The One 2 Ka 4l
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1. Fill 3L bucket FULL. 2. Transfer this water to 5L bucket. It now has 3L.3. Fill 3L bucket FULL again.4. Transfer to 5L bucket till it fills completely. Now 3L bucket has 1L.5. Throw away 5L of water. 6. Transfer 1L water from 3L to 5L7. Fill 3L bucket FULL again.8. Transfer to 5L bucket to get 4L in the bucket.
its too complicated...the easy one..1.fill 5 lt., bucket.2.Take out 3 lt. from it using another bucket of 3 lt.3.5-3=2 lt. left in 5 lt. bucket4.Empty the 3 lt. bucket and transfer this 2 lt. in 3 lt. bucket.5.now in last fill the 5 lt. bucket full and fill the 3 lt. bucketleft in 5 lt.bucket=5-1=4:)
Sorry thats in hindi translating to englishFill 5 L bucket -> Pour it in 3L bucket --> 5L bucket has 2L liquid--> Empty 3L bucket and pour the 2L liquid from 5 L bucket to 3L bucket ---> Now fill 5L bucket---> empty 1 into 3L bucket
A = 5L bucket, B = 3L bucket1. Fill A with one B.2. Try filling the 2L space in A with another B. We are left with 1L in B.3. Empty A and Fill it with 1L 4. Then fill A with complete B
First fill 5L with 3L two times.After filling the 5L completely 3 will contain 1 lit.Now empty 5L and add this 1L in that.Now fill 3L and pour it in 5L.5L bucket now contains 4L.
1. Fill 3L bucket FULL.2. Transfer this water to 5L bucket. It now has 3L.3. Fill 3L bucket FULL again.4. Transfer to 5L bucket till it fills completely. Now 3L bucket has 1L.5. Throw away 5L of water.6. Transfer 1L water from 3L to 5L7. Fill 3L bucket FULL again.8. Transfer to 5L bucket to get 4L in the bucket.
Without wasting the water.. here is a solution:1. Fill 3L bucket FULL. 2. Transfer this water to 5L bucket. It now has 3L.3. Fill 3L bucket FULL again.4. Transfer to 5L bucket till it fills completely. Now 3L bucket has 1L.5. Pour the remaining 1L to 4L bucket. 6. Fill 3L bucket FULL again and pour to 4L Bucket.
first pour 5 ltr bucket water in 3 ltr bucket then 2 ltr remains in the 5 ltr bucket. Empty the 3 ltr bucket and pour that remaining 3 ltr water in it. and again fill 5 ltr bucket and pour 1 ltr water in the 3 ltr bucket which has currently 2 ltr. so in 5 ltr bucket 4 ltr is remains.2 ans:pour 3 ltr in 5 ltr bucket now there is 3 ltr in 5 ltr bucket again pour 3 ltr bucket water in 5ltr then 1 ltr remains in 3 ltrempty the 5 ltr bucket and then pour the 1 ltr of 3 ltr bucket now 5 ltr bucket has 1 ltrpour 3 ltr bucket in itNow 5 ltr bucket has 4 ltr
two buckets3L and 5L bucket3L bucket fill full in 5L bucket and check in 5L bucket in total 3L water and same process and check in 5L bucket in 5L water and in 3L bucket in 1L water after that empty 5L bucket and add 3L bucket in 1L water and again fill full 3L bucket in 5L bucket so total in 5L bucket in 4L water
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The concentration calculator allows you to quickly calculate the volume, mass or concentration of your vial. Simply enter your mass, volume, or concentration values for your reagent and the calculator will determine the rest.
Suppose you have 8L, 5L and a 3L bucket with you. The buckets have no measurement lines on them and you are asked to measure exactly 4L of water using the three buckets. How could you measure exactly 4L water using only those buckets and provided you have as much extra water as you need
Solution:- 1.1 When we are allowed to throw water out of bucket. Step 1: Fill the 8L bucket full. Step 2: Pour the water from 8L bucket to 3L bucket. Water in 8L bucket is now 5L. Empty 3L bucket. Step 3: Pour the remaining water from 8L bucket to 3L bucket. Water in 8L bucket is now 2L. Empty 3L bucket. Step 4: Pour the remaining 2L water from 8L bucket to 5L bucket. 8L bucket and 3L bucket are now empty and 5L bucket has 2L of water in it.
1.2 When we are allowed to throw water out of bucket. Step 1: Fill the 5L bucket full. Step 2: Pour the water from 5L bucket to 3L bucket. Water in 5L bucket is now 2L. Empty 3L bucket. Step 3: Pour the remaining water from 5L bucket to 8L bucket. Water in 8L bucket is now 2L.
2.1 When we are not allowed to throw water out of bucket. Step 1: First fill the 8L bucket full. Step 2: Pour the water from 8L bucket to 5L bucket. Water remaining in 8L bucket is 3L. Step 3: Pour the water from 5L bucket to 3L bucket. Water remaining in 5L bucket is 2L. Step 4: Pour the water from 3L bucket to 8L bucket. Water in 8L bucket is 6L now and 3L bucket gets empty. Step 5: Pour the water from 5L bucket to 3L bucket. Water in 3L bucket is 2L now and 5L bucket gets empty. Step 6: Pour the water from 8L bucket to 5L bucket. Water remaining in 8L bucket is 1L 5L bucket gets full. Step 7: Pour the water from 5L bucket to 3L bucket. Water remaining in 5L bucket is now 4L as 3L bucket already had 2L of water and when we poured water from 5l bucket to 3L bucket we poured 1L of water from 5L bucket and thus the remaining water in 5L bucket is now 4L.
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Pharmacokinetics focuses on drug movement throughout the human body via the processes of absorption, distribution, and elimination. Upon administration, a drug moves from the site of administration and gets absorbed into the systemic circulation where it will then gets distributed throughout the body. The process of distribution refers to the movement of a drug between the intravascular (blood/plasma) and extravascular (intracellular & extracellular) compartments of the body. Within each compartment of the body, a drug exists in equilibrium between a protein-bound or free form. Over time, drugs within the circulation will then be metabolized and excreted from the body by the liver & kidneys.
Drugs that display single compartment distribution kinetics with a straight line graph on plasma vs. time curves. Because the drug is said to distribute instantaneously, the initial plasma concentration of drug at time = 0 (Co) is difficult to measure and is therefore estimated via extrapolation to time = 0 on a plasma concentration vs. time curve.
Half-life (t1/2) refers to the time required for plasma concentration of a drug to decrease by 50%. t1/2 is dependent on the rate constant (k), which is related to Vd & clearance (CL). Half-life can be expressed using the following equation(s):
Only the drug located in the central compartment can be eliminated from the body because the process of elimination is primarily carried out by the liver and kidneys. Drugs with a high Vd will have a large fraction of drug remaining outside of the central compartment. Meanwhile, the fraction of drug in the plasma will be eliminated, causing a shift of equilibrium resulting in drug located in the peripheral compartment to shift into the central compartment. This shift will cause the plasma concentration to remain at a steady-state concentration despite drug removal from the body. This phenomenon causes plasma concentration to decline more slowly during the elimination phase in the setting of a high Vd.
Therefore, at a constant rate of clearance, a drug with a high Vd will have a longer eliminationhalf-life than a drug with lower Vd.
As previously discussed, multiple values of Vd can be calculated depending on the intrinsic drug kinetics (single vs. multiple compartment models) as well as the phase of drug kinetics following drug administration (distribution phase vs steady state vs terminal elimination phase). However, from the clinical perspective, the single most important utility of Vd is calculating the loading dose of a drug.
The loading dose is best calculated using the Vd at steady state (Vss) as it is the most representative of the specific drugs pharmacokinetic properties at desired steady-state plasma concentration. Therefore, the loading dose can be calculated using the following equation: 153554b96e