A Peek Beyond the Point

Given: $15 \frac{3}{10} \frac{4}{100} - 2 \frac{6}{10} \frac{8}{100}$

Step 1: Convert both numbers to hundredths notation (decimal form).

$15 \frac{3}{10} \frac{4}{100}$ means $15 + \frac{3}{10} + \frac{4}{100} = 15 + \frac{30}{100} + \frac{4}{100} = 15 + \frac{34}{100} = 15.34$

$2 \frac{6}{10} \frac{8}{100}$ means $2 + \frac{6}{10} + \frac{8}{100} = 2 + \frac{60}{100} + \frac{8}{100} = 2 + \frac{68}{100} = 2.68$

Step 2: Subtract.

$$15.34 - 2.68$$

Breaking it down by place value:
$$(10 + 5 + \tfrac{3}{10} + \tfrac{4}{100}) - (2 + \tfrac{6}{10} + \tfrac{8}{100})$$

$$= (15 - 2) + (\tfrac{3}{10} - \tfrac{6}{10}) + (\tfrac{4}{100} - \tfrac{8}{100})$$

Since \frac{3}{10} < \frac{6}{10}, regroup: borrow 1 unit from 15, converting it to $\frac{10}{10}$:

$$= (14 - 2) + (\tfrac{13}{10} - \tfrac{6}{10}) + (\tfrac{4}{100} - \tfrac{8}{100})$$

Since \frac{4}{100} < \frac{8}{100}, regroup: borrow $\frac{1}{10}$ from $\frac{13}{10}$, converting it to $\frac{10}{100}$:

$$= (14 - 2) + (\tfrac{12}{10} - \tfrac{6}{10}) + (\tfrac{14}{100} - \tfrac{8}{100})$$

$$= 12 + \tfrac{6}{10} + \tfrac{6}{100}$$

$$= 12.66$$

Answer: $15.34 - 2.68 = 12.66$

Convert to decimal form:

$12\frac{6}{10}\frac{2}{100} = 12 + \frac{6}{10} + \frac{2}{100} = 12.62$

$\frac{9}{10}\frac{9}{100} = \frac{9}{10} + \frac{9}{100} = 0.99$

Subtract:
$$12.62 - 0.99$$

Regrouping:
$= 12 + 0.62 - 0.99$
$= 11 + 1.62 - 0.99$
$= 11 + 0.63$
$= 11.63$

Answer: $11.63$

Detailed place value computation:

$$84.691 - 77.345$$

$$= (80 + 4 + 0.6 + 0.09 + 0.001) - (70 + 7 + 0.3 + 0.04 + 0.005)$$

Ones place: $4 - 7$: need to regroup. Borrow 10 from tens.
Tens place: $80 \to 70$, ones become $14 - 7 = 7$

Tenths place: $0.6 - 0.3 = 0.3$

Hundredths place: $0.09 - 0.04 = 0.05$

Thousandths place: $0.001 - 0.005$: need to regroup. Borrow $0.01$ from hundredths.
$0.05 \to 0.04$, thousandths become $0.011 - 0.005 = 0.006$

Tens place: $70 - 70 = 0$

Putting it all together:
$$= (70 - 70) + (14 - 7) + (0.6 - 0.3) + (0.04 - 0.04) + (0.011 - 0.005)$$
$$= 0 + 7 + 0.3 + 0.04 + 0.006$$
$$= 7.346$$

Compact (column) form:
$$\begin{array}{r} 84.691 \\ -\;77.345 \\ \hline 7.346 \end{array}$$

Answer: $84.691 - 77.345 = 7.346$

$$\begin{array}{r} 29.19 \\ +\;9.91 \\ \hline 39.10 \end{array}$$

Tenths: $1 + 9 = 10$, write 0 carry 1. Hundredths: $9 + 1 = 10$, write 0 carry 1 (already counted). Let me redo carefully:

Hundredths: $9 + 1 = 10$, write 0, carry 1.
Tenths: $1 + 9 + 1 = 11$, write 1, carry 1.
Ones: $9 + 9 + 1 = 19$, write 9, carry 1.
Tens: $2 + 0 + 1 = 3$.

$$= 39.10$$

Answer: $39.10$

$$\begin{array}{r} 6.236 \\ +\;0.487 \\ \hline 6.723 \end{array}$$

Thousandths: $6 + 7 = 13$, write 3, carry 1.
Hundredths: $3 + 8 + 1 = 12$, write 2, carry 1.
Tenths: $2 + 4 + 1 = 7$.
Ones: $6 + 0 = 6$.

Answer: $6.723$

Write $18$ as $18.0$:
$$\begin{array}{r} 18.0 \\ -\;8.8 \\ \hline 9.2 \end{array}$$

Tenths: $0 - 8$, regroup: borrow 1 from ones. $10 - 8 = 2$.
Ones: $8 - 1 - 8$, regroup: borrow from tens. $18 - 1 - 8 = 9$.
Tens: $1 - 0 = 1$... wait: $18.0 - 8.8$:
Ones: $8 - 8 = 0$ after borrowing. Tens: $1 - 0 = 1$...

Let me redo: $18.0 - 8.8$
Tenths: $0 - 8$, borrow 1 tenth-group from ones: ones become 7, tenths become 10. $10 - 8 = 2$.
Ones: $7 - 8$, borrow from tens: tens become 0 (from 1), ones become 17. $17 - 8 = 9$.
Tens: $0 - 0 = 0$.

Result: $9.2$

Answer: $9.2$

Tenths: $4 - 5$, borrow: ones become 9, tenths become 14. $14 - 5 = 9$.
Ones: $9 - 4 = 5$... wait: $10.4 - 4.5$
Ones digit of 10 is 0, tens digit is 1.
Tenths: $4 - 5$, borrow from ones. Ones: $0 \to$ need to borrow from tens. Tens: $1 \to 0$, ones become 10. Now borrow from ones for tenths: ones $10 \to 9$, tenths $4 + 10 = 14$. $14 - 5 = 9$.
Ones: $9 - 4 = 5$.
Tens: $0 - 0 = 0$.

Result: $5.9$

Answer: $5.9$

Write $17$ as $17.000$:

$$17.000 - 16.198$$

Thousandths: $0 - 8$, borrow chain: $10 - 8 = 2$.
Hundredths: $0 - 1 - 9$ (after borrowing): regroup again: $10 - 1 - 9 = 0$...

Let me do it step by step:
Thousandths: $0 - 8$, borrow from hundredths. Hundredths become $-1$ (need to borrow too).
Hundredths: $0 - 1 - 9$, borrow from tenths. Tenths become $-1$.
Tenths: $0 - 1 - 1$, borrow from ones. Ones become $-1$.
Ones: $7 - 1 - 8$, borrow from tens. Tens become $0$.

More cleanly: $17.000 - 16.198$
$$= 17.000 - 16.198 = 0.802$$

Verification: $16.198 + 0.802 = 17.000$ ✓

Answer: $0.802$

Write $17$ as $17.00$:
$$17.00 - 0.05$$

Hundredths: $0 - 5$, borrow. Tenths: $0 - 0 - 1$ (after borrow), borrow from ones. Ones: $7 \to 6$, tenths $\to 10$, then tenths $10 - 1 = 9$, hundredths $10 - 5 = 5$.
Ones: $6 - 0 = 6$. Tens: $1 - 0 = 1$.

Result: $16.95$

Answer: $16.95$

Write $18.1$ as $18.100$:
$$34.505 - 18.100$$

Thousandths: $5 - 0 = 5$.
Hundredths: $0 - 0 = 0$.
Tenths: $5 - 1 = 4$.
Ones: $4 - 8$, borrow from tens: tens $3 \to 2$, ones $14 - 8 = 6$.
Tens: $2 - 1 = 1$.

Result: $16.405$

Answer: $16.405$

Write $9.9$ as $9.90$:
$$9.90 - 9.09$$

Hundredths: $0 - 9$, borrow from tenths. Tenths: $9 \to 8$, hundredths $10 - 9 = 1$.
Tenths: $8 - 0 = 8$.
Ones: $9 - 9 = 0$.

Result: $0.81$

Answer: $0.81$

$$6.236 - 0.487$$

Thousandths: $6 - 7$, borrow. Hundredths: $3 \to 2$, thousandths $16 - 7 = 9$.
Hundredths: $2 - 8$, borrow. Tenths: $2 \to 1$, hundredths $12 - 8 = 4$.
Tenths: $1 - 4$, borrow. Ones: $6 \to 5$, tenths $11 - 4 = 7$.
Ones: $5 - 0 = 5$.

Result: $5.749$

Answer: $5.749$

Pattern identified: Each term increases by $0.4$.

$$6.0 + 0.4 = 6.4$$
$$6.4 + 0.4 = 6.8$$
$$6.8 + 0.4 = 7.2$$

The next 3 terms are: $6.4, 6.8, 7.2$

$$0.34 = \frac{3}{10} + \frac{4}{100} + \frac{0}{1000}$$

Or simply: $0.34 = 3$ tenths $+ 4$ hundredths $+ 0$ thousandths.

Answer: $0.34 = \frac{3}{10} + \frac{4}{100}$

$$1.02 = 1 + \frac{0}{10} + \frac{2}{100}$$

$= 1$ one $+ 0$ tenths $+ 2$ hundredths $+ 0$ thousandths.

Answer: $1.02 = 1 + \frac{2}{100}$

$$0.8 = \frac{8}{10} + \frac{0}{100} + \frac{0}{1000}$$

$= 8$ tenths.

Answer: $0.8 = \frac{8}{10}$

$$0.362 = \frac{3}{10} + \frac{6}{100} + \frac{2}{1000}$$

$= 3$ tenths $+ 6$ hundredths $+ 2$ thousandths.

Answer: $0.362 = \frac{3}{10} + \frac{6}{100} + \frac{2}{1000}$

Note: The number line image is not available. However, the general method is:

1. Identify the two consecutive whole numbers (or decimal numbers) between which the letter lies.
2. Count the equal divisions between them to determine the scale of each small division.
3. Count how many divisions the letter is from the left endpoint and multiply by the value of each division.

For example, if the number line goes from $0$ to $1$ with 10 equal parts, each part $= 0.1$. If letter $A$ is at the 3rd mark, then $A = 0.3$.

Students should apply this method to the actual number line in their textbook.

Compare the numbers:

First compare the whole number (integer) parts:
- $11.01$ and $11.10$ have integer part $11$.
- $1.011$, $1.101$, $1.01$ have integer part $1$.

Among $11.01$ and $11.10$: tenths digit 0 < 1, so 11.10 > 11.01.

Among $1.011$, $1.101$, $1.01$: tenths digit: $1.101$ has $1$, others have $0$. So $1.101$ is greatest among these.
Now compare $1.011$ and $1.01$: $1.011 = 1.011$, $1.01 = 1.010$. So 1.011 > 1.010.

Descending order: 11.10 > 11.01 > 1.101 > 1.011 > 1.01

All have integer part $2$. Compare tenths:
- $2.768$: tenths $= 7$
- $2.698$: tenths $= 6$
- $2.675$: tenths $= 6$
- $2.567$: tenths $= 5$
- $2.499$: tenths $= 4$

Among $2.698$ and $2.675$: hundredths 9 > 7, so 2.698 > 2.675.

Descending order: 2.768 > 2.698 > 2.675 > 2.567 > 2.499

All have integer part $4$. Compare tenths: all have $6$ in tenths except $4.595$ (tenths $= 5$).

So $4.595$ is smallest among these.

Now compare $4.678$, $4.600$, $4.656$, $4.666$ (all tenths $= 6$):
Hundredths: $7, 0, 5, 6$ respectively.
$4.678$: hundredths $= 7$ (largest)
$4.666$: hundredths $= 6$
$4.656$: hundredths $= 5$
$4.600$: hundredths $= 0$

Descending order: 4.678\text{ g} > 4.666\text{ g} > 4.656\text{ g} > 4.600\text{ g} > 4.595\text{ g}

All have integer part $33$. Compare tenths:
- $33.31$, $33.331$, $33.313$: tenths $= 3$
- $33.13$, $33.133$: tenths $= 1$

So $33.13$ and $33.133$ are smaller.

Among $33.31$, $33.331$, $33.313$: hundredths: $1, 3, 1$.
$33.331$: hundredths $= 3$ (largest)
Among $33.31$ and $33.313$: $33.310$ vs $33.313$: thousandths 0 < 3, so 33.313 > 33.310.

Among $33.13$ and $33.133$: $33.130$ vs $33.133$: thousandths 0 < 3, so 33.133 > 33.130.

Descending order: 33.331\text{ m} > 33.313\text{ m} > 33.31\text{ m} > 33.133\text{ m} > 33.13\text{ m}

Given digits: $1, 4, 0, 8, 6$ (each used once).

We need a number close to $30$. The integer part should be close to $30$.

Possible two-digit integers using these digits: $10, 14, 16, 18, 40, 41, 46, 48, 60, 61, 64, 68, 80, 81, 84, 86$...

Closest to $30$: $28$ is not possible. Let's try $28$... digits available are $1,4,0,8,6$.
- Integer part $= 28$: not possible (no $2$).
- Integer part $= 40$: difference from $30$ is $10$.
- Integer part $= 18$: difference from $30$ is $12$.
- Integer part $= 16$: difference $= 14$.
- Integer part $= 41$: difference $= 11$.

Wait — we can also use a decimal point. Try $30$-something:
- $30$ itself: we need digits $3$ and $0$, but $3$ is not available.

Try making numbers like $\mathbf{28.}$... not possible.

Best approach: use digits to form $\mathbf{30.}$... not possible since $3$ is unavailable.

Try $\mathbf{29.}$... not possible.

Try numbers just above or below $30$:
- $\mathbf{40.618}$: difference $= 10.618$
- $\mathbf{18.640}$: difference $= 11.36$
- $\mathbf{18.064}$: difference $= 11.936$

Actually, let's try $\mathbf{30}$ using decimal: e.g., $\mathbf{28.}$... no $2$.

Using all five digits with a decimal point:
- $\mathbf{28.}$... no.
- $\mathbf{30.}$... no $3$.
- $\mathbf{31.}$... no $3$.

Best candidates near $30$:
- $\mathbf{40.618}$: $|40.618 - 30| = 10.618$
- $\mathbf{18.640}$: $|18.640 - 30| = 11.36$
- $\mathbf{41.608}$: $|41.608 - 30| = 11.608$
- $\mathbf{40.168}$: $|40.168 - 30| = 10.168$
- $\mathbf{40.186}$: $|40.186 - 30| = 10.186$
- $\mathbf{40.618}$: $10.618$

Smallest difference so far: $\mathbf{40.168}$ with difference $10.168$.

Can we do better? Try $\mathbf{41.068}$: $|41.068-30|=11.068$. No.
Try $\mathbf{40.816}$: $|40.816-30|=10.816$. No.

Actually, try numbers with integer part in $20$s: no digit $2$.
Try integer part $= 1$-digit: $8.0146$: $|8.0146-30|=21.99$. Worse.

So the closest is $\mathbf{40.168}$ (difference $10.168$) or $\mathbf{40.186}$ (difference $10.186$).

Among all arrangements, $\mathbf{40.168}$ is closest to $30$.

Answer: $40.168$ (using digits $4, 0, 1, 6, 8$) is the decimal number closest to $30$.

Given digits: $1, 4, 0, 8, 6$ (each used once).

We need a number between $100$ and $1000$, so it must be a 3-digit integer part (hundreds digit $\neq 0$).

To make the smallest such number:
- The hundreds digit should be as small as possible (but $\neq 0$): use $1$.
- The tens digit should be as small as possible: use $0$.
- The ones digit: next smallest available: $4$.
- Decimal part: remaining digits $8$ and $6$ → to minimise, place smaller digit first: $0.68$... wait, remaining digits are $8$ and $6$.

So the number is $\mathbf{104.68}$.

But wait — can we make it smaller? Try $\mathbf{104.68}$ vs $\mathbf{104.86}$: 104.68 < 104.86. ✓

Also check $\mathbf{100.68}$... but we only have one $0$ and it's already used in tens place. Digits are $1,4,0,8,6$ — only one $0$.

So: hundreds $= 1$, tens $= 0$, ones $= 4$, tenths $= 6$, hundredths $= 8$ → $\mathbf{104.68}$.

Answer: $104.68$ is the smallest possible decimal number between $100$ and $1000$ using digits $1, 4, 0, 8, 6$.

No, a decimal number with more digits is not necessarily greater than one with fewer digits.

Reason: The value of a number depends on the place value of each digit, not on the count of digits.

Examples:
- $0.009$ has 3 decimal digits but is much smaller than $0.9$ which has 1 decimal digit.
- $1.5$ (2 digits) is greater than $0.999$ (3 decimal digits).
- $100.1$ has more digits than $99.9$ and is indeed greater, but this is because of the hundreds place, not the number of digits.

Conclusion: The number of digits alone does not determine which decimal is greater. We must compare place by place starting from the highest place value.

Given weights:
- Beans: $0.25$ kg
- Carrots: $0.30$ kg
- Potatoes: $0.50$ kg
- Capsicums: $0.20$ kg
- Ginger: $0.05$ kg

Total weight:
$$0.25 + 0.30 + 0.50 + 0.20 + 0.05$$
$$= 0.55 + 0.50 + 0.20 + 0.05$$
$$= 1.05 + 0.20 + 0.05$$
$$= 1.25 + 0.05$$
$$= 1.30 \text{ kg}$$

Answer: Total weight $= 1.30$ kg

Given:
- Milk in first 3 days: $3.79 + 4.2 + 4.25$ L
- Total in 6 days: $25$ L

Step 1: Find total for first 3 days.
$$3.79 + 4.20 + 4.25$$
$$= 7.99 + 4.25$$
$$= 12.24 \text{ L}$$

Step 2: Find total for last 3 days.
$$25 - 12.24 = 12.76 \text{ L}$$

Answer: Pinto supplied $12.76$ L of milk in the last three days.

Given:
- January weight: $35.75$ kg
- February weight: $34.50$ kg

Since 34.50 < 35.75, Tinku has lost weight.

Change in weight:
$$35.75 - 34.50 = 1.25 \text{ kg}$$

Answer: Tinku has lost $1.25$ kg of weight.

Identify the pattern by finding differences between consecutive terms:

$6.4 - 5.5 = +0.9$
$6.39 - 6.4 = -0.01$
$7.29 - 6.39 = +0.9$
$7.28 - 7.29 = -0.01$
$6.18 - 7.28 = -1.1$
$6.17 - 6.18 = -0.01$

Pattern of operations: $+0.9,\ -0.01,\ +0.9,\ -0.01,\ -1.1,\ -0.01,\ +0.9,\ -0.01, \ldots$

Wait, let me re-examine:
$5.5 \xrightarrow{+0.9} 6.4 \xrightarrow{-0.01} 6.39 \xrightarrow{+0.9} 7.29 \xrightarrow{-0.01} 7.28 \xrightarrow{-1.1} 6.18 \xrightarrow{-0.01} 6.17$

The repeating cycle appears to be: $+0.9, -0.01, +0.9, -0.01, -1.1, -0.01$

Next operations after $6.17$: following the cycle, next is $+0.9$, then $-0.01$.

$$6.17 + 0.9 = 7.07$$
$$7.07 - 0.01 = 7.06$$

Answer: The next two terms are $7.07$ and $7.06$.

Given:
- Insurance cost per passenger $= 45$ paise $= ₹0.45$
- Number of passengers $= 1$ lakh $= 1,00,000$

Total insurance fee:
$$= 1{,}00{,}000 \times ₹0.45$$
$$= ₹45{,}000$$

Answer: The total insurance fee paid is ₹$45,000$.

Convert both to the same denominator (or decimal):

$$\frac{10}{1000} = \frac{1}{100} = 0.01$$

$$\frac{1}{10} = 0.1$$

Since 0.1 > 0.01:

Answer: $\frac{1}{10}$ is greater.

Convert to decimals:

One-hundredth $= \frac{1}{100} = 0.01$

90 thousandths $= \frac{90}{1000} = 0.09$

Since 0.09 > 0.01:

Answer: 90 thousandths is greater.

Convert to decimals:

One-thousandth $= \frac{1}{1000} = 0.001$

90 hundredths $= \frac{90}{100} = 0.9$

Since 0.9 > 0.001:

Answer: 90 hundredths is greater.

Given example: 87 ones, 5 tenths and 60 hundredths $= 88.10$

Explanation of the example:
$87 + \frac{5}{10} + \frac{60}{100}$
$= 87 + 0.5 + 0.60$
$= 87 + 1.10$ (since $0.5 + 0.6 = 1.1$)
$= 88.10$ ✓

Answer: $88.10$ (as given in the example).

Compute:

$12$ tens $= 120$

$12$ tenths $= \frac{12}{10} = 1.2$

$$120 + 1.2 = 121.2$$

Answer: $121.2$

Compute:

$10$ tens $= 100$

$10$ ones $= 10$

$10$ tenths $= \frac{10}{10} = 1$

$10$ hundredths $= \frac{10}{100} = 0.1$

$$100 + 10 + 1 + 0.1 = 111.1$$

Answer: $111.1$

Compute:

$25$ tens $= 250$

$25$ ones $= 25$

$25$ tenths $= \frac{25}{10} = 2.5$

$25$ hundredths $= \frac{25}{100} = 0.25$

$$250 + 25 + 2.5 + 0.25 = 277.75$$

Answer: $277.75$

Note: The exact box arrangement is not visible from the image. However, the general approach is:

We need to use digits $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ (each at most once) to fill in an addition problem whose sum is as close to $10.5$ as possible.

Strategy:
- The sum should be close to $10.5$.
- Try to make two numbers that add to approximately $10.5$.
- For example: $9.8 + 0.7 = 10.5$ (uses digits $9, 8, 0, 7$ — all different ✓)
- Or: $7.4 + 3.1 = 10.5$ (uses $7, 4, 3, 1$ ✓)
- Or: $6.3 + 4.2 = 10.5$ (uses $6, 3, 4, 2$ ✓)

A possible answer (depending on the box structure): $\mathbf{6.3 + 4.2 = 10.5}$ (exact), using digits $6, 3, 4, 2$.

Students should apply this strategy to the actual box layout in their textbook to get the exact answer.