Production Pattern for the no.of units (in cubic feet) per day:days No.of units1 1502 1803 1204 2505 1606 120 7 150For a truck that can carry 2000 cubic feet,hiring cost per day is 1000.Storing cost per cubic feet is Rs 5 per day.Any residual material left at the end of the 7th day has to be transferred.Q1. If all units are to be sent to market,then on which days should the trucks be hired to minimize cost :A. 2,4,6,7B.7C.2,4,5,7D.NoneCorrect : AQ2.If the storage cost is reduced to Rs 0.9 per cubic feet per day,then on which day/days should the truck be hired.A.4B.7C.4 and 7D.NoneCorrect : BI want the procedure.Thanks in advance.
No truck after day 1 means 150*5=750 storage cost, which is less than 1000, so no truck today......after day 2, (150+180)*5 which is greater than 1000, so lets get the stuff out by a truck today....similarly day 3, 120*5 is
Q2. day 4 storage cost .9*(150+180+120+250) which is less than 1000 so no truck...6th day storage cost .9(150+180+120+250+160+120)=.9*980 which is still less than 1000, so lets get everything out on the 7th day......
A shopkeeper makes a profit of Q% by selling an object for rs.24/-. Had the cost price and selling price been interchanged ,it would have led to a loss of 62.5Q%.With the latter cost price , what would be the new selling price to get a profit of Q%n
this is q.16 Lod 2 profit and loss can someone help me understand how to get a solution to this.
How many natural numbers less than 100 have exactly 3 prime factors?2458None of theseHow'd you approach this? Do you include 1 as one of the three prime factors?
to have three factors it should be the square of a prime number.. So the numbers falling in this range are 4,9,25,49. Do let me kno the OA..
A shopkeeper makes a profit of Q% by selling an object for rs.24/-. Had the cost price and selling price been interchanged ,it would have led to a loss of 62.5Q%.With the latter cost price , what would be the new selling price to get a profit of Q%nthis is q.16 Lod 2 profit and losscan someone help me understand how to get a solution to this.
let the cost price be "x" and the selling price by 24.. so profit% will be 24-x/x = Q Also if the CP and SP are interchanged then 24 will be the cost price and x will be the selling price. Loss% is given by 24-x/24 = 62.5Q/100
But from the first condition 24-x= xQ/100 substitute for second case xQ/24= 62.5Q/100 ==> x= 1500/100 = 15 So cost price is 15 selling price is 24 therefore profit percentage is 60% Now when the CP is 24 profit percentage 60% substitute and get the answer as 38.4
Well the answer is 5. Can't a number like 2*7*1 be possible? Why only prime squares?
sorry i read the question wrongly.. according to the question your logic is correct So the possible numbers will be (1 is not a prime number ) 2*3*5,2*3*7,2*3*11,2*3*13,2*5*7 Hence 5
sorry i read the question wrongly.. according to the question your logic is correctSo the possible numbers will be (1 is not a prime number )2*3*5,2*3*7,2*3*11,2*3*13,2*5*7Hence 5
Ah right. Thanks man.:)
One more question:
How many divisors/factors of 1200 end with only one €˜zero €™?
Ah right. Thanks man.One more question:How many divisors/factors of 1200 end with only one €˜zero €™?1614121011
answer is 10...look when you factorize it you will get 1200=2^4*3*5^2 So to get a single zero any multiplication of 2 with 5 wil give you single zero. so total 4 possibility next 5^2*2 = 50 5*3*2^x here x can be any of the four values 1,2,3,4. Hence 4 posssibilities next 5^2*3*2 = 150 hence 1 possibility.
answer is 10...look when you factorize it you will get 1200=2^4*3*5^2So to get a single zero any multiplication of 2 with 5 wil give you single zero. so total 4 possibilitynext 5^2*2 = 50 5*3*2^x here x can be any of the four values 1,2,3,4. Hence 4 posssibilitiesnext 5^2*3*2 = 150 hence 1 possibility.is the answer right?
The answer is apparently 12.
Tell me if this is right: Like you said, 1200=2^4*3*5^2
So to make the divisors end in one zero we can write this: (2^3*3*5)*10
So, now we cannot allow 2 and 5 to multiply inside the bracket. So, factors possible with 2 &3 will be (3+1)*(1+1)=8
And factors possible with 3 & 5 will be (1+1)*(1+1)= 4
So 8+4= 12. I think this is right. Not very sure though.
The answer is apparently 12.Tell me if this is right: Like you said, 1200=2^4*3*5^2So to make the divisors end in one zero we can write this: (2^3*3*5)*10So, now we cannot allow 2 and 5 to multiply inside the bracket. So, factors possible with 2 &3 will be (3+1)*(1+1)=8And factors possible with 3 & 5 will be (1+1)*(1+1)= 4So 8+4= 12. I think this is right. Not very sure though.
Lovely. A small permutation doubt:There are 6 pups and 4 cats.In how many ways can they be seated in a row so that no cats sit together?Answer: 6!*P(7, 4)Why is it P(7,4) and not C(7,4)? The cats seem the same, why do we bother with order here?
in that case all the pups are also same why 6! we shud consider all the pups and cat as distinct