### Seniors & juniors algebra word problem

Here's a word problem that someone sent me recently:

This problem gives a lot of information, and it sounds like it can be solved many different ways. But the

To start, you could for example notice the two facts given about the senior class: The senior class

From this it is easy to solve the number of boys and girls in the senior class.

The ratio 5:3 means 5/8 of them are boys and 3/8 of them are girls. Now, 5/8 of 400 is 250, and 3/8 of it is 150.

So seniors are solved... now to juniors. Let

B1 = boys in junior class

G1 = girls in junior class.

The first sentence of the problem gives us an equation:

G1 + 150 = B1 + 250

We also now that G1 + B1 = 300.

So let's solve this

G1 + 150 = B1 + 250

G1 + B1 = 300

I can subtract the bottom one from the top one to get:

150 - B1 = B1 - 50

200 = 2B1

B1 = 100

Since the total was 300, then G1 is 200.

And now the ratio of boys to girls, it is 100:200 or 1:2. All done!

You could also use straightforward algebra and set up four equations originally, using B1, G1, B2, and G2 for the numbers of boys and girls in junior and senior classes. You would get these four equations:

G1 + G2 = B1 + B2

B2 + G2 = 400

B1 + G1 = 300

B2/G2 = 5/3

And from these four you can solve all four unknowns, and then get the ratio of boys to girls in the junior class.

The total number of girls in the combined junior and senior classes is equal to the number of boys in those two classes. If the senior class has 400 students and the junior class has 300 students, and if the ratio of boys to girls in the senior class is 5:3, what is the ratio of boys to girls in junior class?

This problem gives a lot of information, and it sounds like it can be solved many different ways. But the

*first*task is to notice**what we are given**and**what we are asked**.- We are asked about a
__ratio__. - We're given one ratio, and all kinds of totals. There are
**boys**and**girls**,**senior**and**junior**classes. In other words, there are four groups: senior boys, senior girls, junior boys and junior boys.

To start, you could for example notice the two facts given about the senior class: The senior class

**has 400**students and the**ratio of boys to girls**is 5:3.From this it is easy to solve the number of boys and girls in the senior class.

The ratio 5:3 means 5/8 of them are boys and 3/8 of them are girls. Now, 5/8 of 400 is 250, and 3/8 of it is 150.

So seniors are solved... now to juniors. Let

B1 = boys in junior class

G1 = girls in junior class.

The first sentence of the problem gives us an equation:

G1 + 150 = B1 + 250

We also now that G1 + B1 = 300.

So let's solve this

**system of two equations**:G1 + 150 = B1 + 250

G1 + B1 = 300

I can subtract the bottom one from the top one to get:

150 - B1 = B1 - 50

200 = 2B1

B1 = 100

Since the total was 300, then G1 is 200.

And now the ratio of boys to girls, it is 100:200 or 1:2. All done!

You could also use straightforward algebra and set up four equations originally, using B1, G1, B2, and G2 for the numbers of boys and girls in junior and senior classes. You would get these four equations:

G1 + G2 = B1 + B2

B2 + G2 = 400

B1 + G1 = 300

B2/G2 = 5/3

And from these four you can solve all four unknowns, and then get the ratio of boys to girls in the junior class.

## Comments

That seems like a much more efficient way to get the answer.

My brain just started doing it a particular way, and I soon got the answer that way. So, that way was "efficient" for me.

I realize sometimes it is worthwhile to spend some time thinking WHICH way would be the most efficient, before even solving the problem, but this was simple enough that that approach is not absolutely necessary.

Appreciate if you can comment on my approach. Cheers.

--Chi Hung

I think the model is great. Like I said, there are many ways to solve this.

But when I tutor kids for test prep, efficiency and insight become more important because of time constraints, so I tend to go for what strikes me as clever and quick over pretty much everything.

Here's a problem on a recent ACT:

As shown in the standard (x,y) coordinate plane below, P(6,6) lies on the circle with center (2,3) and radius 5 coordinate units). What are the coordinates of the image of P after the circle is rotated 90 degrees clockwise about the center of the circle?

(The graph is given).

A) (2, 3) B) (3, 2) C) (5, -1)

D) (6, 0) E) (7, 3)

Intriguingly, and to my mind all too predictably, students struggle with this problem because they actually try to do calculations or otherwise ascertain an exact answer. (By the way, it's #37 of 60 problems, so above average in difficulty).

But I would hope that it's obvious to students (operant word being "hope") that a 90 degree rotation about the center of a circle by a point in QI will make the image lie in QIV and hence the coordinates must have signs (+, -) respectively. Thus, C is the only plausible answer. I find many students choose D, which suggests many things: that they aren't thinking about quadrants; that they don't visualize 90 degrees very well in this context; and that in general the propensity to calculate first, rather than think before doing anything is a major source of error on both US standardized tests and, likely, international tests with similar problems used to compare nations.

Doing things the hard way sometimes makes a lot of sense, but here we see that not only is it inefficient considering that students only have 60 minutes for 60 problems of increasing difficulty/trickiness, but also more likely to lead to error.

I completely agree that a simple geometric approach is the best bet. As you said, looking at the picture is a lot quicker than solving the geometry, and yields the correct answer.

I disagree, however with what you said "But I would hope that it's obvious to students (operant word being "hope") that a 90 degree rotation about the center of a circle by a point in QI will make the image lie in QIV"

I believe this would only be true the circle was centered at the origin. Since we are rotating the circle off of the origin, its arcs are not equally split up in each quadrant, and the first quadrant contains more than 90 degrees of the circle. (This is evident if you consider the points where the circle meets the positive x and y axis. Clearly the angle from the point on the positive x axis to the center of the circle to the point on the positive y axis is greater than 90 degrees. Thus the point on the circle in the first quadrant just to the right of the y axis, if rotated 90 degrees about (2, 3), would still be in quadrant 1.)