 # RS Aggarwal Class 6 Chapter 1 Number System

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### EXERCISE 1A

Contents
1. Write the numeral for each of the following numbers:
(i)Nine thousand eighteen
(ii)Fifty- four thousand seventy-three
(iii)Three lakh two thousand five hundred six (iv)Twenty
lakh ten thousand eight
(v)Six crore five lakh fifty-seven
(vi)Two crore two Lakh two thousand two hundred two
(vii)Twelve crore twelve lakh twelve thousand twelve
(viii)Fifteen crore fifty lakh twenty thousand sixty-eight Solutions

(i)The numeral form of nine thousand eighteen is: 9018
(ii)The numeral form of fifty four thousand seventy-three is 54073
(iii)The numeral form of three lakh two thousand five hundred six is: 302506
(iv)The numeral form of twenty lakh ten thousand eight is: 2010008 (v)The
numeral form of six crore five lakh fifty-seven is: 60500057
(vi)The numeral form of two crore two lakh two thousand two hundred two is: 20202202
(vii)The numeral form of twelve crore twelve lakh twelve thousand twelve is: 121212012
(viii)The numeral form of fifteen crore fifty lakh twenty thousand sixty-eight is: 155020068

### 2. Write each of the following numbers in words:

(i)63,005 (ii)7,07,075 (iii)34,20,019
(iv)3,05,09,012 (v)5,10,03,604 (vi)6,18,05,008
(vii)19,09,09,900 (viii)6,15,30,807 (ix)6,60,60,060 Solutions

(i)The given number is 63,005
It is written as sixty three thousand and five (ii)The given number is 7,07,075
It is written as seven lakh seven thousand and seventy five
(iii)The given number is 34, 20,019
It is written as thirty four lakh twenty thousand and nineteen
(iv)The given number is 3,05,09,012
It is written as three crore five lakh nine thousand and twelve
(v)The given number is 5,10,03,604
It is written as five crore ten lakh three thousand six hundred and four
(vi)The given number is 6,18,05,008
It is written as six crore eighteen lakh five thousand and eight
(vii)The given number is 19,09,09,900
It is written as nineteen crore nine lakh nine thousand and nine hundred
(viii)The given number is 6,15,30,807
It is written as six crore fifteen lakh thirty thousand eight hundred and seven
(ix)The given number is 6,60,60,060
It is written as six crore sixty lakh sixty thousand and sixty

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

(i)15,768 (ii)3,08,927 (iii)24,05,609
(iv)5,36,18,493 (v)6,06,06,006 (vi)9,10,10,510 Solut
ions

(i)The given number is 15,768
The expanded form of 15,768 = (1×10000) + (5×1000) + (7×100) + (6×10) + (8×1)
(ii)The given number is 3,08,927
The expanded form of 3,08,927 = (3×100000) + (8×1000) + (9×100) + (2×10) + (7×1)
(iii)The given number is 24,05,609
The expanded form of 24,05,609 = (2×1000000) + (4×100000) + (5×1000) + (6×100) + (9×1)
(iv)The given number is 5,36,18,493
The expanded form of
5,36,18,493 = (5×10000000) + (3×1000000) + (6×100000) + (1×10000) + (8×1000) + (4×100) +
(9×10) + (3×1)
(v)The given number is 6,06,06,006
The expanded form of 6,06,06,006 = (6×10000000) + (6×100000) + (6×1000) + (6×1)
(vi)The given number is 9,10,10,510
The expanded form of 9,10,10,510 = (9×10000000) + (1×1000000) + (1×10000) + (5×100) +
(1×10)

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

(i) 6 × 10000 + 2 × 1000 + 5 × 100 + 8 × 10 + 4 × 1
(ii) 5 × 100000 + 8 × 10000 + 1 × 1000 + 6 × 100 + 2 × 10 + 3 × 1
(iii) 2 × 10000000 + 5 × 100000 + 7 × 1000 + 9 × 100 + 5 × 1 (iv) 3 × 1000000 + 4 × 100000 + 6 × 1000 + 5 × 100 + 7 × 1

Solutions
(i)The given expanded form is 6 × 10000 + 2 × 1000 + 5 × 100 + 8 × 10 + 4 × 1
The numeral number is 62584
(ii)The given expanded form is 5 × 100000 + 8 × 10000 + 1 × 1000 + 6 × 100 + 2 × 10 + 3 × 1
The numeral number is 581623
(iii)The given expanded form is 2 × 10000000 + 5 × 100000 + 7 × 1000 + 9 × 100 + 5 × 1
The numeral number is 20507905
(iv)The given expanded form is 3 × 1000000 + 4 × 100000 + 6 × 1000 + 5 × 100 + 7 × 1
The numeral number is 3406507 RS Aggarwal Class 6 Chapter 1 Number System
1. 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
1. 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
1. 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
1. 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
1. 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
1. 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:

1. 63521047, 7354206, 63514759, 7355014, 102345680 Solutions
Here 102345680 is a 9-digit number
63521047 and 63514759 are both 8-digit numbers in which the same numbers namely 6,3, and 5 are
placed at crores, ten lakhs and lakhs place respectively whereas 2 and 1 are placed at ten thousands place
Since 2 > 1
Therefore 63521047 > 63514759
7354206 and 7355014 are both 7-digits numbers
Here the same numbers 7, 3 and 5 are placed at crores, ten lakhs and lakhs place respectively whereas 5
and 4 are placed at ten thousands place
Since 5 > 4
Therefore 7355014 > 7354206
The given numbers in the descending order are:
102345680 > 63521047 > 63514759 > 7355014 > 7354206

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:

1. 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
2. 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

1. 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
2. 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

1. 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
2. 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
3. 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
4. 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
5. 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
1. 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
2. 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
3. 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
4. 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
5. 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

1. 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
2. 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
3. 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
4. (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
1. (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
2. (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
3. (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
4. (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
5. (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
6. (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
7. (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:
8. (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

1. (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
2. (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
3. (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
4. (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
5. (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:
6. (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
1. (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
2. 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
3. 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
4. (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
5. (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
6. 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
7. (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
1. (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
2. (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:
3. (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
4. (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

### 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

1. 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
2. 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
3. 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
4. 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
5. 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:

1. 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
2. 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
3. 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
4. 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

1. 193 ÷ 24 Solution
The estimated quotient of 193 ÷ 24 = 200 ÷ 20 = 10
2. 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

1. 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

1. 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
2. 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
3. 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
4. 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