Get Class 7 Maths Chapter 1 solutions for Large Numbers Around Us with easy step-by-step explanations, number-system concepts, estimation, multiplication tricks and solved Figure It Out questions.
Class 7 Maths Chapter 1 Exercise Solutions
Students looking for Class 7 Maths Chapter 1 Exercise Solutions can use this page to understand the problems from Large Numbers Around Us, the first chapter of the current NCERT Class 7 Mathematics textbook, Ganita Prakash Part I.
NCERT’s current 2026–27 reprint identifies Large Numbers Around Us as Chapter 1. The chapter introduces students to large quantities expressed through lakhs, crores, millions and billions and develops skills involving place value, number systems, approximation, estimation and mathematical patterns.
Unlike the older NCERT Class 7 Mathematics book, the current Ganita Prakash book does not organise this chapter around a conventional sequence such as “Exercise 1.1, Exercise 1.2”. Instead, questions are presented through several Figure it Out sections and activities. Current solution indexes identify sections around pages 3, 6–7, 9, 14 and 19–21.
This guide gives independently written explanations for the chapter questions while keeping the focus on understanding rather than simply copying answers.
Class 7 Maths Chapter 1 – Chapter Overview
| Details | Information |
|---|---|
| Class | 7 |
| Subject | Mathematics |
| Textbook | Ganita Prakash |
| Part | Part I |
| Chapter | 1 |
| Chapter Name | Large Numbers Around Us |
| Academic reference | 2026–27 reprint |
| Main topics | Large numbers, place value, number systems, estimation and patterns |
The official NCERT textbook explains that Chapter 1 introduces students to lakhs, crores, millions and billions through real-life contexts.
Main concepts in the chapter
Students work with:
- Large numbers
- Indian place-value notation
- International/American number notation
- Number names
- Lakhs and crores
- Millions and billions
- Exact and approximate values
- Rounding
- Estimation
- Multiplication patterns
- Number-card puzzles
- Mathematical reasoning
Figure It Out – Page 3 Solutions
The first set of questions uses the population of a town to help students understand the size of one lakh. The three questions involve subtraction and finding an increase.
Question 1
A town has a population of 75,000. How much smaller is this than one lakh?
Given
Population:
[
75,000
]
One lakh:
[
1,00,000
]
Solution
[
1,00,000-75,000=25,000
]
Final Answer
75,000 is 25,000 less than one lakh.
Concept
One lakh is written as:
[
1,00,000
]
To find how much smaller a number is, subtract it from the larger number.
Question 2
The estimated population of the town in 2024 is 1,06,000. Find how much greater this is than one lakh.
Solution
[
1,06,000-1,00,000=6,000
]
Final Answer
1,06,000 is 6,000 more than one lakh.
Common Mistake
Do not confuse 1,06,000 with 1,60,000. The position of each digit matters.
Question 3
Find the increase in population from 75,000 to 1,06,000.
Solution
[
1,06,000-75,000
]
[
=31,000
]
Final Answer
The population increased by 31,000.
Remember
For an increase:
[
\text{Increase}=\text{New value}-\text{Old value}
]
Figure It Out – Pages 6–7 Solutions
This section introduces an interesting idea involving a calculator with buttons that represent place values. It helps students understand how a number can be represented in different ways and why place-value notation is useful.
Question 1
Represent the given numbers in at least two different ways using place-value-style expressions.
(a) 8,300
One possible representation is:
[
8\times1000+3\times100
]
[
=8,000+300
]
[
=8,300
]
Another representation is:
[
83\times100
]
[
=8,300
]
Answer
[
\boxed{8,300}
]
(b) 40,629
One representation:
[
4\times10,000+6\times100+2\times10+9
]
[
=40,000+600+20+9
]
[
=40,629
]
Another:
[
406\times100+29
]
[
=40,600+29
]
[
=40,629
]
Answer
[
\boxed{40,629}
]
(c) 56,354
Using place values:
[
5\times10,000+6\times1,000+3\times100+5\times10+4
]
[
=50,000+6,000+300+50+4
]
[
=56,354
]
Another form is:
[
563\times100+5\times10+4
]
[
=56,300+50+4
]
[
=56,354
]
(d) 66,666
One form:
[
6\times10,000+6\times1,000+6\times100+6\times10+6
]
[
=66,666
]
Another:
[
66\times1,000+6\times100+6\times10+6
]
[
=66,666
]
(e) 3,67,813
Using place values:
[
3\times1,00,000
+6\times10,000
+7\times1,000
+8\times100
+1\times10
+3
]
[
=3,67,813
]
Another possible grouping is:
[
30\times10,000+60\times1,000+70\times100+81\times10+3
]
[
=3,67,813
]
Key Idea
A number can be represented in more than one mathematically correct way. Place value tells us what each digit contributes to the total.
Question 2 – Minimum Button Clicks
The next question asks students to find the smallest number of button presses needed for the numbers above.
The efficient method is to use the largest available place-value button first.
Answers
| Number | Minimum clicks |
|---|---|
| 8,300 | 11 |
| 40,629 | 21 |
| 56,354 | 23 |
| 66,666 | 30 |
| 3,67,813 | 28 |
For example:
[
8,300=(8\times1,000)+(3\times100)
]
So:
[
8+3=11
]
Therefore, the minimum is 11 clicks.
Question 3 – Connection with Place Value
The smallest-click expressions look very similar to the usual place-value expansion of a number.
Explanation
Consider:
[
56,354
]
Its place-value form is:
[
5\times10,000+
6\times1,000+
3\times100+
5\times10+
4
]
The coefficients are exactly the digits of the number.
Therefore, the minimum number of clicks is related to the sum of the digits when each digit is obtained using its largest suitable place-value button.
For example:
[
5+6+3+5+4=23
]
So the number requires 23 minimum clicks under the given button system.
Figure It Out – Page 9 Solutions
This section compares the Indian and International number systems. The Indian system uses lakh and crore, while the International system uses million and billion.
Question 1 – Number Names in Two Systems
(a) 40,50,678
Indian system:
Forty lakh fifty thousand six hundred seventy-eight
International form:
[
4,050,678
]
Four million fifty thousand six hundred seventy-eight
(b) 4,81,21,620
Indian system:
Four crore eighty-one lakh twenty-one thousand six hundred twenty
International form:
[
48,121,620
]
Forty-eight million one hundred twenty-one thousand six hundred twenty
(c) 2,00,22,002
Indian system:
Two crore twenty-two thousand two
International form:
[
20,022,002
]
Twenty million twenty-two thousand two
(d) 24,68,13,579
Indian system:
Twenty-four crore sixty-eight lakh thirteen thousand five hundred seventy-nine
International form:
[
246,813,579
]
Two hundred forty-six million eight hundred thirteen thousand five hundred seventy-nine
(e) 34,50,00,543
Indian system:
Thirty-four crore fifty lakh five hundred forty-three
International form:
[
345,000,543
]
Three hundred forty-five million five hundred forty-three
(f) 1,02,03,04,050
Indian system:
One arab two crore three lakh four thousand fifty
International form:
[
1,020,304,050
]
One billion twenty million three hundred four thousand fifty
Remember
Indian grouping:
[
3,2,2,2,\ldots
]
International grouping:
[
3,3,3,3,\ldots
]
Question 2 – Convert Number Names into Numerals
(a) One crore one lakh one thousand ten
[
1,01,01,010
]
(b) One billion one million one thousand one
First convert to the international value:
[
1,001,001,001
]
Indian notation:
[
\boxed{1,00,10,01,001}
]
(c) Ten crore twenty lakh thirty thousand forty
[
\boxed{10,20,30,040}
]
(d) Nine billion eighty million seven hundred thousand six hundred
International value:
[
9,080,700,600
]
Indian notation:
[
\boxed{9,08,07,00,600}
]
Common Mistake
When converting a billion-based number into Indian notation, do not simply copy the International comma positions.
Question 3 – Compare the Quantities
(a) 30 thousand and 3 lakhs
[
30,000<3,00,000
]
Therefore:
[
\boxed{30\text{ thousand}<3\text{ lakhs}}
]
(b) 500 lakhs and 5 million
[
500\text{ lakhs}=5\text{ crore}=50\text{ million}
]
Therefore:
[
\boxed{500\text{ lakhs}>5\text{ million}}
]
(c) 800 thousand and 8 million
[
800,000<8,000,000
]
Therefore:
[
\boxed{800\text{ thousand}<8\text{ million}}
]
(d) 640 crore and 60 billion
[
640\text{ crore}=6.4\text{ billion}
]
Since:
[
6.4<60
]
we get:
[
\boxed{640\text{ crore}<60\text{ billion}}
]
Figure It Out – Page 14 Solutions
This section develops quick multiplication methods. The main idea is to rearrange factors so that products such as 100, 1,000 or 1,00,000 appear.
Question 1 – Calculate Quickly
(a) (2\times1768\times50)
Rearrange the factors:
[
2\times50\times1768
]
[
=100\times1768
]
[
=\boxed{1,76,800}
]
(b) (72\times125)
Since:
[
125=\frac{1000}{8}
]
Therefore:
[
72\times125
=72\times\frac{1000}{8}
]
[
=9\times1000
]
[
=\boxed{9,000}
]
(c) (125\times40\times8\times25)
Group convenient factors:
[
125\times8=1,000
]
and
[
40\times25=1,000
]
Therefore:
[
1,000\times1,000
]
[
=\boxed{10,00,000}
]
Concept
The trick is not a new formula. It is simply rearranging factors to make easy products.
Question 2 – Quick Products
(a) (25\times12)
[
25\times12
=25\times4\times3
]
[
=100\times3
]
[
=\boxed{300}
]
(b) (25\times240)
[
25\times240
=25\times4\times60
]
[
=100\times60
]
[
=\boxed{6,000}
]
(c) (250\times120)
[
250\times120
=250\times4\times30
]
[
=1,000\times30
]
[
=\boxed{30,000}
]
(d) (2,500\times12)
[
2,500\times12
=2,500\times4\times3
]
[
=10,000\times3
]
[
=\boxed{30,000}
]
(e) Fill the two numbers
We need:
[
___\times___=12,00,00,000
]
One convenient answer is:
[
1,200\times1,00,000
]
Therefore:
[
\boxed{1,200\times1,00,000=12,00,00,000}
]
Common Mistake
Do not lose track of zeroes when multiplying numbers such as 250, 2,500 and 1,00,000.
Figure It Out – Pages 19–21 Solutions
The final section contains 14 mixed reasoning questions. These questions are designed to make students apply place value, number formation, estimation and logical reasoning rather than merely perform routine calculations. Public indexes identify these as the final “Figure it Out” questions of Chapter 1.
Question 1 – Largest Multiple of 5 and Smallest Even Number
Use each digit from 0 to 9 exactly once.
(a) Largest multiple of 5
A number is divisible by 5 when its last digit is 0 or 5.
To make the number as large as possible, put the digits in descending order and place 0 at the end:
[
\boxed{9,876,543,210}
]
(b) Smallest even number
An even number must end in an even digit.
To minimise a 10-digit number, the first digit must be the smallest non-zero digit, namely 1.
Then place 0 next, while reserving a suitable even digit for the final position.
The smallest arrangement is:
[
\boxed{1,023,456,798}
]
Final Answers
- Largest multiple of 5: 9,876,543,210
- Smallest even number: 1,023,456,798
Question 2 – Seven-Digit Number with a Long Number Name
The task is to find a seven-digit number whose English number name contains a large number of letters.
One suitable answer is:
[
\boxed{77,77,777}
]
Its number name is:
Seventy-seven lakh seventy-seven thousand seven hundred seventy-seven.
The important idea is to choose words such as seventy and seven, which contain many letters, repeatedly.
Final Answer
[
\boxed{77,77,777}
]
Note
The question asks for a number with a maximum-length name; the focus is on the reasoning behind the construction rather than memorising one particular number.
Question 3 – A 9-Digit Number with a Special Swapping Property
We need a 9-digit number such that exchanging any two digits makes the number larger.
For this to happen, the digits must increase from left to right.
Consider:
[
1<2<3<4<5<6<7<8<9
]
Therefore:
[
\boxed{123456789}
]
If any two digits are exchanged, a larger digit moves to the left of a smaller digit, making the number larger.
How many such numbers exist?
Using all nine distinct digits, there is only one strictly increasing arrangement:
[
\boxed{123456789}
]
So the answer is:
[
\boxed{1}
]
Question 4 – Removing Digits to Make the Largest Number
The given 20-digit number is formed by repeating:
[
12345
]
four times.
Ten digits have to be removed while keeping the remaining digits in their original order.
To make the resulting 10-digit number as large as possible, we should choose the largest possible digits as early as possible.
A suitable maximum subsequence is:
[
\boxed{5534512345}
]
Why?
The first digit should be as large as possible. By removing suitable earlier digits, we can bring a 5 forward.
Then we maximise the next available digit, and continue this process from left to right.
Final Answer
[
\boxed{5,534,512,345}
]
Question 5 – Consecutive Number Names
The question asks whether we can eventually find two consecutive numbers whose English names have no letter in common.
For example:
- one and two share o
- two and three share t
- three and four share r
- five and six share i
- six and seven share s
- seven and eight share e
The pattern continues.
Final Answer
There is no such pair of consecutive number names under the usual English number naming system.
The activity is mainly designed to encourage students to investigate patterns in number names rather than perform a conventional arithmetic calculation.
Question 6 – Counting Digits in a Long Sequence
Suppose all positive integers are written consecutively:
[
1,2,3,4,\ldots,10,11,12,\ldots
]
The question has three parts.
(a) What is the 1000th digit?
Step 1: One-digit numbers
Numbers 1 to 9 contribute:
[
9\times1=9
]
digits.
Step 2: Two-digit numbers
Numbers 10 to 99 contain 90 numbers.
Therefore:
[
90\times2=180
]
Total so far:
[
9+180=189
]
Step 3: Three-digit numbers
Digits still required:
[
1000-189=811
]
Each three-digit number contributes 3 digits.
[
811=270\times3+1
]
So 270 complete three-digit numbers take us from 100 through 369.
The next number is:
[
370
]
The remaining one digit is the first digit of 370.
Answer
[
\boxed{\text{The 1000th digit is 3, occurring in 370.}}
]
(b) Which number contains the millionth digit?
Digits through 99,999:
[
9+180+2,700+36,000+4,50,000
]
[
=4,88,889
]
Remaining:
[
10,00,000-4,88,889
=5,11,111
]
Each six-digit number contributes 6 digits.
[
5,11,111=85,185\times6+1
]
Thus, after 85,185 complete six-digit numbers, the next number contains the millionth digit.
Starting at 1,00,000:
[
1,00,000+85,185=1,85,185
]
Answer
[
\boxed{1,85,185}
]
The millionth digit is the first digit of this number.
(c) When is the digit 5 written for the 5000th time?
The number of occurrences of digit 5 from 1 through 9,999 is:
[
4,000
]
So another:
[
5,000-4,000=1,000
]
occurrences are needed.
Counting through the five-digit numbers, the 5000th occurrence is reached when writing:
[
\boxed{13,495}
]
Thus, the 5000th occurrence of the digit 5 is written while writing 13,495.
Question 7 – Calculator with +10,000 and +100 Buttons
The calculator has only two useful buttons:
- +10,000
- +100
We express each target using these values.
(a) 20,800
[
2\times10,000+8\times100
]
[
=20,000+800
]
[
=\boxed{20,800}
]
(b) 92,100
[
9\times10,000+21\times100
]
[
=90,000+2,100
]
[
=\boxed{92,100}
]
(c) 1,20,500
[
12\times10,000+5\times100
]
[
=1,20,000+500
]
[
=\boxed{1,20,500}
]
(d) 65,30,000
[
653\times10,000
]
[
=\boxed{65,30,000}
]
(e) 70,25,700
[
702\times10,000+57\times100
]
[
=70,20,000+5,700
]
[
=\boxed{70,25,700}
]
Concept
The coefficients tell us how many times each button needs to be pressed.
Question 8 – Lakhs in One Billion
We know:
[
1\text{ lakh}=1,00,000
]
and
[
1\text{ billion}=1,00,00,00,000
]
Therefore:
[
\frac{1,00,00,00,000}{1,00,000}
]
[
=10,000
]
Final Answer
[
\boxed{10,000\text{ lakhs}}
]
Question 9 – Number Cards
Two sets of cards numbered 1 to 9 are available.
The objective is to arrange them to achieve the required result.
(a) Largest possible sum
To maximise the sum, larger digits should occupy the higher place-value positions.
One maximum arrangement is:
[
9,987,654
]
and
[
87,654
]
Adding:
[
9,987,654+87,654
]
[
=\boxed{10,075,308}
]
Final Answer
One arrangement giving the largest sum is:
[
\boxed{9,987,654+87,654}
]
(b) Smallest possible difference
To make the difference small, the seven-digit number should be as small as possible and the five-digit number as large as possible, subject to the available cards.
A suitable arrangement is:
[
1,122,334
]
and
[
99,887
]
Therefore:
[
1,122,334-99,887
]
[
=\boxed{1,022,447}
]
Key Idea
For a maximum sum, put large digits into high-value positions.
For a minimum difference, make the two resulting numbers as close as the digit restrictions allow.
Question 10 – Number Cards and Closest Values
The available cards are:
[
4,000,;13,000,;300,;70,000,;1,50,000,;20,;5
]
Each card can be used at most once for a particular target.
The aim is to construct a number as close as possible to each target.
(a) Target: 1,10,000
One useful construction is:
[
4,000\times(20+5)+13,000
]
[
=4,000\times25+13,000
]
[
=1,00,000+13,000
]
[
=1,13,000
]
Difference:
[
1,13,000-1,10,000=3,000
]
Answer
A close value is:
[
\boxed{1,13,000}
]
(b) Target: 2,00,000
Use:
[
1,50,000+70,000-(4,000\times5)
]
[
=2,20,000-20,000
]
[
=\boxed{2,00,000}
]
This is exact.
(c) Target: 5,80,000
Use:
[
70,000\times5+1,50,000+4,000\times20
]
[
=3,50,000+1,50,000+80,000
]
[
=\boxed{5,80,000}
]
Again, the target is obtained exactly.
(d) Target: 12,45,000
One close construction is:
[
4,000\times300+70,000-\frac{1,50,000}{5}
]
[
=12,00,000+70,000-30,000
]
[
=\boxed{12,40,000}
]
Difference:
[
12,45,000-12,40,000=5,000
]
Another equally close construction can also be made.
(e) Target: 20,90,800
One construction is:
[
(70,000-4,000)\times20
+1,50,000\times5
+13,000+300
]
[
=66,000\times20+7,50,000+13,000+300
]
[
=13,20,000+7,50,000+13,000+300
]
[
=\boxed{20,83,300}
]
Difference:
[
20,90,800-20,83,300=7,500
]
Key Idea
These questions are not only about arithmetic. They test whether you can choose operations that make a target number or get close to it.
Question 11 – Coins Reaching the Statue of Unity
Assume each coin is 1 mm thick.
The height is approximately:
[
180\text{ m}
]
Convert metres to millimetres:
[
1\text{ m}=1,000\text{ mm}
]
Therefore:
[
180\text{ m}=180\times1,000
]
[
=1,80,000\text{ mm}
]
Each coin is 1 mm thick.
Therefore:
[
\frac{1,80,000}{1}=1,80,000
]
Final Answer
[
\boxed{1,80,000\text{ coins}}
]
The calculation is a good example of converting units before solving a problem.
Question 12 – Albatross Journey
The bird can travel approximately 900–1,000 km per day.
Suppose the journey is about:
[
12,000\text{ km}
]
At 1,000 km per day
[
12,000\div1,000=12
]
So it would take about:
[
12\text{ days}
]
At 900 km per day
[
12,000\div900\approx13.33
]
So it would take about:
[
13.3\text{ days}
]
Final Answer
The trip would take approximately 12 to 13.3 days, depending on the bird’s average daily distance.
Question 13 – Bar-Tailed Godwit
The bird travelled approximately:
[
13,560\text{ km}
]
in about:
[
11\text{ days}
]
Distance per day
[
13,560\div11
]
[
\approx1,232.73
]
So approximately:
[
\boxed{1,233\text{ km/day}}
]
Distance per hour
There are 24 hours in a day.
[
1,232.73\div24
]
[
\approx51.36
]
Final Answer
Approximately:
[
\boxed{1,233\text{ km/day}}
]
and
[
\boxed{51.4\text{ km/hour}}
]
These are approximate values because the journey duration is given as “about 11 days.”
Question 14 – Comparing Heights
Somu’s building has 10 floors, each 4 metres high.
Therefore:
[
10\times4=40\text{ m}
]
So the building is:
[
\boxed{40\text{ m}}
]
high.
Bald eagle
Lower height:
[
4,500\div40=112.5
]
Upper height:
[
6,000\div40=150
]
Therefore, the eagle’s maximum flight height is about:
[
\boxed{112.5\text{ to }150\text{ times the building height}}
]
Mount Everest
[
8,850\div40=221.25
]
Therefore:
[
\boxed{221.25\text{ times}}
]
the height of the building.
Aeroplane
At 10,000 m: [ 10,000\40=250 ]
At 12,800 m: [ 12,800\div40=320 ]
Therefore:
[
\boxed{250\text{ to }320\text{ times}}
]
the height of the building.
The comparison is based on ratios of heights, rather than the everyday use of the phrase “times bigger.”
Important Concepts from Class 7 Maths Chapter 1
1. Indian Number System
The Indian system uses:
- Ones
- Tens
- Hundreds
- Thousands
- Ten thousands
- Lakhs
- Ten lakhs
- Crores
- Ten crores
- Arab
For example:
[
5,43,21,678
]
is read as:
Five crore forty-three lakh twenty-one thousand six hundred seventy-eight.
2. International Number System
The International system uses:
- Ones
- Tens
- Hundreds
- Thousands
- Millions
- Billions
For example:
[
54,321,678
]
is read as:
Fifty-four million three hundred twenty-one thousand six hundred seventy-eight.
3. One Lakh
[
1\text{ lakh}=1,00,000
]
4. One Crore
[
1\text{ crore}=1,00,00,000
]
5. One Million
[
1\text{ million}=10,00,000
]
6. One Billion
[
1\text{ billion}=1,00,00,00,000
]
Therefore:
[
1\text{ billion}=10,000\text{ lakhs}
]
Common Mistakes in Chapter 1
Mistake 1: Mixing Comma Systems
Do not write Indian and International comma patterns interchangeably.
Indian:
[
1,23,45,678
]
International:
[
12,345,678
]
Mistake 2: Confusing Million and Lakh
Remember:
[
1\text{ million}=10\text{ lakhs}
]
Mistake 3: Ignoring Place Value
In:
[
5,43,21,678
]
the digit 5 is in the crore place, while the digit 4 is in the ten-lakh place.
Mistake 4: Rounding Without Checking the Next Digit
When rounding, look at the digit immediately to the right of the place being rounded.
- 0–4 → keep the digit unchanged
- 5–9 → increase it by 1
Mistake 5: Losing Zeroes
Large-number multiplication often produces several zeroes. Write the calculation carefully instead of relying entirely on mental arithmetic.
How to Prepare Chapter 1 for an Exam
A useful preparation sequence is:
Step 1: Revise Number Systems
Learn the difference between Indian and International notation.
Step 2: Practise Number Names
Convert figures into words and words into figures.
Step 3: Practise Comparison
Convert quantities to the same system before deciding which is larger.
Step 4: Practise Estimation
Round numbers correctly and judge whether an answer is reasonable.
Step 5: Practise Multiplication Tricks
Look for factors that can produce 100, 1,000 or 10,000.
Step 6: Attempt Reasoning Questions
Do not skip the puzzle-style questions. They are useful for developing mathematical thinking.
Practice Question 1
A district has a population of 4,86,000 people. After five years, its population becomes 5,32,000.
Find the increase in population and express the result in lakhs.
Concept
To find an increase:
[
\text{Increase}=\text{New population}-\text{Old population}
]
Solution
[
5,32,000-4,86,000
]
[
=46,000
]
Now:
[
46,000=0.46\text{ lakh}
]
Final Answer
The population increased by:
[
\boxed{46,000}
]
or
[
\boxed{0.46\text{ lakh}}
]
Practice Question 2
Calculate quickly:
[
125\times32\times8
]
Concept
Look for factors that make 1,000.
Solution
Group 125 and 8:
[
125\times8=1,000
]
Therefore:
[
125\times32\times8
]
[
=1,000\times32
]
[
=\boxed{32,000}
]
Final Answer
[
\boxed{32,000}
]
Class 7 Maths Chapter 1 Revision Points
- One lakh = 1,00,000.
- One crore = 1,00,00,000.
- One million = 10 lakhs.
- One billion = 10,000 lakhs.
- Indian numbers use the lakh-crore grouping system.
- International numbers use million-billion grouping.
- For an increase, subtract the old value from the new value.
- When rounding, inspect the digit immediately to the right.
- Estimation gives a useful nearby value; it does not have to equal the exact value.
- Rearranging multiplication factors can make calculations much easier.
- In large-number problems, always check the number of zeroes.
- In reasoning questions, explain why an answer works rather than writing only the final number.
Frequently Asked Questions
What is Class 7 Maths Chapter 1 called?
The current NCERT Class 7 Mathematics Chapter 1 in Ganita Prakash Part I is called Large Numbers Around Us.
Does the current Class 7 Maths Chapter 1 have Exercise 1.1?
The current Ganita Prakash Chapter 1 is organised through Figure it Out sections and activities rather than the older-style Exercise 1.1 format. The chapter’s questions occur at several points, including pages 3, 6–7, 9, 14 and 19–21.
What are the main topics of Large Numbers Around Us?
The chapter covers large numbers, Indian and International number systems, place value, exact and approximate values, rounding, estimation and patterns in multiplication.
What is 1 billion in lakhs?
[
1\text{ billion}=10,000\text{ lakhs}
]
How can I study Class 7 Maths Chapter 1?
First understand the Indian and International number systems. Then practise number names, comparisons, estimation, rounding and multiplication patterns. Finally, solve the reasoning-based questions without looking at the solutions.
Should I memorise the solutions?
No. Use solutions to understand the method. After reading an explanation, close the page and solve a similar question independently.
Why are large-number estimation questions important?
Estimation helps students judge the approximate size of an answer and identify unreasonable calculations without performing lengthy exact calculations.
Related Resources for SyllabusStudy.in
Since exact SyllabusStudy.in URLs were not supplied, the following are recommended anchor-text targets only:
- Class 7 Maths Solutions
- Class 7 Maths Chapter 1 Notes
- Class 7 Maths Chapter 1 Important Questions
- Class 7 Maths Chapter 1 MCQs
- Class 7 Maths Chapter 1 Practice Questions
- Class 7 Maths Chapter 1 Practice Test
- Class 7 Maths Revision Notes
- Class 7 Maths Chapter 2 Solutions
- Class 7 Maths Chapter-wise Solutions
- Class 7 Maths Sample Paper
A strong internal-link structure would be:
Class 7 Maths → Chapter 1 → Chapter 1 Solutions → Chapter 1 Notes → Important Questions → MCQs → Practice Test
This gives students a natural learning path instead of creating many nearly identical keyword pages.
Conclusion
Class 7 Maths Chapter 1 – Large Numbers Around Us builds an important foundation for working confidently with large quantities. Students learn how the Indian and International number systems represent numbers, how to compare lakhs, crores, millions and billions, and how estimation can simplify real-world calculations.
The chapter also goes beyond routine arithmetic. The number-card puzzles, digit-counting problems and multiplication activities encourage students to look for patterns and explain their reasoning.
For the best results, students should first attempt each problem themselves, then use the step-by-step solution to identify mistakes, and finally solve a similar problem independently.
NCERT’s current Ganita Prakash textbook specifically aims to encourage mathematical thinking, exploration, problem-solving and connections with everyday situations.