Hello folks, hope you have been tirelessly prepping for IIFT, which is just around the corner. Hope our Live Sessions are helping you out a great deal as far as General Awareness is concerned. We also thought we would share a few tips and tricks in Quant with you along with some facts and figures. Here’s a look at the trend of the questions in the Quantitative Aptitude section for the previous 3 years: (’15, ’14, ’13)
Today we shall check out some of the concepts used in Probability, Permutations and Combinations: (from the previous year question papers '15 and '14)
1. The internal evaluation for Economics course in an Engineering programme is based on the score of four quizzes. Rahul has secured 70, 90 and 80 in the first three quizzes. The fourth quiz has ten True-False type questions, each carrying 10 marks. What is the probability that Rahul’s average internal marks for the Economics course is more than 80, given that he decides to guess randomly on the final quiz?
(1) 12/1024
(2) 11/1024
(3) 11/256
(4) 12/256
What can be immediately deciphered from the options: the total number of ways the quiz can be solved= 2^10=1024
Most Important point to remember from the question: Rahul needs to score MORE than 80 (M>80) and NOT greater than or equal to. Some may commit this mistake and get the answer wrong. Here's the solution:
Average marks of the first three quizzes =80
So, for Rahul to have average internal marks more than 80, he has to score more than 80 marks in the last quiz.
This is possible that he attempts 10 questions or 9 questions correctly.
Number of ways of doing 10 questions correctly: 10C10=1
Number of ways of doing 9 questions correctly: 10C9=10
Adding both we get: 11
The answer hence: 11/1024
2. During the essay writing stage of MBA admission process in a reputed B-school, each group consists of 10 students. In one such group, two students are batchmates from the same IIT department. Assuming that the students are sitting in a row, the numbers of ways in which the students can sit so that the two batchmates are not sitting next to each other, is:
(1) 3540340
(2) 2874590
(3) 2903040
(4) None of the above
Most Important point to remember from the question: The number of ways in which two students can ALWAYS sit together=(n-1)!x2
(Here it becomes 9!x2)
Approach:
Number of ways in which two students can NEVER sit together=No. of ways in which all the students can sit without any restriction-no of ways in which two students can ALWAYS sit together
The answer hence: 10!-9!x2=10x9!-9!x2=9!x8=2903040
Q-Trick 1: If there was NO 'none of the above' option, you could have just gone with Option 3 directly WITHOUT calculation as (3) is the ONLY multiple of 9.
3. In the board meeting of an FMCG Company, everybody present in the meeting shakes hand with everybody else. If the total number of handshakes is 78, the numbers of members who attended the board meeting is:
(1) 7
(2) 9
(3) 11
(4) 13
As this is an easy question we shall directly go with the approach:
nC2=n(n-1)/2=78,
Hence the answer :n=13
4. In a reputed engineering college in Delhi, students are evaluated based on trimesters. The probability that an Engineering student fails in the first trimesters is 0.08. If he does not fail in the first trimester, the probability that he is promoted to the second year is 0.87. The probability that the student will complete the first year in the Engineering college is approximately:
(1) 0.8
(2) 0.6
(3) 0.4
(4) 0.7
Approach and Solution:
We consider that the student fails in the first year if he fails in the first trimester. Therefore, the probability that the student will complete the first year in the Engineering College is approximately = Probability that he passes 1st trimester x probability that he passes 1st semester and is promoted to the second year) = 0.9 x 0.87 = 0.8 (Approx)
Hence, the answer: 0.8
5. The total number of eight – digit landline telephone numbers that can be formed having at least one of their digits repeated is:
(1) 98185600 (2) 97458800 (3) 100000000 (4) None of the above
The total number of 8-digit landline telephone numbers that can be formed having at least one of their digits repeated = The total number of 8-digit landline numbers – The number of8-digit landline numbers in which no digit is repeated.
The total number of 8-digit landline numbers = 10^8(10 Raised to the Power of 8)
The number of 8-digit landline numbers in which no digit is repeated = 10P8= 1814400
Number of required landline numbers = 10^8-181400=98185600 (Hence the answer)
There are TWO answers possible for this question: Why?
Because there is no telephone number which would be 00000000. Hence, a smart student would try to subtract 1 from Option 1 which is the answer. Usually, the question setter DOES NOT give you a 'none of these' as an option. If they do, please go with your original answer. Do not think practically!
6. In the MBA Programme of a B – School, there are two sections A and B. 1/4th of the students in Section A and 4/9th of the students in section B are girls. If two students are chosen at random, one each from section A and Section B as class representative, the probability that exactly one of the students chosen is a girl, is:
(1) 23/72 (2) 11/36 (3) 5/12 (4) 17/36
Approach and Solution:
The probability that out of the two chosen students, one from each section, exactly ONE of the chosen students is a girl
= 1/4 x 5/9 + 4/9 x 3/4 =17/36
The answer hence: 17/36
7. In an Engineering College in Pune, 8 males and 7 females have appeared for Student Cultural Committee selection process. 3 males and 4 females are to be selected. The total number of ways in which the committee can be formed, given that Mr. Raj is not to be included in the committee if Ms. Rani is selected, is:
(1) 1960 (2) 2840 (3) 1540 (4) None of the above
Approach and Solution:
Selecting 3 males from 8 and 4 females from 7 can be done in 8C3 x 7C4 = 1960 ways
The most important part of the problem: Raj and Rani together cannot be in the committee.
Selection of both can be done in 7C2 x 6C3 = 420 ways
Required number of ways = 1960 – 420 = 1540 ways (The answer hence)
Question 8:
Conventional approach:
Alternative Approach/Time saving Approach:
Substitute a=6, (B: Black, R: Red, T-Total)
Bag 1: B:9, R:1, T:10
Bag 2: B:3, R:7, T:10
Bag 3: B:5, R:5, T:10
Bag 4: B:7, R:3, T:10
Probability of selecting one bag=1/4,
Probability of selecting one black ball from one bag=1/4 x 1/10 +1/4 x 3/10 + 1/4 x 5/10 + 1/4 x 7/10 = 2/5
Only choice d gives 2/5
9. Out of 8 consonants and 5 vowels, how many words can be made, each containing 4 consonants and 3 vowels?
(1) 700 (2) 504000
(3) 3528000 (4) 7056000
Approach and Solution:
Number of ways of selecting 4 consonants out of 8: 8C4=70 (A)
Number of ways of selecting 4 vowels out of 5=5C3= 10 (B)
Number of ways of arranging the 7 letters=7!=5040 (C)......Step 3
Hence the answer: A x B x C = 3528000,
Q-Trick 2: You can stop at Step 3 as there is ONLY ONE option that gives you the fourth digit 8 (from the last), you know there are going to be 3 zeroes only, that is also a clue
10. In a sports meet for senior citizens organized by the Rotary Club in Kolkata, 9 married couples participated in Table Tennis mixed double event. The number of ways in which the mixed double team can be made so that no husband and wife play in the same set, is
(1) 1512 (2) 1240
(3) 960 (4) 640
Approach and Solution:
Two men can be selected in 9C2 ways.
After selecting two men, two women can be selected in 7C2 ways from (9 – 2 = 7) women, so that no husband and wife play in the same set.
Also, these selected 4 people can be grouped in 2 ways
Hence the answer:
The total number of mixed double teams = 9C2 x 7C2 = 1512.
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