Quant Boosters - Kamal Lohia - Set 9



  • Q20) There are 'n' points on a circle's boundary. All possible chord are drawn by joining these points. How many triangles are formed, by intersection of these chords, whose all the three vertices lie inside the circle and not circle's boundary?



  • Q21) How many pairs (a, b) are there where a and b are integers in the range from 1 to 100 inclusive, and a^b is a perfect square?



  • Q22) How many right triangle with integral sides exist such that one of the perpendicular side has length 36 units?



  • Q23) What is the sum of all natural numbers which are less than 2012 and coprime with it?



  • Q24) A polygon has all interior angles of 90 or 270 degrees only. If there are 22 interior angles measuring 270 degrees each, then what is the number of sides in the polygon?



  • Q25) Two altitudes of a triangle are of 12 and 20 units repectively. Find the sum of all possible integral lengths of the third altitude.



  • Q26) S is a set of natural numbers formed by digit 1 only such that S = {1, 11, 111, 1111, .....}.
    (i) How many numbers in S are perfect squares?
    (ii) How many numbers in S are perfect cubes?



  • Q27) If N is a four digit number, abcd such that N = (ab)² + (cd)² then find a + b + c + d



  • Q28) Find the maximum value of y/x if (x - 3)² + (y - 3)² = 6.



  • Q29) Two brothers Nimai and Nitai subtarct a positive divisor of 60 (except 60) from 60 alternately starting from Nimai. The brother who subtracts the last number (till resulting number becomes zero) is the winner. For example: If Nimai subtracts 1, then resulting number is 60 - 1 = 59, now Nitai can subtract any positive divisor of 60 less than 60 from the resulting number. Say he subtracts 2 so the the resulting number now is 59 - 2 = 57. Who has the strategy to win assuming both brothers are equally competent and playing to earn a win?



  • Q30) If Team A and Team B have equal chance of winning in each game. Given that Team B won the second game what's the probability that Team B won the first game if Team A won the series where the first team that won three times wins?



  • Let there be n balls of m different colours
    Initial Probability = m * C(n, 2)/C(mn, 2) = (n – 1)/(mn – 1)
    Now, 20 balls of a different colour are added.
    Probability = {m * C(n, 2) + C(20, 2)}/C(mn + 20, 2)
    Equating them we will get,
    (n – 1) = 19(mn – 1)/(2mn + 19) = 9 + (x – 190)/(2x + 19), where x = mn
    Now, possible values for x are 0, -1, -2, -3, -4, -5, -6, -7 and -8, -9 and -10
    Checking for them possible values of (x, n) are (190, 10), (57, 9), (19, 7), (1, 1) and (0, 0)
    Only feasible solution is (190, 10)
    So, initially we had 10 balls of 19 colours
    [credits : Hemant Yadav]



  • f(x) = (x – 2)p(x) + 99
    As, f(2) = 99, we can say that
    Polynomial {f(x) – 99} will have 2 as a root
    Hence, f(x) – 99 = (x – 2)p(x)
    So, f(x) = (x – 2)p(x) + 99
    f(99) = 97p(99) + 99
    So, possible values of f(99) are -95, 2, 196, for p(x) = -2, -1 and 2 respectively
    [credits : Hemant Yadav]



  • 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.


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