Quant Boosters - Quant Mixed Bag - Set 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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