Quant Boosters by VP - Set 2



  • Q26) In an increasing Arithmetic Progression, the product of the 5th term and the 6th term is 300. When the 9th term of this A.P. is divided by the 5th term, the quotient is 5 and the remainder is 4. What isthe first term of the A.P.?



  • Let the 5th term of the A.P. be ‘a’ and the common
    difference be ‘d’.
    The 6th term will be (a + d) and the
    9th term will be (a + 4d).

    So
    a×(a + d) = 300 ------(i)
    And
    5a + 4 = (a + 4d)
    => d = a + 1 ----- (ii)
    Solving (i) and (ii),
    we get a = 12 or -25/2

    If a = -25/2
    => then the value of ‘d’ will also be
    negative, which is not possible in an increasing A.P.

    Therefore,
    a = 12 and d = 13.
    And
    The first term will be
    = (a – 4d)
    = 12 – 52
    = –40.



  • Q27) Of the applicants who passed a certain test, 15 applied to both colleges X and Y. If 20% of the applicants who applied to college X and 25% of the applicants who applied to college Y applied to both college X and Y, how many applicants applied only to college X or college Y?



  • X intersection Y = 15
    20% of the applicants who applied for X applied for both X and Y
    I.e 20% of Total X = 15
    Total X = 75
    Only X = 60
    Similarly Total Y = 15 * 4 = 60
    Only Y = 45
    So 60 +45 = 105



  • Q28) The price of an article reduces to 576 after two successive discounts. The markup is 80% above the cost price of Rs. 500. What is the new profit percentage if instead of two successive discounts the markup price was further increased successively two times by the same percentage?



  • CP = 500
    MP = 1.8 * 500 = 900
    900 * (1- R/100) ^ 2 = 576
    (1- R/100) ^2 = .64
    .8^2
    I.e 20% discount
    Now
    MP = 900 * (1.2)^2 = 1296
    CP = 500
    I.e profit % = 796/500 * 100 = 159.2 %



  • Q29) Sameer and Sumer started running from the same point in opposite directions on a circular track of length 120 m. Their speeds are 20 m/s and 40 m/s respectively. After every second, Sameer increases his speed by 2 m/s whereas Sumer decreases his speed by 2 m/s. How many times would they have met on the track by the time Sumer comes to rest?



  • Sumer speed is 40m/s
    His speed decreased by 2 m/s every second
    And he travels till he comes to rest
    40/2 = 20 seconds
    So he travels only 20 seconds
    Now , Their relative velocity is not changing so in every seconds they will cover a distance of 60 meter and Sumeer stops after 20 seconds
    So in 20 secs the distance covered is 20 * 60 = 1200 and track is of 120m
    So they met = 1200/120 = 10times



  • Q30) In how many ways can 1000 be written as a sum of ‘n’ consecutive natural numbers, where ‘n’ is greater than 1?



  • Concept : numbers of consecutive natural numbers sum to n = (number of odd factors of n ) -1

    1000 = 5^3 * 2^3
    here number of odd factors = 4 and required answer = 4 - 1 = 3


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