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

• A. 12
• B. 14
• C. 13
• D. 15

Answer 1

Answer: (B)

14

Answer 2

Let's break down the steps:

Initial Amounts:

• Vessel A: 10 gms per 100 ml, so in 500 ml, 50 gms.
• Vessel B: 22 gms per 100 ml, so in 500 ml, 110 gms.
• Vessel C: 32 gms per 100 ml, so in 500 ml, 160 gms.

Transfer from A to B (100 ml):

• Amount of salt in B after transfer: 10+110=120 gms in 600 ml.
• New concentration of salt in B: $\frac{120}{600}$=20 gms per 100 ml.
• Amount of salt left in A: 50−10=40 gms in 400 ml.

Transfer from B to C (100 ml):

• Amount of salt in C after transfer: 20+160=180 gms in 600 ml.
• New concentration of salt in C: $\frac{180}{600}$=30 gms per 100 ml.

Transfer from C to A (100 ml):

• Amount of salt in A after transfer: 30+40=70 gms in 500 ml.
• Strength of salt in A: $\frac{70}{500}×100$=14.

Therefore, after these transfers, the final strength of salt in Vessel A is 14.