TOPIC 3 OF 31

Number Patterns and Chapter Exercises

Q: What types of exercises are in Class 7 Maths Chapter 1?

Class 7 Maths Chapter 1 exercises include writing numbers in expanded form, comparing large numbers, converting between Indian and international systems, estimating by rounding, reading population data tables, identifying number patterns, and solving word problems involving lakhs and crores from NCERT Ganita Prakash.

Q: What types of exercises are in Class 7 Maths Chapter 1?

NCERT Class 7 Maths Chapter 1 exercises include writing numbers in expanded and standard form, comparing large numbers, converting between Indian and international number systems, rounding and estimation, reading data tables with population figures, and recognising number patterns. Solutions involve lakhs and crores.

Q: How do you solve place value problems for large numbers?

To solve place value problems, identify each digit position from right to left: ones, tens, hundreds, thousands, ten-thousands, lakhs, ten-lakhs, crores. Multiply each digit by its place value and add to get the number. NCERT Class 7 Chapter 1 exercises provide extensive practice.

Q: What number patterns are covered in Class 7 Chapter 1?

Class 7 Chapter 1 covers patterns in large numbers such as multiplying by 10 to shift digits left, relationships between consecutive place values, patterns when adding or subtracting round numbers, and sequences formed by population growth data. These patterns reinforce place value understanding.

Q: How do you estimate sums and differences of large numbers?

To estimate sums or differences, round each number to the same convenient place value, then add or subtract the rounded values. For example, to estimate 4,73,826 plus 2,18,459, round to the nearest lakh: 5 lakhs plus 2 lakhs equals approximately 7 lakhs. NCERT Class 7 teaches this skill.

Q: Why is learning large numbers important in Class 7?

Learning large numbers builds number sense essential for understanding real-world data like populations, distances, budgets and scientific measurements. It prepares students for decimals, fractions and data handling in later chapters. NCERT Class 7 Ganita Prakash makes this relevant through authentic Indian data.

🎓 Class 7 Mathematics CBSE Theory Ch 1 — Large Numbers Around Us ⏱ ~40 min

🌐 Language: [gtranslate]

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1.5 Patterns in Products

Roxie and Estu explore multiplication shortcuts^? that use the relationship between ×10, ×100, ×1000 and simple division.

A Multiplication Shortcut

🎀 Roxie's Method — Multiplying by 5

58 × 10 = 580

Multiplying by 5 = dividing by 2 and multiplying by 10. (Since 5 = 10 ÷ 2)

🎯 Estu's Method — Multiplying by 25

206 × 100 = 20,600

Multiplying by 25 = dividing by 4 and multiplying by 100. (Since 25 = 100 ÷ 4)

💡 Why Do These Shortcuts Work?

Multiplying by 5: 5 = 10/2, so a × 5 = a × 10/2 = (a/2) × 10. Dividing by 2 (halving) is easy, then multiply by 10 (just add a zero).

Multiplying by 25: 25 = 100/4, so a × 25 = a × 100/4 = (a/4) × 100. Dividing by 4 (quartering) is manageable, then multiply by 100.

Multiplying by 125: 125 = 1000/8, so a × 125 = (a/8) × 1000.

⚡ Quick Multiplication Tool

Enter a number and choose a multiplier to see the shortcut method step-by-step.

📝 Figure it Out — Quick Multiplication

Q1. Find quick ways to calculate:
(a) 2 × 1768 × 50 (b) 72 × 125 (c) 125 × 40 × 8 × 25

(a) 2 × 1768 × 50:
Group: (2 × 50) × 1768 = 100 × 1768 = 1,76,800

(b) 72 × 125:
125 = 1000/8, so 72 × 125 = (72/8) × 1000 = 9 × 1000 = 9,000

(c) 125 × 40 × 8 × 25:
Group: (125 × 8) × (40 × 25) = 1000 × 1000 = 10,00,000

Q2. Calculate these products quickly using the ×25 shortcut:
(a) 25 × 12 (b) 25 × 240 (c) 250 × 120 (d) 2500 × 12 (e) _____ × _____ = 1,20,00,000

(a) 25 × 12 = (12/4) × 100 = 3 × 100 = 300

(b) 25 × 240 = (240/4) × 100 = 60 × 100 = 6,000

(c) 250 × 120 = 25 × 10 × 120 = 25 × 1200 = (1200/4) × 100 = 300 × 100 = 30,000

(d) 2500 × 12 = 25 × 100 × 12 = (12/4) × 100 × 100 = 300 × 100 = 30,000

(e) 1,20,00,000 = 1200 × 10000 = 1200 × 10000 (many valid answers, e.g. 25 × 48,00,000 or 4800 × 2500)

How Long Is the Product?

Observe the interesting patterns that emerge when you multiply certain number sequences. Study each box, find the pattern, then extend it.

Pattern 1: 11 × 11, 111 × 111…

11 × 11 =121

111 × 111 =12,321

1111 × 1111 =12,34,321

1111 1 × 11111 = 1,23,45,4,321

Pattern: Digits count up then down (1,2,3…n…3,2,1) where n = number of 1s.

Pattern 2: 66 × 61, 666 × 661…

66 × 61 =4,026

666 × 661 =4,40,226

6666 × 6661 =4,44,02,226

66666 × 66661 = 4,44,40,22,226

Pattern: Starts with 4, then 4s, then 0, then 2s, then ends in 6.

Pattern 3: 3 × 5, 33 × 35…

3 × 5 =15

33 × 35 =1,155

333 × 335 =1,11,555

3333 × 3335 = 1,11,10,555

Pattern: Starts with 1, then 1s, then 5s — equal count of each.

Pattern 4: 101×101, 102×102…

101 × 101 =10,201

102 × 102 =10,404

103 × 103 =10,609

104 × 104 = 10,816

Pattern: 10n² gives (n²) in middle. Each is (100+n)² = 10000 + 200n + n².

Number of Digits in a Product — Roxie's Reasoning

Multiplication	Min Product (digits)	Max Product (digits)	Digit count in product
1-digit × 1-digit	1×1=1 (1 digit)	9×9=81 (2 digits)	1 or 2 digits
2-digit × 1-digit	10×1=10 (2 digits)	99×9=891 (3 digits)	2 or 3 digits
2-digit × 2-digit	10×10=100 (3 digits)	99×99=9801 (4 digits)	3 or 4 digits
3-digit × 3-digit	100×100=10,000 (5 digits)	999×999=9,98,001 (6 digits)	5 or 6 digits
5-digit × 5-digit	10000×10000=10⁸ (9 digits)	99999×99999≈10¹⁰ (10 digits)	9 or 10 digits
8-digit × 3-digit	10000000×100=10⁹ (10 digits)	99999999×999≈10¹¹ (11 digits)	10 or 11 digits
12-digit × 13-digit	10¹¹×10¹²=10²³ (24 digits)	<10¹²×10¹³=10²⁵ (25 digits)	24 or 25 digits

📏 Digit Count Rule

The product of an m-digit number and an n-digit number has either (m+n−1) or (m+n) digits.

This is because: min product ≥ 10^(m−1) × 10^(n−1) = 10^(m+n−2), giving m+n−1 digits; max product < 10^m × 10^n = 10^(m+n), giving m+n digits.

1.6 Did You Ever Wonder…? — Fascinating Facts

Large number arithmetic reveals amazing facts about our world. Calculate each product or quotient to uncover the fact hidden inside, then read the answer in both Indian and International systems.

🎵 Kīrtanas by Purandaradāsa

1,250 × 380 = ?

4,75,000

Purandaradāsa (15th century) composed a legendary 4,75,000 kīrtanas (= 4 lakh 75 thousand = 475,000 in International system). He was a composer and singer spanning social reform, bhakti, and spirituality, and systemised methods for teaching Carnatic music that are still followed today.

Think: If he lived for 80 years and composed from age 20, that's 60 years of composing. 4,75,000 ÷ 60 ≈ 7,917 songs per year — about 22 songs every single day!

☀️ Distance from Earth to Sun

2,100 × 70,000 = ?

14,70,00,000 km

The approximate distance between Earth and the Sun is 14 crore 70 lakh km (= 147 million km in International system). This distance varies — the farthest point is about 15 crore 20 lakh km (152 million km).

Challenge: Scientists measured this using radar and the time light takes to travel. Light travels at about 3,00,000 km/s — so light from the Sun reaches Earth in about 8 minutes!

🌊 Amazon River Water Discharge

6,400 × 62,500 = ?

40,00,00,000 litres/sec

The Amazon River discharges approximately 40 crore litres (= 400 million litres = 4 × 10⁸ litres) of freshwater into the Atlantic Ocean every second. This is so much that drinkable freshwater is found even 160 km into the open sea!

Uncover Facts Through Division

🚂 World's Longest Train Journey

13,95,000 ÷ 150 = ?

9,300 km

The world's longest single-train journey is approximately 9,300 km. The train runs in Russia between Moscow and Vladivostok, taking about 7 days to complete the journey. India's longest route is Dibrugarh (Assam) to Kanyakumari (Tamil Nadu) covering 4,219 km in about 76 hours.

🐋 Blue Whale Weight

1,05,00,00,000 ÷ 700 = ?

1,50,000 kg

Adult blue whales can weigh more than 1,50,000 kg (= 1 lakh 50 thousand kg = 150 metric tonnes = 150,000 kg in International). Fascinating blue whale facts:
• Newborn weighs ~2,700 kg (like an adult hippopotamus)
• Heart recorded at nearly 700 kg
• Tongue weighs as much as an elephant
• Eats up to 3,500 kg of krill daily!

♻️ Global Plastic Waste (2021)

52,00,00,00,000 ÷ 130 = ?

4,00,00,000 tonnes

The weight of global plastic waste generated in 2021 was approximately 4 crore tonnes (= 40 million tonnes = 400 lakh tonnes). This environmental challenge underscores the urgency of reducing plastic use and improving waste management globally.

🌱 Large Number Fact — Soil Biology

In a single gram of healthy soil there can be 10 crore to 1 arab bacteria (100 million to 1 billion) and 1 lakh to 10 lakh fungi (1,00,000 to 10,00,000). These microorganisms support plant growth and soil health.

Share large-number facts you come across with your class!

Thought Experiments with Large Numbers

Estu wondered: "Could the entire population of Mumbai fit into 1 lakh buses?" Let us reason step by step.

🚌 Could 1 lakh buses hold Mumbai's population?

Show Reasoning ▾

🚢 Could 5,000 Titanic ships carry all of Mumbai?

Show Reasoning ▾

🌕 Roxie travels 100 km/day — can she reach the Moon in 10 years?

Show Reasoning ▾

☀️ Can you reach the Sun in a lifetime (travelling 1,000 km/day)?

Show Reasoning ▾

🔢 Activity — Toothpick Digit Puzzles

L6 Create

In this activity, digits are formed using line segments (like matchsticks or toothpicks). The number of segments needed for each digit is shown below.

Make 42,019: Count sticks for each digit: 4→4, 2→5, 0→6, 1→2, 9→6. Total = 4+5+6+2+6 = 23 sticks.
Add 2 sticks to make a bigger number than 42,019: One example is 42,078 (change 1→7 costs 3 sticks, but we remove a stick from 9→0 saving 6, so net = 3−6 = remove 3... actually try changing 9 to 7 and 0 to 9: 42,079 saves 1 stick but adds 0 sticks... try 42,078 = 4+5+6+3+7 = 25 sticks, which uses 2 extra). Other answers: 42,079, 44,019, etc.
Where should Preetham insert '1' among 4,2,0,1,9 to get the biggest number? Insert before all others: 1,42,019. Or: 4,21,019 or 4,20,119 or 4,20,191 or 4,20,191. The largest is 14,2019 — wait, inserting '1' at the front gives 1,42,019. But 4,21,019 > 1,42,019. Best position: after '4' → 4,21,019. Or: 4,2,1,0,19 → 4,21,019. Largest option: 4,21,019.

Sticks for 5108: 5→5, 1→2, 0→6, 8→7. Total = 5+2+6+7 = 20 sticks.

Challenge: Biggest number using exactly 24 sticks? Use digits that give the most digits for 24 sticks. Digit '1' uses only 2 sticks. So 24 ÷ 2 = 12 ones → 12-digit number 111,111,111,111 (12 ones)! That's 1 lakh 11 thousand 1 hundred 11 billion 1 hundred 11... i.e. 1,11,11,11,11,111.

Smallest number using 24 sticks? Use '8' (7 sticks) as much as possible but minimize digits. 24 ÷ 7 = 3 remainder 3. 3 sticks → digit '7'. So smallest = 788? No — fewest digits means largest digits. 888 = 7+7+7 = 21 sticks, need 3 more → add a '7' before: 7888 = 3+7+7+7 = 24 sticks. Is 7888 smaller than 888? No, it has 4 digits! For fewest digits: Use 8,8 = 14 sticks, one more 8 = 21, plus 3 → 7 → 7888. Or check: 888 with remaining 3 sticks → put '7' in front → 7888. Smallest number = 7888. Alternatively, 8887 is bigger, so 7888 < 8887. Answer: 7888.

Chapter Exercises — Figure it Out

Ex 1. Using all digits from 0–9 exactly once (first digit ≠ 0), create a 10-digit number. Write the:
(a) Largest multiple of 5 (b) Smallest even number

(a) Largest multiple of 5: Must end in 0 or 5. To maximise, end in 5 and arrange remaining digits in descending order:
9,87,64,32,10,5 — wait, we need exactly 10 digits using 0-9 once each. Ending in 5: 9876432105. Ending in 0: 9876543210. Since 9876543210 > 9876432105, the largest multiple of 5 is 9,87,65,43,210. ✔

(b) Smallest even number: Must end in 0, 2, 4, 6, or 8. To minimize, start with the smallest non-zero digit. First digit = 1, second = 0, then arrange remaining in ascending order, ending with an even digit. Try: 1023456789 — ends in 9 (odd). Try: 1023456798 — ends in 8 (even). ✔ Answer: 1,02,34,56,798

Ex 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 with the maximum number of letters.

To maximise letters, each word in the name should be as long as possible. Long number words: "seventy-seven" (12 letters), "thirty-three" (11 letters), "forty-four" (9), "sixty-six" (8), etc. A 7-digit number: __ __ , __ __ , __ __ __.

Example: 77,77,777 = "Seventy-seven lakh seventy-seven thousand seven hundred seventy-seven" = 12+4+6+5+8+5+5+5+6 = many letters. Count: S-e-v-e-n-t-y-s-e-v-e-n (12) + l-a-k-h (4) + ... This is a good candidate. Students should explore systematically — one optimal answer is 77,77,777.

Ex 3. Write a 9-digit number where exchanging any two digits results in a bigger number. How many such numbers exist?

If exchanging any two digits makes the number bigger, the original must already be the smallest possible arrangement — i.e., digits sorted in ascending order from left to right (no leading zero).

A 9-digit number with digits in ascending order: example 1,23,45,678 or 1,12,23,344 (with repeats). Any swap of two positions where the left digit is smaller than the right digit gives a bigger number.

The simplest answer: any number whose digits are in non-decreasing order from left to right (with the first digit ≥ 1). Example: 1,23,45,689. There are many such numbers.

Ex 4. Strike out 10 digits from 12345123451234512345 so that the remaining number is as large as possible.

The original number has 20 digits: 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5. We remove 10, leaving 10 digits. To maximise the remaining 10-digit number, use a greedy approach: from each window pick the largest digit possible as the next kept digit.

Working greedily (keeping 10 digits, removing 10):
We want to keep 10 digits such that the resulting number is largest.
Greedy scan: pick largest digit from positions 1–11 (since we need 10 more from remaining), etc.
One valid large result: 5,54,53,45,45... Applying the standard greedy algorithm carefully, the answer is 5,51,23,45,45. Students should verify by systematic greedy selection.

Ex 5. 'Zero' and 'one' share letters 'e' and 'o'. 'One' and 'two' share 'o'. 'Two' and 'three' share 't'. How far do you count to find two consecutive numbers that share no English letter in common?

Check consecutive pairs systematically:
zero/one: share e, o ✗ one/two: share o ✗ two/three: share t ✗ three/four: share r ✗ four/five: share f ✗ five/six: share i ✗ six/seven: share s, e ✗ seven/eight: share e ✗ eight/nine: share i, e ✗
nine/ten: share n, e ✗ ten/eleven: share e, n ✗ eleven/twelve: share e, l, v ✗ twelve/thirteen: share t, e ✗ ... Continue:
twenty-one/twenty-two: share t, w, e, n, y ✗
This is a research challenge — the answer is that consecutive numbers sharing no common letter turns out to occur first at quite large numbers. This is left as an exploration for the class.

Ex 6. You write 1, 2, 3, 4, …, 9, 10, 11, … The 10th digit written is '1' (from 10) and 11th is '0'.
(a) What is the 1000th digit, and in which number does it occur?
(b) What number contains the millionth digit?
(c) When have you written digit '5' for the 5000th time?

Digit count structure:
1-digit numbers (1–9): 9 numbers × 1 digit = 9 digits
2-digit numbers (10–99): 90 numbers × 2 digits = 180 digits (cumulative: 189)
3-digit numbers (100–999): 900 numbers × 3 digits = 2700 digits (cumulative: 2889)
4-digit numbers (1000–9999): 9000 × 4 = 36,000 digits (cumulative: 38,889)

(a) 1000th digit: After 9 (1-digit) + 180 (2-digit) = 189 digits. Remaining: 1000 − 189 = 811 digits into 3-digit numbers. 811 ÷ 3 = 270 remainder 1. The 270th 3-digit number from 100 is 369. The 1st digit of the next number, 370, is '3'. The 1000th digit is 3 occurring in the number 370.

(b) 1,000,000th digit: After 2-digit: 189. After 3-digit: 2889. After 4-digit: 38,889. After 5-digit (10000–99999): 90,000 × 5 = 4,50,000 (cumulative: 4,88,889). After 6-digit: 9,00,000 × 6 = 54,00,000... Remaining after 5-digit numbers: 10,00,000 − 4,88,889 = 5,11,111 digits into 6-digit numbers. 5,11,111 ÷ 6 = 85,185 remainder 1. The 85,185th 6-digit number from 100000 is 185,184. The 1,000,000th digit is the 1st digit of 185,185 = 1, in the number 1,85,185.

(c) 5000th occurrence of digit '5': This requires counting how often '5' appears in each range — this is an advanced counting problem suitable for exploration.

Ex 7. A calculator has only '+10,000' and '+100' buttons. Write an expression for button clicks to reach:
(a) 20,800 (b) 92,100 (c) 1,20,500 (d) 65,30,000 (e) 70,25,700

(a) 20,800 = 2 × 10,000 + 8 × 100 → Press +10,000 twice, +100 eight times.

(b) 92,100 = 9 × 10,000 + 21 × 100 → Press +10,000 nine times, +100 twenty-one times.

(c) 1,20,500 = 12 × 10,000 + 5 × 100 → Press +10,000 twelve times, +100 five times.

(d) 65,30,000 = 653 × 10,000 + 0 × 100 → Press +10,000 six hundred fifty-three times.

(e) 70,25,700 = 702 × 10,000 + 57 × 100 → Press +10,000 seven hundred two times, +100 fifty-seven times.

Ex 8. How many lakhs make a billion?

1 billion = 1,000,000,000 = 1,00,00,00,000
1 lakh = 1,00,000
Number of lakhs in a billion = 1,00,00,00,000 ÷ 1,00,000 = 10,000 lakhs = 10 thousand lakhs

Ex 9. You have two sets of number cards 1–9. Place a card in each box to get:
(a) Largest possible sum (b) Smallest possible difference
□□□□□ + □□□□□ and □□□□□ − □□□□□

(a) Largest sum: To maximise the sum of two 5-digit numbers using digits 1–9 twice, place the largest digits in the highest place values. Pair digits: 9 and 9 in ten-thousands, 8 and 8 in thousands, etc. → 98,765 + 98,765 = 1,97,530. But each digit from 1–9 appears exactly once per set, so: 97,531 + 97,531? No, we have two sets, so each digit 1–9 appears twice total. Best arrangement: 97,531 + 86,424 or systematically: assign 9,8,7,6,5 to one number and 9,8,7,6,5 — wait, both sets are 1–9 so each digit appears twice. Largest sum: put largest digits in ten-thousands place of both numbers. 98,765 + 94,321... Many valid answers; the key is to interleave large digits across both numbers' highest places.

(b) Smallest difference: To minimise the difference, make the two numbers as close together as possible. Use consecutive arrangements: e.g., 58,413 − 58,412 isn't possible (repeated digits). Best approach: alternate digits between numbers so they are nearly equal. Answer is difference = 1 using adjacent digit arrangements.

Ex 10. Using cards 4000, 13000, 300, 70000, 150000, 20, 5 (each used at most once per target), get as close as possible to:
(b) 2,00,000 (c) 5,80,000 (d) 12,45,000 (e) 20,90,800

(b) 2,00,000: 150000 + 13000 × (20 + 5) + 300... Try: 150000 + 13000 + 70000 − 300 × 20 + ... Try: 70000 + 150000 − 13000 + 4000 − 300 + 20 − 5 = 2,10,715. Better: 4000 × (20+5) + 70000 + 300 + 13000 − 5 = 1,00,000+70,000+300+13,000−5 = 1,83,295. Try: 150000 + 70000 − 20000 = not possible (no 20000 card). Try 150000 + 13000 × (5−4000÷...) — exploration needed. Sample: 150000 + 70000 − 20000 — students explore.

(c) 5,80,000: 4000 × (150000 ÷ ...) — try: 150000 × (13000 ÷ ...). Try: 150000 + 430000 = 150000 + 70000 × (13000÷300)... Exploration: 150000 × 4 − 20000 — but × requires card-based ops. Students explore with +, −, ×, ÷.

(d) 12,45,000: 13000 × (70000 ÷ 4000) + 150000 + ... Students explore combinations.

(e) 20,90,800: 70000 × (13000 ÷ ... ) + ... Students explore. These are open-ended exploration questions — share findings with class.

Ex 11. Find how many coins stacked on top of each other match the height of the Statue of Unity. Each coin is 1 mm thick.

The Statue of Unity (including base) is 240 metres tall (statue alone = 182 m, with base = 240 m).

Height of statue alone = 182 m = 182 × 1000 mm = 1,82,000 mm

Each coin = 1 mm thick, so number of coins = 1,82,000 coins (≈ 1 lakh 82 thousand).

Including base: 240 m = 2,40,000 mm → 2,40,000 coins = 2 lakh 40 thousand coins.

Ex 12. Grey-headed albatrosses cover about 900–1000 km/day. One of the longest single trips recorded is about 12,000 km. How many days would such a trip take? Approximately, how long to cross the Pacific Ocean?

Longest single trip: 12,000 km at 900–1000 km/day
At 1000 km/day: 12,000 ÷ 1000 = 12 days
At 900 km/day: 12,000 ÷ 900 ≈ 13.3 days

Pacific Ocean width ≈ 12,000–15,000 km. At 900 km/day, crossing takes approximately 12,000 ÷ 900 ≈ 13 to 17 days.

Ex 13. A bar-tailed godwit flew 13,560 km from Alaska to Australia non-stop in about 11 days (starting 13 Oct 2022). Find the approximate distance covered every day and every hour.

Total distance: 13,560 km | Total time: 11 days

Per day: 13,560 ÷ 11 ≈ 1,232 km/day

Per hour: 1,232 ÷ 24 ≈ 51.3 km/hour ≈ 51 km/hour

This is remarkable — the bird flew non-stop at ~51 km/h for 11 consecutive days!

Ex 14. Bald eagles fly 4500–6000 m above ground. Mount Everest ≈ 8850 m. Aeroplanes fly 10,000–12,800 m. Somu's building is 15 m high. How many times bigger are these heights compared to Somu's building?

Somu's building = 15 m

Bald eagle: 4500–6000 m ÷ 15 = 300 to 400 times taller

Mount Everest: 8850 ÷ 15 = 590 times taller

Aeroplane: 10,000–12,800 ÷ 15 = 667 to 853 times taller

🏆

Competency-Based Questions — Patterns, Products & Large Number Applications

Context: Students in a CBSE Class 7 school are participating in a Maths olympiad. They encounter problems involving multiplication shortcuts, digit count patterns, and large-number reasoning. Understanding these concepts helps them tackle real-world calculations efficiently — from population analysis to astronomical distances — without always relying on calculators.

Q1. Using the shortcut method, 4 × 125 × 8 × 25 equals:

L3 Apply

(A) 1,00,000
(B) 10,000
(C) 1,000
(D) 10,00,000

Answer: (A) 1,00,000
Group: (4 × 25) × (125 × 8) = 100 × 1000 = 1,00,000. ✔

Q2. Fill in the blanks:
(i) The product of a 4-digit number and a 3-digit number has _____ or _____ digits.
(ii) 72 × 125 = 72 × (1000 ÷ ___) = ___ × 1000 = ___
(iii) Multiplying by 50 is the same as dividing by ___ and multiplying by ___.

L2 Understand

(i) 6 or 7 digits. (min: 1000×100=10⁵=6 digits; max: 9999×999<10⁷=7 digits)
(ii) 72 × 125 = 72 × (1000 ÷ 8) = 9 × 1000 = 9,000
(iii) Dividing by 2 and multiplying by 100. (50 = 100/2)

Q3. Purandaradāsa reportedly composed 4,75,000 kīrtanas. If he composed for 60 years of his life, approximately how many did he compose per day? Express your answer rounded to the nearest ten.

L3 Apply

Solution:
Total kīrtanas = 4,75,000
Years composing = 60
Days composing = 60 × 365 = 21,900 days
Per day = 4,75,000 ÷ 21,900 ≈ 21.7 songs per day
Rounded to nearest ten: approximately 20 kīrtanas per day.
That is an incredible creative output maintained over 60 years!

Q4. Light travels at approximately 3,00,000 km per second. The Sun is about 14,70,00,000 km away. (a) How many seconds does sunlight take to reach Earth? (b) How many minutes is this? (c) If the Sun suddenly "switched off," how long would it take for Earth to go dark?

L4 Analyse

(a) Time = Distance ÷ Speed = 14,70,00,000 ÷ 3,00,000 = 490 seconds
(b) 490 ÷ 60 ≈ 8.17 minutes ≈ 8 minutes 10 seconds
(c) If the Sun switched off, Earth would go dark after approximately 8 minutes and 10 seconds — because that is how long the last light from the Sun would take to reach us.

Q5. (HOT) In the pattern 11×11=121, 111×111=12321, 1111×1111=1234321...
(a) What is the value of 11111×11111?
(b) Can you explain WHY this palindrome pattern occurs using place value expansion?

L6 Create

(a) Following the pattern: 11111 × 11111 = 1,23,45,4,321 = 123,454,321
Digits go: 1,2,3,4,5,4,3,2,1 — count up to 5 (number of 1s) then count down.

(b) Why it works: 111 = 100+10+1. So 111×111 = (100+10+1)² = 10000 + 1000 + 100 + 1000 + 100 + 10 + 100 + 10 + 1
= 10000 + (1000+1000) + (100+100+100) + (10+10) + 1
= 10000 + 2000 + 300 + 20 + 1 = 12,321 ✔
The coefficients 1,2,3,2,1 arise because each place value position can be "hit" by multiple combinations — the middle position (hundreds) is hit by all three terms (100×1, 10×10, 1×100), giving coefficient 3.

🔬 Assertion–Reason Questions

(A) Both true; Reason is the correct explanation (B) Both true; Reason is NOT the correct explanation (C) Assertion true; Reason false (D) Assertion false; Reason true

Assertion (A): Multiplying a 5-digit number by another 5-digit number always gives a 9- or 10-digit product.
Reason (R): The product of an m-digit and an n-digit number has either (m+n−1) or (m+n) digits.

Answer: (A)

Both true. For m=5, n=5: (5+5−1)=9 or (5+5)=10 digits. Min: 10000×10000=10⁸ (9 digits); Max: 99999×99999 < 10¹⁰ (10 digits). The Reason correctly explains the Assertion.

Assertion (A): A quick way to calculate 360 × 25 is: 360 ÷ 4 × 100 = 9000.
Reason (R): 25 = 100 ÷ 4, so multiplying by 25 is equivalent to dividing by 4 and then multiplying by 100.

Answer: (A)

Both true. 360 × 25 = 360 × (100/4) = (360/4) × 100 = 90 × 100 = 9,000. ✔ The Reason correctly explains the shortcut used in the Assertion.

Assertion (A): In the pattern 11×11=121, 111×111=12321, the product of n ones multiplied together has (2n−1) digits.
Reason (R): The product of two n-digit numbers has either (2n−1) or 2n digits.

Answer: (B)

Both statements are true. 11 (2 digits) → 121 (3 = 2×2−1 digits) ✔; 111 (3 digits) → 12321 (5 = 2×3−1 digits) ✔. The Reason is also a true general rule. However, the Reason does not fully explain why specifically the "repunit" products (111...×111...) always give (2n−1) digits rather than 2n — that requires showing the product is always less than 10^(2n). So Reason is true but is not the specific explanation.

📖 Chapter Summary — Large Numbers Around Us

1 Lakh = 1,00,000 (1 followed by 5 zeroes). 1 Crore = 1,00,00,000 (7 zeroes). 1 Arab = 1,00,00,00,000 (9 zeroes).
Indian system uses 3–2–2–2 comma grouping; International system uses 3–3–3 grouping. 1 million = 10 lakhs; 1 billion = 100 crores.
Rounding gives an approximate value — rounding up gives a number greater than the original; rounding down gives a smaller number.
The five nearest neighbours of a number are obtained by rounding to the nearest thousand, ten-thousand, lakh, ten-lakh, and crore.
Multiplication shortcuts: ×5 = ÷2 then ×10; ×25 = ÷4 then ×100; ×125 = ÷8 then ×1000.
The product of an m-digit and n-digit number always has (m+n−1) or (m+n) digits.
Many interesting number patterns emerge in products of repdigit numbers (11×11, 111×111, etc.).
Large numbers help us make sense of real-world data — from city populations and river discharge to distances in space.
Estimation and reasoning in stages (breaking a large calculation into smaller steps) makes working with large numbers manageable.

Frequently Asked Questions

What types of exercises are in Class 7 Maths Chapter 1?

NCERT Class 7 Maths Chapter 1 exercises include writing numbers in expanded and standard form, comparing large numbers, converting between Indian and international number systems, rounding and estimation, reading data tables with population figures, and recognising number patterns. Solutions involve lakhs and crores.

How do you solve place value problems for large numbers?

To solve place value problems, identify each digit position from right to left: ones, tens, hundreds, thousands, ten-thousands, lakhs, ten-lakhs, crores. Multiply each digit by its place value and add to get the number. NCERT Class 7 Chapter 1 exercises provide extensive practice.

What number patterns are covered in Class 7 Chapter 1?

Class 7 Chapter 1 covers patterns in large numbers such as multiplying by 10 to shift digits left, relationships between consecutive place values, patterns when adding or subtracting round numbers, and sequences formed by population growth data. These patterns reinforce place value understanding.

How do you estimate sums and differences of large numbers?

To estimate sums or differences, round each number to the same convenient place value, then add or subtract the rounded values. For example, to estimate 4,73,826 plus 2,18,459, round to the nearest lakh: 5 lakhs plus 2 lakhs equals approximately 7 lakhs. NCERT Class 7 teaches this skill.

Why is learning large numbers important in Class 7?

Learning large numbers builds number sense essential for understanding real-world data like populations, distances, budgets and scientific measurements. It prepares students for decimals, fractions and data handling in later chapters. NCERT Class 7 Ganita Prakash makes this relevant through authentic Indian data.

Frequently Asked Questions — Large Numbers Around Us

What is Number Patterns and Chapter Exercises in NCERT Class 7 Mathematics?

Number Patterns and Chapter Exercises is a key concept covered in NCERT Class 7 Mathematics, Chapter 1: Large Numbers Around Us. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Number Patterns and Chapter Exercises step by step?

To solve problems on Number Patterns and Chapter Exercises, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 7 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 1: Large Numbers Around Us?

The essential formulas of Chapter 1 (Large Numbers Around Us) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Number Patterns and Chapter Exercises important for the Class 7 board exam?

Number Patterns and Chapter Exercises is part of the NCERT Class 7 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Number Patterns and Chapter Exercises?

Common mistakes in Number Patterns and Chapter Exercises include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Number Patterns and Chapter Exercises?

End-of-chapter NCERT exercises for Number Patterns and Chapter Exercises cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 1, and solve at least one previous-year board paper to consolidate your understanding.

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