{"id":170486,"date":"2024-05-24T15:34:42","date_gmt":"2024-05-24T10:04:42","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=170486"},"modified":"2024-05-24T15:35:21","modified_gmt":"2024-05-24T10:05:21","slug":"ml-aggarwal-class-11-maths-solutions-section-a-chapter-1-chapter-test","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-11-maths-solutions-section-a-chapter-1-chapter-test\/","title":{"rendered":"ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Chapter Test"},"content":{"rendered":"

Students appreciate clear and concise ISC Maths Class 11 Solutions<\/a> Chapter 1 Sets Chapter Test that guide them through exercises.<\/p>\n

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

Question 1.
\nDecide, among the following sets, which are subsets of which.
\nA = {x : x is a solution of x2<\/sup> – 8x + 12 = 0},
\nB = {2 ,4, 6},
\nC = {x : x is an even natural number),
\nD = {6).
\nSolution:
\nGiven A = {x : x is a solution of x2<\/sup> – 8x + 12 = 0}
\nNow, x2<\/sup> – 8x + 12 = 0
\n\u21d2 (x – 2) (x – 6) = 0
\n\u21d2 (x – 2) (x – 6) = 0
\n\u21d2 x = 2, 6
\nand B = {2, 4, 6} and A = {2, 6}
\nHence every member of set A be a member of set B
\n\u2234 A be a subset of B.
\nThus A \u2282 B.
\nSince C = {x : x is an even natural number}
\n= {2, 4, 6, 8, ………………….}
\nand D = {6}<\/p>\n

Question 2.
\nLet \u03be = the set of all letters in the word \u2018TAMILNADU\u2019 and X = {x : x is a vowel and x \u2208 \u03be}.
\n(i) Write \u03be and X in the roster form.
\n(ii) Tell n (\u03be) and n(X).
\n(Ill) List all the proper subsets of X.
\n(iv) What is the cardinal number of the power set of X?
\nSolution:
\n(i) In roster form,
\n\u03be = {T, A, M, I, L, N, D, U}
\nand X = {x : x is a vowel and x \u2208 \u03be}
\n= {A, I, U}<\/p>\n

(ii) n (\u03be) = no. of distinct elements in \u03be = 8
\nand n (X) = Number of distinct elements in
\nX = 3<\/p>\n

(iii) The proper subsets of X are given below:
\n\u03a6, {A), {I}, {U}, {A, I}, {A, U}, {I, U)<\/p>\n

(iv) Power set of X = P(X)
\n= {\u03a6, {A}, {I}, {U}, {A, I}, {A, U}, {I, U}, {(A. I, U}}
\n\u2234 n (P (X)) = 23<\/sup> = 8.<\/p>\n

\"ML<\/p>\n

Question 3.
\nLet A be the set of letters in the word “POOR\u201d. Write the power set of A.
\nSolution:
\nA = {P, O, R}
\nP(A) = {\u03a6, {P}, {O}, {R, {P, O}, {O, R}, {P, R}, {P, O, R}}
\n\u2234 Cardinal number of power set of A = n (P (A)) = 8<\/p>\n

Question 4.
\nFind the power set of the following sets:
\n(i) {- 1, 0, 1}
\n(ii) {0, 1, {0, 1}}
\nSolution:
\n(i) Let A = {- 1, 0, 1}
\n\u2234 P (A) = {\u03a6, (- 1), {0}, {1}, {- 1, 0}, {0, 1}, {- 1, 1}, {- 1,0, 1}}<\/p>\n

(ii) {0, 1, {0, 1}
\nLet A = {0, 1, {0, 1}}
\n\u2234 P(A) = {\u03a6, {0}, {1}, {{0, 1}}, {0, 1}, {0, {0, 1}}, {1, {0, 1}}, {0, 1, {0, 1}}}<\/p>\n

Question 5.
\nIf A = {2, 3, 5, 7, 8}, B = {1, 5, 9} and A\u2019 = {1, 4, 6, 9}, verify that
\n(i) (A \u222a B)’ = A’ \u2229 B’
\n(ii) B – A = A’ \u2229 B
\nSolution:
\nGiven A = {2, 3, 5, 7, 8) ;
\nB = {1, 5, 9} and
\nA\u2019 = {1, 4, 6, 9}
\nSince A \u222a A\u2019 = \u03be
\n\u21d2 \u03be = {1, 2, 3, 4, 5, 6, 7, 8, 9}<\/p>\n

(i) A \u222a B = {1, 2, 3, 5, 7, 8, 9}
\n\u2234 (A \u2229 B)’ = \u03be – A \u222a B
\n= {4, 6} ………………(1)
\nB’ = \u03be – B = {2, 3, 4, 6, 7, 8}
\n\u2234 A\u2019 \u2229 B\u2019 = {1, 4, 6, 9} \u2229 {2, 3, 4, 6, 7, 8}
\n= {4, 6} …………………(2)
\n\u2234 from (1) and (2) ; we have
\n(A \u222a B)\u2019 = A\u2019 \u2229 B\u2019<\/p>\n

(ii) L.H.S. = B – A = {1, 9}
\nR.H.S. = A\u2019 \u2229B
\n= {1, 4, 6, 9} \u2229 {1, 5, 9}
\n= {1, 9}
\n\u2234 B – A = A\u2019 \u2229 B<\/p>\n

\"ML<\/p>\n

Question 6.
\nFor all sels A, B and C, is A – (B – C) = (A – B) – C true? Justify your answer.
\nSolution:
\n(i) Let A = {1, 2},
\nB = {2, 3}
\nand C = {1, 3}
\n\u2234 B – C = {2}
\nThus A – (B – C) = {1, 2} – {2} = {1} …………………(1)
\nNow, A – B = {1}
\n\u2234 (A – B) – C = {1} – {1, 3} = \u03a6 ………………………(2)
\n\u2234 from (1) and (2) ; we have
\n\u2234 given statement is false.<\/p>\n

Question 7.
\nIf n (\u03be) = 30, n(A\u2019) = 15, n (B) = 5 and n (A \u2229 B) = 3, find
\n(i) n (A)
\n(ii) n (A \u222a B)
\n(iii) n(A – B).
\nSolution:
\nGiven n (\u03be) = 30,
\nn (A\u2019) = 15,
\nn (B) = 5
\nand n (A \u2229 B) = 3<\/p>\n

(i) n (A) = n(\u03be) – n(A\u2019)
\n= 30 – 15 = 15<\/p>\n

(ii) n (A \u222a B) = n (A) + n (B) – n (A \u2229 B)
\n= 15 + 5 – 3 = 17<\/p>\n

(iii) n (A – B) = n (A) – n (A \u2229 B)
\n= 15 – 3 = 12<\/p>\n

Question 8.
\nIf n (\u03be) = 40, n ((A \u222a B)\u2019)= 12, n (A – B) = 10 and n (B – A) = 14, find
\n(i) n (A)
\n(ii) n (B)
\n(iii) n (A \u2229 B).
\nSolution:
\nGiven n (A) = 40,
\nn ((A \u222a B)\u2019) = 12,
\nn (A – B) = 10
\nand n (B – A) = 14
\nNow n ((A \u222a B)\u2019) = n (\u03be) – n (A \u222a B)
\n\u21d2 12 = 40 – n (A \u222a B)
\n\u21d2 n (A \u222a B) = 40 – 12 = 28
\nwe know that,<\/p>\n

(i) Since n (A – B) = 10
\n\u21d2 n (A) – n (A \u2229 B) = 10
\n\u21d2 n (A) – 4 = 10
\n\u21d2 n(A) = 14<\/p>\n

(ii) Since n (B – A) = 14
\n\u21d2 n (B) – n (A \u2229 B) = 14
\n\u21d2 n(B) = 14 + 4 = 18<\/p>\n

(iii) n (A \u222a B) = n (A – B) + n (A \u2229 B) + n (B – A)
\n\u21d2 28 = 10 + n (A \u2229 B) + 14
\n\u21d2 n (A \u2229 B) = 28 – 24 = 4<\/p>\n

\"ML<\/p>\n

\"ML<\/p>\n

Question 9.
\nTwo sets A and B are such that n (A \u222a B) = 18, n (A\u2019 \u2229 B) = 3 and n (A \u2229 B\u2019) = 5, find n (A \u2229 B).
\nSolution:
\nGiven n (A \u222a B) = 18,
\nn (A’ \u2229 B) = 3
\nand n (A \u2229 B\u2019) = 5<\/p>\n

\"ML<\/p>\n

we know that,
\nn (A\u2019 \u2229 B) = n (B – A)
\nand n (A \u2229 B\u2019) = n (A – B)
\nwe know that,
\nn (A \u222a B) = n (A – B) + n (A \u2229 B) + n (B – A)
\n\u21d2 18 = 5 + n (A \u2229 B) + 3
\n\u21d2 n (A \u2229 B) = 18 – 8 = 10<\/p>\n

Question 10.
\nTwo sets A and B are such that n (A \u222a B) = 21, n (A\u2019 \u2229 B\u2019) = 9 and n (A \u2229 B) = 7, find n ((A \u2229 B)\u2019).
\nSolution:
\nGiven n (A \u222a B) = 21 ;
\nn (A\u2019 \u2229 B\u2019) = 9
\nand n (A \u2229 B) = 7
\nNow n (A\u2019 \u2229 B\u2019) = n ((A \u222a B) = 9
\n[using Demorgan\u2019s law]
\nHere, \u03be = (A \u222a B) \u222a (A \u222a B)\u2019
\n\u21d2 n (\u03be) = n (A \u222a B) + n ((A \u222a B)\u2019)
\n\u21d2 n (\u03be) = 21 + 9 = 30
\n\u2234 n (A \u2229 B)\u2019 = n (\u03be) – n (A \u2229 B)
\n= 30 – 7 = 23<\/p>\n

\"ML<\/p>\n

Question 11.
\nIf n (\u03be) = 50, n (A) = 3x, n (B) = 2x and n (A \u2229 B) = x = n ((A \u222a B)\u2019), find
\n(i) the value of x
\n(ii) n (A – B).
\nSolution:
\nGiven n (\u03be) = 50,
\nn (A) = 3x,
\nn (B) = 2x,
\nn (A \u2229 B) = x = n ((A \u222a B)’)<\/p>\n

(i) Since x = n ((A \u222a B)\u2019)
\n= n (\u03be) – n (A \u222a B)
\n\u21d2 x = 50 – [n (A) + n (B) – n (A \u2229 B)]
\n\u21d2 x = 50 – [3x + 2x – x]
\n\u21d2 x = 50 – 4x
\n\u21d2 5x = 50
\n\u21d2 x = 10<\/p>\n

(ii) n (A – B) = n(A) – n (A \u2229 B)
\n= 3x – x
\n= 2x
\n= 2 \u00d7 10 = 20<\/p>\n

Question 12.
\nIf n (\u03be) = 15, A and B are two sets such that A \u2282 B, n (A) = 8 and n (B) = 12, use Venn diagram to find the following:
\n(i) n (A\u2019)
\n(ii) n (B\u2019)
\n(iii) n (A \u2229 B\u2019)
\n(iv) n (A\u2019 \u2229 B).
\nSolution:
\n(i) Since A \u2282 B,
\nn (A) = 8 ;
\nn (B) = 12
\nand n (\u03be) = 15<\/p>\n

\"ML<\/p>\n

(i) n (A\u2019) = 3 + 47
\n[or n (A\u2019) = n (\u03be) – n (A)
\n= 15 – 8 = 7]<\/p>\n

(ii) n (B\u2019) = 15 – 12 = 3
\n[or n (B\u2019) = n (\u03be) – n(B)
\n= 15 – 12 = 3]<\/p>\n

(iii) n (A \u2229 B)\u2019 = 0
\n[or n (A \u2229 B\u2019) = n (A) – n(A \u2229 B)
\n= 8 – 8 =0]<\/p>\n

(iv) n (A\u2019 \u2229 B) = 4
\n[or n (A\u2019 \u2229 B) = n (B) – n (A \u2229B)
\n= 12 – 8 = 4]<\/p>\n

\"ML<\/p>\n

Question 13.
\nIn an examination, 56 per cent of the candidates failed in English and 48 percent failed in Science. If 18 per cent failed in both English and Science, flnd the percentage of those who passed in both the subjects.
\nSolution:
\nLet \u03be be the set of all candidates.
\nLet E and S be the set of candidates who passed in English and Science respectively.
\nThen n (\u03be) = 100 % ;
\nn (E) = (100 – 56) % = 44 % ;
\nn (S) = (100 – 48) % = 52 %
\nAlso n (E\u2019 \u2229 S\u2019) = 18 %
\n\u21d2 n ((E \u222a S))’ = 18 %
\n\u21d2 n (\u03be) – n (E \u222a S) = 18 %
\n\u21d2 100 % – {n (E) + n (S) – n(E \u2229 S)} = 18 %
\n\u21d2 100% – {44 % + 52 % – n(E \u2229 S)} = 18 %
\n\u21d2 100 – 96 % + n (E \u2229 S) = 18%
\n\u21d2 n (E \u2229 S) = 14%
\nHence, required % of students who passed in both the subjects = 14%<\/p>\n

Question 14.
\nFrom amongst the 6000 literate individuals of a city, 50 % read newspaper A, 45 % read newspaper B and 25 % read neither A nor B. How many individuals read both the newspapers A as well as B?
\nSolution:
\nLet be the total no. of individuals.
\nLetA and B be the set of aIL individuals who read newspaper A and newspaper B.
\n\u2234 n (A) = 50 % ;
\nn (B) = 45 %
\nn (A\u2019 \u2229 B\u2019) = 25 %
\nn ((A \u222a B)\u2019) = 25 %
\n\u21d2 100 % – n (A \u2229 B) = 25 %
\n\u21d2 100 % – [n (A) + n (B) – n (A \u2229 B)] = 25 %
\n\u21d2 100 % – [50 % + 45 % – n (A \u2229 B)] = 25 %
\n\u21d2 5 % + n (A \u2229 B) = 25 %
\n\u21d2 n (A \u2229 B) = 20 %
\nHence total number of individuals who read newspaper A as well as B = 20% of 6000
\n= (\\(\\frac{20}{100}\\) \u00d7 6000) = 1200<\/p>\n

\"ML<\/p>\n

Question 15.
\nIn a beauty contest, half the number of judges voted for Miss A, \\(\\frac{2}{3}\\) of them voted for Miss B, 10 voted for both and 6 did not vote for either Miss A or Miss B. Find how many judges, In all, were present there.
\nSolution:
\nLet the total number of judges in all were present in a beauty contest = x
\nThe no. ofjudges who voted for miss A = n (A) = \\(\\frac{x}{2}\\)
\nand no. ofjudges who noted for miss B = n (B) = \\(\\frac{2}{3}\\) x
\nn (A \u2229 B) = 10
\nand 6 = n ((A \u222a B))\u2019
\n\u21d2 6 = x – n (A \u222a B)
\n\u21d2 6 = x – [n (A) + n (B) – n (A \u2229 B)]
\n\u21d2 6 = x – \\(\\left[\\frac{x}{2}+\\frac{2}{3} x-10\\right]\\)
\n\u21d2 – 4 = \\(\\frac{6 x-3 x-4 x}{6}\\)
\n\u21d2 – 4 = – \\(\\frac{x}{6}\\)
\n\u21d2 x = 4
\nHence the required number of judges were present in beauty contest be 24.<\/p>\n

Question 16.
\nIn a group of 50 students, the number of students studying French, English and Sanskrit were found to be as follows :
\nFrench = 17, English = 13, Sanskrit = 15 ; French and English = 9, English and Sanskrit = 4, French and Sanskrit = 5 ; English, French aiid Sanskrit 3.
\nFind the number of students who study :
\n(i) French only
\n(ii) French and Sanskrit but not English
\n(iii) English only
\n(iv) French and English but not Sanskrit
\n(v) Sanskrit only
\n(vi) English and Sanskrit but not French
\n(vii) atleast one of the three languages
\n(viii) none of the three languages.
\nSolution:
\nLet \u03be be the no. of students in a school.
\nLet F, E and S be the set of students studying French, English and Sanskrit.<\/p>\n

\"ML<\/p>\n

Then the no. of students according to given information in the question are shown in different regions in adjoining venn diagram.<\/p>\n

(i) Reqd. no. of students who study French only = 6<\/p>\n

(ii) Reqd. no. of students who study french and Sanskrit but not English = 2<\/p>\n

(iii) Reqd. no. of students who study English only = 3<\/p>\n

(iv) Reqd. no. of students who study French and English but not Sanskrit = 6<\/p>\n

(v) Reqd. no. of students who study Sanskrit only = 9<\/p>\n

(vi) Reqd. no. of students who study English and Sanskrit but not french = 1<\/p>\n

(vii) Reqd. no. of students who study atleast one of the three languages = 6 + 6 + 3 + 2 + 3 + 1 + 9 = 30<\/p>\n

(viii) Reqd. number of students who study none of the three language = n (\u03be) – no. of students who study atleast one of the three languages
\n= 50 – 30 = 20.<\/p>\n

\"ML<\/p>\n

Question 17.
\nIf A and B are two sets such that n (A) = 10 and n (B) = 7, then find:
\n(i) the least value of n (A \u2229 B)
\n(ii) the greatest value of n (A \u2229 B)
\n(ii,) the greatest value of n (A \u222a B)
\n(iv) the least value of n (A \u222a B).
\nSolution:
\nGiven n (A) = 10 ;
\nn(B) = 7
\nwe know that,
\nn (A \u222a B) = n (A) + n (B) – n (A \u2229 B)
\n\u21d2 n (A \u222a B) = 10 + 7 – n (A \u2229 B)
\n\u21d2 n (A \u222a B) \u2264 17 ………………..(1)
\n[\u2235 n(A \u2229 B) \u2265 0
\n\u2234 – n(A \u2229 B) \u2264 0]
\nNow, equality sign helds in eqn. (1) only when (A \u2229 B) = 0
\ni.e. A \u2229 B = \u03a6
\nNow, A \u2282 A \u222a B and B \u2282 A \u222aB
\n\u21d2 n (A) \u2264 n (A \u222a B) and n(B) \u2264 n (A\u222aB)
\n\u21d2 n (A \u222a B) \u2265 max {n(A), n(B)}
\n\u21d2 n (A \u222a B) \u2265 max {10, 7} = 10
\n\u21d2 10 \u2264 n (A \u222a B) …………………..(2)
\nFrom (1) and (2) ; we have
\n10 \u2264 n (A \u222a B) \u2264 17
\n\u2234 least value of n (A \u222a B) = 10
\ngreatest value of n (A \u222a B) = 17
\nAlso A \u2229 B \u2282 A and A \u2229 B \u2282 B
\n\u21d2 n (A \u2229 B) \u2264 n(A) and n (A \u2229 B) \u2264 n(B)
\n\u21d2 n (A \u2229 B) \u2264 min. (n (A), n(B)}
\n\u21d2 n (A \u2229 B) \u2264 min {10, 7} = 7
\nIt may be noted that equality sign hold when A \u2282 B or B \u2282 A.
\nThus, greatest value of n (A \u2229 B) = 7
\nalso n (A \u2229 B) \u2265 0
\n\u2234 least value of n (A \u2229 B) = 0.
\n[when A \u2229 B = \u03a6].<\/p>\n","protected":false},"excerpt":{"rendered":"

Students appreciate clear and concise ISC Maths Class 11 Solutions Chapter 1 Sets Chapter Test that guide them through exercises. ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Chapter Test Question 1. Decide, among the following sets, which are subsets of which. A = {x : x is a solution of x2 …<\/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\/170486"}],"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=170486"}],"version-history":[{"count":1,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170486\/revisions"}],"predecessor-version":[{"id":170491,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170486\/revisions\/170491"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=170486"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=170486"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=170486"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}