Let the number of students be 100. Number of students who failed in Hindi is 30%.
n(H) = 30
Number of students who failed in English is 45%.
? n(E) = 45
Number of students who failed in both the subjects is 20%. n(H ? E) = 20
Applying the rule,
n(H ? E) = n(H) + n(E) - n(H ? E)
= 30 + 45 - 20 = 55
Percentage of students who failed in Hindi or English or both the subjects = 55%
Number of students who passed in both the subjects = 100 - 55 = 45%
Given in the question,
20% of 80 = 20 * 80 / 100 = 16
Remaining 50%
= (80 ? 16) × 50 / 100 = 32
No. of families not owning any vehicle
= 80 ? (32 + 16) = 80 ? 48 = 32
Draw a figure and calculate.
Number of people speak English only
= 50 ? (10 + 25)
= 15
(A × B) = {1, 2, 3} × {2, 3, 4}
= {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4)}
(C × D) = {1, 3, 4} × {2, 4, 5}
= {(1, 2), (1, 4), (1, 5), (3, 2), (3, 4), (3, 5), (4, 2), (4, 4), (4, 5)}
? (A ? B) ? (C × D)
= {(1, 2), (1, 4), (3, 2), (3, 4)}
Also, (A ? C) = {1, 3} and (B ? D) = {2, 4}. Therefore,
(A ? C) × (B ? D) = {(1, 2), (1, 4), (3, 2), (3, 4)}
Hence, (A × B) ? (C × D) = (A ? C) × (B ? D).
we can calculate below as per given question,
we can calculate below as per given question,
(B ? C) = { b, c, e } ? { b, c, f } = { e }
A × (B ? C) = { a, d } x { e }
(A × B) = { a, d } x { b, c, e }
(A × C) = { a, d } x { b, c, f }
? A × (B ? C) = {(a, e), (d, e)} ...........?(1)
Also, (A × B) ? (A × C) = {(a, e), (d, e)} ..........?(2)
Hence, from (1) and (2), we have
A × (B ? C) = (A × B) ? (A × C).
Given in the question, we can can calculate,
(B ? C) = {b, c, e} ? {b, c, f} = {b, c}
? A × (B ? C) = {a, d} × {b, c}
= {(a, b), (a, c), (d, b), (d, c)}
Also, (A × B) ? (A × C)
= {(a, b), (a, c), (a, b), (d, c)}
? A × (B ? C) = (A × B) ? (A × C).
A new set having those elements which are in A but not in B is said to be the difference of sets A and B and it is denoted by A ? B.
Given that in question, A = {1, 2, 3, 4, 5} and B = {2, 4, 6}
So the numbers which are in A and but not in B = {1, 3, 5}
? A ? B = {1, 3, 5}
NA
(a) 1? A but 1? B and 6 ? B but 6 ? A.
? A and B are not comparable.
(b) A = {x: x ? N and x ? 10} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
Clearly, A ? B ? A and B are comparable.
(c) {4, 5} ? A but {4, 5} ? B and 4 ? B but 4 ? A.
? A and B are not comparable.
Given: n(C) = 50, n(F) = 20, n(C ? F) = 10.
Number of students playing at least one of these two games = n (C ? F) = n (C) + n (F) ? n (C ? F)
= 50 + 20 ? 10 = 60.
NA
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