Quant Boosters - Shashank Prabhu, CAT 100 Percentiler - Set 6

• Q27) Two couples and one single person occupy a row of five chairs at random. What is the probability that neither couple sits together (the husband and the wife should not occupy adjacent seats)?
(A) 1/5
(B) 1/3
(C) 3/8
(D) 2/5
(E) 1/2

• Let A & B be one couple. C & D be the other couple. Let X be the single person.
Case 1: One couple always sits together. Let AB be that couple. Now, we have to make sure that C&D don't sit next to each other. So arrangements will be:
A B _ X _
_ A B X _
_ A B _ X
X _ A B _
_ X A B _
_ X _ A B
For each of these six cases, A&B can exchange their positions. And C&D can also exchange positions.
Hence, total ways = 6 * 2 * 2 = 24
similarly, these 24 cases will be present for CD as the couple sitting together.
So 24 + 24 = 48 ways so far when any one couple sits together.

Case 2: Both the couples sit next to each other.
A B C D X
A B X C D
X A B C D

For each of these cases, A B can be replaced with B A, C D can be replaced with D C. And the positions where we have fixed AB and CD can be exchanged. Hence, we have 3 * 2 * 2 * 2 ways = 24 ways.

In total, we have 72 cases where at least one couple sits together.

Total ways of arranging 5 people = 5! = 120
120 - 72 = 48
48/120 = 2/5

• Q28) In a temple there are some magical bells which tolls 18 times in a day, simultaneously. but every bell tolls at a different interval of time, but not in fraction of minutes. the max number of bells in the temple can be
a) 18
b) 10
c) 24
d) 6

• 18 times in a day. So 18 times in 24 * 60 minutes. 24 * 60/18=80. Number of factors of 80 is 10. So, ten bells tolling after every 1, 2, 4, 5, 8, 10, 16, 20, 40, 80 minutes.

• Q29) A contractor expected to complete a certain task with 120 workers. However, starting from the second day, every four workers dropped out. Due to this, the time took 4 more days than actually scheduled to be. In how many days, the work was completed?

• 120+116+...120-(n+3)4=120n
(n+4)(228-4n)=240n
212n-4n^2+912=240n
4n^2+28n-912=0
n^2+7n-228=0
(n+19)(n-12)=0
n=12
New case is done in (n+4) days so 16.

• Q30) Reminder when 3^3^3 + 7^7^7 is divided by 5

• 3^3^3 mod 5 = 3^27 mod 5 = 3^3 mod 5 = 2
7^7^7 mod 5 = 2^7^7 mod 5
Consider 7^7 mod 4 = -1 = 3
2^3 mod 5 = 3
2+3 mod 5 = 0

Alternatively, 3^3^3 will end in 7 and 7^7^7 will end in 3.. total will end in 0 and so divisible by 5

• @shashank_prabhu part of solution is missing so unable to understand

• @Naman-Jain-0

Number of ways to get a sum greater than 17 = Total - Number of ways to get a sum lesser than or equal to 17

6 dice and each dice can give 6 options. Total cases = 6^6

Now we will find the number of ways to get a sum lesser than or equal to 17
a + b + c + d + e + f ≤ 17
a , b, c ... f can take values from 1 to 6

Because of ≤, add a dummy variable. (say, g)
so we have a + b + c + d + e + f + g = 17

Now we will discuss a very important point. a, b, c .. f are all natural numbers and for g - 0 is possible. Means this is neither in case 1 or case 2. To resolve this let subtract 1 from a, b, c .. f so that they can be given a value of 0 also. Our equation becomes

a + b + c + d + e + f +g = 17 - 6 = 11

Concept : a + b + c ... (k terms) = n has (n + k - 1) C (k - 1) non negative integer solutions

Number of ways = (11 + 7 - 1) C (7 - 1) = 17C6

now we have to remove cases where a > 6.
say a = A + 6
So, A + b + c + d + e + f + g = 11 - 6 = 5
Number of ways which a can be higher than 6 (invalid cases for a) = 11C6 ways.
6 x 11C6 invalid cases if we include for b, c, d, e and f also.

So Final - 17C6 - 6 x 11C6

Now coming back to the original question,
Number of ways to get a sum greater than 17 = Total - Number of ways to get a sum lesser than or equal to 17
= 6^6 - (17C6 - 6 x 11C6)

If you need a clear understanding of this concept, refer the below article. Happy Learning!

Solving Combinatorics Problems Using Stars & Bars Concept

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