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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
(2019)
  • a)
    13
  • b)
    14
  • c)
    12
  • d)
    15
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The strength of a salt solution is p% if 100 ml of the solution contai...
Initial amount of salt in vessel A = 10 gms per 100 ml. solution. Therefore in 500 ml solution in vessel amount of salt = 50 gms
Similarly, initially in 500 ml solution in vessel B amount of salt = 110 gms
and initially in 500 ml solution in vessel C, amount of salt = 160 gms
When 100 ml is transferred from A to B, the amount of salt now in B = 10 + 110 = 120 gms in 600 ml.
The new concentration of salt in B = 120 / 600 x 100
= 20 gms per 100 ml.
Now, the amount of salt in A = 50 – 10 = 40 gms in 400 ml
Now, when 100 ml is transfered from B to C, the amount of salt now in C = 20 + 160 = 180 gms in 600 ml.
The new concentration of salt C = 180 / 600 x 100
= 30 gms per 100 ml
Finally, when 100 ml is transfered from C to A, the amount of salt now in A = 30 + 40 = 70 gms in 500 ml.
∴ Strength of salt in 70 / 500 x 100 = 14
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Most Upvoted Answer
The strength of a salt solution is p% if 100 ml of the solution contai...
Initial amount of salt in vessel A = 10 gms per 100 ml. solution. Therefore in 500 ml solution in vessel amount of salt = 50 gms
Similarly, initially in 500 ml solution in vessel B amount of salt = 110 gms
and initially in 500 ml solution in vessel C, amount of salt = 160 gms
When 100 ml is transferred from A to B, the amount of salt now in B = 10 + 110 = 120 gms in 600 ml.
The new concentration of salt in B = 120 / 600 x 100
= 20 gms per 100 ml.
Now, the amount of salt in A = 50 – 10 = 40 gms in 400 ml
Now, when 100 ml is transfered from B to C, the amount of salt now in C = 20 + 160 = 180 gms in 600 ml.
The new concentration of salt C = 180 / 600 x 100
= 30 gms per 100 ml
Finally, when 100 ml is transfered from C to A, the amount of salt now in A = 30 + 40 = 70 gms in 500 ml.
∴ Strength of salt in 70 / 500 x 100 = 14
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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(2019)a)13b)14c)12d)15Correct answer is option 'B'. Can you explain this answer?
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