A = \u03a6<\/p>\n

When A \u2282 B
\n\u2234 A \u222a B = B
\nand A \u2229 B = A
\nA – B = A – A \u2229 B
\n= A – A = \u03a6
\nand B – A = B – B
\n= B – A \u2260 \u03a6<\/p>\n

\"ML<\/p>\n

Question 11.
\nTwo finite sets have m and n elements respectively. The total number of subsets of the first set is 192 more than the total number of subsets of the second set. The values of in and n respectively are
\n(a) 7, 6
\n(b) 8, 6
\n(c) 8, 5
\n(d) 9, 7
\nAnswer:
\n(b) 8, 6<\/p>\n

The total number of subsets of set containing m elements = 2m<\/sup>
\nand total no. of subsets of set containing n elements = 2n<\/sup>
\nThen according to given condition.
\n2m<\/sup> = 2n<\/sup> + 192
\n\u21d2 2m<\/sup> – 2n<\/sup> = 192
\n\u21d2 2n<\/sup> (2m – n<\/sup> – 1) = 26<\/sup> \u00d7 31<\/sup>
\n2n<\/sup> (2m-n<\/sup> – 1) = 26<\/sup> \u00d7 (22<\/sup> – 1)
\n\u2234 n = 6
\nand m – n = 2
\n\u21d2 m = 8<\/p>\n

Question 12.
\nFor any two sets A and B, A \u2229 (A \u222a B) is equal to
\n(a) A
\n(b) B
\n(c) \u03a6
\n(d) A \u2229 B
\nSolution:
\n(a) A<\/p>\n

A \u2229 (A \u222a B) = A
\n[\u2235 A \u2282 A \u222a B]<\/p>\n

Question 13.
\nThe symmetric difference of A = {0, 1, 2) and B {2, 3, 4} is
\n(a) {0, 1}
\n(b) {3, 4}
\n(c) {0, 1, 3, 4}
\n(d) {0, 1, 2, 3, 4}
\nSolution:
\n(c) {0, 1, 3, 4}<\/p>\n

A = {0, 1, 2)
\nand B = {2, 3, 4}
\n\u2234 A – B = {0, 1} and B – A = {3, 4}
\nThus, A \u2206 B = (A – B) \u222a (B – A)
\n= {0, 1} \u222a {3, 4}
\n= {0, 1, 3, 4}<\/p>\n

\"ML<\/p>\n

Question 14.
\nThe symmetric difference of sets A and B is equal to
\n(a) (A – B) \u222a A
\n(b) (B – A) \u222a B
\n(c) (A \u222a B) – (A \u2229 B)
\n(d) (A \u222a B) (A \u2229 B)
\nSolution:
\n(c) (A \u222a B) – (A \u2229 B)<\/p>\n

A \u2206 B = (A – B) \u222a (B – A)
\n= (A \u2229 B\u2019) \u222a (B \u2229 A\u2019)
\n= {(A \u222a B) \u2229 (B’ \u222a B)} \u2229 {(A \u222a A’) \u2229 (B’ \u222a A’)}
\n= {(A \u222a B) \u2229 U} \u2229 {U \u2229 (B’ \u222a A’)}
\n= (A \u222a B) \u2229 (B\u2019 \u222a A\u2019)
\n= (A \u222a B) \u2229 (B \u2229 A)\u2019 [Demorgan\u2019s law]
\n= (A \u222a B) – (B \u2229 A)<\/p>\n

Question 15.
\nThe symmetric difference of sets A and B is not equal to
\n(a) (A – B) \u2229 (B – A)
\n(b) (B – A) \u222a B
\n(c) (A \u222a B) – (A \u2229 B)
\nd) ((A \u222a B) – B) \u222a ((A \u222a B) – A)
\nSolution:
\n(a) (A – B) \u2229 (B – A)<\/p>\n

A \u2206 B = (A – B) \u222a (B – A)
\nThus A \u2206 B \u2260 (A – B) \u2229 (B – A)
\nFor option (d) ;
\n= ((A \u222a B) – B) \u222a ((A \u222a B) – A)
\n= ((A \u222a B) \u2229 B\u2019) \u222a ((A \u222a B) \u2229 A\u2019)
\n= {(A \u2229 B’) \u222a (B \u2229 B\u2019)) \u222a {(A \u2229 A\u2019) \u222a (B \u2018 \u2229 A\u2019)}
\n= {(A \u2229 B\u2019) \u222a \u03a6) \u222a {\u03a6 \u222a (B \u2229 A\u2019))
\n= (A \u2229 B\u2019) \u222a (B \u2229 A\u2019)
\n= (A – B) \u222a (B – A)
\n= A \u2206 B<\/p>\n

Question 16.
\nFor any two sets X and Y, X \u2229 (X \u222a Y)\u2019 is equal to
\n(a) X
\n(b) Y
\n(c) \u03a6
\n(d) X \u2229 Y
\nSolution:
\n(c) \u03a6<\/p>\n

X \u2229 (X \u222a Y)’ = X \u2229 (X\u2019 \u2229 Y\u2019) [Demorgan\u2019s law]
\n= (X \u2229 X\u2019) \u2229 Y\u2019
\n= \u03a6 \u2229 Y\u2019 = \u03a6<\/p>\n

\"ML<\/p>\n

Question 17.
\nFor any two sets A and B, ((A\u2019 \u222a B\u2019) – A is equal to
\n(a) A
\n(b) B
\n(c) \u03a6
\n(d) A \u2229 B
\nAnswer:
\n(a) A<\/p>\n

((A\u2019 \u222a B\u2019) – A)\u2019 = {(A\u2019 \u222a B\u2019) \u2229 A\u2019)\u2019
\n= (A\u2019 \u222a B\u2019)\u2019 \u222a (A\u2019)\u2019 [Demorgan\u2019s law]
\n= [(A\u2019)\u2019 \u2229 (B\u2019)’} \u222a A
\n= (A \u2229 B) \u222a A = A
\n[\u2235 A \u2229 B \u2282 A]<\/p>\n

Question 18.
\nFor any two sets A and B, (B’ \u222a (B\u2019 – A)]\u2019 is equal to
\n(a) A
\n(b) B
\n(c) \u03a6
\n(d) A \u222a B
\nAnswer:
\n(b) B<\/p>\n

[B\u2019 \u222a (B\u2019 – A)]’ = [B’ \u222a (B’ \u2229 A\u2019)]\u2019
\n= (B\u2019)\u2019 \u2229 [B\u2019 \u2229 A’)’ [Demorgan\u2019s law]
\n= B \u2229 [(B’)’ \u222a (A’)’]
\n= B \u2229 (B \u222a A) = B
\n[\u2235 B \u2282 B \u222a A]<\/p>\n

Question 19.
\nFor any re sets A, B and C, (A – B) \u2229 (C – B) is equal to
\n(a) A – (B \u2229 C)
\n(b) (A – C) \u2229 B
\n(c) (A \u2229 C) – B
\n(d) (A – B) \u2229 C
\nAnswer:
\n(c) (A \u2229 C) – B<\/p>\n

(A – B) \u2229 (C – B) = (A \u2229 B\u2019) \u2229 (C \u2229 B\u2019)
\n= (B\u2019 \u2229 A) \u2229 (B\u2019 \u2229 C)
\n= B’ \u2229 (A \u2229 C)
\n= (A \u2229 C) \u2229 B’
\n= A \u2229 C – B<\/p>\n

\"ML<\/p>\n

Question 20.
\nLet A and B are two disjoint sets and \u03be be the universal set, then A\u2019 \u222a ((A \u222a B) \u2229 B\u2019) is equal to
\n(a) \u03a6
\n(b) \u03be
\n(c) A
\n(d) B
\nSolution:
\n(b) \u03be<\/p>\n

A’ \u222a ((A \u222a B) \u2229 B\u2019) = A\u2019 \u222a ((A \u2229 B\u2019) \u222a (B \u2229 B\u2019))
\n= A\u2019 \u222a ((A \u2229 B\u2019) \u222a \u03a6)
\n= A\u2019 \u222a (A \u2229 B\u2019)
\n= (A\u2019 \u222a A) \u2229 (A\u2019 \u222a B\u2019)
\n= U \u2229 (A\u2019 \u222a B\u2019)
\n= A\u2019 \u222a B\u2019
\n= (A \u2229 B)\u2019
\n[using Demorgan\u2019s law]
\n= \u03a6’ = U = \u03be
\n[\u2235 A and B disjoint sets = A \u2229 B = \u03a6]<\/p>\n

Question 21.
\nLet S = set of points inside the square, T = set of points inside the triangle and C = the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square. Then
\n(a) S \u2229 T \u2229 C = \u03a6
\n(b) S \u222a T \u222a C = C
\n(c) S \u222a T \u222a C = S
\n(d) S \u222a T = S \u2229 C
\nAnswer:
\n(c) S \u222a T \u222a C = S<\/p>\n

Therefore S \u222a T \u222a C = S<\/p>\n

\"ML<\/p>\n

Question 22.
\nLet R be the set of points inside a rectangle of sides a and b (a, b > 1) with two sides along the positive direction of x-axis and y-axis. Then
\n(a) R = {(x, y) : 0 \u2264 x \u2264 a, 0 \u2264 y \u2264 b}
\n(b) R = {(x, y) : 0 \u2264 x < a, 0 \u2264 y \u2264 b}
\n(c) R = {(x, y) : 0 \u2264 x \u2264 a, 0 < y < b}
\n(d) R = {(x, y) : 0 < x < a, 0 < y < b}
\nAnswer:
\n(d) R = {(x, y) : 0 < x < a, 0 < y < b}<\/p>\n

\u2234 R= {(x, y) ; 0 < x < a ; 0 < y < b}<\/p>\n

\"ML<\/p>\n

\"ML<\/p>\n

Question 23.
\nIn a class of 80 students, 39 students play football and 45 students play cricket and 15 students play both the games. Then the number of students who play neither is
\n(a) 11
\n(b) 14
\n(c) 16
\n(d) 18
\nAnswer:
\n(a) 11<\/p>\n

F : set of students who play football
\nC : set of students who play cricket
\nThen n (F) = 39 ;
\nn (C) = 45 ;
\nn (F \u2229 C) = 15
\nand n (\u03be) = 80
\n\u2234 n (F \u222a C) = n (F) + n (C) – n (F \u2229 C)
\n= 39 + 45 – 15 = 69
\nThus. No. of students who play neither = n (F\u2019 \u2229 C\u2019)
\n= n {(F \u222a C)\u2019}
\n= n (\u03be) – n (F \u222a C)
\n= 80 – 69 = 11<\/p>\n

Question 24.
\nIn a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read both. The number of persons who read neither is
\n(a) 210
\n(b) 290
\n(c) 180
\n(d) 260
\nAnswer:
\n(b) 290<\/p>\n

Let us define the following sets as under:
\nH : set of persons who read Hindi
\nE : set of persons who read English
\nThen n (H) = 450 ;
\nn (E) = 300 ;
\nn (H \u2229 E) = 200
\n\u2234 Required no. of persons who read neither = n (H\u2019 \u2229 E\u2019)
\n= n (H \u222a E)\u2019
\n= n (\u03be) – n (H \u222a E)
\n= 840 – {n (H) + n (E)} – n (H \u2229 E)}
\n= 840 – {450 + 300 – 200}
\n= 840 – 550 = 290<\/p>\n

Question 25.
\nIn a group of 70 people, 52 like soft drinks and 37 like tea and each person likes atleast one of the two drinks. Then the number of people who like both the drink
\n(a) 15
\n(b) 19
\n(c) 18
\n(d) 20
\nAnswer:
\n(b) 19<\/p>\n

Let S : set of peoples who like soft drinks
\nT : set of people, who like tea
\nThen n (S) = 52
\nand n (T) = 37
\nand n (S \u2229 T) = 70
\nwe know that,
\nn (S \u222a T) = n (S) + n (T) – n (S \u2229 T)
\n\u21d2 70 = 52 + 37 – n (S \u2229 T)
\n\u21d2 n (S \u2229 T) = 89 – 70 = 19.<\/p>\n

Question 26.
\nA T.V. survey gives the following data for TV. watching 59 % of the people watch program A, 67 % of the people watch program B and x % of the people watch both the program, then
\n(a) x = 26
\n(b) x = 59
\n(c) 26 \u2264 x \u2264 59
\n(d) x \u2265 59
\nAnswer:<\/p>\n

Let P: set of people watch program A
\nQ : set of people watch program B
\nThen n (P) = \\(\\frac{59}{100}\\) ;
\nn (Q) = \\(\\frac{47}{100}\\)
\nand n (P \u2229 Q) = \\(\\frac{x}{100}\\)
\n\u2234 n (P \u222a Q) = n (P) + n (Q) – n (P \u2229 Q)
\n= \\(\\frac{59}{100}+\\frac{67}{100}-\\frac{x}{100}\\)
\nA \u222a B \u2286 \u03be
\n\u21d2 n (A \u222a B) \u2264 n (\u03be)
\n\u21d2 \\(\\frac{59}{100}+\\frac{67}{100}-\\frac{x}{100}\\) \u2264 100 % = \\(\\frac{100}{100}\\) = 1
\n\u21d2 126 – x \u2264 100
\n\u21d2 26 \u2264 x ………………….(1)
\nSince A \u2229 B \u2286 A
\n\u21d2 n (A \u2229 B) \u2264 n (A) = \\(\\frac{59}{300}\\)
\nand A \u2229 B \u2286 B
\n\u21d2 n (A \u2229 B) \u2264 n (B) = \\(\\frac{67}{100}\\)
\n\u21d2 \\(\\frac{x}{100}\\) \u2264 min. \\(\\left\\{\\frac{59}{100}, \\frac{67}{100}\\right\\}\\)
\n\u21d2 \\(\\frac{x}{100} \\leq \\frac{59}{100}\\)
\n\u21d2 x \u2264 59 ……………….(2)
\nFrom (1) and (2) ; we have
\n26 \u2264 x \u2264 59<\/p>\n

\"ML<\/p>\n

Question 27.
\nIf A {(x, y) : y = \\(\\frac{1}{x}\\) \u2260 x \u2208 R},
\nB = {(x, y) : y = – x, x \u2208 R}, then
\n(a) A \u2229 B = A
\n(b) A \u2229 B = B
\n(c) A \u2229 B = \u03a6
\n(d) A \u222a B = A
\nAnswer:
\n(c) A \u2229 B = \u03a6<\/p>\n

Given A = {(x, y) ; y = \\(\\frac{1}{x}\\) ; x \u2260 0 \u2208 R}
\nand B = {(x, y) ; y – x ; x \u2208 R}
\n\u2234 A \u2229 B = {(x, y) ; y = \\(\\frac{1}{x}\\) ; y = – x ; x \u2260 0 \u2208 R}
\nSince y = \\(\\frac{1}{x}\\) = – x
\n\u21d2 1 + x2<\/sup> = 0 which does not gives real values of x.
\n\u21d2 A \u2229 B = \u03a6<\/p>\n

Question 28.
\nIf n (E) = 50, n (A) = 38, n (B) = 30, then the least value of n (A \u2229 B) is
\n(a) 30
\n(b) 38
\n(c) 50
\n(d) 18
\nAnswer:
\n(a) 30<\/p>\n

Given n (\u03be) = 50,
\nn (A) = 38,
\nn (B) = 30
\n\u2234 n (B) = 6
\nThus n (A) – n (B) = 6 – 6 = 0<\/p>\n","protected":false},"excerpt":{"rendered":"

Interactive ISC Mathematics Class 11 Solutions Chapter 1 Sets MCQs engage students in active learning and exploration. ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets MCQs Choose the correct answer from the given four options in questions (1 to 29) : Question 1. Which of the following collection of objects is not …<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170472"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=170472"}],"version-history":[{"count":5,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170472\/revisions"}],"predecessor-version":[{"id":170485,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170472\/revisions\/170485"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=170472"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=170472"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=170472"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}