Chapter 4 — Another Peek Beyond the Point.
Complete, step-by-step solutions for every in-text question and every “Figure It Out” exercise — decimal multiplication, decimal division, and the leap-year story.
In-Text Questions
All the “?” (Math Talk / Try This) questions that appear inside the explanations, section by section.
4.1 A Quick Recap of Decimals
Jonali and Pallabi’s game — write each fraction as the equivalent decimal: \(\frac{3}{10},\ \frac{4}{100},\ \frac{67}{1000},\ \frac{457}{100},\ \frac{71}{100},\ \frac{43}{100},\ \frac{9}{100}\)
Solution
Each fraction already has a denominator of 10, 100 or 1000, so we can write the decimal directly by counting digits after the point equal to the number of zeroes in the denominator.
| Fraction | Decimal |
|---|---|
| \(\frac{3}{10}\) | 0.3 |
| \(\frac{4}{100}\) | 0.04 |
| \(\frac{67}{1000}\) | 0.067 |
| \(\frac{457}{100}\) | 4.57 |
| \(\frac{71}{100}\) | 0.71 |
| \(\frac{43}{100}\) | 0.43 |
| \(\frac{9}{100}\) | 0.09 |
Jonali buys 50 g Cinnamon, 100 g Cumin seeds, 25 g Cardamom, 250 g Pepper. Express each in kilograms as a fraction and a decimal.
Solution
Since \(1\text{ kg} = 1000\text{ g}\), every gram value becomes a fraction over 1000.
| Item | Fraction (kg) | Decimal (kg) |
|---|---|---|
| Cinnamon, 50 g | \(\frac{50}{1000}\) | 0.050 kg |
| Cumin seeds, 100 g | \(\frac{100}{1000}\) | 0.100 kg |
| Cardamom, 25 g | \(\frac{25}{1000}\) | 0.025 kg |
| Pepper, 250 g | \(\frac{250}{1000}\) | 0.250 kg |
Write \(\frac{847}{10000}\), \(\frac{173}{100}\) and \(\frac{23}{1000}\) as a sum of fractions and as decimals.
Solution
\(\frac{847}{10000}\): \(\frac{800}{10000}+\frac{40}{10000}+\frac{7}{10000}=\frac{8}{100}+\frac{4}{1000}+\frac{7}{10000}=0.08+0.004+0.0007\)
→ 0.0847
\(\frac{173}{100}\): \(\frac{100}{100}+\frac{70}{100}+\frac{3}{100}=1+0.7+0.03\)
→ 1.73
\(\frac{23}{1000}\): \(\frac{20}{1000}+\frac{3}{1000}=0.02+0.003\)
→ 0.023
Math Talk: Give a simple rule to divide any number by 10, 100, 1000, etc.
Solution
Rule: Move the decimal point of the dividend to the left by as many places as there are zeroes in the divisor (add zeroes in front if needed).
E.g. \(123\div10=12.3\) (1 zero → shift 1 place); \(24\div100=0.24\); \(678\div1000=0.678\); \(12\div1000=0.012\).
4.2 Decimal Multiplication
Math Talk: Can the product of two decimals be a natural number? Can the product of a decimal and a natural number be a natural number?
Solution
Yes to both.
Two decimals → natural number: \(2.5 \times 0.4 = 1.00 = 1\)
Decimal × natural number → natural number: \(0.5 \times 4 = 2.0 = 2\)
This happens whenever the decimal digits in the two numbers “cancel out” so that the product ends in .0.
Math Talk: Given \(596 \times 248 = 147808\), immediately write down the product of \(5.96 \times 24.8\).
Solution
Count decimal digits: 5.96 has 2, 24.8 has 1 → total = 3 decimal places.
Place the decimal point 3 digits from the right of 147808:
Math Talk: When is the product of two decimals greater than both numbers, and when is it less?
Solution
| Situation | Example | Relationship |
|---|---|---|
| Both numbers > 1 | \(3.4\times6.5=22.1\) | Product > both numbers |
| Both numbers between 0 and 1 | \(0.75\times0.4=0.3\) | Product < both numbers |
| One > 1, one between 0 and 1 | \(0.75\times5=3.75\) | Product is between the two numbers |
Multiply by 10, 100 and 1000: 5.7, 23.02, 0.92, 0.306, 24.67
Solution
Multiplying by 10/100/1000 just shifts the decimal point right by 1/2/3 places.
| Number | × 10 | × 100 | × 1000 |
|---|---|---|---|
| 5.7 | 57 | 570 | 5700 |
| 23.02 | 230.2 | 2302 | 23020 |
| 0.92 | 9.2 | 92 | 920 |
| 0.306 | 3.06 | 30.6 | 306 |
| 24.67 | 246.7 | 2467 | 24670 |
4.3 Decimal Division
Ribbon is cut into 100 equal pieces from 3.9 m — length of each piece? Then, what is 0.039 m in cm and mm?
Solution
\(3.9 \div 100 = \frac{39}{10}\times\frac{1}{100}=\frac{39}{1000}=0.039\) m (already shown in the text).
\(0.039\text{ m} = 0.039\times100 = \) 3.9 cm
\(0.039\text{ m} = 0.039\times1000 = \) 39 mm
Divide by 10, 100, 1000, 10000 to complete the table: 21.1 and 0.13; also find the missing original numbers for the rows giving 2.146 (÷100) and 0.0058 (÷10000).
Solution
Rule: shift the decimal point left by 1 / 2 / 3 / 4 places respectively.
| Decimal | ÷10 | ÷100 | ÷1000 | ÷10000 |
|---|---|---|---|---|
| 18.7 (given) | 1.87 | 0.187 | 0.0187 | 0.00187 |
| 21.1 | 2.11 | 0.211 | 0.0211 | 0.00211 |
| 0.13 | 0.013 | 0.0013 | 0.00013 | 0.000013 |
| 214.6 | 21.46 | 2.146 | 0.2146 | 0.02146 |
| 58 | 5.8 | 0.58 | 0.058 | 0.0058 |
To find a “missing original number”, work backwards: if ÷100 gives 2.146, the original is \(2.146\times100=214.6\). If ÷10000 gives 0.0058, the original is \(0.0058\times10000=58\).
Do you know which month has the extra leap-year day?
Solution
Math Talk: Will the quotient always be greater than the dividend when the divisor is a decimal? Describe the relationship between dividend, divisor and quotient.
Solution
No — it depends on whether the divisor is smaller or larger than 1.
| Divisor | Example | Relationship |
|---|---|---|
| Divisor < 1 | \(128\div0.4=320\) | Quotient > Dividend |
| Divisor = 1 | \(128\div1=128\) | Quotient = Dividend |
| Divisor > 1 | \(128\div4=32\) | Quotient < Dividend |
A “magic number” — 142857 arises from \(1\div7\). Try This: find another cyclic number using \(1\div17\).
Solution
\(\frac{1}{7}=0.\overline{142857}\). Multiplying 142857 by 2,3,4,5,6 just cycles its digits; multiplying by 7 gives 999999.
\(\frac{1}{17}=0.\overline{0588235294117647}\) (a 16-digit repeating block). This repeating block, 0588235294117647, is another cyclic number — multiplying it by 2 through 16 cycles its digits around, just like 142857 did.
4.4 Look Before You Leap!
Math Talk: With one extra day every 4 years, what is the expression and value for the number of days in 100 calendar years? Form a different expression too.
Solution
Number of leap-adjustments in 100 years \( = \frac{100}{4}=25\).
\(\text{Days} = \left(100\times365\right) + \left(\frac{100}{4}\times1\right) = 36500+25 = \) 36,525 days
Alternative expression: \(100\times366 – \left(100 – \frac{100}{4}\right) = 36600 – 75 = 36{,}525\) days (start by assuming every year is 366 days, then subtract the 75 years that are NOT leap years).
The Earth actually needs \(100\times365.2422=36{,}524.22\) days — so this scheme overcompensates by 0.78 days.
Math Talk: With the “no leap day in the 100th year” rule, write an expression for days in 100 calendar years.
Solution
Years divisible by 4 (excluding the 100th year itself) = 25 − 1 = 24 leap years; remaining 76 years are ordinary.
\(\left(\frac{100}{4}-\frac{100}{100}\right)\times366 \;+\; \left(100-\left(\frac{100}{4}-\frac{100}{100}\right)\right)\times365\)
\(= (24\times366)+(76\times365) = 8784+27740 = \) 36,524 days
Compare with actual \(36{,}524.22\) days — very close! Over 1000 years this scheme gives \(36524\times10=365{,}240\) days, while the Earth needs \(1000\times365.2422=365{,}242.2\) days — a shortfall of 2.2 days.
Try This: With the final “leap every 400th year too” scheme, find the number of calendar days in 10,000 years and compare with the actual number of days the Earth takes for 10,000 revolutions. Suggest a fix if the difference is large.
Solution
In 10,000 years:
• Divisible by 400: \(\frac{10000}{400}=25\)
• Divisible by 100 but not 400: \(\frac{10000}{100}-25=100-25=75\)
• Divisible by 4 but not 100: \(\frac{10000}{4}-100=2500-100=2400\)
• Ordinary years: \(10000-(25+75+2400)=7500\)
Total days \(= (7500\times365)+(2400\times366)+(75\times365)+(25\times366)\)
\(= 2{,}764{,}875 + 878{,}400 + 27{,}375 + 9{,}150 = \) 3,652,425 12 days…
Adding carefully: \(2{,}764{,}875+878{,}400=3{,}643{,}275\); \(+27{,}375=3{,}670{,}650\); \(+9{,}150=3{,}652{,}425\)… Grouping ordinary years together (7500+75=7575) and leap years together (2400+25=2425) is cleaner: \(7575\times365=2{,}764{,}875\) and \(2425\times366=887{,}550\); total \(=3{,}652{,}425\) days.
Actual days needed: \(10000\times365.2422=3{,}652{,}420\) — recomputing precisely: \(10000 \times 365.2422 = 3{,}652{,}422\) days.
Fix: astronomers suggest one further correction — skip the leap day in years divisible by 4000 (so a year divisible by 4000 is NOT a leap year even though it’s divisible by 400). This trims the tiny long-term drift.
Try This: Investigate how traditional Indian calendars keep the calendar aligned with astronomical events.
Solution
Traditional Indian calendars are largely lunisolar — months follow the Moon’s phases, while the year is kept aligned to the Sun by periodically inserting an extra month called Adhika Māsa (leap month), roughly once every 2–3 years. Panchang-makers use Drik Ganita (observation/computation based on the actual positions of the Sun and Moon) rather than a fixed rule like “every 4th year,” so the calendar self-corrects using precise astronomical calculation instead of a simple divisibility rule.
Exercise Questions — Figure It Out
All four “Figure It Out” problem sets from the chapter, fully solved with steps.
Figure It Out — Set 1 (Page 73)
Find the products in tenths / hundredths: (a) \(6\times4\) tenths (b) \(7\times0.3\) (c) \(9\times5\) hundredths
Solution
(a) \(6\times4\) tenths \(=24\) tenths \(=\) 2.4 (given as example)
(b) \(0.3 = 3\) tenths, so \(7\times3\) tenths \(=21\) tenths \(=\) 2.1
(c) \(9\times5\) hundredths \(=45\) hundredths \(=\) 0.45
Find the products: (a) \(27.34\times6\) (b) \(4.23\times3.7\) (c) \(0.432\times0.23\)
Solution
(a) Ignore decimals: \(2734\times6=16404\). Decimal places = 2. → 164.04
(b) \(423\times37=15651\). Decimal places = \(2+1=3\). → 15.651
(c) \(432\times23=9936\). Decimal places = \(3+2=5\). → 0.09936
Thejus needs 1.65 m of cloth for a shirt. How many metres for 3 shirts?
Solution
\(1.65\times3=4.95\)
Meenu bought 4 notebooks (₹15.50 each) and 3 erasers (₹2.75 each). Total spend?
Solution
Notebooks: \(4\times15.50=62.00\)
Erasers: \(3\times2.75=8.25\)
Total: \(62.00+8.25=70.25\)
A rupee coin is 1.45 mm thick. What is the height of a stack of 36 coins, in centimetres?
Solution
\(1.45\times36=52.2\) mm
Convert to cm: \(52.2\div10=5.22\) cm
1 kg of oranges costs ₹56.50. What is the price of 2.250 kg? Can we write 56.50 as 56.5 and 2.250 as 2.25, and get the same product?
Solution
\(56.50\times2.250\): drop trailing zeroes → \(56.5\times2.25\).
\(565\times225=127125\); decimal places \(=1+2=3\) → 127.125
Yes — dropping trailing zeroes after the last non-zero decimal digit never changes a number’s value (56.50 = 56.5, 2.250 = 2.25), so the product is exactly the same either way.
Dwarakanath buys notebooks at ₹23.6 (wholesale) and sells at ₹30 each. Profit on selling 50 books in a week?
Solution
Profit per book \(=30-23.6=6.4\)
Profit for 50 books \(=6.4\times50=320\)
Given \(18\times12=216\): (a) \(18\times1.2\) (b) \(18\times0.12\) (c) \(1.8\times1.2\) (d) \(0.18\times0.12\) (e) \(0.018\times0.012\) (f) \(1.8\times12\). Which products are less than 1?
Solution
Place the decimal point in 216 based on total decimal digits of the two factors:
| Expression | Decimal places | Product |
|---|---|---|
| (a) 18 × 1.2 | 1 | 21.6 |
| (b) 18 × 0.12 | 2 | 2.16 |
| (c) 1.8 × 1.2 | 2 | 2.16 |
| (d) 0.18 × 0.12 | 4 | 0.0216 |
| (e) 0.018 × 0.012 | 6 | 0.000216 |
| (f) 1.8 × 12 | 1 | 21.6 |
Which products are less than 1? (a) \(7\times0.6\) (b) \(0.7\times0.6\) (c) \(0.7\times6\) (d) \(0.07\times0.06\). Can you tell without multiplying?
Solution
(a) \(7\times0.6=4.2\) — no
(b) \(0.7\times0.6=0.42\) — yes, < 1
(c) \(0.7\times6=4.2\) — no
(d) \(0.07\times0.06=0.0042\) — yes, < 1
Without multiplying: a product is less than 1 whenever both factors being multiplied are less than 1 (proper decimals). In (b) and (d) both numbers are less than 1; in (a) and (c) one factor (7 or 6) is a whole number ≥ 1, giving a bigger product.
Multiply by 10, 100 and 1000: 5.7, 23.02, 0.92, 0.306, 24.67
Solution
Same rule as before — shift decimal point right by 1 / 2 / 3 places.
| Number | × 10 | × 100 | × 1000 |
|---|---|---|---|
| 5.7 | 57 | 570 | 5700 |
| 23.02 | 230.2 | 2302 | 23020 |
| 0.92 | 9.2 | 92 | 920 |
| 0.306 | 3.06 | 30.6 | 306 |
| 24.67 | 246.7 | 2467 | 24670 |
Figure It Out — Set 2 (Page 83)
Find the quotient by converting to a denominator of 1, 10, 100 or 1000, and verify by long division: (a) \(\frac{18}{5}\) (b) \(\frac{415}{4}\) (c) \(\frac{1217}{2}\) (d) \(\frac{4827}{8}\)
Solution
(a) \(\frac{18}{5}=\frac{18\times2}{5\times2}=\frac{36}{10}=\) 3.6
(b) \(\frac{415}{4}=\frac{415\times25}{4\times25}=\frac{10375}{100}=\) 103.75
(c) \(\frac{1217}{2}=\frac{1217\times5}{2\times5}=\frac{6085}{10}=\) 608.5
(d) \(\frac{4827}{8}=\frac{4827\times125}{8\times125}=\frac{603375}{1000}=\) 603.375
Long-division check for 4827 ÷ 8 = 603.375
Choose the correct answer: (a) \(\frac{1526}{4}=\) (i) 38.15 (ii) 380.15 (iii) 381.5 (iv) 381.05 (b) \(\frac{3567}{8}=\) (i) 4458.75 (ii) 44.5875 (iii) 445.875 (iv) 4458.75
Solution
(a) \(1526\div4=381.5\) → option (iii)
(b) \(3567\div8=445.875\) → option (iii)
What is the quotient? (a) \(132\div4\) (b) \(13.2\div4\) (c) \(1.32\div4\) (d) \(0.132\div4\)
Solution
All share the same digit pattern \(132\div4=33\); only the decimal point shifts.
| (a) 132 ÷ 4 | 33 |
| (b) 13.2 ÷ 4 | 3.3 |
| (c) 1.32 ÷ 4 | 0.33 |
| (d) 0.132 ÷ 4 | 0.033 |
What is the quotient? (a) \(126\div8\) (b) \(12.6\div8\) (c) \(1.26\div8\) (d) \(0.126\div8\) (e) \(0.0126\div8\)
Solution
Base fact: \(126\div8=15.75\)
| (a) 126 ÷ 8 | 15.75 |
| (b) 12.6 ÷ 8 | 1.575 |
| (c) 1.26 ÷ 8 | 0.1575 |
| (d) 0.126 ÷ 8 | 0.01575 |
| (e) 0.0126 ÷ 8 | 0.001575 |
Figure It Out — Set 3 (Page 86–87)
Express as a decimal: (a) \(\frac{2}{5}\) (b) \(\frac{13}{4}\) (c) \(\frac{4}{50}\) (d) \(\frac{5}{8}\)
Solution
(a) \(\frac{2}{5}=\frac{4}{10}=\) 0.4
(b) \(\frac{13}{4}=\frac{325}{100}=\) 3.25
(c) \(\frac{4}{50}=\frac{8}{100}=\) 0.08
(d) \(\frac{5}{8}=\frac{625}{1000}=\) 0.625
Find the quotients: (a) \(24.86\div1.2\) (b) \(5.728\div1.52\)
Solution
Convert each divisor to a whole number first, multiplying dividend the same way.
(a) \(24.86\div1.2=248.6\div12\). Dividing gives \(20.71666\ldots\) — this is a non-terminating, recurring decimal (the digit 6 repeats forever), written \(20.7\overline{6}\), ≈ 20.717 (rounded).
(b) \(5.728\div1.52=572.8\div152\). Dividing gives \(3.76842105\ldots\) — also non-terminating, ≈ 3.768 (rounded).
Both quotients don’t terminate because after converting to whole-number divisors (12 and 152), the divisors carry a prime factor other than 2 or 5 — a preview of the “Does This Ever End?” idea from the chapter.
Using \(156\times12=1872\), evaluate: (a) \(15.6\times1.2\) (b) \(187.2\div1.2\) (c) \(18.72\div15.6\) (d) \(0.156\times0.12\)
Solution
(a) 1+1 = 2 decimal places on 1872 → 18.72
(b) \(187.2\div1.2=1872\div12=\) 156
(c) \(18.72\div15.6=1872\div156=\) 1.2
(d) 3+2 = 5 decimal places on 1872 → 0.01872
Evaluate: (a) \(25\div\_=0.025\) (b) \(25\div\_=250\) (c) \(25\div\_=2.5\) (d) \(25\div10=25\times\_\) (e) \(25\div0.10=25\times\_\) (f) \(25\div0.01=25\times\_\)
Solution
(a) \(25\div0.025=\) needs divisor \(=25/0.025=\) 1000
(b) divisor \(=25/250=\) 0.1
(c) divisor \(=25/2.5=\) 10
(d) dividing by 10 = multiplying by \(\frac{1}{10}\) → 0.1
(e) dividing by 0.10 = multiplying by 10
(f) dividing by 0.01 = multiplying by 100
Find the quotient: (a) \(2.46\div1.5\) (b) \(2.46\div0.15\) (c) \(2.46\div0.015\). Is the quotient of \(24.6\div1.5\) the same as \(2.46\div0.15\)?
Solution
(a) \(2.46\div1.5=246\div150=\) 1.64
(b) \(2.46\div0.15=246\div15=\) 16.4
(c) \(2.46\div0.015=2460\div15=\) 164
\(24.6\div1.5=16.4\), and \(2.46\div0.15=16.4\) as well — yes, they are the same!
Multiplying both dividend and divisor by the same power of 10 doesn’t change the value of a quotient — here both numbers in \(24.6/1.5\) were divided by 10 to get \(2.46/0.15\), so the ratio stays identical.
A 4 m wooden block is cut into 5 equal pieces. Length of each piece?
Solution
\(4\div5=0.8\)
A regular 12-sided polygon has perimeter 208.8 cm. Find the length of one side.
Solution
\(208.8\div12=17.4\)
3 litres of watermelon juice is shared among 8 friends. How much (in ml) does each get?
Solution
\(3\text{ L}=3000\text{ ml}\)
\(3000\div8=375\)
A car covers 234.45 km using 12.6 litres of petrol. Distance per litre?
Solution
\(234.45\div12.6=2344.5\div126\approx18.6071\ldots\) (non-terminating, repeats)
13.5 kg of flour (aata) is shared equally among 15 students. Flour per student?
Solution
\(13.5\div15=0.9\)
Math Talk: \(\frac{1}{2}, \frac{1}{2\times2}, \frac{1}{2\times2\times2}, \ldots\) and \(\frac{1}{5}, \frac{1}{5\times5}, \ldots\) — extend one more step in each and explain the pattern.
Solution
Powers of 2: \(0.5,\ 0.25,\ 0.125,\ 0.0625,\) next \(\frac{1}{32}=\) 0.03125 (each term is exactly half the previous one).
Powers of 5: \(0.2,\ 0.04,\ 0.008,\ 0.0016,\) next \(\frac{1}{3125}=\) 0.00032 (each term is exactly one-fifth of the previous one).
2 and 5 are special because \(10=2\times5\). Any fraction whose denominator is only made of 2’s and/or 5’s can always be converted into an equivalent fraction with denominator 10, 100, 1000, … — so it always terminates as a decimal.
Figure It Out — Set 4 (Page 93–94)
A 210 g packet of peanut chikki costs ₹70.5; a 110 g packet of potato chips costs ₹33.25. Which is cheaper?
Solution
Compare price per 100 g.
Chikki: \(\frac{70.5}{210}\times100=\) ₹33.57 per 100 g
Chips: \(\frac{33.25}{110}\times100=\) ₹30.23 per 100 g
Write the decimal number shown at each arrow mark on the two number lines.
Solution
Between 3.1 and 3.2 there are 10 equal divisions, each worth 0.01. The arrow points to the 5th division from 3.1:
Between 2.15 and 2.17, the gap of 0.02 is split into 10 parts of 0.002 each. The arrow points to the 5th division:
Read from the midpoint marking on the number-line diagram in the textbook; positions may vary slightly by print edition.
Shyamala bought 3 kg bananas at ₹30/kg (35 bananas in all) and sells each banana for ₹5. Find her profit.
Solution
Cost \(=3\times30=₹90\)
Revenue \(=35\times5=₹175\)
Profit \(=175-90=₹85\)
Textbooks 2.5 cm thick are placed on a 160 cm shelf; the teacher wanted to place 80 books. How many actually fit? Is there space left?
Solution
Books that fit \(=160\div2.5=64\)
Space used \(=64\times2.5=160\) cm (the full shelf)
Fill in the blanks: 5.5 km = __ m, 35 cm = __ m, 14.5 cm = __ mm, 68 g = __ kg, 9.02 m = __ mm, 125.5 ml = __ l
Solution
| 5.5 km | 5500 m |
| 35 cm | 0.35 m |
| 14.5 cm | 145 mm |
| 68 g | 0.068 kg |
| 9.02 m | 9020 mm |
| 125.5 ml | 0.1255 l |
(Sridharacharya’s Patiganita) \(6\frac{1}{4}\) divided by \(2\frac{1}{2}\), and \(60\frac{1}{4}\) divided by \(3\frac{1}{2}\). Solve using decimals.
Solution
\(6\frac{1}{4}=6.25\), \(2\frac{1}{2}=2.5\): \(6.25\div2.5=\) 2.5
\(60\frac{1}{4}=60.25\), \(3\frac{1}{2}=3.5\): \(60.25\div3.5=17.2142857\ldots\)
This is a non-terminating, repeating decimal: \(17\frac{3}{14}\approx\) 17.214 (rounded)
Fill the boxes in at least 2 different ways: (a) \(\_\times\_=2.4\) (b) \(\_\times\_=14.5\)
Solution
(a) \(1.2\times2=2.4\) or \(0.8\times3=2.4\) or \(0.6\times4=2.4\)
(b) \(2.9\times5=14.5\) or \(1.45\times10=14.5\) or \(7.25\times2=14.5\)
Given \(756\div36=21\), find: (a) \(75.6\div3.6\) (b) \(7.56\div0.36\) (c) \(756\div0.36\) (d) \(75.6\div360\) (e) \(7560\div3.6\)
Solution
Both numerator and denominator being scaled by the same power of 10 keeps the quotient 21; scaling by different amounts shifts the decimal point of 21.
| (a) 75.6 ÷ 3.6 | 21 |
| (b) 7.56 ÷ 0.36 | 21 |
| (c) 756 ÷ 0.36 | 2100 |
| (d) 75.6 ÷ 360 | 0.21 |
| (e) 7560 ÷ 3.6 | 2100 |
Find the missing cells if each cell represents \(a\div b\) (given some starter values).
Solution
Base fact: \(1517\div37=41\). Every other cell is 41 scaled by powers of 10 based on how much each row/column has shifted its decimal point.
| b ↓ a → | 1517 | 151.7 | 15.17 | 1.517 | 15170 |
|---|---|---|---|---|---|
| 37 | 41 | 4.1 | 0.41 | 0.041 | 410 |
| 3.7 | 410 | 41 | 4.1 | 0.41 | 4100 |
| 0.37 | 4100 | 410 | 41 | 4.1 | 41000 |
| 0.037 | 41000 | 4100 | 410 | 41 | 410000 |
| 370 | 4.1 | 0.41 | 0.041 | 0.0041 | 41 |
Tip: moving a’s decimal point one place right multiplies the quotient by 10; moving b’s decimal point one place right divides the quotient by 10.
Using digits 2, 4, 5, 8, 0 (each once) fill \(\boxed{\ }\boxed{\ }.\boxed{\ }\ \times\ \boxed{\ }.\boxed{\ }\) to get: (a) maximum product (b) minimum product (c) a product greater than 150 (d) product nearest to 100 (e) product nearest to 5
Solution
Format: a 3-digit number “AB.C” times a 2-digit number “D.E”, using each of 0,2,4,5,8 exactly once.
(a) Maximum: put the two biggest digits as the leading (highest place-value) digits, and tuck 0 into the least significant decimal spot.
(b) Minimum: put 0 as the leading digit of the smaller number to shrink it below 1, then keep the other leading digit as small as possible.
(c) Greater than 150: e.g.
(d) Nearest to 100: exploring several combinations:
(e) Nearest to 5: since the smallest possible product overall is 9.16 (see part b), that is also the closest we can get to 5:
These are found by systematically testing arrangements — other combinations may tie or come close; the key strategy is that the digit with the biggest “place-value weight” (the tens digit) has the biggest effect on the product, followed by the units digits, followed by the tenths digits.
Sort in increasing order: (a) \(245.05\times0.942368\) (b) \(245.05\times7.9682\) (c) \(245.05\div7.9682\) (d) \(245.05\div0.942368\) (e) \(245.05\) (f) \(7.9682\)
Solution
Reason about each without computing exactly:
• Multiplying 245.05 by a number less than 1 (0.942368) → result is less than 245.05 → (a) ≈ 230.93
• Multiplying by a number greater than 1 (7.9682) → result is much greater → (b) ≈ 1952.6
• Dividing by a number greater than 1 (7.9682) → result is smaller → (c) ≈ 30.75
• Dividing by a number less than 1 (0.942368) → result is greater than 245.05 → (d) ≈ 260.03
(e) is 245.05 itself; (f) is 7.9682 itself (the smallest value here).
i.e. (f) < (c) < (a) < (e) < (d) < (b)
Puzzle Time — Hidato
The worked Hidato puzzle from the chapter — fill consecutive numbers 1 to 40 so each next number sits in an adjacent (including diagonal) cell.
Complete the Hidato grid from 1 to 40 (some numbers pre-filled, lowest = 1, highest = 40).
Solution — as given in the textbook
| 33 | 35 | 36 | 37 | ||
| 31 | 34 | 24 | 22 | 38 | |
| 30 | 25 | 23 | 21 | 12 | 39 |
| 39 | 26 | 20 | 13 | 40 | 11 |
| 27 | 28 | 14 | 19 | 9 | 10 |
| 15 | 16 | 18 | 8 | 1 | |
| 17 | 7 | 16 | 3 | ||
| 5 | 4 |
Solved grid — the path runs consecutively from 1 (bottom right) up to 40 (middle), moving through adjacent cells including diagonals.
The three extra “Try it yourself” Hidato grids printed at the very end of the chapter are open practice puzzles — grab a pencil and use the same rule: each number’s neighbour (including diagonal) must be one more or one less.