Quant Boosters - Kamal Lohia - Set 9



  • p² - q² = 36²
    (p + q)(p – q) = 4 * 2^2 * 3^4
    So, total (3 * 5 – 1)/2 = 7 possibilities
    [credits : hemant yadav]



  • abcd = ab² + cd²
    => ab(100 - ab) = cd(cd - 1)
    Now, unit digit of RHS can be 0 or 2 or 6,
    When its 0, LHS will have two zeros at the end, so not possible
    When its 2, LHS can never have 2 as unit digit, so not possible
    When its 6, cd can have unit digit as 3 or 8

    Hence, we can write as
    x(x - 1) = p(100 - p) = y, where x = cd, p = ab and y has a unit digit of 6
    => x² - x - y = 0
    D = Discriminant = 4y + 1
    Unit digit of D will be 5, since it is a perfect square its ten's digit should be 2.
    => Tens digit of y should be 5 or 0, so last two digits of y should be 06 or 56.
    Since p has unit digit 2 or 8, one of p and (100 - p) will be q2 and other will be (9 - q)8

    Last two digits of (9 - q)8*q2 = 06 or 56
    => Unit digit of 2(9 - q) + 8q will be 4
    => Unit digit of 18 + 6q will be 4
    => q can be 1 or 6
    => ab = 12(cd = 33) or 62(not possible)
    So, only such number is 1233.

    Couldn't find a better way to get the result.

    [credits : Hemant Yadav]



  • 1, 2, 3, 4, 5, 6 are all factors of 60. So, all of these are winning positions. Next number is 7, which is definitely a losing position as whatever you pick next player will pick everything to win. So everyone will try to leave 7. For 8, 9, 10, 11, 12, 13 one can always pick 1, 2, 3, 4, 5, 6 respectively to leave 7. In case of 14 whatever you pick the next player can either leave 7 for you are pick everything to win. So, we can see that one should always leave a multiple of 7. Also, 7 and 60 are coprime, so we don't have to worry about what if other player subtracts a larger number as he can never leave a multiple of 7.
    [credits : Hemant Yadav]



  • X - 2nd match is won by B and A won the series
    Y - First match is won by B
    => P(Y/X) = P(X and Y)/P(X)
    P(X) = (1/16) + 3 * (1/32) = 5/32
    P(X and Y) = 1/32
    => P(Y/X) = 1/5
    [credits : Hemant Yadav]



  • (x – 3)² + (y – 3)² = 6
    We have to find the maximum value of y/x.
    y/x = (y – 0)/(x – 0), so y/x is slope of a line passing through origin and one of the point on the circle. This slope will be maximum when this line will be tangent to the circle.

    Angle between the line joining the origin and centre of circle = taninv(√6/2√3) = taninv(1/√2)
    Slope of line joining centre of circle and origin = 1
    Maximum value of y/x = (1 + 1/√2)/(1 – 1/√2) = 3 + 2√2
    Similarly minimum value will be 3 - 2√2
    [credits : Hemant Yadav]



  • Area is 25?



  • Number of sucj triangles is nC6.



  • Or we can apply cauchy schwarz ineq as follows:-m=y/x
    ((X-3)^2+(y-3)^2) (m^2+(-1)^2) > =(m(x-3)-1(y-3))^2

    M comes out to be 3+2rt2, 3-2rt2
    Min =3-2rt2



  • 15/2
    8



  • @zabeer sir..

    Why have we considered numbers like 7 and such to proceed further?

    Could you explain the solution a bit more... ?


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