Chapter 4: Another Peek Beyond The Point

To write a fraction whose denominator is $$10, 100, 1000, \ldots$$ as a decimal, we place the decimal point so that the numerator has as many digits after the decimal point as there are zeroes in the denominator.

$$\frac{4}{100}$$: denominator has $$2$$ zeroes, so we need $$2$$ digits after the point $$\Rightarrow 0.04$$.

$$\frac{67}{1000}$$: denominator has $$3$$ zeroes, so we need $$3$$ digits after the point $$\Rightarrow 0.067$$.

$$\frac{457}{100} = \frac{400 + 57}{100} = 4 + \frac{57}{100} = 4.57$$.

$$\frac{71}{100} = 0.71$$.

$$\frac{43}{100} = 0.43$$.

$$\frac{9}{100} = 0.09$$.

We know that $$1\text{ kg} = 1000\text{ g}$$, so $$1\text{ g} = \frac{1}{1000}\text{ kg}$$. To convert grams into kilograms, we divide by $$1000$$.

Cinnamon: $$50\text{ g} = \frac{50}{1000}\text{ kg} = \frac{5}{100}\text{ kg} = 0.05\text{ kg}$$.

Cumin seeds: $$100\text{ g} = \frac{100}{1000}\text{ kg} = \frac{1}{10}\text{ kg} = 0.1\text{ kg}$$.

Cardamom: $$25\text{ g} = \frac{25}{1000}\text{ kg} = 0.025\text{ kg}$$.

Pepper: $$250\text{ g} = \frac{250}{1000}\text{ kg} = \frac{25}{100}\text{ kg} = 0.25\text{ kg}$$.

We split the numerator according to its place values and then group the digits so that each group has a denominator of $$10$$, $$100$$, $$1000$$, $$\ldots$$

Row 2: $$\frac{847}{10000} = \frac{800 + 40 + 7}{10000} = \frac{8}{100} + \frac{4}{1000} + \frac{7}{10000} = 0.08 + 0.004 + 0.0007 = 0.0847$$.

Row 3: $$\frac{173}{100} = \frac{100 + 70 + 3}{100} = 1 + \frac{7}{10} + \frac{3}{100} = 1 + 0.7 + 0.03 = 1.73$$.

Row 4: $$\frac{23}{1000} = \frac{20 + 3}{1000} = \frac{2}{100} + \frac{3}{1000} = 0.02 + 0.003 = 0.023$$.

Look at what happens in the examples we have already seen: $$\frac{50}{1000} = 0.050$$ (decimal moves $$3$$ places left), $$\frac{457}{100} = 4.57$$ (decimal moves $$2$$ places left), $$\frac{3}{10} = 0.3$$ (decimal moves $$1$$ place left).

Rule: To divide a number by $$10, 100, 1000, \ldots$$ (a $$1$$ followed by some zeroes), just shift the decimal point of the number to the left by as many places as the number of zeroes in the divisor. If there are not enough digits on the left, insert zeroes.

Applying the rule:

$$\frac{123}{10} = 12.3$$ (shift $$1$$ place left).

$$\frac{24}{100} = 0.24$$ (shift $$2$$ places left).

$$\frac{678}{1000} = 0.678$$ (shift $$3$$ places left).

The cost of $$5$$ pens $$= 5 \times ₹9.5$$.

Write $$9.5$$ as a sum of its whole and fractional part: $$9.5 = 9 + \frac{5}{10}$$.

So $$5 \times 9.5 = 5 \times \left(9 + \frac{5}{10}\right) = 5 \times 9 + 5 \times \frac{5}{10} = 45 + \frac{25}{10} = 45 + 2.5 = 47.5$$.

Hence Arshad should pay $$₹47.5$$ (that is, $$47$$ rupees and $$50$$ paisa).

Distance $$=$$ (km per litre) $$\times$$ (litres) $$= 12.5 \times 7.5$$ km.

Write each decimal as a fraction over $$10$$: $$12.5 = \frac{125}{10}$$ and $$7.5 = \frac{75}{10}$$.

$$12.5 \times 7.5 = \frac{125}{10} \times \frac{75}{10} = \frac{125 \times 75}{100} = \frac{9375}{100} = 93.75$$.

So the car covers $$93.75$$ km.

Each day Ajay makes a round trip: he walks to school ($$827$$ m) and walks back home ($$827$$ m), so one day he walks $$2 \times 827 = 1654$$ m.

He does this for $$6$$ days a week, so the total distance in a week is $$6 \times 1654 = 9924$$ m.

Convert to kilometres using $$1000\text{ m} = 1\text{ km}$$:

$$9924\text{ m} = \frac{9924}{1000}\text{ km} = 9.924\text{ km}$$.

Area of a rectangle $$=$$ length $$\times$$ breadth $$= 13.3 \times 5.7$$ sq cm.

Write each decimal as a fraction over $$10$$: $$13.3 = \frac{133}{10}$$ and $$5.7 = \frac{57}{10}$$.

$$13.3 \times 5.7 = \frac{133}{10} \times \frac{57}{10} = \frac{133 \times 57}{100}$$.

Compute $$133 \times 57$$: $$133 \times 57 = 133 \times 50 + 133 \times 7 = 6650 + 931 = 7581$$.

So $$13.3 \times 5.7 = \frac{7581}{100} = 75.81$$.

Hence the area is $$75.81\text{ cm}^2$$.

Step 1 — Multiply ignoring the decimal points. Removing the decimals, we get whole numbers $$58$$ and $$124$$.

$$58 \times 124 = 58 \times 100 + 58 \times 24 = 5800 + 1392 = 7192$$.

Step 2 — Count the decimal digits. $$5.8$$ has $$1$$ digit after the decimal point, and $$1.24$$ has $$2$$ digits after the decimal point. Together there are $$1 + 2 = 3$$ decimal digits.

Step 3 — Place the decimal point. Put the decimal point so that the product has $$3$$ digits after the point:

$$5.8 \times 1.24 = 7.192$$.

Check with fractions: $$5.8 \times 1.24 = \frac{58}{10} \times \frac{124}{100} = \frac{7192}{1000} = 7.192$$. $$\checkmark$$

Yes. Even though each factor has digits after the decimal point, their product can turn out to be a whole number.

Example: $$0.5 \times 2.0 = \frac{5}{10} \times \frac{20}{10} = \frac{100}{100} = 1$$.

Another example: $$0.25 \times 4.0 = \frac{25}{100} \times \frac{40}{10} = \frac{1000}{1000} = 1$$.

So the product of two decimals can indeed be a natural number.

Yes. If the natural number is a multiple of the denominator hidden in the decimal, the product will be a whole number.

Example: $$0.5 \times 2 = \frac{5}{10} \times 2 = \frac{10}{10} = 1$$.

Example: $$0.25 \times 4 = \frac{25}{100} \times 4 = \frac{100}{100} = 1$$.

Example: $$1.5 \times 4 = \frac{15}{10} \times 4 = \frac{60}{10} = 6$$.

So the product of a decimal and a natural number can be a natural number.

Count the decimal digits in the factors: $$5.96$$ has $$2$$ digits after the point, and $$24.8$$ has $$1$$ digit after the point, giving a total of $$2 + 1 = 3$$ decimal digits in the product.

Take the whole-number product $$147808$$ and place a decimal point so that there are $$3$$ digits after it:

$$5.96 \times 24.8 = 147.808$$.

Check with fractions: $$5.96 \times 24.8 = \frac{596}{100} \times \frac{248}{10} = \frac{596 \times 248}{1000} = \frac{147808}{1000} = 147.808$$. $$\checkmark$$

Rule for multiplying two decimals:

1. Ignore the decimal points and multiply the two numbers as whole numbers.
2. Count the total number of digits that appear after the decimal points in the two original numbers — call this total $$n$$.
3. In the whole-number product, place the decimal point so that exactly $$n$$ digits lie to the right of it. If needed, put extra zeroes on the left.

Illustration: For $$5.8 \times 1.24$$, ignore the points: $$58 \times 124 = 7192$$. Total decimal digits $$= 1 + 2 = 3$$. So $$5.8 \times 1.24 = 7.192$$.

Recall that multiplying by a number bigger than $$1$$ makes something larger, while multiplying by a number smaller than $$1$$ makes it smaller.

Product greater than both numbers: this happens when both factors are greater than $$1$$.
Example: $$1.2 \times 1.5 = 1.8$$, and $$1.8 > 1.5 > 1.2$$.

Product less than both numbers: this happens when both factors are less than $$1$$.
Example: $$0.6 \times 0.4 = 0.24$$, and $$0.24 < 0.4 < 0.6$$.

In-between case: if one factor is greater than $$1$$ and the other is less than $$1$$, the product lies between the two factors.
Example: $$0.5 \times 2.4 = 1.2$$, and $$0.5 < 1.2 < 2.4$$.

Cloth for $$3$$ shirts $$= 3 \times 1.65$$ m.

Ignore the decimal point: $$3 \times 165 = 495$$.

$$1.65$$ has $$2$$ decimal digits, so the product has $$2$$ decimal digits: $$3 \times 1.65 = 4.95$$.

Hence $$4.95$$ m of cloth is needed.

Cost of $$4$$ notebooks $$= 4 \times ₹15.50$$. Ignoring the point, $$4 \times 1550 = 6200$$, and since $$15.50$$ has $$2$$ decimal digits, $$4 \times 15.50 = ₹62.00$$.

Cost of $$3$$ erasers $$= 3 \times ₹2.75$$. Ignoring the point, $$3 \times 275 = 825$$, and since $$2.75$$ has $$2$$ decimal digits, $$3 \times 2.75 = ₹8.25$$.

Total spent $$= ₹62.00 + ₹8.25 = ₹70.25$$.

Height of $$36$$ coins $$= 36 \times 1.45$$ mm.

Ignoring the decimal point: $$36 \times 145$$.

$$36 \times 145 = 36 \times 100 + 36 \times 45 = 3600 + 1620 = 5220$$.

Since $$1.45$$ has $$2$$ decimal digits, $$36 \times 1.45 = 52.20$$ mm.

Convert to centimetres: $$1\text{ cm} = 10\text{ mm}$$, so

$$52.20\text{ mm} = \frac{52.20}{10}\text{ cm} = 5.22\text{ cm}$$.

Price of $$2.250$$ kg $$= 2.250 \times 56.50$$.

Ignoring decimal points: $$2250 \times 5650$$.

$$2250 \times 5650 = 2250 \times 5000 + 2250 \times 650 = 11250000 + 1462500 = 12712500$$.

Total decimal digits $$= 3 + 2 = 5$$, so $$2.250 \times 56.50 = 127.12500 = ₹127.125$$.

Can we drop the trailing zeros? Yes. $$56.50 = 56.5$$ and $$2.250 = 2.25$$ because trailing zeroes after the decimal point do not change the value of a decimal: $$\frac{5650}{100} = \frac{565}{10}$$ and $$\frac{2250}{1000} = \frac{225}{100}$$.

Check: $$2.25 \times 56.5$$. Ignoring decimals: $$225 \times 565 = 225 \times 500 + 225 \times 65 = 112500 + 14625 = 127125$$. Total decimal digits $$= 2 + 1 = 3$$, so the product is $$127.125$$.

So we get exactly the same answer, $$₹127.125$$, because a trailing zero after the decimal point does not change the number.

Profit on $$1$$ notebook $$=$$ selling price $$-$$ cost price $$= 30 - 23.6 = 6.4$$ rupees.

Profit on $$50$$ notebooks $$= 50 \times 6.4$$. Ignoring the decimal: $$50 \times 64 = 3200$$. Since $$6.4$$ has $$1$$ decimal digit, $$50 \times 6.4 = 320.0 = 320$$.

So Dwarakanath's weekly profit is $$₹320$$.

Rule: Multiplying a decimal by $$10, 100, 1000, \ldots$$ shifts the decimal point to the right by as many places as the number of zeroes in the multiplier. Insert zeros on the right if needed.

$$5.7$$: $$\times 10 = 57$$, $$\times 100 = 570$$, $$\times 1000 = 5700$$.

$$23.02$$: $$\times 10 = 230.2$$, $$\times 100 = 2302$$, $$\times 1000 = 23020$$.

$$0.92$$: $$\times 10 = 9.2$$, $$\times 100 = 92$$, $$\times 1000 = 920$$.

$$0.306$$: $$\times 10 = 3.06$$, $$\times 100 = 30.6$$, $$\times 1000 = 306$$.

$$24.67$$: $$\times 10 = 246.7$$, $$\times 100 = 2467$$, $$\times 1000 = 24670$$.

Length of each piece $$= \frac{3.9}{10}$$ m.

Write $$3.9 = \frac{39}{10}$$, so $$\frac{3.9}{10} = \frac{39}{10 \times 10} = \frac{39}{100} = 0.39$$.

Equivalently, dividing by $$10$$ shifts the decimal point one place to the left: $$3.9 \to 0.39$$.

Hence each piece is $$0.39$$ m long.

The total length $$29$$ m is to be shared equally between $$2$$ people, so each gets $$\frac{29}{2}$$ m.

Divide: $$29 \div 2 = 14$$ remainder $$1$$. The remainder $$1$$ m $$= 10$$ tenths, and $$10 \div 2 = 5$$ tenths $$= 0.5$$ m.

So each gets $$14 + 0.5 = 14.5$$ m.

Use long division by place value.

Thousands: $$1 \div 4 = 0$$ remainder $$1$$. Bring down: $$13$$ hundreds.

Hundreds: $$13 \div 4 = 3$$ remainder $$1$$. Bring down: $$12$$ tens.

Tens: $$12 \div 4 = 3$$ remainder $$0$$. Bring down: $$4$$ units.

Units: $$4 \div 4 = 1$$ remainder $$0$$.

Reading the quotient digits: $$0331$$, i.e. $$331$$.

Check: $$4 \times 331 = 1324$$. $$\checkmark$$

Use long division and, when the whole-number division ends with a remainder, continue into the decimal places.

Hundreds: $$2 \div 8 = 0$$ remainder $$2$$; bring down $$23$$ tens.

Tens: $$23 \div 8 = 2$$ remainder $$7$$; bring down $$77$$ units.

Units: $$77 \div 8 = 9$$ remainder $$5$$.

So $$237 = 8 \times 29 + 5$$, i.e. quotient $$29$$ remainder $$5$$.

Continue into decimals: $$5$$ units $$= 50$$ tenths. $$50 \div 8 = 6$$ remainder $$2$$ (tenth digit $$6$$).

$$2$$ tenths $$= 20$$ hundredths. $$20 \div 8 = 2$$ remainder $$4$$ (hundredth digit $$2$$).

$$4$$ hundredths $$= 40$$ thousandths. $$40 \div 8 = 5$$ remainder $$0$$ (thousandth digit $$5$$).

So $$237 \div 8 = 29.625$$.

Check: $$8 \times 29.625 = 237.000$$. $$\checkmark$$

Weight per bag $$= \frac{9.5}{4}$$ kg.

Write $$9.5 = \frac{95}{10}$$, so $$\frac{9.5}{4} = \frac{95}{40}$$. Divide $$95$$ by $$4$$: $$95 \div 4 = 23$$ remainder $$3$$; the tenths give $$30 \div 4 = 7$$ remainder $$2$$; hundredths give $$20 \div 4 = 5$$ remainder $$0$$. So $$95 \div 4 = 23.75$$.

Hence $$\frac{95}{40} = \frac{23.75}{10} = 2.375$$.

Alternatively, long division of $$9.5$$ by $$4$$: $$9 \div 4 = 2$$ rem $$1$$; $$15\text{ tenths} \div 4 = 3\text{ tenths}$$ rem $$3$$; $$30\text{ hundredths} \div 4 = 7$$ rem $$2$$; $$20\text{ thousandths} \div 4 = 5$$ rem $$0$$. Quotient $$= 2.375$$.

So each bag weighs $$2.375$$ kg.

Write $$0.06 = \frac{6}{100}$$, so $$0.06 \div 5 = \frac{6}{100 \times 5} = \frac{6}{500}$$.

To convert the denominator into $$1000$$, multiply numerator and denominator by $$2$$: $$\frac{6 \times 2}{500 \times 2} = \frac{12}{1000} = 0.012$$.

Alternatively, long division: $$6\text{ hundredths} \div 5 = 1\text{ hundredth}$$ rem $$1\text{ hundredth} = 10\text{ thousandths}$$. $$10\text{ thousandths} \div 5 = 2$$ rem $$0$$. Quotient $$= 0.012$$.

This refers back to Example 6: Anuja's ribbon was $$3.9$$ m long. If it is cut into $$100$$ equal pieces, each piece is

$$\frac{3.9}{100}\text{ m} = \frac{39}{10 \times 100}\text{ m} = \frac{39}{1000}\text{ m} = 0.039\text{ m}$$.

Equivalently, dividing by $$100$$ moves the decimal point two places to the left: $$3.9 \to 0.039$$.

Use the conversions $$1\text{ m} = 100\text{ cm} = 1000\text{ mm}$$.

In centimetres: $$0.039\text{ m} = 0.039 \times 100\text{ cm} = 3.9\text{ cm}$$.

In millimetres: $$0.039\text{ m} = 0.039 \times 1000\text{ mm} = 39\text{ mm}$$.

Apply the rule: to divide by $$10, 100, 1000, \ldots$$ shift the decimal point of the dividend to the left by as many places as there are zeros in the divisor.

$$21.1$$: $$\div 10 = 2.11$$; $$\div 100 = 0.211$$; $$\div 1000 = 0.0211$$; $$\div 10000 = 0.00211$$.

$$0.13$$: $$\div 10 = 0.013$$; $$\div 100 = 0.0013$$; $$\div 1000 = 0.00013$$; $$\div 10000 = 0.000013$$.

Row 4 — the $$\div 100$$ cell is $$2.146$$. So the original decimal is $$2.146 \times 100 = 214.6$$. Then $$\div 10 = 21.46$$; $$\div 1000 = 0.2146$$; $$\div 10000 = 0.02146$$.

Row 5 — the $$\div 10000$$ cell is $$0.0058$$. So the original decimal is $$0.0058 \times 10000 = 58$$. Then $$\div 10 = 5.8$$; $$\div 100 = 0.58$$; $$\div 1000 = 0.058$$.

Continuing Example 7: Neenu has $$29$$ m of ribbon. Shared equally among $$4$$ friends, each gets

$$\frac{29}{4}\text{ m}$$.

Long division: $$29 \div 4 = 7$$ rem $$1$$. Tenths: $$10 \div 4 = 2$$ rem $$2$$. Hundredths: $$20 \div 4 = 5$$ rem $$0$$.

So $$\frac{29}{4} = 7.25$$. Each friend gets $$7.25$$ m of ribbon.

Average speed $$= \dfrac{\text{distance}}{\text{time}} = \dfrac{126}{2.5}$$ km/hr.

To turn the decimal divisor into a whole number, multiply numerator and denominator by $$10$$:

$$\dfrac{126}{2.5} = \dfrac{126 \times 10}{2.5 \times 10} = \dfrac{1260}{25}$$.

Long division: $$126 \div 25 = 5$$ rem $$1$$; bring down $$0$$: $$10 \div 25 = 0$$ rem $$10$$; then $$100\text{ tenths} \div 25 = 4$$ rem $$0$$.

So $$\dfrac{1260}{25} = 50.4$$.

Ravi's average speed was $$50.4$$ km/hr.

To divide by a decimal, first turn the divisor into a whole number by multiplying numerator and denominator by the same power of $$10$$. Here $$1.3$$ has $$1$$ decimal digit, so multiply top and bottom by $$10$$:

$$\dfrac{4.68}{1.3} = \dfrac{4.68 \times 10}{1.3 \times 10} = \dfrac{46.8}{13}$$.

Now divide by $$13$$: $$46 \div 13 = 3$$ rem $$7$$; $$78\text{ tenths} \div 13 = 6$$ rem $$0$$. Quotient $$= 3.6$$.

Check: $$1.3 \times 3.6 = 4.68$$. $$\checkmark$$

Long division of $$10$$ by $$3$$:

$$10 \div 3 = 3$$ rem $$1$$. Bring in a decimal point and a zero: $$10\text{ tenths} \div 3 = 3$$ rem $$1$$. Next: $$10\text{ hundredths} \div 3 = 3$$ rem $$1$$. And so on — the remainder is always $$1$$, and we always bring down another $$0$$.

The digit $$3$$ keeps repeating forever, so

$$10 \div 3 = 3.333\ldots = 3.\overline{3}$$.

No, this process does not end. Since the same remainder $$1$$ shows up over and over, the same digit $$3$$ will keep appearing in the quotient.

$$10 \div 9$$: $$10 \div 9 = 1$$ rem $$1$$. Tenths: $$10 \div 9 = 1$$ rem $$1$$. Every step gives the same remainder $$1$$, so the digit $$1$$ repeats: $$10 \div 9 = 1.111\ldots = 1.\overline{1}$$.

$$100 \div 11$$: $$100 \div 11 = 9$$ rem $$1$$. Tenths: $$10 \div 11 = 0$$ rem $$10$$. Hundredths: $$100 \div 11 = 9$$ rem $$1$$. Thousandths: $$10 \div 11 = 0$$ rem $$10$$… The digits $$09$$ keep repeating, so $$100 \div 11 = 9.0909\ldots = 9.\overline{09}$$.

$$1 \div 7$$: Successive remainders are $$1, 3, 2, 6, 4, 5, 1, 3, 2, \ldots$$ (they cycle), and the corresponding quotient digits are $$1, 4, 2, 8, 5, 7$$ which then repeat. So

$$1 \div 7 = 0.142857142857\ldots = 0.\overline{142857}$$.

No, this too does not end. The remainders can only take values $$1, 2, 3, 4, 5, 6$$ (never $$0$$), so one of them must eventually repeat, forcing the quotient to be periodic.

Compute the six products:

$$1 \times 142857 = 142857$$.

$$2 \times 142857 = 285714$$.

$$3 \times 142857 = 428571$$.

$$4 \times 142857 = 571428$$.

$$5 \times 142857 = 714285$$.

$$6 \times 142857 = 857142$$.

What do you notice? Each product uses the same six digits — $$1, 4, 2, 8, 5, 7$$ — arranged in the same cyclic order. If you write $$142857$$ around a circle, each product is what you read starting from a different digit. That is why $$142857$$ is called a cyclic number.

Now multiply by $$7$$: $$7 \times 142857 = 999999$$.

So $$7 \times 142857 = 999999$$, a string of six nines — remarkable! (This is consistent with $$1/7 = 0.\overline{142857}$$: multiplying by $$7$$ should give $$0.\overline{999999} = 1$$, and indeed $$999999 / 1000000 \approx 1$$.)

Perform long division of $$1$$ by $$17$$. The successive remainders are $$1, 10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12, 1, \ldots$$ — cycling every $$16$$ steps. The corresponding quotient digits give

$$1 \div 17 = 0.\overline{0588235294117647}$$.

So the cyclic number produced is $$0588235294117647$$ (a $$16$$-digit block).

Check that it is cyclic: multiplying $$0588235294117647$$ by any integer from $$1$$ to $$16$$ gives one of the same $$16$$-digit cyclic rearrangements of these digits. For example, $$2 \times 588235294117647 = 1176470588235294$$ (a cyclic shift of the original).

And $$17 \times 588235294117647 = 9999999999999999$$, mirroring the $$142857 \times 7 = 999999$$ observation for $$1/7$$.

No — it is not always true. It depends on whether the divisor is bigger or smaller than $$1$$.

Some examples with dividend $$= 4$$:

Rule:

• If the divisor is $$> 1$$, the quotient is less than the dividend.
• If the divisor is $$= 1$$, the quotient is equal to the dividend.
• If the divisor is between $$0$$ and $$1$$, the quotient is greater than the dividend. The smaller the divisor, the bigger the quotient.

So a decimal divisor gives a quotient bigger than the dividend only when that decimal is less than $$1$$.

Length of each piece $$= \dfrac{4}{5}$$ m.

Convert the denominator to $$10$$: $$\dfrac{4}{5} = \dfrac{4 \times 2}{10} = \dfrac{8}{10} = 0.8$$.

Hence each piece is $$0.8$$ m $$= 80$$ cm.

A regular polygon has all sides equal, so side length $$= \dfrac{\text{perimeter}}{\text{number of sides}} = \dfrac{208.8}{12}$$ cm.

Long division of $$208.8$$ by $$12$$:

$$20 \div 12 = 1$$ rem $$8$$; $$88 \div 12 = 7$$ rem $$4$$; $$48\text{ tenths} \div 12 = 4$$ rem $$0$$. Quotient $$= 17.4$$.

So each side is $$17.4$$ cm long.

Each friend gets $$\dfrac{3}{8}$$ litres.

Convert to a decimal by making the denominator $$1000$$: $$\dfrac{3}{8} = \dfrac{3 \times 125}{1000} = \dfrac{375}{1000} = 0.375$$ litres.

Convert to millilitres using $$1\text{ l} = 1000\text{ ml}$$: $$0.375$$ l $$= 0.375 \times 1000\text{ ml} = 375$$ ml.

Distance per litre $$= \dfrac{234.45}{12.6}$$ km.

Multiply top and bottom by $$10$$: $$\dfrac{234.45}{12.6} = \dfrac{2344.5}{126}$$.

Long division of $$2344.5$$ by $$126$$:

$$234 \div 126 = 1$$ rem $$108$$; bring down $$4$$: $$1084 \div 126 = 8$$ (since $$8 \times 126 = 1008$$), rem $$76$$; bring down $$5$$ (a tenths digit): $$765 \div 126 = 6$$ (since $$6 \times 126 = 756$$), rem $$9$$; bring down $$0$$ (a hundredths digit): $$90 \div 126 = 0$$ rem $$90$$; bring down $$0$$ (thousandths): $$900 \div 126 = 7$$ (since $$7 \times 126 = 882$$), rem $$18$$.

So $$\dfrac{2344.5}{126} \approx 18.607$$ km/l.

Rounded sensibly: about $$18.6$$ km per litre.

Flour per student: $$\dfrac{13.5}{15}$$ kg.

Multiply top and bottom by $$10$$: $$\dfrac{13.5}{15} = \dfrac{135}{150} = \dfrac{9}{10} = 0.9$$ kg.

So each student receives $$0.9$$ kg $$= 900$$ g of flour.

Missing entries in the table.

Left column: $$\dfrac{1}{2^5} = \dfrac{1}{32}$$. Multiply top and bottom by $$5^5 = 3125$$ to get a power-of-ten denominator: $$\dfrac{3125}{100000} = 0.03125$$.

Right column: $$\dfrac{1}{5^5} = \dfrac{1}{3125}$$. Multiply top and bottom by $$2^5 = 32$$: $$\dfrac{32}{100000} = 0.00032$$.

Pattern. Each row on the left, $$\dfrac{1}{2^n}$$, and the corresponding row on the right, $$\dfrac{1}{5^n}$$, have decimal representations with exactly $$n$$ digits after the decimal point, and the two digit-strings multiply to give $$1$$ followed by $$n$$ zeroes.

Row 1: $$0.5 \times 0.2 = 0.1$$. Row 2: $$0.25 \times 0.04 = 0.01$$. Row 3: $$0.125 \times 0.008 = 0.001$$. Row 4: $$0.0625 \times 0.0016 = 0.0001$$. Row 5: $$0.03125 \times 0.00032 = 0.00001$$.

Why 2 and 5 are related this way. Because $$2 \times 5 = 10$$! So $$2^n \times 5^n = 10^n$$, i.e. $$\dfrac{1}{2^n} \times \dfrac{1}{5^n} = \dfrac{1}{10^n}$$. Multiplying $$\dfrac{1}{2^n}$$ by $$5^n$$ or $$\dfrac{1}{5^n}$$ by $$2^n$$ turns the denominator into $$10^n$$, which is what makes these fractions terminate as decimals.

The extra day is added to February. In a normal year February has $$28$$ days, but in a leap year (every fourth year) it has $$29$$ days. That extra $$29$$th day of February is called the leap day.

A normal calendar year has $$365$$ days. If, in every $$4$$th year, we add one extra day, then in $$100$$ years we add $$1$$ extra day for each of the $$25$$ years divisible by $$4$$ ($$4, 8, 12, \ldots, 100$$).

Total number of calendar days in $$100$$ years:

$$100 \times 365 + 25 = 36500 + 25 = 36525\text{ days}$$.

So the expression is $$100 \times 365 + 25 = 36525$$.

The multiples of $$4$$ between $$1$$ and $$100$$ are $$4, 8, 12, \ldots, 100$$.

Their count is $$\dfrac{100}{4} = 25$$.

Yes. Splitting the $$100$$ years into leap years and ordinary years gives many equivalent expressions. There are $$25$$ leap years (of $$366$$ days each) and $$75$$ ordinary years (of $$365$$ days each) in $$100$$ years.

Expression 1: $$100 \times 365 + 25 \times 1 = 36500 + 25 = 36525$$ days.

Expression 2: $$25 \times 366 + 75 \times 365 = 9150 + 27375 = 36525$$ days.

Expression 3: Use the average: $$100 \times 365.25 = 36525$$ days. (Each year contributes $$365.25$$ days on average because $$1$$ extra day every $$4$$ years is $$0.25$$ extra days per year.)

All three give the same answer, $$36525$$ days.

With the earlier scheme we had $$100 \times 365 + 25 = 36525$$ days. The adjustment says: in every $$100$$th year, do not add the leap day. So we subtract $$1$$ from the previous total.

Number of days in $$100$$ years $$= 100 \times 365 + 25 - 1 = 36524$$.

Count of leap days in $$10{,}000$$ years. Every $$4$$th year is a leap year, unless the year is a multiple of $$100$$; but centuries that are multiples of $$400$$ remain leap.

Multiples of $$4$$ in $$10{,}000$$ years: $$\dfrac{10000}{4} = 2500$$.

Multiples of $$100$$ in $$10{,}000$$ years: $$\dfrac{10000}{100} = 100$$. Subtract these from the leap years.

Multiples of $$400$$ in $$10{,}000$$ years: $$\dfrac{10000}{400} = 25$$. Add these back.

Number of leap days $$= 2500 - 100 + 25 = 2425$$.

Calendar days in $$10{,}000$$ years:

$$10000 \times 365 + 2425 = 3650000 + 2425 = 3652425\text{ days}$$.

Actual days for $$10{,}000$$ revolutions: the Earth takes $$365.2422$$ days per revolution, so

$$10000 \times 365.2422 = 3652422\text{ days}$$.

Difference: $$3652425 - 3652422 = 3$$ days. Over $$10{,}000$$ years, the calendar drifts ahead of the seasons by about $$3$$ days.

How to fix it: since the calendar gains about $$3$$ extra days every $$10{,}000$$ years, we could drop $$3$$ leap days every $$10{,}000$$ years — for example, treat $$3$$ specific years that would otherwise be leap years as ordinary $$365$$-day years. Equivalently, skip the leap day in years divisible by, say, $$3200$$ (approximately $$10000/3$$). Small periodic corrections like this keep the calendar aligned with the actual solar year.

People figured it out through careful and patient astronomical observation over centuries.

The most direct way: observe the position of the Sun at noon (or the length of the shadow of a vertical pole) every day for many years. The Sun reaches the same highest/lowest position (a solstice) exactly once per year. By counting the number of days between two solstices — and averaging over many years — one gets a very precise value for the year's length.

Similarly, observing the equinox (when day and night are equal), or the time when a chosen star rises exactly at sunset, gives the same length. Ancient Indian astronomers (Aryabhata, Bhaskara), Babylonians, Egyptians, Greeks and later European astronomers all measured this number, refining it century by century. Modern astronomical instruments (and telescopes, and now satellites) confirm the value is very close to $$365.2422$$ days.

Traditional Indian calendars are luni-solar: they use the Moon to define months and the Sun (and stars) to keep the year aligned with the seasons.

Solar year (Saura māna). The year begins when the Sun enters the sidereal constellation of Mesha (Aries); each solar month begins when the Sun enters the next zodiac sign (saṅkrānti). Because the year is tied to the Sun's actual position among the stars, seasons stay put.

Astronomers such as Āryabhaṭa (5th c. CE), Varāhamihira, Brahmagupta and Bhāskara-II observed the Sun and stars over decades and computed the year length very accurately — Āryabhaṭa's value differs from the modern one only in the third decimal place.

Lunar months (Chāndra māna). Each month runs from one new moon (or full moon) to the next — about $$29.5$$ days. Twelve lunar months make about $$354$$ days, so a lunar year falls short of the solar year by about $$11$$ days.

Adhika-māsa (extra month). To keep the lunar months in step with the seasons, an extra lunar month is inserted about every $$32$$ or $$33$$ months (roughly once in three years). This is the intercalation that keeps festivals like Diwali or Makar Sankranti in the same season year after year.

Kshaya-māsa. Occasionally, when two solar transits fall within a single lunar month, a lunar month is dropped, which finely adjusts the calendar.

Nakṣatras (star groups). The path of the Moon is divided into $$27$$ nakṣatras. Each day the Moon moves through one nakṣatra, so the position of the Moon among the stars is tracked directly, keeping the calendar tied to the actual sky.

Together, these ideas — a sidereal solar year, lunar months, adhika-māsa/kshaya-māsa adjustments, and observation of nakṣatras — let traditional Indian calendars stay accurately aligned with both the seasons and the stars.

Compare the price per gram for the two packets.

Chikki: $$\dfrac{70.5}{210}$$ rupees per gram $$= \dfrac{7050}{21000} = \dfrac{7050 \div 210}{21000 \div 210} = \dfrac{33.57\ldots}{100}$$. Working it out step by step: $$70.5 \div 210 = 0.3357\ldots$$, so about $$₹0.336$$ per gram, i.e. about $$33.6$$ paise per gram.

Chips: $$\dfrac{33.25}{110}$$ rupees per gram $$= 0.3022\ldots$$, so about $$₹0.302$$ per gram, i.e. about $$30.2$$ paise per gram.

Since $$30.2 < 33.6$$, potato chips cost less per gram.

Quicker check via a common weight ($$770$$ g, the LCM). Price of $$770$$ g of chikki $$= \dfrac{770}{210} \times 70.5 = 3.667 \times 70.5 \approx ₹258.5$$. Price of $$770$$ g of chips $$= \dfrac{770}{110} \times 33.25 = 7 \times 33.25 = ₹232.75$$. Chips are cheaper for the same weight.

So the potato chips are cheaper per gram.

(a) The distance from $$3.1$$ to $$3.2$$ is $$0.1$$. Dividing this into $$10$$ equal parts, each part is $$\dfrac{0.1}{10} = 0.01$$.

So the point $$7$$ parts to the right of $$3.1$$ is $$3.1 + 7 \times 0.01 = 3.1 + 0.07 = 3.17$$.

(b) The distance from $$2.15$$ to $$2.17$$ is $$0.02$$. Dividing into $$10$$ equal parts gives $$\dfrac{0.02}{10} = 0.002$$ per part.

The point $$6$$ parts to the right of $$2.15$$ is $$2.15 + 6 \times 0.002 = 2.15 + 0.012 = 2.162$$.

Cost price of $$3$$ kg bananas $$= 3 \times 30 = ₹90$$.

Selling price of $$35$$ bananas $$= 35 \times 5 = ₹175$$.

Profit $$=$$ selling price $$-$$ cost price $$= 175 - 90 = ₹85$$.

Each book takes up $$2.5$$ cm of shelf. Number of books that fit in $$160$$ cm:

$$\dfrac{160}{2.5} = \dfrac{1600}{25} = 64$$.

So $$64$$ books fit on the shelf. The teacher wanted to place $$80$$ books, but only $$64$$ actually fit.

Space used by these $$64$$ books $$= 64 \times 2.5 = 160$$ cm. Space left $$= 160 - 160 = 0$$ cm.

So $$64$$ books can be placed and no space is left. The remaining $$80 - 64 = 16$$ books cannot be accommodated.

Use the given unit conversions and multiply or divide by $$10, 100$$ or $$1000$$ as required.

$$5.5$$ km $$\to$$ m: $$1$$ km $$= 1000$$ m, so $$5.5 \times 1000 = 5500$$ m.

$$35$$ cm $$\to$$ m: $$1$$ m $$= 100$$ cm, so $$35 \div 100 = 0.35$$ m.

$$14.5$$ cm $$\to$$ mm: $$1$$ cm $$= 10$$ mm, so $$14.5 \times 10 = 145$$ mm.

$$68$$ g $$\to$$ kg: $$1$$ kg $$= 1000$$ g, so $$68 \div 1000 = 0.068$$ kg.

$$9.02$$ m $$\to$$ mm: $$1$$ m $$= 1000$$ mm, so $$9.02 \times 1000 = 9020$$ mm.

$$125.5$$ ml $$\to$$ l: $$1$$ l $$= 1000$$ ml, so $$125.5 \div 1000 = 0.1255$$ l.

Convert each mixed fraction into a decimal:

$$6\tfrac{1}{4} = 6 + 0.25 = 6.25$$,   $$2\tfrac{1}{2} = 2 + 0.5 = 2.5$$.

$$60\tfrac{1}{4} = 60 + 0.25 = 60.25$$,   $$3\tfrac{1}{2} = 3 + 0.5 = 3.5$$.

First division: $$\dfrac{6.25}{2.5} = \dfrac{62.5}{25} = 2.5$$.

(Long division: $$62 \div 25 = 2$$ rem $$12$$; $$125\text{ tenths} \div 25 = 5$$ rem $$0$$.)

Second division: $$\dfrac{60.25}{3.5} = \dfrac{602.5}{35}$$.

Long division: $$60 \div 35 = 1$$ rem $$25$$; $$252 \div 35 = 7$$ rem $$7$$; $$75\text{ tenths} \div 35 = 2$$ rem $$5$$; $$50\text{ hundredths} \div 35 = 1$$ rem $$15$$; $$150\text{ thousandths} \div 35 = 4$$ rem $$10$$... so the answer is $$17.214\ldots$$ (a non-terminating decimal).

Actually $$\dfrac{60.25}{3.5} = \dfrac{6025}{350} = \dfrac{241}{14}$$. Since $$14 \times 17 = 238$$, we get $$\dfrac{241}{14} = 17\dfrac{3}{14}$$; as a decimal $$3 \div 14 = 0.2142\overline{857142}$$, so

$$\dfrac{60.25}{3.5} = 17.21428\overline{571428}\ldots \approx 17.214$$.

Each cell is $$a \div b$$. Since $$1517 \div 37 = 41$$, every other cell can be obtained by multiplying (or dividing) $$41$$ by an appropriate power of $$10$$: multiplying the dividend by $$10$$ multiplies the quotient by $$10$$, while multiplying the divisor by $$10$$ divides the quotient by $$10$$.

Row $$b = 37$$: $$\dfrac{1517}{37}=41; \ \dfrac{151.7}{37}=4.1; \ \dfrac{15.17}{37}=0.41; \ \dfrac{1.517}{37}=0.041; \ \dfrac{15170}{37}=410$$.

Row $$b = 3.7$$: divisor is $$10$$ times smaller, so quotients are $$10$$ times larger: $$410, 41, 4.1, 0.41, 4100$$.

Row $$b = 0.37$$: divisor $$100$$ times smaller than row $$1$$: $$4100, 410, 41, 4.1, 41000$$.

Row $$b = 0.037$$: divisor $$1000$$ times smaller than row $$1$$: $$41000, 4100, 410, 41, 410000$$.

Row $$b = 370$$: divisor $$10$$ times larger than row $$1$$, so quotients $$10$$ times smaller: $$4.1, 0.41, 0.041, 0.0041, 41$$.