Exercise 1.6 is the last exercise in Chapter 1 and helps you to practice dealing with large numbers from real life. Questions are from a factual, real-life context to use what you learnt from ´Working with Big Numbers´ including some additional word problems, comparisons, and scenarios concerning data with big numbers.
This exercise contains the most up-to-date NCERT syllabus, and can be a good review of the whole chapter through meaningful practice. This exercise reviews the part about calculation, estimation, and thinking flexibly that are important for both school tests and exams and all CBSE students. The step-by-step NCERT Solutions will highlight the key aspects of each question while solving, giving a clear understanding of the question.
Practicing this exercise can make you feel much more confident while working with large numbers to solve a word problem, and also will provide you a good foundation for competitions like Maths Olympiads.
Exercise 1.6 contains revision-type word problems and real-world numerical application. The NCERT Solutions for Class 7 Maths Chapter 1 provide simple explanations and detailed steps for each one of the questions. Click below to download the free PDF for revision.
This exercise covers:
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
Sol. (a) To form the largest multiple of 5, the last digit must be 0 or 5 .
Arranging the digits in decreasing order, the largest multiple of 5 is 9876543210 (10 digits).
(b) To form the smallest even number, the last digit must be even ( 0,2,4,6, 8), and the digits should be written in ascending order.
The even number is 1023456798 (10 digits).
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.
Sol. 78,78,773 (Seventy-eight lakhs seventyeight thousand seven hundred seventythree). This has 61 letters, making it one of the longest 7-digit numbers.
3. Write a 9-digit number where exchanging any two digits results in a bigger number. How many such numbers exist?
Sol. To ensure swapping any two digits increases the value, the digits must increase from left to right.
So, the arrangement would be: 123456789.
There is only 1 such number.
4. Strike out 10 digits from the number 12345123451234512345 so that the remaining number is as large as possible.
Sol. The given number is 12345123451234512345.
Removing the 10 highlighted digits from left to right: 12345123451234512345 The number we get is 5534512345 , which is the largest possible number.
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?
Sol. Let's list the spellings of some numbers and check for shared letters between consecutive numbers.
1.Zero (Z, E, R, O) and One (O, N, E) - Share ' e ' and ' o '.
2.One ( 0, N,E ) and Two ( T,W,O ) - Share 'o'.
3.Two ( T,W,O ) and Three ( T,H,R,E,E ) Share 't'.
4.Three (T, H, R, E, E) and Four (F, O, U, R) - Share 'r'.
5.Four ( F,O,U,R ) and Five ( F,I,V,E ) Share 'f'.
6.Five (F, I, V, E) and Six (S, I, X) - Share 'i'.
7.Six (S, I, X) and Seven (S, E, V, E, N) - Share 's'.
8.Seven (S, E, V, E, N) and Eight (E, I, G, H, T) - Share 'e'.
9.Eight (E, I, G, H, T) and Nine (N, I, N, E) Share 'e', 'i'.
10.Nine (N, I, N, E) and Ten (T, E, N) - Share ' e′ ', ' n '.
Similarly, we can check for any two consecutive numbers.
Therefore, there are no consecutive numbers that do not share a letter in common.
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?
Sol. (a) 1-digit numbers (1-9): There are 9 numbers. Each number uses 1 digit. Total digits: 9×1=9 digits. The 9th digit is ' 9 ' (from the number 9). 2-digit numbers (10-99): There are 99−10+1=90 numbers. Each number uses 2 digits. Total digits:
90×2=180 digits. Cumulative digits up to 99 : 9+180=189 digits. The 189 th digit is ' 9 ' (from the number 99). 3-digit numbers (100-999): There are 999−100+1=900 numbers. Each number uses 3 digits. Total digits: 900×3=2700 digits. Cumulative digits up to 999 : 189+2700=2889 digits.
To find the 1000th digit:
Digits so far: 9+180=189.
So, the 1000 th digit will lie in the 3 digit numbers range.
Remaining digits: 1000−189=811. Number of 3-digit numbers to reach 811 digits:
811÷3=270, with 1 remaining number.
Thus, first we need to write the first 270 3-digit numbers starting from 100.
So, the 270th 3-digit number =100+270−1=369.
The next number is 370 .
Thus, the 1000th digit is the 1st digit of 370 , which is 3 .
(b) 4-digit numbers (1000-9999): There are 9999−1000+1=9000 numbers. Each number uses 4 digits. Total digits: 9000×4=36,000 digits. Cumulative digits up to 9999: 2889+36000=38,889 digits.
5-digit numbers (10000-99999): There are 99999−10000+1=90,000 numbers. Each number uses 5 digits. Total digits: 90,000×5=450,000 digits. Cumulative digits up to 99999: 38,889+450,000=488,889 digits.
6-digit numbers (100000999999): There are 999999−100000+1=900,000
numbers. Each number uses 6 digits.
Total digits: 900,000×6=5,400,000 digits.
To reach the millionth digit:
Upto 5-digit numbers: 9+180+2700+36,000+4,50,000=4,88,889
Remaining digits in the 6-digit range: 1,000,000 (or ( 10,00,000 ) 4,88,889=5,11,111
The number of 6 -digit numbers required: 5,11,111÷6=85,185, with 1 remaining number.
So, the 85,185 th 6 -digit number is 85,185+1,00,000−1=1,85,184.
The millionth digit occurs in the number 185184+1=1,85,185.
(c) Single-digit numbers (1-9):
5 occurs only 1 time
Two-digit numbers (10-99)
( 15,25,35,…,95 ), 5 occurs 9 times.
50, 51, 52, ..., 59, 5 occurs 10 times.
Thus, 5 occurs 20 times in two digit numbers.
Three-digit numbers (100-999)
(i) Unit's place: In numbers like 105, 115,…..,995,5 occurs 90 times.
(ii) Ten's place: In numbers like 150159, 250-259, ......, 950-959, 5 occurs 90 times.
(iii) Hundred's place: In numbers like 500-599, 5 occurs 100 times.
Thus, 90 (unit's place) +90 (ten's place) + 100 (hundred's place) = 280 occurrences
Total occurrences so far: 20+280=300
Four-digit numbers (1000-9999)
(i) Unit's place: In numbers like 1005, 1015, ..., 9995, 5 occurs 900 times.
(ii) Ten's place: In numbers like 10501059, 1150-1159, ..., 9950-9959, 5 occurs 900 times
(iii) Hundred's place: In number like 1500-1599,2500-2599,..., 95009599, 5 occurs 900 times.
(iv) Thousands position: 5000-5999, 5 occurs 1000 times
Adding these up: 900 (unit's place) + 900 (ten's place) + 900 (hundred's place) + 1000 (thousand's place) = 3700 occurrences
Total occurrences so far:
300+3700=4000
Five-digit numbers (starting from 10000)
(i) In 10001-10999:
For the 5000th number, we require 5000−4000=1000 more numbers that lie in 10001-10999.
Among 10000-10999, one digit 5 appears in 100 numbers (e.g., 10005, 10015,....., 10995).
The digit 5 appears in 100 numbers (e.g., 10050-10059, ..., 1095010959).
The digit 5 appears in 100 numbers (e.g., 10500-10599).
Total occurrences so far:
4000+300=4300
(ii) In 11000-11999
5 at unit's place =100 times
5 at ten's place = 100 times
5 at a hundred's place = 100 times
Total occurrences so far:
4300+300=4600
(iii) In 12000-12999
5 at unit's place =100 times
5 at ten's place = 100 times
5 at a hundred's place = 100 times
Total occurrences so far:
4600+300=4900
(iv) In 13000-13999
5 occurs at unit place in numbers 13005, 13015 .........
So, 5 occurs for 5000th time in 100th number starting from 13005 i.e. 13995
Required number =13995
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
Sol. (a) 20,800=2×10,000+8×100
Number of clicks =2+8=10 clicks
(b) 92,100=9×10,000+21×100
Number of clicks =9+21=30 clicks
(c) 1,20,500=12×10,000+5×100
Number of clicks =12+5=17 clicks
(d) 65,30,000=653×10,000+0×100
Number of clicks =653+0
= 653 clicks
(e) 70,25,700=702×10,000+57×100 Number of clicks =702+57=759 clicks
8. How many lakhs make a billion?
Sol. 1 lakh = 1,00,000
1 billion = 1,000,000,000
1 billion =10,000×100,000
1 billion = 10,000 lakhs
Thus, 10,000 lakhs make a billion.
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.
Sol.
Total Available Digits:
From two sets of digits 1 to 9:
Each digit from 1 to 9 appears twice.
So, total 18 digits to be used.
(a) Greatest 7 digit number using 18 digits = 9988776
Greatest 5 digit number using remaining 11 digits = 65544
Largest possible sum =9988776+65544=10054320
(b) Smallest 7 digit number using 18 digits = 1122334
Greatest 5 digit number using remaining 11 digits = 99887
Smallest possible difference
= 1122334−99887=1022447
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 :
Sol.
(a) 4000×(20+5)+13000
=4000×25+13000
=100000+13000
=113000
This gives us 1,13,000, which is very close to 1,10,000.
(b) 1,50,000+70,000−4000×5
=2,00,000
(c) 70,000×5+4,000×20+1,50,000
= 5,80,000
(d) 70,000×20−300×5−4,000−1,50,000=12,44,500
This gives us 12,44,500, which is very close to 12,45,000.
(e) 13,000×300−70,000(20+5)+4,000−1,50,000=20,04,000
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.
Sol. Approximate height of the Statue of Unity is 182 m=1,82,000 mm
(Since 1 m=1000 mm )
Thickness of one coin =1 mm
Number of coins needed =1,82,000 mm ÷1 mm=1,82,000 coins
Grey-headed albatrosses have a roughly 7-feet wide wingspan. They are known to migrate across several oceans.
Albatrosses can cover about 900-1000 km in a day. One of the longest single trips recorded is about 12,000 km. How many days would such a trip take to cross the Pacific Ocean approximately?
Sol. Time = Speed Dis tance
To estimate the number of days, divide the total distance by the daily distance covered.
If the albatross flies 900 km per day.
900 km/ day 12000 km=13.33 days ≈13 to 14 days
If the albatross flies 1000 km per day.
1000 km/ day 12000 km=12 days
It would take approximately 12 to 14 days for a grey headed albatross to complete a 12,000 km journey across the Pacific Ocean, depending on its flying speed.
12. A bar-tailed godwit holds the record for the longest recorded non-stop flight. It travelled 13,560 km from Alaska to Australia without stopping. Its journey started on 13 October 2022 and continued for about 11 days. Find out the approximate distance it covered every day. Find out the approximate distance it covered every hour.
Sol. Total distance =13,560 km and duration = 11 days
Approximate distance covered every day =13,560÷11∼1,233 km/ day .
Thus, the godwit covers approximately 1,233 km per day.
One day = 24 hours
Approximate distance covered every hour =1,233÷24∼51 km/ hour .
Thus, the godwit covers approximately 51 km per hour.
13. Bald eagles are known to fly as high as 4500-6000 m above the ground level. Mount Everest is about 8850 m high. Aeroplanes can fly as high as 10,00012,800 m. How many times bigger are these heights compared to Somu's building ( 44 m tall)?
Sol. Let's compare these heights to Somu's building:
Bald eagle: 4500-6000 m;
Mount Everest: 8850 m;
Aeroplanes: 10,000−12,800 m
Somu's building is 44 m tall.
Heights compared to Somu's building:
Bald eagle: ( 4500÷44∼102 ) to ( 6000÷44∼136 )
The bald eagle's flying height is 102−136 times higher.
Mount Everest: 8850÷44=201
Mount Everest is 201 times bigger.
Aeroplanes: (10,000÷44∼227) to ( 12,800÷44∼291 )
An aeroplane's flying height is 227-291 times higher.
(Session 2026 - 27)