Seniors & juniors algebra word problem

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

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.

This sounds like it can be solved by setting up some equations and using algebra.

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.

Seniors & juniors algebra word problem

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

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.

This sounds like it can be solved by setting up some equations and using algebra.

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.

Simplify a ratio problem, for your entertainment :)

Simplify the ratio 186:403. The answer is 6:13. How do we get it?

To simplify ratios (or fractions), we need to find COMMON FACTORS of the two numbers. So, one way to do it is to first find the GCF (Greatest Common Factor) of 186 and 403. Then divide 186 and 403 by it.

Alternatively, just find ANY factor of 186 and 403, and divide both by it, to simplify the ratio somewhat, and to get started. Then repeat the process.

Okay, 403 is not divisible by 2, 3, 4, 5, 6. This I know by divisibility tests. Maybe it's divisible by 7... need to try (calculator). No, it isn't.

Maybe by 11? No.

Maybe by 13? YES. My calculator helps. 403 = 13 x 31. I happen to know both of these are primes, so therefore 403 doesn't have any other factors.

Then 186.... is it divisible by 13 or 31?
By 13, no.
By 31, YES!  186 = 31 x 6

So since 186 = 6 x 31 and 403 = 13 x 31, then the ratio 186:403 simplifies to 6:13. Clearly that's as far as we can get, as it's simplified to the lowest terms.

Simplify a ratio problem, for your entertainment :)

Simplify the ratio 186:403. The answer is 6:13. How do we get it?

To simplify ratios (or fractions), we need to find COMMON FACTORS of the two numbers. So, one way to do it is to first find the GCF (Greatest Common Factor) of 186 and 403. Then divide 186 and 403 by it.

Alternatively, just find ANY factor of 186 and 403, and divide both by it, to simplify the ratio somewhat, and to get started. Then repeat the process.

Okay, 403 is not divisible by 2, 3, 4, 5, 6. This I know by divisibility tests. Maybe it's divisible by 7... need to try (calculator). No, it isn't.

Maybe by 11? No.

Maybe by 13? YES. My calculator helps. 403 = 13 x 31. I happen to know both of these are primes, so therefore 403 doesn't have any other factors.

Then 186.... is it divisible by 13 or 31?
By 13, no.
By 31, YES!  186 = 31 x 6

So since 186 = 6 x 31 and 403 = 13 x 31, then the ratio 186:403 simplifies to 6:13. Clearly that's as far as we can get, as it's simplified to the lowest terms.

Ratio word problem solved with block model and algebra

I guess it is time for some more problem solving, since someone sent this question in.

Two numbers are in the ratio of 1:2. If 7 be added to both, their ratio changes to 3:5. What is the greater number?

We can model the two original numbers with blocks. 1 block and 2 blocks makes the ratio to be 1:2.

|-------|

|-------|-------|

Now add the same thing to both (the 7):

          7
|-------|---|

|-------|-------|---|
                 7 

The way I just happened to draw these suggests that I could just split the original block in two, and the problem is solved:

          7
|---|---|---|

|---|---|---|---|---|
                 7 

Here, each little block is 7. The original larger blocks are 14 each.

So the original bigger number, which had two larger blocks, is 28, and the smaller number is 14.

Check:
Their ratio is 28:14 = 2:1. If you add 7 to both, you have 35 and 21, and their ratio is 35:21 = 5:3.


Solving the same problem using algebra

The two numbers in the ratio of 1:2 are x and 2x.

Once 7 is added to both, we have x + 7 and 2x + 7. Their ratio is 3:5, and we can write a proportion using fractions:

x + 7           3
-------   =   ----
2x + 7          5

Cross-multiply to get

5(x + 7) = 3(2x + 7)

5x + 35 = 6x + 21
35 - 21 = x

x = 14

The larger number was 2x or 28. We already checked this earlier.

Ratio word problem solved with block model and algebra

I guess it is time for some more problem solving, since someone sent this question in.

Two numbers are in the ratio of 1:2. If 7 be added to both, their ratio changes to 3:5. What is the greater number?

We can model the two original numbers with blocks. 1 block and 2 blocks makes the ratio to be 1:2.

|-------|

|-------|-------|

Now add the same thing to both (the 7):

          7
|-------|---|

|-------|-------|---|
                 7 

The way I just happened to draw these suggests that I could just split the original block in two, and the problem is solved:

          7
|---|---|---|

|---|---|---|---|---|
                 7 

Here, each little block is 7. The original larger blocks are 14 each.

So the original bigger number, which had two larger blocks, is 28, and the smaller number is 14.

Check:
Their ratio is 28:14 = 2:1. If you add 7 to both, you have 35 and 21, and their ratio is 35:21 = 5:3.


Solving the same problem using algebra

The two numbers in the ratio of 1:2 are x and 2x.

Once 7 is added to both, we have x + 7 and 2x + 7. Their ratio is 3:5, and we can write a proportion using fractions:

x + 7           3
-------   =   ----
2x + 7          5

Cross-multiply to get

5(x + 7) = 3(2x + 7)

5x + 35 = 6x + 21
35 - 21 = x

x = 14

The larger number was 2x or 28. We already checked this earlier.

A simple ratio problem

Problem: If a:b = 1:3 and b:c = 3:4, find a:c.

Two ratios are given, third is to be found. This is very very simple. The picture shows the two given ratios as blocks.

We can see that a is one block and c is four blocks, so the ratio a:c is 1:4.

You don't need an image for that, of course, since the original ratios are so easy. If a:b=1:3 and b:c=3:4, b being the same in both cases, we can write the ratio a:b:c as 1:3:4 right off.

But what if the numbers weren't so friendly? What if it said this way:

If a:b = 1:3 and b:c = 5:7, find a:c.

This is solvable in various ways. I'll use equivalent ratios, in other words change the given ratios to equivalent ratios until we find ones where the b's are the same.

In the first ratio, 1:3, b is 3. In the other ratio, 5:7, it is 5. We can make those to be 15 by changing the ratios to equivalent ratios - which is done in an identical manner as changing fractions to equivalent fractions.

1:3 = 5:15 and 5:7 = 15:21.

Now the ratio of a:b:c is 5:15:21, so the asked ratio a:c is 5:21.

A simple ratio problem

Problem: If a:b = 1:3 and b:c = 3:4, find a:c.

Two ratios are given, third is to be found. This is very very simple. The picture shows the two given ratios as blocks.

We can see that a is one block and c is four blocks, so the ratio a:c is 1:4.

You don't need an image for that, of course, since the original ratios are so easy. If a:b=1:3 and b:c=3:4, b being the same in both cases, we can write the ratio a:b:c as 1:3:4 right off.

But what if the numbers weren't so friendly? What if it said this way:

If a:b = 1:3 and b:c = 5:7, find a:c.

This is solvable in various ways. I'll use equivalent ratios, in other words change the given ratios to equivalent ratios until we find ones where the b's are the same.

In the first ratio, 1:3, b is 3. In the other ratio, 5:7, it is 5. We can make those to be 15 by changing the ratios to equivalent ratios - which is done in an identical manner as changing fractions to equivalent fractions.

1:3 = 5:15 and 5:7 = 15:21.

Now the ratio of a:b:c is 5:15:21, so the asked ratio a:c is 5:21.

A bar diagram to solve a ratio problem

Dave at MathNotations had an interesting ratio problem:

In Virtual HS, the ratio of the number of juniors to seniors is 7:5.
The ratio of (the number of) junior males to junior females is 3:2.
The ratio of senior males to senior females is 4:3.
What is the ratio of junior males to senior females?

He asked if it can be solved using "Singapore" style bar model.

I'm not sure if this is exactly how they'd do it, but this is how I'd do it... so here goes.

After I made the diagrams, I soon saw that Dave's numbers are two awkward; the bar diagram drawing would get too messy because we'd need to divide it into too tiny parts to see anything.

BUT... you probably know about the PROBLEM SOLVING STRATEGY called "solve an easier problem". My agenda is therefore:

• show how to solve a few related simple ratio problems using the bar diagram


• solve a variant of the original problem (with friendly numbers)


• solve the original problem.

1. Here's a bar diagram representing the fact that the ratio of the number of juniors to seniors is 7:5.

As you can see, the "whole" ends up divided into 12 parts (7 + 5).

We can use this diagram to solve problems such as:

If there are 768 juniors and seniors in total, how many juniors are in the school?
(Divide that into 12 parts, and then take 7 of those parts.)

Or...

If there are 95 seniors, how many juniors are there?
(Divide the number of seniors by 5, then multiply by 7.)

Or..
If there are 49 Juniors, how many students are there in all?
(Divide the number of juniors by 7, then multiply by 12.)

The diagram makes all this dividing/multiplying by 5/7/12 all crystal clear.

---

2. Let's change the numbers from the original problem to make them "friendly" for this approach. Let's solve this instead:

In Virtual HS, the ratio of the number of juniors to seniors is 7:5.
The ratio of (the number of) junior males to junior females is 4:3.
The ratio of senior males to senior females is 7:3.
What is the ratio of junior males to senior females?

Our diagram becomes, first of all, like before:

But then we add the additional information into it. The ratio of junior males to females is 4:3, which means juniors as a whole are divided into 7 parts (4 + 3). Similarly, seniors need divided into 10 parts.

Voila! It just so happens (wonder why?) that our original division into 7 and 5 parts works beautifully to give us juniors divided into 7 parts (they already are!) and seniors into 10 parts if we only split each part into 2.

So, what is the ratio of junior males to senior females?

In the diagram, the WHOLE is now divided into 24 parts. Junior males are 8 of those parts, and senior females are 3 of those parts. Their ratio is therefore 8:3.

---

3. Solving the original problem, with those unfriendly numbers.... the diagram looks sort of like this:

It gets messy... I'd prefer using the bar model as a stepping stone and to illustrate the basic situation, but eventually using the "least common denominator" or algebraic methods.

A bar diagram to solve a ratio problem

Dave at MathNotations had an interesting ratio problem:

In Virtual HS, the ratio of the number of juniors to seniors is 7:5.
The ratio of (the number of) junior males to junior females is 3:2.
The ratio of senior males to senior females is 4:3.
What is the ratio of junior males to senior females?

He asked if it can be solved using "Singapore" style bar model.

I'm not sure if this is exactly how they'd do it, but this is how I'd do it... so here goes.

After I made the diagrams, I soon saw that Dave's numbers are two awkward; the bar diagram drawing would get too messy because we'd need to divide it into too tiny parts to see anything.

BUT... you probably know about the PROBLEM SOLVING STRATEGY called "solve an easier problem". My agenda is therefore:

• show how to solve a few related simple ratio problems using the bar diagram


• solve a variant of the original problem (with friendly numbers)


• solve the original problem.

1. Here's a bar diagram representing the fact that the ratio of the number of juniors to seniors is 7:5.

As you can see, the "whole" ends up divided into 12 parts (7 + 5).

We can use this diagram to solve problems such as:

If there are 768 juniors and seniors in total, how many juniors are in the school?
(Divide that into 12 parts, and then take 7 of those parts.)

Or...

If there are 95 seniors, how many juniors are there?
(Divide the number of seniors by 5, then multiply by 7.)

Or..
If there are 49 Juniors, how many students are there in all?
(Divide the number of juniors by 7, then multiply by 12.)

The diagram makes all this dividing/multiplying by 5/7/12 all crystal clear.

---

2. Let's change the numbers from the original problem to make them "friendly" for this approach. Let's solve this instead:

In Virtual HS, the ratio of the number of juniors to seniors is 7:5.
The ratio of (the number of) junior males to junior females is 4:3.
The ratio of senior males to senior females is 7:3.
What is the ratio of junior males to senior females?

Our diagram becomes, first of all, like before:

But then we add the additional information into it. The ratio of junior males to females is 4:3, which means juniors as a whole are divided into 7 parts (4 + 3). Similarly, seniors need divided into 10 parts.

Voila! It just so happens (wonder why?) that our original division into 7 and 5 parts works beautifully to give us juniors divided into 7 parts (they already are!) and seniors into 10 parts if we only split each part into 2.

So, what is the ratio of junior males to senior females?

In the diagram, the WHOLE is now divided into 24 parts. Junior males are 8 of those parts, and senior females are 3 of those parts. Their ratio is therefore 8:3.

---

3. Solving the original problem, with those unfriendly numbers.... the diagram looks sort of like this:

It gets messy... I'd prefer using the bar model as a stepping stone and to illustrate the basic situation, but eventually using the "least common denominator" or algebraic methods.