a certain quantity of 50 % solution is replaced with 25 % solution such that the new concentration is 35 %. what is the fraction of the solution that was replaced?	Answers.
A) 1 / 4	B) 1 / 3	C) 1 / 2	D) 2 / 3	E) 3 / 5
<gadget id="calculator">35 - 25</gadget>
<output>10</output>

<gadget id="calculator">50 - 35</gadget>
<output>15</output>

<gadget id="calculator">10 / 15</gadget>
<output>2/3 = around 0.666667</output>

<gadget id="calculator">(2/3) + 1</gadget>
<output>5/3 = around 1.666667</output>

<gadget id="calculator">1 / (5/3)</gadget>
<output>3/5 = around 0.6</output>

<result>E</result>
a certain quantity of 50 % solution is replaced with 25 % solution such that the new concentration is 35 %. what is the fraction of the solution that was replaced?
inverse(add(divide(subtract(35, 25), subtract(50, 35)), const_1))
"let ' s say that the total original mixture a is 100 ml the original mixture a thus has 50 ml of alcohol out of 100 ml of solution you want to replace some of that original mixture a with another mixture b that contains 25 ml of alcohol per 100 ml . thus , the difference between 50 ml and 25 ml is 25 ml per 100 ml of mixture . this means that every time you replace 100 ml of the original mixture a by 100 ml of mixture b , the original alcohol concentration will decrease by 25 % . the question says that the new mixture , let ' s call it c , must be 35 % alcohol , a decrease of only 15 % . therefore , 15 out of 25 is 3 / 5 and e is the answer ."