  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.9M−1x%M+1  9M+1%M−1  (9M+1)x%M−1  9M+x%M+1

Solution

## The correct option is C (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+4046] % concentration of salted seawater. The value of the increase percentage will be 40046. (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=40046, Only option (b) matches the situation. (9×5+1)40046(5−1)=100  Suggest corrections   