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 solutionsMethod 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;
 when x > 4 x1  x2 + x4 = (x1)(x2)+(x4) = m => x3 = m => m > 1
 when 2 < x < = 4 x1x2+x4 = (x1)(x2)(x4) = m => 5x = m => 1 < = m < 3
 when 1 < x < = 2 x1x2+x4 = (x1)+(x2)(x4) = m => x+1 = m => 1 < m < = 3
 when Infinity < x < = 1 x1x2+x4 = (x1)+(x2)(x4) = m => 3x = 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 > mLooking 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 satisfyHence, 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) = 32f(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