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Resulting concentration after successive transfers

CAT 2019 Slot 2 · Quantitative Ability · Hard · Mixtures

This is a hard Quantitative Ability question from the CAT 2019 Slot 2 paper. It tests Mixtures. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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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

  1. A.

    15

  2. B.

    12

  3. C.

    13

  4. D.

    14

Answer

D

Explanation

Initial salt amounts:

  • A: 500 ml500\text{ ml} at 10%=50 g10\% = 50\text{ g}
  • B: 500 ml500\text{ ml} at 22%=110 g22\% = 110\text{ g}
  • C: 500 ml500\text{ ml} at 32%=160 g32\% = 160\text{ g}

Step 1: Transfer 100 ml100\text{ ml} (10 g10\text{ g} salt) from A to B.

  • A has 400 ml400\text{ ml} (40 g40\text{ g} salt left).
  • B has 600 ml600\text{ ml} containing 110+10=120 g110 + 10 = 120\text{ g} salt. Strength of B =120600=20%= \frac{120}{600} = 20\%.

Step 2: Transfer 100 ml100\text{ ml} (20 g20\text{ g} salt) from B to C.

  • B has 500 ml500\text{ ml} left.
  • C has 600 ml600\text{ ml} containing 160+20=180 g160 + 20 = 180\text{ g} salt. Strength of C =180600=30%= \frac{180}{600} = 30\%.

Step 3: Transfer 100 ml100\text{ ml} (30 g30\text{ g} salt) from C to A.

  • A now has 400+100=500 ml400 + 100 = 500\text{ ml} containing 40+30=70 g40 + 30 = 70\text{ g} salt.
  • Final strength of A =70500=14%= \frac{70}{500} = 14\%.

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