Quant Boosters - Gaurav Sharma - Set 3



  • Q26) Find the missing term: 4, 6, 12, 18, 30, 42, 60, 72, 102, 108, ?



  • 4 ( 3 and 5 are prime )
    6 ( 5 and 7 are prime )
    12 ( 11 and 13 are prime )
    18 ( 17 and 19 are prime )
    30 ( 29 and 31 are prime )
    42 ( 41 and 43 are prime )

    108 ( 107 and 109 are prime )
    138 ( 137 and 139 are prime )



  • Q27) If a, b, c, d, e, f are non negative real numbers such that a + b + c + d + e + f = 1, then the maximum value of ab + bc + cd + de + ef is
    a) 1/6
    b) 1
    c) 6
    d) 1/4



  • If we take a = b = 12 and c = d = e = f = 0 then the given equation holds so the maximum value is 1/4
    Option b and c do not hold as they are integers
    option a is also rejected as 1/6 is less than 1/4
    Hence option d is correct.



  • Q28) The number of integral values of x satisfying root(-x^2 + 10x – 16) < x – 2 is
    a) 0
    b) 1
    c) 2
    d) 3



  • root(-x^2 + 10x – 16) < x – 2
    -x^2 + 10x – 16 ≥ 0
    x^2 - 10x + 16 ≤ 0
    2 ≤ x ≤ 8 … (1)
    Also, -x^2 + 10x – 16 < x^2 – 4x + 4
    2x^2 – 14x + 20 > 0
    x^2 – 7x + 10 > 0
    x > 5 or x < 2 … (2)
    From (1) and (2), 5 < x ≤ 8
    x = 6, 7, 8



  • Q29) Find the sum of (1 x 2 x 3)/5^1 + (2 x 3 x 5)/5^2 + (3 x 4 x 5)/5^3 + (4 x 5 x 6)/5^4 + …
    a) 375/128
    b) 125/64
    c) 625/128
    d) 875/64



  • S = (1 x 2 x 3)/5^1 + (2 x 3 x 5)/5^2 + (3 x 4 x 5)/5^3 + (4 x 5 x 6)/5^4 + …
    S = 6/5 + 24/5^2 + 60/5^3 + 120/5^4 + …
    S/5 = 6/5^2 + 24/5^3 + 60/5^4 + 120/5^5 + …
    S – S/5 = 4S/5 = 6/5 + 18/5^2 + 36/5^3 + 60/5^4 + 90/5^5 + …
    4S/25 = 6/5^2 + 18/5^3 + 36/5^4 + 60/5^5 + …
    4S/5 – 4S/25 = 16S/25 = 6/5 + 12/5^2 + 18/5^3 + 24/5^4 + 30/5^5 + …
    16S/125 = 6/5^2 + 12/5^3 + 18/5^4 + 24/5^5 + 30/5^6 + …
    64S/125 = 6/5 + 6/5^2 + 6/5^3 + …
    S = (6/5 / ( 1 – 1/5) ) X 125/64 = 375/128



  • Q30) Using the vertices of a regular hexagon as vertices, how many triangles can be formed with distinct areas?
    a) 24
    b) 20
    c) 18
    d) 28



  • Let us number the vertices from 1 – 6, there are three different types of triangles that can be formed.
    Type 1 : (1, 2, 3) [ this is same as (2, 3, 4), (3, 4, 5) … etc)
    Type 2 : (1, 2, 4) [ this is same as (1, 2, 5) (2, 3, 5) (3, 4, 6)]
    Type 3 : (1, 3, 5) and (2, 4, 6) have same shape and area

    So there will be three triangles of different areas that can be formed from a regular hexagon.
    There are 6 triangles of Type 1, 12 of type 2 and 2 of type 3.
    Adding these we get 6 + 12 + = 20 triangles.

    Total number of triangles possible = 6C3 = 20


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