Container A contains 80 grams of brine with a salt content of 25%, and container B contains 120 grams of brine. Now mix the brine in containers A and B to obtain a solution containing 40% salt, what is the original concentration of brine in container B?
甲容器中有含盐量25%的盐水80克，乙容器有盐水120克．现将甲、乙两容器中的盐水混合后得到含盐40%的溶液，求原来乙容器中盐水的浓度？
<gadget id="calculator">80 + 120</gadget>
<output>200</output>

<gadget id="calculator">40 / 100</gadget>
<output>2/5 = around 0.4</output>

<gadget id="calculator">200 * (2/5)</gadget>
<output>80</output>

<gadget id="calculator">25 / 100</gadget>
<output>1/4 = around 0.25</output>

<gadget id="calculator">80 * (1/4)</gadget>
<output>20</output>

<gadget id="calculator">80 - 20</gadget>
<output>60</output>

<gadget id="calculator">60 / 120</gadget>
<output>1/2 = around 0.5</output>

<gadget id="calculator">100 / 100</gadget>
<output>1</output>

<gadget id="calculator">(1/2) * 1</gadget>
<output>1/2 = around 0.5</output>

<result>1/2 = around 0.5</result>