### EXERCISE 1A

**Contents**hide

- 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

#### ANSWER 1A

(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**i

**ons**

(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

**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

**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

**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

**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

**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

**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 <:

**1003467 ☐ 987965 Solution**

1003467 > 987965**3572014 ☐ 10235401**

Solution

3572014 < 10235401**3254790 ☐ 3260152**

Solution

3254790 < 3260152- 1
**0357690 ☐ 11243567**

Solution

10357690 < 11243567 **27596381 ☐ 7965412**

Solution

27596381 > 7965412**47893501 ☐ 47894021**

Solution

47893501 < 47894021

#### Arrange the following numbers in descending order:

**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:

**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**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

**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**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

**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**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**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**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**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

**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**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**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**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**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

**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**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**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**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**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

**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**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

**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**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**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**(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

**(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**(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**(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**(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**(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**(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**(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:- (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

**(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**(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**(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**(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**(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:**(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

**(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**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**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**(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**(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**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**(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

**(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**(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:**(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**(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 × 63SolutionGiven**

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

**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**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**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**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- 15 × 34

**SolutionGiven15 × 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:

**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**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**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**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

- 193 ÷ 24 Solution

The estimated quotient of 193 ÷ 24 = 200 ÷ 20 = 10 - 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

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

**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) 486Solution**

Here the place value of 6 is 6 lakhs = (6 × 100000) = 600000

Hence option (c) is the correct answer

**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**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**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**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