# IPMAT Question Paper 2020 | IPM Indore Quants

###### IPMAT Sample Paper | IPMAT Question Paper | Question 23

IPMAT 2020 Question Paper IPM Indore Quantitative Ability. Solve questions from IPMAT 2020 Question Paper from IPM Indore and check the solutions to get adequate practice. The best way to ace IPMAT is by solving IPMAT Question Paper. To solve other IPMAT Sample papers, go here: IPM Sample Paper

Question 23 : A 2 × 2 matrix is filled with four distinct integers randomly chosen from the set {1,2,3,4,5,6} Then the probability that the matrix generated in such a way is singular is

1. $$frac{2}{45}$ 2. $\frac{1}{45}$ 3. $\frac{4}{15}$ 4. $\frac{1}{15}$ ## Best CAT Online Coaching Try upto 40 hours for free Learn from the best! #### 2IIM : Best Online CAT Coaching. ## Best CAT Coaching in Chennai #### CAT Coaching in Chennai - CAT 2022Limited Seats Available - Register Now! ### Explanatory Answer A matrix is singular if and only if the determinant of the matrix is 0. In a 2 × 2 matrix, $\left|\begin{array}{cc}a & c \\d & b\end{array}\right| \\$ = ab - cd = 0 Or ab = cd If one of a,b,c or d is 5, then ab ≠ cd, because in the set {1,2,3,4,5,6} there is only one multiple of 5. So, we just have the set {1,2,3,4,6} to deal with and out of these we'll have select 4 integers. We can't pick either 3 or 6. We'll have to pick them both. That is 3 and 6 have to stay on either side of the equation ab = cd So, the equation now becomes, 3b = 6d Clearly, b > d {b, d} ⊂ {1, 2, 4} If b = 4, d = 2 If b = 2, d = 1 So, only two such combinations exist. {a, b, c, d} = {1, 2, 3, 6} or {a, b, c, d} = {4, 2, 3, 6} We can select four numbers from the six in 6C4 = 15 ways After selecting the 4 integers, we can arrange them in order in 4! ways. So, the total number of matrices that can be constructed = 15 × 4! Once we select {a, b, c, d} = {1, 2, 3, 6} or {a, b, c, d} = {4, 2, 3, 6}, To construct a singular matrix, for example, 2 & 3 should be on the same diagonal. Similarly, 4 & 3 too in case of {a, b, c, d} = {4, 2, 3, 6}. Only one-third of the arrangments yeild a singular matrix and the others dont. But the probabilty of getting a singular matrix = $\frac{2 × 4! ÷ 3}{15 × 4!}$ = $\frac{2}{45}$ The question is " A 2 × 2 matrix is filled with four distinct integers randomly chosen from the set {1,2,3,4,5,6} Then the probability that the matrix generated in such a way is singular is " ##### Hence, the answer is $\frac{2}{45}$ Choicve A is the correct answer. ###### Best Indore IPM & Rohtak IPM CoachingSignup and sample 9 full classes for free. Register now! ###### Already have an Account? ##### 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 / 94459 38484
WhatsApp: WhatsApp Now
Email: info@2iim.com