Quant Boosters - Vikas Saini - Set 4



  • Q26) There are 3 consecutive natural numbers such that square of the second is 19 more than 9 times of sum of first & third number. Find second number.



  • Let 3 consecutive numbers are n-1, n and n+1.
    => n^2 – 19 = 9(n -1+n+1)
    => n^2 – 19 = 9 x 2n
    => n^2 – 18n + 19 = 0.
    => (n-19) (n+1) = 0.
    => n = 19, -1.
    Second number is 19.



  • Q27) Let N = 999999 ... 36 times. How many 9’s are there in N^2 ?



  • 99^2 = 9801 (one 9)
    999^2 = 998001 (two 9)
    9999^2 = 99980001 (three 9)
    Therefore (99999……….36 time)^2 should have 35 nines.



  • Q28) How many positive numbers are there which is equal to 12 times sum of their digits ?



  • For 2 digit number
    Let the number is 10a+b.
    10a + b = 12(a+b)
    = > 10a + b = 12a + 12b
    = > -2a = 11b
    None two digit number exist here.

    For 3 digit numbers
    Let the number is 100a+10b+c.
    100a + 10b + c = 12(a+b+c)
    = > 100a + 10b + c = 12a + 12b + 12c
    = > 88a – 2b = 11c.
    b = 0.
    = > 88 a = 11 c.
    = > 8a = c.
    a = 1, c = 8.
    Number 108.

    For 4 digit number
    1000a+100b+10c+d
    1000a+100b+10c+d = 12(a+b+c+d)
    = > 1000 a + 100 b + 10 c + d = 12 a + 12 b + 12 c + 12 d
    = > 988 a + 88b = 2c+11d
    Not possible.

    Hence only positive number is 108.



  • Q29) If we arrange 4 different color balls like
    1 red ball, 2 black balls, 3 green balls, 4 white balls
    5 red balls, 6 black balls, 7 green balls, 8 white balls etc...
    How many balls would be required to reach the 100th black ball ?



  • Number\BallsRedBlackGreenWhite
    11234
    25678
    39101112
    413141516
    517181920
    621222324
    725262728
    8292--
    Total120100105112

    Total = 120 + 100 + 105 + 112 = 437.

    437th ball is the 100th black ball.



  • Q30) If A, B, C, D, E represent distinct digit from 1 to 5.
    ABE = E x BE.
    CDA = CA x AA.
    BCA + CCD = EEE.
    Find the value of AB x CE.



  • Here given E x E = E and A x A =A.
    So E = 1 or 5, A = 5 or 1.
    Let’s put E = 1 and A = 5.
    5B1 = 1 x B1
    Not possible.
    Let’s put E = 5 and A = 1.
    1B5 = 5 x B5
    So B = 2.
    BBA + CCD = EEE.
    221 + CCD = 555.
    1 + D = 5.
    D = 4.
    2 + C = 5.
    C = 3.
    AB x CE = 12 x 35 = 420.


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