Quant Capsules by Shashank Prabhu - Set 18
CAT 100%iler, 5 times AIR 1, Director - Learningroots, Ex ITC, Pagalguy, TAS
A watch loses 2/3% time during the 1st week and gains 1/3% time during next week. If on a Sunday noon, it showed the right time, what time will it show at noon on the Sunday after the next?
(A) 11:26:24 a.m
(B) 10:52:48 a.m
(C) 10:52:18 a.m
(D) 11:36:24 a.m
86400 * 7 * 2/300= 14 * 288=loss of 4032 seconds
86400 * 7 * 1/300= 7 * 288=gain of 2016 seconds
Overall loss of 2021 seconds. So, 2016/60=33 minutes and 36 seconds
From the word INSTITUTE. How many words can be formed using 5 alphabets?
I(2), N, S, T(3), U, E
5d-6c5-6 ways-6 * 5!-720 cases
3d2s-2c1 * 5c3-20 ways-20 * 5!/2!-1200 cases
2d3s-5c2-10 ways-10 * 5!/3!-200 cases
2s2s1d-1 * 4-4 ways-4 * 5!/2!2!-120 cases
3s2s-1 way-5!/2!3!-10 cases
Total 2250 cases
Find the remainder when 209! divided by 422
Chinese remainder theorem.
If the divisor N can be expressed in the form of a*b where a and b are coprime, then the smallest non negative integral solution to the equation obtained by dividing the original dividend by a and b should give you the remainder when the dividend is divided by N.
209! Mod 211 is 1
209! Mod 2 is 0
Remainder is 212
In WIMWI, each student is on two student committees and any two student committees have exactly one student in common. There are six student committees. How many students are there in WIMWI?
2 committees from 6 can be selected in 6c2 = 15 ways and so, that would be the number of students common to exactly two committees.
Find the sum of the last three digits of s where s=871 * 873 * 875 * 878 * 881 * 883
Last 3 digits are nothing but the remainder when divided by 1000. So basically everything gets cancelled out and you have to find the remainder when divided by 4 which comes to be 1. But, as you had cancelled out 250 from the numerator and the denominator, you have to multiply it back with the original remainder. So, the final remainder is 250.
A can do a piece of work in 10 days; B in 15 days. They work for 5 days. The rest of the work was finished by C in 2 days. If they get 1500 Rs for the whole work, the daily wages of B and C are?
Let the total work be 150 units. A does 15 units/day and B does 10 units/day.
In 5 days, together they will do 5*(15+10) = 125 units. Remaining units are done by C = 25 units.
They will get paid in the ratio of work done = A:B:C = 75:50:25 = 750:500:250 in terms of Rs.
B gets 500 for 5 days = 100/day
C gets 250 for 2 days = 125/day
Total = 225/day
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)?
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
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.
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?
New case is done in (n+4) days so 16.
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.