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Exponents and Roots: Chapter Test

20 multiple-choice questions and 10 free-response questions, drawn from across the chapter and mixed together.

Free response

10 questions in parts, 152 points in total. Work them out on paper. There are no hints here: reveal each question's answer, worked solution, and rubric when you are ready to mark that one.

Free response · work it on paper

Question 1 of 10

1. How far the exponent reaches . 14 points. Question 1 of 10.

Each pair of expressions below is built from the same digits and the same exponent, and the members of a pair differ only by a set of parentheses. These parts evaluate two such pairs and then ask what the parentheses were doing in each.
1. Part A.
  Evaluate $8 + 3 \times 5^{3}$ and $(8 + 3) \times 5^{3}$ . Report both values, and for each one name the tier of the order of operations you settled first.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Evaluate $(-6)^{2}$ and $-6^{2}$ , showing the factors you multiplied in each case, and say which number the exponent is attached to in each.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Using one expression from each of the earlier parts as your cases, explain what a pair of parentheses decides about which numbers an exponent, and a minus sign, are attached to. Then say what $9^{1}$ equals, and what exponent a number written with no exponent at all is carrying.
  
  Carry your own answer forward Argue from the two expressions you evaluated in parts A and B, whatever values you reached for them.
  
  Explain why it works A sentence or two. Reasons, not steps. 6 points
The answer

Part A

$8 + 3 \times 5^{3} = 383$ and $(8 + 3) \times 5^{3} = 1375$ . The first settles the power first, the second settles the bracket first.

Part B

$(-6)^{2} = 36$ and $-6^{2} = -36$ . The first squares the base $-6$ ; the second squares $6$ alone and negates the result afterwards.

Part C

Parentheses do two jobs. In $(8 + 3) \times 5^{3}$ they group a neighbouring operation, leaving the cube on $5$ alone; in $(-6)^{2}$ they widen the base to take in the minus sign. Without them the exponent takes the number beside it and a leading minus stays outside. And $9^{1} = 9$ , so a bare $9$ carries exponent $1$ .

Worked solution

Part A

In the first expression nothing is bracketed, so the power goes first, then the multiplication, then the addition.

$5^{3} = 125, \quad 3 \times 125 = 375, \quad 8 + 375 = 383$

In the second the bracket is resolved ahead of everything, and only then does the power act on its own base.

$8 + 3 = 11, \quad 11 \times 125 = 1375$

The exponent never reached the $8$ or the $3$ in either case: the bracket changed which numbers were multiplied, not which number was cubed.

Part B

With parentheses the base is the single number $-6$ , and two negative factors cancel.

$(-6)^{2} = (-6) \times (-6) = 36$

Without them the exponent reaches only the $6$ , and the minus sign is applied to the finished square.

$-6^{2} = -(6 \times 6) = -36$

The same digits and the same exponent give answers $72$ apart, and the parentheses are the only difference between the two expressions.

Part C

An exponent always acts on exactly one base, and parentheses are the only thing that can widen what counts as that base. But not every pair of parentheses is doing that job.

In part A the bracket in $(8 + 3) \times 5^{3}$ never touches the cube: it groups the addition so that the multiplication meets $11$ rather than $3$ , while the exponent still acts on $5$ alone. Remove it and the same digits give $383$ instead of $1375$ .

In part B the bracket does widen the base, and there the two values part company.

$(-6)^{2} = 36, \qquad -6^{2} = -36$

On the left both factors carry the sign and the two negatives cancel; on the right a single minus sign waits outside a positive square. That the exponent is even is what makes them differ. At an odd exponent they coincide, since $(-6)^{3}$ and $-6^{3}$ both come to $-216$ : three negative factors leave one minus sign standing, which is exactly what the unbracketed version has anyway.

Finally, an exponent of $1$ asks for a single copy of the base, and one copy multiplied by nothing else is the base itself.

$9^{1} = 9$

So a number written with no exponent is not missing one; it carries the exponent $1$ silently, which is why $9 \times 9^{3}$ is a product of two powers of $9$ .

In one line

$8 + 3 \times 5^{3} = 383$ while $(8 + 3) \times 5^{3} = 1375$ , and $(-6)^{2} = 36$ while $-6^{2} = -36$ . The two pairs of parentheses are doing different jobs. In the first they group a neighbouring operation, so the multiplication meets $11$ rather than $3$ while the cube still acts on $5$ alone. In the second they widen the base itself, pulling the minus sign in so that both factors carry it; without them the exponent takes the single number beside it and a leading minus waits outside the completed power. That second contrast shows up only at an even exponent: $(-6)^{3}$ and $-6^{3}$ both come to $-216$ . An exponent of $1$ asks for one copy of the base, so $9^{1} = 9$ and a number written bare already carries the exponent $1$ .

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

Resolves the power before the multiplication and the addition in the unbracketed expression, and resolves the bracket first in the other. . Worth 2 points.

Reaches both values, $383$ and $1375$ . . Worth 1 point.

Names which tier was settled first in each expression. . Worth 1 point.

Part B 4 points

Squares $-6$ with both factors carrying the sign, and squares $6$ alone in the other expression. . Worth 2 points.

Reaches $36$ and $-36$ . . Worth 1 point.

Says which number the exponent is attached to in each expression. . Worth 1 point.

Part C 6 points

Says that parentheses decide what the exponent's base is, and distinguishes a bracket that widens the base from one that merely groups a neighbouring operation. . Worth 3 points. needs an explanation, not just an answer

Says that a leading minus sign stays outside the power unless parentheses capture it, and that at an even exponent this changes the value. . Worth 2 points. needs an explanation, not just an answer

Gives $9^{1} = 9$ and says a number with no written exponent carries an exponent of $1$ . . Worth 1 point.
2. One value, chosen on purpose . 13 points. Question 2 of 10.

A radical sign gets written inside calculations as freely as any other number, which means it has to name one definite value. These parts evaluate several roots, compare them with and without a minus sign in front, and then examine what had to be settled before the symbol could be used that way.
1. Part A.
  Evaluate $\sqrt{529}$ and $\sqrt{841}$ , giving for each the check that confirms it.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Which is larger, $\sqrt{841}$ or $\sqrt{784}$ ? Which is larger, $-\sqrt{841}$ or $-\sqrt{784}$ ? Give all four values, and state the rule your two answers follow.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Explain why the radical sign is defined to return just one of the values that square to the radicand, say which one it returns, and say what would go wrong if a single symbol named two numbers at once. Then say what, if anything, $\sqrt{-64}$ names.
  
  Explain why it works A sentence or two. Reasons, not steps. 6 points
The answer

Part A

$\sqrt{529} = 23$ and $\sqrt{841} = 29$ . The checks are $23^{2} = 529$ and $29^{2} = 841$ .

Part B

$\sqrt{841} = 29$ is larger than $\sqrt{784} = 28$ , but $-\sqrt{784} = -28$ is larger than $-\sqrt{841} = -29$ . A bigger radicand gives a bigger root, and negating reverses an order.

Part C

Two numbers square to any positive radicand, so a radical naming both would not name a number and could not be carried through a calculation. Fixing on the non-negative one makes it single-valued, and the other is written with a minus sign in front. And $\sqrt{-64}$ is not real: no real number squares to a negative.

Worked solution

Part A

Both radicands sit between $400$ and $900$ , so their roots sit between $20$ and $30$ , and the last digit narrows the search further. A square ending in $9$ comes from a number ending in $3$ or in $7$ , and a square ending in $1$ comes from a number ending in $1$ or in $9$ .

$23^{2} = 529, \qquad 29^{2} = 841$

Of the candidates in the twenties, only $23$ ends the first radicand correctly, and only $29$ ends the second one correctly. So $\sqrt{529} = 23$ and $\sqrt{841} = 29$ . Squaring each answer returns the radicand it came from, which is the check, and both roots are whole numbers because both radicands are perfect squares.

Part B

Both radicands are perfect squares, so read the roots off directly.

$\sqrt{841} = 29, \qquad \sqrt{784} = 28$

A bigger radicand always has a bigger root, since squaring a bigger non-negative number gives a bigger square, so $\sqrt{841}$ wins the first comparison.

Putting a minus sign in front turns each root into its opposite, and on the number line the opposite of the larger is the further left.

$-29 < -28$

So $-\sqrt{784}$ wins the second comparison. The minus sign in front of a radical is not part of the root: the radical is worked out first and then negated.

Part C

Ask which numbers square to a given positive radicand. They always come in a matched pair, one positive and one negative.

$29^{2} = 841 \qquad \text{and} \qquad (-29)^{2} = 841$

If $\sqrt{841}$ stood for both at once, it would not name a number, and it could not be written inside a calculation: a reader meeting $10 + \sqrt{841}$ would have no way to decide between $39$ and $-19$ . So one member of the pair is singled out by a fixed rule, and the non-negative one is chosen. That single agreed value is what "principal square root" means, and the other member has not vanished: it is written $-\sqrt{841} = -29$ whenever it is the one wanted.

A negative radicand is a different matter. Squaring a positive number gives a positive result, squaring a negative number gives a positive result, and $0^{2} = 0$ , so no real number squares to $-64$ and the symbol $\sqrt{-64}$ has nothing real to name.

In one line

$\sqrt{529} = 23$ and $\sqrt{841} = 29$ , each checked by squaring back. Of the roots, $\sqrt{841} = 29$ beats $\sqrt{784} = 28$ , but once both are negated the order reverses and $-\sqrt{784} = -28$ beats $-\sqrt{841} = -29$ . The radical returns the non-negative value because two numbers square to any positive radicand, and a symbol naming both would not name a number at all; the other member of the pair is written with a minus sign in front of the radical. A negative radicand has no real root, since squaring any real number gives zero or a positive result, so $\sqrt{-64}$ names no real number.

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

Identifies the whole number that squares to each radicand. . Worth 2 points.

Squares each answer back to the radicand as the check. . Worth 1 point.

Part B 4 points

Gives all four values, $29$ , $28$ , $-29$ and $-28$ . . Worth 2 points.

Picks the larger member of each pair, including reversing the order once the values are negated. . Worth 1 point.

States the rule the two answers follow, that a bigger radicand gives a bigger root and that negating reverses an order. . Worth 1 point.

Part C 6 points

Says that two numbers square to any positive radicand, and that a symbol naming both could not be used inside a calculation. . Worth 3 points. needs an explanation, not just an answer

Names the chosen value as the non-negative one and shows how the other is written when it is wanted. . Worth 2 points.

Says $\sqrt{-64}$ is not a real number because no real number squares to a negative. . Worth 1 point. needs an explanation, not just an answer
3. One base, three rules . 15 points. Question 3 of 10.

Powers of one shared base combine under three rules, and each rule is a statement about how many copies of the base end up in the answer. These parts apply the rules on their own and in combination, and then ask where one of them comes from.
1. Part A.
  Write $12^{2} \times 12^{5}$ and $\dfrac{12^{9}}{12^{4}}$ each as a single power of $12$ , and say for each what happened to the count of factors and what happened to the base.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Simplify $\left(\dfrac{6^{7}}{6^{3}}\right)^{2} \times 6$ to a single power of $6$ , working from the inside out and naming the law you use at each step.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  Explain why dividing powers of one base subtracts the exponents rather than dividing them. Use your explanation to say how many factors survive in the quotient from part A, and say what becomes of the base on the way.
  
  Carry your own answer forward Count the surviving factors in whichever quotient you simplified in part A, using your own answer for it.
  
  Explain why it works A sentence or two. Reasons, not steps. 6 points
The answer

Part A

$12^{2} \times 12^{5} = 12^{7}$ and $\dfrac{12^{9}}{12^{4}} = 12^{5}$ . The counts add and then subtract; the base is $12$ throughout and never changes.

Part B

$6^{9}$ , from the quotient rule inside the bracket, then the power rule on the bracket, then the product rule with the final factor of $6 = 6^{1}$ .

Part C

Each factor below cancels one factor above, so the number cancelled is the count below and what survives is the difference of the counts, never their quotient. In part A that leaves $9 - 4 = 5$ factors of $12$ . The base cannot change because cancelling removes copies of a factor and never alters what that factor is.

Worked solution

Part A

Two powers side by side put their factors into one long product, so the counts add.

$12^{2} \times 12^{5} = 12^{2+5} = 12^{7}$

In a quotient each factor below cancels one factor above, since $\tfrac{12}{12} = 1$ , so four of the nine cancel and five are left standing.

$\frac{12^{9}}{12^{4}} = 12^{9-4} = 12^{5}$

In both lines the factor being counted is still $12$ , so the base comes through untouched: what changes is only how many copies of it there are.

Part B

Settle the innermost grouping first. Inside the bracket the base is shared, so the quotient rule subtracts the exponents.

$\frac{6^{7}}{6^{3}} = 6^{7-3} = 6^{4}$

The bracket now carries an outer exponent, and a power of a power multiplies the exponents.

$\left(6^{4}\right)^{2} = 6^{4 \times 2} = 6^{8}$

The last factor is $6 = 6^{1}$ , standing side by side with $6^{8}$ , so the product rule adds the exponents.

$6^{8} \times 6^{1} = 6^{9}$

Part C

Expand both parts of a quotient into their factors. The numerator holds one run of copies of the base and the denominator holds another.

$\frac{12^{9}}{12^{4}} = \frac{12 \times \cdots \times 12}{12 \times \cdots \times 12}$

Each factor in the denominator cancels one factor in the numerator, because a nonzero number over itself is $1$ . There are four factors below, so exactly four of the nine above are cancelled, and what is left is the nine reduced by four.

$9 - 4 = 5 \quad \text{factors of } 12 \text{ survive}$

That is a subtraction and not a division because cancelling removes factors one for one; dividing the counts, $9 \div 4$ , would answer a question nobody asked. The base is untouched for a separate reason: the operation removes copies of a factor and never changes what the factor is, so a run of twelves is still a run of twelves however many are struck out.

In one line

$12^{2} \times 12^{5} = 12^{7}$ and $\dfrac{12^{9}}{12^{4}} = 12^{5}$ , the counts adding and then subtracting while the base stays $12$ . Working outward, $\left(\dfrac{6^{7}}{6^{3}}\right)^{2} \times 6 = \left(6^{4}\right)^{2} \times 6^{1} = 6^{8} \times 6^{1} = 6^{9}$ . Division subtracts the exponents because each factor below cancels one factor above, so four of the nine copies of $12$ are struck out and five survive; dividing the counts would answer a different question. The base is untouched throughout, since cancelling removes copies of a factor without changing what that factor is.

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

Adds the exponents for the product and subtracts them for the quotient, reaching $12^{7}$ and $12^{5}$ . . Worth 2 points.

Says what happened to the count of factors in each case, and that the base is unchanged. . Worth 2 points.

Part B 5 points

Works from the innermost grouping outward rather than combining exponents in the order they appear. . Worth 2 points.

Reads the bare factor of $6$ as $6^{1}$ and reaches $6^{9}$ . . Worth 2 points.

Names the quotient rule, the power rule and the product rule at the steps that use them. . Worth 1 point.

Part C 6 points

Argues that each factor in the denominator cancels one in the numerator, so the surviving count is the difference of the two counts. . Worth 3 points. needs an explanation, not just an answer

Says why the counts are subtracted rather than divided. . Worth 2 points. needs an explanation, not just an answer

Reports the surviving count for the part A quotient and says why the base cannot change. . Worth 1 point.
4. A year of washers . 15 points. Question 4 of 10.

A factory ships $3{,}600{,}000$ washers in a year, and one washer weighs $0.00045$ kilograms. Numbers of that shape are awkward on paper in ordinary form: one is long enough to miscount and the other is small enough to lose a zero in, which is exactly the situation scientific notation was built for.
1. Part A.
  Write $3{,}600{,}000$ and $0.00045$ in scientific notation. For each one, say how many places the decimal point moved and in which direction.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Find the total weight of a year's washers, in kilograms. Give it in proper scientific notation and also as an ordinary number.
  
  Carry your own answer forward Multiply the two scientific-notation figures you wrote in part A, whatever they came to.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  The washers are packed into crates of $2.5 \times 10^{4}$ . How many crates does a year's output fill? Give the count in proper scientific notation and as an ordinary number, and say whether this answer needed the tidying step that part B's did, with your reason.
  
  Carry your own answer forward Divide by the crate size using your own scientific-notation figure for the year's output from part A, whatever it came to.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 6 points
The answer

Part A

$3{,}600{,}000 = 3.6 \times 10^{6}$ , the point moving $6$ places left, and $0.00045 = 4.5 \times 10^{-4}$ , the point moving $4$ places right.

Part B

$1.62 \times 10^{3}$ kilograms, which written out is $1620$ kilograms.

Part C

$1.44 \times 10^{2}$ crates, which is $144$ crates. No tidying was needed: dividing the coefficients gave $3.6 \div 2.5 = 1.44$ , which already sits in $1 \le a < 10$ , whereas part B's coefficients multiplied to $16.2$ , which did not.

Worked solution

Part A

Park the decimal point so exactly one nonzero digit stands in front of it, then count the hops. In the large number the point starts after the final zero and travels left.

$3{,}600{,}000 \;\longrightarrow\; 3.6 \quad (6 \text{ places left})$

Trailing zeros after the last nonzero digit are dropped from the coefficient, so it is $3.6$ and not $3.600000$ , and moving left makes the exponent positive: $3.6 \times 10^{6}$ .

In the small number the point travels right until it clears the leading zeros.

$0.00045 \;\longrightarrow\; 4.5 \quad (4 \text{ places right})$

Moving right makes the exponent negative, so the weight is $4.5 \times 10^{-4}$ kilograms. Both coefficients sit in $1 \le a < 10$ , as the form requires.

Part B

Handle the coefficients and the powers of ten separately, since multiplication can be reordered freely.

$3.6 \times 4.5 = 16.2, \qquad 10^{6} \times 10^{-4} = 10^{2}$

That gives $16.2 \times 10^{2}$ , which is the right value in the wrong form: the coefficient $16.2$ is not below $10$ . Peel a factor of ten off it and hand that factor to the power.

$16.2 \times 10^{2} = 1.62 \times 10^{1} \times 10^{2} = 1.62 \times 10^{3}$

The exponent $3$ sends the point three places right, so the total is $1620$ kilograms.

Part C

Division splits into the same two jobs as multiplication, but each one runs the other way: divide the coefficients, and subtract the exponents.

$3.6 \div 2.5 = 1.44, \qquad \frac{10^{6}}{10^{4}} = 10^{2}$

So the count is $1.44 \times 10^{2}$ crates, and moving the point two places right gives $144$ crates. The unit is a plain count here, not a weight, because dividing washers by washers-per-crate leaves crates.

Whether a tidying step is needed is decided after the arithmetic, by looking at where the new coefficient landed. Here $1.44$ is already at least $1$ and below $10$ , so the form is finished as it stands. In part B the coefficients gave $16.2$ , which is at least $10$ , so a factor of ten had to be peeled off and handed to the power. Nothing about division guarantees either outcome: a quotient of coefficients can fall below $1$ just as easily, and then a factor of ten has to be borrowed back the other way.

In one line

In scientific notation the year's output is $3.6 \times 10^{6}$ washers and one washer weighs $4.5 \times 10^{-4}$ kilograms, the decimal point having moved $6$ places left and $4$ places right. Multiplying gives $16.2 \times 10^{2}$ , which renormalises to $1.62 \times 10^{3}$ kilograms, or $1620$ kilograms. Dividing the output by the crate size of $2.5 \times 10^{4}$ gives $3.6 \div 2.5 = 1.44$ and $10^{6-4} = 10^{2}$ , so $1.44 \times 10^{2}$ crates, that is $144$ crates. That answer needed no tidying, because $1.44$ already sits in $1 \le a < 10$ , while part B's coefficient reached $16.2$ and had to hand a factor of ten to the power.

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

Places the decimal point after the first nonzero digit in each number, giving the coefficients $3.6$ and $4.5$ . . Worth 2 points.

Gets both exponents right, positive for the large number and negative for the small one. . Worth 1 point.

Reports how many places the point moved and in which direction for each number. . Worth 1 point.

Part B 5 points

Multiplies the coefficients and adds the exponents, rather than mixing the two jobs. . Worth 2 points.

Renormalises $16.2 \times 10^{2}$ by raising the exponent, reaching $1.62 \times 10^{3}$ . . Worth 2 points.

Reports the total as a weight in kilograms and gives it as an ordinary number too. . Worth 1 point.

Part C 6 points

Divides the coefficients and subtracts the exponents, rather than mixing the two jobs or reversing either. . Worth 2 points.

Reaches $1.44 \times 10^{2}$ crates and gives $144$ as the ordinary number, labelled as a count of crates. . Worth 2 points.

Decides the tidying question by where the new coefficient landed, and contrasts that with the coefficient reached in part B. . Worth 2 points. needs an explanation, not just an answer
5. Below the first power . 15 points. Question 5 of 10.

A power began as a count of copies of the base, which needs an exponent of at least $1$ . An exponent of $0$ , or a negative one, is not a count of that kind, so for a nonzero base the meaning of such a power has to be settled separately. These parts evaluate several of them and then weigh two rival answers to one.
1. Part A.
  Write $8^{-2}$ and $8^{0}$ as plain numbers, and write $\dfrac{1}{8^{3}}$ as a single power of $8$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Write $\dfrac{12^{-3}}{12^{2} \times 12^{-4}}$ as an equivalent expression with no negative exponent, and give its value as a fraction.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  Two students disagree about $\left(\dfrac{1}{5}\right)^{-2}$ . One answers $\dfrac{1}{25}$ and the other answers $25$ . Decide which is right, say what a negative exponent does to a fraction base, and confirm your verdict by treating the fraction as a quotient of powers.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer

Part A

$8^{-2} = \tfrac{1}{64}$ and $8^{0} = 1$ , while $\dfrac{1}{8^{3}} = 8^{-3}$ .

Part B

$\dfrac{1}{12}$ , since the expression collapses to $12^{-1}$ .

Part C

The second student is right: $\left(\tfrac{1}{5}\right)^{-2} = 25$ . A negative exponent asks for the reciprocal, and the reciprocal of a fraction turns it over, so the base becomes $5$ and the exponent becomes $+2$ . Reading it as $\dfrac{1^{-2}}{5^{-2}} = \dfrac{1}{5^{-2}} = 5^{2}$ confirms it.

Worked solution

Part A

A negative exponent asks for the reciprocal of the matching positive power, so evaluate that power and put it under $1$ .

$8^{-2} = \frac{1}{8^{2}} = \frac{1}{64}$

A zero exponent asks for no factors of the base at all, and a multiplication with nothing in it stands at $1$ , so $8^{0} = 1$ .

The third one runs the same rule backwards. A reciprocal power can always be rewritten with the minus sign moved into the exponent.

$\frac{1}{8^{3}} = 8^{-3}$

All three values are positive; nothing about a zero or a negative exponent puts a minus sign on the value.

Part B

Combine the denominator first with the product rule, adding the exponents and carrying the minus sign.

$12^{2} \times 12^{-4} = 12^{2 + (-4)} = 12^{-2}$

Now the expression is a quotient of one base, so subtract the denominator's exponent from the numerator's, again carrying the signs.

$\frac{12^{-3}}{12^{-2}} = 12^{-3 - (-2)} = 12^{-1}$

A negative exponent moves the power across the fraction bar, so with a positive exponent the answer reads $\tfrac{1}{12^{1}}$ , whose value is $\tfrac{1}{12}$ .

Part C

Take the two jobs in order: the minus sign in the exponent calls for a reciprocal, and only then is a positive power evaluated.

$\left(\frac{1}{5}\right)^{-2} = \frac{1}{\left(\tfrac{1}{5}\right)^{2}} = \frac{1}{\tfrac{1}{25}} = 25$

So the second student is right. The shortcut is to turn the fraction over and drop the minus sign, which is the same calculation with the reciprocal taken first.

$\left(\frac{1}{5}\right)^{-2} = \left(\frac{5}{1}\right)^{2} = 25$

The check treats the fraction as a quotient of powers, since an exponent spreads across a quotient.

$\frac{1^{-2}}{5^{-2}} = \frac{1}{5^{-2}} = 5^{2} = 25$

The first student's $\tfrac{1}{25}$ is the value of $\left(\tfrac{1}{5}\right)^{2}$ , which is what you get by ignoring the minus sign in the exponent. A base below $1$ raised to a negative power comes out above $1$ , which is a quick sanity check on the verdict.

In one line

$8^{-2} = \tfrac{1}{64}$ , $8^{0} = 1$ , and $\dfrac{1}{8^{3}} = 8^{-3}$ . The compound expression collapses in two steps, $12^{2} \times 12^{-4} = 12^{-2}$ and then $\dfrac{12^{-3}}{12^{-2}} = 12^{-1}$ , whose value is $\tfrac{1}{12}$ . On the disputed power the second student is right: $\left(\tfrac{1}{5}\right)^{-2} = 25$ , because a negative exponent asks for the reciprocal, which turns a fraction base over and leaves a positive exponent, and reading the fraction as $\dfrac{1^{-2}}{5^{-2}} = 5^{2}$ confirms it. The answer $\tfrac{1}{25}$ belongs to $\left(\tfrac{1}{5}\right)^{2}$ instead.

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

Evaluates the negative power as the reciprocal of the positive one, reaching $\tfrac{1}{64}$ . . Worth 2 points.

Gives $8^{0} = 1$ rather than $0$ . . Worth 1 point.

Runs the rule backwards to write $\dfrac{1}{8^{3}}$ as $8^{-3}$ . . Worth 1 point.

Part B 5 points

Combines the denominator's two powers by adding their exponents, keeping the minus sign on the $-4$ . . Worth 2 points.

Subtracts a negative exponent correctly, reaching $12^{-1}$ . . Worth 2 points.

Rewrites the result so that no negative exponent is left, and gives its value as $\tfrac{1}{12}$ . . Worth 1 point.

Part C 6 points

Decides for $25$ and shows the reciprocal being taken before the square is evaluated. . Worth 3 points. needs an explanation, not just an answer

States what a negative exponent does to a fraction base, turning it over and leaving a positive exponent. . Worth 2 points. needs an explanation, not just an answer

Confirms the verdict by spreading the exponent across the quotient and reaching $5^{2}$ . . Worth 1 point.
6. Tidying a length and pinning it down . 15 points. Question 6 of 10.

A cutting list gives two lengths in centimetres as $\sqrt{175}$ and $\sqrt{288}$ . These parts tidy each one, pin the first against whole numbers, and then weigh up how the second is best recorded before anything is ordered.
1. Part A.
  Write $\sqrt{175}$ and $\sqrt{288}$ in simplest radical form, naming the largest perfect-square factor you removed each time.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Trap $\sqrt{175}$ between two consecutive whole numbers of centimetres, naming the perfect squares that do the trapping and saying which end the length leans toward.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Set the two exact records of the second length side by side, $\sqrt{288}$ and the simplified form you gave for it in part A. Say whether either is more exact than the other, and say why no decimal, however many places it runs to, can record that length exactly. Then decide, without evaluating any root, whether $\sqrt{288}$ is above or below $17$ centimetres.
  
  Carry your own answer forward Set $\sqrt{288}$ beside the simplified form you wrote for the second length in part A, whatever it came to.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 6 points
The answer

Part A

$\sqrt{175} = 5\sqrt{7}$ centimetres, removing the square factor $25$ , and $\sqrt{288} = 12\sqrt{2}$ centimetres, removing the square factor $144$ .

Part B

$13 < \sqrt{175} < 14$ centimetres, trapped by $169 = 13^{2}$ and $196 = 14^{2}$ , and it leans toward $13$ .

Part C

Neither is more exact: $\sqrt{288}$ and $12\sqrt{2}$ are one value written two ways. No decimal records it exactly, because $288$ is not a perfect square, so its root is irrational and its decimal neither ends nor repeats. And $\sqrt{288}$ is below $17$ , because $17^{2} = 289$ overshoots $288$ .

Worked solution

Part A

Split each radicand into its largest perfect-square factor times whatever is left, then take the root of the square outside.

$\sqrt{175} = \sqrt{25 \times 7} = \sqrt{25} \times \sqrt{7} = 5\sqrt{7}$

Nothing further comes out, since $7$ is prime and has no square factor above $1$ , so the first length is $5\sqrt{7}$ centimetres. The second radicand hides a larger square.

$\sqrt{288} = \sqrt{144 \times 2} = \sqrt{144} \times \sqrt{2} = 12\sqrt{2}$

So the second length is $12\sqrt{2}$ centimetres. Had you taken $288 = 4 \times 72$ instead, the form $2\sqrt{72}$ would still hide the square $36$ , so it would not be finished. Always take the largest square available.

Part B

Find the perfect squares just below and just above the radicand, and take roots across the inequality, which preserves the order.

$169 < 175 < 196$

Since $169 = 13^{2}$ and $196 = 14^{2}$ , taking roots gives the trap.

$13 < \sqrt{175} < 14$

The radicand sits $6$ above $169$ and $21$ below $196$ , so it is much nearer the lower end and the length leans toward $13$ centimetres.

Part C

The two records are the same number written two ways, since the product rule moved a factor out from under the radical without changing the value.

$\sqrt{288} = \sqrt{144 \times 2} = 12\sqrt{2}$

Neither is more exact than the other. The simplified form is usually preferred because the size of the length is easier to judge from it, but that is a matter of readability rather than of accuracy.

A decimal is a different kind of record altogether. A whole number that is not a perfect square has an irrational root, whose decimal neither ends nor repeats, so every decimal anyone could write down stops somewhere and is therefore slightly wrong. Rounding is useful for cutting material and useless for an exact statement.

The last comparison needs no root at all. Squaring is what undoes a root, so square the number being compared against and set it beside the radicand.

$17^{2} = 289 > 288$

A larger square belongs to a larger non-negative number, so $17$ is larger than $\sqrt{288}$ . The margin is thin: the two squares differ by $1$ , so judging the root by eye would not have settled it, and squaring the comparison value is what does.

In one line

The two lengths are $\sqrt{175} = 5\sqrt{7}$ centimetres and $\sqrt{288} = 12\sqrt{2}$ centimetres, the largest perfect-square factors removed being $25$ and $144$ . Since $169 < 175 < 196$ , the first length satisfies $13 < \sqrt{175} < 14$ centimetres and leans toward $13$ , being only $6$ above $169$ but $21$ below $196$ . Of the records of the second length, $\sqrt{288}$ and $12\sqrt{2}$ are one exact value written two ways, neither more exact than the other, and no decimal captures it at all, since $288$ is not a perfect square and its root is irrational. Squaring the comparison value settles the last question without a root: $17^{2} = 289$ is above $288$ , so $\sqrt{288} < 17$ centimetres, by a margin too thin to judge by eye.

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 5 points

Splits each radicand at its largest perfect-square factor, $25$ and $144$ . . Worth 2 points.

Reaches $5\sqrt{7}$ and $12\sqrt{2}$ , with nothing left under either radical that could still come out. . Worth 2 points.

Keeps the results labelled as lengths in centimetres. . Worth 1 point.

Part B 4 points

Names $169$ and $196$ as the bracketing perfect squares and their roots as $13$ and $14$ . . Worth 2 points.

States the trap as $13 < \sqrt{175} < 14$ and says it leans toward $13$ , with the distances that decide it. . Worth 2 points.

Part C 6 points

Identifies the radical and its simplified form as two spellings of one exact value, neither more exact than the other. . Worth 2 points.

Says no decimal can record the length exactly, and gives the reason, that a non-perfect-square radicand has an irrational root. . Worth 2 points. needs an explanation, not just an answer

Settles the comparison by squaring $17$ to $289$ and setting that against the radicand $288$ . . Worth 2 points.
7. Backwards and forwards through a radical . 17 points. Question 7 of 10.

The product rule for radicals reads in both directions. Forwards it splits a radicand into a perfect square and a leftover; backwards it gathers a number standing outside a radical back under it. These parts use both readings and then test a claim about when the result comes out whole.
1. Part A.
  Write $\sqrt{50} \times \sqrt{18}$ and $\dfrac{\sqrt{405}}{\sqrt{5}}$ each as a whole number, naming the rule you used at each step.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Gather $6\sqrt{3}$ under a single radical, then check your answer by simplifying that radical back again.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  A classmate claims that $\sqrt{a} \times \sqrt{b}$ is a whole number whenever $a$ and $b$ are whole numbers. Decide the claim, giving one pair of whole numbers from this question where it holds and one where it fails, and state what has to be true for it to hold.
  
  Carry your own answer forward Use the pair of whole numbers from the first product in part A as one of your cases, whatever value you found for it.
  
  Justify your claim State the claim, then give the reason it has to be true. 7 points
The answer

Part A

$\sqrt{50} \times \sqrt{18} = 30$ by the product rule, and $\dfrac{\sqrt{405}}{\sqrt{5}} = 9$ by the quotient rule.

Part B

$6\sqrt{3} = \sqrt{108}$ . Simplifying back, $\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}$ .

Part C

The claim fails. It holds at $a = 50$ and $b = 18$ , where $\sqrt{900} = 30$ , and fails at $a = 50$ and $b = 5$ , where $\sqrt{250}$ is irrational. The product is a whole number exactly when $a \times b$ is a perfect square, which is a condition on the product and not on $a$ and $b$ separately.

Worked solution

Part A

The product rule joins two radicals into one, so multiply the radicands and then look for a whole-number root.

$\sqrt{50} \times \sqrt{18} = \sqrt{900} = 30$

Neither radicand was a perfect square on its own, yet their product is, which is why joining them first is the shorter route.

The quotient rule works the same way with a division underneath.

$\frac{\sqrt{405}}{\sqrt{5}} = \sqrt{\frac{405}{5}} = \sqrt{81} = 9$

Both answers can be checked by squaring: $30^{2} = 900 = 50 \times 18$ , and $9^{2} = 81 = 405 \div 5$ .

Part B

Run the product rule backwards: the number standing outside a radical goes under it as its square.

$6\sqrt{3} = \sqrt{36} \times \sqrt{3} = \sqrt{36 \times 3} = \sqrt{108}$

So the single radical is $\sqrt{108}$ . It is $36$ and not $6$ that goes underneath, because $36$ is the radicand the outside $6$ came from.

The check runs the rule forwards again, taking the largest perfect-square factor of $108$ .

$\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}$

That returns the form we started from, so the two directions agree.

Part C

Test the claim on the pair from part A, where it does hold.

$\sqrt{50} \times \sqrt{18} = \sqrt{900} = 30$

Neither radicand is a perfect square, so a whole-number answer is not a sign that the pieces were tidy. Now change one of the numbers and the claim collapses.

$\sqrt{50} \times \sqrt{5} = \sqrt{250}$

The radicand $250$ lies between $225 = 15^{2}$ and $256 = 16^{2}$ , so its root is not a whole number, and since $250$ is not a perfect square that root is irrational and no decimal captures it. One such pair is enough to sink a claim made about every pair.

What the two cases show is where the condition really sits. The product rule turns the question into a single radical, $\sqrt{a \times b}$ , so the answer is a whole number exactly when the radicand $a \times b$ is a perfect square. That is a condition on the product alone: $50$ and $18$ are individually untidy and multiply to $900$ , while $50$ and $5$ are no worse individually and multiply to $250$ .

In one line

Joining the radicals first gives $\sqrt{50} \times \sqrt{18} = \sqrt{900} = 30$ and $\dfrac{\sqrt{405}}{\sqrt{5}} = \sqrt{81} = 9$ . Running the product rule backwards, $6\sqrt{3} = \sqrt{36 \times 3} = \sqrt{108}$ , and simplifying $\sqrt{108}$ forwards returns $6\sqrt{3}$ . The classmate's claim fails: $\sqrt{50} \times \sqrt{5} = \sqrt{250}$ is irrational, while $\sqrt{50} \times \sqrt{18} = 30$ is whole even though neither radicand is a perfect square. The product $\sqrt{a} \times \sqrt{b}$ is a whole number exactly when $a \times b$ is a perfect square.

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 5 points

Joins the two radicals before rooting, rather than trying to evaluate each radical on its own. . Worth 2 points.

Reaches $30$ and $9$ , and names the product rule and the quotient rule at the steps that use them. . Worth 2 points.

Reports each answer as a whole number, as asked, rather than leaving a radical standing. . Worth 1 point.

Part B 5 points

Squares the outside factor before moving it under the radical, using $36$ rather than $6$ . . Worth 2 points.

Reaches the single radical $\sqrt{108}$ . . Worth 2 points.

Checks by simplifying $\sqrt{108}$ forwards and recovering $6\sqrt{3}$ . . Worth 1 point.

Part C 7 points

Rejects the claim and supplies a pair of whole numbers for which $\sqrt{a} \times \sqrt{b}$ is not a whole number. . Worth 3 points. needs an explanation, not just an answer

Supplies a pair for which it does hold, and notes that neither radicand needed to be a perfect square. . Worth 2 points. needs an explanation, not just an answer

States the condition as $a \times b$ being a perfect square, a condition on the product rather than on the two numbers separately. . Worth 2 points.
8. Which of two small powers is the larger . 15 points. Question 8 of 10.

The powers below carry negative exponents, and one of them has a fraction for its base. These parts evaluate them, put three of them in order of size, and then test a rule someone might reach for when ranking powers of this kind.
1. Part A.
  Evaluate $3^{-4}$ , $\left(\dfrac{1}{3}\right)^{4}$ and $\left(\dfrac{1}{3}\right)^{-4}$ , giving each as a fraction or a whole number, and say which two of the three are equal.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Put $2^{-5}$ , $5^{-2}$ and $10^{-1}$ in order from smallest to largest, giving each as a fraction.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Set your order from part B beside the pair $2^{-1}$ and $10^{-3}$ . Decide whether the size of the base on its own can settle which of two negative powers is the larger, and state what does settle it.
  
  Carry your own answer forward Argue from the order you produced in part B, whatever order that was.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 7 points
The answer

Part A

$3^{-4} = \tfrac{1}{81}$ , $\left(\tfrac{1}{3}\right)^{4} = \tfrac{1}{81}$ and $\left(\tfrac{1}{3}\right)^{-4} = 81$ . The first two are equal.

Part B

$2^{-5} = \tfrac{1}{32}$ , $5^{-2} = \tfrac{1}{25}$ and $10^{-1} = \tfrac{1}{10}$ , so the order is $2^{-5} < 5^{-2} < 10^{-1}$ .

Part C

It cannot. The part B order happens to run with the bases, but $2^{-1} = \tfrac{1}{2}$ is far larger than $10^{-3} = \tfrac{1}{1000}$ despite the much smaller base. What settles it is the reciprocal: $a^{-n} = \tfrac{1}{a^{n}}$ , so whichever positive power $a^{n}$ is larger gives the smaller negative power.

Worked solution

Part A

A negative exponent asks for the reciprocal of the matching positive power.

$3^{-4} = \frac{1}{3^{4}} = \frac{1}{81}$

Raising a fraction to a positive power raises the top and the bottom separately, which lands on the same value from the other side.

$\left(\frac{1}{3}\right)^{4} = \frac{1^{4}}{3^{4}} = \frac{1}{81}$

So those two agree, which is the reciprocal rule and the power-of-a-quotient rule saying the same thing. The third turns the fraction over before the power acts.

$\left(\frac{1}{3}\right)^{-4} = 3^{4} = 81$

A base below $1$ raised to a negative power comes out above $1$ , so none of the three values is negative and only the first pair coincide.

Part B

Turn each one into a fraction, which needs only the matching positive power.

$2^{-5} = \frac{1}{32}, \qquad 5^{-2} = \frac{1}{25}, \qquad 10^{-1} = \frac{1}{10}$

All three are fractions with numerator $1$ , so the one with the largest denominator is the smallest, and the one with the smallest denominator is the largest.

$\frac{1}{32} < \frac{1}{25} < \frac{1}{10}$

So the order is $2^{-5} < 5^{-2} < 10^{-1}$ , and every value is a positive number below $1$ .

Part C

In part B the smallest base carried the smallest value and the largest base the largest, so the base looks as though it decides the ranking. Test that against a pair where the exponents are further apart.

$2^{-1} = \frac{1}{2}, \qquad 10^{-3} = \frac{1}{1000}$

Here the smaller base gives by far the larger value, so the base alone settles nothing. What was really doing the work in part B was the size of the positive power underneath: $2^{5} = 32$ , $5^{2} = 25$ and $10^{1} = 10$ , and those run in the opposite order to the values.

$a^{-n} = \frac{1}{a^{n}}$

So comparing two negative powers means comparing the two positive powers under the line and then reversing the result, since a bigger denominator makes a smaller unit fraction. The base is only half of what fixes that denominator; the exponent is the other half.

In one line

$3^{-4} = \tfrac{1}{81}$ and $\left(\tfrac{1}{3}\right)^{4} = \tfrac{1}{81}$ are equal, while $\left(\tfrac{1}{3}\right)^{-4} = 81$ . Ordering the three small powers gives $2^{-5} = \tfrac{1}{32}$ , then $5^{-2} = \tfrac{1}{25}$ , then $10^{-1} = \tfrac{1}{10}$ . That order runs with the bases, but the base alone settles nothing: $2^{-1} = \tfrac{1}{2}$ is far larger than $10^{-3} = \tfrac{1}{1000}$ . Since $a^{-n} = \tfrac{1}{a^{n}}$ , the comparison is decided by the positive powers under the line, with the order reversed, and the exponent has as much say in those as the base does.

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

Evaluates each power as a reciprocal where the exponent is negative, reaching $\tfrac{1}{81}$ , $\tfrac{1}{81}$ and $81$ . . Worth 2 points.

Identifies the first two as the equal pair, and keeps every value positive. . Worth 2 points.

Part B 4 points

Turns each negative power into the reciprocal of its positive power, reaching $\tfrac{1}{32}$ , $\tfrac{1}{25}$ and $\tfrac{1}{10}$ . . Worth 2 points.

Ranks unit fractions by their denominators, giving the order $2^{-5} < 5^{-2} < 10^{-1}$ . . Worth 2 points.

Part C 7 points

Rejects the base-alone rule, using the pair given to show a smaller base with a larger value. . Worth 3 points. needs an explanation, not just an answer

Notes that the part B order agreeing with the bases does not establish the rule. . Worth 2 points. needs an explanation, not just an answer

States the rule that does settle it, comparing the positive powers under the line and reversing the result. . Worth 2 points.
9. An exponent over a whole product . 17 points. Question 9 of 10.

The last two laws are about a base that is itself built from pieces. These parts spread an exponent across a product and across a quotient, combine what results with the same-base rules, and then examine a claim about a base built from a sum instead.
1. Part A.
  Write $\dfrac{(3 \times 5)^{6}}{3^{4} \times 5^{2}}$ in the form $3^{a} \times 5^{b}$ , naming the law that produced each exponent.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Write $\left(\dfrac{2^{5}}{7}\right)^{2} \times 7^{3}$ as a product of powers with no fraction left standing.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  A classmate writes $\left(3^{2} + 5^{2}\right)^{2} = 3^{4} + 5^{4}$ , defending it by saying the outer exponent lands on each piece just as it does in a product. Decide the claim by evaluating both sides, and state exactly which structures an exponent may be spread across and which it may not, with the reason.
  
  Justify your claim State the claim, then give the reason it has to be true. 7 points
The answer

Part A

$3^{2} \times 5^{4}$ . The power of a product spreads the $6$ onto each factor, and the quotient rule then subtracts $4$ and $2$ from the two sixes.

Part B

$2^{10} \times 7$ .

Part C

The claim fails: the left side is $34^{2} = 1156$ and the right side is $81 + 625 = 706$ . An exponent spreads across a product or a quotient, because copies of each factor can be gathered separately. It does not spread across a sum: squaring a sum leaves cross terms, which vanish only when one of the terms is $0$ .

Worked solution

Part A

Spread the outer exponent across the product first, since the bracket is the innermost grouping.

$(3 \times 5)^{6} = 3^{6} \times 5^{6}$

The expression is now a quotient with two shared bases, so the quotient rule applies once to each base, subtracting the denominator's exponent from the numerator's.

$\frac{3^{6} \times 5^{6}}{3^{4} \times 5^{2}} = 3^{6-4} \times 5^{6-2} = 3^{2} \times 5^{4}$

The powers of $3$ and the powers of $5$ are handled apart from each other here, because no SAME-BASE rule reaches across two different bases: the product and quotient rules each require one shared base. That is not a ban on unlike bases ever combining. The law that relates them is the power of a product, and read backwards it can gather them even when the exponents differ, since $3^{2} \times 5^{4} = 3^{2} \times \left(5^{2}\right)^{2} = 75^{2}$ .

Part B

An exponent on a quotient lands on the top and the bottom alike, and a power of a power multiplies the exponents.

$\left(\frac{2^{5}}{7}\right)^{2} = \frac{\left(2^{5}\right)^{2}}{7^{2}} = \frac{2^{10}}{7^{2}}$

Now the factor of $7^{3}$ meets a $7^{2}$ below the line, and those share a base, so the quotient rule subtracts.

$\frac{2^{10}}{7^{2}} \times 7^{3} = 2^{10} \times 7^{3-2} = 2^{10} \times 7$

No fraction is left, because the three factors of $7$ on top outnumbered the two below. The $2^{10}$ sits out the quotient rule entirely, since no other factor here carries the base $2$ .

Part C

Evaluate the two sides rather than arguing about the shape.

$\left(3^{2} + 5^{2}\right)^{2} = (9 + 25)^{2} = 34^{2} = 1156$

$3^{4} + 5^{4} = 81 + 625 = 706$

They differ, so the claim is false, and one case is enough to sink a rule offered for every case.

The reason the product version works is that $n$ copies of $a \times b$ hold $n$ copies of $a$ and $n$ copies of $b$ , which multiplication lets you reorder into $a^{n} \times b^{n}$ . Nothing of the kind happens with a sum: squaring $9 + 25$ multiplies the whole of $34$ by the whole of $34$ , and that product contains cross terms which $81 + 625$ has no room for. The general position is that an exponent spreads across a product and across a quotient, and not across a sum or a difference: with a sum, the grouping is resolved first and the power acts on the single number that results. The one escape is the trivial one, where a term is $0$ and there is nothing for the cross terms to be built from.

In one line

$\dfrac{(3 \times 5)^{6}}{3^{4} \times 5^{2}} = \dfrac{3^{6} \times 5^{6}}{3^{4} \times 5^{2}} = 3^{2} \times 5^{4}$ , and $\left(\dfrac{2^{5}}{7}\right)^{2} \times 7^{3} = \dfrac{2^{10}}{7^{2}} \times 7^{3} = 2^{10} \times 7$ . The classmate's claim fails, since $\left(3^{2} + 5^{2}\right)^{2} = 34^{2} = 1156$ while $3^{4} + 5^{4} = 706$ . An exponent spreads across a product or a quotient, because the copies of each factor can be reordered and gathered, and not across a sum or a difference, where the grouping has to be resolved before the power acts; squaring a sum leaves cross terms, and only a term of $0$ makes them disappear.

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 5 points

Spreads the outer exponent onto both factors of the bracket before dividing. . Worth 2 points.

Applies the quotient rule to each base separately, reaching $3^{2} \times 5^{4}$ . . Worth 2 points.

Names the power of a product and the quotient rule at the steps that use them. . Worth 1 point.

Part B 5 points

Sends the outer exponent to the numerator and the denominator alike, and multiplies exponents on the power of a power. . Worth 2 points.

Cancels the powers of $7$ with the quotient rule and reaches $2^{10} \times 7$ , with no fraction left. . Worth 2 points.

Leaves the powers of $2$ and $7$ side by side, since no same-base rule reaches across two different bases. . Worth 1 point.

Part C 7 points

Evaluates both sides, reaching $1156$ and $706$ , and rejects the claim on that evidence. . Worth 3 points. needs an explanation, not just an answer

Says why the product version works, that copies of each factor may be reordered and gathered. . Worth 2 points. needs an explanation, not just an answer

States the boundary: products and quotients split, while a sum does not, apart from the trivial case where one of the terms is $0$ . . Worth 2 points.
10. The digit a power ends on . 16 points. Question 10 of 10.

A power grows too fast to write out for long, but the digit it ends on is a much smaller question than its value. These parts list the first few powers of $7$ , use that list to reach far larger exponents, and then look at why the list behaves as it does.
1. Part A.
  Evaluate $7^{1}$ , $7^{2}$ , $7^{3}$ , $7^{4}$ and $7^{5}$ , and list the ones digit of each.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Using the pattern in your list, give the ones digit of $7^{12}$ and of $7^{23}$ , saying in each case how the pattern gave it.
  
  Carry your own answer forward Read the repeating block off the list of ones digits you produced in part A, whatever digits it held.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Multiply $26$ by $7$ in two pieces: split $26$ into $20$ and $6$ , multiply each piece by $7$ , and add the two products. Use them to explain which part of $26$ decided the ones digit of the answer. Then say what happens to your list in part A from the point where one entry's ones digit matches an earlier entry's, naming entries by their position rather than by their value.
  
  Carry your own answer forward Read the repeat off the list of ones digits you wrote in part A, whatever digits it held.
  
  Explain why it works A sentence or two. Reasons, not steps. 7 points
The answer

Part A

$7$ , $49$ , $343$ , $2401$ and $16{,}807$ , whose ones digits are $7$ , $9$ , $3$ , $1$ , $7$ .

Part B

The ones digit of $7^{12}$ is $1$ and the ones digit of $7^{23}$ is $3$ . The list repeats every $4$ exponents, and $12$ is a multiple of $4$ while $23$ leaves remainder $3$ .

Part C

$7 \times 20 = 140$ ends in $0$ , because a whole number of tens stays one, so nothing of it reaches the ones column; $7 \times 6 = 42$ ends in $2$ , and the total $182$ takes its ones digit from the $6$ alone. In part A the fifth entry's ones digit matches the first entry's, so from there the same four digits repeat for ever.

Worked solution

Part A

Bring one new copy of the base into a running product at each step, starting from the first power, which is one copy of the base and so the base itself.

$7^{1} = 7, \quad 7^{2} = 49, \quad 7^{3} = 343$

Carrying on the same way gives the next two.

$7^{4} = 2401, \quad 7^{5} = 16{,}807$

Reading the last digit off each value gives the list $7$ , $9$ , $3$ , $1$ , $7$ . The fifth entry has come back to the first, so four steps have taken the ones digit all the way round.

Part B

The ones digits run $7$ , $9$ , $3$ , $1$ and then start again, so the pattern has a block of length $4$ and the exponent's remainder on division by $4$ picks the entry.

$12 = 4 \times 3 \quad (\text{remainder } 0)$

A remainder of $0$ lands on the last entry of a block, which is the entry belonging to $7^{4}$ , so the ones digit of $7^{12}$ is $1$ .

$23 = (4 \times 5) + 3$

A remainder of $3$ lands on the third entry of a block, which belongs to $7^{3}$ , so the ones digit of $7^{23}$ is $3$ . Neither answer needed the value of the power, which is the point of following the ones digit alone.

Part C

Split the number at its ones column and multiply each piece separately.

$7 \times 20 = 140, \qquad 7 \times 6 = 42$

Adding those gives $140 + 42 = 182$ , which is $7 \times 26$ . Now look at where each piece landed. The $140$ ends in $0$ , and it was bound to: $20$ is a whole number of tens, so multiplying it by anything leaves a whole number of tens, and a whole number of tens contributes nothing to the ones column. Only the $42$ can put a digit there, and it ends in $2$ , which is exactly the ones digit of $182$ .

So the ones digit of the product came from the ones digit of $26$ and from nothing else. That split works whatever number you start from, so at every step of the part A list the new ones digit is settled by the previous ones digit alone, and the list can be carried on without ever writing out a value.

That is what traps the list. The fifth entry's ones digit turned out to match the first entry's, and since each entry is decided by the one before it, the entry after the fifth has to repeat whatever followed the first, the one after that has to repeat whatever followed the second, and so on without end. The four digits lying between those two matching entries therefore run round for ever, and the list can never wander off somewhere new.

In one line

The first five powers of $7$ are $7$ , $49$ , $343$ , $2401$ and $16{,}807$ , whose ones digits run $7$ , $9$ , $3$ , $1$ , $7$ , so the block has length $4$ . Dividing the exponent by $4$ then picks the entry: $12$ leaves remainder $0$ and lands on the $1$ , while $23$ leaves remainder $3$ and lands on the $3$ . Splitting $26$ into $20$ and $6$ shows why the shortcut works: $7 \times 20 = 140$ ends in $0$ , because a whole number of tens stays a whole number of tens and reaches no further than the tens column, while $7 \times 6 = 42$ ends in $2$ , which is the ones digit of $182$ . Each ones digit is therefore decided by the one before it, so once the fifth entry came back to the first entry's digit, the block $7$ , $9$ , $3$ , $1$ had to repeat from there on.

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

Evaluates all five powers correctly, taking $7^{1}$ as $7$ . . Worth 2 points.

Lists the ones digit of each value as $7$ , $9$ , $3$ , $1$ , $7$ . . Worth 2 points.

Part B 5 points

Identifies the repeating block and its length of $4$ from the list in part A. . Worth 2 points.

Divides each exponent by the block length and uses the remainder to pick the entry, reaching $1$ and $3$ . . Worth 2 points.

Reports each answer as a ones digit rather than as a value of the power. . Worth 1 point.

Part C 7 points

Multiplies both pieces, reaching $7 \times 20 = 140$ and $7 \times 6 = 42$ , and adds them back to $182$ . . Worth 3 points.

Says that $140$ is a whole number of tens and so contributes nothing to the ones column, leaving the ones digit of $26$ to decide the ones digit of the product. . Worth 2 points. needs an explanation, not just an answer

Observes that the fifth entry of the part A list repeats the ones digit of the first, and that everything after it must therefore repeat the same block of four. . Worth 2 points. needs an explanation, not just an answer

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. How far the exponent reaches 14 points

Part A 4

Resolves the power before the multiplication and the addition in the unbracketed expression, and resolves the bracket first in the other. 2
Reaches both values, $383$ and $1375$ . 1
Names which tier was settled first in each expression. 1

Part B 4

Squares $-6$ with both factors carrying the sign, and squares $6$ alone in the other expression. 2
Reaches $36$ and $-36$ . 1
Says which number the exponent is attached to in each expression. 1

Part C 6

Says that parentheses decide what the exponent's base is, and distinguishes a bracket that widens the base from one that merely groups a neighbouring operation. 3
Says that a leading minus sign stays outside the power unless parentheses capture it, and that at an even exponent this changes the value. 2
Gives $9^{1} = 9$ and says a number with no written exponent carries an exponent of $1$ . 1

2. One value, chosen on purpose 13 points

Part A 3

Identifies the whole number that squares to each radicand. 2
Squares each answer back to the radicand as the check. 1

Part B 4

Gives all four values, $29$ , $28$ , $-29$ and $-28$ . 2
Picks the larger member of each pair, including reversing the order once the values are negated. 1
States the rule the two answers follow, that a bigger radicand gives a bigger root and that negating reverses an order. 1

Part C 6

Says that two numbers square to any positive radicand, and that a symbol naming both could not be used inside a calculation. 3
Names the chosen value as the non-negative one and shows how the other is written when it is wanted. 2
Says $\sqrt{-64}$ is not a real number because no real number squares to a negative. 1

3. One base, three rules 15 points

Part A 4

Adds the exponents for the product and subtracts them for the quotient, reaching $12^{7}$ and $12^{5}$ . 2
Says what happened to the count of factors in each case, and that the base is unchanged. 2

Part B 5

Works from the innermost grouping outward rather than combining exponents in the order they appear. 2
Reads the bare factor of $6$ as $6^{1}$ and reaches $6^{9}$ . 2
Names the quotient rule, the power rule and the product rule at the steps that use them. 1

Part C 6

Argues that each factor in the denominator cancels one in the numerator, so the surviving count is the difference of the two counts. 3
Says why the counts are subtracted rather than divided. 2
Reports the surviving count for the part A quotient and says why the base cannot change. 1

4. A year of washers 15 points

Part A 4

Places the decimal point after the first nonzero digit in each number, giving the coefficients $3.6$ and $4.5$ . 2
Gets both exponents right, positive for the large number and negative for the small one. 1
Reports how many places the point moved and in which direction for each number. 1

Part B 5

Multiplies the coefficients and adds the exponents, rather than mixing the two jobs. 2
Renormalises $16.2 \times 10^{2}$ by raising the exponent, reaching $1.62 \times 10^{3}$ . 2
Reports the total as a weight in kilograms and gives it as an ordinary number too. 1

Part C 6

Divides the coefficients and subtracts the exponents, rather than mixing the two jobs or reversing either. 2
Reaches $1.44 \times 10^{2}$ crates and gives $144$ as the ordinary number, labelled as a count of crates. 2
Decides the tidying question by where the new coefficient landed, and contrasts that with the coefficient reached in part B. 2

5. Below the first power 15 points

Part A 4

Evaluates the negative power as the reciprocal of the positive one, reaching $\tfrac{1}{64}$ . 2
Gives $8^{0} = 1$ rather than $0$ . 1
Runs the rule backwards to write $\dfrac{1}{8^{3}}$ as $8^{-3}$ . 1

Part B 5

Combines the denominator's two powers by adding their exponents, keeping the minus sign on the $-4$ . 2
Subtracts a negative exponent correctly, reaching $12^{-1}$ . 2
Rewrites the result so that no negative exponent is left, and gives its value as $\tfrac{1}{12}$ . 1

Part C 6

Decides for $25$ and shows the reciprocal being taken before the square is evaluated. 3
States what a negative exponent does to a fraction base, turning it over and leaving a positive exponent. 2
Confirms the verdict by spreading the exponent across the quotient and reaching $5^{2}$ . 1

6. Tidying a length and pinning it down 15 points

Part A 5

Splits each radicand at its largest perfect-square factor, $25$ and $144$ . 2
Reaches $5\sqrt{7}$ and $12\sqrt{2}$ , with nothing left under either radical that could still come out. 2
Keeps the results labelled as lengths in centimetres. 1

Part B 4

Names $169$ and $196$ as the bracketing perfect squares and their roots as $13$ and $14$ . 2
States the trap as $13 < \sqrt{175} < 14$ and says it leans toward $13$ , with the distances that decide it. 2

Part C 6

Identifies the radical and its simplified form as two spellings of one exact value, neither more exact than the other. 2
Says no decimal can record the length exactly, and gives the reason, that a non-perfect-square radicand has an irrational root. 2
Settles the comparison by squaring $17$ to $289$ and setting that against the radicand $288$ . 2

7. Backwards and forwards through a radical 17 points

Part A 5

Joins the two radicals before rooting, rather than trying to evaluate each radical on its own. 2
Reaches $30$ and $9$ , and names the product rule and the quotient rule at the steps that use them. 2
Reports each answer as a whole number, as asked, rather than leaving a radical standing. 1

Part B 5

Squares the outside factor before moving it under the radical, using $36$ rather than $6$ . 2
Reaches the single radical $\sqrt{108}$ . 2
Checks by simplifying $\sqrt{108}$ forwards and recovering $6\sqrt{3}$ . 1

Part C 7

Rejects the claim and supplies a pair of whole numbers for which $\sqrt{a} \times \sqrt{b}$ is not a whole number. 3
Supplies a pair for which it does hold, and notes that neither radicand needed to be a perfect square. 2
States the condition as $a \times b$ being a perfect square, a condition on the product rather than on the two numbers separately. 2

8. Which of two small powers is the larger 15 points

Part A 4

Evaluates each power as a reciprocal where the exponent is negative, reaching $\tfrac{1}{81}$ , $\tfrac{1}{81}$ and $81$ . 2
Identifies the first two as the equal pair, and keeps every value positive. 2

Part B 4

Turns each negative power into the reciprocal of its positive power, reaching $\tfrac{1}{32}$ , $\tfrac{1}{25}$ and $\tfrac{1}{10}$ . 2
Ranks unit fractions by their denominators, giving the order $2^{-5} < 5^{-2} < 10^{-1}$ . 2

Part C 7

Rejects the base-alone rule, using the pair given to show a smaller base with a larger value. 3
Notes that the part B order agreeing with the bases does not establish the rule. 2
States the rule that does settle it, comparing the positive powers under the line and reversing the result. 2

9. An exponent over a whole product 17 points

Part A 5

Spreads the outer exponent onto both factors of the bracket before dividing. 2
Applies the quotient rule to each base separately, reaching $3^{2} \times 5^{4}$ . 2
Names the power of a product and the quotient rule at the steps that use them. 1

Part B 5

Sends the outer exponent to the numerator and the denominator alike, and multiplies exponents on the power of a power. 2
Cancels the powers of $7$ with the quotient rule and reaches $2^{10} \times 7$ , with no fraction left. 2
Leaves the powers of $2$ and $7$ side by side, since no same-base rule reaches across two different bases. 1

Part C 7

Evaluates both sides, reaching $1156$ and $706$ , and rejects the claim on that evidence. 3
Says why the product version works, that copies of each factor may be reordered and gathered. 2
States the boundary: products and quotients split, while a sum does not, apart from the trivial case where one of the terms is $0$ . 2

10. The digit a power ends on 16 points

Part A 4

Evaluates all five powers correctly, taking $7^{1}$ as $7$ . 2
Lists the ones digit of each value as $7$ , $9$ , $3$ , $1$ , $7$ . 2

Part B 5

Identifies the repeating block and its length of $4$ from the list in part A. 2
Divides each exponent by the block length and uses the remainder to pick the entry, reaching $1$ and $3$ . 2
Reports each answer as a ones digit rather than as a value of the power. 1

Part C 7

Multiplies both pieces, reaching $7 \times 20 = 140$ and $7 \times 6 = 42$ , and adds them back to $182$ . 3
Says that $140$ is a whole number of tens and so contributes nothing to the ones column, leaving the ones digit of $26$ to decide the ones digit of the product. 2
Observes that the fifth entry of the part A list repeats the ones digit of the first, and that everything after it must therefore repeat the same block of four. 2

Multiple choice

Free response