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 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. 15
2. 13
3. 14
4. 12

Answer 1

Correct Answer - Option 3 : 14

Calculation:

The initial amount of salt in each vessel = 50 ml, 110 ml, 160 ml

100 ml of solution is transferred from A to B

A would have 400 ml, B would have 600 ml of solution

Amount of salt from A which is transferred to B = 100 × 1/10

⇒ 10 ml

So, Total salt in B = 110 + 10 = 120 grams (After first transfer)

Total salt in A = 40 grams (After first transfer)

Now, 100 ml from Vessel B is transferred to Vessel C

So similarly salt would transfer from B to C = 1/6th total salt

Amount of salt from B which is transferred to C = 120 × 1/6

⇒ 20 ml

Total Salt in B = 120 – 20 = 100 grams (After second transfer)

Total Salt in C = 160 + 20 = 180 grams (After second transfer)

Now, 100 ml from Vessel C is transferred to Vessel A

So similarly of salt would transfer from C to A = 1/6th total salt

Amount of salt from C which is transferred to A = 180 × 1/6

⇒ 30 ml

Total Salt in C = 180 – 30 = 150 grams (After third transfer)

Total Salt in A = 40 + 30 = 70 grams (After third transfer)

So, Vessel A contains 70 grams Salt in 500 ml solution

Strength of Salt Solution in Vessel A = 14%

∴ The strength in the percentage of the resulting solution in vessel A is 14%