CAT 2017 Question Paper | Quants Slot 1

CAT Previous Year Paper | CAT Permutation and Combination Questions | Question 25

The best questions to practice for CAT Exam are the actual CAT Question Papers. 2IIM offers you exactly that, in a student friendly format to take value from this. If you would like to take the same inside a testing engine (for Free) head out here: CAT Official Question Mocks. In CAT 2017 we saw some beautiful questions that laid emphasis on Learning ideas from basics and being able to comprehend more than remembering gazillion formulae and shortcuts. Original CAT Question paper is the best place to start off your CAT prep practice. This page provides exactly that. To check out about 1000 CAT Level questions with detailed video solutions for free, go here: CAT Question Bank

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

2IIM : Best Online CAT Coaching


2IIM's Online CAT Coaching
Get CAT Last Mile Prep Course for 799 /-
CAT Online Coaching


Video Explanation


Best CAT Coaching in Chennai


CAT Coaching in Chennai - CAT 2020
Starts Sat, November 2nd, 2019


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

 

CAT Questions | CAT Quantitative Aptitude

CAT Questions | CAT DILR

CAT Questions | Verbal Ability for CAT


Where is 2IIM located?

2IIM Online CAT Coaching
A Fermat Education Initiative,
58/16, Indira Gandhi Street,
Kaveri Rangan Nagar, Saligramam, Chennai 600 093

How to reach 2IIM?

Phone: (91) 44 4505 8484
Mobile: (91) 99626 48484
WhatsApp: WhatsApp Now
Email: prep@2iim.com