Quant Marathon by Gaurav Sharma - Set 5


  • Director, Genius Tutorials, Karnal ( Haryana ) & Delhi | MSc (Mathematics)


    Solve the inequality x^3 – 8x^2 + 21x – 18 > 0

    1. (3,infinity)
    2. (- infinity, 2) U (3, infinity)
    3. (2,3) U (3,infinity)
    4. (- infinity, 2)

    x^3 – 8x^2 + 21x – 18 > 0

    x^3 – 8x^2 + 21x – 18 = (x – 2) (x – 3) (x – 3)

    (x – 2) (x – 3) (x – 3) > 0

    Since (x – 3) is repeated, the sign on the number line won’t change when we will go from 3 to 2

    So, our answer will be (2,3) U ( 3, infinity)

    A sequence is defined as 2 = an – a(n-1) = a(n+1) – an.  Sn is the sum up to n terms of the sequence, a4 = 7. How many values of x, y exist such that Sx – Sy = 33

    1. 2
    2. 3
    3. 4
    4. More than 4

    Putting n = 4, we get 2 = a4 – a3 = > a3 = 7 – 2 = 5

    Similarly, a1 = 1, a2 = 3, a3 = 5, a4 = 7 etc

    S1 = 1, S2 = ( 1 + 3 ) = 2^2, S3 = ( 1 + 3 + 5) = 3^2

    So Sx = x^2 and Sy = y^2

    x^2 – y^2 = 33

    Number of ways in which 33 can be written as difference between two perfect squares

    = number of factors of 33/2 = 4/2 = 2

    In an art gallery, paintings are put on two walls facing each other and they are numbered 1, 2, 3 etc. The painting opposite to painting 6 is 19. If there are an even number of paintings, how many paintings are there in total?

    1. 22
    2. 24
    3. 26
    4. Can’t be determined

    Let say there are n paintings ( n even )

    Opposite painting 2 is (n-1)th painting

    So sum of painting = n + 1

    6 + 19 = n + 1 = > n = 24

    But this is the answer if it is given that the order or placing the painting number is reverse in one row compared to other. Because this information is not given and we assume that order is increasing in both the rows then the total number of paintings comes out to be 26. Hence option d is correct

    Consider (AB )^2 = CDA, where A, B, C and D represent distinct digits. Then D equals

    1. 1
    2. 6
    3. 3
    4. 9

    Unit’s digit of squares of numbers can only be 1,4,5,6 and 9

    A can be 1 or 4 or 5 or 6

    Now AB will be a three digit number only when A is 1

    Hence B can be 1 or 9

    But if B = 1 = > AB = 11 = > (AB )^2 = 121, where C and A are not distinct.

    A = 1 and B = 9

    19^2 = 361, D = 6

    a = 1^2, b = 2^3, c = 3^4  ... z = 26^27. How many zeroes will the product of all the alphabets end in

    1. 100
    2. 104
    3. 80
    4. 106

    Number of zeroes in this product will be equal to the number of 5s

    = 5^6 x 10^11 x 15^16 x 20^21 x 25^26

    = 5^(6 + 11 + 16 + 21 + 52)

    = 5^106

    Number of zeroes will be 106

    The graph of y = f(x) is symmetrical about both x = 2 and x = 4 then f (4 + x) =

    1. f (x)
    2. f (2 – x)
    3. f (x – 2)
    4. f (x – 3)

    y = f(x) is symmetrical about both x = 2 and x = 4

    Means, f (2 + x) = f (2 –x) and

    f ( 4 + x ) = f ( 4 – x) = f [ 2 + ( 2 – x) ]

    = f [2 – (2 – x)]

    = f(x)

    So, f (4 + x) = f(x)

    If the coefficients a, b, c of quadratic equation ax^2 + bx + c = 0 are chosen at random with replacement from the set S = { 1, 2, 3, 4, 5, 6}. Find out the probability that the roots of the quadratic equation are real and distinct.

    For roots of ax^2 + bx + c = 0 to be real and distinct, b^2 – 4ac > 0

    Total ways = 6 x 6 x 6 = 216

    Favorable cases = 38

    Required probability = 38/216 = 19/108

    If the equations ax^2 + bx + c = 0 and x^2 + 2x + 3 = 0, have a common root then find a : b : c

    1. 1 : 2 : 3
    2. 2 : 1 : 3
    3. 3 : 2 : 4
    4. 1 : 4 : 5

    Since roots of x^2 + 2x + 3 = 0 are imaginary, both roots will be common.

    Hence both equations are identical.

    a : b : c = 1 : 2 : 3

    1. 631
    2. 730
    3. 731
    4. 733

    S = (3/2)^2 + (5/2)^2 + (7/2)^2 + … + (25/2)^2

    = (3^2 + 5^2 + 7^2 + … + 25^2)/4 = [ ( 1^2 + 3^2 + … + 25^2 ) – 1) / 4

    = [ n (2n – 1) ( 2n + 1) ] / 3 = [ ( 13 x 25 x 27 ) / 3 – 1] / 4 = 2924/4 = 731

    Find the nth term of the series 2 + 5 + 12 + 31 + 86 + …

    Given series : 2, 5, 12, 31, 86 …

    First consecutive difference : 3, 7, 19, 55

    Second consecutive difference are in GP with common ration 3 : 4, 12, 36 …

    Tn = a +  bn + c x 3^(n – 1)

    T1 = a + b + c = 2

    T2 = a + 2b + 3c = 5

    T3 = a + 3b + 9c = 12

    We get, a = 0, b = 1 and c = 1

    Tn = n + 3^(n – 1)

     


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