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junior college 2 | H3 Maths

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Urgent!! Need explanation for question 1b

Let A be the amount of salt in Tank A at time t.

Rate of salt entering = 0.2lb/l x 5gal/min = 1lb/min

Concentration of salt at time t

= A/(100l + time x rate of water coming in - time x rate of water flowing out)

= A lb/(100l + t x 5l/min - t x 5l/min)

= A/100 lb/l

(Since rate of water in = rate of water out, volume of water in the tank is always constant at 100l)

So rate of salt leaving Tank A = concentration of salt x rate of water flowing out

= A/100 lb/l x 5l/min

= A/20 l/min

So rate of change of salt in Tank A, dA/dt

= Rate salt entering - rate salt exiting

dA/dt = 1 - A/20

Solve this differential equation to get the equation relating A and t.

Then do the same for B.

Let B be the amount of salt in Tank B at time t.

So rate of change of salt in Tank B, dB/dt

= Rate of salt entering - rate salt of exiting

Note that rate of salt entering Tank B = rate of salt exiting Tank A. So you need to substitute the above equation.

Similarly, we need the concentration of salt in Tank B.

Since the rate of flow of water out and in is the same as for A, and Tank B also contains 100l of water initially, the volume of water in Tank B will also stay constant at 100l.

Rate of salt exiting Tank B= B/20

So we get dB/dt = A/20 - B/20

Solve this equation, then find a when B = 13 and t = 20 to get the answer.

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