Quant Boosters - Quant Mixed Bag - Set 1



  • [Credits : Tanveer Iqbal]
    As per the question,
    10^(z−1) < = N < 10^z = > 10^(3z−3) < = N^3 < 10^3z.
    So, y can be {3z − 2, 3z − 1, 3z} and y+z can be {4z-2 ,4z-1,4z)
    Clearly all the natural numbers except those of type 4k+1 satisfy the value for y+z.
    So, 35 cannot be the value.
    = > Choice (d) is the right answer



  • Q22) A real number x is chosen at random in the interval [-21/2, 21/2]. What is the probability that the closest integer to x is odd?
    a) 1/21
    b) 5/11
    c) 10/21
    d) 11/21
    e) 1/2



  • [Credits : Tanveer Iqbal]
    From [-10.5,10.5] There can be 21 intervals [-10.5,-9.5],[-9.5,-8.5].....[9.5,10.5]..and there are 10 intervals [-9.5,-8.5]....[8.5,9.5]..in which any x selected in those intervals is closest to an odd integer.
    So the probability is 10/21
    Choice (c) is the right answer



  • Q23) A positive integer p is called "CAT number" if p^3 + 7p = q^3 + 133 for some positive integer q. The sum of all such "CAT numbers" is
    (a) 24
    (b) 26
    (c) 30
    (d) 34
    (e) 38



  • [Credits : Taveer Iqbal]
    We consider the size of p relative to q. If p = q, then q is a CAT number if and only if
    7q = 133. In this case, we get 19 is a good number. Now, suppose that p > = q + 1. Then
    q^3 + 7q − 133 = p^3 > = (q + 1)^3 = q^3 + 3q^2 + 3q + 1
    so that 3q^2 − 4q + 134 < = 0. For positive integers q, it is easy to see that this is impossible. It remains to consider p < = q − 1. For such p, we have q^3 + 7q − 133 = p^3 < = (q − 1)^3 = q^3 − 3q^2 + 3q − 1 which implies 3q^2 + 4q − 132 < 0. It follows here that q < 6. One checks that 6^3 + 7 · 6 − 133 = 5^3 and 5^3 + 7 · 5 − 133 = 3^3,
    so 6 and 5 are CAT numbers. For q < = 4, we see that q^3 +7q−133 < 0 and, hence, q^3 +7q−133 cannot equal p^3 for a positive integer p. Therefore, the sum of all CAT numbers is 19 + 6 + 5 = 30.



  • Q24) If P : Q = 1 / 2 : 3 /8 , Q : R = 1 / 3 : 5 / 9 and R : S = 5 / 6 : 3 / 4 , Then P : Q : R : S = ?



  • [Credits : Tanveer Iqbal]

    P : Q = 1 / 2 : 3 / 8 = 4 : 3 , Q : R = 1 / 3 : 5 / 9 = 3 : 5 and R : S = 5 / 6 : 3 / 4 = 10 : 9
    P : Q = 4 : 3 , Q : R = 3 : 5 , R : S = 10 / 2 : 9 / 2 = 5 : 9 / 2
    P : Q : R : S = 4 : 3 : 5 : 9 / 2 = 8 : 6 : 10 : 9



  • Q25) What digit does “a” represent, if 35! = 10333147966386144929a66651337523200000000?
    (a) 4
    (b) 6
    (c) 2
    (d) 1



  • [Credits : Tanveer Iqbal]
    35!= 1 * 2 * 3 * 4 * 5 *... * 35
    so the above number will also be divisible by 3,9,11( such number which uses the sum of number to check the divisibility)
    Check by 3:- you will get the different values of a which also have to satisfy for 9,11.
    Check by 9:- (Sum of digits +a) has to be divisible by 9.
    Check by 11 directly,
    Sum of odd places- sum of even places= 11 n or 0.
    (66 + a) - 72=0, as a cant be greater than 9. to give the difference multiple of 11.
    so a = 6



  • Q26) In base 7, 65324 + 50231 - 133 × P = 130462. What is the value of 'P'?
    a) 110
    b) 101
    c) 1001
    d) 111



  • [Credits : Ashutosh Mishra]
    Going by last digits
    4 + 1 - 3 = 2
    Now second last digit
    2 + 3 - x = 6
    5 - x = 6
    Only possible when you would have carried over other wise 6 won't be the result
    12 - 6 = 6
    So x = 6
    Now only multiplying 133 by 111 gives second last digit as 6.



  • Q27) Sum of (1/101)+ (1/102)+ (1/103) + ... +(1/200) is
    a. 1
    b. 0.5
    c. 1.85
    d. None of these



  • [Credits : Ashutosh Mishra]
    Method 1:
    Sum of series of 1/n is ln(n). Ln is log to the base e.
    Ln 200-ln100
    Ln(200/100)=ln(2)
    E=2.3
    Ln e base e=1
    So answer will be less than 1 but can it be 0.5??
    Rt 2=1.414
    So rt e=1.7 approx
    But 2 is greater than 1.7
    So answer will lie between 0.5 to 1 so d

    Method 2:
    S = 1/200 + 1/2000... = 100 / 200 = 0.5
    but we have terms 1/101 , 1/102..likewise which will increase the sum
    and if we substitute S = 1/101 + 1/101 ... = 100 / 101 = 0.99.
    S = 0.5 < S < 1

    Method -3
    S = {(1/101) + (1/102) + .. + (1/150)} + {(1/151) + (1/152) + ... + (1/200)}
    1/100 > 1/101 so if we find the sum of below it wil be greater than the asked one
    S < {(1/100) + (1/100) + .. + (1/100)} + (1/150) + (1/150) + ... + (1/150)
    = 50/100+ 50/150 = (1/2) + (1/3) = 5/6 ~ 0.83
    also S > {(1/150) + (1/150) + .. + (1/150)} + {(1/200) + (1/200) + ... + (1/200)} = {50/150} + {50/200} = (1/3) + (1/4) ~ 0.58
    So 0.58 < S < 0.83



  • Q28) 1 + 2 x 2 + 3 x 2^2 + 4 x 2^3 + ... + 100 x 2^99 = ?
    a. 99 x 2^100
    b. 99 x 2^100 + 1
    c. 99 x 2^100 - 1
    d. 100 x 2^100



  • [Credits : Ashutosh Mishra]
    Method 1: (that I wouldn't recommend)
    S = 1 + 2.2 + 3.2² + 4.2³ .... 100.2^99
    2S = 2 + 2.2² + 3.2³ .... 99.2^99 + 100.2^100
    S - 2S = (1-0) + (2.2-2) +( 3.2² - 2.2²) ..... (100.2^99 - 99.2^99) - 100.2^100
    -S = 1 + 2 + 4 + 8 ..... 2^99 - 2^100.
    S = 100.2^100 - ( 1 + 2 + 4 + 8+...... 2^99)
    S = 100.2^100 - 1(2^100 - 1)/(2-1).
    S = 100.2^100 - 2^100 + 1
    S = 99.2^100 + 1

    Method 2 : Generalize
    Take first 2 terms
    1 + 2 × 2 = 5
    So if answer would have been for only 2 terms then answers would have been
    1 × 2^2
    1 × 2^2 + 1
    1 × 2^2 - 1
    2 × 2^2
    Now you have the answer and this is a CAT problem :)



  • Q29) Consider the set S = (1, 2, 3, …, 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?
    (a) 3
    (b) 4
    (c) 6
    (d) 7
    (e) 8



  • [Credits : Ashutosh Mishra]
    Let number of terms in the arithmetic progression be n, then
    1000 = 1 + (n–1) d
    =>(n–1) d = 999
    => n – 1 = 999/d
    Since n is an integer, so n – 1 is also an integer. This means that ‘d’ is a factor of 999.
    Now 999 = 33 × 37. so the total factors of 999 are 4 × 2 = 8.
    Out of these 8 factors one factor is 999 and we will reject it as in that case there will be only two terms in the
    A.P. i.e. 1 and 1000, which is not possible.
    Hence, ‘d’ can take 7 different values.
    So, in total, 7 APs are possible



  • Q30) If 3x + y + 4 = 2xy, where x and y are natural numbers, then find the ratio of the sum of all possible values of x to the sum of all possible values of y.
    a) 2/3
    b) 15/19
    c) 17/21
    d) 7/9



  • 2xy - 3x - y = 4
    2x(y - 3/2) - y = 4
    add 3/2on both sides
    2x(y - 3/2) - y + 3/2 = 4 + 3/2
    2x(y - 3/2) - 1(y - 3/2) = 11/2
    (2x - 1)(y - 3/2) = 11/2
    (2x - 1)(2y - 3) = 11
    11 = 1 * 11
    Case-1) 2x - 1 = 1 => x=1
    2y - 3 = 11, y = 7
    Case-2) 2x - 1 = 11
    x = 6
    2y - 3 = 1, y = 2
    so 7/9


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