The Other Side of Zero

Given: Starting Floor = +2, Movement = -3.

Using the formula: Starting Floor + Movement = Target Floor

$$( +2) + (-3) = -1$$

You will reach Floor -1.

Using the concept Starting Floor + Movement = Target Floor:

a. $(+1) + (+4) = \mathbf{+5}$

Starting at Floor +1, moving up 4 floors → Floor +5.

b. $(+4) + (+1) = \mathbf{+5}$

Starting at Floor +4, moving up 1 floor → Floor +5.

c. $(+4) + (-3) = \mathbf{+1}$

Starting at Floor +4, moving down 3 floors → Floor +1.

d. $(-1) + (+2) = \mathbf{+1}$

Starting at Floor -1, moving up 2 floors → Floor +1.

e. $(-1) + (+1) = \mathbf{0}$

Starting at Floor -1, moving up 1 floor → Floor 0 (Ground).

f. $0 + (+2) = \mathbf{+2}$

Starting at Ground Floor, moving up 2 floors → Floor +2.

g. $0 + (-2) = \mathbf{-2}$

Starting at Ground Floor, moving down 2 floors → Floor -2.

Using: Movement needed = Target Floor − Starting Floor = $(-5) - \text{Starting Floor}$

Several examples:

1. Starting Floor = $0$: Movement = $(-5) - 0 = -5$
$$0 + (-5) = -5$$

2. Starting Floor = $+3$: Movement = $(-5) - (+3) = -8$
$$(+3) + (-8) = -5$$

3. Starting Floor = $-1$: Movement = $(-5) - (-1) = -4$
$$(-1) + (-4) = -5$$

4. Starting Floor = $-3$: Movement = $(-5) - (-3) = -2$
$$(-3) + (-2) = -5$$

5. Starting Floor = $+5$: Movement = $(-5) - (+5) = -10$
$$(+5) + (-10) = -5$$

6. Starting Floor = $-5$: Movement = $(-5) - (-5) = 0$
$$(-5) + (0) = -5$$

In general, any starting floor $S$ requires a movement of $(-5 - S)$ to reach Floor $-5$.

Combining button presses: pressing +1 and then +4 means moving up 1 floor and then up 4 more floors.

$$(+1) + (+4) = \mathbf{+5}$$

Pressing +4 and then +1 means moving up 4 floors and then up 1 more floor.

$$(+4) + (+1) = \mathbf{+5}$$

Pressing +4, then -3, then -2:

Step 1: $(+4) + (-3) = +1$

Step 2: $(+1) + (-2) = -1$

$$(+4) + (-3) + (-2) = \mathbf{-1}$$

Pressing -1, then +2, then -3:

Step 1: $(-1) + (+2) = +1$

Step 2: $(+1) + (-3) = -2$

$$(-1) + (+2) + (-3) = \mathbf{-2}$$

The inverse (additive inverse) of a number $n$ is the number which when added to $n$ gives 0.

| Number | Inverse |
|--------|--------|
| $+4$ | $-4$ |
| $-4$ | $+4$ |
| $-3$ | $+3$ |
| $0$ | $0$ |
| $+2$ | $-2$ |
| $-1$ | $+1$ |

Verification:
- $(+4) + (-4) = 0$ ✓
- $(-4) + (+4) = 0$ ✓
- $(-3) + (+3) = 0$ ✓
- $0 + 0 = 0$ ✓
- $(+2) + (-2) = 0$ ✓
- $(-1) + (+1) = 0$ ✓

Note: The inverse of 0 is 0 itself. Lines should connect each number to its inverse as listed above.

Based on the Building of Fun described in the chapter (standard layout):

1. Jay is in the Art Centre → Floor +2
2. Asin is in the Sports Centre → Floor -1 (Sports Centre is below ground)
3. Binnu is in the Cinema Centre → Floor -2 (Cinema is further below)
4. Aman is in the Toys Store → Floor +3 (Toys Store is above Art Centre)

*(Note: Exact floor numbers depend on the building diagram. The key concept is that lower floor numbers mean lower positions.)*

The person on the lowest floor is the one with the smallest (most negative) floor number.

Concept: On the number line (or building), the number to the left (lower floor) is smaller.

a. -2 \ \boxed{<} \ +5

(Floor -2 is below Floor +5; all negative numbers are less than positive numbers)

b. -5 \ \boxed{<} \ +4

(Floor -5 is below Floor +4)

c. -5 \ \boxed{<} \ -3

(Floor -5 is lower than Floor -3; among negative numbers, the one with larger absolute value is smaller)

d. +6 \ \boxed{>} \ -6

(Floor +6 is above Floor -6)

e. 0 \ \boxed{>} \ -4

(Floor 0 is above all negative floors; all negative numbers are less than 0)

f. 0 \ \boxed{<} \ +4

(Floor 0 is below Floor +4; all positive numbers are greater than 0)

a. -10 \ \boxed{>} \ -12

(Floor -10 is higher than Floor -12; -10 is closer to 0)

b. +17 \ \boxed{>} \ -10

(All positive numbers are greater than all negative numbers)

c. 0 \ \boxed{>} \ -20

(0 is greater than all negative numbers)

d. +9 \ \boxed{>} \ -9

(Positive numbers are greater than negative numbers)

e. -25 \ \boxed{<} \ -7

(Floor -25 is much lower than Floor -7)

f. +15 \ \boxed{>} \ -17

(Positive numbers are greater than negative numbers)

Given information: Floor A = -12, Floor D = -1, Floor E = +1.

The floors are equally spaced on the number line. Between A (-12) and D (-1) there are floors B and C, dividing the interval into equal parts.

From A to D: $-1 - (-12) = 11$ units over 3 intervals → each interval = not equal.

Assuming the floors are marked at consecutive integers or at equal intervals based on the figure (which shows a vertical number line):

If the spacing between consecutive marked floors is consistent:
- A = -12, and moving upward by equal steps to reach D = -1 with B and C in between.
- 3 gaps from A to D: step = $\frac{-1-(-12)}{3} = \frac{11}{3}$ — not integer.

More likely the floors are at integer values and the labels A through H mark specific floors:
- A = -12, B = -9 (or similar), C = -6, D = -1 (or as per diagram spacing).

*Note: The exact answer depends on the figure. Based on a typical equal-spacing assumption with 7 labeled points (A to G) and given A = -12, D = -1, E = +1:*

From D to E: $+1 - (-1) = 2$ units, 1 gap → step between D and E = 2.

Assuming uniform spacing throughout: step = 2 (since D = -1 and E = +1 differ by 2).

Working backwards from D = -1 with step 2:
- C = -1 - 2 = -3
- B = -3 - 2 = -5
- A = -5 - 2 = -7 ≠ -12 (contradiction)

Alternative: step between D and E = 2, but different step elsewhere. Most likely the figure shows a number line where each unit is 1:
- A = -12, B = -8, C = -4, D = -1 (not uniform)

*Since the exact figure is not available, the general method is:*

Use the given anchor points to determine the scale/step size, then apply it to find the remaining floors. With A = -12, D = -1, E = +1:
- Step from D to E = 2 units per division
- F = E + 2 = +3
- G = F + 2 = +5
- H = G + 2 = +7
- C = D - 2 = -3
- B = C - 2 = -5

(Students should read off values directly from their figure.)

To mark these floors on the number line (building shown as a vertical line):

- -7: Mark 7 units below 0 (below ground level).
- -4: Mark 4 units below 0.
- +3: Mark 3 units above 0.
- -10: Mark 10 units below 0.

On the number line, the order from bottom to top is:
-10 < -7 < -4 < 0 < +3

Students should locate and label these points on their building/number line diagram accordingly.

Using: Target Floor − Starting Floor = Movement needed

a. $(+1) - (+4)$: To go from Floor +4 to Floor +1, move down 3.
$$(+1) - (+4) = \mathbf{-3}$$

b. $(0) - (+2)$: To go from Floor +2 to Floor 0, move down 2.
$$(0) - (+2) = \mathbf{-2}$$

c. $(+4) - (+1)$: To go from Floor +1 to Floor +4, move up 3.
$$(+4) - (+1) = \mathbf{+3}$$

d. $(0) - (-2)$: To go from Floor -2 to Floor 0, move up 2.
$$(0) - (-2) = \mathbf{+2}$$

e. $(+4) - (-3)$: To go from Floor -3 to Floor +4, move up 7 (3 to reach 0, then 4 more).
$$(+4) - (-3) = \mathbf{+7}$$

f. $(-4) - (-3)$: To go from Floor -3 to Floor -4, move down 1.
$$(-4) - (-3) = \mathbf{-1}$$

g. $(-1) - (+2)$: To go from Floor +2 to Floor -1, move down 3.
$$(-1) - (+2) = \mathbf{-3}$$

h. $(-2) - (-2)$: To go from Floor -2 to Floor -2, no movement needed.
$$(-2) - (-2) = \mathbf{0}$$

i. $(-1) - (+1)$: To go from Floor +1 to Floor -1, move down 2.
$$(-1) - (+1) = \mathbf{-2}$$

j. $(+3) - (-3)$: To go from Floor -3 to Floor +3, move up 6 (3 to reach 0, then 3 more).
$$(+3) - (-3) = \mathbf{+6}$$

Using: Missing addend = Target − Starting; and subtraction rules.

a. $(+40) + \_ = +200$

Missing = $200 - 40 = +160$
$$(+40) + \mathbf{(+160)} = +200$$

b. $(+40) + \_ = -200$

Missing = $-200 - 40 = -240$
$$(+40) + \mathbf{(-240)} = -200$$

c. $(-50) + \_ = +200$

Missing = $200 - (-50) = 200 + 50 = +250$
$$(-50) + \mathbf{(+250)} = +200$$

d. $(-50) + \_ = -200$

Missing = $-200 - (-50) = -200 + 50 = -150$
$$(-50) + \mathbf{(-150)} = -200$$

e. $(-200) - (-40) = ?$

Subtracting a negative = adding its positive:
$$(-200) - (-40) = (-200) + (+40) = \mathbf{-160}$$

f. $(+200) - (+40) = ?$

$$(+200) - (+40) = \mathbf{+160}$$

g. $(-200) - (+40) = ?$

$$(-200) - (+40) = -200 - 40 = \mathbf{-240}$$

Using the infinite lift concept: Starting Level + Movement = Target Level, and Target − Starting = Movement.

a. $-125 + (-30)$

Start at -125, move down 30:
$$-125 + (-30) = \mathbf{-155}$$

b. $+105 - (-55)$

Target = +105, Starting = -55. Movement = $105 - (-55) = 105 + 55$:
$$+105 - (-55) = \mathbf{+160}$$

c. $+105 + (+55)$

Start at +105, move up 55:
$$+105 + (+55) = \mathbf{+160}$$

*(Note: (b) and (c) give the same answer — subtracting a negative equals adding a positive.)*

d. $+80 - (-150)$

$= 80 + 150$:
$$+80 - (-150) = \mathbf{+230}$$

e. $+80 + (+150)$

$$+80 + (+150) = \mathbf{+230}$$

*(Note: (d) and (e) give the same answer.)*

f. $-99 - (-200)$

$= -99 + 200 = 200 - 99$:
$$-99 - (-200) = \mathbf{+101}$$

g. $-99 + (+200)$

$$-99 + (+200) = \mathbf{+101}$$

*(Note: (f) and (g) give the same answer.)*

h. $+1500 - (-1500)$

$= 1500 + 1500$:
$$+1500 - (-1500) = \mathbf{+3000}$$

On the number line, mark:

Positive numbers (to the right of 0): For example, $+1, +3, +5$

Negative numbers (to the left of 0): For example, $-1, -3, -5$

These are placed at their respective positions on the number line, with positive numbers to the right of 0 and negative numbers to the left of 0.

Based on the example in Question 1, the three negative numbers marked are:

$$\boxed{-1} \quad \boxed{-3} \quad \boxed{-5}$$

(Students should write the three negative numbers they actually marked on their number line.)

Is 2 > -3?

Yes, 2 > -3.

Reason: On the number line, 2 lies to the right of -3. All positive numbers are greater than all negative numbers. Since 2 is positive and -3 is negative, 2 > -3.

Is -2 < 3?

Yes, -2 < 3.

Reason: On the number line, -2 lies to the left of 3. Since -2 is negative and 3 is positive, and all negative numbers are less than all positive numbers, -2 < 3.

a. $-5 + 0$

Adding 0 to any number gives the same number:
$$-5 + 0 = \mathbf{-5}$$

b. $7 + (-7)$

A number plus its additive inverse equals 0:
$$7 + (-7) = \mathbf{0}$$

c. $-10 + 20$

Start at -10, move up 20: $20 - 10 = 10$ (positive, since 20 > 10):
$$-10 + 20 = \mathbf{+10}$$

d. $10 - 20$

Start at 10, target is found by going back 20: $10 - 20 = -(20-10)$:
$$10 - 20 = \mathbf{-10}$$

e. $7 - (-7)$

Subtracting a negative = adding its positive:
$$7 - (-7) = 7 + 7 = \mathbf{14}$$

f. $-8 - (-10)$

Subtracting a negative = adding its positive:
$$-8 - (-10) = -8 + 10 = \mathbf{+2}$$

Using the token method: positive tokens (+) and negative tokens (-); zero pairs cancel out.

a. $(+6) + (+4)$

Place 6 positive tokens and 4 positive tokens. No zero pairs. Total = 10 positive tokens.
$$(+6) + (+4) = \mathbf{+10}$$

b. $(-3) + (-2)$

Place 3 negative tokens and 2 negative tokens. No zero pairs. Total = 5 negative tokens.
$$(-3) + (-2) = \mathbf{-5}$$

c. $(+5) + (-7)$

Place 5 positive and 7 negative tokens. Cancel 5 zero pairs. Remaining: 2 negative tokens.
$$(+5) + (-7) = \mathbf{-2}$$

d. $(-2) + (+6)$

Place 2 negative and 6 positive tokens. Cancel 2 zero pairs. Remaining: 4 positive tokens.
$$(-2) + (+6) = \mathbf{+4}$$

*(Note: The exact token images are not visible, but the method is described below.)*

Method: Count the positive tokens (green/+) and negative tokens (red/-). Cancel equal numbers of positive and negative tokens (zero pairs). The remaining tokens give the floor.

a. After cancelling zero pairs, if $p$ positive tokens remain → Floor $+p$; if $n$ negative tokens remain → Floor $-n$.

For example, if there are 5 positive and 3 negative tokens:
- Cancel 3 zero pairs → 2 positive tokens remain
- Floor = +2
- Addition statement: $(+5) + (-3) = +2$

b. Similarly, if there are 2 positive and 6 negative tokens:
- Cancel 2 zero pairs → 4 negative tokens remain
- Floor = -4
- Addition statement: $(+2) + (-6) = -4$

*(Students should apply this method to their actual token images.)*

Using the token method for subtraction: place tokens for the first number, then remove tokens of the second number (adding zero pairs if needed).

a. $(+10) - (+7)$

Place 10 positive tokens. Remove 7 positive tokens. Remaining: 3 positive.
$$(+10) - (+7) = \mathbf{+3}$$

b. $(-8) - (-4)$

Place 8 negative tokens. Remove 4 negative tokens. Remaining: 4 negative.
$$(-8) - (-4) = \mathbf{-4}$$

c. $(-9) - (-4)$

Place 9 negative tokens. Remove 4 negative tokens. Remaining: 5 negative.
$$(-9) - (-4) = \mathbf{-5}$$

d. $(+9) - (+12)$

Place 9 positive tokens. Need to remove 12 positive but only have 9. Add 3 zero pairs (3 positive + 3 negative). Now have 12 positive and 3 negative. Remove 12 positive. Remaining: 3 negative.
$$(+9) - (+12) = \mathbf{-3}$$

e. $(-5) - (-7)$

Place 5 negative tokens. Need to remove 7 negative but only have 5. Add 2 zero pairs. Now have 7 negative and 2 positive. Remove 7 negative. Remaining: 2 positive.
$$(-5) - (-7) = \mathbf{+2}$$

f. $(-2) - (-6)$

Place 2 negative tokens. Need to remove 6 negative but only have 2. Add 4 zero pairs. Now have 6 negative and 4 positive. Remove 6 negative. Remaining: 4 positive.
$$(-2) - (-6) = \mathbf{+4}$$

Using the rule: subtracting a number = adding its additive inverse.

a. $(-5) - (-7) = (-5) + (+7) = \mathbf{+2}$

b. $(+10) - (+13) = 10 - 13 = \mathbf{-3}$

c. $(-7) - (-9) = (-7) + (+9) = \mathbf{+2}$

d. $(+3) - (+8) = 3 - 8 = \mathbf{-5}$

e. $(-2) - (-7) = (-2) + (+7) = \mathbf{+5}$

f. $(+3) - (+15) = 3 - 15 = \mathbf{-12}$

Given: $-3 - (+5)$

Step 1: Place 3 negative tokens to represent $-3$.

Step 2: We need to remove 5 positive tokens, but we have none.

Step 3: Add 5 zero pairs (5 positive + 5 negative tokens). This does not change the value.

Now we have: 5 positive tokens and $(3 + 5) = 8$ negative tokens.

Step 4: Remove 5 positive tokens.

Remaining: 8 negative tokens.

$$-3 - (+5) = \mathbf{-8}$$

We had to put in 5 zero pairs.

Using the rule: $a - (+b) = a + (-b)$ and $a - (-b) = a + (+b)$.

a. $(-3) - (+10) = (-3) + (-10) = \mathbf{-13}$

*Token method: Place 3 negative. Add 10 zero pairs. Remove 10 positive. Left with 13 negative.*

b. $(+8) - (-7) = (+8) + (+7) = \mathbf{+15}$

*Token method: Place 8 positive. Add 7 zero pairs. Remove 7 negative. Left with 15 positive.*

c. $(-5) - (+9) = (-5) + (-9) = \mathbf{-14}$

*Token method: Place 5 negative. Add 9 zero pairs. Remove 9 positive. Left with 14 negative.*

d. $(-9) - (+10) = (-9) + (-10) = \mathbf{-19}$

e. $(+6) - (-4) = (+6) + (+4) = \mathbf{+10}$

f. $(-2) - (+7) = (-2) + (-7) = \mathbf{-9}$

Given:
- Starting balance = ₹0
- Credits (positive): ₹30, ₹40, ₹50
- Debits (negative): ₹40, ₹50, ₹60

Total credits = $30 + 40 + 50 = ₹120$

Total debits = $40 + 50 + 60 = ₹150$

Final balance = Starting balance + Total credits − Total debits
$$= 0 + 120 - 150 = \mathbf{-₹30}$$

The bank account balance is −₹30 (i.e., ₹30 in debt/overdraft).

Given:
- Starting balance = ₹0
- Debits: ₹1, 2, 4, 8, 16, 32, 64, 128
- Credit: ₹256

Total debits = $1 + 2 + 4 + 8 + 16 + 32 + 64 + 128$

This is a geometric series: $= 2^0 + 2^1 + 2^2 + \cdots + 2^7 = 2^8 - 1 = 256 - 1 = 255$

Final balance = $0 - 255 + 256 = \mathbf{+₹1}$

The bank account balance is ₹1 (positive).

Why maintain a positive balance:

1. A positive balance means you have money available to spend without borrowing.
2. Banks typically charge interest or fees when your balance goes negative (overdraft charges), making you owe more money.
3. A negative balance means you are in debt to the bank, which can lead to financial stress.
4. Maintaining a positive balance ensures financial security and good credit standing.

Circumstances where a temporary negative balance may be worthwhile:

1. Business investment: If you make a large purchase (like equipment or stock) that will generate more income than the cost, a temporary negative balance can be justified — as shown in the chapter example where a ₹150 purchase led to ₹200 earnings.
2. Emergency expenses: Urgent medical or repair costs may require spending beyond your current balance.
3. Education or skill development: Investing in education that leads to higher future earnings.

In general, a negative balance should only be temporary and planned, with a clear path to returning to positive.

*(Note: The exact figure is not visible in the OCR. The following is based on a typical geographical cross-section used in NCERT Class 6 textbooks.)*

In a geographical cross-section, heights above sea level are positive and heights below sea level are negative.

Based on the standard figure in the textbook:
- a ≈ +8000 m (high mountain peak, e.g., Himalayan range)
- b ≈ +500 m (plateau or hill)
- c ≈ 0 m (sea level)
- d ≈ -100 m (shallow ocean floor)
- e ≈ -3000 m (deep ocean)
- f ≈ +200 m (coastal plain)
- g ≈ -500 m (ocean trench)

*Students should read the actual heights from their textbook figure and fill in accordingly.*

Highest point: The point with the greatest positive height above sea level (the tallest peak in the cross-section).

Lowest point: The point with the most negative height (deepest below sea level, such as an ocean trench).

*Students should identify these from their actual figure. In general:*
- The highest point is the mountain peak (largest positive value).
- The lowest point is the deepest ocean floor (most negative value).

Method: Read the height of each point (A through G) from the cross-section figure.

- Decreasing order (highest to lowest): Arrange points from the largest positive height to the most negative height.
Example: A > B > C > D > E > F > G (if A is highest and G is lowest)

- Increasing order (lowest to highest): Reverse of the above.
Example: G < F < E < D < C < B < A

*Students should substitute the actual letters and their heights from the figure to write the correct sequences.*

The highest point above sea level on Earth is Mount Everest (also called Sagarmatha or Chomolungma), located in the Himalayas on the border of Nepal and Tibet (China).

$$\text{Height} = +8848.86 \text{ m above sea level}$$

(approximately +8849 m or +8.85 km)

The lowest point on Earth with respect to sea level is the Challenger Deep in the Mariana Trench in the Pacific Ocean.

$$\text{Depth} \approx -10{,}994 \text{ m} \approx -11{,}000 \text{ m below sea level}$$

The lowest point on land (dry land below sea level) is the Dead Sea shore, located on the border of Jordan and Israel:

$$\text{Height} \approx -430 \text{ m below sea level}$$

Places in India where temperatures go below 0°C:

1. Leh, Ladakh — temperatures can drop to -20°C or lower in winter.
2. Drass, Jammu & Kashmir — one of the coldest inhabited places in India, temperatures can reach -45°C.
3. Shimla, Himachal Pradesh — temperatures occasionally drop below 0°C in winter.
4. Manali, Himachal Pradesh — sub-zero temperatures in winter.
5. Srinagar, Jammu & Kashmir — temperatures below 0°C in winter.
6. Spiti Valley, Himachal Pradesh — extreme cold in winter.

What is common among these places:
- All these places are located at high altitudes in the Himalayan or sub-Himalayan mountain ranges in northern India.
- They are in the northern/north-western part of India.

Why it becomes colder there:
- High altitude: As altitude increases, atmospheric pressure decreases and the air becomes thinner, holding less heat. Temperature drops by approximately 6.5°C for every 1000 m increase in altitude.
- Distance from the sea: These places are far from the moderating influence of the ocean.
- Cold winds: Cold, dry winds from Central Asia and the Himalayas blow through these regions in winter.
- Snow cover: Snow reflects sunlight (high albedo), preventing the ground from absorbing heat.

Reasoning: During a day in November in Leh:
- The warmest time is in the afternoon (around 2:00 p.m.) when the sun is highest.
- The temperature is moderately warm in the late morning (11:00 a.m.) as it is still warming up.
- The temperature is cold late at night (11:00 p.m.) after sunset.
- The coldest time is in the early morning (2:00 a.m.) just before dawn.

Matching:

| Temperature | Time |
|-------------|------|
| 14°C | 02:00 p.m. (warmest — afternoon) |
| 8°C | 11:00 a.m. (warming up — late morning) |
| -2°C | 11:00 p.m. (cooling down — late night) |
| -4°C | 02:00 a.m. (coldest — early morning) |

Given second grid:
$$\begin{array}{|c|c|c|}\hline 5 & -3 & -5 \\ \hline 0 & \text{[blank]} & -5 \\ \hline -8 & -2 & 7 \\ \hline \end{array}$$

Calculating the border rows and columns:

Top row: $5 + (-3) + (-5) = 5 - 3 - 5 = \mathbf{-3}$

Bottom row: $(-8) + (-2) + 7 = -8 - 2 + 7 = \mathbf{-3}$

Left column: $5 + 0 + (-8) = 5 + 0 - 8 = \mathbf{-3}$

Right column: $(-5) + (-5) + 7 = -5 - 5 + 7 = \mathbf{-3}$

All four border sums equal -3.

$$\text{Border sum of the second grid} = \mathbf{-3}$$

*(Note: The partial grids in the images are not fully visible. The method for completing them is described below.)*

Method: In a 3×3 hollow grid, the top row, bottom row, left column, and right column must each sum to the required border sum. The centre cell is not part of any border sum.

For Border sum = +4:

One possible grid:
$$\begin{array}{|c|c|c|}\hline 3 & -1 & 2 \\ \hline 1 & \text{[any]} & 0 \\ \hline 0 & 2 & 2 \\ \hline \end{array}$$
Check: Top: $3+(-1)+2=4$ ✓, Bottom: $0+2+2=4$ ✓, Left: $3+1+0=4$ ✓, Right: $2+0+2=4$ ✓

For Border sum = -2:

One possible grid:
$$\begin{array}{|c|c|c|}\hline -1 & 0 & -1 \\ \hline 0 & \text{[any]} & -2 \\ \hline -1 & -2 & 1 \\ \hline \end{array}$$
Check: Top: $-1+0+(-1)=-2$ ✓, Bottom: $-1+(-2)+1=-2$ ✓, Left: $-1+0+(-1)=-2$ ✓, Right: $-1+(-2)+1=-2$ ✓

For Border sum = -4:

One possible grid:
$$\begin{array}{|c|c|c|}\hline -2 & 0 & -2 \\ \hline 0 & \text{[any]} & -4 \\ \hline -2 & -4 & 2 \\ \hline \end{array}$$
Check: Top: $-2+0+(-2)=-4$ ✓, Bottom: $-2+(-4)+2=-4$ ✓, Left: $-2+0+(-2)=-4$ ✓, Right: $-2+(-4)+2=-4$ ✓

*Students should fill in the given partial grids using these principles.*

Border sum = -4

The constraint is: top row sum = bottom row sum = left column sum = right column sum = -4. The centre cell is free (it doesn't affect any border sum).

Way 1:
$$\begin{array}{|c|c|c|}\hline -2 & 0 & -2 \\ \hline 0 & 5 & -4 \\ \hline -2 & -4 & 2 \\ \hline \end{array}$$

Way 2:
$$\begin{array}{|c|c|c|}\hline -4 & 2 & -2 \\ \hline 2 & 0 & -6 \\ \hline -2 & -6 & 4 \\ \hline \end{array}$$

Way 3:
$$\begin{array}{|c|c|c|}\hline 0 & -1 & -3 \\ \hline -1 & 7 & -3 \\ \hline -3 & -3 & 2 \\ \hline \end{array}$$

In each case, the centre cell can be any integer — it does not affect the border sums. This is why there are infinitely many ways to fill the grid.

Answer: Any grid where the centre cell is free (not part of any border sum constraint) can be filled in multiple ways.

Reason: In the 3×3 hollow grid, the centre cell does not belong to the top row, bottom row, left column, or right column. Therefore, it can be filled with any integer without affecting the border sums. This gives infinitely many solutions.

Additionally, even the border cells have some freedom — as long as the row/column sums equal the required border sum, individual values can vary. For example, in the top row, if the sum must be -4, we could have $(-1) + (-1) + (-2)$ or $0 + (-2) + (-2)$ or $1 + (-3) + (-2)$, etc.

Conclusion: Grids with a free centre cell (like the 3×3 hollow grid) always have multiple solutions. The more constraints are removed, the more ways the grid can be filled.

Example puzzle (make your own):

Fill in the missing numbers so that the border sum is +3:

$$\begin{array}{|c|c|c|}\hline 2 & ? & -1 \\ \hline ? & \text{[free]} & ? \\ \hline -1 & ? & 3 \\ \hline \end{array}$$

Solution:
- Top row: $2 + ? + (-1) = 3 \Rightarrow ? = 2$
- Bottom row: $-1 + ? + 3 = 3 \Rightarrow ? = 1$
- Left column: $2 + ? + (-1) = 3 \Rightarrow ? = 2$
- Right column: $-1 + ? + 3 = 3 \Rightarrow ? = 1$

$$\begin{array}{|c|c|c|}\hline 2 & 2 & -1 \\ \hline 2 & \text{any} & 1 \\ \hline -1 & 1 & 3 \\ \hline \end{array}$$

*Students should create their own puzzles with different border sums and challenge classmates.*

Observation: No matter which numbers you circle (following the game rules — circle one number per row and per column, then strike out the rest of that row and column), the sum of the circled numbers is always the same.

Reason: The grid is constructed so that each number can be written as $r_i + c_j$ where $r_i$ is a row value and $c_j$ is a column value. When you pick one number from each row and each column, the sum equals $(r_1 + r_2 + r_3 + r_4) + (c_1 + c_2 + c_3 + c_4)$, which is always the same regardless of which specific numbers are chosen.

Conclusion: The sum will always be the same, no matter how many times you try or which numbers you pick.

Grid 1:
$$\begin{array}{|c|c|c|c|}\hline 7 & 10 & 13 & 16 \\ \hline -2 & 1 & 4 & 7 \\ \hline -11 & -8 & -5 & -2 \\ \hline -20 & -7 & -14 & -11 \\ \hline \end{array}$$

Let us verify with one selection — pick 7 (row 1, col 1), 1 (row 2, col 2), -5 (row 3, col 3), -11 (row 4, col 4):
$$7 + 1 + (-5) + (-11) = 7 + 1 - 5 - 11 = -8$$

Another selection — pick 16 (row 1, col 4), -2 (row 2, col 1), -8 (row 3, col 2), -14 (row 4, col 3):
$$16 + (-2) + (-8) + (-14) = 16 - 2 - 8 - 14 = -8$$

The answer for Grid 1 is always $\mathbf{-8}$.

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Grid 2:
$$\begin{array}{|c|c|c|c|}\hline -11 & -10 & -9 & -8 \\ \hline -7 & -6 & -5 & -4 \\ \hline -3 & -2 & -1 & 0 \\ \hline 1 & 2 & 3 & 4 \\ \hline \end{array}$$

Selection — pick -11 (row 1, col 1), -6 (row 2, col 2), -1 (row 3, col 3), 4 (row 4, col 4):
$$-11 + (-6) + (-1) + 4 = -11 - 6 - 1 + 4 = -14$$

Another selection — pick -8 (row 1, col 4), -7 (row 2, col 1), -2 (row 3, col 2), 3 (row 4, col 3):
$$-8 + (-7) + (-2) + 3 = -8 - 7 - 2 + 3 = -14$$

The answer for Grid 2 is always $\mathbf{-14}$.

The magic is in BOTH the numbers AND the way they are arranged.

Why it works:

Each entry in the grid can be expressed as:
$$\text{entry}_{ij} = r_i + c_j$$
where $r_i$ is a value associated with row $i$ and $c_j$ is a value associated with column $j$.

When you pick exactly one entry from each row and each column (like a permutation), the sum is:
$$\sum_{i=1}^{4}(r_i + c_{\sigma(i)}) = \sum_{i=1}^{4} r_i + \sum_{i=1}^{4} c_{\sigma(i)} = \sum_{i=1}^{4} r_i + \sum_{j=1}^{4} c_j$$

This sum is always the same regardless of the permutation $\sigma$, because you always add all row values and all column values exactly once.

Making your own such grid:

Choose any row values $r_1, r_2, r_3, r_4$ and column values $c_1, c_2, c_3, c_4$. Fill entry $(i,j)$ as $r_i + c_j$.

Example: $r = (0, 1, 2, 3)$, $c = (0, 2, 4, 6)$:
$$\begin{array}{|c|c|c|c|}\hline 0 & 2 & 4 & 6 \\ \hline 1 & 3 & 5 & 7 \\ \hline 2 & 4 & 6 & 8 \\ \hline 3 & 5 & 7 & 9 \\ \hline \end{array}$$
The constant sum = $(0+1+2+3)+(0+2+4+6) = 6+12 = 18$.

Note: 'Between' means strictly between (not including the endpoints).

a. Between 0 and -7 (in increasing order):
$$-6, -5, -4, -3, -2, -1$$

b. Between -4 and 4 (in increasing order):
$$-3, -2, -1, 0, 1, 2, 3$$

c. Between -8 and -15 (in increasing order):
$$-14, -13, -12, -11, -10, -9$$

d. Between -30 and -23 (in increasing order):
$$-29, -28, -27, -26, -25, -24$$

We need three integers whose sum is $-8$.

Example 1: $-1 + (-3) + (-4) = -8$ ✓

Example 2: $0 + (-3) + (-5) = -8$ ✓

Example 3: $2 + (-4) + (-6) = -8$ ✓

Example 4: $-10 + 5 + (-3) = -8$ ✓

Many answers are possible. Any three integers $a, b, c$ such that $a + b + c = -8$.

Given: Each die has faces: $-1, 2, -3, 4, -5, 6$.

All possible sums when rolling two such dice (each die shows one of: -5, -3, -1, 2, 4, 6):

Let the two values be $a$ and $b$ where $a, b \in \{-5, -3, -1, 2, 4, 6\}$.

Possible sums (listing unique values):

$-5+(-5)=-10$, $-5+(-3)=-8$, $-5+(-1)=-6$, $-5+2=-3$, $-5+4=-1$, $-5+6=1$

$-3+(-3)=-6$, $-3+(-1)=-4$, $-3+2=-1$, $-3+4=1$, $-3+6=3$

$-1+(-1)=-2$, $-1+2=1$, $-1+4=3$, $-1+6=5$

$2+2=4$, $2+4=6$, $2+6=8$

$4+4=8$, $4+6=10$

$6+6=12$

Set of all possible sums:
$$\{-10, -8, -6, -4, -3, -2, -1, 1, 3, 4, 5, 6, 8, 10, 12\}$$

All integers from -10 to +12:
$$-10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$$

Numbers NOT possible (in the range -10 to +12):
$$\boxed{-9, -7, -5, 0, 2, 7, 9, 11}$$

These 8 values cannot be obtained as the sum of two faces of these dice.

Solving each expression:

Row 1:

$8 - 13 = \mathbf{-5}$

$(-8) - (13) = -8 - 13 = \mathbf{-21}$

$(-13) - (-8) = -13 + 8 = \mathbf{-5}$

$(-13) + (-8) = \mathbf{-21}$

Row 2:

$8 + (-13) = 8 - 13 = \mathbf{-5}$

$(-8) - (-13) = -8 + 13 = \mathbf{+5}$

$(13) - 8 = \mathbf{+5}$

$13 - (-8) = 13 + 8 = \mathbf{+21}$

Summary table:

| Expression | Result |
|---|---|
| $8 - 13$ | $-5$ |
| $(-8) - (13)$ | $-21$ |
| $(-13) - (-8)$ | $-5$ |
| $(-13) + (-8)$ | $-21$ |
| $8 + (-13)$ | $-5$ |
| $(-8) - (-13)$ | $+5$ |
| $(13) - 8$ | $+5$ |
| $13 - (-8)$ | $+21$ |

*(Taking present year as 2025 CE)*

a. 150 years ago from 2025:
$$2025 - 150 = \mathbf{1875 \text{ CE}}$$

b. 2200 years ago from 2025:

Since there was no year 0 (the calendar goes from 1 BCE directly to 1 CE), we must account for this:
$$2025 - 2200 = -175$$
Since there is no year 0, we adjust: $-175$ corresponds to 176 BCE (i.e., 175 + 1 = 176 BCE).

$$\mathbf{176 \text{ BCE}}$$

c. 320 years after 680 BCE:

680 BCE means $-680$ on the integer timeline (with no year 0).

Moving 320 years forward from 680 BCE:
$$680 - 320 = 360 \text{ BCE}$$

Since 680 - 320 = 360 > 0, we are still in BCE:
$$\mathbf{360 \text{ BCE}}$$

*(The year is still before the common era, so it is 360 BCE.)*

a. Sequence: $-40, -34, -28, -22, \ldots$

Common difference: $-34 - (-40) = +6$. Each term increases by 6.

$$-22 + 6 = -16, \quad -16 + 6 = -10, \quad -10 + 6 = -4$$

Answer: $(-40), (-34), (-28), (-22), \mathbf{-16}, \mathbf{-10}, \mathbf{-4}$

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b. Sequence: $3, 4, 2, 5, 1, 6, 0, 7, \ldots$

Observe two interleaved sequences:
- Odd positions: $3, 2, 1, 0, \ldots$ (decreasing by 1)
- Even positions: $4, 5, 6, 7, \ldots$ (increasing by 1)

Next terms:
- 9th term (odd position): $0 - 1 = -1$
- 10th term (even position): $7 + 1 = 8$
- 11th term (odd position): $-1 - 1 = -2$

Answer: $3, 4, 2, 5, 1, 6, 0, 7, \mathbf{-1}, \mathbf{8}, \mathbf{-2}$

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c. Sequence: $\_, \_, 12, 6, 1, -3, -6, \_, \_, \_$

Differences between consecutive known terms:
$6 - 12 = -6$, $1 - 6 = -5$, $-3 - 1 = -4$, $-6 - (-3) = -3$

The differences are: $-6, -5, -4, -3, \ldots$ (increasing by 1 each time, i.e., decreasing in magnitude).

Continuing forward: next difference = $-2$, then $-1$, then $0$.
$$-6 + (-2) = -8, \quad -8 + (-1) = -9, \quad -9 + 0 = -9$$

Working backwards from 12:
- Difference before 12 to 6 is -6, so difference before the term before 12 should be -7:
- Term before 12: $12 - (-7) = 12 + 7 = 19$
- Term before 19: $19 - (-8) = 19 + 8 = 27$

Answer: $\mathbf{27}, \mathbf{19}, 12, 6, 1, -3, -6, \mathbf{-8}, \mathbf{-9}, \mathbf{-9}$

*(Note: The pattern of differences is $\ldots, -8, -7, -6, -5, -4, -3, -2, -1, 0, \ldots$)*

Given cards: $+1, +7, +18, -5, -2, -9$

Goal: Make an expression as close to $-30$ as possible.

Strategy: To get a very negative result, we want to subtract large positive numbers and add negative numbers.

Attempt:
$$(-9) + (-5) + (-2) - (+18) - (+7) + (+1)$$
$$= -9 - 5 - 2 - 18 - 7 + 1 = -40$$

Too negative. Let's try:
$$(-9) + (-5) + (-2) - (+18) - (+7)$$
$$= -9 - 5 - 2 - 18 - 7 = -41$$

Try:
$$(-9) + (-5) - (+18) + (+1) + (+7) - (-2)$$
$$= -9 - 5 - 18 + 1 + 7 + 2 = -22$$

Try:
$$(-9) + (-5) + (-2) - (+18) + (+7) - (+1)$$
$$= -9 - 5 - 2 - 18 + 7 - 1 = -28$$

Try:
$$(-9) + (-5) + (-2) - (+18) + (+7) - (-1)$$
— but +1 card, not -1.

Try:
$$(-9) + (-5) + (-2) - (+18) + (+1) + (+7)$$
$$= -9 - 5 - 2 - 18 + 1 + 7 = -26$$

Try:
$$(-9) + (-5) + (-2) - (+18) - (+1) + (+7)$$
$$= -9 - 5 - 2 - 18 - 1 + 7 = -28$$

Try:
$$(-9) - (+18) - (+7) + (+1) - (-5) - (-2)$$
$$= -9 - 18 - 7 + 1 + 5 + 2 = -26$$

Closest result found:
$$(-9) + (-5) + (-2) - (+18) + (+7) - (+1) = -28$$

or

$$(-9) + (-5) + (-2) - (+18) - (+1) + (+7) = -28$$

Expression giving $-28$ (closest to $-30$ using all six cards):
$$(-9) + (-5) + (-2) - (+18) - (+1) + (+7) = \mathbf{-28}$$

*(Students may find other combinations; the key is to get as close to -30 as possible.)*