Quant Boosters - Vikas Saini - Set 3



  • Number of Questions - 30
    Topic - Quant Mixed Bag
    Solved ? - Yes
    Source -



  • Q1) xy + yx in base 8 = abc in base 10. How many values are possible for abc



  • 8x + y + 8y + x = abc.
    9x + 9y = abc.
    9(x+y) = abc.
    x = 5, y = 7 then abc = 108.
    x = 6, y =7 then abc = 117.
    but abc are distinct digits.
    so only 1 value 108.



  • Q2) A two digit number in base 7 is equal to the two digit number formed by reversing it digits but in base 13. How many such numbers are possible ?



  • (xy)_7 = (yx)_13
    7x+y = 13y+x
    6x = 12y
    x = 2y.
    x = 1,2,3.
    y = 2,4,6.
    three numbers 12, 24 and 36



  • Q3) A 99 digit number is formed by writing first 54 natural numbers in front of each other as 123 ... 54. Find the remainder when this number is divided by 8.
    (a) 4
    (b) 7
    (c) 2
    (d) 0 (CAT 1998)



  • For the divisibility by 8, we check only for the last three digits (because 8 = 2^3) of the number. If the last three digits of a number is divisible by 8, then the number will be divisible otherwise not.
    So, here the number is 1234567891011121314…….525354.
    Last three digits of the number = 354
    Remainder when the number is divided by 8 = Rem(354/8) = 2



  • Q4) If log_10(x) – log_10(x^1/3) = 6log_x (10) then find the value of x
    (a) 10
    (b) 30
    (c) 100
    (d) 1000. (IIFT 2013)



  • log_10(x) – (1/3)log_10(x) = 6log_x(10).
    = > (2/3)log_10(x)= 6 / log_10 (x)
    Put log_10 (x) = m.
    = > (2/3) m = 6 / m
    = > m^2 = 9.
    = > m = 3, -3.
    We need to remove m = -3.
    m = 3
    log_10 (x) = 3.
    x = 1000.
    Hence option d.



  • Q5) The sum of the possible values of x in the equation |x + 7| + |x-8| = 16.
    (a) 0
    (b) 1
    (c) 2
    (d) 3
    (e) none of the above. (XAT 2014)



  • First put extreme value of x = 8.
    At x = 8, equation = 15.
    At x = 8.5, equation = 16.
    Now put extreme negative value x = -7.
    At x = -7, equation = 15.
    At x = -7.5, equation = 16.
    Sum = 8.5 – 7.5 = 1.



  • Q6) Sum of 6 terms of an AP is 6. If 3rd term of AP is 0 then find the 6th term of it.



  • Suppose 3rd term of AP is a.
    2nd & 1st term are a-d & a-2d respectively.
    4th, 5th & 6th terms are a+d,a+2d,a+3d respectively.
    So (a-d) + (a-2d) + a + (a+d) + (a+2d) + (a+3d) = 6.
    = > 6a + 3d = 6.
    = > 0 + 3d = 6.
    = > d = 2.
    So a+3d = 6.
    6th term is 6.



  • Q7) Sum of 3 terms of an AP is 30. If we add 6 in product of these 3 terms then it becomes a perfect square. Find all 3 terms of AP.



  • Let three terms are a-d, a, a+d.
    = > (a-d)+a+(a+d) = 30.
    = > 3a = 30.
    = > a = 10.
    (10 – d) x 10 x (10+d) + 6 = k^2.
    = > 10(100 – d^2)+6 = k^2.
    d = 1,2,3,4,5,6,7,8,9.
    d^2 = 1,4,9,16,25,36,49,64,81.
    100 – d^2 = 99,96,91,84,75,64,51,36,19.
    10(100 – d^2) + 6 = 996,966,916,846,756,646,516,366,196.
    196 is the only number here which is perfect square.
    So d = 9.
    3 terms of AP are = (10-9), 10, (10+9)
    = 1, 9, 19.



  • Q8) Sum of first 4 terms of an AP is 42. Sum of square of 4 terms are 566. Find all the 4 terms



  • Let terms are a-2d, a-d , a+d , a+2d.
    (a-3d)+(a-d)+(a+d)+(a+3d) = 42.
    = > 4 a = 42.
    a = 10.5.
    (a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 566.
    = > a^2 + 9d^2 – 6ad + a^2 + d^2 – 2ad + a^2 + d^2 + 2ad + a^2 + 9d^2 + 6ad = 566.
    = > 4a^2 + 20 d^2 = 566.
    = > 4 x (10.5)^2 + 20 d^2 = 566.
    = > 441 + 20d^2 = 566.
    = > 20 d^2 = 125.
    = > d^2 = 6.25
    = > d = 2.5 .
    Numbers are 3, 8, 13, 18.



  • Q9) From the first 20 natural numbers how many Arithmetic Progressions of five terms can be formed such that the common difference is a factor of the fifth term ?



  • For d=1, total = 16.
    (1,2,3,4,5), (2,3,4,5,6) ... (16,17,18,19,20)

    For d=2, total = 6.
    (2,4,6,8,10), (4,6,8,10,12) ... (12,14,16,18,20).

    For d=3, total = 2.
    (3,6,9,12,15), (6,9,12,15,18 )

    For d=4, total = 1.
    (4,8,12,16,20)

    Total = 16 + 6 + 2 + 1 = 25.



  • Q10) Ratio of sum of two APs is (n + 1) : 2n. Find the ratio of the 6th terms of both series.


 

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