a can do a piece of work in 18days and B in 9 days and C in 6days.A and B start working together and after 2 days c join them.in how many days will the job be completed??
plz give the proper solution
my take
1 day's work for A = 1/18 1 day's work for B = 1/ 9 1 day's work for C = 1/ 6
work done in 1st 2 days :-
= (wrk of A for 1 day + wrk of B for 1 day ) * 2 = (1/18 + 1/9 ) * 2 = 6/18
Let total work be 1
so work left = 1- 6/18 = 12/18
Now C joins A+B :-
work done by all 3 in a day = (wrk of A + wrk of B + wrk of C) = 1/18 + 1/ 9 + 1/ 6 = 6/18
so number of days in which the remaining wrk will be completed = (12/18 )/(6/18 ) = 2 days
So, total number of days consumed = 2 + 2 = 4 days = Ans
Gyaan I am interpreting your variables as followed: Aw=Average weight = Total weight/Total Quantity-----------------(1) A1=Weight of 1 n1=Quantity of 1 A2=Weight of 2 n2=Quantity of 2
Assume
A1=Cost of cow, Say 'C', which is its fundamental weight n1=Profit on cow = 0.25 A2=Cost of calf = (1300-C) n2=Profit on calf =0.20
Using (1) above: Aw= Total Cost/Total profit = Total SP = (1+(23 1/13%))*1300
Start Solving (1+(23 1/13%))*1300=(0.25C+260-0.2C)/(0.25+0.20) Which will give C= 9200
Hence Solved
Bolbacchans First verify the answer and let me know if it is right or not. To apply the formula anywhere, just remember (1) and replace the variables.
when cost of calf+cow=1300 how come cow be 9200 bro...i got the same answer when i used allegations method...but answer is 800...what is bolbacchans bro??...i asked for allegation method bcs author had given that it can be used..i wanted to know how he uses it ...good that u dint buy that book...theory part is good but when it comes to problems part !! not that good as many tough problems are left unsolved...in examples only naive problems are solved
17.A dishonest grocer professes to sell pure butter at CP but he mixes it with adulterated fat and gains 25%.find the %age of adultr fat in the mixture which is freely available.
I wished to use the A1,n1.... method but thought of something simpler. Assume: Grocer has x% of pure butter in the mixture. So cost of x+25% profit =100% SP x+0.25x=1
Start Solving x=4/5 that is amount of pure butter So 1/5 is amount of the adultrating fat Hence Solved
22.there are 2 mixtures of honey and water, the quantity of honey in them being 25% and 75% of the mixture.if 2 gallons of the first are mixed with the 3 gallons of the second what will be the ratio of the honey to water in the new mixture?????
Here we'll use your A1,n1... method Assume A1=weight of honey in mixture 1 = 0.25 n1=Amount of mixture 1 = 2 gallons A2=weight of honey in mixture 1 = 0.25 n1=Amount of mixture 2 = 3 gallons
Aw=weight of honey in mixture/Gallon
Start Solving Aw=(2*0.25+3*0.75)/5 = 2.75 But wait! this is the weight of honey in mixture, weight of water = 5-2.75= 2.25 ratio of honey:water = 2.75/2.25
Hence Solved
cat2011 Says
hello frinds.
Hi cat2011,nice to see you r here early : pls contribute with your QnA wrt Quant.
when cost of calf+cow=1600(when selling) how come cow be 9200 bro...i got the same answer when i used allegations method...answer is 800...what is bolbacchans bro??...i asked for allegation method bcs author had given that can be used..i wanted to know how he uses it ...good that u dint buy that book...theory part is good but when it comes to problems part !! not that good as many tough problems are left unsolved...in examples only naive problems are solved
:splat: on me. you are right mate. How come I didn't notice it!!!!!!! Will give the method another try.
Bolbacchans are the quotes by losers that they give to the whole world. See, how perfectly it fits for me.
And I am really happy I didn't buy that book. Arun Sharma Will give method a try though.
Three runners A,B and C run a race, with runner A finishing 24 mts ahead of B and 36 mts ahead of C,while runner B finishes 16 mts ahead of C.Each runner travels the entire distance at constant speed. What was the lenght of the race? a) 72 mts b)96 mts c)120 mts d)144 mts
Three runners A,B and C run a race, with runner A finishing 24 mts ahead of B and 36 mts ahead of C,while runner B finishes 16 mts ahead of C.Each runner travels the entire distance at constant speed. What was the lenght of the race? a) 72 mts b)96 mts c)120 mts d)144 mts
In the time that B runs 24 metres, C runs 20 metres. Relative distance in this case : 4 metres.
In creating a relative distance of 4 metres, B runs 24 metres. When A finishes the race, B and C have a relative distance of 12 metres. So, to create this relative distance, B must have run 24*3 = 72 metres. After having run this distance, the finish line is still 24 metres away. Therefore, race length = 24 + 72 = 96 metres 😁
A dealer buys dry fruitRs 100,Rs 80, Rs 60 per kilogram. he mixes them in the ratio 4:5:6 by weight ,and sells at a profit of 50 %. At what price per kilogram does he sells the dry fruit??? a)Rs 116 b)Rs 106 c)Rs 115 d) none of these.
Suppose weights are 4 ,5 and 6 kgs respectively Total CP = 400 + 400 + 360 = 1160 [ for 15 kg ]
SP = 1.5 * 1160 = 1740 [ for 15 kgs ] SP per kg = 1740/15 = 116
I ve a problem.. Q. M is a two digit number which has d property tht : product of factorial of its digits > sum of factorial of its digits. How many values of M exist?? Ans.. 56, 64, 63, n none of these
Let the number M=ab where a and b are the digits. so, we have a!*b! > a!+b! in other words, 1/a! + 1/b! thus we eliminate these cases.... a=0 b=0 a=1 b=1 and a=2=b thus we r left with 63 numbers..... hope that helps....
I ve a problem.. Q. M is a two digit number which has d property tht : product of factorial of its digits > sum of factorial of its digits. How many values of M exist?? Ans.. 56, 64, 63, n none of these
Let the number M=ab where a and b are the digits. so, we have a!*b! > a!+b! in other words, 1/a! + 1/b! thus we eliminate these cases.... a=0 b=0 a=1 b=1 and a=2=b thus we r left with 63 numbers..... hope that helps....
My take on it: Begin with basics
For any real positive integers n and k, it is evident that n*k which eliminates 01,02,03,04,05,06,07,08,09 and 10,20,30,40,50,60,70,80,90 2. either n or k =1 which eliminates 11,12,13,14,15,16,17,18,19 and 21,31,41,51,61,71,81,91
And n*k=n+k if: n=k=2 which eliminates 22
So seems you were right in elimination: Total numbers =99 99-36 =63
Hey Guys...I want to know that are all the answers of the book "arun sharma" correct??...actually i have just started with it n found some answers in numbers systems not matching with mine.... Please Help!!!!!
Hey Guys...I want to know that are all the answers of the book "arun sharma" correct??...actually i have just started with it n found some answers in numbers systems not matching with mine.... Please Help!!!!!
There are quite a few mistakes in the answers provided. It's better if you clarify your doubts here than trying to prove those answers right.
Hence Solved 
...theory part is good but when it comes to problems part !! not that good as many tough problems are left unsolved...in examples only naive problems are solved
: pls contribute with your QnA wrt Quant.
Arun Sharma
f(1) = 1 + a +b =5 .......(1)