RS Aggarwal Class 6 Chapter 1 Number System

EXERCISE 1A

ANSWER 1A

2. Write each of the following numbers in words:

3. Write each of the following numbers in expanded form:

4.Write the corresponding numeral for each of the following:

5. Find the difference between the place values of the two nines in 79520986
   Solution
   The given number is 79520986
   Here the place value of nine at lakhs is 9000000 and the place value of nine at hundreds is 900
   Therefore required difference is 9000000 – 900 = 8999100

6. Find the difference between the place value and face value of 7 in 27650934
   Solution
   The given number is 27650934
   Here the place value of 7 in 27650934 is 70 lakhs = 70,00,000
   The face value of 7 in 27650934 is 7, Therefore required difference is 70,00,000 – 7 = 69,99,993

7. How many 6-digit numbers are there in all?
   Solution
   The largest 6-digit number is 999999
   The smallest 6-digit number is 100000
   Therefore total number of 6-digit numbers = (999999-100000) + 1
   =899999 + 1
   =900000
   =9 lakhs

8. How many 7-digit numbers are there in all?
   Solution
   The largest 7-digit number is 9999999
   The smallest 7-digit number is 1000000
   Therefore total number of 7-digit numbers = (9999999-1000000) + 1
   =8999999+1
   =9000000
   =Ninety lakhs

9. How many thousands make a lakh?
   Solution
   1 lakh means 1,00,000 which is equal to one hundred thousand
   Therefore 1,00,000 = (100×1000)
   Hence one hundred thousand make a lakh

10. How many thousands make a crore? Solution
    One crore=1,00,00,000 which is equal to one hundred lakh
    Therefore 1,00,00,000=(10,000×1000)
    Hence ten thousand thousands make a crore

EXERCISE 1B

Fill in each of the following boxes with the correct symbol > or <:

1. 1003467 ☐ 987965 Solution
   1003467 > 987965
2. 3572014 ☐ 10235401
   Solution
   3572014 < 10235401
3. 3254790 ☐ 3260152
   Solution
   3254790 < 3260152
4. 10357690 ☐ 11243567
   Solution
   10357690 < 11243567
5. 27596381 ☐ 7965412
   Solution
   27596381 > 7965412
6. 47893501 ☐ 47894021
   Solution
   47893501 < 47894021

Arrange the following numbers in descending order:

8.5032786, 23794206, 5032790, 23756819, 987876 Solution

Here 23794206 and 23756819 are both 8-digits numbers in which the same numbers 2,3 and 7 are
placed ten lakhs, lakhs, ten thousands, thousands and hundreds places respectively whereas 9 and 5 are
placed at tens place respectively.
Since 9 > 5
Therefore 23794206 > 23756819
Here 5032786 and 5032790 are both 7-digits numbers in which the same numbers 5,0,3,2 and 7 are
Placed at ten lakhs, lakhs, ten thousands, thousands and hundreds places respectively whereas 9 and 8
are placed at tens place respectively.
Since 9 > 8
Therefore 5032790 > 5032786
987876 is a 6-digits number
The given numbers in the descending order are:
23794206 > 23756819 > 5032790 > 5032786 > 987876

9.190909, 1808088, 16060666, 16007777, 181888, 1808090 Solution

Here 16060666 and 16007777 are 8-digits numbers in which the same numbers 1, 6 and 0 are placed at
crores, ten lakhs and lakhs respectively whereas 6 and 0 are placed at ten thousands place respectively
Since 6 > 0
Therefore 16060666 > 16007777
1808088 and 1808090 are both 7-digits numbers in which the same numbers 1, 8, 0, 8 and o are placed
at ten lakhs, lakhs, ten thousands, thousands and hundreds places respectively whereas 9 and 8 are
placed at tens place respectively
Since 9 > 8
Therefore 1808090 > 1808088
190909 and 181888 are both 6-digits numbers
In both the numbers the same digit 1 is at lakhs place and the digits at ten thousands place are 9 and 8
Since 9 > 8
Therefore 190909 > 181888
The given numbers in the descending order are:
16060666 > 16007777 > 1808090 > 1808088 > 190909 > 181888

10.199988, 1704382, 200175, 1702497, 201200, 1712040 Solution

Here 1712040, 1704382 and 1702497 are all 7-digits numbers in which the same numbers 1 and 7 are
placed at ten lakhs and lakhs places respectively whereas the digits in ten thousands place are 1, 0 and 0
Hence 1712040 is the largest number and the digits in the remaining two numbers at thousands place are

4 and 2 Since 4 > 2
Therefore 1704382 > 1702497
201200, 200175 and 199988 are 6-digits numbers in which 2 and 1 are at lakhs place respectively
Since 2 is greater than 1
Therefore 201200 > 200175
Thus 199988 is the smallest number
The given numbers in the descending order are:
1712040 > 1704382 > 1702497 > 201200 > 200175 >199988

Arrange the following numbers in ascending order:

11. 9873426, 24615019, 990357, 9874012, 24620010 Solution
    9873426 and 9874012 are both 7-digits numbers in which the same numbers 9, 8 and 7 are placed at ten
    lakhs, lakhs and ten thousands places respectively and the digits 3 and 4 are placed at thousands places
    Since 3 < 4
    Hence 9873426 < 9874012
    24615019 and 24620010 are both 8-digits numbers in which the same numbers 2, 4 and 6 are placed at
    crores, ten lakhs and lakhs respectively, since 2 and 1 are ten thousands place
    Hence 24615019 < 24620010
    The given numbers in ascending order are:
    990357 < 9873426 < 9874012 < 24615019 < 24620010
12. 56943201, 5694437, 56944000, 5695440, 56943300 Solution
    5694437 and 5695440 are both 7-digits numbers in which the same numbers 5, 6 and 9 are placed at ten
    lakhs lakhs and ten thousands places respectively and the digits 4 and 5 are at thousand places
    Since 4 < 5
    Hence 5694437 < 5695440
    56943201, 56943300 and 56944000 are all 8-digits numbers
    The same numbers 5, 6, 9 and 4 are placed at crores, ten lakhs, lakhs and ten thousands place
    The digits 3 and 4 are at thousands place
    Since 3 < 4
    56943300 < 56944000
    The digits 2 and 3 are at hundreds place
    Since 2 < 3
    56943201 < 56943300 ,

The given numbers in ascending order are:
5694437 < 5695440 < 56943201 < 56943300 < 56944000

13. 700087, 8014257, 8015032, 10012458, 8014306
    Solution
    700087 is 6-digit number
    8014257, 8015032 and 8014306 are all 7-digit numbers
    The same numbers 8, 0 and 1 are placed at ten lakhs, lakhs and ten thousands place respectively
    At thousands place one number has 5 while the other two numbers have 4
    Hence 8015032 is the largest number
    The digits 2 and 3 are at hundreds place in the remaining two numbers
    Hence 8014257 < 8014306
    10012458 is an 8 digit number
    The given numbers in ascending order are:
    700087 < 8014257 < 8014306 < 8015032 < 10012458
14. 1020304, 893245, 980134, 1021403, 893425,
    1020216 Solution
    893245, 893425 and 980134 are 6-digits number
    980134 is the largest number
    The same numbers in the remaining two numbers are 8, 9 and 3 are placed at lakhs, ten thousands and
    thousands places respectively and the digits 2 and 4 are at hundreds place
    Hence 893245 < 893425
    1020304, 1021403 and 1020216 are all 7-digits number
    Since they have the same digits at ten lakhs, lakhs and ten thousand places namely 1,0 and 2 respectively
    1021403 has 1 at thousand places and the remaining two numbers have the digits at hundred places
    namely 2 and 3
    Since 2 < 3
    ∴ 1020216 < 1020304
    Hence, the given numbers in ascending order are
    893245 < 893425 < 980134 < 1020216 < 1020304 < 1021403

EXERCISE 1C

1. The number of persons who visited the holy shrine of Mata Vaishno Devi during last two
   consecutive years was 13789509 and 12976498 respectively. How many persons visited the shrine
   during these two years?
   Solution
   Persons visited the shrine in first year = 13789509
   Persons visited the shrine in second year = 12976498
   Therefore number of persons visited the shrine in these two years = 13789509+12976498
   =26766007
2. Last year, three sugar factories in a town produced 24809565 bags, 18738576 bags and 9564568
   bags of sugar respectively. How many bags were produced by all the three factories during last
   year
   Solution
   Number of sugar bags produced by first factory in last year = 24809565
   Number of sugar bags produced by second factory in last year = 18738576
   Number of sugar bags produced by third factory in last year = 9564568
   ∴Total number of sugar bags produced by three factories during last year
   = 24809565 + 18738576 + 9564568
   = 53112709
3. A number exceeds 37684955 by 3615045.What is that number?
   Solution
   Given numbers are 37684955 and 3615045
   ∴Sum of both the numbers = 37684955 + 3615045 = 41300000
4. There were three candidates in an election. They received 687905 votes, 495086 votes and 93756
   votes respectively. The number of invalid votes was 13849.If 25467persons did not vote, find how
   many votes were registered
   Solution
   Number of votes received by three candidates = 687905 + 495086 + 93756 = 1276747
   Number of invalid votes = 13849
   Number of persons who did not vote = 25467
   ∴Total number of registered votes = 1276747 + 13849 + 25467
   = 1316063
5. A survey conducted on an Indian state shows that 1623564 people have only primary
   education; 9768678 people have secondary education;6837954 people have higher education and
   2684536 people are illiterate. If the number of children below the age of school admission is
   698781, find the total population of the state.
   Solution
   People having primary education = 1623564
   People having secondary education = 9768678

People having higher education = 6837954
People who are illiterate = 2684536
Children below the age of school admission = 698781
∴Total population in the state = 1623564 + 9768678 + 6837954 + 2684536 + 698781
= 21613495

6. In a particular year a company produced 8765435 bicycles. Next year, the number of bicycles
   produced was 1378689 more than those produced in the preceding year. How many bicycles were
   produced during the second year? How many bicycles were produced during these two years?
   Solution
   Number of bicycles produced in first year = 8765435
   Number of bicycles produced in second year = 8765435 + 1378689
   = 10144124
   Total number of bicycles produced during these two years = 8765435 + 10144124
   = 18909559
7. The sale receipt of a company during a year was ₹ 20956480.Next year; it increased by ₹
   6709570.What was the total sale receipt of the company during these two
   years? Solution
   Company sale receipts in first year = ₹20956480
   Company sale receipts in second year = ₹20956480 + ₹6709570
   = ₹27666050
   Total number of company sale receipts during these two years = ₹20956480 + ₹27666050
   = ₹48622530
8. The total population of a city is 28756304.If the number of males is 16987059, find the
   of females in the city. number
   Solution
   Total number of population in the city = 28756304
   Total number of males in the city = 16987059
   Hence number of females in the city = 28756304 – 16987059
   = 11769245
9. By how much is 13246510 larger than 4658642?
   Solution
   Given numbers are 13246510 and 4658642
   Required number = 13246510 – 4658642
   = 8587868
   Hence 13246510 is larger than 4658642 by the number 8587868
10. By how much is 5643879 smaller than one crore?
    Solution
    Given number is 5643879
    Required number=1 crore – 5643879
    =10000000 – 5643879
    =4356121, Hence the number 5643879 is smaller than one crore by the number 4356121

11. What number must be subtracted from 11010101 to get 2635967?
    Solution
    Given number = 11010101
    Required number = 2635967
    11010101 – Required number= 2635967
    Required number = 11010101 – 2635967
    =8374134
    Hence the number 8374134 must be subtracted from 11010101 to get 2635967
12. The sum of two numbers is 10750308.If one of them is 8967519, what is the other number?
    Solution
    The sum of two numbers = 10750308
    Given number = 8967519
    Other number = 10750308 – 8967519 =1782789
    Hence 1782789 is the other number
13. A man had ₹ 20000000 with him. He spent ₹13607085 on buying a school building. How much
    money is left with him?
    Solution
    Money already having with him = ₹20000000
    Money spent on buying a school building = ₹13607085
    Total amount of money left = ₹20000000 – ₹13607085
    =6392915
    Therefore total amount of money left with him is ₹6392915
14. A society needed ₹ 18536000 to buy a property .It collected ₹7253840 as membership fee, took
    a loan of ₹5675450 from a bank and collected ₹2937680 as donation. How much is the society still
    short of?
    Solution
    Money needed by the society to buy a property = ₹18536000
    Amount collected as membership fee = ₹7253840
    Amount taken as loan from bank = ₹5675450 Amount
    collected as donation from bank = ₹2937680
    Total amount short = ₹18536000 – (₹7253840 + ₹5675450 + ₹2937680)
    = ₹18536000 – ₹15866970 = ₹2669030
    Hence total amount short = ₹2669030
15. A man had ₹ 10672540 with him. He gave ₹ 4836980 to his wife, ₹ 3964790 to his son and the
    rest to his daughter. How much money was received by the daughter?
    Solution
    Amount already with him = ₹10672540
    Amount given to his wife = ₹4836980

Amount given to his son = ₹3964790
Amount received by the daughter = ₹10672540 – (₹4836980 – ₹3964790)
= ₹10672540 – ₹8801770
= ₹1870770
Hence total amount received by the daughter is ₹1870770

EXERCISE 1D

1. Round each of the following numbers to the nearest ten:
   (a) 36 (b) 173 (c) 3869 (d) 16378
   Solution
   (a) 36
   Here 6 is in once digits place
   6 > 5
   ∴ required rounded number is 40
   (b) 173
   Here 3 is in once digits place
   3 < 5 ∴ required rounded number is 170 (c) 3869 Here 9 is in once digits place 9 > 5
   ∴ required rounded number is 3870
   (d) 16378
   Here 8 is in once digits place 8 > 5
   ∴ required rounded number is 16380
2. Round each of the following numbers to the nearest hundred:
   (a) 814 (b) 1254 (c) 43126(d) 98165
   Solution
   (a) 814
   Here tens digits place is 1
   1 < 5
   ∴ required rounded number is 800
   (b) 1254
   Here tens digits place is 5
   5=5
   ∴ required rounded number is 1300
   (c) 43126
   Here tens digits place is 2
   2 < 5
   ∴ required rounded number is 43100
   (d) 98165
   Here tens digits place is 6,

6 > 5
∴ required rounded number is 98200

3. Round each of the following numbers to the nearest thousand:
   (a) 793 (b) 4826 (c) 16719(d) 28394
   Solution
   (a) Here 7 is in hundred digits place
   7 > 5
   ∴ required rounded number is 1000
   (b) Here 8 is in hundred digits place
   8 > 5
   ∴ required rounded number is 5000
   (c)Here 7 is in hundred digits place
   7 > 5
   ∴ required rounded number is 17000
   (d) Here 3 is in hundred digits place
   3 < 5
   ∴ required rounded number is 28000
4. Round each of the following numbers to the nearest ten thousand:
   (a) 17514 (b) 26340 (c) 34890(d) 272685
   Solution
   (a) Thousand digits number is 7 > 5
   ∴ required rounded number is 20000
   (b) Thousand digits number is 6 > 5
   ∴ required rounded number is 30000
   (c)Thousand digit number is 4 < 5 ∴
   required rounded number is 30000
   (d) Thousand digit number is 2 < 5 ∴
   required rounded number is 270000
5. Estimate each sum to the nearest
   ten: 5. (57 + 34)
   Solution
   Estimated value to the nearest ten of 57 = 60
   Estimated value to the nearest ten of 34 = 30
   Total required estimation = (60 + 30)
   = 90
6. (43 +78)
   Solution
   Estimated value to the nearest ten of 43 = 40
   Estimated value to the nearest ten of 78 = 80, Total required estimation = (40 + 80) = 120

7. (14 + 69)
   Solution
   Estimated value to the nearest ten of 14 = 10
   Estimated value to the nearest ten of 69 = 70
   Total required estimation = (10 + 70) = 80
8. (86 + 19)
   Solution
   Estimated value to the nearest ten of 86 = 90
   Estimated value to the nearest ten of 19 = 20
   Total required estimation = (90 + 20) = 110
9. (95 + 58)
   Solution
   Estimated value to the nearest ten of 95 = 100
   Estimated value to the nearest ten of 58 = 60
   Total required estimation = (100 + 60) = 160
10. (77 + 63)
    Solution
    Estimated value to the nearest ten of 77 = 80
    Estimated value to the nearest ten of 63 = 60
    Total required estimation = (80 +60) = 140
11. (356 + 275)
    Solution
    Estimated value to the nearest ten of 356 = 360
    Estimated value to the nearest ten of 275 = 280
    Total required estimation = (360 + 280) = 640
12. (463 + 182)
    Solution
    Estimated value to the nearest ten of 463 = 460
    Estimated value to the nearest ten of 182 = 180
    Total required estimation = (460 +180) = 640
13. (538 + 276)
    Solution
    Estimated value to the nearest ten of 538 = 540
    Estimated value to the nearest ten of 276 = 280
    Total required estimation = (540 + 280) =820
    Estimate each sum to the nearest hundred:
14. (236 + 689)

Solution
Estimated value to the nearest hundreds of 236 = 200
Estimated value to the nearest hundreds of 689 = 700
Total required estimation = (200 +700)
=900

15. (458 + 324)
    Solution
    Estimated value to the nearest hundreds of 458 = 500
    Estimated value to the nearest hundreds of 324 = 300
    Total required estimation = (500 + 300)
    = 800
16. (170 + 395)
    Solution
    Estimated value to the nearest hundreds of 170 =200
    Estimated value to the nearest hundreds of 395 = 400
    Total required estimation = (200 + 400) = 600
17. (3280 + 4395)
    Solution
    Estimated value to the nearest hundreds of 3280 = 3300
    Estimated value to the nearest hundreds of 4395 = 4400
    Total required estimation = (3300 + 4400) = 7700
18. (5130 + 1410)
    Solution
    Estimated value to the nearest hundreds of 5130 = 5100
    Estimated value to the nearest hundreds of 1410 = 1400
    Total required estimation = (5100 + 1400)
    = 6500
19. (10083 + 29380)
    Solution
    Estimated value to the nearest hundreds of 10083 = 10080
    Estimated value to the nearest hundreds of 29380 = 29400
    Total required estimation = (10080 + 29400) = 39500
    Estimate each sum to the nearest thousand:
20. (32836 + 16466)
    Solution
    Estimated value to the nearest thousands of 32836 = 33000
    Estimated value to the nearest thousands of 16466 = 16000
    Total required estimation = (33000 + 16000)= 49000

21. (46703 + 11375)
    Solution
    Estimated value to the nearest thousands of 46703 = 47000
    Estimated value to the nearest thousands of 11375 =
    11000 Total required estimation = (47000 + 11000) = 58000
22. There are 54 balls in box A and 79 balls in box Estimate the total number of balls in both the
    boxes taken together
    Solution
    Total number of balls in box A = 54
    Total number of balls in box B = 79
    Total number of estimation of balls in box A = 50
    Total number of estimation of balls in box B = 80
    Total number of estimation of balls in both the boxes = (50 + 80)
    =130
23. Estimate each difference to the nearest
    ten: 23. (53 – 18)
    Solution
    Estimated value to the nearest ten of 53 = 50
    Estimated value to the nearest ten of 18 = 20
    ∴ required estimation = (50 – 20) = 30
24. (97 – 38)
    Solution
    Estimated value to the nearest ten of 97 = 100
    Estimated value to the nearest ten of 38 = 40
    ∴ required estimation = (100 – 40) = 60
25. (409 – 148)
    Solution
    Estimated value to the nearest ten of 409 = 410
    Estimated value to the nearest ten of 148 = 150
    ∴ required estimation = (410 – 150)= 260
26. Estimate each difference to the nearest
    hundred: 26. (678 – 215)
    Solution
    Estimated value to the nearest hundreds of 678 = 700
    Estimated value to the nearest hundreds of 215 = 200
    ∴ required estimation = (700 – 200) = 500
27. (957 – 578)
    Solution
    Estimated value to the nearest hundreds of 957 = 1000
    Estimated value to the nearest hundreds of 578 = 600, ∴ required estimation = (1000 – 600) = 400

28. (7258 – 2429)
    Solution
    Estimated value to the nearest hundreds of 7258 =7300
    Estimated value to the nearest hundreds of 2429 = 2400
    ∴ required estimation = (7300 – 2400) = 4900
29. (5612 – 3095)
    Solution
    Estimated value to the nearest hundreds of 5612 = 5600
    Estimated value to the nearest hundreds of 3095 = 3100
    ∴required estimation = (5600 – 3100) = 2500
    Estimate each difference to the nearest thousand:
30. (35863 – 27677)
    Solution
    Estimated value to the nearest thousands of 35863 = 36000
    Estimated value to the nearest thousands of 27677 = 28000
    ∴ required estimation = (36000 – 28000) = 8000
31. (47005 – 39488)
    Solution
    Estimated value to the nearest thousands of 47005 = 47000
    Estimated value to the nearest thousands of 39488 = 39000
    ∴ required estimation = (47000 – 39000) = 8000

EXERCISE 1E

Estimate each of the following products by rounding off each number to the nearest ten:

38 × 63
SolutionGiven
38 × 63
Estimated value to the nearest ten of 38 = 40
Estimated value to the nearest ten of 63 = 60
∴ required estimation = (40 × 60) = 2400

2. 54 × 47
   SolutionGiven
   54 × 47
   Estimated value to the nearest ten of 54 = 50
   Estimated value to the nearest ten of 47 = 50
   ∴ required estimation = (50 × 50) = 2500
3. 28 × 63
   SolutionGiven
   28 × 63
   Estimated value to the nearest ten of 28 = 30
   Estimated value to the nearest ten of 63 = 60
   ∴ required estimation = (30 × 60) = 1800
4. 42 × 75
   SolutionGiven
   42 × 75
   Estimated value to the nearest ten of 42 = 40
   Estimated value to the nearest ten of 75 = 80
   ∴ required estimation = (40 × 80)
   = 3200
5. 64 × 58
   SolutionGiven
   64 × 58
   Estimated value to the nearest ten of 64 = 60
   Estimated value to the nearest ten of 58 = 60
   ∴ required estimation = (60 × 60)
   = 3600
6. 15 × 34

SolutionGiven
15 × 34
Estimated value to the nearest ten of 15 = 20
Estimated value to the nearest ten of 34 = 30
∴ required estimation = (20 × 30)
= 600
Estimate each of the following products by rounding off each number to the nearest hundred:

7. 376 × 123
   SolutionGiven
   376 × 123
   Estimated value to the nearest hundreds of 376 = 400
   Estimated value to the nearest hundreds of 123 = 100
   ∴ required estimation = (400 × 100)
   = 40000
8. 264 × 147
   SolutionGiven
   264 × 147
   Estimated value to the nearest hundreds of 264 = 300
   Estimated value to the nearest hundreds of 147 = 100
   ∴ required estimation = (300 × 100)
   = 30000
9. 423 × 158
   SolutionGiven
   423 × 158
   Estimated value to the nearest hundreds of 423 = 400
   Estimated value to the nearest hundreds of 158 = 200
   ∴ required estimation = (400 × 200)
   = 80000
10. 509 × 179
    SolutionGiven
    509 × 179
    Estimated value to the nearest hundreds of 509 = 500
    Estimated value to the nearest hundreds of 179 = 200
    ∴ required estimation = (500 × 200)
    = 100000

EXERCISE 1F

Find the estimated quotient for each of the following:
1.87 ÷ 28 Solution
The estimated quotient of 87÷ 28 = 90 ÷30 = 3
2.83 ÷ 17 Solution
The estimated quotient of 83 ÷ 17 = 80÷ 20 = 4
3.75 ÷ 23 Solution
The estimated quotient of 75 ÷ 23 = 80 ÷ 20 = 4

4. 193 ÷ 24 Solution
   The estimated quotient of 193 ÷ 24 = 200 ÷ 20 = 10
5. 725 ÷ 23 Solution
   The estimated quotient of 725 ÷ 23 = 700 ÷ 20 = 35

EXERCISE 1G

Express each of the following as a Roman numeral:
(i) 2 (ii) 8 (iii) 14 (iv) 29
(v) 36 (vi) 43 (vii) 54 (viii) 61
(ix) 73 (x) 81 (xi) 91 (xii) 95
(xiii) 99 (xiv) 105 (xv) 114
Solutions
(i) Roman number of 2 is written as II
(ii) Roman number of 8 is written as (5 + 3) = VIII
(iii) Roman number of 14 is written as (10 + 4) = XIV
(iv) Roman number of 29 is written as (10 + 10 + 9) = XXIX
(v) Roman number of 36 is written as (10 + 10 + 10 + 6) = XXXVI
(vi) Roman number of 43 is written as (50 – 10) + 3 = XLIII
(vii) Roman number of 54 is written as (50 + 4) = LIV
(viii) Roman number of 61 is written as (50 + 10 + 1) = LXI
(ix) Roman number of 73 is written as (50 + 10 + 10 + 3) =LXXIII
(x) Roman number of 81 is written as (50 + 10 + 10 + 10+ 1) = LXXXI
(xi) Roman number of 91 is written as (100 – 10) + 1 = XCI
(xii) Roman number of 95 is written as (100 – 10) + 5 = XCV
(xiii) Roman number of 99 is written as (100 – 10) + 9 = XCIX
(xiv) Roman number of 105 is written as (100 + 5) = CV
(xv) Roman number of 114 is written as (100 + 10) + 4 = CXIV

2. Express each of the following as a Roman numeral
   (i) 164 (ii) 195 (iii) 226 (iv) 341
   (v) 475 (vi) 596 (vii) 611 (viii) 759
   Solutions
   (i) Roman numeral of 164 = (100 + 50 + 10 +4) = CLXIV
   (ii) Roman numeral of 195 = (100) + (100 – 10) + 5 = CXCV
   (iii) Roman numeral of 226 = (100 + 100 + 10 + 10 + 6) = CCXXVI
   (iv) Roman numeral of 341 = 100 + 100 + 100 + (50 – 10) + 1 = CCCXLI
   (v) Roman numeral of 475 = (500 – 10) + 50 + 50 + 10 + 10 + 5 = CDLXXV
   (vi) Roman numeral of 596 = 500 + (100 – 10) + 6 = DXCVI
   (vii) Roman numeral of 611 = 500 + 100 + 11 = DCXI
   (viii) Roman numeral of 759 = 500 + 100+ 100 + 59 = DCCLIX

EXERCISE 1H

The place value of 6 in the numeral 48632950 is
(a) 6 (b) 632950 (c) 600000 (d) 486
Solution
Here the place value of 6 is 6 lakhs = (6 × 100000) = 600000
Hence option (c) is the correct answer

2. The face value of 4 in the numeral 89247605 is
   (a) 4 (b) 40000 (c) 47605 (d) 8924
   Solution
   Here the face value remains same irrespective of its place in the place value chart.
   Hence the face value of 4 will remain same Hence (a) is correct answer
3. The difference between the place value and the face value of 5 in the numeral 78653421 is
   (a) 53416 (b) 4995 (c) 49995 (d) none of these
   Solution
   Here place value of 5 = 5 × 10000 = 50000
   Face value of 5 = 5
   ∴ required difference = (place value – face value)
   = 50000 – 5
   = 49995
   Hence option (c) is the correct answer
4. The smallest counting number is
   (a) 0 (b) 1 (c) 10 (d) none of these
   Solution
   The smallest counting number is 1
   Hence option (b) is correct answer
5. How many 4 – digit numbers are there?
   (a) 8999 (b) 9000 (c) 8000 (d) none of these
   Solution
   The largest 4-digit number = 9999
   The smallest 4-digit number = 1000
   Total number of 4-digit numbers = (9999-1000) + 1
   = 8999 + 1
   = 9000
   Hence option (b) is correct answer