Actual readings depend on the printed scale, but following the pattern shown for the worked example (\(2\tfrac{7}{10}\) cm), typical readings for the three screws in the right‑hand column are:

Method: To read the length, see where the tip of the screw lines up on the scale. Since a whole unit (1 cm) is divided into 10 equal parts, count how many of those small parts (tenths) the tip crosses past the last whole‑cm mark.

The scale on which each 1 cm unit is divided into 10 equal smaller parts (millimetre markings) helps measure the screws more accurately, because it lets us read lengths like \(2\tfrac{7}{10}\) cm exactly instead of only estimating “between 2 cm and 3 cm”. Finer sub‑divisions always give more precise, accurate measurements.

Because the two screws looked almost the same length, a whole centimetre unit was too big to show the tiny difference between them. Splitting each unit into 10 equal smaller parts (tenths) let Sonu see and record the small but important difference in their lengths — this is exactly why decimals/fractional units are needed for accurate measurement.

This is a hands‑on activity — measure your own objects; your readings may differ slightly.

Convert all to tenths and compare: (h) \(0.4\), (a) \(0.9\), (b) \(1.7\), (g) \(6.7\), (f) \(7.6\), (e) \(10.5\), (c) \(13.0\), (d) \(13.1\).

\(\tfrac{4}{10}=0.4\); \(4\tfrac{1}{10}=4.1\); \(\tfrac{41}{10}=4.1\); \(41\tfrac{1}{10}=41.1\)

1. \((2+3) + \left(\dfrac{7}{10}+\dfrac{6}{10}\right) = 5 + \dfrac{13}{10}\)
2. \(\dfrac{13}{10} = \dfrac{10}{10}+\dfrac{3}{10} = 1+\dfrac{3}{10}\)
3. \(5 + 1 + \dfrac{3}{10} = 6\dfrac{3}{10}\)

1. Whole numbers: \(2+5+7=14\)
2. Tenths: \(\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}=\dfrac{12}{10}=1\dfrac{2}{10}\)
3. Total \(=14+1\dfrac{2}{10}=15\dfrac{2}{10}\)

Finger length \(= \left(12+\dfrac{4}{10}\right) – \left(6+\dfrac{7}{10}\right)\)

1. \((12-6) + \left(\dfrac{4}{10}-\dfrac{7}{10}\right) = 6 – \dfrac{3}{10}\)
2. \(6-\dfrac{3}{10} = 5+1-\dfrac{3}{10} = 5+\dfrac{10}{10}-\dfrac{3}{10} = 5+\dfrac{7}{10}\)

\(3\tfrac{5}{10}=\dfrac{35}{10}\) and \(2\tfrac{7}{10}=\dfrac{27}{10}\)

\(2\tfrac{4}{10}=\dfrac{24}{10}\); difference \(=\dfrac{24}{10}-\dfrac{9}{10}=\dfrac{15}{10}=1\dfrac{5}{10}\)

Each smaller part \(=\dfrac{1}{100}\) of a unit (a “hundredth”). Since \(10\) hundredths make \(1\) tenth, and there are \(10\) tenths in a unit, there are \(10\times 10=100\) such smaller parts in \(1\) unit.

\(\dfrac{1}{10}=\dfrac{10}{100}\), so 10 one‑hundredths make one‑tenth.

Continuing the pattern of hundredths from 0 to past 2 units:

No — both methods give the same result, \(18\tfrac{2}{100}\). Method 1 adds whole numbers, tenths, and hundredths separately and regroups; Method 2 does the same regrouping using the standard column format. They are just two ways of writing the identical reasoning.

Yes. In both cases we (1) split the number by place value, (2) add matching place values separately, and (3) regroup whenever a sum in one place value reaches 10 or more, carrying \(1\) over to the next higher place value. In \(483+268\), \(11\) ones become \(1\) ten \(+1\) one; in decimals, \(13\) hundredths become \(1\) tenth \(+3\) hundredths — the underlying carrying logic is identical.

1. \(25\tfrac{9}{10}=\tfrac{2590}{100}\); \(6\tfrac{4}{10}\tfrac{7}{100}=\tfrac{647}{100}\)
2. \(\tfrac{2590}{100}-\tfrac{647}{100}=\tfrac{1943}{100}\)
3. \(\tfrac{1943}{100}=19\tfrac{43}{100}=19\tfrac{4}{10}\tfrac{3}{100}\)

\(15\tfrac{3}{10}\tfrac{4}{100}=\tfrac{1534}{100}\); \(2\tfrac{6}{10}\tfrac{8}{100}=\tfrac{268}{100}\)

Just like \(653-268=385\) requires borrowing a ten to subtract in the ones place, subtracting hundredths here needed borrowing a tenth (\(9\to14\) hundredths) — the same regrouping idea used for whole numbers applies to decimals.

Yes, a unit can be split into any number of equal parts (e.g. 4 quarters, 16 sixteenths). But we usually split into 10 because the Indian place value system itself is built on powers of 10 — each place value is exactly 10 times the one to its right (ones, tens, hundreds…). Dividing by 10 lets fractional parts (tenths, hundredths, thousandths…) extend this same place‑value pattern smoothly below one, so all numbers — whole or fractional — can be written and compared using one consistent system.

Yes — the chart can be extended indefinitely in both directions (bigger place values to the left: lakhs, ten‑lakhs…; smaller place values to the right: ten‑thousandths…).

More questions you could ask: “How many hundredths make 1 hundred?” (10,000); “How many thousandths make 1 hundred?” (1,00,000).

No, we cannot simply write \(4\tfrac{2}{10}\) as “42” — it would be impossible to tell whether “42” means \(4\) tens \(+2\) ones (\(=42\)) or \(4\) units \(+2\) tenths (\(=4.2\)). The two quantities are very different, so we need a distinct symbol — the decimal point — to separate the whole‑number part from the fractional part. This is exactly why decimal notation (like \(4.2\)) was introduced.

1. \(234\) hundredths \(=\dfrac{234}{100}=\dfrac{200}{100}+\dfrac{30}{100}+\dfrac{4}{100}=2.34\)
2. \(105\) tenths \(=\dfrac{105}{10}=\dfrac{100}{10}+\dfrac{5}{10}=10.5\)

Since \(1\ cm=10\ mm\), \(1\ mm=\dfrac{1}{10}\) cm.

Since \(1\ m=100\ cm\), \(1\ cm=\dfrac{1}{100}m=0.01\ m\).

(a) \(10\ cm=\dfrac{10}{100}m=0.1\ m\)   (b) \(15\ cm=\dfrac{15}{100}m=0.15\ m\)

Since \(1\ kg=1000\ g\), \(1\ g=\dfrac{1}{1000}kg=0.001\ kg\).

\(5\ g = \dfrac{5}{1000}kg=0.005\ kg\);   \(10\ g=\dfrac{10}{1000}kg=\dfrac{1}{100}kg=0.010\ kg\)

Since \(1\) rupee \(=100\) paise, \(1\) paisa \(=\dfrac{1}{100}\) rupee \(=₹0.01\).

Sonu is right for trailing zeros: \(0.2=0.20=0.200\), because all three represent the same quantity — 2 tenths (0 hundredths and 0 thousandths add nothing). Zara’s caution is correct only for zeros placed before the significant digits after the decimal point (e.g. \(0.2\neq0.02\neq0.002\)), since those do change the place value of the 2.

(a) Zoom \(9\to10\), then \(9.8\to9.9\), then \(9.87\to9.88\); \(9.876\) sits \(0.006\) past \(9.87\).

(b) Zoom \(0\to1\), then \(0.4\to0.5\), then \(0.40\to0.41\); \(0.407\) sits \(0.007\) past \(0.40\).

Both numbers have 6 units and 4 tenths (equal so far). At the hundredths place, \(6.456\) has \(5\) hundredths while \(6.465\) has \(6\) hundredths. Since \(5<6\):

Yes, we can be sure. Once two numbers match in every place value down to some point and then differ at a place, the digit that’s larger there already contributes more than the maximum possible total of every place value after it combined. For example, in the hundredths place, the biggest possible sum of everything from the thousandths place onward is just under \(0.01\) (e.g. \(0.00999\ldots<0.01\)) — smaller than the value of one extra hundredth. So no combination of later digits can ever overturn the difference already created at an earlier (higher) place value.

1. (a) \(1.32 > 1.23\) (tenths: 3 > 2)
2. (b) \(13.800 > 3.81\) (13 units > 3 units)
3. (c) \(1.090 > 1.009\) (tenths: 0 vs 0, hundredths: 9 > 0)

Closest to 1.09: Distances → \(|0.9-1.09|=0.19\); \(|1.1-1.09|=0.01\); \(|1.01-1.09|=0.08\); \(|1.11-1.09|=0.02\). Smallest is \(0.01\).

Closest to 4: \(|3.56-4|=0.44\); \(|3.65-4|=0.35\); \(|3.099-4|=0.901\).

Closest to 1: \(|0.8-1|=0.2\); \(|0.69-1|=0.31\); \(|1.08-1|=0.08\).

The best overall choice is 25.148 — using the digits 2 and 5 as the whole‑number part (exactly 25) makes the number as close to 25 as possible.

Compact form:

Detailed place-value view (with regrouping):

1. Thousandths: \(1-5\) — borrow 1 hundredth (=10 thousandths): \(11-5=6\)
2. Hundredths: \(9-1(\text{borrowed})-4=4\)
3. Tenths: \(6-3=3\)
4. Ones: \(4-7\) — borrow 1 ten: \(14-7=7\)
5. Tens: \(8-1(\text{borrowed})-7=0\)

Each term increases by 0.4.

\(25.936+8.202=34.138\). Whole‑number parts sum \(=25+8=33\). Check: \(33 < 34.138 < 35\;(33+2)\) ✔ — the claim holds.

This works for any two decimal numbers, because each decimal (fractional) part is at least \(0\) and strictly less than \(1\); so the sum of the two fractional parts is at least \(0\) and strictly less than \(2\). Hence the total sum always lies between the sum of whole parts and \(2\) more than that sum.

If two decimals have whole‑number parts \(A\) and \(B\) (with \(A>B\)), their difference always lies strictly between \((A-B)-1\) and \((A-B)+1\). This is because each fractional part lies between \(0\) (inclusive) and \(1\) (exclusive), so subtracting one fractional part from another can shift the result by up to just under \(1\) unit in either direction — never a full unit.

None of these — because hours and minutes are not a decimal (base‑10) system; 1 hour = 60 minutes, not 100. \(0.5\) hours means splitting 1 hour into 10 equal parts and taking 5, i.e. \(\dfrac{5}{10}\times60=30\) minutes.

Since \(1\) over \(=6\) balls (not 10), \(5.5\) overs means \(5\dfrac{5}{6}\) overs \(=5\) overs and \(5\) balls — not a true decimal.

Other examples of “non-decimal” decimal-looking notation: time written as hours.minutes (e.g. “2.30” for 2 hrs 30 min, where 1 hour = 60 min); book/document section numbers (e.g. 3.2 for “Chapter 3, Section 2” — not a fraction at all); shooting/archery scorecards; and rupee amounts occasionally written loosely as “5.50” meaning ₹5 and 50 paise (this one actually IS a true decimal, since 1 rupee = 100 paise).

1. (a) 11.01, 1.011, 1.101, 11.10, 1.01 → 11.10 > 11.01 > 1.101 > 1.011 > 1.01
2. (b) 2.567, 2.675, 2.768, 2.499, 2.698 → 2.768 > 2.698 > 2.675 > 2.567 > 2.499
3. (c) 4.678g, 4.595g, 4.600g, 4.656g, 4.666g → 4.678 > 4.666 > 4.656 > 4.600 > 4.595
4. (d) 33.13m, 33.31m, 33.133m, 33.331m, 33.313m → 33.331 > 33.313 > 33.31 > 33.133 > 33.13

1. (a) The two‑digit whole part closest to 30 achievable from these digits is 40. Arranging the rest as decimals gives 40.168.
2. (b) For the smallest 3‑digit whole‑part number, place smallest usable digits first: hundreds=1, tens=0, ones=4, decimals=6,8 → 104.68.

Not necessarily. The number of digits doesn’t decide the size — the place value of each digit does.

\(0.25+0.3+0.5+0.2+0.05\)

1. \(0.25+0.30=0.55\)
2. \(0.55+0.50=1.05\)
3. \(1.05+0.20=1.25\)
4. \(1.25+0.05=1.30\)

1. Milk in first 3 days \(=3.79+4.2+4.25=12.24\) L
2. Milk in last 3 days \(=25-12.24=12.76\) L

Since \(35.75>34.50\), Tinku has lost weight.

The pattern alternates: add 0.9, then subtract 0.01, repeatedly.

\(5.5\xrightarrow{+0.9}6.4\xrightarrow{-0.01}6.39\xrightarrow{+0.9}7.29\xrightarrow{-0.01}7.28\xrightarrow{+0.9}8.18\xrightarrow{-0.01}8.17\xrightarrow{+0.9}9.07\xrightarrow{-0.01}9.06\)

\(1\ km=1000\ m\), and \(1\ m=1000\ mm\), so \(1\ km=1000\times1000\ mm\).

Fee per passenger \(=45\) paise \(=₹0.45\)

Total \(=₹0.45\times1,00,000 = ₹45,000\)

1. (a) \(\tfrac{10}{1000}=0.01\); \(\tfrac{1}{10}=0.1\) → \(\tfrac{1}{10}\) is greater
2. (b) one-hundredth = 0.01; 90 thousandths = 0.090 → 90 thousandths is greater
3. (c) one-thousandth = 0.001; 90 hundredths = 0.90 → 90 hundredths is greater

One good choice: \(9.476+1.025\)

1. \(9.476+1.025=10.501\)
2. Digits used: 9,4,7,6,1,0,2,5 — all different, each used at most once ✔
3. \(|10.501-10.5|=0.001\) — extremely close