Quant Boosters - Kamal Lohia - Set 4



  • Total milk to be filled is 4 cups and total capacity available is 2 + 4 = 6 cups i.e. in all we can fill 4/6 = 2/3 of both vessels. Also as each vessel should be filled in equal proportion, that means both the vessel contains 2/3 of their capacity. So 4 cup vessel contains (2/3) * 4 = 8/3 cups of milk.



  • Q28) Find the sum of first n terms of the series 1.2.3 + 2.3.4 + 3.4.5 + ..... ?



  • See the pattern:
    1×2×3 = (1/4) × [1×2×3×4 - 0×1×2×3]
    2×3×4 = (1/4) × [2×3×4×5 -1×2×3×4]
    ...
    n(n + 1)(n + 2) = (1/4) × [n(n + 1)(n + 2)(n + 3) - (n - 1)n(n + 1)(n + 2)]
    => The required sum is = (1/4) × [n(n + 1)(n + 2)(n + 3)] = [(n² + 3n + 1)² - 1]/4



  • Q29) What is the remainder when 128 ^ 1000 is divided by 153.



  • This is simply application of Chinese Remainder Theorem.

    For two divisors D1 and D2 such that HCF(D1 , D2) = 1, first find integers x, y so that xD1 + yD2 = 1
    Next let N ≡ r1 mod D1
    and also N ≡ r2 mod D2
    Then according to Chinese Remainder Theorem (CRT):
    N ≡ (xD1r2 + yD2r1) mod D1D2.

    153 = 3^2 * 17 = 9 * 17
    and 128^1000 = 2^1000mod9 = 2^(3 * 333 + 1)mod9 = -1 * 2mod9 = 7mod9.
    also 128^1000 = (-8)^1000mod17 = 2^3000mod17 = 2^(4 * 750)mod17 = 1mod17.
    Also 9 * 2 - 17 * 1 = 1
    Using Chinese Remainder Theorem
    128^1000 = (9 * 2 * 1 - 17 * 1 * 7)mod153 = -101mod153 = 52mod153



  • Q30) Find the number of integer solution to the equation 1/x + 1/y = 1/48



  • Let x = 48 + a
    and y = 48 + b
    so 1/x + 1/y = 1/48
    => 1/(48 + a) + 1/(48 + b) = 1/48
    => 48(48 + a) + 48(48 + b) = (48 + a)(48 + b) = 48² + 48a + 48b + ab
    => ab = 48² = 2^8 3^2
    Now we just need to find the number of ways in which RHS can be written as product of two integers.
    See, number of (positive) factors of RHS = (8 + 1)(2 + 1) = 27.....so number of (ordered) pairs of positive factors possible = 27.
    And including the negative values for a, b we get total 2 × 27 = 54 pairs.



  • @kamal_lohia why is that one of x,y,z is not greater than 3 ?
    @zabeer



  • @Naman-Jain-0

    atleast one of x, y and z is not more than 24
    as HCF is given as 8, we know x, y and z would be a multiple of 8. So to ensure atleast one among them is not greater than 24, one of x, y, and z is not greater than 3.
    [just for example, if x = 4 then a = 4x = 32 > 24]



  • @kamal_lohia thank you @zabeer

    also , has anyone posted memory based LRDI questions from 2015 and 2016 ?


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