x/(x+35) = 30/100 so 10x = 3x + 105 7X = 105 X = 15 LT quantity of milk initially 15 lt
milk = 15 lt water = 35 lt add 40 lt water to this mixture
milk = 15
water = 35 +40 = 75
now we add x lt water to make concentration of milk = 8%
15 /(90+x) = 8/100
1500 = 720 +8x 780 = 8x
x = 780 /8 x = 97.5 need add 97.5 lt of water in mixture ????????????????????????????
The soln is dilute. hence,it contains some quantity of water initially. M-milk W-water M/(M+W+30)=.3------(1) M/(M+W+75)=.2------(2) solve above 2 equations and get the values of M & W. and calculate quantity of water to be added=300-(M+W+75)
Quote: Originally Posted by shashankjha View Post 1-> total work = 10*7 = 70 womandays similarly , total work = 14*10 = 140 childdays so, 1 womanday = 2 childdays Now, 5w+10c = 20c i.e 20 childdays. since , 10 children could complete it in 14 days, so 20 children could complete it in 7 days. . From Ajayjr Hi Shashank Can u plz explain how u arrived at the boldened part? --------------------------------------------------------------------------------
Ajay, As calculated in the solution total work = 70 womandays of effort = 140 childdays of effort this shows that 1 womanday of effort = 2 childday of effort. Now, we need to calculate for 5woman and 10child effort to do the same amount of work thus, 5W+10C = 5womanday of effort + 10childday of effort = (5*2)childdays of effort + 10 childdays of effort = 20 childdays of effort.
@ajeetaryans -- please explain i have seen it for the first time
let the number be x. then let 1% of x = 1.5 = 3/2. => 1/100 * x = 3/2 => 1/100 *x * 1/(3/2) = 1 => 1/100 * x * 2/3 = 1 => (2/3)/100 * x = 1 or 2/3% of x = 1.
So, it can be concluded that if m% of x = n, then, 1/n% of x = m. :clap: I hope its clear...you can put any random number in place of x and check for youself.
A contractor employed 30 men to complete the project in 100 days.but later on he realized that just after 25 days only 20% of the work had been completed. a) how many extra days, than the scheduled time are required? b)to compute the work on the scheduled time how many men he has to increase? c)if the amount of work is also increased by 20% of the actual work ,then how many extra days are required but the number of men remain constant d)how many men should be increased so that work will be completed in 25 days less than the scheduled time.?
For questions of this type, always remember that no. of men is inversely proportional to no. of days required as well as to amount of work done.
Now, a) let total no. of days required to complete the work = x days. 30 men completes 20% of work i.e 1/5th part of work in 25 days. => (30 * 25) / (30 * x) = (1/5)/1. => x = 125 days. so extra days = 125-100 = 25 days.
b) let extra men = x. so, (30*25) / ((30+x)*75) = (20/100) / (80/100) = 1/4 => x = 10.
c) Now, work is increased by 20% , so total work is now 120 or 1.2 part. total no. of days required to complete the work = x days => (30*25) / (30*x) = (1/5) / 1.2 => x = 150. so, extra days = 150-100 = 50 days.
d) here ,total days = 100-25 =75 days, let extra men = x. thus, (30*25) / ((30+x)*50) = (20/100) / (80/100) = 1/4 =>x = 30 men. :
For questions of this type, always remember that no. of men is inversely proportional to no. of days required as well as to amount of work done.
Now, a) let total no. of days required to complete the work = x days. 30 men completes 20% of work i.e 1/5th part of work in 25 days. => (30 * 25) / (30 * x) = (1/5)/1. => x = 125 days. so extra days = 125-100 = 25 days.
b) let extra men = x. so, (30*25) / ((30+x)*75) = (20/100) / (80/100) = 1/4 => x = 10.
c) Now, work is increased by 20% , so total work is now 120 or 1.2 part. total no. of days required to complete the work = x days => (30*25) / (30*x) = (1/5) / 1.2 => x = 150. so, extra days = 150-100 = 50 days.
d) here ,total days = 100-25 =75 days, let extra men = x. thus, (30*25) / ((30+x)*50) = (20/100) / (80/100) = 1/4 =>x = 30 men. :
Salute ur patience . Bhai1doubts hai how dd u arrive at the equation
10 years ago the average age of all the 25 teachers of college was 45 yrs. 4yr ago,the principal has retired from her post at the age of 60 yr. so after one yr a new principal whose age was 54 yrs recruited from outside.The present average age of all the teachers is ,if principal is also considered as a teacher..?
10 years ago the average age of all the 25 teachers of college was 45 yrs. 4yr ago,the principal has retired from her post at the age of 60 yr. so after one yr a new principal whose age was 54 yrs recruited from outside.The present average age of all the teachers is ,if principal is also considered as a teacher..?
10 years ago the average age of all the 25 teachers of college was 45 yrs. 4yr ago,the principal has retired from her post at the age of 60 yr. so after one yr a new principal whose age was 54 yrs recruited from outside.The present average age of all the teachers is ,if principal is also considered as a teacher..?
Sol : There are 25 teachers and average age is 45. Total/25 = 45 Total = 45*25 = 1125 this was before 10 years. Now presently the sum of ages will be 1125 + 25(10) = 1125 + 250 = 1375. But he says a principal of age 60 retires and after 1 year a principal of age 54 joins. This means age reduced by 60-54 = 6. And in between for 1 year there was no principal so reduce 1 year from total. This means from total we must subtract 7 to get present sum of ages. 1375 - 7 = 1368 Average = 1368/25 = 54.72 :)
Salute ur patience . Bhai1doubts hai how dd u arrive at the equation
d - (30*25) / ((30+x)*50)
Now, as total no of days = 75.Out of these , for 25 days 30 men worked and completed 20% of work. So, for the remaining 75-25 = 50 days (30+x) no of men worked and completed remaining 80% of work. thus the ratio (30*25) / ((30+x)*50) = (20/100) / (80/100).
Got these in my first AIMCAT 1) if x = 3+1 find x^4 + 6(x^2) + 8x 2) 3^256 divided by 13 gives what remainder? Can someone solve with proper steps! Thanks in advance
Got these in my first AIMCAT 2) 3^256 divided by 13 gives what remainder? Can someone solve with proper steps! Thanks in advance
The ans should be 3 Method: Euler No. of 13 = 12 (For a prime No. PIts Euler No. is (P-1 ) ) Thus 3^12 would give remainder 1 with 13. Thus3^56 can be written as = 3^12 * 3^12 * 3^12.3^4 Now,for 3^4find the remainder with 13..which is 3!!
Got these in my first AIMCAT 1) if x = 3+1 find x^4 + 6(x^2) + 8x 2) 3^256 divided by 13 gives what remainder? Can someone solve with proper steps! Thanks in advance
let the number be x. then let 1% of x = 1.5 = 3/2. => 1/100 * x = 3/2 => 1/100 *x * 1/(3/2) = 1 => 1/100 * x * 2/3 = 1 => (2/3)/100 * x = 1 or 2/3% of x = 1.
So, it can be concluded that if m% of x = n, then, 1/n% of x = m. :clap: I hope its clear...you can put any random number in place of x and check for youself. :cheers:
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