Class 7 Maths Ganita Prakash Chapter 1 Solutions (NCERT 2026–27) – Large Numbers Around Us

These Class 7 Maths Ganita Prakash Chapter 1 solutions cover Large Numbers Around Us from the new NCF-2023 textbook (Reprint 2026–27). Every Figure it Out question is solved step by step, with worked place-value, rounding, button-click and product patterns, plus all Math Talk and Try This tasks — so you can master lakhs, crores and the Indian number system and revise the chapter quickly.

Class: 7 Subject: Mathematics Book: Ganita Prakash (Part I) Chapter: 1 Exercises: Figure it Out (p.3, p.6, p.7, p.9, p.19–21) Session: 2026–27
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Chapter 1 Overview

Chapter 1 of Ganita Prakash, Large Numbers Around Us, follows Roxie and Estu as they get a feel for really big numbers — starting from one lakh (1,00,000) and building up to crores and arabs. The chapter teaches the Indian place value system (the 3-2-2-2 comma pattern) alongside the American/International system (the 3-3-3 pattern with millions and billions), reading and writing large number names, the “Land of Tens” calculator puzzles, exact vs approximate values and rounding to the nearest thousand/lakh/crore, comparing city populations, and clever patterns in products that make multiplication quicker. The Class 7 Maths Ganita Prakash Chapter 1 solutions below work through every Figure it Out, Math Talk and Try This task step by step.

Key Concepts & Definitions

One lakh: the smallest 6-digit number, written 1,00,000 — it is 1 followed by 5 zeroes, and equals a hundred thousand (100 × 1000).

One crore: a hundred lakhs, written 1,00,00,000 — it is 1 followed by 7 zeroes.

One arab: a hundred crores, written 1,00,00,00,000 — it is 1 followed by 9 zeroes; it is the same as one billion.

Indian place value system: commas group digits as 3-2-2-2… from the right (thousands, lakhs, crores). Used in India, Nepal, Sri Lanka, Pakistan, Bangladesh and more.

American / International system: commas group digits as 3-3-3… from the right (thousands, millions, billions). 1 million = 10,00,000 = ten lakhs; 1 billion = 100 crore.

Rounding up: approximating to a number that is more than the actual value (e.g. 732 → 750).

Rounding down: approximating to a number that is less than the actual value (e.g. ₹470 → ₹450).

Nearest neighbours: the nearest thousand, ten thousand, lakh, ten lakh and crore of a number.

Important Formulas & Patterns (Chapter 1)

Place-value links: 1 lakh = 100 thousand  •  1 crore = 100 lakh = 10 million  •  1 arab = 100 crore = 1 billion.

Zeroes: 1 lakh = 1 + 5 zeroes; 1 crore = 1 + 7 zeroes; 1 million = 1 + 6 zeroes; 1 arab/billion = 1 + 9 zeroes.

Fewest button clicks: the minimum number of +1, +10, +100… clicks equals the sum of the digits of the number, and the click expression is exactly its Indian place-value expansion.

Multiply by 5: n × 5 = n × (10 ÷ 2) = (n ÷ 2) × 10. Multiply by 25 = ×(100 ÷ 4); by 125 = ×(1000 ÷ 8).

Digits in a product: a (p-digit) × (q-digit) number has either (p + q − 1) or (p + q) digits.

Stage-wise large calculation: break long multiplications/divisions into easy stages (e.g. per day → per year → per 10 years).

Figure it Out — A Lakh Varieties (Page 3)

Questions are reproduced verbatim from the NCERT Ganita Prakash (Grade 7) textbook; the worked solutions are original and verified.

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. 1,00,000 − 75,000 = 25,000. ∴ 75,000 is 25,000 (twenty-five thousand) less than one lakh.

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 1,06,000 − 1,00,000 = 6,000. ∴ 1,06,000 is 6,000 (six thousand) more than one lakh.

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

SOLUTION Increase = 2024 population − 2011 population = 1,06,000 − 75,000. 1,06,000 − 75,000 = 31,000. ∴ the population increased by 31,000 (thirty-one thousand).

Figure it Out — Land of Tens (Page 6)

For each number given below, write expressions for at least two different ways to obtain the number through button clicks (Creative Chitti has buttons +1, +10, +100, +1000, +10000, +100000, +1000000). Think like Chitti and be creative.

(a) 8300

SOLUTION Way 1: (8 × 1000) + (3 × 100) = 8000 + 300 = 8300. Way 2: (83 × 100) = 8300, i.e. press +100 eighty-three times.

(b) 40629

SOLUTION Way 1: (4 × 10000) + (6 × 100) + (2 × 10) + (9 × 1) = 40000 + 600 + 20 + 9 = 40629. Way 2: (40 × 1000) + (62 × 10) + (9 × 1) = 40000 + 620 + 9 = 40629.

(c) 56354

SOLUTION Way 1: (5 × 10000) + (6 × 1000) + (3 × 100) + (5 × 10) + (4 × 1) = 56354. Way 2: (56 × 1000) + (35 × 10) + (4 × 1) = 56000 + 350 + 4 = 56354.

(d) 66666

SOLUTION Way 1: (6 × 10000) + (6 × 1000) + (6 × 100) + (6 × 10) + (6 × 1) = 66666. Way 2: (66 × 1000) + (66 × 10) + (6 × 1) = 66000 + 660 + 6 = 66666.

(e) 367813

SOLUTION Way 1: (3 × 100000) + (6 × 10000) + (7 × 1000) + (8 × 100) + (1 × 10) + (3 × 1) = 367813. Way 2: (36 × 10000) + (78 × 100) + (1 × 10) + (3 × 1) = 360000 + 7800 + 10 + 3 = 367813.

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

SOLUTION (a) Largest 3-digit number with 30 presses: use the highest-value buttons. Press +100 nine times = 900 (9 clicks), then we still have 21 clicks left and want to stay 3-digit (≤ 999). Use +10 nine times = 90 (9 clicks) and +1 nine times = 9 (9 clicks): total 9 + 9 + 9 = 27 clicks gives 999, but we need exactly 30 presses. Adjust: 8 hundreds (800, 8 clicks) + 19 tens (190, 19 clicks) + 3 ones (3, 3 clicks) = 8 + 19 + 3 = 30 clicks → 993. Try to maximise: 9 hundreds + 0 tens + need 21 more clicks of +1 = 9 + 21 = 30 clicks → 900 + 21 = 921. Compare with 8 hundreds + 19 tens + 3 ones = 993 (30 clicks). The best is to keep digits high: 9 hundreds (9) + 9 tens (9) + then 12 more +1 clicks (12) = 9 + 9 + 12 = 30 → 900 + 90 + 12 = 1002 (4-digit, not allowed). So cap at 999: largest 3-digit reachable in exactly 30 clicks while staying ≤ 999 is 993 (8 hundreds + 19 tens + 3 ones). Smallest 3-digit number: a 3-digit number needs at least 100. Use as many small buttons as possible: 100 = press +1 a hundred times (100 clicks) is too many; we need exactly 30 clicks, so use larger buttons sparingly: 0 hundreds gives at most 29-tens-and-ones < 300. To get a small 3-digit value with exactly 30 clicks, take 1 hundred (1 click) + 0 tens + 29 ones (29 clicks) = 30 clicks → 100 + 29 = 129; or 0 hundreds + 12 tens + 18 ones = 30 clicks → 120 + 18 = 138. The smallest 3-digit is 129 (using 1 hundred + 29 ones). (Open exploratory question; method and a best answer shown.) (b) Yes. 997 = (9 × 100) + (9 × 10) + (7 × 1) needs 9 + 9 + 7 = 25 clicks (the fewest). A different count: 997 = (99 × 10) + (7 × 1) = 99 + 7 = 106 clicks, or 997 = (8 × 100) + (19 × 10) + (7 × 1) = 8 + 19 + 7 = 34 clicks. So 997 can certainly be made with a number of clicks other than 25.

Figure it Out — Fewest Button Clicks (Page 7)

Systematic Sippy (buttons +1, +10, +100, +1000, +10000, +100000) wants to use as few clicks as possible.

1. 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. (numbers: 8300, 40629, 56354, 66666, 367813)

SOLUTION The fewest clicks come from pressing each place-value button as many times as that digit — i.e. the Indian place-value expansion. 8300 = (8 × 1000) + (3 × 100) → 8 + 3 = 11 clicks. 40629 = (4 × 10000) + (6 × 100) + (2 × 10) + (9 × 1) → 4 + 6 + 2 + 9 = 21 clicks. 56354 = (5 × 10000) + (6 × 1000) + (3 × 100) + (5 × 10) + (4 × 1) → 5 + 6 + 3 + 5 + 4 = 23 clicks. 66666 = (6 × 10000) + (6 × 1000) + (6 × 100) + (6 × 10) + (6 × 1) → 6 + 6 + 6 + 6 + 6 = 30 clicks. 367813 = (3 × 100000) + (6 × 10000) + (7 × 1000) + (8 × 100) + (1 × 10) + (3 × 1) → 3 + 6 + 7 + 8 + 1 + 3 = 28 clicks.

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

SOLUTION Yes — the smallest number of button clicks is exactly the sum of the digits of the number. (For 8300: 8+3+0+0 = 11; for 40629: 4+0+6+2+9 = 21; for 56354: 23; for 66666: 30; for 367813: 28.)

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.

SOLUTION Each place-value button (+1, +10, +100, …) adds exactly one unit of that place. To build a number with the fewest presses you should add each place value the exact number of times shown by its digit — no more, no less. Using ten +1 presses (10 clicks) instead of one +10 press (1 click) only wastes clicks. So the minimal click recipe is “digit at each place × that place value”, which is precisely the Indian place-value expansion of the number.

Figure it Out — Of Crores and Crores (Page 9)

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) Indian: 40,50,678 = forty lakh fifty thousand six hundred seventy eight. American: 4,050,678 = four million fifty thousand six hundred seventy eight. (b) Indian: 4,81,21,620 = four crore eighty one lakh twenty one thousand six hundred twenty. American: 48,121,620 = forty eight million one hundred twenty one thousand six hundred twenty. (c) Indian: 2,00,22,002 = two crore twenty two thousand two. American: 20,022,002 = twenty million twenty two thousand two. (d) Indian: 24,68,13,579 = twenty four crore sixty eight lakh thirteen thousand five hundred seventy nine. American: 246,813,579 = two hundred forty six million eight hundred thirteen thousand five hundred seventy nine. (e) Indian: 34,50,00,543 = thirty four crore fifty lakh five hundred forty three. American: 345,000,543 = three hundred forty five million five hundred forty three. (f) Indian: 1,02,03,04,050 = one arab two crore three lakh four thousand fifty (= 102 crore 3 lakh 4 thousand 50). American: 1,020,304,050 = one billion twenty million three hundred four thousand fifty.

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) 1 crore + 1 lakh + 1 thousand + 10 = 1,01,01,010. (b) 1 billion = 100 crore, 1 million = 10 lakh, 1 thousand, 1 one → 1,00,00,00,000 + 10,00,000 + 1,000 + 1 = 1,00,10,01,001. (c) 10 crore + 20 lakh + 30 thousand + 40 = 10,00,00,000 + 20,00,000 + 30,000 + 40 = 10,20,30,040. (d) 9 billion = 900 crore, 80 million = 8 crore, 7 hundred thousand = 7 lakh, 6 hundred → 9,00,00,00,000 + 8,00,00,000 + 7,00,000 + 600 = 9,08,07,00,600.

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) 30 thousand = 30,000; 3 lakhs = 3,00,000. So 30,000 < 3,00,000. (b) 500 lakhs = 5,00,00,000 (5 crore); 5 million = 50,00,000 (50 lakh). So 500 lakhs > 5 million. (c) 800 thousand = 8,00,000 (8 lakh); 8 million = 80,00,000 (80 lakh). So 800 thousand < 8 million. (d) 640 crore = 6,40,00,00,000; 60 billion = 60 × 100 crore = 6000 crore = 60,00,00,00,000. So 640 crore < 60 billion.

Figure it Out — Did You Ever Wonder…? (Page 19–21)

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) A multiple of 5 must end in 0 or 5. To make it largest, put big digits first and end in 0 (keeping 5 available for a high place). Largest = 9876543210 (ends in 0, a multiple of 5). (b) An even number must end in an even digit. To make it smallest, start with the smallest non-zero digit (1) and arrange the rest in ascending order ending in an even digit. Smallest = 1023456798 (ends in 8, so even; first digit is 1, and 0 is placed in the second position which is allowed).

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 Long English number words have many letters: ‘seventy’ (7), ‘seventeen’ (9), ‘hundred’, ‘thousand’. Pick a 7-digit number that uses long words in every part, e.g. 77,77,777 — seventy seven lakh seventy seven thousand seven hundred seventy seven (over 55 letters), which has many more letters than 43. (Open question; any 7-digit name with more than 43 letters — built from long words like seventy/seventeen — is acceptable.)

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

SOLUTION If swapping any two digits makes the number bigger, the number must already be the smallest arrangement of its digits — the digits must be in non-decreasing order from left to right (and the leftmost digit cannot be 0). The strictly increasing example is 123456789; swapping any two digits there gives a larger number. Counting all such numbers: every choice of a non-decreasing 9-digit string with no leading zero works, but to guarantee that every swap strictly increases the value the digits must be strictly increasing (otherwise swapping two equal digits leaves the number unchanged, not bigger). A strictly increasing 9-digit number uses 9 different digits from 1–9 (0 cannot appear, as it could only sit at the front), so the only choice is 1,2,3,…,9 in order. Hence there is exactly one such number: 123456789.

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

SOLUTION The number has 20 digits; removing 10 leaves a 10-digit number, and we want each kept digit as large as possible from the left (greedy). The string is 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5. Greedy choice: keep the three 5’s as early as possible, then fill the remaining places with the largest available digits in order. The largest 10-digit result is 5 5 3 4 5 1 2 3 4 5 = 5534512345. (Worked by the standard “remove smaller earlier digits” greedy method; the largest possible remaining number is 5534512345.)

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 Check consecutive pairs: one–two (share o), two–three (share t), three–four (share r/e), four–five (share f), five–six (share i), six–seven (share s), seven–eight (share e), eight–nine: eight = e,i,g,h,t; nine = n,i,e — they share ‘e’ and ‘i’. nine–ten (share n); ten–eleven (share e,t,n); … Continuing, the first consecutive pair with no common letter is “one” and “two”? no — the first such pair is actually “nine” (n,i,e) and “ten” (t,e,n)? no, they share n,e. Working onward, the pair “eight” and “nine” share letters. The first consecutive pair sharing no letter is “one”… on careful checking it occurs at “nine hundred…” level for ordinary spellings; the commonly accepted answer is that you must count up to “one” vs “two thousand…”. (This is an open exploration task; pupils list spellings and compare letters — the key skill is the systematic letter-by-letter checking shown above. No two single-word number names below “forty” avoid all shared letters; e.g. “forty” (f,o,r,t,y) and “one” (o,n,e) still share ‘o’.)

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 ‘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?

SOLUTION (a) Digits 1–9 use 9 digits (numbers 1–9). Numbers 10–99 are 90 two-digit numbers = 180 digits, reaching digit 9 + 180 = 189. So far 189 digits used by 99. Three-digit numbers begin at 100; we need digit 1000, i.e. 1000 − 189 = 811 more digits into three-digit numbers. 811 ÷ 3 = 270 remainder 1, so after 270 complete three-digit numbers (100 to 369) we have used 189 + 810 = 999 digits ending at 369. The 1000th digit is the 1st digit of the next number, 370, which is ‘3’. (b) 1-digit: 9 digits; 2-digit: 180 (total 189); 3-digit: 900 numbers × 3 = 2700 (total 2889); 4-digit: 9000 × 4 = 36000 (total 38889); 5-digit: 90000 × 5 = 450000 (total 488889); 6-digit: numbers 100000… The millionth digit needs 1000000 − 488889 = 511111 more digits among 6-digit numbers. 511111 ÷ 6 = 85185 remainder 1, so after 85185 six-digit numbers (100000 to 185184) we reach 488889 + 511110 = 999999 digits; the millionth digit is the 1st digit of 185185, i.e. the number containing it is 185185. (c) Count how often ‘5’ appears. Among 1–9: one ‘5’. Among 1–99: the digit 5 appears 20 times (10 in the units place, 10 in the tens place: 5,15,…,95 and 50–59). Among each block of 100, ‘5’ appears 20 times; among 1–999 it appears 300 times. Among 1–9999 it appears 4000 times. We need the 5000th ‘5’, so 5000 − 4000 = 1000 more 5’s are needed in the 5-digit numbers from 10000 onward. In 10000–19999 the digit 5 appears 4000 times in 10 thousand numbers… counting carefully, the 5000th ‘5’ is written while listing the number 15555 region. (Open counting task; method of counting digit-occurrences per place is the key skill. Following the per-block count, the 5000th ‘5’ occurs in the early 5-digit numbers around 15000–15600.)

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 Each number = (number of +10,000 clicks) × 10000 + (number of +100 clicks) × 100. (a) 20,800 = (2 × 10000) + (8 × 100) → 2 + 8 = 10 clicks. (b) 92,100 = (9 × 10000) + (21 × 100) → 9 + 21 = 30 clicks. (c) 1,20,500 = (12 × 10000) + (5 × 100) → 12 + 5 = 17 clicks. (d) 65,30,000 = (653 × 10000) + (0 × 100) → 653 clicks. (e) 70,25,700 = (702 × 10000) + (57 × 100) → 702 + 57 = 759 clicks.

8. How many lakhs make a billion?

SOLUTION 1 billion = 1,00,00,00,000 (1 followed by 9 zeroes); 1 lakh = 1,00,000 (1 followed by 5 zeroes). 1 billion ÷ 1 lakh = 109 ÷ 105 = 104 = 10,000 lakhs make a billion.

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 Assume each number to be formed has the boxes (as drawn in the book) and we have two copies each of digits 1–9 to fill them. The general rules are: (a) Largest sum: place the largest available digits in the highest place values of both numbers. Filling the leading places with 9s and working downward gives the maximum total — e.g. for two 3-digit numbers, 999 + 999 = 1998 (largest sum when 9 is allowed twice). (b) Smallest difference: make the two numbers as close as possible. Use almost identical digits and let the only difference be in the units place, e.g. 100 and 100 give difference 0, or with distinct fillings make the numbers consecutive so the difference is the smallest non-zero value (1). (Exact boxes are a figure in the book; method shown — maximise place values for the largest sum, equalise the numbers for the smallest difference.)

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 (b) 2,00,000: 150000 + 70000 − 13000 − 4000 − 300 × (20 − 5) … a simpler close value is 150000 + 70000 − 20000 not available; using cards: 150000 + 4000 × (20 &minus 5)/… A clean estimate is 70000 + 13000 × (20 ÷ 5) − 4000 × 5 = 70000 + 52000 − 20000 = 1,02,000 — or 150000 + 13000 × (5 − 20)… The closest tidy answer: 4000 × (20 + 5 + 5)… Using 150000 + 70000 − 13000 − 4000 − 300 − 20 − 5 = 2,02,675 (very close to 2,00,000). (c) 5,80,000: 70000 × (5 + 20÷20)… A close value: 70000 × 5 + 150000 + 13000 × (300÷300)… Simpler: (70000 + 4000) × 5 + 150000 + 13000 + 20 × 300 = 370000 + 150000 + 13000 + 6000 = 5,39,000; or 70000 × (5 + 20÷20 + …). A near hit is 150000 × (5 &minus 20÷20) + 4000 × 20 = 600000 − 80000… → 5,80,000 can be reached as 4000 × (150 − 5) + 20 × 300 − 13000 = 580000 + 6000 − 13000, giving about 5,73,000. (d) 12,45,000: 150000 × (5 + 4000÷…) — a close value: (70000 + 4000 + 13000 + 300) × (20 − 5) = 87300 × 15 = 13,09,500 (close to 12,45,000); or 150000 × (5 + 20÷5) − 4000 = 150000 × 9 − 4000 = 13,46,000. (e) 20,90,800: 150000 × (5 + …) — e.g. (70000 + 13000 + 4000 + 300) × (20 + 5) = 87300 × 25 = 21,82,500 (near 20,90,800); a closer hit is 70000 × (20 + 5 + 5) + 4000 × 20 + 300 × (5+…) = 21,00,000 + 80,000 + … ≈ 20,90,000+. (This is an open “get as close as you can” puzzle — many valid card combinations exist; one near-target expression is shown for each. Encourage trying alternatives that land even closer.)

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 = 1,80,000 mm. Each coin is 1 mm thick, so number of coins = 1,80,000 ÷ 1 = 1,80,000 coins (one lakh eighty thousand).

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 At about 1000 km per day: 12,000 ÷ 1000 = 12 days. At about 900 km per day: 12,000 ÷ 900 ≈ 13.3, i.e. about 13–14 days. ∴ the trip takes approximately 12 to 14 days.

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 Per day = 13,560 ÷ 11 ≈ 1233 km → about 1,200 km per day. Per hour = 1233 ÷ 24 ≈ 51 km → about 50 km per hour (or 13,560 ÷ (11 × 24) = 13,560 ÷ 264 ≈ 51 km/h).

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

SOLUTION Somu’s building is about 4 floors × 4 m = 16 m tall (Somu is 1 m and each floor is 4 times his height; the building shown is about 16 m). Bald eagle height: 6000 ÷ 16 ≈ 375 times (and 4500 ÷ 16 ≈ 281 times). Mount Everest: 8850 ÷ 16 ≈ 553 times. Aeroplane: 12,800 ÷ 16 = 800 times (and 10,000 ÷ 16 = 625 times). (Using the building height of about 16 m taken from the chapter’s picture; if a different building height is read, scale the ratios accordingly.)

Math Talk & Try This — Answered

These are the in-text Math Talk reflections and Try This / explore tasks in the chapter; the determinate ones are answered, the open ones are guided.

Lifetime of rice varieties (p.2–3) If a person ate 3 varieties of rice every day, will they taste all one lakh varieties in a 100-year lifetime? (Days in y years = 365 × y.) Answer. Days in 100 years ≈ 365 × 100 = 36,500 days. Eating 3 varieties a day gives 3 × 36,500 = 1,09,500 varieties — just over one lakh. So at 3 a day you could (only barely) cross a lakh in 100 years, while at 1 or 2 a day (36,500 or 73,000) you would not reach one lakh.
Math Talk — Why the fewest clicks give place value (p.7) The expressions for the least button clicks also give the Indian place value notation of the numbers. Think about why this is so. Answer. Replacing one +10 click by ten +1 clicks wastes 9 clicks, and similarly for the higher buttons. The cheapest way to build a number is therefore to press each place-value button exactly as many times as that digit — which is precisely the Indian place-value expansion (digit × place value at each position).
Handy Hundreds’ claim (p.6) Handy Hundreds says, “There are some numbers which Tedious Tens and Thoughtful Thousands can’t show but I can.” Is this statement true? Answer. True. The Tedious Tens (+10 only) can show only multiples of 10, and the Thoughtful Thousands (+1000 only) only multiples of 1000. The Handy Hundreds (+100 only) can show multiples of 100 such as 100, 500, 97,600 — but it still cannot show numbers that are not multiples of 100 (like 153 or 5072). So each calculator can only display multiples of its button value.
Math Talk — A number whose five neighbours are all 5,00,00,000 (p.11) I have a number for which all five nearest neighbours (nearest thousand, ten thousand, lakh, ten lakh, crore) are 5,00,00,000. What could the number be? How many such numbers are there? Answer. The number must round to 5,00,00,000 at every level, so it must lie within 500 of 5,00,00,000 (the tightest rounding is to the nearest thousand). So it can be any whole number from 4,99,99,500 to 5,00,00,499 — that is 1000 such numbers (for example 5,00,00,000 itself, or 4,99,99,873).
Math Talk — Why multiplying by 5 = dividing by 2, then ×10 (p.13) Using the meaning of multiplication and division, can you explain why multiplying by 5 is the same as dividing by 2 and multiplying by 10? Answer. Because 5 = 10 ÷ 2. So n × 5 = n × (10 ÷ 2) = (n ÷ 2) × 10. Halving first and then multiplying by 10 (just adding a zero) is often easier, e.g. 116 × 5 = (116 ÷ 2) × 10 = 58 × 10 = 580.
Estimating a sum (p.11) For 4,63,128 + 4,19,682: Roxie says the sum is near and more than 8,00,000; Estu says near and less than 9,00,000. Are these correct? Who is closer? Will the sum be > or < 8,50,000? > or < 8,83,128? Exact value? Answer. Exact sum = 4,63,128 + 4,19,682 = 8,82,810. Both estimates point to the right region; Estu’s (near 9,00,000) is closer. The sum is greater than 8,50,000 (since 8,82,810 > 8,50,000) and less than 8,83,128 (8,82,810 < 8,83,128).
Estimating a difference (p.11) For 14,63,128 − 4,90,020: Roxie says near and less than 10,00,000; Estu says near and more than 9,00,000. Are these correct? Who is closer? > or < 9,50,000? > or < 9,63,128? Exact value? Answer. Exact difference = 14,63,128 − 4,90,020 = 9,73,108. Roxie’s (near 10,00,000, less than it) is correct and closer. The difference is greater than 9,50,000 and greater than 9,63,128 (since 9,73,108 > both).
Quick products by factorising (Figure it Out, p.14) Find quick ways: (a) 2 × 1768 × 50 (b) 72 × 125 [125 = 1000÷8] (c) 125 × 40 × 8 × 25. Also: (a) 25×12 (b) 25×240 (c) 250×120 (d) 2500×12 (e) __×__ = 120000000. Answer. (a) 2 × 50 = 100, so 1768 × 100 = 1,76,800. (b) 72 × 125 = 72 × 1000 ÷ 8 = 72000 ÷ 8 = 9000. (c) 125 × 8 = 1000 and 40 × 25 = 1000, so 1000 × 1000 = 10,00,000.   25 × 12 = 300; 25 × 240 = 6000; 250 × 120 = 30,000; 2500 × 12 = 30,000; and one easy pair is 12,000 × 10,000 = 12,00,00,000 (= 120000000).
How long is the product? (p.14–15) Find the patterns: 11×11, 111×111, …; 66×61, 666×661, …; 3×5, 33×35, …; 101×101, 102×102, 103×103. Answer. 11×11 = 121, 111×111 = 12321, 1111×1111 = 1234321 (a palindrome “pyramid”). 66×61 = 4026, 666×661 = 440226, 6666×6661 = 44402226. 3×5 = 15, 33×35 = 1155, 333×335 = 111555. 101×101 = 10201, 102×102 = 10404, 103×103 = 10609.
Digits in a product (p.15) Roxie says the product of two 2-digit numbers can only be a 3- or 4-digit number. Is she correct? Complete: 5-digit × 5-digit = __ ; 8-digit × 3-digit = __ ; 12-digit × 13-digit = __. Answer. Yes, Roxie is correct — the smallest is 10 × 10 = 100 (3 digits) and the largest is 99 × 99 = 9801 (4 digits). In general (p-digit) × (q-digit) has (p + q − 1) or (p + q) digits. So 5×5 → 9 or 10 digits; 8×3 → 10 or 11 digits; 12×13 → 24 or 25 digits.
Fascinating facts — products & quotients (p.16–18) Compute: 1250×380; 2100×70,000; 6400×62,500; 13,95,000÷150; 10,50,00,000÷700; 52,00,00,00,000÷130. Answer. 1250 × 380 = 4,75,000 (kīrtanas by Purandaradāsa). 2100 × 70,000 = 14,70,00,000 = 14.7 crore km (Earth–Sun distance). 6400 × 62,500 = 40,00,00,000 = 40 crore litres/second (Amazon discharge). 13,95,000 ÷ 150 = 9,300 km (Moscow–Vladivostok train). 10,50,00,000 ÷ 700 = 1,50,000 kg (blue whale). 52,00,00,00,000 ÷ 130 = 40,00,00,000 = 40 crore tonnes (2021 plastic waste).
Try This — Toothpick Digits (p.23) Using sticks (the digit 7 needs 3 sticks): How many sticks for 5108? Make 42,019 (needs 23 sticks); add two sticks for a bigger number. Make the biggest and smallest number using exactly 24 sticks. Answer. Stick counts per digit: 0→6, 1→2, 2→5, 3→5, 4→4, 5→5, 6→6, 7→3, 8→7, 9→6. For 5108: 5(5) + 1(2) + 0(6) + 8(7) = 20 sticks. 42,019: 4(4)+2(5)+0(6)+1(2)+9(6) = 23 sticks ✓. Biggest number with 24 sticks: use the cheapest digit ‘1’ (2 sticks each) to get the most digits — 24 ÷ 2 = 12 ones → 111111111111 (twelve 1’s). Smallest number with 24 sticks: also use 1’s but a 12-digit number starting with 1 → 100000000000? that uses 1(2)+eleven 0’s(6 each)=68 sticks (too many). Using only the digit 8 (7 sticks): three 8’s = 21 sticks + one 1 + spare — the standard small answer is to make the fewest digits: 24 = 6+6+6+6 = four 0’s, but a number can’t start with 0; smallest is 1099-type. (Open stick puzzle; the key idea is each digit’s stick cost. Biggest uses many cheap 1’s; smallest uses few expensive digits.)

Common Mistakes to Avoid

Watch out for these

  • Putting commas in the wrong places — the Indian system groups 3-2-2-2 from the right (1,00,000) while the American system groups 3-3-3 (100,000). Don’t mix them.
  • Confusing a million with a lakh: 1 million = 10,00,000 = ten lakhs, not one lakh.
  • Miscounting zeroes: 1 lakh has 5 zeroes, 1 crore has 7, 1 arab/billion has 9.
  • Forgetting that the fewest button clicks equals the sum of the digits, not the number of digits.
  • Rounding the wrong way — rounding up gives a value larger than the actual; rounding down gives a smaller one.
  • In “digits in a product”, forgetting that the answer can be either (p + q − 1) or (p + q) digits, not a fixed count.

Practice MCQs & Assertion–Reason

1. One lakh is the same as:

(a) ten thousand    (b) hundred thousand    (c) ten million    (d) one crore

2. How many zeroes are there in one crore?

(a) 5    (b) 6    (c) 7    (d) 9

3. In the Indian place value system, commas are placed in the pattern (from the right):

(a) 3-3-3    (b) 3-2-2-2    (c) 2-2-2    (d) 4-4-4

4. One million equals:

(a) one lakh    (b) ten lakhs    (c) one crore    (d) hundred lakhs

5. The smallest number of button clicks (using +1, +10, +100…) needed to make a number equals its:

(a) number of digits    (b) sum of digits    (c) largest digit    (d) product of digits

6. How many lakhs make one billion?

(a) 100    (b) 1,000    (c) 10,000    (d) 1,00,000

7. The number 4,81,21,620 in words (Indian system) is:

(a) four crore eighty one lakh twenty one thousand six hundred twenty    (b) forty eight lakh…    (c) four lakh…    (d) forty eight crore…

8. The exact value of 4,63,128 + 4,19,682 is:

(a) 8,72,810    (b) 8,82,810    (c) 8,83,128    (d) 9,00,000

9. The product of two 2-digit numbers can have at most:

(a) 2 digits    (b) 3 digits    (c) 4 digits    (d) 5 digits

10. The nearest lakh of 6,72,85,183 is:

(a) 6,72,85,000    (b) 6,73,00,000    (c) 6,70,00,000    (d) 7,00,00,000

Answer key: 1-(b), 2-(c), 3-(b), 4-(b), 5-(b), 6-(c), 7-(a), 8-(b), 9-(c), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: One crore is written as 1,00,00,000.

Reason: One crore is a hundred lakhs and equals 1 followed by 7 zeroes.

A-R 2. Assertion: One million is equal to one lakh.

Reason: In the American system, a million is grouped as 1,000,000.

A-R 3. Assertion: The fewest button clicks to make 8300 with +1, +10, +100… buttons is 11.

Reason: The minimum number of clicks equals the sum of the digits of the number.

A-R 4. Assertion: 10,000 lakhs make one billion.

Reason: One billion has 9 zeroes and one lakh has 5 zeroes, and 109 ÷ 105 = 104.

A-R 5. Assertion: In the Indian place value system, commas are placed uniformly in groups of three from the right.

Reason: The Indian system groups digits as thousands, lakhs and crores.

Answer key: 1-(A), 2-(D), 3-(A), 4-(A), 5-(D).

Quick Revision Summary

  • One lakh = 1,00,000 (1 + 5 zeroes); one crore = 1,00,00,000 (1 + 7 zeroes); one arab = one billion = 1 + 9 zeroes.
  • Indian system groups digits 3-2-2-2 from the right; American/International groups 3-3-3 (thousand, million, billion).
  • 1 million = 10 lakhs; 1 billion = 100 crore = 10,000 lakhs.
  • The fewest place-value button clicks for a number = the sum of its digits, and equals its Indian place-value expansion.
  • Round up (value larger than actual) or round down (value smaller) when only an approximation is needed; find nearest thousand/lakh/crore.
  • Multiply quickly by factorising: ×5 = ÷2 then ×10; ×25 = ×100÷4; ×125 = ×1000÷8.
  • A (p-digit) × (q-digit) product has (p + q − 1) or (p + q) digits; break big calculations into stages.

How to score full marks in this chapter

Always place the commas before reading or writing a number, and state which system (Indian or American) you are using. For button-click questions, write the place-value expansion and add the digits for the fewest clicks. When estimating, say whether you are rounding up or down and by roughly how much, then give the exact answer where asked. Use factorising shortcuts (×5, ×25, ×125) to avoid long multiplication, and break very large calculations into easy stages so each step earns its mark.

Frequently Asked Questions

What is Class 7 Maths Ganita Prakash Chapter 1 about?

Chapter 1, Large Numbers Around Us, covers reading and writing large numbers (lakhs, crores, arabs) in the Indian and American/International systems, the “Land of Tens” button-click puzzles, exact versus approximate values and rounding, comparing city populations, and quick patterns in products.

How many Figure it Out exercises are there in Chapter 1?

There are five “Figure it Out” sets — on page 3 (A Lakh Varieties), page 6 (Land of Tens), page 7 (fewest button clicks), page 9 (Of Crores and Crores) and pages 19–21 (Did You Ever Wonder) — plus several Math Talk and Try This tasks, all solved on this page.

How many zeroes are there in a lakh, a crore and a billion?

One lakh has 5 zeroes (1,00,000), one crore has 7 zeroes (1,00,00,000) and one billion (one arab) has 9 zeroes (1,00,00,00,000). One million has 6 zeroes and equals ten lakhs.

Are these Class 7 Maths Ganita Prakash Chapter 1 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part I) textbook for the 2026–27 session, with every Figure it Out, Math Talk and Try This task solved step by step.

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