Quant Boosters - Nitin Gupta - Set 1



  • a/72 + b/63 + c/56 = 4
    a/56 + b/63 + c/72 = 14/3
    Add them to get:-
    2a/63 + 2b/63 + 2c/63 = 26/3
    => a + b + c = 21 * 13 = 273



  • Q27) India and Brazil play the Soccer World Cup final in which India defeats Brazil 4 – 2. In how many different ways could the goals have been scored given that Brazil never had a lead over India during the match?
    (a) 9
    (b) 10
    (c) 8
    (d) None of these



  • If Brazil never gets a lead over India, the first goal of the match should be scored by India. At this stage India leads by 1 – 0, and the only way in which Brazil can take a lead is by scoring the next two goals i.e. a scoring pattern like IBBIII. In all other cases Brazil would never be able to lead India. Total cases where India scores the first goal
    5!/ 3!2! = 10
    Different scoring patterns possible = 10 – 1 = 9.



  • Q28) N girls and 2N boys played a chess tournament. Every player played every other player exactly once. The boys won 7/5 times as many matches as the girls (and there were no draws). Then which among the following is definitely false? (Assume 1 point for a win and 0 for a loss)
    (a) Boys pocketed prime number of points against girls
    (b) Girls always won twice or more matches than boys won against them
    (c) The sum of the scores of top 3 individual players was not between 25 and 33
    (d) The sum of the scores of top 3 individual players was 69
    (e) none of the foregoing



  • Total number of matches among boys were 2nC2, among girls were nC2 and between boys and girls were n*2n. Please note that 2nC2 + nC2 + 2n^2 = 3nC2. Assume 1 point for a win and 0 for a loss.

    => Girls pocketed nC2 points amongst themselves and boys pocketed 2nC2 points among themselves. Let boys take k points from their matches against girls => girls take 2n^2 - k from their matches gainst boys.

    => 2nC2 + k = 7/5*(nC2 + 2n^2 - k), solving we get 8k = n(5n+1).
    For n = 3, k = 6.
    For n = 8, k = 41
    For n = 11, k = 77.

    (a) can be true as for n = 8, k = 41.
    (b) can be true as can be seen for for n = 3, 8, 11, ...
    (c) is true as for n = 3, top 3 can score 8+7+6 = 21 points and for n = 11, when 33 matches are played top 3 will always score more than 16+15+14 = 45.
    (d) is true, For n = 11, we can have the top 3 score as 23+23+23 = 69.

    Hence, choice (e) is the right answer



  • Q29) Find the number of positive integral solutions for a + b + c = 20 if a ≤ b ≤ c



  • Method 1 :
    a + b + c = 20
    So, C(19, 2) = 171 solutions
    When all are equal, no solution
    When two of them are equal, 3*9 = 27 solutions
    So, answer will be (171 - 27)/6 + 9 = 24 + 9 = 33 solutions

    Method 2 :
    Number of ways in which a + b + c = n , where a ≤ b ≤ c (a,b,c are positive integers) = {n^2/12}
    where {.} means nearest integer; here the answer is {20^2/12} = {33.33} = 33



  • Q30) The equation |x - 1| - |x - 2| + |x - 4| = m has exactly n real solutions for some real m. Then which among the following relations between m and n can not be true?
    a) m = n
    b) m/n = 3/5
    c) m/n = 5/3
    d) m = n - 1
    e) m/n = 3/2



  • Since 1, 2, 4 are the critical point, we divide the domain into 4 regions;

    1. when x > 4 |x-1| - |x-2| + |x-4| = (x-1)-(x-2)+(x-4) = m => x-3 = m => m > 1
    2. when 2 < x < = 4 |x-1|-|x-2|+|x-4| = (x-1)-(x-2)-(x-4) = m => 5-x = m => 1 < = m < 3
    3. when 1 < x < = 2 |x-1|-|x-2|+|x-4| = (x-1)+(x-2)-(x-4) = m => x+1 = m => 1 < m < = 3
    4. when -Infinity < x < = 1 |x-1|-|x-2|+|x-4| = -(x-1)+(x-2)-(x-4) = m => 3-x = m => 2 < = m < = Infinity

    Clearly
    for n = 4 we have 2 < m < 3
    for n = 3 we have m = 2, 3
    for n = 2 we have 1 < m < 2
    for n = 1 we have m = 1
    for n = 0 we have 1 > m

    Looking at the choices
    (a) m = 2.4 and n = 4 satisfy
    (b) m = n = 3 satisfy
    (d) m = 10/3 and n = 2 satisfy
    (e) m = -1 and n = 0 satisfy

    Hence, choice (c) is the right answer.



  • Warning - There might be an easier method!

    f(1) + f(6) = 0
    f(1) + f(4) + 24 = 0
    f(1) + f(2) + 8 + 24 = 0
    f(1) + f(0) = -32

    f(1) + f(2) + f(3) + f(4) + f(5) + f(6)
    = f(1) + f(6) + f(0) + f(1) + 3 + f(2) + 8 + f(3) + 15
    = 0 + f(0) + f(1) + 3 + f(0) + 8 + f(1) + 3 + 15
    = 2 (f(0) + f(1)) + 29
    = 2 (-32) + 29
    = -64 + 29 = -35
    Option D


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