@iLoveTorres said:loss of ~26%
loss of RS 1000
@Asfakul said:unfortunately I dont have OA . can you please explain your steps ?
my bad. made a mistake. it CBD.
1/m = a , 1/w = b
let 3 men and 1 women can do the work in k days ( k is less than 20)
2a + 3b = 1/20
3a + 1b = 1/k (say)
multiply 1st by 2 and subtract the equations
a + 5b = 1/10 - 1/k
>20
@vbhvgupta said:Labor allocation is a very important process. A particular weaving section has 20 hours looms and with five laborer €™s loom efficiency is 75% and production of a loom at 100% efficiency is 10 meter per hour. Salary of a laborer is Rs. 11000 per month. I removed one laborer due to which efficiency came down to 70%. How much did I gain or lose due to this action? Assume that profit on one meter cloth is Rs. 4 and looms are working for 30 days in a month and 10 hours per day.
bhai mera approach mein ya gadbadi hai batado..
20 hour loom, 5 worker, 10 hours per day, at 100% efficiency 10mts/hr so at 75% efficiency 7.5mts/hour, 30 days, profit = 4/mt
case 1) 20*5*10*7.5*30*4=900000-5*11000=845000
case 2) 20*4*10*7*30*4=672000-4*11000=628000
difference is no where 1000rs
20 hour loom, 5 worker, 10 hours per day, at 100% efficiency 10mts/hr so at 75% efficiency 7.5mts/hour, 30 days, profit = 4/mt
case 1) 20*5*10*7.5*30*4=900000-5*11000=845000
case 2) 20*4*10*7*30*4=672000-4*11000=628000
difference is no where 1000rs
@maroof10 said:Ram has to weigh 6 distinct packets.He weighs them 4 at a time,weighing all the possible combinations of the packets from the six.The average weight of all the weighing combinations is found to be 500 gm. What is the combined weight of all the six packets?
is it 1500 ?? by any chance .
@iLoveTorres said:bhai mera approach mein ya gadbadi hai batado..20 hour loom, 5 worker, 10 hours per day, at 100% efficiency 10mts/hr so at 75% efficiency 7.5mts/hour, 30 days, profit = 4/mtcase 1) 20*5*10*7.5*30*4=900000-5*11000=845000case 2) 20*4*10*7*30*4=672000-4*11000=628000difference is no where 1000rs
let the units of work be 20
in a day men do m units
women do w units
2m+3w=1.........................................................................(1)
3m+w>1............................................................................(2)
2*(1) - (2)
m+5w=2- (>1)
Now u dont now how much big " greater than 1" is
OA CBD
@techgeek2050 said:my bad. made a mistake. it CBD.1/m = a , 1/w = blet 3 men and 1 women can do the work in k days ( k is less than 20)2a + 3b = 1/203a + 1b = 1/k (say)multiply 1st by 2 and subtract the equationsa + 5b = 1/10 - 1/k it CBD.
bro take the total units of work as 180.
So 2M + 3W = 9 units/day
W can be 1 or 3. when W = 1, M = 3; W=3 , M = 0
Now 3M + 1W should do more than 9 unit of work according to the question.
When W=1 and M=3 the total unit of work done would be 10 which satisfies the condition. the second set of values dont satisfy the given condition.
Now check for M + 5W = 3+5=8 which is less than 9. hence it will take more than 20 days
So 2M + 3W = 9 units/day
W can be 1 or 3. when W = 1, M = 3; W=3 , M = 0
Now 3M + 1W should do more than 9 unit of work according to the question.
When W=1 and M=3 the total unit of work done would be 10 which satisfies the condition. the second set of values dont satisfy the given condition.
Now check for M + 5W = 3+5=8 which is less than 9. hence it will take more than 20 days
@maroof10 said:Ram has to weigh 6 distinct packets.He weighs them 4 at a time,weighing all the possible combinations of the packets from the six.The average weight of all the weighing combinations is found to be 500 gm. What is the combined weight of all the six packets?
750 gm?
@iLoveTorres said:bro take the total units of work as 180.So 2M + 3W = 9 units/dayW can be 1 or 3. when W = 1, M = 3; W=3 , M = 0Now 3M + 1W should do more than 9 unit of work according to the question.When W=1 and M=3 the total unit of work done would be 10 which satisfies the condition. the second set of values dont satisfy the given condition.Now check for M + 5W = 3+5=8 which is less than 9. hence it will take more than 20 day
bhai, in the end u'll get
M + 5W = 0.1 - (>0.05)
(if M = 1/m and W = 1/w)
answer should be >20
@iLoveTorres said:1500 toh ho hi nahi sakta.. i guess 625 hona chahiye
may be . How many times a particular box will be repeated if we take 4 box out of 6?
all total 15 combinations are possible .
all total 15 combinations are possible .
@techgeek2050 said:bhai, in the end u'll get M + 5W = 0.1 - (>0.05) (if M = 1/m and W = 1/w)so we cannot say if it will be greater or less than 0.05
you are subtracting a value greater than .05 from .1 toh obviously it will be less than .05..
@Asfakul said:may be . How many times a particular box will be repeated if we take 4 box out of 6? all total 15 combinations are possible .
i guess 12 times
@iLoveTorres said:you are subtracting a value greater than .05 from .1 toh obviously it will be less than .05..
ya ryt! 
@maroof10 said:@piyushrohella12 yes...what is the approach?
takin 4 out of 6 means 6c4=15
now when u make all the poss comb of say
A B C D E F ( u r taking 4 out of 6)
how many A/B/C/D/E/F is repeated
5c3= 10
so every item is repaeted 10 tymes that mean 10 * sum of all items/15= 500
sum of all= 750
@maroof10 said:Ram has to weigh 6 distinct packets.He weighs them 4 at a time,weighing all the possible combinations of the packets from the six.The average weight of all the weighing combinations is found to be 500 gm. What is the combined weight of all the six packets?
every packets will come 10 times in the combination with other packets.
and there will be 15 total combination.
so , 10(p1+p2+....+p6)/15 = 0.5
OA = 750 gm
@vbhvgupta said:Arun has to pay off his debt by paying Rs. 2000 in the first month and subsequently, has to keep repaying 80% of the amount he paid in previous month until he clears the total debt. The approximate amount of his debt is:
shdnt it b CBD
2000 + 2000(0.8) + 2000(0.8)^2 +.....i can say 1 month mei hi clear ho gya debt may b 2 month
do u have a solution given in the buk
@piyushrohella12 said:takin 4 out of 6 means 6c4=15now when u make all the poss comb of sayA B C D E F ( u r taking 4 out of 6)how many A/B/C/D/E/F is repeated5c3= 10so every item is repaeted 10 tymes that mean 10 * sum of all items/15= 500sum of all= 750
If we generalize it - is it something like C(n-1,r-1) n= total number of box and r= number of box taken at a time ?