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The Cartesian product X × Y of sets X , Y is the set of all ordered pairs ( x , y ).
Two ordered pairs ( x1 , y1 ), ( x2 , y2 ) .
Given that P = { 1, 2, 3 } and Q = { 4 , 5 } then,
P x Q = { 1, 2, 3 } x { 4 , 5 }
? P x Q = { 1 } x { 4 , 5 } , { 2 } x { 4 , 5 } , { 3 } x { 4 , 5 }
? P x Q = { 1 , 4 } , { 1 , 5 } , { 2 , 4 } , { 2 , 5 } , { 3 , 4 }, { 3 , 5 }
? P x Q = { (1 , 4 ) , ( 1 , 5 ) , ( 2 , 4 ) , ( 2 , 5 ) , ( 3 , 4 ), ( 3 , 5) }
Here, P(1 + 10/100)t > 2P
? (11/10)t > 2
When t = 8 ? (11/10)8 = 2.14358
t =7 ? (11/10)7 = 1.9487
By trial, [11 x 11 x 11 x 11 x 11 x 11 x 11 x 11] / [10 x 10 x 10 x 10 x 10 x 10 x 10 x 10] >2
Hence, the first years in which sum of money will become more than double in amount is 8th year
Given, P = ? 1750, R = 8%,
n = 2 and a/b = 1/2
According to the formula,
Amount = P(1 + R/100)n x [1 + {(a/b) x R}/100]
=1750 (1 + 8/100 )2 [1 + {(1/2) x 8} /100]
=1750 (27/25)2 x 26/25
= 1750 x (27/25) x (27/25) x (26/25)
=? 2122.848
= ? 2122.85
Let the population at the beginning of the first year be N .
Then, according to the question,
N x [1+ 5/100] x [1 - 5/100] = 47880
? N x (105/100) x (95/100) = 47880
? N = 47880 x (100/105) x (100/95) = 48000
CI = 8000 x (1 + 10/100)2 - 8000
= 8000 x (11/10) x (11/10) - 8000
= ? 9680 - 8000
= ? 1680
? Sum = (840 x 100)/(3 x 8) = ? 3500
[ ? SI is half of CI ]
? SI = 1680/2 = ? 840
A ? B = {a, b, c} ? {c, d, e, f}
A ? B = { c }
A ? C = { a, b, c } ? { c, d, e }
A ? C = { c }
? (A ? B) ? (A ? C) = { c }.
U ? A = {a, b, c, d, e, f} ? {a, b, c} = {a, b, c, d, e, f} = U
(U ? A)? = ?.
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Let, A = {0}. The possible subsets of this set A are ? and {0}, so the power set of the given set A is P(A) = {f, {0}}.
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