CAT 2017 Question Paper | Quants Slot 1

This is a question from CAT permutation and combination. This question is more like a puzzle than a question. But an interesting puzzle nevertheless. We have to find the positive integral solutions of a set of variables. But there's a constraint that governs these variables. CAT Permutation and Combination and Probability is an important topic in theCAT Exam. One can expect to get 2-3 questions from this topic in CAT exam.

Question 25 : The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is

1. 101
2. 99
3. 87
4. 105

Video Explanation

Explanatory Answer

Method of solving this CAT Question from Permutation and Combination

We have to find the number of solutions (x, y, z) to the given equation.
x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12.
x – y – z = 25, so the maximum value y and z can take is 12 and x can take is 40.
From 40 we can subtract some values to get 25 i.e.15 can be subtracted,
If x = 40 then y + z = 15
y ≤ 12 so y can take all the corresponding values 12,11,10,......3 it is a set of 10 values.
z ≤ 12, so z can take corresponding other values to get with 15 when added with x
Similarly,
If x = 39, then y + z = 14 here y can take all the value from 12 ,11,...till 2
So there are 11 values.
If x = 38, then y + z = 13 here y can take all values from 12,11,10,....till 1
So there are 12 values.
If x = 37 then y + z = 12 here y can take 11,10,9,.....,1 so there are 11 values
If x = 36 then y + z = 11 here y can take 10,9,......,1 so there are 10 values
If x = 35 then y + z = 10 here y can take 9,8,.......,1 so there are 9 values
If x = 34 then y + z = 9 here y can take 8,.......,1 so there are 8 values
If x = 33 then y + z = 8 here y can take 7,........,1 so there are 7values
If x = 32 then y + z = 7 here y can take 6,........,1 so there are 6 values
If x = 31 then y + z = 6 here y can take 5.........,1 so there are 5 values
If x = 30 then y + z = 5 here y can take 4,......,1 so there are 4 values
If x = 29 then y + z = 4 here y can take 3,...,1 so there are 3values
If x = 28 then y + z = 3 here y can take 2,1 so there are 2 values
If x = 27 then y + z = 2 here y can take 1 so there is 1 value
x cannot be less than 27 because it is given that y and z are positive integers so it has to be at least one cannot be less than one
So, \\frac{12 × 13}{2}) + 21 = 78 + 21 = 99.
The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is 99.

The question is "The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is"

Hence, the answer is 99

Choice B is the correct answer

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