Counting things is easy for us now. We can count objects in large numbers, eg: no. of students in the school, and represent them through numerals. we can also communicate large numbers using suitable number names. A number is a mathematical object, also it's used to count, measure, and label.
How to Comparing numbers
Let us look at some examples:
Example: 1
Compare 4678 and 5543
5543 is greater as the digit at the thousands place in 5543 is greater than that in 4678.
Example: 2
Compare 9264,8695,8402 and 7824
9264 is the greatest as it has the greatest digit at the thousands place in all the numbers.
Whereas 7824 is the smallest as it has the smallest digit at the thousands place in all the numbers.
Example: 3 – Very Special Case
Compare 56311 and 56833
Here, we will start by checking the thousands place.As the digit 5 at ten thousand place is same so we will move forward and see the thousands place. The digit 6 is also same so we will still move on further to check the hundreds place.
The digit at the hundreds place in 56833 is greater than that in 56311
Thus 56833 is greater than 56311
Let Us See Next Ascending and Descending Order:
Ascending Order:
It means arrangement from the smallest to the greatest.
Ascending order = 80 < 150 < 180 < 195 < 210
Descending Order:
It means arrangement from the greatest to the smallest.
Descending order = 240 >195 >180 > 150 > 70
How many numbers can be formed using a certain number of digits
If a certain number of digits are given, we can make different numbers having the same number of digits by interchanging positions of digits.
Example:
Consider 4 digits: 4, 0, 9, 5. Using these four digits,
Largest number possible = 9540
Smallest number possible = 4059 (Since 4 digit number cannot have 0 as the leftmost number, as the number then will become a 3 digit number)
Shifting Digits:
Take any 4 or 3 digit number with different digits and interchange the digit at the thousands place and hundreds place and the digit at the ones place.
Example:
Consider a number 678, If we swap the hundredths place digit with the digit at units place, we will get 876 which is greater than 678.
Similarly, if we exchange the tenths place with the units place, we get 687, which is greater than 678.
Place value
It refers to the positional notation which defines a digits position.
7542
Here, 2 is at one's place, 4 is at tens place, 5 is at hundreds place and 7 is at thousands place.
Indian Method:
Periods 
Crores 
Lakhs 
Thousands 
Ones 

Place Value 
Ten Crores 
Crore 
Ten Lakhs 
Lakh 
Ten Thousands 
Thousand 
Hundreds 
Tens 
Ones 










Indian Method:
Ones = 1
Tens = 10
Hundreds = 100
Thousands = 1,000
Ten Thousands = 10,000
One Lakh = 1,00,000
Ten Lakhs = 10,00,000
One Crore = 1,00,00,000
International Method:
Periods 
Billions 
Millions 
Thousands 
Ones 

Place Value 
Hundred Billions 
Ten Billions 
Billions 
Hundred Millions 
Ten Millions 
Millions 
100 Thousands 
10 Thousands 
Thousand 
Hundreds 
Tens 
Ones 













International Method:
Ones = 1
Tens = 10
Hundreds = 100
Thousands = 1,000
Hundred Thousands = 100,000
One Million = 1,000,000
Ten Millions = 10,000,000
Hundred Million = 100,000,000
Formation of large numbers:
Greatest Number 
Add 
Equals 
Smallest Number 
Number Name 
Greatest 1 digit no = 9 
1 
10 
Smallest 2 digit no. = 10 
Ten 
Greatest 2 digit no = 99 
1 
100 
Smallest 3 digit no. = 100 
Hundred 
Greatest 3 digit no. = 999 
1 
1000 
Smallest 4 digit no. = 1000 
Thousand 
Greatest 4 digit no. = 9999 
1 
10000 
Smallest 5 digit no. = 10000 
Ten Thousand 
Greatest 5 digit number = 99999 
1 
100000 
Smallest 6 digit no. = 100000 
1 lakh 
Greatest 6 digit number = 999999 
1 
1000000 
Smallest 7 digit no = 1000000 
Ten Lakhs 
Greatest 7 digit number = 9999999 
1 
10000000 
Smallest 8 digit no = 10000000 
1 Crore 
Define Estimation
There are a number of situations in which we don't need the correct quantity but need only an estimate of this quantity. It means mostly approximating a quantity to the desired accuracy.
Estimating to the nearest tens by rounding off
The estimation is done by rounding off the numbers to the nearest tens. Thus, 18 is estimated as 20 to the nearest tens; 11 is estimated as 10 to the nearest tens.
Estimating to the nearest hundreds by rounding off
Numbers 1 to 49 are closer to 0 than to 100. So they are rounded off to 0. Numbers 51 to 99 are closer to 100 than to 0, and so are rounded off to 100. Number 50 is equidistant from 0 and 100 both. It is customary to round it off as 100.
Estimating to the nearest thousands by rounding off
Numbers 1 to 499 are nearer to 0 than 1000, so these numbers are rounded off as 0. The numbers 501 to 999 are nearer to 1000 than 0, so they are rounded off as 1000. Number 500 is customarily rounded off as 1000.
Roman Numerals:
Digits 09 in Roman are represented as : I, II, III, IV, V, VI, VII, VIII, IX, X
Some other Roman numbers are : I = 1, V = 5 , X = 10 , L = 50 , C = 100 , D = 500 , M = 1000
Example:
If a symbol is repeated, its value is added as many times as it occurs.
XV = 10 + 5 = 15
NCERT SOLUTIONS
EXERCISE – 1.1
1. Fill in the blanks:
(a) 1 lakh = 10 ten thousand.
(b) 1 million = 10 hundred thousand.
(c) 1 crore = 10 ten lakh.
(d) 1 crore = 10 million.
(e) 1 million = 10lakh.
2. Place commas correctly and write the numerals:
(a) Seventy three lakh seventy five thousand three hundred seven.
Ans: 73,75,307
(b) Nine crore five lakh forty one.
Ans: 9,05,00,041
(c) Seven crore fifty two lakh twenty one thousand three hundred two.
Ans: 7,52,21,302
(d) Fifty eight million four hundred twenty three thousand two hundred two.
Ans: 5,84,23,202
(e) Twenty three lakh thirty thousand ten.
Ans: 23,30,010
3. Insert commas suitably and write the names according to Indian System of Numeration:
(a) 87595762 (b) 8546283 (c) 99900046 (d) 98432701
(a) 87595762
Ans: 8,75,95,762
Eight crore seventyfive lakh ninetyfive thousand seven hundred sixty two
(b) 8546283
Ans: 85,46,283
Eightyfive lakh fortysix thousand two hundred eightythree
(c) 99900046
Ans: 9,99,00,046
Nine crore ninetynine lakh fortysix
(d) 98432701
Ans: 9,84,32,701
Nine crore eightyfour lakh thirtytwo thousand seven hundred one
4. Insert commas suitably and write the names according to International System of Numeration:
(a) 78921092 (b) 7452283 (c) 99985102 (d) 48049831
(a) 78921092
Ans: 78,921,092
Seventyeight million nine hundred twentyone thousand ninetytwo
(b) 7452283
Ans: 7,452,283
Seven million four hundred fifty two thousand two hundred eightythree
(c) 99985102
Ans: 99,985,102
Ninetynine million nine hundred eightyfive thousand one hundred two
(d) 48049831
Ans: 48,049,831
Fortyeight million fortynine thousand eight hundred thirtyone
EXERCISE – 1.2
1. A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.
Ans:
No. of tickets sold on the first day – 1094
No. of tickets sold on the second day – 1812
No. of tickets sold on the third day – 2050
No. of tickets sold on the fourth day – 2751
Total no. of tickets sold on all the four days = (1094 + 1812 + 2050 + 2751) = 7,707
2. Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?
Ans:
Shekhar has so far scored = 6980 runs
He wishes to complete = 10,000 runs.
∴ Total number of runs needed by him = 10,000 – 6980 = 3020 runs
3. In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?
Ans:
Number of votes secured by the successful candidate = 5,77,500
Number of votes secured by his nearest rival = 3,48,700
∴ Margin of votes to win the election = 5,77,500 – 3,48,700 = 2,28,800
4. Kirti bookstore sold books worth Rs 2,85,891 in the first week of June and books worth Rs 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?
Ans:
Books sold in 1st week of June worth = 2,85,891
Books sold in 2nd week of the month worth = 4,00,768
∴, total sale of books in the two weeks together = = 2,85,891 + 4,00,768 = 6,86,659
In the 2nd week of the month, the sale of books was greater.
Difference of the sale of books = 4,00,768 – 2,85,891 = 1,14,877
Hence, in 2nd week of June, the sale of books was more by ₹1,14,877.
5. Find the difference between the greatest and the least 5digit number that can be written using the digits 6, 2, 7, 4, 3 each only once.
Ans:
Given digits are = 6, 2, 7, 4, 3
Greatest number = 76432
Least number = 23467
∴ Difference = 76432 – 23467 = 52965
6. A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
Ans:
No. of screws manufactured in a day = 2,825
No. of screws manufactured in January month (2006) = 31 x 2825 = 87,575
7. A merchant had Rs 78,592 with her. She placed an order for purchasing 40 radio sets at Rs 1200 each. How much money will remain with her after the purchase?
Ans:
Amount of money with the merchant = 78,592
No. of radio sets = 40
Price of one radio set = 1200
∴ cost of 40 radio sets = 1200 x 40 = 48,000
Remaining money with the merchant = 78,592 – 48000 = ₹30,592
Hence, amount of ₹30,592 will remain with her after purchasing the radio sets.
8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?
Ans:
Difference between 65 and 56 i.e (65 – 56) = 9
∴ Difference between the correct and incorrect answer = 7236 × 9 = 65124
Hence, by 65124, the answer was greater than the correct answer.
9. To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?
Ans:
Total length of the cloth = 40 m
= 40 x 100 cm = 4000 cm.
Cloth needed to stitch a shirt = 2 m 15 cm
= 2 x 100 + 15 cm = 215 cm
Therefore, the number of shirts stitched = 4000/215
So, the no. of shirts stitched = 18
Remaining cloth = 130 cm = 1 m 30 cm
10. Medicine is packed in boxes, each weighing 4 kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Ans:
Weight of one box = 4 kg 500 g
= 4 x 1000 + 500 = 4500 g
800 kg = 800 x 1000 = 800000 g
= 800000/4500
∴ 177 boxes can only be loaded in the van.
11. The distance between the school and a student’s house is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days.
Ans:
Distance between school and house = 1 km 875 m = (1000 + 875) m = 1875 m.
Distance travelled by the student in both ways = 2 x 1875 = 3750 m
Distance travelled in 6 days = 3750 m x 6
= 22500 m
= 22 km 500 m.
Hence, total distance covered in six days = 22 km 500 m.
12. A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Ans:
Quantity of curd in a vessel = 4 1 500 ml
= (4 x 1000 + 500) ml
= 4500 ml.
Capacity of 1 glass = 25 ml
Therefore, number of glasses = 4500/25 = 180
EXERCISE – 1.3
1. Estimate each of the following using general rule:
(a) 730 + 998 (b) 796 – 314 (c) 12904 + 2888 (d) 28292 – 21496
Make ten more such examples of addition, subtraction and estimation of their outcome.
(a) 730 + 998
Ans:
Rounding off 730 nearest to hundreds = 700
Rounding off 998 nearest to hundreds = 1,000
∴ 730 + 998 = 700 + 1000 = 1700
(b) 796 – 314
Ans:
Rounding off 796 nearest to hundreds = 800
Rounding off 314 nearest to hundreds = 300
∴ 796 – 314 = 800 – 300 = 500
(c) 12,904 + 2,888
Ans:
Rounding off 12,904 nearest to thousands = 13000
Rounding off 2888 nearest to thousands = 3000
∴ 12,904 + 2,888 = 13000 + 3000 = 16000
(d) 28,292 – 21,496
Ans:
Rounding off 28,292 nearest to thousands = 28,000
Rounding off 21,496 nearest to thousands = 21,000
∴ 28,292 – 21,496 = 28,000 – 21,000 = 7,000
Make ten more such examples of addition, subtraction and estimation of their outcome.
Example 1:
1220 + 2385 = 1200 + 2400 = 3600
Example 2:
3873 + 6535 = 4000 + 7000 = 11,000
Example 3:
8772 – 3654 = 9,000 – 4,000 = 5,000
Example 4:
4548 – 2965 = 5,000 – 3,000 = 2,000
Example 5:
1957 + 3185 = 2000 + 3,000 = 5,000
Example 6:
3269 – 1698 = 3000 – 2000 = 1,000
Example 7:
8765 + 6232 = 9000 + 6000 = 15,000
Example 8:
1078 – 1028 = 1000 – 1000 = 0
Example 9:
6382 + 5830 = 6,000 + 6,000 = 12,000
Example 10:
9894 – 6389 = 10,000 – 6000 = 4,000
2. Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):
(a) 439 + 334 + 4317 (b) 108734 – 47599 (c) 8325 – 491 (d) 489348 – 48365
(a) 439 + 334 + 4317
Ans:
Rough estimate (Rounding off to nearest hundreds)
439 + 334 + 4,317 = 400 + 300 + 4300 = 5,000
Closer estimate (Rounding off to nearest tens)
439 + 334 + 4317 = 440 + 330 + 4320 = 5090.
(b) 108734 – 47599
Ans:
Rough estimate (Rounding off to nearest hundreds)
1,08,734 – 47,599 = 1,08,700 – 47,600 = 61,100
Closer estimate (Rounding off to nearest tens)
1,08,734 – 47,599 = 1,08,730 – 47,600 = 61,130.
(c) 8325 – 491
Ans:
Rough estimate (Rounding off to nearest hundreds)
8325 – 491 = 8300 – 500 = 7800
Closer estimate (Rounding off to nearest tens)
8325 – 491 = 8330 – 490 = 7840.
(d) 489348 – 48365
Ans:
Rough estimate (Rounding off to nearest hundreds)
4,89,348 – 48,365 = 4,89,300 – 48,400 = 4,40,900
Closer estimate (Rounding off to nearest tens)
4,89,348 – 48,365 = 4,89,350 – 48,370 = 4,40,980
Make four more such examples.
Example 1:
364 + 562
Solution:
Rough estimate (Rounding off to nearest hundreds)
364 + 562 = 400 + 600
= 1,000
Closer estimate (Rounding off to nearest tens)
364 + 562 = 360 + 560
= 920
Example 2:
8785 – 3820
Solution:
Rough estimate (Rounding off to nearest hundreds)
8785 – 3820 = 8800 – 3900
= 4900
Closer estimate (Rounding off to nearest tens)
8785 – 3820 = 8790 – 3820
= 4970
Example 3:
6643 – 8265
Solution:
Rough estimate (Rounding off to nearest hundreds)
6643 + 8265 = 6700 + 8300
= 15,000
Closer estimate (Rounding off to nearest tens)
6643 + 8265 = 6640 + 8270
= 14910
Example 4:
3836 – 1262
Solution:
Rough estimate (Rounding off to nearest hundreds)
3836 – 1262 = 3800 – 1300
= 2500
Closer estimate (Rounding off to nearest tens)
3836 – 1262 = 3840 – 1260
= 2580
3. Estimate the following products using general rule:
(a) 578 × 161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
Ans:
(a) 578 x 161 = 600 x 200 = 1,20,000
(b) 5281 x 3491 = 5000 x 3000 = 1,50,00,000
(c) 1291 x 592 = 1300 x 600 = 7,80,000
(d) 9250 x 29 = 9000 x 30 = 2,70,000
Make four more such examples.
Example 1.
385 x 1063 = 400 x 1000 = 4,00,000
Example 2.
6851 x 1290 = 7000 x 1000 = 70,00,000
Example 3.
3888 x 9360 = 4000 x 9000 = 3,60,00,000
Example 4.
3406 x 7503 = 3000 x 8000 = 2,40,00,000