Experiencing the Power Play
AActivity — Take a sheet of paper, as large a sheet as you can find. Fold it once. Fold it again, and again. How many times can you fold it over and over? Try it with different types of paper and see what happens.
Every fold doubles the thickness and halves the length of the folded stack.
After a few folds the stack is thick and short, so it cannot be bent any more.
An ordinary sheet of paper can be folded about 7 times.
A thinner or much larger sheet (newspaper, tissue) allows a few more folds.
Even with kilometre-long rolls of paper, people have managed only about 12 folds.
MTMath Talk — Now, what do you think the thickness would be after 30 folds? 45 folds? Make a guess.
Fill the table below.
Fill the table below.
Take the thickness of the sheet as 0.001 cm.
The thickness doubles after every fold, so the thickness after \(n\) folds is \(0.001 \times 2^{n}\) cm.
For example, after 18 folds: \(0.001 \times 2^{18} = 0.001 \times 262144 = 262.144\) cm \(\approx 262\) cm.
Convert to metres when the value passes 100 cm, and to kilometres when it passes 1 km (100000 cm).
| Fold | Thickness | Fold | Thickness | Fold | Thickness |
|---|---|---|---|---|---|
| 18 | ≈ 262 cm | 27 | ≈ 1.3 km | 36 | ≈ 687 km |
| 19 | ≈ 524 cm | 28 | ≈ 2.7 km | 37 | ≈ 1,374 km |
| 20 | ≈ 10.5 m | 29 | ≈ 5.4 km | 38 | ≈ 2,749 km |
| 21 | ≈ 21.0 m | 30 | ≈ 10.7 km | 39 | ≈ 5,498 km |
| 22 | ≈ 41.9 m | 31 | ≈ 21.5 km | 40 | ≈ 10,995 km |
| 23 | ≈ 83.9 m | 32 | ≈ 42.9 km | 41 | ≈ 21,990 km |
| 24 | ≈ 168 m | 33 | ≈ 85.9 km | 42 | ≈ 43,980 km |
| 25 | ≈ 336 m | 34 | ≈ 172 km | 43 | ≈ 87,961 km |
| 26 | ≈ 671 m | 35 | ≈ 344 km | 44 | ≈ 1,75,922 km |
| 45 | ≈ 3,51,844 km |
After 30 folds the thickness is about 10.7 km, which is the height at which aeroplanes fly.
After 45 folds the thickness is about 3,51,844 km.
After 46 folds it is \(0.001 \times 2^{46}\) cm \(\approx 7.04 \times 10^{5}\) km, which is more than the distance to the Moon (3,84,400 km).
Note: the textbook table gives about 10.4 m for 20 folds. \(0.001\times2^{20} = 1048.576\) cm \(\approx 10.5\) m.
?In-text Question — Notice the change in thickness after two folds. By how much does it increase? After any 3 folds the thickness increases 8 times (= 2 × 2 × 2). Check if that is true.
Take fold 4 and fold 6 (two folds apart): \(0.064 \div 0.016 = 4\).
Take fold 9 and fold 11: \(2.048 \div 0.512 = 4\).
Each fold doubles the thickness, so two folds multiply it by \(2 \times 2 = 4\).
Check for 3 folds: fold 4 to fold 7 is \(0.128 \div 0.016 = 8\).
Fold 5 to fold 8 is \(0.256 \div 0.032 = 8\).
So the statement for 3 folds is true.
Exponential Notation and Operations
?In-text Question — Which expression describes the thickness of a sheet of paper after it is folded 10 times? The initial thickness is represented by the letter-number v.
(i) 10v
(ii) 10 + v
(iii) 2 × 10 × v
(iv) 210
(v) 210v
(vi) 102v
Each fold multiplies the thickness by 2.
After 10 folds the thickness is \(v \times 2 \times 2 \times \dots \times 2\) (ten 2s), which is \(v \times 2^{10} = 2^{10}v\).
(i) \(10v\) is 10 times the thickness, so it is wrong.
(ii) \(10 + v\) adds 10 to the thickness, so it is wrong.
(iii) \(2 \times 10 \times v = 20v\) is only 20 times the thickness, so it is wrong.
(iv) \(2^{10}\) has no \(v\), so it is only a number and not a thickness. So it is wrong.
(v) \(2^{10}v\) is exactly \(v\) multiplied by ten 2s, so it is correct.
(vi) \(10^{2}v = 100v\) is 100 times the thickness, so it is wrong.
?In-text Question — What is (−1)5? Is it positive or negative? What about (−1)56? Is (−2)4 = 16? Verify. What is 02, 05? What is 0n?
\((-1)^{5} = (-1)\times(-1)\times(-1)\times(-1)\times(-1) = -1\).
It is negative.
Pairs of \((-1)\) multiply to \(+1\).
The exponent 5 is odd, so one \((-1)\) is left over.
\((-1)^{56}\): the exponent 56 is even, so all the \((-1)\)s pair up.
\((-1)^{56} = 1\).
It is positive.
\((-2)^{4} = (-2)\times(-2)\times(-2)\times(-2) = 4 \times 4 = 16\).
Yes, it is true.
\(0^{2} = 0 \times 0 = 0\).
\(0^{5} = 0 \times 0 \times 0 \times 0 \times 0 = 0\).
In general \(0^{n} = 0\) for every counting number \(n\).
Figure it Out — Set 1 (Pages 22–23)
Three questions from pages 22–23 of the textbook.
1Express the following in exponential form:
(i) 6 × 6 × 6 × 6
(ii) y × y
(iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7
(v) 2 × 2 × a × a
(vi) a × a × a × c × c × c × c × d
(i) \(6\times6\times6\times6 = 6^{4}\) [6 is repeated 4 times]
(ii) \(y\times y = y^{2}\)
(iii) \(b\times b\times b\times b = b^{4}\)
(iv) \(5\times5\times7\times7\times7 = 5^{2}\times7^{3}\)
(v) \(2\times2\times a\times a = 2^{2}\times a^{2}\)
(vi) \(a\times a\times a\times c\times c\times c\times c\times d = a^{3}\times c^{4}\times d\) [\(d\) appears once, so \(d = d^{1}\)]
(ii) \(y^{2}\)
(iii) \(b^{4}\)
(iv) \(5^{2}\times7^{3}\)
(v) \(2^{2}\times a^{2}\)
(vi) \(a^{3}c^{4}d\)
2Express each of the following as a product of powers of their prime factors in exponential form.
(i) 648
(ii) 405
(iii) 540
(iv) 3600
(i) Divide 648 repeatedly by primes.
\(648 \div 2 = 324\)
\(324 \div 2 = 162\)
\(162 \div 2 = 81\)
\(81 \div 3 = 27\)
\(27 \div 3 = 9\)
\(9 \div 3 = 3\)
\(3 \div 3 = 1\)
\(648 = 2^{3}\times3^{4}\)
(ii)
\(405 \div 5 = 81\)
\(81 = 3\times3\times3\times3 = 3^{4}\)
\(405 = 3^{4}\times5\)
(iii) Divide 540 repeatedly by primes.
\(540 \div 2 = 270\)
\(270 \div 2 = 135\)
\(135 \div 3 = 45\)
\(45 \div 3 = 15\)
\(15 \div 3 = 5\)
\(5 \div 5 = 1\)
\(540 = 2^{2}\times3^{3}\times5\)
(iv)
\(3600 = 36 \times 100\)
\(= (2^{2}\times3^{2})\times(2^{2}\times5^{2})\)
\(3600 = 2^{4}\times3^{2}\times5^{2}\) [\(2^{2}\times2^{2} = 2^{2+2} = 2^{4}\)]
(ii) \(3^{4}\times5\)
(iii) \(2^{2}\times3^{3}\times5\)
(iv) \(2^{4}\times3^{2}\times5^{2}\)
3Write the numerical value of each of the following:
(i) 2 × 103
(ii) 72 × 23
(iii) 3 × 44
(iv) (−3)2 × (−5)2
(v) 32 × 104
(vi) (−2)5 × (−10)6
(i) \(2\times10^{3} = 2\times1000 = 2000\)
(ii) \(7^{2}\times2^{3} = 49\times8 = 392\)
(iii) \(3\times4^{4} = 3\times256 = 768\)
(iv) \((-3)^{2}\times(-5)^{2} = 9\times25 = 225\) [even powers are positive]
(v) \(3^{2}\times10^{4} = 9\times10000 = 90000\)
(vi) \((-2)^{5} = -32\) [odd power] and \((-10)^{6} = 1000000\) [even power].
\((-2)^{5}\times(-10)^{6} = -32\times1000000 = -3,20,00,000\)
(ii) 392
(iii) 768
(iv) 225
(v) 90000
(vi) \(-3,20,00,000\)
The Stones that Shine
?In-text Question — 37 can also be written as 32 × 35. Can you reason out why?
\(3^{7} = 3\times3\times3\times3\times3\times3\times3\) [seven 3s]
Group the first two 3s and the remaining five 3s.
\(3^{7} = (3\times3)\times(3\times3\times3\times3\times3) = 3^{2}\times3^{5}\)
The exponents add: \(2 + 5 = 7\).
MTMath Talk — Use the observation \(n^a \times n^b = n^{a+b}\) to compute the following.
(i) 29
(ii) 57
(iii) 46
(i) \(2^{9} = 2^{4}\times2^{5} = 16\times32 = 512\)
(ii) \(5^{7} = 5^{3}\times5^{4} = 125\times625 = 78125\)
(iii) \(4^{6} = 4^{3}\times4^{3} = 64\times64 = 4096\)
(ii) 78125
(iii) 4096
?In-text Question — Write the following expressions as a power of a power in at least two different ways:
(i) 86
(ii) 715
(iii) 914
(iv) 58
Use \((n^{a})^{b} = n^{ab}\).
Split the exponent into two factors.
(i) \(8^{6} = (8^{2})^{3} = (8^{3})^{2}\) [\(2\times3 = 6\)]
Also \(8^{6} = (2^{3})^{6} = 2^{18}\).
(ii) \(7^{15} = (7^{3})^{5} = (7^{5})^{3}\) [\(3\times5 = 15\)]
(iii) \(9^{14} = (9^{2})^{7} = (9^{7})^{2}\) [\(2\times7 = 14\)]
Also \(9^{14} = (3^{2})^{14} = 3^{28}\).
(iv) \(5^{8} = (5^{2})^{4} = (5^{4})^{2}\) [\(2\times4 = 8\)]
(ii) \((7^{3})^{5}\), \((7^{5})^{3}\)
(iii) \((9^{2})^{7}\), \((9^{7})^{2}\)
(iv) \((5^{2})^{4}\), \((5^{4})^{2}\)
Magical Pond
?In-text Question — Write the number of lotuses (in exponential form) when the pond was —
(i) fully covered
(ii) half covered
The number of lotuses doubles every day.
Assume the pond started with 1 lotus.
After \(n\) days there are \(2^{n}\) lotuses.
(i) The pond is fully covered on day 30, so the number of lotuses is \(2^{30}\).
(ii) The pond is half covered on day 29, so the number of lotuses is \(2^{29}\). [half of \(2^{30}\) is \(2^{29}\)]
(ii) \(2^{29}\)
?In-text Question — Use the observation \(m^a \times n^a = (mn)^a\) to compute the value of 25 × 55.
\(2^{5}\times5^{5} = (2\times5)^{5}\) [same exponent, so multiply the bases]
\(= 10^{5} = 100000\)
?In-text Question — Simplify \(\dfrac{10^4}{5^4}\) and write it in exponential form.
The exponents are the same, so \(\dfrac{10^{4}}{5^{4}} = \left(\dfrac{10}{5}\right)^{4}\) [\(\dfrac{m^{a}}{n^{a}} = \left(\dfrac{m}{n}\right)^{a}\)]
\(= 2^{4} = 16\)
How Many Combinations
?In-text Question — Roxie has 7 dresses, 2 hats, and 3 pairs of shoes. How many different ways can Roxie dress up?
For each dress there are 2 choices of hat.
That gives \(7\times2 = 14\) dress-and-hat pairs.
For each of these 14 pairs there are 3 choices of shoes.
Total ways \(= 7\times2\times3 = 42\).
?In-text Question — Estu says, “Next time, I will buy a lock that has 6 slots with the letters A to Z. I feel it is safer.” How many passwords are possible with such a lock?
Each of the 6 slots can hold any of the 26 letters.
Number of passwords \(= 26\times26\times26\times26\times26\times26 = 26^{6}\).
\(26^{3} = 17576\), so \(26^{6} = 17576\times17576 = 308915776\).
TTTry This — Think about how many combinations are possible in different contexts:
(i) Pincodes of places in India,
(ii) Mobile numbers,
(iii) Vehicle registration numbers. Try to find out how these numbers or codes are allotted/generated.
(i) A Pincode has 6 digits.
The first digit gives the region (zone) of the country.
The first two digits together give the sub-region, and the first three digits give the sorting district.
The last three digits identify the individual post office.
For example, in 464001 (Vidisha) the first digit 4 is the zone, and 001 is the post office.
There are at most \(10^{6} = 10,00,000\) six-digit codes, but not all are in use.
(ii) An Indian mobile number has 10 digits.
The first digit is 6, 7, 8 or 9, so there are at most \(4\times10^{9}\) numbers.
(iii) A vehicle number has a state code (two letters), a district/RTO code (two digits), a series of letters and a four-digit number.
For example, MP 04 AB 1234.
The four-digit number alone gives \(10^{4}\) choices for each series.
Each new letter series gives \(26^{2} = 676\) more series of two letters.
The Other Side of Powers
?In-text Question — What is 2100 ÷ 225 in powers of 2?
Use \(n^{a} \div n^{b} = n^{a-b}\).
\(2^{100} \div 2^{25} = 2^{100-25} = 2^{75}\)
MTMath Talk — Why can’t n be 0?
The rule is \(n^{a} \div n^{b} = n^{a-b}\).
If \(n = 0\), then \(n^{b} = 0^{b} = 0\), so we would be dividing by 0.
Division by 0 is not defined, so the rule cannot be used.
MTMath Talk — We had required a and b to be counting numbers. Can a and b be any integers? Will the generalised forms still hold true?
Test the three rules with negative exponents.
Product rule: \(2^{3}\times2^{-5} = 8\times\dfrac{1}{32} = \dfrac{1}{4}\), and \(2^{3+(-5)} = 2^{-2} = \dfrac{1}{4}\).
It holds.
Power of a power: \((2^{-2})^{3} = \left(\dfrac{1}{4}\right)^{3} = \dfrac{1}{64}\), and \(2^{(-2)\times3} = 2^{-6} = \dfrac{1}{64}\).
It holds.
Quotient rule: \(2^{-3} \div 2^{-5} = \dfrac{1}{8}\times32 = 4\), and \(2^{-3-(-5)} = 2^{2} = 4\).
It holds.
?In-text Question — Write equivalent forms of the following.
(i) 2−4
(ii) 10−5
(iii) (−7)−2
(iv) (−5)−3
(v) 10−100
Use \(n^{-a} = \dfrac{1}{n^{a}}\).
(i) \(2^{-4} = \dfrac{1}{2^{4}} = \dfrac{1}{16}\)
(ii) \(10^{-5} = \dfrac{1}{10^{5}} = \dfrac{1}{100000}\)
(iii) \((-7)^{-2} = \dfrac{1}{(-7)^{2}} = \dfrac{1}{49}\)
(iv) \((-5)^{-3} = \dfrac{1}{(-5)^{3}} = \dfrac{1}{-125} = -\dfrac{1}{125}\)
(v) \(10^{-100} = \dfrac{1}{10^{100}}\)
(ii) \(\dfrac{1}{10^{5}}\)
(iii) \(\dfrac{1}{(-7)^{2}}\)
(iv) \(\dfrac{1}{(-5)^{3}}\)
(v) \(\dfrac{1}{10^{100}}\)
?In-text Question — Simplify and write the answers in exponential form.
(i) 2−4 × 27
(ii) 32 × 3−5 × 36
(iii) p3 × p−10
(iv) 24 × (−4)−2
(v) 8p × 8q
(i) \(2^{-4}\times2^{7} = 2^{-4+7} = 2^{3}\)
(ii) \(3^{2}\times3^{-5}\times3^{6} = 3^{2-5+6} = 3^{3}\)
(iii) \(p^{3}\times p^{-10} = p^{3-10} = p^{-7}\)
(iv) \((-4)^{-2} = \dfrac{1}{(-4)^{2}} = \dfrac{1}{16} = 4^{-2} = 2^{-4}\)
\(2^{4}\times(-4)^{-2} = 2^{4}\times2^{-4} = 2^{4-4} = 2^{0} = 1\)
(v) \(8^{p}\times8^{q} = 8^{p+q}\)
(ii) \(3^{3}\)
(iii) \(p^{-7}\)
(iv) \(2^{0} = 1\)
(v) \(8^{p+q}\)
Power Lines
?In-text Question — How many times larger than 4−2 is 42?
Number of times \(= 4^{2} \div 4^{-2}\).
\(4^{2} \div 4^{-2} = 4^{2-(-2)} = 4^{4}\)
Check: \(4^{2} = 16\), \(4^{-2} = \dfrac{1}{16}\), and \(16 \div \dfrac{1}{16} = 16\times16 = 256 = 4^{4}\).
?In-text Question — Use the power line for 7 to answer the following questions.2,401 × 49 =493 =343 × 2,401 =\(\dfrac{16{,}807}{49}=\)\(\dfrac{7}{343}=\)\(\dfrac{16{,}807}{8{,}23{,}543}=\)\(1{,}17{,}649\times\dfrac{1}{343}=\)\(\dfrac{1}{343}\times\dfrac{1}{343}=\)
Write each number as a power of 7 using the power line, then use the exponent rules.
| Question | Working | Answer |
|---|---|---|
| 2,401 × 49 | \(7^{4}\times7^{2} = 7^{4+2}\) | \(7^{6}\) |
| 49³ | \((7^{2})^{3} = 7^{2\times3}\) | \(7^{6}\) |
| 343 × 2,401 | \(7^{3}\times7^{4} = 7^{3+4}\) | \(7^{7}\) |
| \(\dfrac{16{,}807}{49}\) | \(7^{5}\div7^{2} = 7^{5-2}\) | \(7^{3}\) |
| \(\dfrac{7}{343}\) | \(7^{1}\div7^{3} = 7^{1-3}\) | \(7^{-2}\) |
| \(\dfrac{16{,}807}{8{,}23{,}543}\) | \(7^{5}\div7^{7} = 7^{5-7}\) | \(7^{-2}\) |
| \(1{,}17{,}649\times\dfrac{1}{343}\) | \(7^{6}\times7^{-3} = 7^{6-3}\) | \(7^{3}\) |
| \(\dfrac{1}{343}\times\dfrac{1}{343}\) | \(7^{-3}\times7^{-3} = 7^{-3-3}\) | \(7^{-6}\) |
Powers of 10 and Scientific Notation
?In-text Question — Write these numbers in the same way (using powers of 10):
(i) 172,
(ii) 5642,
(iii) 6374.
Write each digit times its place value as a power of 10.
(i) \(172 = 1\times10^{2} + 7\times10^{1} + 2\times10^{0}\)
(ii) \(5642 = 5\times10^{3} + 6\times10^{2} + 4\times10^{1} + 2\times10^{0}\)
(iii) \(6374 = 6\times10^{3} + 3\times10^{2} + 7\times10^{1} + 4\times10^{0}\)
(ii) \(5\times10^{3} + 6\times10^{2} + 4\times10^{1} + 2\times10^{0}\)
(iii) \(6\times10^{3} + 3\times10^{2} + 7\times10^{1} + 4\times10^{0}\)
?In-text Question — Write the large-number facts we read just before in this form.
(i) The Sun is located 30,00,00,00,00,00,00,00,00,000 m from the centre of our Milky Way galaxy.
(ii) The number of stars in our galaxy is 1,00,00,00,00,000.
(iii) The mass of the Earth is 59,76,00,00,00,00,00,00,00,00,00,000 kg.
(i) The Sun is located 30,00,00,00,00,00,00,00,00,000 m from the centre of our Milky Way galaxy.
(ii) The number of stars in our galaxy is 1,00,00,00,00,000.
(iii) The mass of the Earth is 59,76,00,00,00,00,00,00,00,00,00,000 kg.
Write the number as (a number between 1 and 10) \(\times\) (a power of 10).
The exponent is the number of places the decimal point moves.
(i) The number has 21 digits: 3 followed by 20 zeros.
\(30,00,00,00,00,00,00,00,00,000 = 3\times10^{20}\) m
(ii) The number has 12 digits: 1 followed by 11 zeros.
\(1,00,00,00,00,000 = 1\times10^{11}\)
(iii) The number has 25 digits: 5976 followed by 21 zeros.
The decimal point moves 24 places, so the mass is \(5.976\times10^{24}\) kg.
(ii) \(1\times10^{11}\)
(iii) \(5.976\times10^{24}\) kg
?In-text Question — The distance between the Sun and Saturn is 14,33,50,00,00,000 m = 1.4335 × 1012 m. The distance between Saturn and Uranus is 14,39,00,00,00,000 m = 1.439 × 1012 m. The distance between the Sun and Earth is 1,49,60,00,00,000 m = 1.496 × 1011 m.
Can you say which of the three distances is the smallest?
On the number line, mark the relative position of the Earth.
Can you say which of the three distances is the smallest?
On the number line, mark the relative position of the Earth.
Write all three with the same power of 10.
\(1.4335\times10^{12} = 14.335\times10^{11}\)
\(1.439\times10^{12} = 14.39\times10^{11}\)
\(1.496\times10^{11}\) stays as it is.
\(1.496 < 14.335 < 14.39\), so the Sun–Earth distance is the smallest.
On the number line the Earth is at \(\dfrac{1.496}{14.335}\approx0.10\) of the way from the Sun to Saturn.
This is about one tenth of the distance.
?In-text Question — Express the following numbers in standard form.
(i) 59,853
(ii) 65,950
(iii) 34,30,000
(iv) 70,04,00,00,000
(i) \(59853 = 5.9853\times10^{4}\) [the decimal point moves 4 places]
(ii) \(65950 = 6.595\times10^{4}\) [4 places]
(iii) \(34,30,000 = 3.43\times10^{6}\) [6 places]
(iv) \(70,04,00,00,000 = 7.004\times10^{10}\) [10 places]
(ii) \(6.595\times10^{4}\)
(iii) \(3.43\times10^{6}\)
(iv) \(7.004\times10^{10}\)
Did You Ever Wonder?
?In-text Question — Roxie wonders, “Instead of jaggery if we use 1-rupee coins, how many coins are needed to equal my weight?” Would the number of coins be in hundreds, thousands, lakhs, crores, or even more? Make an instinctive guess, then find the answer by making reasonable assumptions.
Guess: a typical guess is “thousands”.
Relationship: number of coins \(=\) Roxie’s weight \(\div\) weight of one coin.
Assumptions: Roxie’s weight is 45 kg \(= 45000\) g.
A 1-rupee coin weighs about 5 g (weigh one on a kitchen scale to check).
Compute: \(45000 \div 5 = 9000\) coins.
The coins are worth \(9000\times1 = 9000\) rupees.
?In-text Question — Estu asks, “What if we use 5-rupee coins or 10-rupee notes instead? How much money could it be?” Make an instinctive guess first. Then find out (make necessary and reasonable assumptions about the unknown details and find the answers).
Guess: tens of thousands for coins and more for notes.
Assumptions: Estu’s weight is 50 kg \(= 50000\) g.
A 5-rupee coin weighs about 6 g.
A 10-rupee note weighs about 1 g.
5-rupee coins: number of coins \(= 50000 \div 6 \approx 8333\).
Money \(\approx 8333\times5 \approx 41,667 \approx 4.2\times10^{4}\) rupees.
10-rupee notes: number of notes \(= 50000 \div 1 = 50000\).
Money \(= 50000\times10 = 5,00,000 = 5\times10^{5}\) rupees.
MTMath Talk — Estu says, “When I become an adult, I would like to donate notebooks worth my weight every year”. Roxie says, “When I grow up, I would like to do annadāna (offering grains or meals) worth my weight every year”. How many people might benefit from each of these offerings in a year? Again, guess first before finding out.
Assumptions: an adult weighs 60 kg.
A notebook weighs about 150 g.
A student uses about 6 notebooks in a year.
A person eats about 0.5 kg of grain in a day.
Notebooks: number of notebooks \(= 60000 \div 150 = 400\).
Students helped for a year \(= 400 \div 6 \approx 67\).
Annadāna: grain \(= 60\) kg.
People fed for one day \(= 60 \div 0.5 = 120\).
The same 60 kg feeds one person for 120 days, which is about 4 months.
So worth one year of her weight, grain feeds about 120 people for a day, or one person for about 4 months.
?In-text Question — Roxie and Estu overheard someone saying — “We did pādayātra for about 400 km to reach this place! We arrived early this morning.” How long ago would they have started their journey? Find answers by making necessary assumptions and approximations. Do guess first before calculating.
Guess: about two weeks.
Assumptions: a pilgrim walks about 5 km in an hour, for about 5 hours a day.
So the distance in a day is \(5\times5 = 25\) km.
Compute: number of days \(= 400 \div 25 = 16\).
?In-text Question — Can you come up with some examples of linear growth and of exponential growth?
Linear growth adds the same amount each time.
Examples: the steps of a ladder that rise 20 cm each;
saving ₹100 every week;
a taxi fare that increases by a fixed amount for every km.
Exponential growth multiplies by the same number each time.
Examples: the thickness of a folded paper;
the lotuses in the pond that double every day;
a message that every person forwards to 2 new people.
Getting a Sense for Large Numbers
?In-text Question — Fill in the blanks in the table of populations:
(i) The estimated global population of starlings is around 1.3 arab / 1.3 billion (_____).
(ii) The estimated mosquito population worldwide is 11 neel / 110 trillion (_____).
(i) \(1.3\) billion \(= 1.3\times10^{9}\) [a billion is \(10^{9}\)].
(ii) \(110\) trillion \(= 110\times10^{12} = 1.1\times10^{2}\times10^{12} = 1.1\times10^{14}\) [a trillion is \(10^{12}\)].
(ii) \(1.1\times10^{14}\)
?In-text Question — With a global human population of about 8 × 109 and about 4 × 105 African elephants, can we say that there are nearly 20,000 people for every African elephant?
People for each elephant \(= \dfrac{8\times10^{9}}{4\times10^{5}}\).
\(= \dfrac{8}{4}\times10^{9-5} = 2\times10^{4}\)
\(2\times10^{4} = 20000\).
?In-text Question — Fill in the blanks:
(i) A fossil of Kelenken guillermoi, a type of terror bird, is dated to 15 million years ago (≈ _____ seconds).
(ii) Plants on land started 47 crore / 470 million years ago (≈ _____ seconds).
(i) A fossil of Kelenken guillermoi, a type of terror bird, is dated to 15 million years ago (≈ _____ seconds).
(ii) Plants on land started 47 crore / 470 million years ago (≈ _____ seconds).
Take 1 year \(\approx 3.16\times10^{7}\) seconds.
1 year \(= 365\times24\times60\times60 = 31,536,000\) seconds \(\approx 3.16\times10^{7}\) seconds.
(i) 15 million years \(= 1.5\times10^{7}\) years.
Seconds \(= 1.5\times10^{7}\times3.16\times10^{7} = 4.74\times10^{14}\).
(ii) 470 million years \(= 4.7\times10^{8}\) years.
Seconds \(= 4.7\times10^{8}\times3.16\times10^{7} = 14.85\times10^{15} \approx 1.5\times10^{16}\).
(ii) about \(1.5\times10^{16}\) seconds
?In-text Question — Calculate and write the answer using scientific notation:
(i) How many ants are there for every human in the world?
(ii) If a flock of starlings contains 10,000 birds, how many flocks could there be in the world?
(iii) If each tree had about 104 leaves, find the total number of leaves on all the trees in the world.
(iv) If you stacked sheets of paper on top of each other, how many would you need to reach the Moon?
Use these values from the chapter: ants \(= 2\times10^{16}\), humans \(\approx 8\times10^{9}\), starlings \(= 1.3\times10^{9}\), trees \(= 3\times10^{12}\).
(i) Ants for every human \(= \dfrac{2\times10^{16}}{8\times10^{9}} = 0.25\times10^{7} = 2.5\times10^{6}\).
(ii) Flocks \(= \dfrac{1.3\times10^{9}}{10^{4}} = 1.3\times10^{9-4} = 1.3\times10^{5}\).
(iii) Leaves \(= 3\times10^{12}\times10^{4} = 3\times10^{16}\).
(iv) Take the thickness of a sheet as 0.001 cm \(= 10^{-3}\) cm (as in the folding problem).
Distance to the Moon \(= 3,84,400\) km \(= 3.844\times10^{5}\times10^{5}\) cm \(= 3.844\times10^{10}\) cm.
Number of sheets \(= \dfrac{3.844\times10^{10}}{10^{-3}} = 3.844\times10^{13}\).
(ii) \(1.3\times10^{5}\) flocks
(iii) \(3\times10^{16}\) leaves
(iv) \(3.844\times10^{13}\) sheets
?In-text Question — “How old are you?” asked Estu. Roxie said, “I’m 4840 days old today! … I’m _____ hours old!” Make an estimate before finding this number.
Estimate: 4840 days is about 5000 days, and a day has about 24 hours.
So the answer is about \(5000\times24 = 1,20,000\) hours.
Compute: hours \(= 4840\times24 = 116160\).
?In-text Question — Estu: “I am 4070 days old today. Can you find out my date of birth?”
Count back 4070 days from today.
Take today as 9 October 2026 (use the date on which you solve it).
11 years back from 9 October 2026 is 9 October 2015.
Days in these 11 years \(= 11\times365 + 3 = 4018\) [3 leap days: 29 February 2016, 2020 and 2024].
Remaining days \(= 4070 - 4018 = 52\).
Go back 52 days from 9 October 2015: 9 days back is 30 September, 30 days more is 31 August, and 13 days more is 18 August.
?In-text Question — What could the number “69,70,710 …” in the speech bubble mean? Find out!
The bubble says “I am 69,70,710 … old”.
Test whether it is Roxie’s age in minutes.
Roxie’s age in minutes \(= 4840\times24\times60 = 69,69,600\).
This is very close to 69,70,710 (the difference is only 1110 minutes, less than one day).
In days, \(69,70,710 \div 1440 \approx 4840.8\).
?In-text Question — If you have lived for a million seconds, how old would you be?
1 million \(= 10^{6}\) seconds.
1 day \(= 24\times60\times60 = 86400\) seconds.
Days \(= \dfrac{10^{6}}{86400} \approx 11.57\).
?In-text Question — 105 seconds ≈ 1.16 days and 106 seconds ≈ 11.57 days. Think of some events or phenomena whose time is of the order of
(i) 105 seconds and
(ii) 106 seconds. Write them in scientific notation.
(i) Order of \(10^{5}\) seconds (1 to 10 days):
A two-day weekend trip: \(2\times86400 = 1.7\times10^{5}\) seconds.
One week: \(7\times86400 = 6.05\times10^{5}\) seconds.
(ii) Order of \(10^{6}\) seconds (10 to 100 days):
A fortnight’s holiday: \(14\times86400 = 1.2\times10^{6}\) seconds.
The Moon going once around the Earth (27.3 days): \(27.3\times86400 = 2.4\times10^{6}\) seconds.
(ii) a fortnight \(1.2\times10^{6}\) s; one Moon cycle \(2.4\times10^{6}\) s
TTTry This — Calculate and write the answer using scientific notation:
(i) If one star is counted every second, how long would it take to count all the stars in the universe? Answer in terms of the number of seconds.
(ii) If one could drink a glass of water (200 ml) every 10 seconds, how long would it take to finish the entire volume of water on Earth?
(i) The number of stars is \(2\times10^{23}\).
One star is counted per second.
Time \(= 2\times10^{23}\) seconds.
(ii) The water on Earth is \(2\times10^{25}\) drops, with 16 drops in a millilitre.
Volume in ml \(= \dfrac{2\times10^{25}}{16} = 0.125\times10^{25} = 1.25\times10^{24}\) ml.
Number of glasses \(= \dfrac{1.25\times10^{24}}{200} = \dfrac{1.25\times10^{24}}{2\times10^{2}} = 0.625\times10^{22} = 6.25\times10^{21}\).
Time \(= 6.25\times10^{21}\times10 = 6.25\times10^{22}\) seconds.
(ii) \(6.25\times10^{22}\) seconds
A Pinch of History
?In-text Question — Observe the names million (106), billion (109), trillion (1012), quadrillion (1015), quintillion (1018), sextillion (1021), septillion (1024), octillion (1027), nonillion (1030), decillion (1033). What does the first part of each name denote?
Look at the exponents: 6, 9, 12, 15, 18, 21, 24, 27, 30, 33.
Each is 3 more than the one before.
Write each as \(10^{3k} = (1000)^{k}\).
| Name | Power of 10 | Power of 1000 | First part |
|---|---|---|---|
| million | \(10^{6}\) | \(1000^{2}\) | mi (1) |
| billion | \(10^{9}\) | \(1000^{3}\) | bi (2) |
| trillion | \(10^{12}\) | \(1000^{4}\) | tri (3) |
| quadrillion | \(10^{15}\) | \(1000^{5}\) | quadri (4) |
| decillion | \(10^{33}\) | \(1000^{11}\) | dec (10) |
The first part (bi, tri, quadri, …) tells how many times 1000 is multiplied with 1000.
For example, a billion is \(1000\times1000\times1000\): two 1000s are multiplied with 1000, and “bi” means 2.
Tremendous in Ten!
AActivity — Round 2 of the game: Roxie wrote 101000 + 101000 + 101000 + 101000 and Estu wrote (101000000) × 9000. Can you say which is greater?
Roxie: \(10^{1000}+10^{1000}+10^{1000}+10^{1000} = 4\times10^{1000}\).
This is a number with 1001 digits.
Estu: \(10^{1000000}\times9000 = 9\times10^{3}\times10^{1000000} = 9\times10^{1000003}\).
This is a number with 1000004 digits.
Estu’s number has many more digits, so it is much greater.
AActivity — Below are some conditions that you may consider for different rounds of “Tremendous in Ten!”.
(i) Exponents are not allowed. Only addition is allowed.
(ii) Exponents are not allowed. Only addition and multiplication are allowed.
(iii) Exponents are allowed. Only addition is allowed.
(iv) Exponents are allowed. Any arithmetic operation is allowed.
You can create your own conditions and/or involve more people to play together.
This is a game, so there is no single answer. Here is one sample round for each condition.
(i) Only addition, no exponents
Roxie writes 9999999999, a number with 10 digits.
Estu writes \(99999+99999+99999 = 299997\), a number with 6 digits.
Roxie’s number is greater. Adding a few small numbers cannot match one long number.
(ii) Addition and multiplication, no exponents
Roxie writes 99999999, a number with 8 digits.
Estu writes \(9999\times9999 = 99980001\), also a number with 8 digits.
Compare digit by digit: \(99999999 > 99980001\).
Roxie’s number is greater. Multiplying does not gain digits, because two 4-digit numbers give at most 8 digits.
(iii) Addition, with exponents
Roxie writes \(9^{99999}\).
\(9^{99999} > 9^{10} > 10^{9}\), and \(9^{99999}\) is a number with more than 95,000 digits.
Estu writes 99999999999 + 99999999999, a number with 11 digits.
Roxie’s number is greater. One exponent beats any number made of the same digits.
(iv) Any operation, with exponents
Roxie writes \(9^{(9^{9})}\). Here \(9^{9} = 387420489\), so the number is \(9^{387420489}\).
Estu writes \(9^{99999999}\).
Both have base 9, and \(387420489 > 99999999\), so \(9^{387420489} > 9^{99999999}\).
Roxie’s number is greater, and she wrote fewer digits.
Figure it Out — Set 2 (Pages 44–45)
Fourteen questions from pages 44–45 of the textbook.
1Find out the units digit in the value of 2224 ÷ 432? [Hint: 4 = 22]
\(4^{32} = (2^{2})^{32} = 2^{64}\) [power of a power]
\(2^{224} \div 4^{32} = 2^{224} \div 2^{64} = 2^{224-64} = 2^{160}\)
The units digits of powers of 2 repeat in a cycle of 4: 2, 4, 8, 6 (for \(2^{1}, 2^{2}, 2^{3}, 2^{4}\)).
\(160 \div 4 = 40\) with remainder 0, so \(2^{160}\) ends in the same digit as \(2^{4}\), which is 6.
2There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would be there after 40 days?
One container is brought in every day, and each has 5 bottles.
In 40 days, 40 containers are brought in.
Number of bottles \(= 40\times5 = 200\).
\(200 = 2\times10^{2}\)
This is linear growth, since the same number (5) is added every day.
3Write the given number as the product of two or more powers in three different ways. The powers can be any integers.
(i) 643
(ii) 1928
(iii) 32−5
(i) \(64^{3} = (2^{6})^{3} = 2^{18}\).
Split the exponent 18:
\(2^{18} = 2^{10}\times2^{8}\)
\(2^{18} = 4^{5}\times4^{4}\) [\(4^{9} = 2^{18}\)]
\(2^{18} = 8^{3}\times8^{3}\) [\(8^{6} = 2^{18}\)]
(ii) \(192 = 64\times3 = 2^{6}\times3\), so \(192^{8} = 2^{48}\times3^{8}\) [power of a product].
\(192^{8} = 2^{48}\times3^{8}\)
\(192^{8} = 2^{40}\times2^{8}\times3^{8}\)
\(192^{8} = 2^{40}\times(2\times3)^{8} = 2^{40}\times6^{8}\)
(iii) \(32^{-5} = (2^{5})^{-5} = 2^{-25}\).
Split the exponent \(-25\):
\(2^{-25} = 2^{-10}\times2^{-15}\)
\(2^{-25} = 2^{-5}\times2^{-20}\)
\(2^{-25} = 4^{-12}\times2^{-1}\) [\(4^{-12} = 2^{-24}\)]
(ii) \(2^{48}\times3^{8}\), \(2^{40}\times2^{8}\times3^{8}\), \(2^{40}\times6^{8}\)
(iii) \(2^{-10}\times2^{-15}\), \(2^{-5}\times2^{-20}\), \(4^{-12}\times2^{-1}\)
Other correct splits are possible.
4Examine each statement below and find out if it is ‘Always True’, ‘Only Sometimes True’, or ‘Never True’. Explain your reasoning.
(i) Cube numbers are also square numbers.
(ii) Fourth powers are also square numbers.
(iii) The fifth power of a number is divisible by the cube of that number.
(iv) The product of two cube numbers is a cube number.
(v) q46 is both a 4th power and a 6th power (q is a prime number).
(i) Only Sometimes True.
\(64 = 4^{3} = 8^{2}\) and \(729 = 9^{3} = 27^{2}\) are both cubes and squares.
But \(8 = 2^{3}\) and \(27 = 3^{3}\) are cubes that are not squares.
\(n^{3}\) is a square when \(n\) is a square.
If \(n = m^{2}\), then \(n^{3} = m^{6} = (m^{3})^{2}\).
(ii) Always True.
\(n^{4} = (n^{2})^{2}\), which is a square.
(iii) Always True.
\(\dfrac{n^{5}}{n^{3}} = n^{2}\), which is a whole number for a non-zero whole number \(n\).
So \(n^{3}\) divides \(n^{5}\).
(iv) Always True.
\(a^{3}\times b^{3} = (ab)^{3}\), which is a cube.
(v) Never True.
\(q\) is prime, so a number equal to \(q^{46}\) can be made only from \(q\)s.
So if \(q^{46}\) were a 4th power, it would be \((q^{k})^{4} = q^{4k}\) for some whole number \(k\).
This needs \(4k = 46\).
But \(46 \div 4\) is not a whole number.
In the same way, \(46 \div 6\) is not a whole number, so \(q^{46}\) is not a 6th power either.
(ii) Always True
(iii) Always True
(iv) Always True
(v) Never True
5Simplify and write these in the exponential form.
(i) 10−2 × 10−5
(ii) 57 ÷ 54
(iii) 9−7 ÷ 94
(iv) (13−2)−3
(v) m5n12(mn)9
(i) \(10^{-2}\times10^{-5} = 10^{-2+(-5)} = 10^{-7}\)
(ii) \(5^{7}\div5^{4} = 5^{7-4} = 5^{3}\)
(iii) \(9^{-7}\div9^{4} = 9^{-7-4} = 9^{-11}\)
(iv) \((13^{-2})^{-3} = 13^{(-2)\times(-3)} = 13^{6}\)
(v) \((mn)^{9} = m^{9}n^{9}\)
\(m^{5}n^{12}\times m^{9}n^{9} = m^{5+9}\times n^{12+9} = m^{14}n^{21}\)
(ii) \(5^{3}\)
(iii) \(9^{-11}\)
(iv) \(13^{6}\)
(v) \(m^{14}n^{21}\)
6If 122 = 144 what is
(i) (1.2)2
(ii) (0.12)2
(iii) (0.012)2
(iv) 1202
(i) \((1.2)^{2} = \left(\dfrac{12}{10}\right)^{2} = \dfrac{144}{100} = 1.44\)
(ii) \((0.12)^{2} = \left(\dfrac{12}{100}\right)^{2} = \dfrac{144}{10000} = 0.0144\)
(iii) \((0.012)^{2} = \left(\dfrac{12}{1000}\right)^{2} = \dfrac{144}{1000000} = 0.000144\)
(iv) \(120^{2} = (12\times10)^{2} = 144\times100 = 14400\)
(ii) 0.0144
(iii) 0.000144
(iv) 14400
7Circle the numbers that are the same —
24 × 36 64 × 32 610 182 × 62 624
Write each number using prime factors.
\(2^{4}\times3^{6}\) is already in prime-factor form.
\(6^{4}\times3^{2} = (2\times3)^{4}\times3^{2} = 2^{4}\times3^{4}\times3^{2} = 2^{4}\times3^{6}\)
\(6^{10} = 2^{10}\times3^{10}\)
\(18^{2}\times6^{2} = (2\times3^{2})^{2}\times(2\times3)^{2} = 2^{2}\times3^{4}\times2^{2}\times3^{2} = 2^{4}\times3^{6}\)
\(6^{24} = 2^{24}\times3^{24}\)
\(2^{4}\times3^{6}\), \(6^{4}\times3^{2}\) and \(18^{2}\times6^{2}\) have the same prime factors \(2^{4}\times3^{6}\).
8Identify the greater number in each of the following —
(i) 43 or 34
(ii) 28 or 82
(iii) 1002 or 2100
(i) \(4^{3} = 64\) and \(3^{4} = 81\).
Since \(81 > 64\), \(3^{4}\) is greater.
(ii) \(2^{8} = 256\) and \(8^{2} = 64\).
Since \(256 > 64\), \(2^{8}\) is greater.
(iii) \(100^{2} = 10^{4} = 10000\).
\(2^{100} = (2^{10})^{10} = 1024^{10}\).
Since \(1024 > 1000\), \(1024^{10} > 1000^{10} = 10^{30}\).
\(10^{30} > 10^{4}\), so \(2^{100}\) is greater.
(ii) \(2^{8}\)
(iii) \(2^{100}\)
9A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID (identifier) code for each packet. If they choose to use the digits 0–9, how many digits should the code consist of?
8.5 billion \(= 8.5\times10^{9} = 8,50,00,00,000\) packets.
An \(n\)-digit code using the digits 0–9 gives \(10^{n}\) different codes.
We need \(10^{n} \ge 8.5\times10^{9}\).
\(10^{9} = 1,00,00,00,000\), which is less than \(8.5\times10^{9}\).
So 9 digits are not enough.
\(10^{10} = 10\times10^{9}\), which is more than \(8.5\times10^{9}\).
So 10 digits are enough.
10Math Talk — 64 is a square number (82) and a cube number (43). Are there other numbers that are both squares and cubes? Is there a way to describe such numbers in general?
Yes.
Look at the prime factorisation of such a number.
A square has every prime with an even exponent.
A cube has every prime with an exponent divisible by 3.
So every exponent is divisible by both 2 and 3, that is, by 6.
Hence the number is \(n^{6} = (n^{3})^{2} = (n^{2})^{3}\), which is a square and a cube.
Examples: \(1^{6} = 1\), \(2^{6} = 64\), \(3^{6} = 729\), \(4^{6} = 4096\), \(5^{6} = 15625\).
11A digital locker has an alphanumeric (it can have both digits and letters) passcode of length 5. Some example codes are G89P0, 38098, BRJKW, and 003AZ. How many such codes are possible?
Each place can be a digit (10 choices) or a letter (26 choices).
Choices for each place \(= 10 + 26 = 36\).
Assume the letters are all capital letters.
There are 5 places, so the number of codes \(= 36\times36\times36\times36\times36 = 36^{5}\).
\(36^{2} = 1296\), \(36^{3} = 46656\), \(36^{4} = 1679616\), \(36^{5} = 60466176\).
12The worldwide population of sheep (2024) is about 109, and that of goats is also about the same. What is the total population of sheep and goats?
(i) 209
(ii) 1011
(iii) 1010
(iv) 1018
(v) 2 × 109
(vi) 109 + 109
Total \(= 10^{9} + 10^{9} = 2\times10^{9}\).
(v) \(2\times10^{9}\) is the same number, so (v) is correct.
(vi) \(10^{9}+10^{9}\) is the same sum, so (vi) is correct.
(i) \(20^{9} = 2^{9}\times10^{9} = 512\times10^{9}\), which is far too large.
(ii) \(10^{11} = 100\times10^{9}\), which is 100 times \(10^{9}\) and not 2 times.
(iii) \(10^{10} = 10\times10^{9}\), which is 10 times \(10^{9}\) and not 2 times.
(iv) \(10^{18} = 10^{9}\times10^{9}\), which is far too large.
Note: the textbook prints the label (ii) twice in this question. The first option, 209, is (i).
(vi) \(10^{9}+10^{9}\) (both are the same number).
13Calculate and write the answer in scientific notation:
(i) If each person in the world had 30 pieces of clothing, find the total number of pieces of clothing.
(ii) There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees.
(iii) The human body has about 38 trillion bacterial cells. Find the bacterial population residing in all humans in the world.
(iv) Total time spent eating in a lifetime in seconds.
(i) World population \(\approx 8.2\times10^{9}\) (from the chapter).
Pieces of clothing \(= 8.2\times10^{9}\times30 = 246\times10^{9} = 2.46\times10^{11}\).
(ii) 100 million \(= 10^{8}\) colonies, and \(50000 = 5\times10^{4}\) bees in each colony.
Bees \(= 10^{8}\times5\times10^{4} = 5\times10^{12}\).
(iii) 38 trillion \(= 38\times10^{12} = 3.8\times10^{13}\).
Bacteria \(= 3.8\times10^{13}\times8.2\times10^{9} = 31.16\times10^{22} = 3.116\times10^{23}\).
(iv) Assumptions: a lifetime of 70 years, and 1 hour of eating each day.
1 hour \(= 3600\) seconds.
Time \(= 70\times365\times3600 = 9,19,80,000 = 9.198\times10^{7}\) seconds.
(ii) \(5\times10^{12}\) bees
(iii) \(3.116\times10^{23}\) bacterial cells
(iv) \(9.198\times10^{7}\) seconds (assuming 70 years and 1 hour a day)
14Try This — What was the date 1 arab / 1 billion seconds ago?
1 billion seconds \(= 10^{9}\) seconds.
Days \(= \dfrac{10^{9}}{24\times60\times60} = \dfrac{10^{9}}{86400} \approx 11574\) days.
In years, \(11574 \div 365.25 \approx 31.7\) years.
Count back 11574 days from today.
Take today as 9 October 2026.
31 years back from 9 October 2026 is 9 October 1995.
This covers \(31\times365 + 8 = 11323\) days [8 leap days: 1996, 2000, 2004, 2008, 2012, 2016, 2020, 2024].
Remaining days \(= 11574 - 11323 = 251\).
251 days before 9 October 1995 is 31 January 1995.
Continue with Chapter 3
Move on to the Story of Numbers next.