Quant Boosters - Ravi Handa - Set 1



  • From the leaves that are torn off, the first page number should be odd.
    Let us assume it is 2a + 1.
    Since there are total of 62 pages which are torn off, the last page number will be 2a + 62
    Sum of 'n' terms in an Arithmetic Progression = n/2[First term + Last Term]
    => Sum of page numbers = 62/2 [2a + 1 + 2a + 62]
    => Sum of page numbers = 31(4a + 63)

    Now, let's look at the options
    Option (a) 1955 is ruled out because it is not divisible by 31

    Option (b) 2201
    => 31(4a + 63) = 2201
    => 4a + 63 = 2201/31 = 71
    => 4a = 71 - 63 = 8
    => a = 2
    => 2201 is a valid value

    Option (c) 2079 is ruled out because it is not divisible by 31



  • Q26) How many times is a key of a typewriter pressed in order to type the first 299 natural numbers by pressing the space bar once between any two successive natural numbers?



  • Number of 1 digit natural numbers {1, 2, 3 ... 9}= 9
    Keystrokes required to type them = 9*1 = 9

    Number of 2 digit natural numbers {10, 11, 12 ... 99} = 90
    Keystrokes required to type them = 90*2 = 180

    Number of 3 digit natural numbers {100, 101, 102 ... 299} = 200
    Keystrokes required to type them = 200*3 = 600

    Number of spaces between the numbers = 298
    Keystrokes required to type them = 298*1 = 298

    Total keystrokes required = 9 + 180 + 600 + 298 = 1087



  • Q27) Which number is greater, 80^50 or 60^54?



  • To solve this, it would help if you know Log(2) = 0.3010 and Log(3) = 0.4771
    If we have to compare a and b, we can compare Log(a) and Log(b).

    Log (60^54) = 54 Log (60) = 54 [Log (2) + Log(3) + Log(10)] = 54[0.3010 + 0.4771 + 1] = 54 * 1.771 = 95.634

    Log (80^50) = 50 Log(80) = 50 [Log(8) + Log(10)] = 50[3 Log(2) + 1] = 50[3*0.3010 + 1] = 50 * 1.9030 = 95.15

    So, Log (60^54) > Log (80^50) => 60^54 > 80^50



  • Q28) How many 6 digit numbers can be made using digits 1, 2, 3, 4, 5, and 6 without repetition such that the hundred digit is greater than the ten digit, and the ten is greater than the one digit?



  • Let us assume that the number that we have is abcdef
    The condition given is d > e > f

    Let us first select three digits for a, b, and c.
    From the given 6 digits (1,2,3,4,5,6), 3 digits for a,b, and c can be selected in 6C3 = 6!/3!3! = 20 ways.
    These three selected digits can be arranged on three positions of a, b, and c in 3! = 6 ways

    The remaining 3 digits are automatically selected for d, e, and f.
    So, we have 1 way of selecting them.

    The biggest digit will be allocated to 'd'.
    The second biggest digit will be allocated to 'e'.
    The third biggest digit will be allocated to 'f'
    So, we have 1 way of arranging them on the three positions of d, e, and f.

    Total ways = 20 * 6 * 1 * 1 = 120



  • Q29) A and B started at a time towards each other. After crossing each other, they took 16hrs, 25hrs respectively to reach their destinations. If A had 40kmph speed, what is the speed of B?



  • Let us say that A and B started from point P and point Q respectively and met at point R in the middle after time 't'.

    Speed of A is given to us as 40 kmph.

    Let us assume the speed of B to be 's'.

    If we consider the distance PR, it was covered by A in 't' hours and B in 25 hours.
    => PR = 40 * t = s * 25

    If we consider the distance RQ, it was covered by B in 't' hours and A in '16' hours.
    => RQ = s * t = 40 * 16

    Let us divide the two equations. We will get

    40t/st = 25s/640
    => 40/s = 5s/128
    => s^2 = 40 * 128/5 = 8 * 128 = 1024
    => Speed of B = s = 32 kmph



  • Q30) What degree occurs at a time 3:30 in the clock?



  • At 3:30, the hour hand would be exactly between 3 and 4
    => 15 deg from both 3 and 4
    At 3:30, the minute hand would be exactly at 6
    The gap between the hour hand and the minute hand will be
    30 deg (between 5 and 6) + 30 deg (between 4 and 5) + 15 deg (between hour hand and 4) = 75 degrees


Log in to reply
 

  • 51
  • 61
  • 61
  • 61
  • 42
  • 64