NCERT Solutions Class 7 Maths Chapter 1 Large Numbers Around Us
In Chapter 1 – Large Numbers Around Us, students will build upon the understanding of numbers and be introduced to how these large numbers are used in the world around us. Students will also learn the ways to read, write, compare, and use some calculations with very large numbers - into the crores and more.
By examining real-world examples such as population figures, distances between planets, and national income, students will discover how large numbers appear in our everyday lives. The NCERT Solutions for Class 7 Maths Chapter 1 offers the complete and structured answers to the textbook questions step-by-step all the way to the practical use of large numbers.
1.0NCERT Solutions Class 7 Maths Chapter 1 Large Numbers Around Us: Download PDF
Download our FREE PDF of NCERT Solutions Class 7 Maths Chapter 1 – Large Numbers Around Us, exclusively crafted by experts at ALLEN.
2.0Key Concepts in Chapter 1: Large Numbers Around Us
The first chapter of Class 7 Maths begins with real-world examples and builds up to numerical operations involving large values. Below are the key topics and concepts covered in this chapter:
1. Reading and Writing Large Numbers
- Numbers beyond six digits (lakh, crore).
- Indian and International Place Value System.
- Writing numbers in words and numerals.
- Example: 7,28,69,002 is written as “Seven crore twenty-eight lakh sixty-nine thousand and two” in the Indian system.
2. Comparing Large Numbers
- Determining the greater or smaller among two large numbers using place values.
- Useful in interpreting real-world data like populations or revenues.
- Example: Comparing population of states using crores and lakhs.
3. Estimation
- Estimating sums, differences, products, and quotients using rounding off.
- Helps in mental math and checking the reasonableness of answers.
- Example: Estimating total cost of multiple products in lakhs or crores.
4. Operations Using Large Numbers
- Performing addition, subtraction, multiplication and division on 6-digit and 7-digit numbers.
- Application-based questions involving cost, budget, distance, and more.
5. Use of Large Numbers in Daily Life
- Data interpretation involving census, sports statistics, scientific data, etc.
- Students are encouraged to explore examples around them (e.g., cricket scores, mobile users, metro city population).
6. Roman Numerals (Brief Introduction)
- Conversion between Roman numerals and Hindu-Arabic numerals.
- Application in daily life: Clock faces, event names, etc.
3.0NCERT Solutions for Class 7 Chapter 1 Large Numbers Around Us: All Exercises
Exercise 1.1
Class 7 Maths NCERT Solutions Chapter 1 – Exercise 1.1 introduces students to the concept of large numbers through real-life examples such as population data and census statistics. Students learn to compare numbers with benchmark values like one lakh and determine differences or increases in quantities. The exercise strengthens understanding of the Indian place value system and helps learners interpret numbers used in everyday contexts.
Key Concepts Covered:
- Comparing large numbers
- Understanding lakh as a benchmark value
- Real-life applications of large numbers
- Basic subtraction with large numbers
Exercise 1.2
In this exercise, the focus is on expressing numbers using place values and mathematical expressions. Here, students break down numbers into thousands, hundreds, tens, and ones, and represent them using different combinations. The exercise also encourages logical thinking by exploring multiple ways to obtain the same number using place-value operations.
Key Concepts Covered:
- Place value representation
- Expanded form of numbers
- Multiple ways to express numbers
- Understanding thousands, hundreds, tens, and ones
Exercise 1.3
Class 7 Maths NCERT Solutions Chapter 1 – Exercise 1.3 builds on place-value understanding by exploring efficient ways to construct numbers with the least number of operations. Students can learn to analyse the positions of digits and determine how many steps or operations are required to form numbers. This exercise will help learners to see how place value can affect number construction and efficiency in calculations.
Key Concepts Covered:
- Relationship between digits and place value
- Constructing numbers using place-value operations
- Minimising steps in calculations
- Understanding number structure
Exercise 1.4
Reading and writing very large numbers in both the Indian and International number systems is the main focus ofthis exercise. In this exercise, students learn to convert number names to numerals, write numbers in place-value notation, and also compare quantities across the two systems. Learning this exercise helps in strengthening the understanding of crores, millions, billions, and number naming conventions.
Key Concepts Covered:
- Indian place value system
- International place value system
- Writing number names
- Converting between number systems
- Comparing large numbers
Exercise 1.5
Exercise 1.5 encourages students to use smart calculation techniques in order to simplify problems on multiplications. Helpful numerical patterns are recognised by regrouping numbers and helps the learners to discover quicker ways to perform calculations. This exercise highlights the importance of strategic thinking and factor grouping in arithmetic operations.
Key Concepts Covered:
- Quick multiplication strategies
- Factor regrouping methods
- Simplifying calculations
- Pattern-based computation
Exercise 1.6
Exercise 1.6 is another important section in the first chapter where it presents logical number puzzles and real-life numerical problems that involve reasoning with large numbers. Students explore digit arrangements, patterns in numbers, counting occurrences, and problem-solving situations related to distances, measurements, and estimations. The exercise strengthens analytical thinking and real-world mathematical reasoning.
Key Concepts Covered:
- Logical number puzzles
- Digit arrangement and number patterns
- Counting occurrences of digits
- Real-life applications of large numbers
- Mathematical reasoning and estimation
4.0NCERT Class 7 Maths Chapter 1 Large Numbers Around Us: Detailed Solutions
FIGURE IT OUT-01
- 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 ?
Sol. One lakh =1,00,000
Now, 1,00,000−75,000=25,000
Thus, the population of Chintamani in 2011 was 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 ?
Sol. 1,06,000−1,00,000=6,000
Thus, the population in 2024 is 6,000 more than one lakh.
- By how much did the population of Chintamani increase from 2011 to 2024?
Sol. Increase in population of Chintamani from 2011 to 2024 will be
=1,06,000−75,000=31,000
FIGURE IT OUT-02
- 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
Sol. Chitti is a calculator like smart matrix and it has the following buttons: +1,+10, +100, +1000, +10000, +100000 and +1000000 .
(a) Way1: (8×1000)+(3×100)
=8000+300=8300
Way2: (83×100)=8300
(b) Way1: (4×10000)+(6×100)+(2
×10)+(9×1)
=40000+600+20+9
=40629
Way2: (40×1000)+(6×100)+(29×1)
=40000+600+29
= 40629
(c) Way1: (5×10000)+(6×1000)+(3
×100)+(54×1)
=50000+6000+300+54
=56354
Way2: (56×1000)+(35×10)+(4×1)
=56000+350+4
= 56354
(d) Way1: (6×10000)+(6×1000)+
(6×100)+(66×1)
=60000+6000+600+66
= 66666
Way2: (66×1000)+(66×10)+(6×1)
=66000+660+6
= 66666
(e) Way1: (3×100000)+(6×10000)+
(7×1000)+(8×100)+(13×1)
=300000+60000+7000+800+13
=367813
Way2: (36×10000)+(78×100)+
( 13×1 )
=360000+7800+13
= 367813
FIGURE IT OUT-03
- 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.
Sol. (a) 8300
(8×1000)+(3×100)=8300
8×(+1000)=8000 ( 8 clicks)
3×(+100)=300 ( 3 clicks)
Total clicks: 8+3=11
(b) 40629
(4×10000)+(6×100)+(2×10)+
(9×1)=40629
4×(+10000)=40000 (4 clicks)
6×(+100)=600 ( 6 clicks)
2×(+10)=20 ( 2 clicks)
9×(+1)=9 ( 9 clicks)
Total clicks: 4+6+2+9=21
(c) 56354
(5×10000)+(6×1000)+(3×100)
+(5×10)+(4×1)=56354
5×(+10000)=50000 ( 5 clicks)
6×(+1000)=6000 ( 6 clicks)
3×(+100)=300 ( 3 clicks)
5×(+10)=50 ( 5 clicks)
4×(+1)=4 ( 4 clicks)
Total clicks: 5+6+3+5+4=23
(d) 66666
(6×10000)+(6×1000)+(6×100)
+(6×10)+(6×1)=66666
6×(+10000)=60000 ( 6 clicks)
6×(+1000)=6000 ( 6 clicks)
6×(+100)=600 ( 6 clicks)
6×(+10)=60 ( 6 clicks)
6×(+1)=6 ( 6 clicks)
Total clicks: 6+6+6+6+6=30
(e) 367813
(3×100000)+(6×10000)+(7×
1000)+(8×100)+(1×10)+(3×1)
=367813
3×(+100000)=300000 ( 3 clicks)
6×(+10000)=60000 ( 6 clicks)
7×(+1000)=7000 ( 7 clicks)
8×(+100)=800 ( 8 clicks)
1×(+10)=10 ( 1 click)
3×(+1)=3 ( 3 dicks)
Total clicks: 3+6+7+8+1+3=28
- Do you see any connection between each number and the corresponding smallest number of button clicks?
Sol. The smallest number of button clicks depends on the place value of the digit in the number. If the digit in a particular place is small, then less number of clicks are required and if it is big, then more number of clicks are required.
- 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.
Sol. When we write a number using the least button clicks, we divide it into parts like lakh, thousand, hundred, ten, and one. These are the same place values we use in the Indian place value system, so both look the same.
FIGURE IT OUT-04
- 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
Sol. (a) Indian System 40,50,678
Forty lakh fifty thousand six hundred seventy-eight.
American System 4,050,678
Four million fifty thousand six hundred seventy-eight.
(b)
Indian System 4,81,21,620
Four crore eighty-one lakh twentyone thousand six hundred twenty.
American System
48,121,620
Forty-eight million one hundred twenty-one thousand six hundred twenty.
(c)
Two crore twenty-two thousand two.
American System:
20,022,002
Twenty million thousand two.
(d)
Twenty-four crore sixty-eight lakh
thirteen thousand five hundred
American System
246,813,579
Two hundred forty-six million eight hundred thirteen thousand five hundred seventy-nine
(e) Indian System:
34,50,00,543
Thirty-four crore fifty lakh five hundred forty-three.
American System:
345,000,543
Three hundred forty-five million five hundred forty-three.
(f) Indian System:
1,02,03,04,050
One arab two crore three lakh four thousand fifty.
American System:
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.
Sol. (a) In Indian place value notation 1,01,01,010
(b) One billion one million one thousand one - 1,001,001,001 In Indian place value notation 1,00,10,01,001
(c) In Indian place value notation 10,20,30,040
(d) Nine billion eighty million seven hundred thousand six hundred 9,080,700,600
In Indian place value notation 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
Sol. (a) 30 thousand =30,000
3 lakhs = 3,00,000
30 thousand < 3 lakh
(b) 500 lakhs =50000000
5 million =5000000
500 lakhs > 5 million
(c) 800 thousand = 800000
8 million = 8000000
800 thousand < 8 million
(d) 640 crore =6400000000
60 billion = 60000000000
640 crore < 60 billion
FIGURE IT OUT-05
- Find quick ways to calculate these products:
(a) 2×1768×50
(b) 72×125 [Hint: 125=81000 ]
(c) 125×40×8×25
Sol. (a) 2×1768×50=2×1768×2100
=1768×100=176800
(b) 72×125=72×81000
=9×1000=9000
(c) 125×40×8×25
=81000×40×8×4100=10,00,000
- Calculate these products quickly.
(a) 25×12= ____
(b) 25×240= ____
(c) 250×120= ____
(d) 2500×12= ____
(e) ___ × ___ =120000000
Sol. (a) 25×12=25×4×3=100×3=300
(b) 25×240=25×4×60
=100×60=6000
(c) 250×120=250×4×30
=1000×30=30000
(d) 2500×12=2500×4×3
=10000×3=30000
(e) 25000×4800=25000×4×1200
=100000×1200=120000000
FIGURE IT OUT-06
- 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
Sol. (a) To form the largest multiple of 5, the last digit must be 0 or 5 .
Arranging the digits in decreasing order, the largest multiple of 5 is 9876543210 (10 digits).
(b) To form the smallest even number, the last digit must be even ( 0,2,4,6, 8), and the digits should be written in ascending order.
The even number is 1023456798 (10 digits).
- 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.
Sol. 78,78,773 (Seventy-eight lakhs seventyeight thousand seven hundred seventythree). This has 61 letters, making it one of the longest 7-digit numbers.
- Write a 9-digit number where exchanging any two digits results in a bigger number. How many such numbers exist?
Sol. To ensure swapping any two digits increases the value, the digits must increase from left to right.
So, the arrangement would be: 123456789.
There is only 1 such number.
- Strike out 10 digits from the number 12345123451234512345 so that the remaining number is as large as possible.
Sol. The given number is 12345123451234512345.
Removing the 10 highlighted digits from left to right: 12345123451234512345 The number we get is 5534512345 , which is the largest possible number.
- 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?
Sol. Let's list the spellings of some numbers and check for shared letters between consecutive numbers.
1.Zero (Z, E, R, O) and One (O, N, E) - Share ' e ' and ' o '.
2.One ( 0, N,E ) and Two ( T,W,O ) - Share 'o'.
3.Two ( T,W,O ) and Three ( T,H,R,E,E ) Share 't'.
4.Three (T, H, R, E, E) and Four (F, O, U, R) - Share 'r'.
5.Four ( F,O,U,R ) and Five ( F,I,V,E ) Share 'f'.
6.Five (F, I, V, E) and Six (S, I, X) - Share 'i'.
7.Six (S, I, X) and Seven (S, E, V, E, N) - Share 's'.
8.Seven (S, E, V, E, N) and Eight (E, I, G, H, T) - Share 'e'.
9.Eight (E, I, G, H, T) and Nine (N, I, N, E) Share 'e', 'i'.
10.Nine (N, I, N, E) and Ten (T, E, N) - Share ' e′ ', ' n '.
Similarly, we can check for any two consecutive numbers.
Therefore, there are no consecutive numbers that do not share a letter in common.
- 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?
Sol. (a) 1-digit numbers (1-9): There are 9 numbers. Each number uses 1 digit. Total digits: 9×1=9 digits. The 9th digit is ' 9 ' (from the number 9). 2-digit numbers (10-99): There are 99−10+1=90 numbers. Each number uses 2 digits. Total digits:
90×2=180 digits. Cumulative digits up to 99 : 9+180=189 digits. The 189 th digit is ' 9 ' (from the number 99). 3-digit numbers (100-999): There are 999−100+1=900 numbers. Each number uses 3 digits. Total digits: 900×3=2700 digits. Cumulative digits up to 999 : 189+2700=2889 digits.
To find the 1000th digit:
Digits so far: 9+180=189.
So, the 1000 th digit will lie in the 3 digit numbers range.
Remaining digits: 1000−189=811. Number of 3-digit numbers to reach 811 digits:
811÷3=270, with 1 remaining number.
Thus, first we need to write the first 270 3-digit numbers starting from 100.
So, the 270th 3-digit number =100+270−1=369.
The next number is 370 .
Thus, the 1000th digit is the 1st digit of 370 , which is 3 .
(b) 4-digit numbers (1000-9999): There are 9999−1000+1=9000 numbers. Each number uses 4 digits. Total digits: 9000×4=36,000 digits. Cumulative digits up to 9999: 2889+36000=38,889 digits.
5-digit numbers (10000-99999): There are 99999−10000+1=90,000 numbers. Each number uses 5 digits. Total digits: 90,000×5=450,000 digits. Cumulative digits up to 99999: 38,889+450,000=488,889 digits.
6-digit numbers (100000999999): There are 999999−100000+1=900,000
numbers. Each number uses 6 digits.
Total digits: 900,000×6=5,400,000 digits.
To reach the millionth digit:
Upto 5-digit numbers: 9+180+2700+36,000+4,50,000=4,88,889
Remaining digits in the 6-digit range: 1,000,000 (or ( 10,00,000 ) 4,88,889=5,11,111
The number of 6 -digit numbers required: 5,11,111÷6=85,185, with 1 remaining number.
So, the 85,185 th 6 -digit number is 85,185+1,00,000−1=1,85,184.
The millionth digit occurs in the number 185184+1=1,85,185.
(c) Single-digit numbers (1-9):
5 occurs only 1 time
Two-digit numbers (10-99)
( 15,25,35,…,95 ), 5 occurs 9 times.
50, 51, 52, ..., 59, 5 occurs 10 times.
Thus, 5 occurs 20 times in two digit numbers.
Three-digit numbers (100-999)
(i) Unit's place: In numbers like 105, 115,…..,995,5 occurs 90 times.
(ii) Ten's place: In numbers like 150159, 250-259, ......, 950-959, 5 occurs 90 times.
(iii) Hundred's place: In numbers like 500-599, 5 occurs 100 times.
Thus, 90 (unit's place) +90 (ten's place) + 100 (hundred's place) = 280 occurrences
Total occurrences so far: 20+280=300
Four-digit numbers (1000-9999)
(i) Unit's place: In numbers like 1005, 1015, ..., 9995, 5 occurs 900 times.
(ii) Ten's place: In numbers like 10501059, 1150-1159, ..., 9950-9959, 5 occurs 900 times
(iii) Hundred's place: In number like 1500-1599,2500-2599,..., 95009599, 5 occurs 900 times.
(iv) Thousands position: 5000-5999, 5 occurs 1000 times
Adding these up: 900 (unit's place) + 900 (ten's place) + 900 (hundred's place) + 1000 (thousand's place) = 3700 occurrences
Total occurrences so far:
300+3700=4000
Five-digit numbers (starting from 10000)
(i) In 10001-10999:
For the 5000th number, we require 5000−4000=1000 more numbers that lie in 10001-10999.
Among 10000-10999, one digit 5 appears in 100 numbers (e.g., 10005, 10015,....., 10995).
The digit 5 appears in 100 numbers (e.g., 10050-10059, ..., 1095010959).
The digit 5 appears in 100 numbers (e.g., 10500-10599).
Total occurrences so far:
4000+300=4300
(ii) In 11000-11999
5 at unit's place =100 times
5 at ten's place = 100 times
5 at a hundred's place = 100 times
Total occurrences so far:
4300+300=4600
(iii) In 12000-12999
5 at unit's place =100 times
5 at ten's place = 100 times
5 at a hundred's place = 100 times
Total occurrences so far:
4600+300=4900
(iv) In 13000-13999
5 occurs at unit place in numbers 13005, 13015 .........
So, 5 occurs for 5000th time in 100th number starting from 13005 i.e. 13995
Required number =13995
- 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
Sol. (a) 20,800=2×10,000+8×100
Number of clicks =2+8=10 clicks
(b) 92,100=9×10,000+21×100
Number of clicks =9+21=30 clicks
(c) 1,20,500=12×10,000+5×100
Number of clicks =12+5=17 clicks
(d) 65,30,000=653×10,000+0×100
Number of clicks =653+0
= 653 clicks
(e) 70,25,700=702×10,000+57×100 Number of clicks =702+57=759 clicks
- How many lakhs make a billion?
Sol. 1 lakh = 1,00,000
1 billion = 1,000,000,000
1 billion =10,000×100,000
1 billion = 10,000 lakhs
Thus, 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.
Sol.
Total Available Digits:
From two sets of digits 1 to 9:
Each digit from 1 to 9 appears twice.
So, total 18 digits to be used.
(a) Greatest 7 digit number using 18 digits = 9988776
Greatest 5 digit number using remaining 11 digits = 65544
Largest possible sum =9988776+65544=10054320
(b) Smallest 7 digit number using 18 digits = 1122334
Greatest 5 digit number using remaining 11 digits = 99887
Smallest possible difference
= 1122334−99887=1022447
- 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 :
Sol.
(a) 4000×(20+5)+13000
=4000×25+13000
=100000+13000
=113000
This gives us 1,13,000, which is very close to 1,10,000.
(b) 1,50,000+70,000−4000×5
=2,00,000
(c) 70,000×5+4,000×20+1,50,000
= 5,80,000
(d) 70,000×20−300×5−4,000−1,50,000=12,44,500
This gives us 12,44,500, which is very close to 12,45,000.
(e) 13,000×300−70,000(20+5)+4,000−1,50,000=20,04,000
- Find out how many coins should be stacked to match the height of the Statue of Unity. Assume each coin is 1 mm thick.
Sol. Approximate height of the Statue of Unity is 182 m=1,82,000 mm
(Since 1 m=1000 mm )
Thickness of one coin =1 mm
Number of coins needed =1,82,000 mm
÷1 mm=1,82,000 coins
- 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?
Sol. Time = Speed Dis tance
To estimate the number of days, divide the total distance by the daily distance covered.
If the albatross flies 900 km per day.
900 km/ day 12000 km=13.33 days ≈13 to 14 days
If the albatross flies 1000 km per day.
1000 km/ day 12000 km=12 days
It would take approximately 12 to 14 days for a grey headed albatross to complete a 12,000 km journey across the Pacific Ocean, depending on its flying speed.
- 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.
Sol. Total distance =13,560 km and duration = 11 days
Approximate distance covered every day =13,560÷11∼1,233 km/ day .
Thus, the godwit covers approximately 1,233 km per day.
One day = 24 hours
Approximate distance covered every hour =1,233÷24∼51 km/ hour .
Thus, the godwit covers approximately 51 km per 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 10,00012,800 m. How many times bigger are these heights compared to Somu's building ( 44 m tall)?
Sol. Let's compare these heights to Somu's building:
Bald eagle: 4500-6000 m;
Mount Everest: 8850 m;
Aeroplanes: 10,000−12,800 m
Somu's building is 44 m tall.
Heights compared to Somu's building:
Bald eagle: ( 4500÷44∼102 ) to ( 6000÷44∼136 )
The bald eagle's flying height is 102−136 times higher.
Mount Everest: 8850÷44=201
Mount Everest is 201 times bigger.
Aeroplanes: (10,000÷44∼227) to ( 12,800÷44∼291 )
An aeroplane's flying height is 227-291 times higher.
5.0Key Features of NCERT Solutions Class 7 Maths Chapter 1 : Large Numbers Around Us
- Conceptual Clarity: Each concept, from understanding 'lakh' to mastering the Indian place value system, is explained in a clear and concise manner.
- Step-by-Step Explanations: Solutions to all exercises are provided with detailed, easy-to-follow steps, making it simple for students to understand the reasoning behind each answer.
- Real-World Relevance: The solutions connect mathematical concepts to real-life situations, helping students appreciate the practical application of large numbers.
- Adherence to NCERT Syllabus: The solutions strictly follow the updated NCERT curriculum, ensuring that students are well-prepared for their examinations and aligned with academic standards.
- Practice-Oriented Approach: Numerous examples and practice questions, derived directly from the textbook's activities, are included to reinforce learning and build confidence.
- Downloadable PDF Format: Conveniently available as a PDF, allowing students to access the solutions offline anytime, anywhere.
