Practicing Ganita Prakash Class 7 Solutions and RBSE Class 7 Maths Chapter 1 Large Numbers Around Us Solutions Question Answer helps develop logical thinking and accuracy.
RBSE Class 7 Maths Chapter 1 Large Numbers Around Us Solutions
Ganita Prakash Class 7 Chapter 1 Solutions
Class 7 Maths Ganita Prakash Part 1 Chapter 1 Solutions
In-Text Questions
Page 4
Question 1.
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
Solution:
(a) Three lakh six hundred
(b) Five Iakh four thousand eighty five
(c) Twenty seven lakh thirty thousand
(d) Seventy lakh fifty three thousand one hundred thirty eight
Page 5
Question 1.
Write the corresponding number in the Indian place value system for each of the following:
(a) One Iakh twenty three thousand four hundred and fifty six
(b) Four lakh seven thousand seven hundred and four
(c) Fifty Iakhs five thousand and fifty
(d) Ten Iakhs two hundred and thirty five
Solution:
(a) 1,23,456 .
(b) 4,07,704
(e) 50,05,050
(d) 10,00,235
Pages 8-9
Question 1.
How many zeroes does a thousand lakh have’
Solution:
A thousand Iakh has 8 zeroes.
Question 2.
How many zeroes does a hundred thousand have?
Solution:
A hundred thousand has 5 zeroes.
Page 11
Question 1.
Write the five nearest neighbours for these numbers:
(a) 3,87,69,957
(b) 29,05,32,481
Solution:
(a)
| Nearest thousand | 3,87,70,000 |
| Nearest ten thousand | 3,87,70,000 |
| Nearest lakh | 3,88,00,000 |
| Nearest ten lakh | 3,90,00,000 |
| Nearest crore | 4,00,00,000 |
(b)
| Nearest thousand | 29,05,32,000 |
| Nearest ten thousand | 29,05,30,000 |
| Nearest lakh | 29,05,00,000 |
| Nearest ten lakh | 29,10,00,000 |
| Nearest crore | 36,00,00,000 |
Question 2.
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?
Solution:
The number could be any number within the range of 4,99,99,995 to 5,00,00,004. There are 10 such numbers.
Page 14 (Math Talk)
Question 1.
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?
Solution:
The explanation is given by the following examples:
(1) 110 × 5 = 110 × \(\frac {10}{2}\) [∵ 5 = \(\frac {10}{2}\)]
= 55 × 10 = 550
(2) 820 × 25 = 820 × \(\frac {100}{4}\) [∵ 25 = \(\frac {100}{4}\)]
= 205 × 100 = 20500
Class 7 Maths Ganita Prakash Chapter 1 Solutions
Figure it Out (Page 3)
Question 1.
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?
Solution:
One lakh = 1,00,000
So, 75,000 is less than one lakh by 25,000.
Question 2.
The estimated population of Chintamani in the year 2024 is 1,06,000. How much more than one lakh is 1,06,000?
Solution:
One lakh = 1,00,000
So, 1,06,000 is more than one lakh by 6,000.
Question 3.
By how much did the population of Chintamani increase from 2011 to 2024?
Solution:
Population of Chintamani in 2011 = 75,000
Population of Chintamani in 2024 = 1,06,000
So, the population of Chintamani increased from 2011 to 2024 by 31,000.
Figure it Out (Pages 6-7)
Question 1.
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
Solution:
(a) (8 × 1000)+ (3 × 100) = 8300
(83 × 100) = 8300
(5 × 1000)+ (33 × 100) = 8300
(b) (4 × 10000) + (6 × 100) + (2 × 10) + (9 × 1) = 40629
(40 × 1000) + (6 × 100) + (2 × 10) + (9 × 1) = 40629
(30 × 1000) + (100 × 100) + (6 × 100) + (2 × 10) + (9 × 1) = 40629
(c) (5 × 10000) + (6 × 1000) + (3 × 100) + (5 × 10) +(4 × l) = 56354
(5 × 10000) + (6 × 1000) + (2 × 100) + (15 × 10) +(4 × 1) = 56354
(5 × 10000) + (6× 1000) + (1 × 100) + (25 × 10) +(4 × 1) = 56354
(d) (6 × 10000) + (6 × 1000) + (6 × 100) + (6 × 10) +(6 × 1) = 66666
(5 × 10000) + (16 × 1000) + (6 × 100) + (6 × 10) +(6 × 1) = 66666
(4 × 10000) + (26 × 1000) + (5 × 100) + (16 × 10) +(6 × 1) = 66666
(e) (3 × 100000) + (6 × 10000) + (7 × 1000) + (8 × 100) + (1 × 10) + (3 × 1) = 367813
(3 × 100000) +(6 × 10000) + (7 × 1000) +(7 × 100)+ (11 × 10) + (3 × 1) = 367813
(3 × 100000) + (6 × 10000) + (5 × 1000)+ (28 × 100) + (1 × 10) + (3 × 1) = 367813
Question 2.
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.
Solution:
(i) 8300 = (8 × 1000) + (3 × 100)
(ii) 40629 = (4 × 10000) + (6 × 100) + (2× 10)+ (9× 1)
(iii) 56354 = (5 × 10000) + (6 × 1000) + (3 × 100) + (5 × 10) + (4 × 1)
(iv) 66666 = (6×l 0000) + (6 ×l 000) + (6 × 100) + (6 × 10) + (6 × 1)
(v) 367813 = (3 × 100000) + (6 × 10000) + (7 × 1000) + (8 × 100) + (1 × 10) + (3 × 1)
Question 2.
Do you see any connection between each number and the corresponding smallest number of button clicks?
Solution:
The expression for the least button clicks gives the Indian place value notation of the numbers.
Question 3.
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.
What if we press the +10,00,000 button ten times? What number will come up? How many zeroes will it have? What should we call it?
Solution:
This is on account of the place value of each digit in the given number.
If we press the +10,00,000 button ten times, the number 100 lakhs will come up. It will have 1 followed by seven zeros. We call it a crore.
Thus, 1 crore is written as 1,00,00,000.
Figure it Out (Page 9)
Question 1.
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
Solution:
(a)4050678
= 40,50,678 (in Indian place value notation)
= Forty lakh fifty thousand six hundred seventy eight (in Indian system)
= 4,050,678 (in American system)
= Four million fifty thousand six hundred seventy eight (in American system)
(b) 48121620
= 4,81,21,620 (in Indian place value notation)
= Four crore eighty one lakh twenty one thousand six hundred twenty (in Indian system) = 48,121,620 (in American system)
= Forty eight million one hundred twenty one thousand six hundred twenty (in American system)
(c) 20022002
= 2,00,22,002 (in Indian place value notation)
= Two crore twenty two thousand two (in Indian system)
= 20,022,002 (in American system)
= Twenty million twenty two thousand two (in American system)
(d) 246813579
= 24,68,13,579 (in Indian place value notation)
= Twenty four crore sixty eight lakh thirteen thousand five hundred seventy nine (in Indian system)
= 246,813,579 (in American system)
= Two hundred forty six million eight hundred thirteen thousand five hundred seventy nine (in American system) .
(e) 345000543
= 34,50,00,543 (in Indian place value notation)
= Thirty four crore fifty lakh five hundred forty three (in Indian system)
= 345,000,543 (in American system)
= Three hundred forty five million five hundred forty three (in American system)
(f) 1020304050
= 1,02,03,04,050 (in Indian place value notation)
= One arab two crore three lakhs four thousand fifty (in Indian system)
= 1,020,304,050 (inAmerican system) = One billion twenty million three hundred four thousand fifty (inAmerican system)
Question 2.
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
Solution:
(a) One crore one lakh one thousand ten = 1,01,01,010 (in Indian place value notation)
(b) One billion one million one thousand one = 1,001,001,001 (inAmerican system) = 1,00,10,01,001 (in Indian system notation)
(c) Ten crore twenty lakh thirty thousand forty = 10,20,30,040 (in Indian place value notation)
(d) Nine billion eighty million seven hundred thousand six hundred
= 9,080,700,600 (inAmerican system)
= 9,08,07,00,600 (in Indian place value notation)
Question 3.
Compare and write ‘>’ or‘=’:
(a) 30 thousand …………… 3 lakhs
(b) 500 lakhs ……………….. 5 million
(c) 800 thousand ……………….. 8 million
(d) 640 crore ………………… 60 billion
Solution:
(a) < (b) >
(c) <
(d) <
Figure it Out (Page 14)
Question 1.
Find quick ways to calculate these products:
(a) 2 × 1768 ×50
Solution:
2 × 1768 ×50
= 2 × 1768 × \(\frac {100}{2}\) [∵ 50 = \(\frac {100}{2}\)]
= 1768 × 100 = 176800
(b) 72 × 125 [Hint: 125 = \(\frac {1000}{8}\)]
Solution:
72 × 125 = 72 × \(\frac {1000}{8}\)
= 9 × 1000 = 9000
(c) 125 × 40 × 8 × 25
Solution:
125 × 40 × 8 × 25 [∵ 25 = \(\frac {100}{4}\)]
= 125 × 40 × 8 × \(\frac {100}{4}\)
= 125 × 40 × 200 = 125 × 8000
= \(\frac {1000}{8}\) × 8000 [∵ 125 = \(\frac {1000}{8}\)]
= 10,00,000
Question 2.
Calculate these products quickly.
(a) 25 × 12 = ……………..
Solution:
25 × 12 = \(\frac {100}{4}\) × 12 [∵ 25 = \(\frac {100}{4}\)]
= 100 × 3 = 300
(b) 25 × 240 = …………………
Solution:
25 × 240 = \(\frac {100}{4}\) × 240 [∵ 25 = \(\frac {100}{4}\)]
= 100 × 60 = 6,000
(c) 250 × 120 = ………………….
Solution:
250 × 120 = \(\frac {1000}{4}\) × 120 [∵ 250 = \(\frac {1000}{4}\)]
= 1000 × 30 = 30,000
(d) 2500 × 12 = ………………….
Solution:
2500 × 12 = 25 × 100 × 12
\(\frac {10}{4}\) × 100 × 12 [∵ 25 = \(\frac {100}{4}\)]
= 100 × 100 × 3
= 10,000 × 3 = 30,000
(e) …………… × ………………… = 120000000
Solution:
10000000 × 12 = 120000000
Figure it Out (Pages 19-21)
Question 1.
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
Solution:
(a) Largest multiple of 5 = 9876543210
(b) Smallest even number = 1023456798
Question 2.
The number 10,30,285 in words is Ten lakhs thirty thousand two hundred eighty five, which has 43 letters. Give a 7-digit number name which has the maximum number of letters.
Solution:
Required 7-digit number = 77,77,777
= Seventy seven lakh seventy seven thousand seven hundred seventy seven
This number in words has 60 letters.
Question 3.
Write a 9-digit number where exchanging any two digits results in a bigger number. How many such numbers exist?
Solution:
Required 9-digit number = 12,34,56,789
There exist only one such number.
Question 4.
Strike out 10 digits from the number 12345123451234512345 so that the remaining number is as large as possible.
Solution:
The largest possible number is 5534512345.
Question 5.
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?
Solution:
There exist no two such consecutive numbers.
Question 6.
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 ‘O’, 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?
Solution:
Numbers 1-9 contribute 9 digits (1 digit each)
Numbers 10-99 contribute 90 × 2 = 180 digits (2 digits each)
Numbers 100-999 contribute 900 × 3 = 2700 digits (3 digits each)
(a) Total digits upto 99 = 9 + 180 = 189
So, the 1000th digit will lie in the 3 digit numbers range.
Remaining digits = 1000 – 189 = 811
Number of 3 digit numbers to reach 811 digits
= \(\frac {811}{3}\) = 270, with 1 remaining number.
So, first we need to write the first 270 3-digit numbers starting from 100.
So, 270th 3-digit number = 100 + 270 -1 = 369
So next number is 370.
Hence, the 1000th digit is the’Ist digit of 370, which is 3 and the 1000th digit occurs within the number 370.
(b) Numbers 1 -9 contribute 9 digits (1 digit each)
Numbers 10-99 contribute 90 × 2
= 180 digits (2 digits each)
Numbers 100-999 contribute 900 × 3 = 2700 digits (3 digits each)
Numbers 1000-9999 contribute 9000 × 4 = 36,000 digits (4 digits each)
Numbers 10,000-99999 contribute
90,000 × 5 = 4,50,000 digits (5 digits each)
Numbers 1,00,000 – 9,99,999 contribute 9,00,000 × 6 = 54,00,000 digits (6 digits each)
To reach the millionth digit: upto 5-digit numbers = 9 + 180 + 2700 + 36,000 + 4,50,000 = 4,88,889
Remaining numbers in the 6-digit range = 10,00,000 – 4,88,889 = 5,11,111
The number of 6 digit numbers required = 5,11,111 ÷ 6 = 81,185, with 1 remaining number.
So, the 85,185th 6-digit number = 85,185 + 1,00,000 – 1 = 1,85,184
The millionth digit occurs in the number = 185,184 + 1 = 1,85,185
(c) Digit 5 in the range of 1 – 9 = 1
Digit 5 in the range of 10 – 99 = 19
Digit 5 in the range of 100 – 999 = 280
Digit 5 in the range of1000 – 9999 = 3700
Total 1 + 19 + 280 + 3700 = 4000
For the 5000th number, we need 5000 – 4000 = 1000 more numbers which come in 10000 – 99999.
Thus the last number is 13995. Which has the digit 5 and the 5000th number.
Question 7.
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
Solution:
(a) 20,800 = 20000 + 800
= (2 × 10,000) + (8 × 100)
(b) 92,100 =90,000+ 2100
= (9 × 10,000) + (21 × 100)
(c) 1,20,500= 1,20,000 + 500
= (12 × 10,000) + (5 × 100)
(d) 65,30,000 = (653 × 10,000)
(e) 70,25,700 = 70,20,000 + 5700
= (702 × 10,000) + (57 × 100)
Question 8.
How many lakhs make a billion?
Solution:
10000 lakhs make a billion.
Question 9.
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.
Solution:
To get the largest possible sum, we have to take the largest numbers in boxes :
Now to get the smallest possible difference:
Question 10.
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 :
Solution:
(a) Closest number
= (25 × 4000) +(1 × 13000)
= 1,00,000 + 13,000= 1,13,000
(b) Closest number
= (1 × 1,50,000) +(4 × 13,000)
= 1,50,000 + 52,000 = 2,02,000
(c) Closest number
= (8 × 70,000) +(5 × 4,000)
= 560000 + 20000 = 580000
(d) Closest number
= (8 × 150000) + (3 × 13000) + (20 × 300)
= 1200000 + 39000 + 6000 = 12,45,000
(e) Closest number
= (500 × 4000) + (1 × 70000) + (1 × 13000)+ (26 × 300)
= 20,00,000 + 70,000 + 13,000 + 7800
= 2090800
Question 11.
Find out how many coins should be stacked to match the height of the Statue of Unity. Assume each coin is 1 mm thick.
Solution:
Height of the statue of unity = 180 metres
= 180 × 1000 mm = 180000 mm
Thickness of one coin = 1 mm
Number of coins that should be stacked
= \(\frac {180000}{1}\) = 180000
Question 12.
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?
Solution:
Distance for the trip = 12,000 km
Albatrosses can cover about 900 to 1000 km/ day.
Time taken by the Albatrosses to cross the pacific ocean (using their fastest speed)
= \(\frac{\text { Distance }}{\text { Speed }}\)
= \(\frac{12,000 \mathrm{~km}}{1000 \mathrm{~km} / \text { day }}\) = 12 days (about)
Time taken by the Albatrosses to cross the pacific ocean (using their slowest speed)
= \(\frac{12,000 \mathrm{~km}}{900 \mathrm{~km} / \text { day }}\) = 13.3 days
So, Albatrosses would take approximately between 12 days to 13.3 days to cross the pacific ocean.
Question 13.
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.
Solution:
Approximate distance it covered every day
= \(\frac {13560}{11}\)
= 1233 km approximately
1 day = 24 hours
∴ Approximate distance it covered every hour
= \(\frac {1233}{24}\)
= 51 km approximately
Question 14.
Bald eagles are known to fly as high as 4500-6000 m above the ground level. Mount Everest is about 8850 pi high. Aeroplanes can fly as high as 10,000-12,800 m. How many times bigger are these heights compared to Somu’s building?
Solution:
Height of Somu’s building = 40 m
\(\frac {4500}{40}\) = 112 times
\(\frac {6000}{40}\) = 150 times
\(\frac {8850}{40}\) = 221 times
\(\frac {10000}{40}\) = 250 times
\(\frac {12800}{40}\) = 320 times
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