Directions: Refer the following data to answer the questions given below.
Solution
Weight of the student = 50 x 5 – 52.3 x 4 = 41.8 kg.
Directions: Refer the following data to answer the questions given below.
Directions: Refer the following data to answer the questions given below.
Directions: Refer the following data to answer the questions given below.
Solution:
Situation Before Boiling
We have to mix river water, lake water, and sea water in the ratio of 4:2:1.
Let us assume we are mixing 4x, 2x and x unit of each.
This will give us 7x units of the total mixture.
In this 7x units, amount of salt from riveris 0.5% of 4x, i.e. units of salt
Similarly, in 7x units, amount of salt from lake is 1% of 2x, i.e. units of salt
And, in 7x units, amount of salt from sea is 5% of x, i.e. units of salt
Adding all the units of salt in 7x units of water, we get overall salt in mixture, which is:
Concentration of salt in 7x units of water
This means, in 7x units of mixture we have unit of salt, which also means the amount of water in this mixture is
After Boiling
When the above mixture is boiled, the concentration of salt doubles (i.e., the salt does not change in amount, but the water evaporates in the mixture and salt remains such that it has double concentration in the mixture now).
That is,
Concentration of salt in the (remaining) mixture becomes
Since, the concentration of salt is doubling without any change in quantity of salt in the mixture, it is imperative that the total volume of the mixture is getting halved…this is possible only when the units of water in the mixture are evaporating in such a way that the volume of mixture remaining after boiling is exactly half the volume of the original mixture…in other words, as water evaporates on boiling 7x units of the mixture, the remaining volume of the mixture is now 3.5x only.
Therefore, water remaining in the mixture is units
Therefore, the amount of water evaporated = 50%
Directions: Refer the following data to answer the questions given below.
Solution:
A check may be carried out as follows:
Say, we have 500 ml of alcohol in beaker 1and 500 ml of water in beaker 2.
Now, assume 3 cups hold 300 ml of any liquid — alcohol or water, so if we take 3 cups of alcohol from beaker 1 (the cup will have 300 ml of alcohol) and pour it into beaker 2…
In Beaker 1:
200 ml of alcohol is left
In Beaker 2:
…we will have 500 ml of water and 300 ml of alcohol as a mixture, 800 ml mixture.
In this mixture, alcohol is 300/800 parts and water is 500/800 parts.
Now, if we fill 3 cups from beaker 2…
In Beaker 2 :
500 ml mixture remains, the constituents of the mixture are:
{3/8}*500 =187.5ml alcohol and{5/8}*500=312.5 ml water
In Beaker 1:
We were holding 200 ml of alcohol to which 300 ml of mixture is added from beaker 2.
The constituents of 300 ml of mixture that is added are:
{3/8}*300=112.5ml of and {5/8}*300ml water
Therefore, final situation of constituents in Beaker 1 = 200 + 112.5 = 312.5 ml alcohol
…and187.5 ml water
Comparison:
Percentage of water in beaker 1 = {187.5/500}*100
Percentage of alcohol in beaker 2 = { 187.5/500}*100