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Order of Operations: Free Response

5 questions in parts, 57 points in total. Work each one out on paper, taking a hint if you get stuck. When you have an answer, reveal the answer to check it, and the full solution only if you still want it. The rubric is there so you can mark your own work.

Free response · work it on paper Question 1 of 5

1. What a bare line does not decide . Foundational, 10 points. Question 1 of 5.

A notice board carries the expression $7 + 6 \times 2$ and nothing else. Two readers work it out and get different answers. Neither of them has made a mistake in arithmetic. This question asks what the symbols decide on their own, and what had to be agreed.
1. Part A.
  Work out $7 + 6 \times 2$ twice. First take the operations in the order they are written. Then take the multiplication first. Report both values, say which reading gave each one, and say which value the order of operations selects.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Write one expression that forces the addition to happen first, and one that forces the multiplication to happen first. Use grouping symbols, so that no reader has to guess. Then say which of your two pairs of grouping symbols the convention already makes unnecessary.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  A basic calculator works each step out as you press it, so keying in $7 + 6 \times 2$ shows $26$ . Your textbook prints the same line and calls it $19$ . Neither has made an arithmetic slip. Say what the calculator and the textbook are doing differently. Then rewrite the notice board's line so that it names $19$ both ways: worked a step at a time as the keys arrive, and read as a finished line under the tiers.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
Hints
One at a time, each one a step further than the last. Take only as many as you need.
1. Hint 1 of 3
  None of this needs a hard calculation. It turns on one point: a line of symbols is not yet a set of instructions until an agreement says how to read it.
2. Hint 2 of 3 · Part B
  Grouping symbols do not change the numbers. They change which operation happens first. Once you have written both lines, check each one against the two values you already found in part A.
3. Hint 3 of 3 · Part C
  Before deciding who is wrong, check whether both values can be reached by correct arithmetic. If they can, the disagreement is about reading, not about calculating, and the repair is something you write rather than something you work out.
That is every hint for this question.
Answer and solution
Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.
The answer

Part A

Taking the operations in written order gives $26$ . Taking the multiplication first gives $19$ , and $19$ is the value the order of operations selects.

Part B

$(7 + 6) \times 2$ forces the addition, and $7 + (6 \times 2)$ forces the multiplication. The second pair is the unnecessary one, since the convention already reads the bare string that way.

brackets or braces read exactly as round parentheses do, so $[\,7 + 6\,] \times 2$ is the same expression

Part C

The calculator never reads the whole line. It works each step out as the keys arrive, so it settles $7 + 6$ first and shows $26$ . The textbook reads the finished line under the tiers, so it settles the multiplication first and gets $19$ . Written as $6 \times 2 + 7$ , the multiplication comes first either way, so both report $19$ .

Not what you got? Look for the slip on your own paper before you open the solution. Finding it yourself is worth more than reading it.
Worked solution

Part A

Each reading is a complete calculation, so carry both out in full.

Taking the operations in the order they appear, the addition happens first and the sum is then doubled:

$7 + 6 = 13, \qquad 13 \times 2 = 26.$

Taking the multiplication first, the product is formed and the $7$ is added at the end:

$6 \times 2 = 12, \qquad 7 + 12 = 19.$

The two readings give $26$ and $19$ . Both calculations are correct arithmetic, so the disagreement is not about arithmetic. It is about what the written line asks for. The order of operations picks one reading, and it puts multiplication on a higher tier than addition, so the line names $19$ .

Part B

A pair of grouping symbols says: settle what is inside me first. So enclosing the operation you want first removes every doubt.

To force the addition, enclose it:

$(7 + 6) \times 2 = 13 \times 2 = 26.$

To force the multiplication, enclose that instead:

$7 + (6 \times 2) = 7 + 12 = 19.$

Each of these says exactly one thing, and between them they account for both values from part A.

The second pair tells a reader nothing they did not already have. The convention puts multiplication on the higher tier, so $7 + 6 \times 2$ and $7 + (6 \times 2)$ are read the same way, and those brackets may be dropped. The first pair cannot be dropped, because without it the line says something else. That is what the agreement buys. The reading we want most often needs no brackets at all, so an ordinary formula does not fill up with them.

Part C

Both values are already on the table from part A:

$(7 + 6) \times 2 = 26, \qquad 7 + (6 \times 2) = 19.$

Neither line is an error. Each one correctly works out a different quantity.

The calculator is not choosing between them, because it never sees the whole line. It acts on each key as it arrives, so by the time the times key is pressed it is already holding $13$ , and it doubles that. The textbook does something else. It reads a finished line under the order of operations, which puts multiplication on the higher tier and gives $19$ .

Brackets settle it for a human reader, since $7 + (6 \times 2)$ leaves nothing to choose. They do nothing for this calculator, which acts before it could ever see the closing one.

What works for both is to move the multiplication to the front:

$6 \times 2 + 7 = 12 + 7 = 19.$

Now the multiplication is the first thing either reader meets. The calculator is holding $12$ before the $+$ key arrives, and the tiers put the same product first, so the notice board names $19$ whoever reads it.

In one line

In written order $7 + 6 \times 2$ comes to $26$ , and with the multiplication taken first it comes to $19$ . The order of operations selects $19$ . Both values can be written without doubt as $(7 + 6) \times 2 = 26$ and $7 + (6 \times 2) = 19$ , and the second pair of brackets is the one the convention already supplies. A calculator that works each key out as it arrives never reads the whole line at all, so the repair for the notice board is $6 \times 2 + 7$ , which names $19$ either way.
Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Carries both readings out in full, rather than working one out and describing the other. . Worth 2 points.

Labels each of the two values with the reading that produced it, and states which one the convention selects. . Worth 1 point.

Part B 3 points

Produces one expression for each of the two values, with the grouping placed around the operation that is meant to go first. . Worth 2 points.

Singles out the pair of grouping symbols the convention already supplies, rather than treating both pairs as equally necessary. . Worth 1 point.

Part C 4 points

Locates the difference in how each one reads the line, a step at a time as the keys arrive against the finished line under the tiers, rather than in a calculation error. . Worth 2 points. needs an explanation, not just an answer

Gives a line that names the same value both ways, worked a step at a time and read under the tiers, and says which value that is. . Worth 2 points.
2. The tier that goes first, and why that ranking is useful . Reasoning, 11 points. Question 2 of 5.

The tier list can look arbitrary, as though someone wrote it down and everyone else memorised it. This question compares the two readings of $6 + 4 \times 5$ and then asks why the standard ranking is useful, and how it saves parentheses.
1. Part A.
  The product $4 \times 5$ is a short way of writing a repeated addition of equal groups. Rewrite $6 + 4 \times 5$ with that repeated addition written out in full. Keep the repeated copies together as one quantity, and give the total.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Now read the rival version, $(6 + 4) \times 5$ . Say in words what collection of equal groups it describes, give its total, and say what has happened to the $6$ that stood on its own in part A.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points
3. Part C.
  Use your two readings to say why the standard ranking is useful, rather than only which answer it gives. Then write down what a reader would have to put on the page to name your part A total, if the ranking went the other way.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
Hints
One at a time, each one a step further than the last. Take only as many as you need.
1. Hint 1 of 3
  Quoting the rule will not settle this, because the rule is the thing being explained. Go back to what a product means before anyone works it out: a count of equal groups, written short.
2. Hint 2 of 3 · Part A
  Write the copies out and then enclose them. Decide first how many copies there are and how big each one is, and take care over which of the two numbers plays which role.
3. Hint 3 of 3 · Part C
  Look at the brackets in your part A line. Ask what they are holding together, and what would happen to those copies if the addition were settled first instead.
That is every hint for this question.
Answer and solution
Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.
The answer

Part A

$6 + (5 + 5 + 5 + 5)$ , and the total is $26$ .

$6 + (4 + 4 + 4 + 4 + 4)$ is equally correct: nothing in the question pins which of the two numbers counts the groups. Keep whichever reading you choose for part B as well, where the rival candidate then describes five copies of $10$ instead.

Part B

It describes ten copies of $5$ , which is $50$ . The $6$ is now part of the count of copies, instead of standing on its own.

Part C

The two readings describe different collections, so the symbols do not decide and the ranking is a choice. The standard choice is useful because it lets the product stay inside the sum with no brackets around it. Under the other ranking, the same total would have to be written $6 + (4 \times 5)$ .

Not what you got? Look for the slip on your own paper before you open the solution. Finding it yourself is worth more than reading it.
Worked solution

Part A

Written out in full, $4 \times 5$ is four copies of $5$ :

$4 \times 5 = 5 + 5 + 5 + 5.$

Put that in place of the product. Keep the four copies together, because they are what the single symbol group $4 \times 5$ stands for:

$6 + 4 \times 5 = 6 + (5 + 5 + 5 + 5).$

The copies come to $20$ , and the lone $6$ is added at the end:

$6 + 20 = 26.$

So this reading counts four copies of $5$ , with one separate $6$ standing beside them.

Part B

Inside the brackets the sum settles first, so the expression becomes

$(6 + 4) \times 5 = 10 \times 5,$

which stands for ten copies of $5$ :

$10 \times 5 = 50.$

If you matched the two numbers the other way round in part A, read this as five copies of $10$ throughout. The total and the comparison below are unchanged.

Now compare the two collections. Part A held four copies of $5$ and one separate $6$ beside them. This one holds ten copies of $5$ and nothing standing beside them. The $6$ has stopped being a quantity of its own and is now part of the count of groups.

So the two readings are not two routes to one number. They describe different collections.

Part C

Set the two readings side by side:

$6 + (5 + 5 + 5 + 5) = 26, \qquad (6 + 4) \times 5 = 50.$

The first counts four copies of $5$ and a separate $6$ . The second counts ten copies of $5$ and nothing separate. They are different collections, so the symbols on their own cannot pick one. The ranking is a choice.

What the standard choice buys is short writing. It leaves the product alone inside the sum, so $6 + 4 \times 5$ needs no brackets around $4 \times 5$ , and neither does any other line of that shape.

The other ranking would work too, as long as everyone used it. Readers there would take $6 + 4 \times 5$ to mean $50$ , so to name $26$ they would have to write

$6 + (4 \times 5) = 26$

every time. That is the cost the standard ranking avoids.

In one line

Written out, $6 + 4 \times 5 = 6 + (5 + 5 + 5 + 5) = 26$ : four copies of $5$ and one separate $6$ . The rival reading $(6 + 4) \times 5 = 50$ describes ten copies of $5$ , with the $6$ counted as extra groups, so the two readings describe different collections rather than two routes to one number. The symbols cannot choose between them, so the ranking is a choice, and the standard one is useful because the product then needs no brackets. Under the other ranking, $26$ would have to be written $6 + (4 \times 5)$ .

Another way: Count the pile instead of quoting the rule

The two readings can be settled with counters. Lay out four rows of five counters for the product, and put six loose counters beside them. Count everything on the table:

$5 + 5 + 5 + 5 + 6 = 26.$

Now try to build the rival reading from the same counters. It calls for ten rows of five, which needs $50$ counters, so it is not another way of counting this pile. It is a different pile. The counters cannot tell you which ranking to adopt. What they show is that the two readings count different piles, so something outside the symbols has to choose.

When it is worth it Whenever a rule feels arbitrary. Building both quantities by hand shows what the rule is choosing between.
Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Writes the product out as the correct number of equal copies, matching each of its two numbers to the count of groups or the size of a group. . Worth 2 points.

Keeps the copies together as one quantity, so that the lone term is added to the whole bundle rather than to a single copy. . Worth 2 points.

Part B 3 points

Describes the collection in words as a number of equal groups, rather than only reporting the total. . Worth 2 points.

Says what has happened to the term that stood on its own in part A. . Worth 1 point.

Part C 4 points

Says that the two readings describe different collections, so the ranking is a choice, and names what the standard choice saves: no brackets around the product. . Worth 3 points. needs an explanation, not just an answer

Writes the expression a reader would need for the same total under the other ranking. . Worth 1 point.
3. Same tier, and the reading that settles it . Foundational, 10 points. Question 3 of 5.

Multiplication and division share one tier, and addition and subtraction share another. Within a tier the agreed reading is strictly left to right. This question puts that reading to work on two chains, then asks which chains actually need it.
1. Part A.
  Work out $30 - 12 - 6$ twice, once with the leftmost subtraction settled first and once with the rightmost settled first. Report both values, say which grouping produced each, and state the value the convention names.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  A classmate says that in $54 \div 9 \times 3$ the multiplication has to be done before the division, because multiplication comes earlier in the mnemonic they memorised. Work out the value their belief produces and the value the convention produces, and report both.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  The chain $8 + 5 + 7$ also has two operations from one tier. Settle it both ways, decide whether the left to right reading is needed for that chain to name a single number, and say how that compares with the chain in part A.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
Hints
One at a time, each one a step further than the last. Take only as many as you need.
1. Hint 1 of 3
  A reading rule only earns its keep where two groupings of the same chain disagree. So test a chain for that first, and only then ask what the rule is doing there.
2. Hint 2 of 3 · Part B
  Put a bracket around whichever pair each reader reaches for first, and the two beliefs become two different expressions on paper: the classmate is working out $54 \div (9 \times 3)$ , while the convention starts from $54 \div 9$ . Evaluate each of those in full.
3. Hint 3 of 3 · Part C
  Once you hold the two values for the new chain, compare them with each other before you compare anything else. A reading rule can only choose between values that differ.
That is every hint for this question.
Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Evaluates both groupings in full, rather than one of them plus a description of the other. . Worth 1 point.

Labels the two values by the grouping that produced them, and states which one the left to right reading names. . Worth 2 points.

Part B 3 points

Works the classmate's belief out in full, rather than dismissing it without a calculation. . Worth 2 points.

Attaches each value to the reading that produced it, and states which of the two readings the convention selects. . Worth 1 point.

Part C 4 points

Settles the new chain both ways and reaches a verdict from those two values rather than from the rule itself. . Worth 2 points. needs an explanation, not just an answer

Sets that verdict beside the chain in part A, saying what the rule settles in each case. . Worth 2 points. needs an explanation, not just an answer

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

Evaluate $40 - 15 - 5$ under the convention, and show that grouping the last two numbers instead would give a different value. Then evaluate $36 \div 6 \times 2$ under the convention, and give the value a reader would reach who believed multiplication always outranks division.
4. Five households, and where the bracket closes . Application, 12 points. Question 4 of 5.

A community centre charges $9$ dollars for an adult ticket and $6$ dollars for a child ticket. Five households book together for one visit. Each household brings $2$ adults and $4$ children, and each household holds a member credit of $10$ dollars, which comes off that household's own tickets. The treasurer has to write the group's total as a single expression before paying it.
1. Part A.
  Write a single expression for the amount the five households pay in total, using grouping symbols so that each household's credit comes off that household's own tickets. Do not evaluate it.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Evaluate your expression one operation per line, naming at each step which group or which tier you are settling, and state the group's total with its unit.
  
  Carry your own answer forward Work from the expression you wrote in part A, whatever form it took. If that did not come out, settle the innermost group first and work outward, so that the total is reached one operation at a time.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  A volunteer writes the group's total as $5 \times (2 \times 9 + 4 \times 6) - 10$ . Work out what that line comes to, say what claim it makes about where the credit is applied, and identify what has changed about the grouping.
  
  Carry your own answer forward Judge the volunteer's line against the total you reached in part B, whatever it came to. If that did not come out, settle the group's total by a route you trust before you compare.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 5 points
Hints
One at a time, each one a step further than the last. Take only as many as you need.
1. Hint 1 of 3
  Price one household completely before you think about five of them. A grouping symbol is how you say that a household's total is one quantity, and where it closes is what decides whether the credit belongs to a household or to the whole group.
2. Hint 2 of 3 · Part B
  Nothing outside a bracket may touch its contents until they have become one number, and inside the bracket the tiers still hold, so the two products are formed before anything is added on or taken away.
3. Hint 3 of 3 · Part C
  The volunteer's arithmetic may well be perfect. Ask instead how many credits their line hands out, and set that against how many the centre promised.
That is every hint for this question.
Answer and solution
Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.
The answer

Part A

$5 \times [\,(2 \times 9 + 4 \times 6) - 10\,]$

$5 \times (2 \times 9 + 4 \times 6 - 10)$ is also correct: the addition and the subtraction inside share a tier and run left to right, so the innermost pair may be dropped
the two ticket products may be written in either order, and brackets or braces read as parentheses do

Part B

$160$ dollars.

Part C

It comes to $200$ dollars. Its bracket encloses only the tickets, which leaves the credit outside the multiplication, so one credit reaches the group as a whole instead of one reaching each household.

Not what you got? Look for the slip on your own paper before you open the solution. Finding it yourself is worth more than reading it.
Worked solution

Part A

Build one household first, then account for all five.

A household's tickets are $2$ adults at $9$ dollars and $4$ children at $6$ dollars, which is one quantity made of two products:

$2 \times 9 + 4 \times 6.$

Inside that quantity the tiers already do the right thing. Each product is formed before the addition, which is exactly what pricing two kinds of ticket and then combining them needs, so no grouping has to be added.

The credit comes off that household's own tickets, so the subtraction has to be enclosed with them:

$(2 \times 9 + 4 \times 6) - 10.$

Five households pay five times that amount, and the outer bracket is what keeps the household total together as the thing being multiplied:

$5 \times [\,(2 \times 9 + 4 \times 6) - 10\,].$

The inner pair could be dropped, since the addition and the subtraction inside it share a tier and run left to right. The outer pair could not. It is the only thing saying that the credit belongs to a household rather than to the group.

Part B

Work from the inside out, settling one thing at a time.

The innermost group holds two products, a sum and a subtraction. The products are on the higher tier, so they go first:

$(2 \times 9 + 4 \times 6) - 10 = (18 + 4 \times 6) - 10.$

$(18 + 4 \times 6) - 10 = (18 + 24) - 10.$

Now the addition and the subtraction, which share a tier and run left to right:

$(18 + 24) - 10 = 42 - 10.$

$42 - 10 = 32.$

That $32$ is one household's share in dollars. The group is now a single multiplication:

$5 \times 32 = 160.$

The five households pay $160$ dollars in total.

Part C

Evaluate the volunteer's line under the ordinary tiers. The bracket settles first:

$5 \times (2 \times 9 + 4 \times 6) - 10 = 5 \times 42 - 10.$

Multiplication outranks subtraction, so the product is formed before the credit is taken off:

$5 \times 42 - 10 = 210 - 10 = 200.$

So the volunteer's line comes to $200$ dollars, against the $160$ dollars the households owe.

The arithmetic in that line is faultless. What is wrong is what it says. Its bracket closes straight after the tickets, which leaves the credit outside the multiplication, and an operation written outside a multiplication happens once. So the line charges five households' tickets in full and then hands back a single credit of $10$ dollars. Four of the five promised credits never happen, and the gap between the two totals is exactly those four:

$200 - 160 = 40.$

The repair is to move the closing bracket to after the credit, so that the subtraction sits inside the quantity being multiplied by five. Where a grouping symbol closes decides what the expression claims.

In one line

The group's total is $5 \times [\,(2 \times 9 + 4 \times 6) - 10\,]$ , which comes to $160$ dollars. The volunteer's line $5 \times (2 \times 9 + 4 \times 6) - 10$ comes to $200$ dollars, because closing the bracket straight after the tickets leaves the credit outside the multiplication: one credit reaches the group instead of one reaching each household, and the gap of $40$ dollars is the four credits that never happened.
Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Represents every quantity the situation states: both ticket prices, both ticket counts, the credit, and the number of households. . Worth 2 points.

Uses grouping alone to show which amounts belong to a single household and which apply to the group, without evaluating anything. . Worth 2 points.

Part B 3 points

Settles the innermost group completely before the multiplication that surrounds it. . Worth 1 point.

Inside the group, forms each ticket product before combining them, and applies the credit after that combining rather than to one product. . Worth 1 point.

Reports the result as an amount of money, with the unit attached. . Worth 1 point.

Part C 5 points

Evaluates the volunteer's line correctly under the tiers, so that the comparison rests on what that line actually says rather than on a guess. . Worth 2 points.

Reads the line back into the situation, accounting for the credits it applies against the credits the centre promised. . Worth 2 points.

Locates the difference in where a grouping symbol closes, and states the change that would make the line say what the centre promised. . Worth 1 point. needs an explanation, not just an answer

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

A sports club orders kit for $4$ teams. A shirt costs $12$ dollars and a cap costs $7$ dollars. Each team takes $3$ shirts and $2$ caps, and each team holds a voucher for $15$ dollars that comes off its own order. Write one expression for the club's total and evaluate it, then work out what the line $4 \times (3 \times 12 + 2 \times 7) - 15$ would come to instead.
5. The grouping symbol that is not written . Reasoning, 14 points. Question 5 of 5.

A stacked fraction such as $\dfrac{20 + 16}{4 + 5}$ contains no brackets at all, and yet readers do not disagree about what it says. The bar is doing grouping work that nobody writes down. This question makes that work explicit and then tests how far it reaches.
1. Part A.
  Evaluate $\dfrac{20 + 16}{4 + 5}$ , settling the top and the bottom separately before dividing, and state the single number the whole expression names.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Rewrite that fraction on one line, using a division sign and grouping symbols, so that it still names the same number. Then evaluate the string $20 + 16 \div 4 + 5$ , which is what is left when the bar's division is written out but its grouping is not.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  Consider $\dfrac{24}{2 \times 3}$ . Decide whether its one-line form may be written as $24 \div 2 \times 3$ , and support the decision by evaluating both.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
4. Part D.
  Describe one family of fractions whose one-line form needs no grouping symbols at all, and say why that family needs none.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
Hints
One at a time, each one a step further than the last. Take only as many as you need.
1. Hint 1 of 4
  The fraction has no brackets in it, so decide first what a reader supplies without being told. Whatever that turns out to be, you will have to write it down by hand once the bar is gone.
2. Hint 2 of 4 · Part B
  Writing the bar's division on one line, with no grouping added, leaves a string in which division outranks addition. Work out what that string divides, and by what, then check those against what the fraction was dividing.
3. Hint 3 of 4 · Part C
  The bottom of the new fraction is built out of an operation and the top is not. Evaluate the fraction and the proposed line separately, then ask which piece of the bottom the line never brought under the division.
4. Hint 4 of 4 · Part D
  Go back through parts B and C and ask, each time, what the brackets were holding together. Then ask what a top or a bottom would have to look like for there to be nothing to hold.
That is every hint for this question.
Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Settles the whole top and the whole bottom before dividing, rather than dividing a term of one by a term of the other. . Worth 2 points.

Reports one number as the value of the whole expression. . Worth 1 point.

Part B 3 points

Produces a one-line expression that names the same number as the stacked fraction, rather than a string that merely reuses the same numbers in the same order. . Worth 2 points.

Evaluates the unbracketed string under the tiers and reports its value, rather than assuming it must come to the same thing. . Worth 1 point.

Part C 5 points

Evaluates both the fraction and the proposed one-line form, and reaches the verdict from those two values rather than from an impression. . Worth 3 points. needs an explanation, not just an answer

Says what the proposed form does to the leftover factor, rather than only that the move is not allowed, and gives the repaired one-line form. . Worth 2 points.

Part D 3 points

Describes one family of fractions that needs no grouping when flattened, with an example. . Worth 1 point.

Gives the reason, that a bracket is needed only where an operation on one side of the bar would otherwise be reached by the tiers. . Worth 2 points. needs an explanation, not just an answer

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

Evaluate $\dfrac{15 + 9}{3 + 1}$ , write its one-line form with grouping symbols, and evaluate the string that keeps the bar's division but none of its grouping. Then decide whether $\dfrac{30}{5 \times 2}$ may be written on one line as $30 \div 5 \times 2$ .

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. What a bare line does not decide 10 points

Part A 3

Carries both readings out in full, rather than working one out and describing the other. 2
Labels each of the two values with the reading that produced it, and states which one the convention selects. 1

Part B 3

Produces one expression for each of the two values, with the grouping placed around the operation that is meant to go first. 2
Singles out the pair of grouping symbols the convention already supplies, rather than treating both pairs as equally necessary. 1

Part C 4

Locates the difference in how each one reads the line, a step at a time as the keys arrive against the finished line under the tiers, rather than in a calculation error. 2
Gives a line that names the same value both ways, worked a step at a time and read under the tiers, and says which value that is. 2

2. The tier that goes first, and why that ranking is useful 11 points

Part A 4

Writes the product out as the correct number of equal copies, matching each of its two numbers to the count of groups or the size of a group. 2
Keeps the copies together as one quantity, so that the lone term is added to the whole bundle rather than to a single copy. 2

Part B 3

Describes the collection in words as a number of equal groups, rather than only reporting the total. 2
Says what has happened to the term that stood on its own in part A. 1

Part C 4

Says that the two readings describe different collections, so the ranking is a choice, and names what the standard choice saves: no brackets around the product. 3
Writes the expression a reader would need for the same total under the other ranking. 1

3. Same tier, and the reading that settles it 10 points

Part A 3

Evaluates both groupings in full, rather than one of them plus a description of the other. 1
Labels the two values by the grouping that produced them, and states which one the left to right reading names. 2

Part B 3

Works the classmate's belief out in full, rather than dismissing it without a calculation. 2
Attaches each value to the reading that produced it, and states which of the two readings the convention selects. 1

Part C 4

Settles the new chain both ways and reaches a verdict from those two values rather than from the rule itself. 2
Sets that verdict beside the chain in part A, saying what the rule settles in each case. 2

4. Five households, and where the bracket closes 12 points

Part A 4

Represents every quantity the situation states: both ticket prices, both ticket counts, the credit, and the number of households. 2
Uses grouping alone to show which amounts belong to a single household and which apply to the group, without evaluating anything. 2

Part B 3

Settles the innermost group completely before the multiplication that surrounds it. 1
Inside the group, forms each ticket product before combining them, and applies the credit after that combining rather than to one product. 1
Reports the result as an amount of money, with the unit attached. 1

Part C 5

Evaluates the volunteer's line correctly under the tiers, so that the comparison rests on what that line actually says rather than on a guess. 2
Reads the line back into the situation, accounting for the credits it applies against the credits the centre promised. 2
Locates the difference in where a grouping symbol closes, and states the change that would make the line say what the centre promised. 1

5. The grouping symbol that is not written 14 points

Part A 3

Settles the whole top and the whole bottom before dividing, rather than dividing a term of one by a term of the other. 2
Reports one number as the value of the whole expression. 1

Part B 3

Produces a one-line expression that names the same number as the stacked fraction, rather than a string that merely reuses the same numbers in the same order. 2
Evaluates the unbracketed string under the tiers and reports its value, rather than assuming it must come to the same thing. 1

Part C 5

Evaluates both the fraction and the proposed one-line form, and reaches the verdict from those two values rather than from an impression. 3
Says what the proposed form does to the leftover factor, rather than only that the move is not allowed, and gives the repaired one-line form. 2

Part D 3

Describes one family of fractions that needs no grouping when flattened, with an example. 1
Gives the reason, that a bracket is needed only where an operation on one side of the bar would otherwise be reached by the tiers. 2

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