Question

10% of salty sea water contained in a flask was poured out into a beaker. After this, a part of the water contained in the beaker was vapourised by heating and due to this, the percentage of salt in the beaker increased M times. If it is known that after the content of the beaker was poured into the flask. the percentage of salt in the flask increased by x%.
Find the original quantity of seawater in the flask.

• A. $\frac{9M+1\%}{M-1}$
  
• B. $\frac{(9M+1)x\%}{M-1}$
  
• C. $\frac{9M-1x\%}{M+1}$
  
• D. $\frac{9M+x\%}{M+1}$

Answer 1

The correct option is B $\frac{(9M+1)x\%}{M-1}$


Let the initial percentage of salt be 10% in 100 litres of seawater in the flask. 10% of this is poured out (i.e.,10 litres are poured out) and the water heated so as to increase the percentage of salt in the beaker 5 times (we have assumed M as 5 here.)
This means that there will be 50% salt in the beaker. Since the salt concentration is increased by only evaporating water, the amount of salt remains the same.
Initially, the salt was 10% of 10 litres ( = worth 1 litre). Hence, the water must have been worth 9 litres.
Now, since this amount of salt becomes worth 50% of the total solution, the amount of water left alter evaporation would have been 1 litre and the total would be 2 litres.
When the 2 litres are mixed back again: The new concentration of salt in seawater would go up. In this specific case by alligation we would get the following alligation situation:
Mix 90 litres of 10% salted seawater with 2 litres of 50% salted seawater.
The result using alligation will be: $[10+\frac{40}{46}]$ % concentration of salted seawater. The value of the increase percentage will be $\frac{400}{46}$. (this will be the value of x)
Now, try to use the given options in order to match the fact that originally the flask contained 100 litres of seawater.
Use $M=5,x=\frac{400}{46},$
Only option (b) matches the situation.
$\frac{(9\times 5+1)\frac{400}{46}}{(5-1)}=100$