The Other Side of Zero — NCERT Solutions

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.

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:

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 $\mathbf{-8}$.

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:

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

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

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