Quant Boosters - Vikas Saini - Set 5



  • first we take extreme value at x-13 = 0
    = > x = 13
    at x = 13.
    |x-1|+|x-5|+|x-9|+|x-13| = 24.
    if we inverse value of x by 1 then total value will increase by 4.
    we need to increase value of 18,we will have to increase value by 4.5.
    To increase by 4 = +1.
    To increase by 1 = +1/4.
    To increase by 18 = 18/4 = 4.5.
    at x = 13+4.5 = 17.5
    |x-1|+|x-5|+|x-9|+|x-13|
    = 16.5 + 12.5 + 8.5 + 4.5
    = 42.
    max value of x = 17.5
    for minimum value of x-1 = 0
    = > x = 1.
    at x = 1
    |x-1|+|x-5|+|x-9|+|x-13|
    = 24.
    again to increase by 18,we need to reduce value of x by 4.5.
    x = 1 - 4.5 = -3.5
    |x-1|+|x-5|+|x-9|+|x-13|
    = 4.5 + 8.5 + 12.5 + 16.5
    = 42.
    x € [-3.5,17.5]



  • Q18) What was the day on 15th August 1947 ?



  • How to find the day of any century
    In a non leap year 365 days.
    365 = 7 x 52 + 1.
    It means one odd day in a non leap year.
    In a year every single day of week repeats by 52 times at least, whereas 1 odd day will be remain in a year itself.
    For a leap year 2 odd days will be remain.
    In a century we find 76 non leap years and 24 leap years.
    So total no of odd days in a century = 76 x 1 + 24 x 2 = 124 days.
    124 mod 7 = 5 days.

    In a century there is 5 odd days.
    100 years = 5 odd days.
    200 years = 5 x 2 mod 7 = 3 odd days
    300 years = 5 x 3 mod 7 = 1odd day.
    400 years = (5 x 4 +1) mod 7 = 0 odd.

    It means after 400 years no odd days.

    For month odd days code
    January - 0
    February - 3
    March - 3
    April - 6
    May - 1
    June - 4
    July - 6
    August - 2
    September - 5
    October - 0
    November - 3
    December - 5

    And code for century
    100 years = +4.
    200 years = +2.
    300 years = 0.
    400 years = -1.

    An awesome formula to find the day if you know the date :
    Day = ( Date + Month code + year + [year/4]+ century code ) mod 7

    Given question,
    Date = 15
    Month code = 2
    Year = 47
    century code = 0(1900 = +300)
    = (15 + 2 + 47 + [47/4] + 0 ) mod 7
    = (15 + 2 + 47 + 11) mod 7
    = 75 mod 7
    = 5.
    5th day of week is Friday.

    Some more examples

    What was the day 26th January 1950.
    Day = 26
    Month code = 0
    Year = 50
    century code = 0.
    Day = (26+0+50+[50/4]+0) mod 7
    = (26+50+12+0) mod 7
    = 88 mod 7
    = 4.
    4th day of week = Thursday.

    What was the day on 12 December 1729.
    Date = 12.
    Month code = 5
    Year = 29.
    Century code = +4 (+100 year)
    Day = (12+5+29+[29/4]+4) mod 7
    = (12+5+29+7+4) mod 7
    = 57 mod 7
    = 1
    1st day of week is Monday

    What will be the day on 30th September 2016.
    Date = 30.
    Month code = 5.
    Year = 16.
    Century code = -1 ( +400)
    Day = (30 + 5 + 16 + [16/4] -1) mod 7
    = (30+5+16+4-1) mod 7
    = 5
    5th day of week is Friday.



  • Q19) What is the angle between Hour hand and minute hand at 4:45 PM ?



  • Finding angle between Hands
    12 h = 360°
    1 h = 30°
    60 m = 30°
    1 m = 0.5°
    Angle = | 5.5 M - 30 H |
    M = minute
    H = Hour

    Given question,
    5 hours = 5 x 30° = 150°
    45 min = 45 x 0.5° = 22.5°
    Total = 150° - 22.5° = 127.5°

    using Shortcut :-
    Angle = | 5.5 M - 30 H |
    = | 5.5 x 45 - 30 x 4 |
    = 127.5 °

    One more example -

    To find the angle between hour hand and minute hand at 3:30
    3 Hours = 30° x 3 = 90°
    30 min = 0.5° x 30 = 15°
    Total = 90° - 15° = 75°.
    Shortcut :-
    Angle = | 5.5 M - 30 H |
    = | 5.5 x 30 - 30 x 3 |
    = 75°



  • Q20) A watch gains 5 sec in every 3 minutes and was set right at 8 AM. So what time will it show at 10 PM on the same day ?



  • Some concepts before the solution

    Relative speed of 2 clock hands
    12 H = 60 div
    1 H = 5 div.
    SH = 60 M = 50 div.
    SH = 5/60 div/min.
    SH = 1/12 div/min.
    SM = 60 div / 60 min = 1div/1min
    Relative speed = 1 - 1/12 = 11/12 div/min.
    T = 60 / (11/12)
    T = 65 + (5/11)
    T = 1:05:5/11
    Both hand will coincide after every 1:05:5/11.

    Examples

    By what time both hands will coincide in between 5-6 PM ?
    (1:05:5/11) x 5
    = 5:25:25/11
    = 5:27:3/11

    By what time both hand will coincide in between 9 - 10 PM ?
    (1:05:5/11) x 9
    = 9:45:45/11
    = 9:49:1/11.

    By what time in between 2-3 PM both hand will make angle of 30°.
    angle = | 5.5 M - 30 H |
    30° = | 5.5 x M - 30 x 2 |
    30° = | 5.5 M - 60° |
    90° = 5.5 M
    M = 900 / 55 = 180/11 = 16 (4/11).
    Time = 2:16:4/11.

    By what time in between 5-6 PM both hands will make angle of 60°.
    angle = | 5.5 M - 30 H |
    60° = | 5.5 M - 30 x 5 |
    60° = 5.5 M - 150
    210° = 5.5 M
    M = 2100/55 = 420/11 = 38(2/11)
    Time = 5:38:2/11.

    Given question

    3 Min = 5 sec
    1 min = 5/3 sec.
    60 min = 100 sec.
    1 Hour = 100 sec.
    14 Hour = 1400 sec.
    1400 sec = 23 min 20 seconds.
    Time = 10:23:20 sec.



  • Q21) If all the roots of the equation (x – m)^2 * (x – 10) + 4 = 0 are integers, find the number of distinct values that ‘m’ can have ?



  • (x – m)^2 * (x-10) = -4.
    -4 = 4 x (-1) , 1 x (-4),

    (x – m)^2 (x -10) = 4 x (-1)
    x – 10 = -1.
    x = 9.
    (9 – m)^2 = 4.
    (9 – m) = 2, -2.
    m = 7, 11.

    (x – m)^2 (x – 10) = 1 x (-4)
    x – 10 = -4.
    x = 6.
    ( 6 – m)^2 = 1.
    6 – m =1, -1.
    m = 5, 7.

    Hence m can take three values.



  • Q22) Find the number of integer solutions of the equation x^2/y = 4x – 3, where x and y are non zero real numbers



  • Suppose x^2 / y = t.
    t = 4x – 3.
    x = 1, t = 1, y = 1.
    x = 3, t = 9, y = 1.
    Only two integer solutions.



  • Q23) If a, b are integers then how many ordered pairs (a, b) satisfy the equation a^2 + ab + b^2 = 1 ?



  • a^2 + ab + ab + b^2 – ab = 1
    (a+b)^2 = 1 + ab.
    a = 0, b = 1.
    a = 1, b = 0.
    a = 0, b = -1.
    a = -1, b = 0.
    a = 1, b = -1.
    a = -1, b=1.
    6 ordered pairs.



  • Q24) x^2 * y^3 = 8, where x, y > 0. What is the minimum value of 4x + 3y ?



  • 4x + 3y = P.
    4x = 2k
    x = k/2.
    3y = 3k.
    y = k.
    (k/2)^2 k^3 = 8.
    K^5/4 = 8.
    K = 2.
    4x + 3y = 2k + 3k = 5k = 10.



  • Q25) [ log (1) + log (1+3) + log (1+3+5) ... log(1+3+5…….19) - 2 (log 1 + log 2 ... log 7) = m + nx + ay. If log 2 = x and log 3 = y, then find the value of m, n & a ?



  • (log 1 + log 4 + log 9……..log 100) – 2(log 1 + log 2…….log 7)
    = 2(log 1 + log 2 + log 3……..log 10) – 2( log 1 + log 2…………log 7)
    = 2 (log 8 +log 9 + log 10)
    = 2(3log2 + 2log3 + 1)
    = 2 + 6log 2 + 4log 3
    = 2 + 6x + 4y
    m = 2, x =6, y =4.



  • Q26) All three roots of the cubic equation x^3 - 10x^2 + 31x - k = 0 are prime numbers. What is the value of K ?



  • Let’s three roots l,m and n.
    l + m + n = 10.
    lm + mn + nl = 31.
    lmn = k.
    all roots are prime.
    Hence l = 2, m = 3, n = 5.
    lmn = 30.




  • {x} = x – [x]

    (a) 0
    (b) 1
    (c) 2
    (d) 3


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