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Mathematics Test - 18

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Mathematics Test - 18
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  • Question 1
    1 / -0

    The probability that A speaks truth is4/5, while this probability for B is3/4. The probability that they contradict each other when asked to speak on a fact is

    Solution
    In short the event described here is P(A) P(B) +P(A)P(B)
    where P(A) and P(B) are the probabilities of A and B speaking the truth
    respectively.
    P(E)=P(A)P(B)+P(A)P(B)=15×34+45×14
    P(E)=320+420=720=1440
  • Question 2
    1 / -0

    The minimum number of elements that must be added to the relation R= {(1,2)(2,3)} on the set of natural numbers so that it is an equivalence is

    Solution
    For R to be equivalence relation, it should be reflexive, symmetric and transitive Now, R is reflexive (x,x)R for all x{1,2,3}
    (1,1),(2,2),(3,3)R
    R is symmetric (1,2),(2,3)R
    (2,1),(3,2)R
    R is transitive
    (1,2),(2,3)R
    (1,3)R
    Also (3,1)R as R is symmetric. So, the total number of elements is 9
    Hence, minimum 7 elements must be added.
  • Question 3
    1 / -0

    If the circles x2 + y2 = 9 and x2 + y2 − 2αx + 2y + 1 = 0 touch each other internally, then α=

    Solution
    Circle 1:x2+y2=32
    radius is r1=3 and centre is (0,0) Circle 2:x2+y22αx+(α)2+21y+1(α)2+11=0
    (xα)2+(y+1)21(α)2+1=0
    (xα)2+(y+1)2=(α)2
    radius is r2=α and centre is (α,1)
    Distance between 2 centres is diff between their radii
    r2r1=C1C2
    C1C2=(α0)2+(10)2=α2+12
    α3=α2+12
    (α3)2=α2+12
    α2+96α=α2+1
    6α=8
    α=43
  • Question 4
    1 / -0
    The value of (0.2)log5(14+18+116+.10n) is
    Solution
    Let's calculate the sum of infinite series 14+18+116+
    Let's consider it to be S
    The series is a GP with first term (a)=14 and common ratio (r)=12
    S=a1r
    S=12
    To find (15)log312
    By Properties of log, we can write it as (12)2log515
    =(12)2log55
    =(12)2
    =22=4
    Hence, option A is correct.
  • Question 5
    1 / -0

    If one root of the equation ax2 + bx + c = 0 be n times the other root, then

    Solution
    Let the roots be α and nα
    Sum of roots, α+nα=baα=ba(n+1)(i)
    and product, α.n.α=caα2=cna. (ii)
    From (i) and (ii), we get
    [ba(n+1)]2=cnab2a2(n+1)2=cnanb2=ac(n+1)2
    Note : Students should remember this question as a fact.
  • Question 6
    1 / -0

    If n(A) + n(B) = m, then the number of possible bijections from A to B is

    Solution
    A bijection from A to B is a function which maps every element of A to uniqu element of B i.e. injective. n(B)n(A)
    Also, it ensures that every element of B is an image of some element of A n(A)n(B)
    n(A)=n(B)
    n(A)=m2=n(B)
    Let A={a1,a2,,am2} and
    B={b1,b2,,bm2}
    Let f:AB defined by f(ai)=bi is a bijection. Any and all images of some fixed ai appears in at least one such f. And each f unique for each permutation (b1,b2,bm2) Hence, the number of functions is exactly equal to the number of such permutations, which is (m2)!
  • Question 7
    1 / -0

    Solution

  • Question 8
    1 / -0

    If the7thterm of a harmonic progression is 8 and the8thterm is 7, then its15thterm is

    Solution
    Obviously, 7th term of corresponding A.P is 18 and 8th term will be 17a+6d=18 and a+7d=17
    Solving these, we get d=156 and a=156
    Therefore 15th term of this A.P.
    =156+14×156=1556
    Hence the required 15th term of the H.P. is 5615
  • Question 9
    1 / -0

    Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is

    Solution
    For a particular house being selected Probability =13 Prob(all the persons apply for the same house) =(13×13×13)×3=19 or 218
  • Question 10
    1 / -0

    In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then number of families which buy A only is

    Solution

    n(A) = 40% of 10,000 = 4,000

    n(B) = 20% of 10,000 = 2,000

    n(C) = 10% of 10,000 = 1,000

    n(A∩ B) = 5% of 10,000 = 500

    n(B∩ C) = 3% of 10,000 = 300

    n(C∩ A) = 4% of 10,000 = 400

    n(A∩ B∩ C) = 2% of 10,000 = 200

    We want to find the number of families which buy only A= n(A) - [n(A ∩ B) + n(A ∩ C) - n(A ∩ B∩ C)]

    =4000 - [500 + 400 - 200] = 4000 - 700 = 3300

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