A container contains 40 litres of milk. From this container, 4 litres of milk were taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
29.16 liters (Option B)If you are taking out 4 liters out of a 40 liter solution and replacing it with water, you are effectively replacing 1/10th of the solution with water. => Fraction of milk will become (9/10) of the original If the process is repeated 'n' times, fraction of milk will become (9/10)^n of the original. We are given the process was done thrice=> Final quantity of milk = (9/10)^3 * 40 = 0.729*40 = 29.16
A container is 4/5 full of water. After taking out 8 liters, it became 2/3 full. how many liters of water have to be added to make it full?
Easy math:Bring every thing to the same unit:4/5 = 12/15 and 2/3 = 10/15the difference of those 2 volume levels is 8 liter.→ 12/15 - 10/15 = 8 liter ==> 1/15 = 4 liter.A full tank is 15/15 or 4 x 15 = 60 liters.
The length of a cubical water container is 50 cm. How many litres of water at the most can it contain?
Let consider length of cube as a (given =a=50) now formula of volume of cubical container =a^3 (cube of a )Vol=50×50×50= 125000 unit^3For unit of cubical container units may from mks and cgs and or in other system of unitsThanks for visiting
What formula can I use to calculate how much hot and cold water to add in order to hit a desired target volume and temperature?
Tim’s answer is helpful but ignores the math.By using thermodynamics we can derive a formula.q_lost = q_gain (enthalpy conservation)q = mass * specific heat * temp changeAssuming M2 is the mass of the cooler body of water:m1(T1 - Tfinal)*C = m2(Tfinal - T2)*CThus, m2 = m1(T1-Tfinal)/(Tfinal-T2)Since mass is proportional to volume, you could easily convert this into a volumetric calculation.
How do I calculate the storage capacity of a tank in liters?
In any calculator, use the decimeter unit to calculate.1 liters = 1 dm x 1 dm x 1 dmFormula volume of a rectangular tank:V = A * B * HA - lengthB - widthH - heightFormulas volume of a cylinder:The simplest calculator for rectangular tank calculation - Volume of a cube | Cube Volume CalculatorThe simplest calculator for cylinder tank calculation - Volume of a tank | Tank Volume Calculator
There are 81 litres of pure milk in a container. One third of the milk is replaced by water in the container. Again one-third of the mixture is replaced with an equal amount of water. What is the ratio of milk to water in the new mixture?
4 : 581l milk is in containerone third (27l) is replaced by water. So there is 54l milk and 27l milk is in container (2 parts milk + 1 part water -> 2:1 ratio)this is the tricky part of solution… since milk is miscible in water, the 2:1 ratio is maintained through out the container. So if some amount of mixture is removed from any place in container, removed mixture is in 2:1 ratio & also remaining mixture maintains same 2:1 ratio… Now, as one third is removed from 81l mixture, 54l mixture is remaining which has 2:1 ratio. So there is 36l milk & 18l water in container.Now 27l water is added as replacement for the removed 1/3rd part of mixture. So there is 36l milk and 18+27=45l water now36:45=4:5
A container contains 5 litres of milk. From this container, 1 litre of milk is taken out and replaced with water. This process is repeated twice more. How much milk is now in the container?
This question can be solved quickly using the formula: Final Concentration = Initial Concentration x [1 - (amount being replaced / total amount)]^(no. of times the replacement is made)Initial Concentration = 100%, Amount being replaced = 1 L, Total amount = 5 L, No. of times the replacement is made = 3.Final Concentration = 100 x [1 - (1/5)]^3 = 100 x (4/5)^3 = 100 x 64 / 125 = 51.2%The amount of milk in the container now = 5L x 0.512 = 2.56L.Following are some other tips to solving aptitude questions on Mixtures & Alligations quickly:-Volume of a substance in the final solution = Sum of volumes of the substance in the ingredientsFrom a solution of volume V containing x percent of a substance, if y liters are replaced by water, the new percentage of the substance in the solution = x (V-y)/ VFor more, you may refer to the following info-graphic:-You might also want to solve Online Aptitude Test Questions in order to evaluate and improve your Quantitative Aptitude skills.Hope this helps :)
How can you convert LITERS into GRAMS?
Otis - - Not like that, you idiot. None of this will be used like that. We have to have expert witnesses to testify in this cause. This is for our own personal benefit at the current time. The expert witness will testify as to whether or not this process can be completed. In the meantime, some 3 months prior to trial, we are getting our act together to know what is and is not possible.
A vessel contains 60 litres of milk. 12 litres of milk was taken out from it and replaced with water. From the mixture, 12 litres is taken out and replaced with water. What is the ratio of milk and water in the resultant mixture?
Let's solve it in an uncomplicated way.Step 1Milk=60 litreOnly Milk was there.Step 2Milk=60-12 =48 litreWater =12 litreNow,a mixture is created with 4:1 ratio of milk and water.Means 4/5 part Milk and 1/5 part of Water.Step 3Removing 12 litres of mixture meansremoving Milk=12×4/5 part=48/5 partAnd removing water=12×1/5 part =12/5 part.So,removing 12 litre of mixture and addition of 12 litre water gives the result as—Milk=48 - 48/5 = 48×4/5=192/5 litre.Water=12 - 12/5 + 12 = 24 - 12/5 =12×9/5 =108/5 litre.Here,ratio =192/5 : 108/5 =192/108=16/9 Ans.
Is my answer for the empirical formula correct?
A xenon fluoride compound can be prepared by heating a mixture of Xe and F2 gases to a high temperature in a pressure proof container. Assume that xenon gas was added to a 0.25 L container until its pressure was 0.12 atm at 0.0°C. Fluorine gas was then added until the total pressure was 0.72 atm at 0.0°C. After the reaction was complete, the xenon was consumed completely and the pressure of the F2 remaining in the container was 0.36 atm at 0°C. What is the empirical formula of the xenon fluoride compound? I got XeF4, is this correct? Thanks for your help!!