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

Updated On: Aug 20, 2024
  • 12
  • 14
  • 13
  • 15
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The Correct Option is B

Solution and Explanation

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.

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