6. Applied Verbal Problems

Rough estimate: The amount coming out of the 2 pipes is about 300 L/min so if each pipe has the same width, then the rate will be about 150 L/min for each pipe. But one pipe's rate is 50 L/min more than the other, so we expect the answers to be around 125 L/min and 175 L/min.

Solution: Let x be the rate of the slower pipe (in L/min).

The rate of the faster pipe will be

`(x + 50)` L/min.

Together they release: x + (x + 50) = 330

So `2x + 50 = 330`

`2x = 280`

`x = 140`

So the drainage rates of the 2 pipes are `140` L/min and `190` L/min.

[This is close to our earlier estimate. Together they give a rate of `330` L/min and they differ by 50 L/min, so we can be confident our answer is correct.]

Let A be the rate of the faster pipe and B be the rate of the slower pipe.

We have A − B = 50 and A + B = 330.

Solving these simultaneous equations, we have:

`A - B = 50`

`A + B = 330`

`2A = 380`

So `A = 190`. Since `B` is `50` L/min slower, then `B = 140`.

CHECK:

`190 + 140 = 330`

`190 - 50 = 140.`

So the drainage rate of the faster pipe is `190` L/min and the rate of the slower pipe is `140` L/min.

Here is our system. Note the off-center weight acting on the beam.

Rough estimate

We expect the three forces to each be less than 10 N, based on some quick trial and error. If the first upwards force is 2 N, the second will be about 8 N (that's 6 N more) and 2 + 8 = 10. Our downward force is 5 times the first, somewhere near the 4 times that what we need, so it's a pretty good guess.

Solution

Let the first upward force be F, as shown in the diagram.

The statement "the downwards force is four times the first upward force" gives us:

• The downwards force is `4F`, and

The statement "the second of the two upward forces is `6.4\ "N"` more than the first" gives us

• The second upward force is `F+6.4`.

The second sentence in the question tells us that the sum of the 2 upwards forces equals the downwards force. So

So `F = 3.2`

The second force is `F + 6.4 = 3.2 + 6.4 = 9.6`

The third (downward) force is `4F = 4 xx 3.2 = 12.8`

So the 3 forces are: `3.2` N, `9.6` N and `12.8` N.

This is close to our earlier estimate and all statements in the question are true with these forces.

Let the 3 forces be F and G (upwards) and H (downwards).

We are told F + G = H .....(1)

Also, the second force is `6.4` more than the first, `G = F + 6.4` .....(2)

For the third (downwards) force, `H = 4 xx F` .....(3)

If we substitute equations (2) and (3) into (1), we have:

So `F = 3.2`

This gives `G = 9.6` and `H = 12.8`.

Checking:

(1) `F + G = 3.2 + 9.6 = 12.8 = H` (OK)

(2) `F + 6.4 = 3.2 + 6.4 = 9.6` (OK)

(3) `4F = 4 xx 3.2 = 12.8` (OK)

So the three forces are `3.2` N, `9.6` N and `12.8` N, respectively.

Yes, Dale, these problems often seem to have funny English...

Look at the second sentence:

If the second of the two forces is 6.4 N more than the first and the third is four times the first, what are the forces?

To illustrate what this means, let me just take a guess value for the first force.

Force 1: 10 N (this is my guess value)

Force 2: 16.4 N (this is because it says the second force has to be 6.4 N more than the first)

Force 3: 40 N (this is because the third force has to be four times the first).

[Note: My guess of 10 N for the first force is not the right answer, because the first condition fails ("the sum of two forces must equal the third force") since 10 + 16.4 ≠ 40.]

It may have been easier if the question said:

"The second of the two forces is 6.4 N more than the first. The third force is four times the first. What are the forces?"

We need to use consistent units throughout. The 2 g of drug is equivalent ot 2000 mg.

Rough estimate: We expect the dosages to be around 700 mg and 1300 mg, based on similar thinking to Example 1 above.

Let the amount the adult gets be d mg.

The child will get d − 660.

Together, the two dosages must be: d + (d − 660) = 2000

That is 2d − 660 = 2000

2d = 2660

d = 1330

And d − 660 = 1330 − 660 = 670

So the adult gets 1330 mg and the child gets 670 mg.

Let the first amount be x and the second y.

We have: `x + y = 2000` and

`x - y = 660`

Adding gives `2x = 2660` so `x = 1330`, then `y = 670`.

Checking: `x + y = 1330 + 670 = 2000` (OK)

`x - y = 1330 - 670 = 660` (OK)

So the amounts of dosage should be `1330` mg for the adult and `670` mg for the child.

In this question, we must consider the percentage of gasoline in each mixture.

Let A be the amount of 100% gasoline required. So the amount of the 75% gasoline-oil mixture will be `8.0 − A`.

The amount of gasoline in the gasoline-oil mixture is

`75% xx (8.0 − A)` ` = 0.75 xx (8.0 − A)` ` = 6.0 − 0.75A`

A ratio of 15:1 means that out of 16 parts, 15 parts must be gasoline. So, the amount of gasoline required in the outboard engine mix is

`(15/16) xx 8 = 7.5`

So considering the amount of gasoline in each part, we have

`A + (6.0 − 0.75A) = 7.5`

`0.25A = 1.5`

`A = 6`

Therefore, we need to add `6` L of pure gasoline and `2` L of the 75% gasoline-oil mix to make the correct fuel for the scooter engine.

We will let the amount of 100% gasoline be G, the amount of 75% mix be M and the final amount is 8.0 L.

The final mixture is in the ratio 15:1 which means 15/16 is gasoline.

So `G + M = 8` ...(1)

Considering the amount of gasoline in each one,

`100%G + 75%M = 15/16 xx 8`

That is:

`G + 0.75M = 7.5` ...(2)

We subtract equation (2) from equation (1) and get:

`0.25M = 0.5`

and so

`M = 2`.

CHECK: G = 6 (since G + M = 8) and

`6 + 0.75× 2 = 6 + 1.5 = 7.5` OK

So the amount of pure gasoline required is `6` L.