Quant Boosters - CatOMania - Set 1


  • CatOMania - coz we bell it !


    what is the probability of always having x + y > 0 ,where y < 0
    a)1/2
    b)1/8
    c)3/4
    d)1/4

    4 cases
    X negative
    X zero
    X positive and greater than y
    X positive and less than y
    We want only 1 case so 1/4

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

    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.

    Aseem writes three single-digit numbers, starting from 1, on a black board. The numbers written by him form an arithmetic progression. Ruchika, who is a friend of Aseem, erases one of the numbers written on the black board. If the average of the remaining numbers is 5.5. The number erased by Ruchika is
    a 7
    b 1
    c 4
    d Cannot be determined

    a - d, a, a + D
    a + a + D = 11
    2a + D = 11
    a - d = 1
    3a = 12
    A = 4

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

    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

    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

    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 :)

    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

    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.

    What is the remainder when 12345678987654321 is divided by 1001 ?

    111111111^2mod 1001
    111111mod 1001=0
    (111111 * 1000+111)^2mod 1001
    111^2mod1001
    12321
    12 × 1000 + 321 mod 1001
    12 × -1 + 321 = 309

    The least number which must be added to 3775 * 3789 to make the result the result a perfect square is?
    a)64
    b)49
    c)121
    d) 9

    Method 1
    Take y = 3782
    (y - 7)(y + 7) = y^2 - 49
    So if we add 49 it will be a perfect square.

    Method - 2
    Don't take x,
    Take 1
    1×15=15
    1
    Minimum number from options - 15 + 9 = 24, not a square
    Next 15 + 49 = 64 => done

    Similar questions came this year in CAT
    See, if you generalize above like may be it is possible for 1 not for the number given
    Taking the question first
    2 × 16 = 32 + 49 = 81
    3 × 17 = 51 + 49 = 100
    So you can see it is a perfect square for every number.

    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

    Two joggers run around an oval track in opposite directions. One jogger runs around the track in 56 seconds. The joggers meet every 24 seconds. How many seconds does it take for the second jogger to run around the track?

    As they run in opposite direction and they meet in 24 seconds 1st runner would run as much as 2nd runner would do in 32 seconds ( 56 - 24 = 32)
    So ratio of time= 32:24 = 8:6 = 4:3
    So total time will also be, 56:42


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