Question:
The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively. Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A. The strength, in percentage, of the resulting solution in vessel A is
- 12
- 14
- 13
- 15
Correct Answer: Option: 2
Initial amount of salt in vessel A=10 gms per 100 ml, therefore in 500 ml the amount of salt =50 gms
Initial amount of salt in vessel B=22 gms per 100 ml, therefore in 500 ml the amount of salt =110 gms
Initial amount of salt in vessel C=32 gms per 100 ml, therefore in 500 ml the amount of salt =160 gms
When 100 ml is trasfered from A to B, the amount of salt now in B = 10+110=120 gms in 600ml.
The new concentration of salt in B = 120/600=20 gms per 100 ml.
Also, the amount of salt lef in A =50-10 =40 gms in 400ml.
Now, when 100 ml is trasfered from B to C, the amount of salt now in C = 20+160=180 gms in 600ml.
The new concentration of salt in C= 180/600=30 gms per 100 ml.
Finally, when 100 ml is trasfered from C to A, the amount of salt now in A = 30+40=70 gms in 500ml.
Therefore, the strength of salt in A = $\frac{70}{500}\times 100=14%$
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