Large Numbers Around Us

Estu has tasted only 3 varieties of rice so far. If you tried a new variety of rice each day, would you even come close to tasting all the (about one lakh) varieties in a lifetime of 100 years? What do you think? Guess.

Answer

A year has about 365 days (ignoring leap years).

Number of days in 100 years = 365 × 100

= 36,500 days

If one new variety of rice is tasted each day, then only 36,500 varieties can be tasted in 100 years.

Since 36,500 < 1,00,000 (one lakh), this is much less than the total number of rice varieties.

Hence, a person would not come close to tasting all one lakh varieties of rice in a lifetime of 100 years.

Observe the pattern and fill in the boxes given below.

• The largest 3-digit number is 999
• The smallest 4-digit number is ............... (+1)
• The largest 4-digit number is ...............
• The smallest 5-digit number is ............... (+1)
• The largest 5-digit number is ...............
• The smallest 6-digit number is 1,00,000 (+1)

(1,00,000 is read as "One Lakh")

Answer

Adding 1 to the largest number with a certain number of digits gives the smallest number with one more digit. Using this pattern:

• The largest 3-digit number is 999
• The smallest 4-digit number is 1,000 (999 + 1)
• The largest 4-digit number is 9,999
• The smallest 5-digit number is 10,000 (9,999 + 1)
• The largest 5-digit number is 99,999
• The smallest 6-digit number is 1,00,000 (99,999 + 1)

Fill in the missing numbers in the sequence below:

99,995 → 99,996 → ............... → 99,998 → ............... → ............... → ............... → ............... → ...............

Answer

Filling the sequence by counting on by 1 each time:

99,995 → 99,996 → 99,997 → 99,998 → 99,999 → 1,00,000 → 1,00,001 → 1,00,002 → 1,00,003

What if a person ate 3 varieties of rice every day? Will they be able to taste all the lakh varieties in a 100 year lifetime? Find out.

Answer

The number of days in a lifetime of 100 years (ignoring leap years) is:

365 × 100

= 36,500 days

If a person eats 3 varieties of rice every day, then the number of varieties tasted in 100 years would be:

3 × 36,500

= 1,09,500 varieties

Since 1,09,500 > 1,00,000 (one lakh), the person can taste all one lakh varieties within a 100-year lifetime.

Hence, yes, a person eating 3 varieties of rice every day would be able to taste all one lakh varieties in 100 years.

According to the 2011 Census, the population of the town of Chintamani was about 75,000. How much less than one lakh is 75,000?

Answer

One lakh = 1,00,000.

To find how much less 75,000 is than one lakh, we subtract:

1,00,000 - 75,000

= 25,000

Hence, 75,000 is 25,000 less than one lakh.

The estimated population of Chintamani in the year 2024 is 1,06,000. How much more than one lakh is 1,06,000?

Answer

To find how much more 1,06,000 is than one lakh, we subtract:

1,06,000 - 1,00,000

= 6,000

Hence, 1,06,000 is 6,000 more than one lakh.

By how much did the population of Chintamani increase from 2011 to 2024?

Answer

Population in 2011 = 75,000

Population in 2024 = 1,06,000.

Increase in population = Population in 2024 - Population in 2011

= 1,06,000 - 75,000

= 31,000

Hence, the population of Chintamani increased by 31,000 from 2011 to 2024.

• The world's tallest statue is the 'Statue of Unity' in Gujarat depicting Sardar Vallabhbhai Patel. Its height is about 180 metres.
• Kunchikal waterfall in Karnataka is said to drop from a height of about 450 metres.

(i) Somu is 1 metre tall. If each floor is about four times his height, what is the approximate height of the building?

(ii) Which is taller — The Statue of Unity or this building? How much taller? ............... m.

(iii) How much taller is the Kunchikal waterfall than Somu's building? ............... m.

(iv) How many floors should Somu's building have to be as high as the waterfall? ............... .

Answer

(i)

Somu is 1 metre tall, and each floor is about four times his height:

Height of one floor = 4 × 1 = 4 m

From the picture, Somu's building has about 10 floors. So its approximate height is:

10 × 4

= 40 m

Approximate height of the building = 40 m.

(ii)

Height of Statue of Unity = 180 m

Height of building = 40 m

∴ Statue of Unity is taller by:

180 - 40

= 140 m

The Statue of Unity is taller by about 140 m.

(iii)

Height of waterfall = 450 m

Height of building = 40 m

The waterfall is taller by:

450 - 40

= 410 m

The Kunchikal waterfall is taller than Somu's building by about 410 m.

(iv)

Height of each floor = 4 m

Height of waterfall = 450 m

To be as high as the waterfall, the number of floors needed = $\dfrac{\text{Height of waterfall}}{\text{Height of each floor}}$

= $\dfrac{450}{4}$

= 112.5

Since the number of floors must be a whole number, the building should have about 113 floors to be as high as the waterfall.

Hence, the building would need about 113 floors to be as high as the waterfall.

Write each of the numbers given below in words:
(a) 3,00,600
(b) 5,04,085
(c) 27,30,000
(d) 70,53,138

Answer

Using the Indian place value system (grouped as crores–lakhs–thousands–hundreds):

(a) 3,00,600 — Three lakh six hundred

(b) 5,04,085 — Five lakh four thousand eighty five

(c) 27,30,000 — Twenty seven lakh thirty thousand

(d) 70,53,138 — Seventy lakh fifty three thousand one hundred thirty eight

Write the corresponding number in the Indian place value system for each of the following:

(a) One lakh twenty three thousand four hundred and fifty six
(b) Four lakh seven thousand seven hundred and four
(c) Fifty lakhs five thousand and fifty
(d) Ten lakhs two hundred and thirty five

Answer

Writing each number in the Indian place value system (with commas in the 3-2-2 pattern from the right):

(a) One lakh twenty three thousand four hundred and fifty six = 1,23,456

(b) Four lakh seven thousand seven hundred and four = 4,07,704

(c) Fifty lakhs five thousand and fifty = 50,05,050

(d) Ten lakhs two hundred and thirty five = 10,00,235

In the Land of Tens, there are special calculators with special buttons.

The Thoughtful Thousands only has a +1000 button. How many times should it be pressed to show:
(a) Three thousand? 3 times
(b) 10,000? ...............
(c) Fifty three thousand? ...............
(d) 90,000? ...............
(e) One Lakh? ...............
(f) ...............? 153 times
(g) How many thousands are required to make one lakh?

Answer

Each press of the button adds 1000.

So the number of presses needed is the required number divided by 1000.

(a) Three thousand = $\dfrac{3,000}{1,000} = 3$ times. (given)

(b) $\dfrac{10,000}{1,000} = 10$ times.

(c) Fifty three thousand = $\dfrac{53,000}{1,000} = 53$ times.

(d) $\dfrac{90,000}{1,000} = 90$ times.

(e) One lakh = $\dfrac{1,00,000}{1,000} = 100$ times.

(f) Pressing the button 153 times shows 153 × 1000 = 1,53,000 (One lakh fifty three thousand).

(g) Since one lakh needs 100 presses of +1000, 100 thousands are required to make one lakh.

The Tedious Tens only has a +10 button. How many times should it be pressed to show:
(a) Five hundred? ...............
(b) 780? ...............
(c) 1000? ...............
(d) 3700? ...............
(e) 10,000? ...............
(f) One lakh? ...............
(g) ...............? 435 times

Answer

Each press adds 10.

So the number of presses is the required number divided by 10.

(a) Five hundred = $\dfrac{500}{10} = 50$ times.

(b) $\dfrac{780}{10} = 78$ times.

(c) $\dfrac{1,000}{10} = 100$ times.

(d) $\dfrac{3,700}{10} = 370$ times.

(e) $\dfrac{10,000}{10} = 1,000$ times.

(f) One lakh = $\dfrac{1,00,000}{10} = 10,000$ times.

(g) Pressing the button 435 times shows 435 × 10 = 4,350.

The Handy Hundreds only has a +100 button. How many times should it be pressed to show:
(a) Four hundred? ............... times
(b) 3,700? ...............
(c) 10,000? ...............
(d) Fifty three thousand? ...............
(e) 90,000? ...............
(f) 97,600? ...............
(g) 1,00,000? ...............
(h) ...............? 582 times
(i) How many hundreds are required to make ten thousand?
(j) How many hundreds are required to make one lakh?
(k) Handy Hundreds says, "There are some numbers which Tedious Tens and Thoughtful Thousands can't show but I can." Is this statement true? Think and explore.

Answer

Each press adds 100.

So the number of presses is the required number divided by 100.

(a) Four hundred $= \dfrac{400}{100} = 4$ times.

(b) $\dfrac{3,700}{100} = 37$ times.

(c) $\dfrac{10,000}{100} = 100$ times.

(d) Fifty three thousand $= \dfrac{53,000}{100} = 530$ times.

(e) $\dfrac{90,000}{100} = 900$ times.

(f) $\dfrac{97,600}{100} = 976$ times.

(g) $\dfrac{1,00,000}{100} = 1,000$ times.

(h) Pressing the button 582 times shows 582 × 100 = 58,200.

(i) Ten thousand = $\dfrac{10,000}{100} = 100$,

∴ 100 hundreds are required to make ten thousand.

(j) One lakh = $\dfrac{1,00,000}{100} = 1,000$,

∴ 1,000 hundreds are required to make one lakh.

(k) The statement is not true. Every number shown by Handy Hundreds is a multiple of 100, and every multiple of 100 can also be shown by Tedious Tens. Therefore, there is no number that Handy Hundreds can show but Tedious Tens cannot.

Creative Chitti is a different kind of calculator. It has the following buttons: +1, +10, +100, +1000, +10000, +100000 and +1000000. It always has multiple ways of doing things. "How so?", you might ask. To get the number 321, it presses +10 thirty two times and +1 once. Will it get 321? Alternatively, it can press +100 two times and +10 twelve times and +1 once.

Answer

Checking the first way — pressing +10 thirty two times and +1 once:

(32 × 10) + (1 × 1)

= 320 + 1

= 321

This gives 321.

Checking the alternative way — pressing +100 two times, +10 twelve times and +1 once:

(2 × 100) + (12 × 10) + (1 × 1)

= 200 + 120 + 1

= 321

This also gives 321.

So the same number can be reached in more than one way.

Hence, yes, Creative Chitti gets 321 by both methods, showing that a number can be made in many different ways.

Two of the many different ways to get 5072 are shown below:

These two ways can be expressed as:
(a) (50 × 100) + (7 × 10) + (2 × 1) = 5072
(b) (3 × 1000) + (20 × 100) + (72 × 1) = 5072

Find a different way to get 5072 and write an expression for the same.

Answer

One more way is to use the place value of each digit of 5072 — that is,

press +1000 five times, +10 seven times and +1 two times:

(5 × 1000) + (7 × 10) + (2 × 1)

= 5000 + 70 + 2

= 5072

Hence, one different way is (5 × 1000) + (7 × 10) + (2 × 1) = 5072.

For each number given below, write expressions for at least two different ways to obtain the number through button clicks. Think like Chitti and be creative.
(a) 8300
(b) 40629
(c) 56354
(d) 66666
(e) 367813

Answer

For each number, two possible ways are shown below (many others are also correct).

(a) 8300

Way 1:

(8 × 1000) + (3 × 100)

= 8000 + 300

= 8300

Way 2:

83 × 100

= 8300

(b) 40629

Way 1:

(4 × 10,000) + (6 × 100) + (2 × 10) + (9 × 1)

= 40,000 + 600 + 20 + 9

= 40,629

Way 2:

(40 × 1000) + (6 × 100) + (29 × 1)

= 40,000 + 600 + 29

= 40,629

(c) 56354

Way 1:

(5 × 10,000) + (6 × 1000) + (3 × 100) + (5 × 10) + (4 × 1)

= 50,000 + 6000 + 300 + 50 + 4

= 56,354

Way 2:

(56 × 1000) + (35 × 10) + (4 × 1)

= 56,000 + 350 + 4

= 56,354

(d) 66666

Way 1:

(6 × 10,000) + (6 × 1000) + (6 × 100) + (6 × 10) + (6 × 1)

= 60,000 + 6000 + 600 + 60 + 6

= 66,666

Way 2:

(66 × 1000) + (66 × 10) + (6 × 1)

= 66,000 + 660 + 6

= 66,666

(e) 367813

Way 1:

(3 × 1,00,000) + (6 × 10,000) + (7 × 1000) + (8 × 100) + (1 × 10) + (3 × 1)

= 3,00,000 + 60,000 + 7000 + 800 + 10 + 3

= 3,67,813

Way 2:

(36 × 10,000) + (7 × 1000) + (813 × 1)

= 3,60,000 + 7000 + 813

= 3,67,813

Hence, each number can be obtained in more than one way through button clicks, as shown above.

Creative Chitti has some questions for you —
(a) You have to make exactly 30 button presses. What is the largest 3-digit number you can make? What is the smallest 3-digit number you can make?
(b) 997 can be made using 25 clicks. Can you make 997 with a different number of clicks?

Answer

(a) If all 30 presses were of the +1 button, the number obtained would be 30.

Replacing one +1 press by a +10 press increases the number by 9, while replacing one +1 press by a +100 press increases it by 99. Thus, every possible total differs from 30 by a multiple of 9.

The largest 3-digit value of this form is 993. It can be made as follows:

(8 × 100) + (19 × 10) + (3 × 1)

= 800 + 190 + 3

= 993

Number of presses = 8 + 19 + 3 = 30

Hence, the largest 3-digit number is 993.

The smallest 3-digit value of this form is 102. It can be made as follows:

(8 × 10) + (22 × 1)

= 80 + 22

= 102

Number of presses = 8 + 22 = 30

Hence, the smallest 3-digit number is 102.

(b)

Yes. One way to make 997 is:

(8 × 100) + (19 × 10) + (7 × 1) = 800 + 190 + 7 = 997

8 + 19 + 7 = 34 clicks

So yes, 997 can be made with a different number of clicks (such as 34, 43, … clicks).

Hence, 997 can be made with a different number of clicks, such as 34 clicks.

Systematic Sippy is a different kind of calculator. It has the following buttons: +1, +10, +100, +1000, +10000, +100000. It wants to be used as minimally as possible.

How can we get the numbers
(a) 5072,
(b) 8300
using as few button clicks as possible?

Here is one way to get the number 5072. This method uses 23 button clicks in total.

Answer

To use the fewest button clicks, we use the largest place-value buttons possible.

(a) For 5072:

(5 × 1000) + (7 × 10) + (2 × 1)

= 5000 + 70 + 2

= 5072

This uses 5 + 7 + 2 = 14 clicks, which is less than 23 clicks.

Hence, 5072 can be obtained in 14 clicks.

(b) For 8300:

(8 × 1000) + (3 × 100)

= 8000 + 300

= 8300

This uses 8 + 3 = 11 clicks.

Hence, 8300 can be obtained in 11 clicks.

For the numbers in the previous exercise, find out how to get each number by making the smallest number of button clicks and write the expression.

Answer

The smallest number of clicks is obtained by pressing each place-value button as many times as the corresponding digit of the number.

(a) 8300 = (8 × 1000) + (3 × 100)

smallest number of clicks = 8 + 3 = 11.

(b) 40629 = (4 × 10000) + (6 × 100) + (2 × 10) + (9 × 1)

smallest number of clicks = 4 + 0 + 6 + 2 + 9 = 21.

(c) 56354 = (5 × 10000) + (6 × 1000) + (3 × 100) + (5 × 10) + (4 × 1)

smallest number of clicks = 5 + 6 + 3 + 5 + 4 = 23.

(d) 66666 = (6 × 10000) + (6 × 1000) + (6 × 100) + (6 × 10) + (6 × 1)

smallest number of clicks = 6 + 6 + 6 + 6 + 6 = 30.

(e) 367813 = (3 × 100000) + (6 × 10000) + (7 × 1000) + (8 × 100) + (1 × 10) + (3 × 1)

smallest number of clicks = 3 + 6 + 7 + 8 + 1 + 3 = 28.

Do you see any connection between each number and the corresponding smallest number of button clicks?

Answer

Yes. In the least-click method, each place-value button is pressed as many times as the digit in that place. Therefore, the total number of clicks is equal to the sum of the digits of the number.

For example,

56354 = (5 × 10000) + (6 × 1000) + (3 × 100) + (5 × 10) + (4 × 1)

So, the number of clicks = 5 + 6 + 3 + 5 + 4 = 23,

which is the sum of its digits.

Hence, the smallest number of button clicks required for a number is equal to the sum of its digits.

If you notice, the expressions for the least button clicks also give the Indian place value notation of the numbers. Think about why this is so.

Answer

To use the least number of clicks, each place-value button (+1, +10, +100, +1000, …) is pressed as many times as the digit in that place. Therefore, the expression becomes:

(digit × its place value) + (digit × its place value) + ...

This is precisely the expanded form of the number.

For example:

40629 = (4 × 10,000) + (0 × 1000) + (6 × 100) + (2 × 10) + (9 × 1)

Hence, the least-clicks expression is the same as the place value expansion because each digit is taken exactly the number of times equal to its value at its own place.

How many zeros does a thousand lakh have? ...............

Answer

A thousand lakh means 1000 × 1,00,000.

1000 × 1,00,000

= 10,00,00,000

This number is 1 followed by 8 zeros (it is the same as ten crore).

Hence, a thousand lakh has 8 zeros.

How many zeros does a hundred thousand have? ...............

Answer

A hundred thousand means 100 × 1000.

100 × 1000

= 1,00,000

This number is 1 followed by 5 zeros (it is the same as one lakh).

Hence, a hundred thousand has 5 zeros.

Read the following numbers in Indian place value notation and write their number names in both the Indian and American systems:
(a) 4050678
(b) 48121620
(c) 20022002
(d) 246813579
(e) 345000543
(f) 1020304050

Answer

Placing commas in the Indian system (3-2-2 pattern from the right) and in the American system (3-3-3 pattern from the right), we get:

(a) 4050678

• Indian: 40,50,678 — Forty lakh fifty thousand six hundred seventy eight
• American: 4,050,678 — Four million fifty thousand six hundred seventy eight

(b) 48121620

• Indian: 4,81,21,620 — Four crore eighty one lakh twenty one thousand six hundred twenty
• American: 48,121,620 — Forty eight million one hundred twenty one thousand six hundred twenty

(c) 20022002

• Indian: 2,00,22,002 — Two crore twenty two thousand two
• American: 20,022,002 — Twenty million twenty two thousand two

(d) 246813579

• Indian: 24,68,13,579 — Twenty four crore sixty eight lakh thirteen thousand five hundred seventy nine
• American: 246,813,579 — Two hundred forty six million eight hundred thirteen thousand five hundred seventy nine

(e) 345000543

• Indian: 34,50,00,543 — Thirty four crore fifty lakh five hundred forty three
• American: 345,000,543 — Three hundred forty five million five hundred forty three

(f) 1020304050

• Indian: 1,02,03,04,050 — One arab two crore three lakh four thousand fifty
• American: 1,020,304,050 — One billion twenty million three hundred four thousand fifty

Write the following numbers in Indian place value notation:
(a) One crore one lakh one thousand ten
(b) One billion one million one thousand one
(c) Ten crore twenty lakh thirty thousand forty
(d) Nine billion eighty million seven hundred thousand six hundred

Answer

Building each number place by place (and recalling that 1 million = 10 lakh and 1 billion = 100 crore):

(a) One crore one lakh one thousand ten

1,00,00,000 + 1,00,000 + 1,000 + 10

= 1,01,01,010

(b) One billion one million one thousand one

1,00,00,00,000 + 10,00,000 + 1,000 + 1

= 1,00,10,01,001

(c) Ten crore twenty lakh thirty thousand forty

10,00,00,000 + 20,00,000 + 30,000 + 40

= 10,20,30,040

(d) Nine billion eighty million seven hundred thousand six hundred

9,00,00,00,000 + 8,00,00,000 + 7,00,000 + 600 = 9,08,07,00,600

Compare and write '<', '>' or '=':
(a) 30 thousand ............... 3 lakhs
(b) 500 lakhs ............... 5 million
(c) 800 thousand ............... 8 million
(d) 640 crore ............... 60 billion

Answer

Converting both sides to ordinary numbers and comparing:

(a)

30 thousand = 30,000 and 3 lakhs = 3,00,000.

Since 30,000 < 3,00,000

∴ 30 thousand < 3 lakhs.

(b)

500 lakhs = 5,00,00,000 and 5 million = 50,00,000.

Since 5,00,00,000 > 50,00,000

∴ 500 lakhs > 5 million.

(c)

800 thousand = 8,00,000 and 8 million = 80,00,000.

Since 8,00,000 < 80,00,000

∴ 800 thousand < 8 million.

(d)

640 crore = 6,40,00,00,000 and 60 billion = 60,00,00,00,000 (which is 6000 crore).

Since 6,40,00,00,000 < 60,00,00,00,000,

∴ 640 crore < 60 billion.

Think and share situations where it is appropriate to (a) round up, (b) round down, (c) either rounding up or rounding down is okay and (d) when exact numbers are needed.

Answer

Different situations call for different ways of approximating. Some illustrative examples are:

(a) Round up — When we should not fall short of what is needed. For example, while buying sweets for a school function, we may round the number up and buy a few extra sweets.

(b) Round down — when a smaller, safe estimate is useful. For example, while judging how much time is left before leaving to catch the school bus, it is better to assume slightly less time than is actually available, so we round the time down and stay punctual.

(c) Either rounding up or down is okay — when only a rough sense of size matters. For example, while estimating the distance between two far-off places, saying "about 450 km" instead of an exact figure is good enough, and rounding either way is acceptable.

(d) Exact numbers are needed — When accuracy is important. For example, while counting votes, recording marks, or handling money in a bank, exact numbers are required.

With large numbers it is useful to know the nearest thousand, lakh or crore. For example, the nearest neighbours of the number 6,72,85,183 are shown in the table below.

Similarly, write the five nearest neighbours for these numbers:
(a) 3,87,69,957
(b) 29,05,32,481

Answer

For each level, we keep the digits above that place and round off using the digit just below it.

(a) 3,87,69,957

(b) 29,05,32,481

Hence, the five nearest neighbours of each number are as listed in the tables above.

I have a number for which all five nearest neighbours are 5,00,00,000. What could the number be? How many such numbers are there?

Answer

For all five nearest neighbours to be 5,00,00,000, the number must round to 5,00,00,000 even when rounded to the nearest thousand.

A number rounds to 5,00,00,000 to the nearest thousand if it lies between:

4,99,99,500 to 5,00,00,499

For example:

• 4,99,99,750
• 5,00,00,000
• 5,00,00,321

all have:

• Nearest thousand = 5,00,00,000
• Nearest ten thousand = 5,00,00,000
• Nearest lakh = 5,00,00,000
• Nearest ten lakh = 5,00,00,000
• Nearest crore = 5,00,00,000

The number of whole numbers in this range is:

5,00,00,499 − 4,99,99,500 + 1

= 1,000

Hence, the number can be any whole number from 4,99,99,500 to 5,00,00,499, and there are 1000 such numbers.

Roxie and Estu are estimating the values of simple expressions.

4,63,128 + 4,19,682
Roxie: "The sum is near 8,00,000 and is more than 8,00,000."
Estu: "The sum is near 9,00,000 and is less than 9,00,000."

(a) Are these estimates correct? Whose estimate is closer to the sum?
(b) Will the sum be greater than 8,50,000 or less than 8,50,000? Why do you think so?
(c) Will the sum be greater than 8,83,128 or less than 8,83,128? Why do you think so?
(d) Exact value of 4,63,128 + 4,19,682 = ...............

Answer

The exact sum is:

4,63,128 + 4,19,682 = 8,82,810

Distance from 8,00,000 = 8,82,810 − 8,00,000 = 82,810

Distance from 9,00,000 = 9,00,000 − 8,82,810 = 17,190

Since 17,190 < 82,810, the sum is much closer to 9,00,000.

(a)

Both estimates are correct. The sum is more than 8,00,000 and less than 9,00,000. The exact sum is closer to 9,00,000.

Hence, Estu's estimate is closer.

(b)

The sum is greater than 8,50,000. Because,

4,63,128 > 4,50,000 and 4,19,682 > 4,00,000.

So their sum is more than 4,50,000 + 4,00,000 = 8,50,000.

Hence, the sum is greater than 8,50,000.

(c)

The sum is less than 8,83,128. Because,

4,19,682 < 4,20,000

4,63,128 + 4,19,682 < 8,83,128

Hence, the sum is less than 8,83,128.

(d) Adding exactly:

4,63,128 + 4,19,682

= 8,82,810

Hence, the exact value is 8,82,810.

14,63,128 – 4,90,020
Roxie: "The difference is near 10,00,000 and is less than 10,00,000."
Estu: "The difference is near 9,00,000 and is more than 9,00,000."

(a) Are these estimates correct? Whose estimate is closer to the difference?
(b) Will the difference be greater than 9,50,000 or less than 9,50,000? Why do you think so?
(c) Will the difference be greater than 9,63,128 or less than 9,63,128? Why do you think so?
(d) Exact value of 14,63,128 – 4,90,020 = ...............

Answer

The exact difference is:

14,63,128 − 4,90,020 = 9,73,108

Distance from 10,00,000 = 10,00,000 − 9,73,108 = 26,892

Distance from 9,00,000 = 9,73,108 − 9,00,000 = 73,108

Since 26,892 < 73,108, the difference is much closer to 10,00,000.

(a)

Both estimates are correct. The difference is more than 9,00,000 and less than 10,00,000. The exact difference is closer to 10,00,000.

Hence, Roxie's estimate is closer.

(b)

The difference is greater than 9,50,000. Because,

14,63,128 − 5,00,000 = 9,63,128 and 9,63,128 > 9,50,000.

Since 4,90,020 < 5,00,000, we are subtracting a smaller number, so the actual difference is even larger.

Hence, the difference is greater than 9,50,000.

(c)

The difference is greater than 9,63,128. Because,

14,63,128 − 5,00,000 = 9,63,128 and 4,90,020 < 5,00,000

Since we subtract less than 5,00,000, the actual difference is greater than 9,63,128.

Hence, the difference is greater than 9,63,128.

(d) Subtracting exactly:

14,63,128 − 4,90,020

= 9,73,108

Hence, the exact value is 9,73,108.

Observe the populations of some Indian cities in the table below.

From the information given in the table, answer the following questions by approximation:

1. What is your general observation about this data? Share it with the class.
2. What is an appropriate title for the above table?
3. How much is the population of Pune in 2011? Approximately, by how much has it increased compared to 2001?
4. Which city's population increased the most between 2001 and 2011?
5. Are there cities whose population has almost doubled? Which are they?
6. By what number should we multiply Patna's population to get a number/population close to that of Mumbai?

Answer

1.

Most cities show an increase in population from 2001 to 2011. Some cities, such as Bengaluru, Hyderabad, Surat and Vadodara, have grown very rapidly. Kolkata shows a slight decrease in population.

2.

An appropriate title is: "Population of Some Indian Cities in 2001 and 2011"

3.

Population of Pune in 2011 = 31,15,431

Population of Pune in 2001 = 25,38,473

The increase in population = Population of Pune in 2011 - Population of Pune in 2001

31 lakh − 25 lakh

= 6 lakh

Hence, Pune's population increased by about 6 lakh.

4.

Comparing the increase for each city, the largest rise is for Bengaluru:

84,25,970 − 43,01,326

= 41,24,644

The largest increase is for Bengaluru, by about 41 lakh.

5.

Yes. A city's population has almost doubled when the 2011 figure is close to twice the 2001 figure. They are:

• Bengaluru
• Hyderabad
• Surat
• Vadodara

6.

Patna's population in 2011 = 16,84,222

Mumbai's population in 2011 = 1,24,42,373.

The multiplying factor = $\dfrac{\text{Mumbai's population}}{\text{Patna's population}}$

= $\dfrac{1,24,42,373}{16,84,222} \approx 7.39$

We need to multiply Patna’s population by 7.39 to get a number close to the population of Mumbai.

Using the meaning of multiplication and division, can you explain why multiplying by 5 is the same as dividing by 2 and multiplying by 10?

Answer

Since 5 = $\dfrac{10}{2}$,

multiplying a number by 5 is the same as multiplying it by $\dfrac{10}{2}$.

$\therefore n × 5 = n × \dfrac{10}{2} \\[1em] = \dfrac{n \times 10}{2}$

This means we can first divide the number by 2 and then multiply by 10.

For example,

116 × 5

= $\dfrac{116}{2} \times 10$

= 58 × 10

= 580

Hence, multiplying a number by 5 is the same as dividing it by 2 and then multiplying by 10, because 5 = $\mathbf {\dfrac{10}{2}}$.

Find quick ways to calculate these products:
(a) 2 × 1768 × 50
(b) 72 × 125 [Hint: $125 = \dfrac{1000}{8}$]
(c) 125 × 40 × 8 × 25

Answer

The idea is to regroup the factors so that they form easy round numbers like 100 or 1000.

(a)

Pairing 2 and 50 first gives a round 100:

2 × 1768 × 50

= (2 × 50) × 1768

= 100 × 1768

= 1,76,800

Hence, the product is 1,76,800.

(b)

Using the hint $125 = \dfrac{1000}{8}$:

72 × 125

= $72 \times \dfrac{1000}{8}$

= $\dfrac{72}{8} \times 1000$

= 9 × 1000

= 9000

Hence, the product is 9000.

(c)

Pairing 125 × 8 and 40 × 25 gives two thousands:

125 × 40 × 8 × 25

= (125 × 8) × (40 × 25)

= 1000 × 1000

= 10,00,000

Hence, the product is 10,00,000.

Calculate these products quickly.
(a) 25 × 12 = ...............
(b) 25 × 240 = ...............
(c) 250 × 120 = ...............
(d) 2500 × 12 = ...............
(e) ............... × ............... = 120000000

Answer

Here we use $25 = \dfrac{100}{4}$, $250 = \dfrac{1000}{4}$ and $2500 = \dfrac{10,000}{4}$ to turn each multiplication into a simple division followed by multiplication by a round number.

(a) 25 × 12

$= \dfrac{100}{4} \times 12 \\[1em] = \dfrac{12}{4} \times 100 \\[1em] = 3 \times 100 \\[1em] = 300$

Hence, the product is 300.

(b) 25 × 240

$= \dfrac{100}{4} \times 240 \\[1em] = \dfrac{240}{4} \times 100 \\[1em] = 60 \times 100 \\[1em] = 6000$

Hence, the product is 6000.

(c) 250 × 120

$= \dfrac{1000}{4} \times 120 \\[1em] = \dfrac{120}{4} \times 1000 \\[1em] = 30 \times 1000 \\[1em] = 30,000$

Hence, the product is 30,000.

(d) 2500 × 12

$= \dfrac{10000}{4} \times 12 \\[1em] = \dfrac{12}{4} \times 10000 \\[1em] = 3 \times 10000 \\[1em] = 30,000$

Hence, the product is 30,000.

(e)

We need two numbers whose product is 12,00,00,000.

One convenient pair is:

1200 × 1,00,000

= 12,00,00,000

Hence, the product is 12,00,00,000.

In each of the following boxes, the multiplications produce interesting patterns. Evaluate them to find the pattern. Extend the multiplications based on the observed pattern.

Answer

Evaluating each multiplication and then continuing the pattern with one more line:

The product climbs up from 1 to the number of 1s being multiplied and then comes back down — a palindrome. With k ones in each factor, the digits run 1, 2, ..., k, ..., 2, 1

With k sixes in the first number, the product is (k-1) fours, then a 0, then (k-1) twos, and finally a 6.

With k threes in the first number, the product is k ones followed by k fives.

Each line is a square: 101², 102², 103², ... The next square continues the list as 104² = 10816.

Observe the number of digits in the two numbers being multiplied and their product in each case. Is there any connection between the numbers being multiplied and the number of digits in their product?

Answer

Observe 11 × 11 = 121

Here, a 2-digit number × a 2-digit number gives a 3-digit product.

111 × 111 = 12321

Here, a 3-digit number × a 3-digit number gives a 5-digit product.

101 × 101 = 10201

Here, a 3-digit number × a 3-digit number gives a 5-digit product.

From these examples, we see that if one number has m digits and the other has n digits, then their product has either (m + n - 1) digits or (m + n) digits.

Hence, the product of an m-digit number and an n-digit number always has (m + n − 1) digits or (m + n) digits.

(i) Roxie says that the product of two 2-digit numbers can only be a 3- or a 4-digit number. Is she correct?

(ii) Should we try all possible multiplications with 2-digit numbers to tell whether Roxie's claim is true? Or is there a better way to find out?

Answer

(i)

Yes, Roxie is correct.

The smallest 2-digit number is 10, so the smallest product is:

10 × 10 = 100 $\quad$ (a 3-digit number)

The largest 2-digit number is 99, so the largest product is:

99 × 99 = 9801 $\quad$ (a 4-digit number)

Every product of two 2-digit numbers lies between 100 and 9801. Therefore, the product can only have 3 digits or 4 digits.

Hence, Roxie is correct.

(ii)

No, we do not need to try all possible multiplications.

The smallest 2-digit number is 10 and the largest 2-digit number is 99.

10 × 10 = 100 $\quad$ (a 3-digit number)

99 × 99 = 9801 $\quad$ (a 4-digit number)

Every product of two 2-digit numbers lies between 100 and 9801. Therefore, every such product must have either 3 digits or 4 digits.

Hence, there is no need to try all possible multiplications. Checking the smallest and largest possible products is enough to confirm Roxie's claim.

Can multiplying a 3-digit number with another 3-digit number give a 4-digit number?

Answer

No.

The smallest 3-digit number is 100.

So the smallest possible product of two 3-digit numbers is:

100 × 100 = 10,000 $\quad$ (a 5-digit number)

Since even the smallest product of two 3-digit numbers has 5 digits, every product of two 3-digit numbers must have 5 digits or 6 digits.

Therefore, a 4-digit product is not possible.

Hence, multiplying a 3-digit number by another 3-digit number can never give a 4-digit number.

Can multiplying a 4-digit number with a 2-digit number give a 5-digit number?

Answer

Yes.

The smallest 4-digit number is 1000 and the smallest 2-digit number is 10.

So the smallest possible product is:

1000 × 10 = 10,000 $\quad$ (a 5-digit number)

Therefore, a 5-digit product is possible.

Also, the largest possible product is:

9999 × 99 = 9,89,901 $\quad$ (a 6-digit number)

So the product of a 4-digit number and a 2-digit number can have either 5 digits or 6 digits.

Hence, yes — multiplying a 4-digit number by a 2-digit number can give a 5-digit number.

Observe the multiplication statements below. Do you notice any patterns? See if this pattern extends for other numbers as well.

1-digit × 1-digit = 1-digit or 2-digit
2-digit × 1-digit = 2-digit or 3-digit
2-digit × 2-digit = 3-digit or 4-digit
3-digit × 3-digit = 5-digit or 6-digit
5-digit × 5-digit = ............... or ...............
8-digit × 3-digit = ............... or ...............
12-digit × 13-digit = ............... or ...............

Answer

The pattern observed is:

If one number has m digits and the other has n digits, then their product has either (m + n − 1) digits or (m + n) digits.

Applying the rule:

• 5-digit × 5-digit:

(m + n − 1) = 5 + 5 - 1 = 9

(m + n) = 5 + 5 = 10

Hence, 5-digit × 5-digit = 9-digit or 10-digit.

• 8-digit × 3-digit

(m + n − 1) = 8 + 3 - 1 = 10

(m + n) = 8 + 3 = 11

Hence, 8-digit × 3-digit = 10-digit or 11-digit.

• 12-digit × 13-digit

(m + n − 1) = 12 + 13 - 1 = 24

(m + n) = 12 + 13 = 25

Hence, 12-digit × 13-digit = 24-digit or 25-digit.

The RMS Titanic ship carried about 2500 passengers, the population of Mumbai is 1,24,42,373. Can the population of Mumbai fit into 5000 such ships?

Answer

Each ship carries about 2500 passengers.

So, 5000 such ships can carry:

5000 × 2500

= 1,25,00,000 passengers

Population of Mumbai = 1,24,42,373

Comparing the two numbers:

1,25,00,000 > 1,24,42,373

Therefore, 5000 ships can carry more people than the population of Mumbai.

Hence, yes, the population of Mumbai can fit into 5000 such ships.

Roxie wondered, "If I could travel 100 kilometers every day, could I reach the Moon in 10 years?" (The distance between the Earth and the Moon is 3,84,400 km.)
(i) How far would she have travelled in a year?
(ii) How far would she have travelled in 10 years?

Answer

(i)

Roxie travels 100 km every day.

A year has about 365 days.

So the distance in one year is:

100 × 365

= 36,500 km

Hence, Roxie would travel 36,500 km in a year.

(ii)

In 10 years, the distance travelled is:

36,500 × 10

= 3,65,000 km

Hence, Roxie would travel 3,65,000 km in 10 years.

The distance between the Earth and the Moon is 3,84,400 km.

Comparing:

3,65,000 < 3,84,400

She falls short of the Moon by:

3,84,400 - 3,65,000

= 19,400 km

Hence, Roxie would be 19,400 km short of the Moon. So she would not quite reach the Moon in 10 years.

Find out if you can reach the Sun in a lifetime, if you travel 1000 kilometers every day. (The distance between the Earth and the Sun = 2100 × 70,000 = 14,70,00,000 km).

Answer

Distance between the Earth and the Sun:

2100 × 70,000

= 14,70,00,000 km (about 14 crore 70 lakh km)

Taking a lifetime to be 100 years, the number of days is:

365 × 100

= 36,500 days

If a person travels 1000 km every day, then the distance travelled in a lifetime is:

1000 × 36,500

= 3,65,00,000 km (about 3 crore 65 lakh km)

Comparing this with the distance to the Sun:

3,65,00,000 < 14,70,00,000

So, the distance travelled in a lifetime is much less than the distance to the Sun.

In fact, the number of days needed to cover 14,70,00,000 km at 1000 km a day is:

$\dfrac{14,70,00,000}{1000}$

= 1,47,000 days

In years, this is:

$\dfrac{1,47,000}{365} \approx 402.7$ years

which is about 403 years — much more than a single lifetime.

Hence, no, you cannot reach the Sun in a lifetime even if you travel 1000 km every day.

Make necessary reasonable assumptions and answer the questions below:
(a) If a single sheet of paper weighs 5 grams, could you lift one lakh sheets of paper together at the same time?
(b) If 250 babies are born every minute across the world, will a million babies be born in a day?
(c) Can you count 1 million coins in a day? Assume you can count 1 coin every second.

Answer

(a)

Weight of one sheet of paper = 5 g

Weight of one lakh sheets:

1,00,000 × 5

= 5,00,000 grams

Converting to kilograms:

1 kg = 1000 g

5,00,000 g = $\dfrac{5,00,000}{1000}$

= 500 kg

A person cannot lift 500 kg at once.

Hence, you cannot lift one lakh sheets of paper together.

(b)

Number of hours in a day = 24

Number of minutes in a day:

24 × 60

= 1440 minutes

Number of babies born every minute = 250

Number of babies born in a day = Number of babies per minute × Number of minutes in a day

= 250 × 1440

= 3,60,000 babies

Comparing:

3,60,000 < 10,00,000,

Hence, a million babies will not be born in a day.

(c)

Number of coins counted per second = 1 coin

Number of seconds in a day:

24 × 60 × 60

= 86,400 seconds

∴ Number of coins counted in a day = (Number of coins counted per second) × (Number of seconds in a day)

= 86,400 × 1

= 86,400

Comparing:

86,400 < 10,00,000

Hence, you cannot count 1 million coins in a day.

Using all digits from 0 – 9 exactly once (the first digit cannot be 0) to create a 10-digit number, write the —
(a) Largest multiple of 5
(b) Smallest even number

Answer

(a)

A multiple of 5 must end in 0 or 5.

To make the number as large as possible, the bigger digits should come first, so we arrange the digits in descending order:

9876543210

This is divisible by 5.

Hence, the largest multiple of 5 is 9,87,65,43,210.

(b)

An even number must end in 0, 2, 4, 6 or 8.

To make the number as small as possible, the smaller digits should come first. The first digit cannot be 0, so it must be 1, and the next digit should be 0.

Arranging the rest in increasing order gives 1, 0, 2, 3, 4, 5, 6, 7, 8, 9 — but this ends in 9, which is odd. So we swap the last two digits (8 and 9) to make the last digit even, keeping the number as small as possible:

1, 0, 2, 3, 4, 5, 6, 7, 9, 8

This gives 1,02,34,56,798, which ends in 8 and is therefore even.

Hence, the smallest even number is 1,02,34,56,798.

The number 10,30,285 in words is Ten lakhs thirty thousand two hundred eighty five, which has 42 letters. Give a 7-digit number name which has the maximum number of letters.

Answer

One such number is 77,77,777, which is read as:

Seventy seven lakhs seventy seven thousand seven hundred seventy seven

This number name has 61 letters when spaces are ignored.

Hence, 77,77,777 is a 7-digit number whose number name has the maximum number of letters.

Write a 9-digit number where exchanging any two digits results in a bigger number. How many such numbers exist?

Answer

For swapping any two digits to always produce a larger number, every digit must be smaller than all the digits to its right. Therefore, the digits must be in strictly increasing order from left to right.

A 9-digit number needs 9 distinct digits chosen from 0 – 9.

• If 0 is included, then it must be the first digit (because the digits are in increasing order), which is not allowed in a 9-digit number.
• Therefore, 0 must be excluded.

The only possible digits are 1, 2, 3, 4, 5, 6, 7, 8, 9

giving the number:

12,34,56,789

Checking it: in 123456789, every digit is smaller than all digits to its right, so swapping any two digits always pushes a larger digit leftward and makes the number bigger.

Hence, the required number is 12,34,56,789, and exactly one such number exists.

Strike out 10 digits from the number 12345123451234512345 so that the remaining number is as large as possible.

Answer

The number 12345123451234512345 has 20 digits. Striking out 10 leaves a 10-digit number, and we want it to be as large as possible.

So we keep the largest available digits as far left as possible while preserving their order.

After striking out 10 digits from 12345123451234512345

the largest 10-digit number that can be formed is 5534512345.

In Indian notation: 5,53,45,12,345

Hence, the required number is 5,53,45,12,345.

The words 'zero' and 'one' share letters 'e' and 'o'. The words 'one' and 'two' share a letter 'o', and the words 'two' and 'three' also share a letter 't'. How far do you have to count to find two consecutive numbers which do not share an English letter in common?

Answer

No such pair of consecutive numbers exists under the usual English number-naming system.

For numbers within the same ten, hundred, thousand or larger place-value block, the two consecutive number names retain a common word or common part. At a boundary, the two names still share at least one letter. For example:

• nine and ten share 'n' and 'e';
• nineteen and twenty share 't' and 'e';
• ninety nine and one hundred share 'n';
• nine hundred ninety nine and one thousand share several letters.

The same pattern continues at larger place-value boundaries.

Hence, no matter how far we count, two consecutive number names always share at least one English letter.

Suppose you write down all the numbers 1, 2, 3, 4, …, 9, 10, 11, ... The tenth digit you write is '1' and the eleventh digit is '0', as part of the number 10.
(a) What would the 1000th digit be? At which number would it occur?
(b) What number would contain the millionth digit?
(c) When would you have written the digit '5' for the 5000th time?

Answer

First, count how many digits are used by numbers of different lengths:

1-digit numbers (1 – 9):

9 × 1 = 9 digits

2-digit numbers (10 – 99):

90 × 2 = 180 digits

Total: 9 + 180 = 189

3-digit numbers (100 – 999):

900 × 3 = 2700 digits

Total: 189 + 2700 = 2889

(a)

The 1000th digit lies among the 3-digit numbers.

Digits already used up to 99 = 189

So within the 3-digit numbers, it is the

1000 − 189 = 811^th digit.

Since each 3-digit number contributes 3 digits,

810 ÷ 3 = 270

Thus, 270 complete numbers (100 to 369) use 810 digits.

The next digit is the first digit of 370.

Therefore, the 1000th digit is 3, and it occurs in 370.

(b)

Continue the digit count:

4-digit numbers:

9000 × 4 = 36,000

Total: 2889 + 36,000 = 38,889

5-digit numbers:

90,000 × 5 = 4,50,000

Total: 38,889 + 4,50,000 = 4,88,889

The millionth digit lies among the 6-digit numbers.

Digits needed after 4,88,889:

10,00,000 − 4,88,889 = 5,11,111

Each 6-digit number contributes 6 digits.

5,11,111 ÷ 6 = 85,185 remainder 1

So 85,185 complete 6-digit numbers use 5,11,110 digits.

The next digit is the first digit of the next number:

100000 + 85,185 = 185185

The millionth digit occurs in the number 1,85,185.

(c)

From 1 to 9999, the digit 5 occurs 4000 times.

From 10,000 to 12,999, it occurs 900 more times. Therefore, up to 12,999, it has occurred:

4000 + 900 = 4900 times

From 13,000 to 13,399, it occurs 80 more times:

4900 + 80 = 4980 times

From 13,400 to 13,494, it occurs 19 more times:

4980 + 19 = 4999 times

The number 13,495 contains one digit 5. Therefore, that digit is the 5000th occurrence.

Hence, the digit 5 is written for the 5000th time at 13,495.

A calculator has only '+10,000' and '+100' buttons. Write an expression describing the number of button clicks to be made for the following numbers:
(a) 20,800
(b) 92,100
(c) 1,20,500
(d) 65,30,000
(e) 70,25,700

Answer

Each press of '+10,000' adds ten thousand and each press of '+100' adds one hundred. So we write each number as a multiple of 10,000 plus a multiple of 100, using as many '+10,000' presses as possible. The total number of clicks is the sum of the two multipliers.

(a) 20,800

20,800 = (2 × 10,000) + (8 × 100)

Total clicks = 2 + 8 = 10

(b) 92,100

92,100 = (9 × 10,000) + (21 × 100)

Total clicks = 9 + 21 = 30

(c) 1,20,500

1,20,500 = (12 × 10,000) + (5 × 100)

Total clicks = 12 + 5 = 17

(d) 65,30,000

65,30,000 = (653 × 10,000)

Total clicks = 653 + 0 = 653

(e) 70,25,700

70,25,700 = (702 × 10,000) + (57 × 100)

Total clicks = 702 + 57 = 759

How many lakhs make a billion?

Answer

A billion is 1 followed by 9 zeroes:

1 billion = 1,00,00,00,000

In the Indian system:

1,000,000,000 = 1,00,00,00,000 $\quad$ [1 billion is the same as 100 crore]

Also, 1 crore = 100 lakh and 1 billion = 100 crore

∴ 1 billion = 100 × 100 lakh

= 10,000 lakh

We can also check this by division:

$\dfrac{1,00,00,00,000}{1,00,000} = 10,000$

Hence, 10,000 lakhs make a billion.

You are given two sets of number cards numbered from 1 – 9. Place a number card in each box below to get the (a) largest possible sum (b) smallest possible difference of the two resulting numbers.

Answer

From the figure,

• Top row has 7 boxes → a 7-digit number.
• Bottom row has 5 boxes, aligned under the last 5 digits of the top number → a 5-digit number.
• There are two sets of cards 1–9, so each digit can be used at most twice.

(a) Largest possible sum

To make the sum as large as possible, the largest digits should sit in the highest place values. The two leftmost places (the ten-lakhs and lakhs places) belong only to the 7-digit number, so the two 9s go there. Each of the remaining five places (ten-thousands down to ones) receives a digit from both numbers, so the next-largest digits are placed there in pairs:

7-digit number = 99,87,654 and 5-digit number = 87,654.

99,87,654 + 87,654

= 1,00,75,308

Hence, the largest possible sum is 1,00,75,308.

(b) Smallest possible difference

The 7-digit number is always greater than the 5-digit number, so the difference is (7-digit number) − (5-digit number). To make it as small as possible, the 7-digit number should be as small as possible (so the two 1s go in the two leftmost places), while in each shared place a small digit is placed on top and a large digit below:

7-digit number = 11,22,334 and 5-digit number = 99,887.

11,22,334 - 99,887

= 10,22,447

Hence, the smallest possible difference is 10,22,447.

You are given some number cards; 4000, 13000, 300, 70000, 150000, 20, 5. Using the cards get as close as you can to the numbers below using any operation you want. Each card can be used only once for making a particular number.
(a) 1,10,000: Closest I could make is 4000 × (20 + 5) + 13000 = 1,13,000
(b) 2,00,000:
(c) 5,80,000:
(d) 12,45,000:
(e) 20,90,800:

Answer

We combine the given cards using the operations +, −, × and ÷, using each card at most once for a particular number. One set of close (or exact) values is shown below.

(a) 1,10,000 — as given:

4000 × (20 + 5) + 13000

= 4000 × 25 + 13000

= 1,00,000 + 13,000

= 1,13,000

(b) 2,00,000:

70,000 × 5 − 1,50,000

= 3,50,000 − 1,50,000

= 2,00,000 (exact)

(c) 5,80,000:

$\phantom{=} 4000 \times \dfrac{300 + \dfrac{13000}{5}}{20} \\[1em] = 4000 \times \dfrac{300 + 2600}{20} \\[1em] = 4000 \times \dfrac{2900}{20} \\[1em] = 4000 \times 145 \\[1em] = 5,80,000 \text{ exact}$

(d) 12,45,000:

(13,000 + 70,000) × (20 − 5)

= 83,000 × 15

= 12,45,000 (exact)

(e) 20,90,800:

One very close value is:

$70,000 + \left(\dfrac{1,50,000 - 13,000}{20}\right)(300 - 5) \\[1em] = 70,000 + 6850 \times 295 \\[1em] = 70,000 + 20,20,750 \\[1em] = 20,90,750.$

This is only 50 less than 20,90,800.

Hence, using the given cards, the targets in (b), (c) and (d) can be obtained exactly. For (a), we get 1,13,000, which is 3,000 away from the target, and for (e), we get 20,90,750 which is a very close value.

Find out how many coins should be stacked to match the height of the Statue of Unity. Assume each coin is 1 mm thick.

Answer

The height of the Statue of Unity is about 180 m.

Converting to millimetres:

1 m = 1000 mm

180 m = 180 × 1000

= 1,80,000 mm

Each coin is 1 mm thick.

So the number of coins needed is:

$\dfrac{1,80,000}{1} = 1,80,000$

Hence, about 1,80,000 coins should be stacked to match the height of the Statue of Unity.

Grey-headed albatrosses have a roughly 7-feet wide wingspan. They are known to migrate across several oceans. Albatrosses can cover about 900 – 1000 km in a day. One of the longest single trips recorded is about 12,000 km. How many days would such a trip take to cross the Pacific Ocean approximately?

Answer

The albatross travels about 900–1000 km in a day.

The distance of the trip is about 12,000 km.

If it travels 1000 km per day:

$\dfrac{12,000}{1000} = 12 \text{ days}$

So, it would take about 12 days.

If it travels 900 km per day:

$\dfrac{12,000}{900} \approx 13.3 \text{ days}$

So, it would take about 13 to 14 days.

Hence, such a 12,000 km trip would take approximately 12 to 14 days.

A bar-tailed godwit holds the record for the longest recorded non-stop flight. It travelled 13,560 km from Alaska to Australia without stopping. Its journey started on 13 October 2022 and continued for about 11 days. Find out the approximate distance it covered every day. Find out the approximate distance it covered every hour.

Answer

The bar-tailed godwit travelled about 13,560 km in 11 days.

Distance covered every day:

$\dfrac{13,560}{11} \approx 1233 \text{ km}$

Each day has 24 hours, so the distance covered every hour:

$\dfrac{1233}{24} \approx 51 \text{ km}$

Hence, the bar-tailed godwit covered approximately 1233 km every day and about 51 km every hour.

Bald eagles are known to fly as high as 4500 – 6000 m above the ground level. Mount Everest is about 8850 m high. Aeroplanes can fly as high as 10,000 – 12,800 m. How many times bigger are these heights compared to Somu's building?

Answer

Somu's building is approximately 40 m high.

To find how many times greater a height is than Somu's building, divide the height by 40.

(i)

Bald eagles' flight height = 4500 – 6000 m

$\dfrac{4500}{40} = 112.5$

$\dfrac{6000}{40} = 150$

Hence, bald eagles fly about 112 to 150 times as high as Somu's building.

(ii)

Mount Everest's height = 8850 m

$\dfrac{8850}{40} = 221.25$

Hence, Mount Everest is about 221 times as tall as Somu's building.

(iii)

Aeroplanes' flight height = 10,000 – 12,800 m

$\dfrac{10{,}000}{40} = 250$

$\dfrac{12{,}800}{40} = 320$

Hence, aeroplanes fly about 250 to 320 times as high as Somu's building.