Quant Boosters - Shashank Prabhu, CAT 100 Percentiler - Set 1
shashank_prabhu last edited by
45 is correct. Let's say all parcels are of different weights such that a>b>c>d. So, there could be 6 possible combinations ab, ac, ad, bc, bd, cd. Out of them, there should be 2 pairs such that in each pair, addition of two quantities gives us the same result. If we look at it carefully, ab, cd, ac, bd cannot be equal to anything else. That leaves us with ad being equal to bc which defeats the given condition. So, there are 2 weights that are equal to each other and so, yield 4 cases.
Addition of these 2 weights would also give a result. As all the weights are integral in nature, the sum should be an even number. So, either the equal weights are 47 kg each or 52 kg each.
47, 47 will give us the other weights to be 50 and 54
52, 52 will give us the other wights to be 49 and 45
Shivaraman R last edited by
@shashank_prabhu whats the logic behind 13c7
zabeer last edited by zabeer
Number of non-negative integral solutions = (n + r - 1)C(r - 1)
Here, a + b + c + d + e + f + g + h = 6
so r = 8 and n = 6
Number of non-negative integral solutions = (8 + 6 - 1) C (8 - 1) = 13c7
Read the below article if you want to refer more on this concept.
Sairam Akella last edited by zabeer
@shashank_prabhu sir but here only case is a > b > c but u have taken 6 cases??
zabeer last edited by zabeer
Hi Sairam, what is the confusion here ?
Total solution of a + b + c = 60 is 62C2 = 1891.
Now we need to remove cases with repetition.
When a = b or b = c or c = a case.
when a = b,
2a + c = 60 --> 31 cases
Similarly when b = c,
a + 2b = 60 --> 31 - 1 = 30 cases (as a = b = c = 20 case is already counted in the previous case)
Also, when c = a
2a + b = 30 --> 31 - 1 = 30 cases (as a = b = c = 20 case is already counted)
Total = 31 + 30 + 30 = 91 cases
So we have 1891 - 91 = 1800 cases where a, b and c are different.
Now we need to find only a > b > c ≥ 0 cases, the best method is already explained in shashank's solution.
Will give something traditional to find the cases satisfying a > b > c (reference - math.stackexchange)
let say d = a - c and e = b - c
as a > b > c, we can say d > e > 0
d + e = a + b - 2c = a + b + c - 3c = 60 - 3c
d + e = 60 - 3c
So the number of cases possible are [(60 - 3c - 1)/2]
c can be anything from 0 to 19 (19 = [60/3] - 1).
(a, b, c) cases would be (31, 29, 0) to (21, 20, 19)
So our required answer is Summation of [(60 - 3c - 1)/2] where c = 0 to 19
=> 29 + 28 + 26 + 25 + 23 + 22 + .... + 2 + 1
which is basically 1 + 2 + 3 + 4 ... + 30 - (3 + 6 + 9 + .... + 30)
= 30 x 31/2 - 5 x (3 + 30) = 15 x 31 - 15 x 11 = 15 x 20 = 300
As you can see, traditional method will take away a good chunk of your time. Hence it's best to go with the method explained by shashank or else you can mug up the below formula.
Number of possible solutions for (a, b, c) when a + b + c = N and a > b > c ≥ 0 is [(n^2 + 6)/12]
So here N = 60, so possible cases for a > b > c ≥ 0 is [(3606/12)] = [300.5] = 300
Also, if we need to find out for a ≥ b ≥ c ≥ 0, then just add [N/2] + 1 to the above equation.
Find the possible solutions for a + b + c = 100 if
a) a > b > c ≥ 0
solution : [(100^2 + 6)/12] = 833
b) a ≥ b ≥ c ≥ 0
solution : 833 + [100/2] + 1 = 884
Sonam Priya last edited by zabeer
@shashank_prabhu Sir how do we get this 3 * 3 = 9 * x
zabeer last edited by
@Sonam-Priya it is called chord theorem.
This theorem states that A × B is always equal to C × D no matter where the chords are.