its 1/2
Hey guys !1 i am new to this thread !!
@Tusharrr said:@chillfactor 15 IS not answer
Oops, we can select the person who has not been asked to speak in 5 ways
So probability will become 5/14
=> a + b = 19
This one from the vault, CAT2001
If a,b,c,d are four positive real numbers such that abcd=1, what is the minimum value of (1+a)(1+b)(1+c)(1+d)?
1.4
2.1
3.16
4.18
1.4
2.1
3.16
4.18
Do you need the OA?
@junefever I think the answer is 16 . Is it correct ?
7 Indian 4 chinese finish a job in 5 days. 7 japanese and 3 chinese finish the same job in 7 days. Given that the efficiency of each person of a particular nationality is same but different from others. 1 Indian 1 chinese and 1 japanese will complete the work in
@junefever said:This one from the vault, CAT2001If a,b,c,d are four positive real numbers such that abcd=1, what is the minimum value of (1+a)(1+b)(1+c)(1+d)?1.4 2.13.164.18Do you need the OA?
(1+a) >= 2_/a
Similarly for others..
So, (1+a)(1+b)(1+c)(1+d) >= 2^4 _/abcd = 16 ..
@Tusharrr
the answer is 5/14..used a very simple visualization to solve it...
Basically they are divided into groups of 2 with no repetition....and there are 5c2 such combinations of people, each of the 4 weeks can be allocated any one of these pairs, but the pairs cant be reused.
so total possibilities are 10*9*8*7 = denominator.
Now we see that every person would be repeated in 4c1, groups after which we have 6 groups to choose from, and the person to be excluded can be chosen in 5 ways. Note that we cannot exclude 2 people as that would only give us 3c2=3 combinations and ..we need at least 5.
So basically choose the person to be excluded, then allocated the remaining groups to each week, ensuring that groups cannot be reused. Hence we get:
(5*6*5*4*3)/(10*9*8*7) = 5/14 as the answer.
the answer is 5/14..used a very simple visualization to solve it...
Basically they are divided into groups of 2 with no repetition....and there are 5c2 such combinations of people, each of the 4 weeks can be allocated any one of these pairs, but the pairs cant be reused.
so total possibilities are 10*9*8*7 = denominator.
Now we see that every person would be repeated in 4c1, groups after which we have 6 groups to choose from, and the person to be excluded can be chosen in 5 ways. Note that we cannot exclude 2 people as that would only give us 3c2=3 combinations and ..we need at least 5.
So basically choose the person to be excluded, then allocated the remaining groups to each week, ensuring that groups cannot be reused. Hence we get:
(5*6*5*4*3)/(10*9*8*7) = 5/14 as the answer.
@vbhvgupta said:7 Indian 4 chinese finish a job in 5 days. 7 japanese and 3 chinese finish the same job in 7 days. Given that the efficiency of each person of a particular nationality is same but different from others. 1 Indian 1 chinese and 1 japanese will complete the work in
245/12 days imo
ScareCrow28
can u solve this
There are N ordered integer quadruples (a,b,c,d) subject to 0≤a,b,c,d≤99990 such that ad−bc≡1 (mod 99991). What are the last three digits of N?
@vbhvgupta said:7 Indian 4 chinese finish a job in 5 days. 7 japanese and 3 chinese finish the same job in 7 days. Given that the efficiency of each person of a particular nationality is same but different from others. 1 Indian 1 chinese and 1 japanese will complete the work in
W = n(I + J + C) ; n = ?
W = 5(7I + 4C) = 7(7J + 3C)
=> 35I + 20C = W
=> 49J + 21C = W
From here, C = 35I - 49J
W = 5(7I + 4*35I - 4*49J) = 35(I + 20I - 28J)
= 245(3I - 4J) = n(36I - 48J) = 12n(3I - 4J)
We can see that 12n = 245
Hence, n = 245/12
I hope calc mistakes didn't creep in
a piece of work can be completed by 10 men and 6 women in 18 days. Men works 9 hours per day while women work 7.5 hours per day. Per hour efficiency of a woman is 2/3 of man efficiency. In how many days 10 men and 9 women complete the workk.
@vbhvgupta said:a piece of work can be completed by 10 men and 6 women in 18 days. Men works 9 hours per day while women work 7.5 hours per day. Per hour efficiency of a woman is 2/3 of man efficiency. In how many days 10 men and women complete the workk.
W = 18(10m + 6w) = 18(10*9 + 6*7.5*2/3) = 18(90 + 30) = 18*120
I am assuming that to be 10 men and 10 women since its not clear
No of days required = 18*120/(90+50) = 18*120/140 = 108/7 days
Edit: Question changed so the answer..
No of days acc to 10 men and 9 women = 18*120/(90+45) = 18*120/135 = 16..
@vbhvgupta said:a piece of work can be completed by 10 men and 6 women in 18 days. Men works 9 hours per day while women work 7.5 hours per day. Per hour efficiency of a woman is 2/3 of man efficiency. In how many days 10 men and women complete the workk.
108/7...
___/\___
@ScareCrow28 said:W = 18(10m + 6w) = 18(10*9 + 6*7.5*2/3) = 18(90 + 30) = 18*120I am assuming that to be 10 men and 10 women since its not clearNo of days required = 18*120/(90+50) = 18*120/140 = 108/7 days
yar its 9 women
@erm said:What is the remainder if the sum 2012^2013+2013^2012 is divided by 2012.2013?
2013+2012= 4025
@vbhvgupta said:a piece of work can be completed by 10 men and 6 women in 18 days. Men works 9 hours per day while women work 7.5 hours per day. Per hour efficiency of a woman is 2/3 of man efficiency. In how many days 10 men and 9 women complete the workk.
16 days
regards
scrabbler
@vbhvgupta said:a piece of work can be completed by 10 men and 6 women in 18 days. Men works 9 hours per day while women work 7.5 hours per day. Per hour efficiency of a woman is 2/3 of man efficiency. In how many days 10 men and 9 women complete the workk.
My approach was different.
1 woman = 2/3 * 7.5/9 = 5/9 man. So 5 men = 9 women
So in first case 10 men + 6 women = 18 + 6 = 24 women take 18 days
In second case 10 men + 9 women = 18 + 9 = 27 women take 8/9 of 18 = 16 days
regards
scrabbler