Figure It Out — Questions & Solutions

Given

Find the ordered pair of integers for each sum and difference: (a) 27, 9; (b) 4, 12; (c) 0, 10; (d) 0, −10; (e) −7, −1; (f) −7, −13.

To Find

First and second integer in each pair

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. If the numbers are x and y, add x+y = S and x−y = D to obtain 2x = S+D

   then y = S−x.

2. (a) x = 36 ÷ 2 = 18

   y = 9.

   Check: 18+9 = 27 and 18−9 = 9.

3. (b) x = 16 ÷ 2 = 8

   y = −4.

   (c) x = 10 ÷ 2 = 5

   y = −5.

4. (d) x = −10 ÷ 2 = −5

   y = 5.

   (e) x = −8 ÷ 2 = −4

   y = −3.

5. (f) x = −20 ÷ 2 = −10

   y = 3.

   Check: −10+3 = −7 and −10−3 = −13.

Result

(a) $\left(18,9\right)$; (b) $\left(8,-4\right)$; (c) $\left(5,-5\right)$; (d) $\left(-5,5\right)$; (e) $\left(-4,-3\right)$; (f) $\left(-10,3\right)$.

Given

Use tokens to evaluate (a) 3 × (−2), (b) (−5) × (−2), (c) (−4) × (−1), (d) (−7) × 3.

To Find

The four products

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. (a) Add two negative tokens three times: −6.

2. (b) Removing two negative tokens five times has the same effect as adding ten positive tokens: 10.

   Add zero pairs first if needed.

3. (c) Removing one negative token four times leaves value 4.

4. (d) Removing three positive tokens seven times leaves value −21.

   Zero pairs allow removal without changing the starting value.

Result

(a) −6; (b) 10; (c) 4; (d) −21.

Given

Given 123 × 456 = 56088, evaluate (a) (−123) × 456, (b) (−123) × (−456), (c) 123 × (−456).

To Find

Signed products without repeating the long multiplication

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. The magnitude remains 56088 in all three cases.

2. One negative factor makes (a) and (c) negative.

   Two negative factors make (b) positive.

Result

(a) −56088; (b) 56088; (c) −56088.

Given

Frame a simple rule for multiplying two integers.

To Find

A multiplication rule covering positive, negative and zero factors

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. Multiply the magnitudes.

   For two nonzero factors, give the result a positive sign when their signs match and a negative sign when they differ.

2. If either factor is zero, the product is zero.

Result

Same signs: positive; opposite signs: negative; a zero factor: zero.

Given

Find (a) 4 × (−3), (b) (−6) × (−3), (c) (−5) × (−1), (d) (−8) × 4, (e) (−9) × 10, (f) 10 × (−17). Parts (c)–(f) continue on page 34.

To Find

The six products

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. (a) Opposite signs: −(4 × 3) = −12.

   (b) Same signs: 6 × 3 = 18.

2. (c) Same signs: 5 × 1 = 5.

   (d) Opposite signs: −(8 × 4) = −32.

3. (e) −(9 × 10) = −90.

   (f) −(10 × 17) = −170.

Result

(a) −12; (b) 18; (c) 5; (d) −32; (e) −90; (f) −170.

Given

Evaluate (a) 14 × (−15), (b) (−16) × (−5), (c) 36 ÷ (−18), (d) (−46) ÷ (−23).

To Find

The four values

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. (a) −(14 × 15) = −210.

   (b) 16 × 5 = 80.

2. (c) −(36 ÷ 18) = −2.

   (d) 46 ÷ 23 = 2.

Result

(a) −210; (b) 80; (c) −2; (d) 2.

Given

A room starts at 32°C and cools by 5°C each hour for 10 hours.

To Find

Room temperature after 10 hours

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. The signed change is 10 × (−5) = −50°C.

2. Final temperature = 32+(−50) = −18°C.

Result

−18°C.

Given

White cement earns ₹8 per bag; grey cement loses ₹5 per bag. (a) Sell 3000 white and 5000 grey bags. (b) With 6400 grey bags, how many white bags give neither profit nor loss?

To Find

Net profit/loss in (a) and break-even white-bag count in (b)

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. (a) Net = 3000 × 8+5000 × (−5) = 24000−25000 = −1000 rupees.

2. (b) Loss on grey bags = 6400 × 5 = 32000 rupees.

   Let w be the white-bag count: 8w−32000 = 0.

3. Thus w = 32000 ÷ 8 = 4000 bags.

Result

(a) ₹1000 loss. (b) 4000 white bags.

Given

Fill the blanks: (a) (−3) × □ = 27; (b) 5 × □ = −35; (c) □ × (−8) = −56; (d) □ × (−12) = 132; (e) □ ÷ (−8) = 7; (f) □ ÷ 12 = −11.

To Find

The missing integers

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. (a) 27 ÷ (−3) = −9.

   (b) −35 ÷ 5 = −7.

2. (c) −56 ÷ (−8) = 7.

   (d) 132 ÷ (−12) = −11.

3. (e) 7 × (−8) = −56.

   (f) −11 × 12 = −132.

Result

(a) −9; (b) −7; (c) 7; (d) −11; (e) −56; (f) −132.

Given

Evaluate (a) (−5) × (18+(−3)), (b) (−7) × 4 × (−1), (c) (−2) × (−1) × (−5) × (−3).

To Find

The three expression values

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. (a) First 18−3 = 15

   then −5 × 15 = −75.

2. (b) −7 × 4 = −28

   multiplying by −1 gives 28.

3. (c) Pair the negatives: 2 × 15 = 30.

Result

(a) −75; (b) 28; (c) 30.

Given

Evaluate (a) (−27) ÷ 9, (b) 84 ÷ (−4), (c) (−56) ÷ (−2).

To Find

The three quotients

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. (a) −(27 ÷ 9) = −3.

   (b) −(84 ÷ 4) = −21.

   (c) 56 ÷ 2 = 28.

Result

(a) −3; (b) −21; (c) 28.

Given

Find the integer whose product with −1 is (a) 27, (b) −31, (c) −1, (d) 1, (e) 0.

To Find

The five original integers

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. Multiplication by −1 takes the additive inverse, so negate each given result.

2. Check: (−27)(−1) = 27

   31(−1) = −31

   1(−1) = −1

   (−1)(−1) = 1

   0(−1) = 0.

Result

(a) −27; (b) 31; (c) 1; (d) −1; (e) 0.

Given

Given 47−56+14−8+2−8+5 = −4, find −47+56−14+8−2+8−5 without evaluating each term.

To Find

Value of the sign-reversed expression

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. Every term is the opposite of its original term.

   Factor out −1 from the entire new expression.

2. The new value is (−1) × (−4) = 4.

Result

4.

Given

Even n → n ÷ 2; odd n → (−3) × n + 1.

To Find

Investigate the three suggested starts without asserting a universal outcome.

Formula / Property / Theorem / Reasoning

Compute each signed step; repetition establishes a cycle for that particular sequence.

Solution

1. From −7: −7 → 22 → 11 → −32 → −16 → −8 → −4 → −2 → −1 → 4 → 2 → 1 → −2.

2. From −21: −21 → 64 → 32 → 16 → 8 → 4, then 2 → 1 → −2 → −1 → 4.

3. From −6: −6 → −3 → 10 → 5 → −14 → −7, then follow the first sequence.

4. These starts reach the cycle −2, −1, 4, 2, 1.

   Testing examples is not a proof for all integer starts

   0, for instance, stays 0.

Result

The three tested starts enter the same five-term cycle. No universal claim is made.

Given

A test awards +4 for each correct answer and −2 for each wrong answer. (a) Anita answers all questions, scores 40 and has 15 correct. (b) Anil scores −10 with 5 correct. Find wrong and unanswered counts.

To Find

Incorrect answers and total/unanswered questions

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. (a) Correct answers earn 15 × 4 = 60.

   The wrong answers contribute 40−60 = −20, so there are (−20) ÷ (−2) = 10 wrong answers.

   Total questions: 15+10 = 25.

2. (b) Correct answers earn 5 × 4 = 20.

   Wrong answers contribute −10−20 = −30, so there are 15 wrong answers.

3. Anil answered 5+15 = 20 of the same 25 questions, leaving 5 unanswered.

   This uses the test convention that unanswered questions score zero.

Result

Anita: 10 wrong, 25 questions. Anil: 15 wrong, 5 unanswered.

Given

The machine maps (4,8,−3)→28, (6,9,6)→−48, (2,3,−2)→8, (−9,5,−8)→31 and (7,−4,−6)→−17. Find a matching rule and the output for (−16,−6,−9).

To Find

A rule fitting every displayed row and its missing output

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. Try a−b × c, multiplying before subtracting.

2. 4−8 × (−3) = 28

   6−9 × 6 = −48

   2−3 × (−2) = 8.

3. −9−5 × (−8) = 31 and 7−(−4) × (−6) = −17, so the rule fits all five rows.

4. For the last row, −16−(−6) × (−9) = −16−54 = −70.

Result

A matching rule is a−bc; missing output −70. Finite examples alone do not make a rule unique.

Given

The temperature starts at 8°C and drops 5°C per hour for 4 hours. Write an expression.

To Find

Expression for the final temperature

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. Each hour contributes −5°C.

   Four hours contribute 4 × (−5)°C.

2. The required expression is 8+4 × (−5)

   its value is −12.

Result

8+4 × (−5) = −12°C.

Given

Find three consecutive integers with product (a) −6, (b) 120.

To Find

The consecutive-integer triples

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. (a) (−3) × (−2) × (−1) = 6 × (−1) = −6.

2. (b) 4 × 5 × 6 = 20 × 6 = 120.

Result

(a) −3,−2,−1. (b) 4,5,6.

Given

Coins are worth +13 or −9 pibs. Make +85 pibs, then the page-44 totals (a) 20, (b) 40, (c) −50, (d) 8, (e) 10, (f) −2, (g) 1, (h) 1568.

To Find

Non-negative coin counts for every requested total

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. With x positive-value and y negative-value coins, the total is 13x−9y.

   Counts must be non-negative integers.

2. Example: 13 × 10−9 × 5 = 85.

   (a) 13 × 5−9 × 5 = 20.

   (b) 13 × 10−9 × 10 = 40.

3. (c) 13 × 1−9 × 7 = −50.

   (d) 13 × 2−9 × 2 = 8.

4. (e) 13 × 7−9 × 9 = 10.

   (f) 13 × 4−9 × 6 = −2.

5. (g) 13 × 7−9 × 10 = 1.

   (h) 13 × 122−9 × 2 = 1586−18 = 1568.

Result

Coin pairs $\left(x,y\right)$: example $\left(10,5\right)$; (a) $\left(5,5\right)$; (b) $\left(10,10\right)$; (c) $\left(1,7\right)$; (d) $\left(2,2\right)$; (e) $\left(7,9\right)$; (f) $\left(4,6\right)$; (g) $\left(7,10\right)$; (h) $\left(122,2\right)$. These are valid examples, not unique solutions.

Given

Evaluate (a) (32 × (−18)) ÷ (−36), (b) 32 ÷ ((−36) × (−18)), (c) (25 × (−12)) ÷ (45 × (−27)), (d) (280 × (−7)) ÷ ((−8) × (−35)).

To Find

The four exact quotients

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. (a) −576 ÷ (−36) = 16.

2. (b) Denominator = 648, so $\frac{32}{648}$ = $\frac{4}{81}$.

3. (c) Numerator = −300 and denominator = −1215.

   Thus $\frac{300}{1215}$ = $\frac{20}{81}$.

4. (d) Numerator = −1960 and denominator = 280.

   Thus $\frac{-1960}{280}$ = −7.

Result

(a) 16; (b) $\frac{4}{81}$; (c) $\frac{20}{81}$; (d) −7.

Given

Arrange in increasing order: (a) (−348)+(−1064), (b) (−348)−(−1064), (c) 348−(−1064), (d) (−348) × (−1064), (e) 348 × (−1064), (f) 348 × 964.

To Find

Increasing order of the six labelled expressions

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. Evaluate: (a) −1412

   (b) 716

   (c) 1412

   (d) 370272

   (e) −370272

   (f) 335472.

2. The negative number with larger magnitude comes first

   order the remaining positive numbers by magnitude.

Result

(e) < (a) < (b) < (c) < (f) < (d).

Given

Given (−548) × 972 = −532656, find (a) (−547) × 972, (b) (−548) × 971, (c) (−547) × 971.

To Find

The three neighbouring products

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. (a) (−548+1) × 972 = −532656+972 = −531684.

2. (b) (−548) × (972−1) = −532656+548 = −532108.

3. (c) Use (a): (−547) × (972−1) = −531684+547 = −531137.

Result

(a) −531684; (b) −532108; (c) −531137.

Given

Given 207 × (−33+7) = −5382, find −207 × (33−7).

To Find

Value of −207 × (33−7)

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. Since −33+7 = −(33−7), the given expression is exactly −207 × (33−7).

Result

−5382.

Given

Four fixed signed numbers and three operations, each used exactly once; brackets allowed.

To Find

Find and verify the maximum and minimum.

Formula / Property / Theorem / Reasoning

Compare all valid orders and parenthesisations; do not insert an extra unary minus or reuse an operation.

Solution

1. (−6) × ((−2) − (5 + 3)) = (−6) × (−10) = 60.

2. ((3 + 5) − (−2)) × (−6) = 10 × (−6) = −60.

3. A systematic complete check considers 24 number orders, 6 operation orders and 5 binary bracket structures: 720 expressions.

4. The five shapes are ((ab)c)d, (a(bc))d, (ab)(cd), a((bc)d), a(b(cd))

   each adjacency stands for one of the three operations.

   Exact integer evaluation of these cases gives no value outside −60 to 60.

Result

Maximum 60; minimum −60. The examples attain the bounds and the complete enumeration checks optimality.

Given

Fill the boxes with integers in at least five ways: (a) □+□ × □ = −36; (b) (□−□) × □ = 12; (c) □−(□−□) = −1.

To Find

Five valid ordered triples of entries for each expression

Reasoning

Apply the integer sign rules, order of operations and inverse operations.

Solution

1. (a) Put the entries (−36−k,k,1), for k = 0,1,2,3,4: (−36,0,1), (−37,1,1), (−38,2,1), (−39,3,1), (−40,4,1).

   Each gives −36−k+k = −36.

2. (b) Use (k+3,k,4), for k = 0,1,2,3,4: (3,0,4), (4,1,4), (5,2,4), (6,3,4), (7,4,4).

   Each gives 3 × 4 = 12.

3. (c) Use (k,k+1,0), for k = 0,1,2,3,4: (0,1,0), (1,2,0), (2,3,0), (3,4,0), (4,5,0).

   Each gives k−(k+1) = −1.

Result

Five verified fillings for each part; many others are possible.