The elementary grades 1-4 laid the foundations, the basics of arithmetic: addition, subtraction, multiplication, division, and fractions. In grade 5, students are expected to master most aspects of fraction math and begin working with the rest of the Math Monsters: decimals, ratios, and percents (all of which are specialized fractions).
Word problems grow ever more complex as well, and learning to explain (justify) multi-step solutions becomes a first step toward writing proofs.
This installment of my elementary problem solving series is based on the Singapore Primary Mathematics, Level 5A.
For your reading pleasure, I have translated the problems into the world of J.R.R. Tolkien’s classic, The Hobbit.
UPDATE: Problems have been genericized to avoid copyright issues.
The hobbit had 3 times as many apple tarts as mince pies in his larder. If he had 24 more apple tarts than mince pies, how many of the pastries (both tarts and pies) did he have altogether?
Long before ratios are specifically introduced, students get plenty of practice with the informal ratios twice as many and three times as many. In each case, we draw the smaller group as one unit, and the larger group as whatever number of units make the “times as many.”
We know there are 24 more tarts than pies, so:
2 units = 24
1 unit = 24 2 = 12
And finally, we count how many units we have in all:
4 units = 4 12 = 48
There are 48 pastries altogether.
The three trolls had 123 pieces of gold. Tom had 15 pieces of gold more than Bert. Bert had 3 pieces fewer than William. How many pieces of gold did William have?
This is a comparison problem, where they tell us how much more or less one thing is than another. To model it, we will draw a bar for each thing we’re comparing. The left edges of the bars line up, making it easy to compare which is larger or smaller.
In this case, we will need three bars — one for each troll’s amount of gold. We’ll label each treasure with the troll’s name. (My kids would usually just put the troll’s initial.) Notice that Bert has the smallest number of pieces. Tom has Bert’s amount plus 15 more, and William has Bert’s amount plus 3 more.
We could make an algebra equation:
But we won’t. Remember, this is a 5th grade problem! Instead, we’ll use the “stealth algebra” of our diagram to help us think through the numbers.
All the gold together is 123 pieces. We can imagine taking away the extra bits, reducing each of the other trolls’ loot to match Bert’s stash:
3 units = 123 – 18 = 105
1 unit = 105 3 = 35
And now, because I forgot to put a question mark in my drawing, I have to go back and read the problem again. Oh, yes: we need to find William’s amount.
1 unit + 3 = 35 + 3 = 38
William had 38 gold coins.
The Great Goblin had twice as many goblin soldiers as his cousin, the Gross Goblin. How many soldiers must the Great Goblin send to his cousin so that they will each have 1200 goblin soldiers?
What a cool problem! It almost seems like they haven’t given enough information, doesn’t it?
Students who are not used to bar diagrams often get confused by transfer problems, which have a beginning situation and then something is moved from one person to another to set up the end of the story. Desperate children will grab any number and guess at the answer: “Send half of them. 1200 2…”
Let’s see what the bar diagrams tell us. First, the starting ratio:
And this is what we want to end up with:
But how can we get there? A useful problem-solving tool is to work backwards. In this case, we’ll start with the end of the story, since that’s where they gave us a number to work with. We know how many goblins will be in each army, which means we can find the total number of soldiers:
All goblin soldiers = 1200 2 = 2400
Aha! Moving some of the soldiers around wouldn’t change the total number, so there must have been 2400 soldiers in the first diagram, too:
3 units = 2400
1 unit = 2400 3 = 800
But how many soldiers need to move? The diagram makes the answer clear: We need to move half a unit:
1/2 unit = 800 2 = 400
And to double-check:
1 unit + 400 = 800 + 400 = 1200
The Great Goblin sent 400 soldiers to his cousin.
There are usually many ways to approach any word problem. Kitten chose a method similar to, but shorter than, the way I did it. She started with the simple 2:1 ratio, and then figured out that the Great Goblin would have to send 1/4 of his soldiers:
Since the “extra” unit had to be divided in half, she divided all the other units in half, too. Finally, she realized that the three smaller units remaining in the Great Goblin’s army must be 1200 soldiers:
3 units = 1200
1 unit = 1200 3 = 400
The Great Goblin sent 400 soldiers to his cousin.
The cave creature caught 10 small fish. He divided the fish to make 4 equal meals. How many fish did he eat at each meal?
The 5th grade unit on fractions begins by making explicit something that in earlier grades has only been implied — the connection between fractions and division. I tell my students, “The division symbol looks like a little fraction, with dots for the numbers. Let that remind you: Every fraction is a division problem, top divided by bottom. And any division problem can be written as a fraction.”
He had 10 fish, split into 4 equal groups. Each group is 1/4 of the whole 10:
1 unit = 10 4 = 10/4
And putting that into simplest form:
10/4 = 5/2 = 2
The creature ate fish at each meal.
The bear-shifter baked a large loaf of wholegrain bread. He ate 1/3 of the loaf himself (with plenty of honey!), and he sliced 1/2 of the same loaf to feed the dwarves and their friend. What fraction of the loaf was left?
Bar diagrams help students see the need for a common denominator. Unless all the pieces are the same size, it doesn’t make much sense to say, “We have one piece left.”
The bear-shifter cut 1/3 of the loaf from one side, and 1/2 of the loaf from the other:
What size is the white piece in the middle? It is clearly half of the middle third. Let’s get a common denominator by cutting all of the thirds in half.
Aha! Half of a third is the same as a sixth.
There was 1/6 of his loaf of bread left.
Again, Kitten doesn’t think the way I do. She finds adding fractions much easier than subtracting them, so she reasoned, “First I’ll find out how much they ate. Then I can see what is left.”
There was 1/6 of his loaf of bread left.
The king of the elves had a barrel of fine wine. His butler poured 3/4 gallon of it into a small keg. He drank 1/2 of the keg and gave the other half to his friend, the chief of the guards. How much wine did the butler drink?
I teach my students to connect the idea of multiplication with the preposition “of” (see If It Ain’t Repeated Addition, What is It? and A Mathematical Trauma). In this problem, we need to find 1/2 of 3/4 of a gallon, which means:
This could be done on a single bar, but I find it easier to draw two:
Notice that the line that splits the keg in half also bisects the unit above it (the second fourth of the gallon). Let’s make an equivalent fraction by splitting all of the units in half:
The butler drank 3/8 of a gallon of wine.
2/3 of the items in the dragon’s treasure were made of gold. 1/4 of the remaining part was precious gems. What fraction of treasure was precious gems?
As students gain skill in working with fractions, the problems grow correspondingly more complex. Now we introduce the fraction of the remaining part problem. In this type of problem, what was a part of the original whole becomes a new “whole thing” to be cut up into parts of its own.
Here, the treasure is divided into thirds, and one of these thirds is our remaining part. We show that it is being treated separately by drawing a new bar below the first, connected with lines to its original position.
We are interested in 1/4 of the remaining part, and we want to know what fraction of the original bar (the whole treasure) it would be:
Well, if our remaining third is cut into fourths, we could make a common denominator by cutting all of the thirds the same way. What size pieces would we have then?
1/12 of the treasure was precious gems.
When the king of the elves heard the dragon had been killed, he set out to claim a share of the treasure. 2/5 of his army were archers. 1/2 of the remainder fought with spears, and the rest carried swords. If 300 soldiers carried swords, how many elves marched out with the king of the elves?
Here is another remaining part puzzle. In this case, our original bar (representing the army) is divided into fifths, two of which are archers. The other three fifths become our remainder bar, of which 1/2 fight with spears:
The last chunk is the swordsmen (swords-elves?), and there are 300 of them. That means we can find the size of our remaining part:
1 unit = 300
2 units = 300 2 = 600
If the remaining part of the army is 600, then on the original bar:
3 units = 600
1 unit = 600 3 = 200
5 units = 200 5 = 1000
There were 1,000 elves marching with their king.
The master bowman gathered a small army of 600 survivors from the town the dragon had destroyed. The ratio of archers to swordsmen was 2:3. How many archers followed the bowman to the last battle?
Now our students have matured, graduating from the simple three times as many situations to full-fledged ratios. Without bar diagrams (or similar pictorial methods), middle school students find ratios an abstract and difficult subject, but a diagram makes it easy to see relationships.
The ratio tells how many parts (units) to draw for each group. 2:3 means 2 units of archers and 3 units of swordsmen:
5 units = 600
1 unit = 600 5 = 120
2 units = 120 2 = 240
240 archers followed the bowman to the last battle.
The dwarves rewarded their hobbit friend with two chests of gold, silver, and small gems — 6000 pieces of treasure altogether. There were twice as many pieces of gold as there were gems. There were 600 more pieces of silver than gems. How much of each type of treasure did the hobbit receive?
At the end of our textbook, we meet another comparison problem, like the earlier one with the trolls. Of course, we hope our students have learned something in the interim, so this problem will take a few extra steps — for example, we are asked to find the amount for all three types of treasure, not just for one of them.
We are comparing the number of pieces of gold, gems, and silver, so we will need three bars. The fewest are the gems. There are twice as many (notice the ratio?) pieces of gold. And the pieces of silver match the gems plus 600 more:
As we did earlier in the troll problem, we first remove the “extra” pieces. This will let us work with the unknown units by themselves:
4 units = 6000 600 = 5400
1 unit = 5400 4 = ?
Excuse me while I do the long division…
1 unit = 1350
2 units = 1350 2 = 2700
1 unit + 600 = 1950
And on complex problems, it’s always a good idea to double-check:
1350 + 2700 + 1950 = 6000
(Yes, it really is important to check! I made a mental math error as I was typing this post, which I caught only because I took the time to add up my answers and see if they made sense.)
The hobbit received 2,700 pieces of gold; 1,350 gems; and 1,950 pieces of silver.
Practice and Learn
To get more practice creating bar diagrams, your students may enjoy these online tutorials:
And for some fun practice with fractions, geometry, decimals, percentages, averages, and more: