29 is the smallest as it contains only 2 digits. 1234 is the greatest as it contains four digits, which is the maximum number of digits amongst all.

769 is the smallest as it contains minimum number of digits amongst all and 741965 is the greatest as it contains six digits.

Both 97632 and 96543 have same number of digits i.e. five. Now, compare them by looking at the digits at ten thousand place, which is 9 for both the numbers.
Next, compare the second digit from the left. Here, it is 7 in 97632 and 6 in 96543.
Since, 7> 6
So, 97632> 96543

Both 67454 and 78465 are five-digit numbers. So, we compare them by comparing the digits at ten thousand places. Since, 7> 6. So, 78465> 67454.

Given, numbers are 847, 9754, 8320 and 571.
To convert numbers into ascending numbers, we arrange numbers smallest to the greatest.
Now, ascending order is 571 < 847 < 8320 < 9754.

Given, numbers are 819, 5240, 890 and 6249.
To convert numbers into descending order, we arrange numbers greatest to the smallest.
Now, descending order is 6249 > 5240> 890> 819.

For getting the smallest number, firstly we convert the numbers into ascending order.
The smallest 4-digit number is 2478.
For getting greatest number, firstly we convert the numbers into descending order.
The greatest 4-digit number is 8742.

For getting the smallest number, firstly we convert the numbers into ascending order.
The smallest 4-digit number is 1479.
For getting greatest number, firstly we convert the numbers into descending order.
The greatest 4-digit number is 9741.

Given, number is 97621.
Hence, place value of 7 is 7000 and face value of 7 is 7.

59785 is the greatest and 18 is the smallest.

89423 is the greatest number, because it has highest number of digits i.e. 5 and 100 is the smallest number, because it has least number of digits i.e. 3.

75284 is the greatest number, because it has highest number of digits i.e. 5 and III is the smallest number, because it has least number of digits i.e. 3.

12002 is the greatest number, because it has highest number of digits i.e. 5 and 124 is the smallest number, because it has least number of digits i.e. 3. Yes, it was easy.We just identify the number of digits and locate the greatest and smallest numbers, because greatest number has more digits.

Here, descending order of the given digits is 8, 7, 4, 2.
∴ Greatest number = 8742
and ascending order of the given digits is 2, 4, 7, 8.
∴ Smallest number = 2478.

Here, descending order of the given digits is 9, 7, 4,1.
∴ Greatest number = 9741
and ascending order of the given digits is 1,4,7,9.
∴ Smallest number = 1479.

Here, descending order of the given digits is 7, 5,4, 0.
∴ Greatest number = 7540
and ascending order of the given digits is 0, 4, 5, 7.
But, 0457 is a three-digit number, so we have to interchange the place of 4 and 0 i.e. 4057.
∴ Smallest number = 4057.

Here, descending order of the given digits is 7, 6, 2, 1
∴ Greatest number = 7621
and ascending order of the given digits is 1,2,6, 7.
∴ Smallest number = 1267.

Here, descending order of the given digits is 5,4, 3, 0.
∴ Greatest number = 5430
and ascending order of the given digits is 0,3,4, 5.
But 0345 is a three-digit number. So, we have to interchange the place of 3 and 0 i.e 3045.
∴ Smallest number = 3045.

10 - 1 = 9

100 - 1= 99

10000 -1 = 9999

100000 -1 = 99999

10000000 -1 = 9999999

Numbers after commas are as follows: 73, 75, 307.

Numbers after commas are as follows: 9, 05, 00, 041.

Numbers after commas are as follows: 7, 52, 21, 302.

Numbers after commas are as follows: 58, 423, 202.

Numbers after commas are as follows: 23, 30, 010.

984671

2369

6 \(\times\) 10⁷ or 60000000

7

808

9 \(\times\)10000 + 6\(\times\)1000+ 2\(\times\)100 + 4\(\times\)10 + 2\(\times\)1

8621

12356

36436

Production of milk in one day = 75678 L                                  75678
Supply to milk depot in one day = 67689 L                             -67689
So, the milk left in the dairy = (75678 - 67689) = 7989 L           7989

We know that, 1 km = 1000 m = 1000 \(\times\) 100 cm [∵ 1 m= 100 cm]
=100000 cm

We know that, 1 kg = 1000 g = (1000 X 1000) mg [∵ 1 g = 1000 mg]
= 1000000 mg
∴ 1000000 milligrams make one kilogram

Rs.10636875

292 kg 500 g

62 km 625 m per hour

Rs.1642

Given number is 2936. Here, hundred digit is 9, which is more than 5 i.e. 9 > 5.
∴ The required rounded number 2936 = 3000

In 4235, hundred digit is 2 < 5.
∴ The required rounded number 4235 = 4000

We have, 27 \(\times\) 105 = 27 \(\times\) (100 + 5) = 27 \(\times\) 100 + 27 \(\times\) 5 = 2700 + 135 = 2835 [∵ 105 =100+5]

We have, 27 \(\times\) 99 = 27 \(\times\) (100 -1) = 2700 - 27 = 2673  [∵ 99 =100-1]

We have, 73 = LXXIII [∵73=50+10+10+1+1+1 i.e. 50=L, 10=X, 1 =I]

We have, 92 = XCII  [∵ 92=100-10+1+1 i.e.100=C,10=X, 1 = I]

596 = 500 +90 + 6= DXCVI [∵ 500= 0, 90=100-10=XC, 6=5 +1= VI]

479 = 400 + 70 + 9 = CDLXXIX  [∵ 400=500-100=CD, 70=50+10+10=LXX, 1 =I]

∵ 87⟶ 90 [rounding off to nearest tens]
313⟶300 [rounding off to nearest hundreds]
∴ Estimated product = 90 \(\times\) 300 = 27,000

∵ 9 ⟶ 10 [rounding off to nearest tens]
795 ⟶ 800 [rounding off to nearest hundreds]
∴ Estimated product = 10 \(\times\) 800 = 8,000

∵ 898 ⟶ 900 [rounding off to nearest hundreds]
785 ⟶ 800 [rounding off to nearest hundreds].
∴ Estimated product = 900 \(\times\) 800 = 7,20,000

∵ 958 ⟶ 1000 [rounding off to nearest hundreds]
387 ⟶ 400 [rounding off to nearest hundreds]
∴ Estimated product = 1000 \(\times\) 400 = 4,00,000

∵ 730 ⟶ 700 [rounding off to nearest hundreds]
998 ⟶1000 [rounding off to nearest hundreds]
∴ Estimated sum = 700 + 1000 = 1700.

∵ 796 ⟶ 800 [rounding off to nearest hundreds]
314 ⟶ 300 [rounding off to nearest hundreds]
∴ Estimated difference = 800 - 300 = 500

∵12904 ⟶ 13000[rounding off to nearest thousands]
2888 ⟶ 3000 [rounding off to nearest thousands]
∴ Estimated sum = 13000 + 3000 = 16000

∵ 28,292 ⟶ 28,000[rounding off to nearest thousands]
21,496 ⟶ 21,000 [rounding off to nearest thousands]
∴ Estimated difference = 28000 - 21000 = 7000

We have, 578 \(\times\) 161
578 ⟶ 600 [rounding off to nearest hundreds]
161 ⟶ 200 [rounding off to nearest hundreds]
∴ Estimated product = 600 \(\times\) 200 = 1,20,000

We have, 5281\(\times\)3491
5281 ⟶ 5000 [rounding off to nearest thousands]
3491 ⟶ 3500 [rounding off to nearest hundreds]
∴ Estimated product = 5060 \(\times\) 3500 = 1,75,00,000

We have, 1291\(\times\)592
1291⟶ 1300 [rounding off to nearest hundreds]
592 ⟶  600 [rounding off to nearest hundreds]
∴ Estimated product = 1300\(\times\) 600
=7,80,000

We have, 9250\(\times\)29
9250 ⟶ 9000 [rounding off to nearest thousands]
29 ⟶ 30 [rounding off to nearest tens]
∴ Estimated product = 9000 \(\times\) 30
= 2,70,000

According to the question, sum of 9 and 2 = 9 + 2
Now, mutiplying it by 4, we have 4 \(\times\) (9 + 2)
which is the required expression.

According to the question,
Difference of 18 and 6 = (18 - 6)
Now, dividing this difference by 4, we have (18 - 6) ÷ 4
which is the required expression.

According to the question, sum of 3 and 2 = (3 + 2)
Now, multiplying it by 3, we get 3 \(\times\) (3 + 2)
Again, divide 45 by this result, we have 45 ÷ 3 \(\times\) (3 + 2)
which is the required expression.

90000

200

80000

2943

1170000

CCCXLI

CCCLXXXIX

8000

8000

60000

DCCCLXVIII

CMLXI

199

76

Here, ten thousands digit is 4 in both numbers.
But thousands digit in 47645 is 7 and thousands digit in 48740 is 8. So, 8 > 7
Hence, 48740 > 47645

In 15896, ten thousands digit is 1 and ten thousands digit in 26760 is 2. So, 2 > 1.
Hence, 26760> 15896

In the given numbers, we see that 28706 is the smallest and 87604 is the greatest.

Here, given digits are 3, 5, 4, 6. For the smallest number, we write the digits in ascending order. So, the smallest four-digit number is 3456.

The ascending order of the given numbers are as follows: 1462 <1562 < 2605 < 3164

Given, digits are 1, 2, 7, 9, 4.
For greatest number, we write the digits in descending order.
So, the greatest five-digit number is 97421.

According to Indian system of numeration, 
24,64,056

According to  Indian system of numeration, 
68,96,462

Expanded form of given numbers as
76496 = 7 \(\times\) 10000 + 6 \(\times\) 1000 + 4 \(\times\) 100 + 9 \(\times\) 10 + 6

Expanded form of given numbers as
986256 =9 \(\times\) 100000 + 8 \(\times\) 10000 + 6 \(\times\) 1000 + 2 \(\times\) 100 + 5 \(\times\) 10 + 6

According to International system of numeration 2,546,726

According to International system of numeration 7,869,420

The numeral form is 20,502,632.

The place value of 6 in 762540 = 6 \(\times\)10000
=60000

We know that, 1 kg =1000 g, 5 kg =5000 g.
So, 5 kg 290 g =(5000+ 290) g
=5290 g

We know that,
1 km =1000 m
and 1 m =100 cm
∴ 1 km =1000 m=1000 \(\times\) 100 cm
=100000 cm

We know that,
1 m =100 cm and 1 cm =10 mm
∴ 1 m =100 cm =100\(\times\) 10 mm
=1000 mm

In 3945, we know that, ones digit is 5,
Then, rounded form of 3945 = 3950

In 4942, ones digit is 2 < 5.
Then, rounded form of 4942 = 4940

Here, rounded form of 945 = 900
and rounded form of 623 = + 600
So, estimated sum = 1500

Here, 945 is rounded off = 900
560 is rounded off = - 600
Estimated difference = 300

Here: 39 is rounded off = 40
and 42 is rounded off = \(\times\) 40
Estimated product = 1600

Here, given numbers are 137 and 198.
137= 100+ 30+ 7 = CXXXVIl
198= 100+ 90+ 8 = CXCVIII

XXIV = 20+ (5-1) = 24

XLVI =(50 -10) + 5 + 1 = 46

10000023

60000

96700

CXIV

90,000

65 km/h

400

Given, number is 475320, according to Indian system of numeration,

We read it as, four lakhs seventy-five thousands three hundreds twenty.
According to International system of numeration, using commas it is written as 475,320

We have, 4536, 4892, 4370, 4452
Here, each of the given numbers is containing 4-digits and their digits at thousand places are also same. So, we compare the hundreds place digit.

∵ 8>5>4>3
∴ The greatest number is 4892 and the smallest number is 4370.

We have, 15623, 15073, 15189, 15800
Here, the two leftmost digits 15 i.e. digits at thousand and ten thousands place are same in each numbers. So, we compare the hundreds place digit

∴ The greatest number is 15800 and the smallest number is 15073.

We have, 25286, 25245, 25270, 25210
Here, the three leftmost digits 252 are same in each number. So, we compare the tens place digit.

∴The greatest number is 25286 and the smallest number is 25210.

Given digits are 3, 8 and 7.
Here, descending order of the given digits is 8, 8, 7, 3. [∵ greatest digit 8 is taken twice]
∴ Greatest 4-digit number = 8873
and ascending order of the given digits is 3, 3, 7, 8. [∵ smallest digit 3 is taken twice]
∴ Smallest 4-digit number = 3378

Given digits are 9, 0 and 5.
Here, descending order of the given digits is 9, 9, 5, 0. [∵ greatest digit 9 is taken twice]
∴ Greatest 4-digit number = 9950
and ascending order of the given digits is 0,0, 5, 9. [∵ smallest digit 0 is taken twice]
But 0059 is a 2-digit number, because 0 at first two leftmost places is meaningless. So, keeping 5 (instead of 0) at the leftmost place and taking the smallest digit o twice i.e. 5009.
∴ Smallest 4-digit number = 5009

Given digits are 0, 4 and 9.
Here, descending order of the given digits is 9, 9, 4, 0. [∵ greatest digit 9 is taken twice]
∴ Greatest 4-digit number = 9940
and ascending order of the given digit is 0, 0, 4, 9. [∵ smallest digit 0 is taken twice]
But 0049 is a 2-digit number, because 0 at first two leftmost places is meaningless. So, keeping 4 (instead of 0) at the leftmost place and taking the smallest digit o twice, we get 4009. 
∴ Smallest 4-digit number = 4009.

Given digits are 8, 5 and 1.
Here, descending order of the given digits is 8, 8, 5, 1. [∴ greatest digit 8 is taken twice]
∴ Greatest 4-digit number = 8851
and ascending order of the given digits is 1, 1, 5, 8. [∴ smallest digit 1 is taken twice]
∴ Smallest 4-digit number = 1158.

Digits in ascending order are 0,1,2,3,4,5,6,7,8,9 and digits in descending order are 9,8,7,6,5,4,3,2,1, 0. Also, we know that 0 can never be taken at the leftmost place.
Keeping the digit 7 at ones place, we can fill other places tens, hundreds and thousands by remaining numbers.
Greatest number 

-[Keeping the greatest digits i.e. 9, 8 and 6 at thousands, hundreds and tens place]
Smallest number

[Keeping the smallest digits i.e. 0, 1, 2 at hundreds, thousands and tens place. If keep 0 at thousands place, then number will be equal to 3-digit number]

Digits in ascending order are 0,1,2,3,4,5,6,7,8,9 and digits in descending order are 9,8,7,6,5,4,3,2,1, 0. Also, we know that 0 can never be taken at the leftmost place.
Keeping the digit 4 at tens place, we can fill other places thousands, hundreds and ones place by remaining number.
Greatest number

[Keeping the greatest digits i.e. 9, 8 and 7 at thousands, hundreds and tens place]
Smallest number

[Keeping the smallest digits i.e. 9, 8 and 7 at thousands, hundreds and tens place].

Digits in ascending order are 0,1,2,3,4,5,6,7,8,9 and digits in descending order are 9,8,7,6,5,4,3,2,1, 0. Also, we know that 0 can never be taken at the leftmost place.
Keeping the digit 9 at hundreds place, we can fill thousands, tens and ones place by remaining digits.
Greatest number

[Keeping the greatest digit i.e. 8, 7 and 6 at thousands, tens and ones place]
Smallest number 

[Keeping the smallest digit l.e. 1, 0 and 2 at thousands, tens and ones place]

Digits in ascending order are 0,1,2,3,4,5,6,7,8,9 and digits in descending order are 9,8,7,6,5,4,3,2,1, 0. Also, we know that 0 can never be taken at the leftmost place.
Keeping the digit 1 at thousands place, we can fill hundreds, tens and ones place digit by remaining digits.
Greatest number 

[Keeping the greatest digit i.e, 9, 8 and 7 at hundreds, tens and ones place]
Smallest number

[Keeping the smallest digit i.e. 0, 2 and 3 at hundreds, tens and ones place]

Two given digits are 2 and 3. We want to make 4-digit numbers using both the digit equal number of times.
The possible 4-digit numbers are 2233, 3322, 2332, 3223, 2323 and 3232.
(i) The greatest number is 3322.
(ii) The smallest number is 2233.
(iii) We can make 6 different numbers in all.

(a) Yes, prices can be arranged in increasing order as follows:
Rs.1788 < Rs.1897 < Rs. 2635 < Rs. 2854 < Rs. 3975
(b) Yes, prices can be arranged in decreasing order as follows:
Rs.3975 > Rs. 2854 > Rs. 2635 > Rs. 1897 > Rs. 1788.

Five more situations may be
(i) Prices of 4 different kinds of branded shoes are Rs.4225, Rs.3835, Rs. 2620 and Rs. 2230, respectively.
(ii) The weight of 4 students of Class VI are 33 kg, 38 kg, 35 kg and 37 kg, respectively.
(iii) Marks obtained by 4 students in Mathematics out of 100 are 87, 88, 90 and 78, respectively.
(iv) The tution fee paid by 4 students of different classes are Rs. 170, Rs. 190, Rs. 150 and Rs. 200, respectively.
(v) The runs scored by 4 players A, B, C and D in an innings are 75, 60, 25 and 82, respectively.

(a) We have, 847, 9754, 8320, 571
Ascending order of given numbers is as follows:
571 < 847 < 8320 < 9754
(b) We have, 9801, 25751, 36501, 38802
Ascending order of given numbers is as follows:
9801 < 25751 < 36501 < 38802

(a) We have, 5000, 7500, 85400, 7861
Descending order of given numbers is as follows:
85400> 7861 > 7500 > 5000
(b) We have, 1971,45321,88715,92547
Descending order of given numbers is as follows:
92547> 88715 > 45321 > 1971

The least 6-digit number is 100000 (one lakh). Now, five examples, where the number of things counted would be more than 6-digit number are as follows:
(i) The number of people in a big city.
(ii) The number of stars in a clear dark night.
(iii) The number of motorcycles in a big town.
(iv) The number of grains in a sack full of wheat.
(v) The number of pages of notebooks of all students in a big town.

The greatest 6-digit number = 999999
1st previous number =999999 -1 = 999998
2nd previous number = 999999 - 2 = 999997
3rd previous number = 999999 - 3 = 999996
4th previous number =999999 - 4 = 999995
5th previous number = 999999 - 5 = 999994
Now, descending order of these numbers is as follows:
999998 > 999997 > 999996 > 999995 > 999994.

The smallest 8-digit number = 10000000
∴ 1 st next number = 10000000 + 1 = 10000001
2nd next number = 10000000 + 2 = 10000002
3rd next number =10000000 + 3 = 10000003
4th next number =10000000 + 4 = 10000004
5th next number = 10000000 + 5 = 10000005
Now, ascending order of these numbers is as follows:
10000001 < 10000002 < 10000003 < 10000004 < 10000005.

Given digits are 4,5,6,7,8 and 9.
We can make many numbers by using these digits, three of them are as follows :
98877456,98877465 and 98877654
Using commas, numbers can be rewritten as
(i) 9,88,77,456
(ii) 9,88,77,465
(iii) 9,88,77,654.

Given digits are 3, 0 and 4.
We can make many numbers by using these digits, five of them are as follows:
(i) 330443 (ii) 343404 (iii) 344430 (iv) 434043 (v) 344034
Using commas, these numbers can be rewritten as :
(i) 3,30,443 (ii) 3,43,404 (iii) 3,44,430 (iv) 4,34,043 (v) 3,44,034

We can draw a table to show these operations:

We can draw a table to show these operations:

Given, weight of one tablet = 20 mg
Total weight of 200000 tablets in the box = 200000 \(\times\) 20 mg
=4000000 mg
Now, weight in grams =\(\frac{4000000}{1000}g=4000g\) \(​​\left[ \because \quad 1mg=\frac { 1 }{ 1000 } g \right] \)
Now, weight in kilograms =\(\frac{4000}{1000}kg=4kg\) \(​​\left[ \because \quad 1g=\frac { 1 }{ 1000 } kg \right] \).

Given, number of tickets sold on the first day = 1094
Number of tickets sold on the second day = 1812
Number of tickets sold on the third day = 2050
Number of tickets sold on the fourth day = 2751
∴ Total number of tickets sold on all the four days
= 1094 + 1812 + 2050 + 2751 = 7707
Hence, there are 7707 tickets sold on all the four days.

Given, number of runs scored by Shekhar = 6,980
Target of runs to be scored = 10,000
∴  Number of runs needed more
= Target of runs - Run scored yet
= 10,000 - 6,980 = 3,020

Given, votes registered by the successful candidate =5,77,500
Votes registered by the nearest rival = 3,48,700
∴ Margin, by which the successful candidate won
= Votes of successful candidate - Votes of nearest rival
= 5,77,500 - 3,48,700 = 2,28,800

Given, sale of books in Ist week = Rs. 2,85,891
Sale of books in IInd week = Rs. 4,00,768
Total sales of two weeks =Sale of Ist week + Sale of IInd week
= 2,85,891 + 4,00,768 = Rs. 6,86,659
Since, 4,00,768 > 2,85,891
Thus, sale in the second week is greater.
∴ Difference in the sales amount = 4,00,768 - 2,85,891
= Rs. 1,14,877
Hence, sale in the second week was greater by Rs. 1,14,877.

Using the digits 6, 2, 7, 4 and 3, we get
Greatest number = 76,432 and least number = 23,467
∴ Difference = Greatest number - Least number
=76,432 - 23,467 = 52,965.

Number of screws manufactured by the machine in one day = 2825
We know that, number of days in the month of January =31
∴ Number of screws produced by the machine in the month of January, 2006 
= 2,825\(\times\) 31 = 8,7575
Hence, number of screws produced by the machine in the month of january, 2006 are 87,575.

Given, cost of one radio set = Rs. 1200
Number of radio sets to be purchased = 40
∴ Cost of 40 radio sets = Cost of one radio set \(\times\) Number of radio sets
= 40 \(\times\) 1200 = Rs. 48,000
∵ Total money with the merchant = Rs. 78,592
∴ Money left with the merchant after purchase of
40 radio sets = Total money - Cost of 40 radio sets
= 78,592 - 48,000 = Rs.30,592
Hence, Rs. 30592 will remain with her after the purchase.

If a student multiplies 7,236 by 65, he will get
= 7,236\(\times\)65
= 4,70,340
According to the question, Student multiplies 7236 by 56 and he will get
= 7,236\(\times\)56= 4,05,216
Now, required difference = Actual answer - Wrong answer
= 4,70,340 - 4,05,216
= 65,124

Total length of cloth = 40 m = 40 \(\times\) 100 cm [∵1 m =100 cm]
= 4000 cm
Cloth needed for one shirt = 2 m 15 cm
= (2\(\times\)100) cm +15 cm

= (200+15) cm = 215 cm 
∴ Number of shirts stitched out of total cloth
\(=\frac { Total\quad length\quad of\quad cloth }{ Cloth\quad needed\quad for\quad stitched\quad shirt } \)
\(=\frac { 4000 }{ 215 } =18\frac { 130 }{ 215 } \)
Hence, 18 shirts can be stitched and cloth left over is 130 cm i.e. 1 m 30 cm.

Given, distance between the school and the house
= 1 km 875 m =1\(\times\)1000+875
= 1000 + 875 = 1875 m [∵ 1 km = 1000 m]
∴ Distance covered by the student in one day
= 2\(\times\) 1875 = 3750 m [∵ She walks both ways in one day]
∴ Distance covered in 6 days = 6 x 3750
= 22500 m = (22 \(\times\) 1000 + 500) m
= 22 km 500 m    [∵1 km = 1000 m]

Given, capacity of vessel = 4 L 500 mL
= 4 \(\times\)1000 + 500 = 4000 + 500 = 4500 mL. [1L= 1000mL]
and capacity of one glass = 25 mL
∴ Number of glasses of curd filled out of vessel
\(=\frac { Capacity\quad of\quad a\quad vessel }{ Capacity\quad of\quad a\quad glass } =\frac { 4500 }{ 25 } =180\)
Hence, curd can be filled in 180 glasses.

Rs.112338

152

51975

Rs. 16570

We can arrange it as a table

On checking the rounding off in respect of given numbers, we find that, the rounding off of 286 and 9870 are incorrect. Their correct result should be 300 and 9900, respectively.

On checking the rounding off in respect of given numbers, we find that the rounding off of 7805, 53552 and 3499 are incorrect.
Their correct results should be 7800, 54000 and 3000, respectively.

The complete table is shown as below:

We have, 439 + 334 + 4317
\(\therefore \begin{matrix} 439\rightarrow 400 \\ 334\rightarrow 300 \\ 4317\rightarrow 4300 \end{matrix}\} [rounding\quad off\quad to\quad nearest\quad hundreds]\)
∴ Rough estimate = 439 + 334 + 4317
= 400 + 300 + 4300 = 5000
\(Again\begin{matrix} 439\rightarrow 440 \\ 334\rightarrow 330 \\ 4317\rightarrow 4320 \end{matrix}\} [rounding\quad off\quad to\quad nearest\quad tens]\)
and closer estimate = 439 +334 +4317
= 440 + 330 + 4320 = 5090

We have, 1,08,734 - 47599
\(\because \begin{matrix} 1,08,734\rightarrow 1,08,700 \\ 47,599\rightarrow 47,600 \end{matrix}\} [rounding\quad off\quad to\quad nearest\quad hundreds]\)
∴  Rough estimate = 1,08,734 - 47,599
= 1,08,700 - 47,600 = 61,100
\(\because \begin{matrix} 1,08,734\rightarrow 1,08,730 \\ 47,599\rightarrow 47,600 \end{matrix}\} [rounding\quad off\quad to\quad nearest\quad tens]\)
and closer estimate = 1,08,734 - 47,599
=1,08,730 - 47,600 = 61,130

We have, 8325 - 491
\(\because \begin{matrix} 8325\rightarrow 8300 \\ 491\rightarrow 500 \end{matrix}\} [rounding\quad off\quad to\quad nearest\quad hundreds]\)
∴ Rough estimate = 8325 - 491 = 8300 - 500 = 7800
\(Again\begin{matrix} 8325\rightarrow 8330 \\ 491\rightarrow 490 \end{matrix}\} [rounding\quad off\quad to\quad nearest\quad tens]\)
and closer estimate = 8325 - 491 = 8330 - 490 = 7840

We have, 4,89,348 - 48,365
\(\because \begin{matrix} 4,89,348\rightarrow 4,89,300 \\ 48,365\rightarrow 48,400 \end{matrix}\} [rounding\quad off\quad to\quad nearest\quad hundreds]\)
∴  Rough estimate = 4,89,348 - 48,365
= 4,89,300 - 48,400 = 4,40,900
\(Again\begin{matrix} 4,89,348\rightarrow 4,89,350 \\ 48,365\rightarrow 48,370 \end{matrix}\} [rounding\quad off\quad to\quad nearest\quad tens]\)
and closer estimate = 4,89,348 - 48,365
= 4,89,350 - 48,370 = 4,40,980

We can make many situations of this kind, three situations of them are as follows:
Situation 1. Prabha read 5 pages of a novel in the evening and 8 pages in the morning. How many pages did Prabha read in 6 days?
Situation 2. Meena and Reena work for 6 days. Meena earns Rs. 5 per day and Reena earns Rs.8 per day. What do they earn together in 6 days?
Situation 3. Ram and Shyam walk for 6 days. Ram walks 5 km in a day and Shyam walks 8 km in a day. How many kilometres do both of them walk in 6 days?

We can make many situations of this kind, five situations of them are as follows:
Situation 1. Marry is a maid (servant). She charges Rs. 8 per day to clean a house but gives back Rs. 3 to the house owner for saving. What amount does she carry in her hand from 7 houses?
Situation 2. Ankit's pocket money is Rs. 8 per day. He spents Rs. 3 daily. Find his total saving in 7 days.
Situation 3. What will be the seven times the difference of eight and three?
Situation 4. A clerk works 8 h daily in an office, but he remains absent for 3 h daily. In a week, for how many hours did he work?
Situation 5. Seven children with Rs. 8 each went to market. Each of them bought pencils costing Rs. 3. What is total money left with them?

Here, we have
The greatest five-digit number is 76542.
The smallest five-digit number is 24567.

Here, we have
The greatest five-digit number is 87320.
The smallest five-digit number is 20378.

According to the Indian and International system of numeration, 3748763

According to the Indian and International system of numeration, 32660505

According to the Indian and International system of numeration, 2004004

Given, number is 6523689.
Place value of first 6 = 6 \(\times\) 1000000 = 6000000
Place value of second 6 = 6 \(\times\) 100 = 600
∴ Difference between the place values of two 6's
= 6000000 - 600 = 5999400

(i) 5981 rounded off to nearest tens = 5980
4428 rounded off to nearest tens = 4430
∴ Estimated product = 5980 \(\times\) 4430 = 26491400
(ii) 5981 rounded off to nearest hundreds = 6000
4428 rounded off to nearest hundreds = 4400
∴ Estimated product = 6000 \(\times\) 4400 = 26400000

Monthly saving = Rs. 250
We know that, 1 yr = 12 months, 2 yr = 24 months
Total saving in 2 yr = 24 x 250 = Rs. 6000

Given, mass of a cylinder = 14 kg 250 g
Mass of 20 cylinders = 20 \(\times\) (14 kg + 250 g)
= 280 kg + 5000 g
= 280 kg + 5 kg
= 285 kg

Distance covered in one day = 365 m
Number of days = 120
So, total distance covered by Tony in 120 days
= 365 \(\times\) 120 m = 43800 m
∴ Estimated product = 44000 m = 44 km

We have, 8739 \(\times\)102 = 8739 \(\times\) (100 + 2)
= 8739 \(\times\) 100 + 8739 \(\times\) 2
= 873900 + 17478 = 891378

37\(\times\)108=37\(\times\)(100+8) =37\(\times\)100+37\(\times\)8
= 3700 + 296 = 3996

54 \(\times\) 105 = 54 \(\times\) (100 + 5)
= 5400 + 54 \(\times\) 5 = 5400 + 270 = 5670

Cloth required for 1 shirt =3 m 05 cm
Cloth required for 14 shirts = (3 m 05 cm) \(\times\) 14
=14\(\times\)3+14\(\times\)05=42 m 70 cm

Total cost of 20 flats = 348642060
Cost of each flat = 348642060 + 20 = Rs. 17432103

Distance cover in 18 h = 1126 km
Distance cover in 1 h = 1126 ÷ 18 = 62 km 555 m
Hence, the speed of the car is 62 km 555 m per hour.

Total number of students in school = 1261
Contribution of money by each student =Rs. 520
∴ Total money =1261 \(\times\) 520 =Rs. 655720

Rs.124,250

Rs. 27666050, ~Rs. 48622530

1325 g or 1 kg 325 g

2 m²

29000 (nearest thousand)

143000 (nearest thousand)

1234524

We have, 6895, 23787, 24569, 24659
Here, 6895 is a 4-digit number and other numbers are 5 digits, so it is clear that 6895 is the smallest number. Now, in remaining three numbers, the digit at ten thousands place is same in each number. So, we compare the thousands place digit.

∵ 4>3
Again, in two numbers, digit at thousands place is same. So, we compare the hundreds place digit.

∵ 6>5
∴ The greatest number is 24659 and smallest number is 6895.

(i) From the given figure, it is clear that,
160> 159> 158> 154
∴ Ramhari is the tallest, because he has the height of 160 cm.
(ii) From the given figure, it is clear that,
154 < 158 < 159 < 160
∴ Dolly is the shortest, because she has the height of 154 cm.
We can arrange height from two types:
(a) Yes, we can arrange their heights in increasing order as follows: Dolly (154 cm) < Mohan (158 cm) < Shashi (159 cm) < Ramhari (160 cm).
(b) Yes, we can arrange their heights in decreasing order as follows: Ramhari (160 cm) > Shashi (159 cm) > Mohan (158 cm) > Dolly (154 cm).

The remaining blanks are shown as below with appropriate number, name and expansion:

The remaining blanks are shown as below with appropriate number, name and expansion

After reading these numbers, we have
(i) 475320 Four lakhs seventy five thousands three hundred twenty.
(ii) 9847215 Ninety eight lakhs forty seven thousands two hundreds fifteen.
(iii) 97645310 Nine crores seventy six lakhs forty five thousands three hundreds ten.
(iv) 30458094 Three crores four lakhs fifty eight thousands ninety four.
The placement boxes is shown as below:

Expanded forms of given numbers are as follow:
(i) 475320 = 4 \(\times\)100000 + 7 \(\times\) 10000 + 5\(\times\) 1000 + 3 \(\times\) 100 +2\(\times\)10+ 0 \(\times\) 1
(ii) 9847215 = 9 \(\times\) 1000000 + 8 \(\times\) 100000 + 4 \(\times\) 10000 + 7 \(\times\) 1000 + 2 \(\times\) 100 + 1 \(\times\)10 + 5 \(\times\) 1
(iii) 97645310 = 9 \(\times\) 10000000 + 7 \(\times\) 1000000 + 6 \(\times\) 100000 + 4 \(\times\) 10000 + 5 \(\times\) 1000 + 3 \(\times\) 100 + 1 \(\times\) 10 + 0 \(\times\)1
(iv) 30458094 = 3\(\times\) 10000000 + 4 \(\times\) 100000 + 5 \(\times\) 10000 + 8 \(\times\) 1000 + 0 \(\times\) 100 + 9\(\times\) 10 + 4 \(\times\) 1
(a) The smallest number = 475320
(b) The greatest number = 97645310
(c) Ascending order of given numbers are as follow:
475320 < 9847215 < 30458094 < 97645310
Descending order of given numbers are as follow:
97645310> 30458094 > 9847215> 475320.

After reading these number, we have
(i) 527864 Five lakhs twenty seven thousands eight hundreds sixty four.
(ii) 95432 Ninety five thousands four hundreds thirty two.
(iii) 18950049 One crore eighty nine lakhs fifty thousands forty nine.
(iv) 70002509 Seven crores two thousand five hundreds nine.
(a) By USIng placement boxes, gIven number can be written as
By using the commas, given numbers can be written as :

(b) Ascending order of these numbers are as follow:
95432 < 527864 < 18950049 < 70002509
Descending order of these numbers are as follow:
70002509> 18950049> 527864 > 95432.

Given digits are 4, 5, 6, 0, 7 and 8.
We can make many numbers by using these digits, five of them are as follows:
876540, 867540, 876450, 876045 and 867405
(a) After putting commas, numbers are as follows:
(i) 8,76,540 (ii) 8,67,540
(iii) 8,76,450 (iv) 8,76,045
(v) 8,67,405
(b) Ascending order of numbers is as follows:
867405 < 867540 < 876045 < 876450 < 876540
Descending order of numbers is as follows:
876540 > 876450 > 876045 > 867540 >  867405.

The names of five large cities in India are Delhi, Bengaluru, Kolkata, Hyderabad and Chennai.
Process of making the pair of cities are given below:
We have, Delhi, Bengaluru, Kolkata, Hyderabad and Chennai
First, we take Delhi as base city
So,

Now, we take Bengaluru as base city.

Now, we take Kolkata as base city.
So,

Now, we take Hyderabad as base city.
So,

Their respective population can be taken from a book of general awareness. However, the distance (in kilometres) between each pair of these cities is given below:

These are the national distances between the cities.

From the given figure,
(i) Total distance covered by bus from A to D
= AB + BC + CD =4170 +3410 + 2160 =9740 km
(ii) Total distance covered by bus from D to G
= DE + EF + FG = 8140 + 4830 + 2550 = 15520 km
(iii) Total distance covered by bus = Distance from A to D + Distance from D to G + Distance from G to A
=9740+15520+1290 [from (i) and (ii) parts)
=26550 km
(iv) Distance from C to D = 2160 km
Distance from D to E = 8140 km
∴ Difference of distances from C to D and D to E
=8140 - 2160 = 5980 km
(v) To find out the time for individual places, we will use the formula of time i.e.
\(Time=\frac { Distance }{ Speed } \)
(a) Time taken by bus to reach from A to B
\(=\frac { Distance\quad from\quad AtoB }{ Speed } =\frac { 4170 }{ 60 } =\frac { 139 }{ 2 } =69\frac { 1 }{ 2 } h\)
(b) Time taken by bus to reach from C to D
\(=\frac { Distance\quad from\quad CtoD }{ Speed } =\frac { 2160 }{ 60 } =36h\)
(c) Time taken by bus to reach from E to G
\(=\frac { Distance\quad from\quad EtoG }{ Speed } =\frac { 4830+2550 }{ 60 } =\frac { 7380 }{ 60 } =123h\)
(d) Time taken by bus for total journey
\(=\frac { Total\quad distance\quad covered }{ Speed } =\frac { 26550 }{ 60 } =\frac { 885 }{ 2 } =442\frac { 1 }{ 2 } h\).

(a) Weight of apples sold during the last year = 2457 kg
Weight of oranges sold during the last year = 3004 kg
∴ Total weight of apples and oranges Raman sold, during the last year = 2457 + 3004= 5461 kg
(b) Total money Raman got by selling apples
=Total number of apples \(\times\) Cost per kg
= 2457 \(\times\) 40 = Rs. 98280
(c) Total money Raman got by selling oranges
=Total number of oranges \(\times\) Cost per kg
=3004 \(\times\) 30 = Rs. 90120
∴ Total money Raman got by selling apples and oranges together =98280 + 90120 = Rs. 188400
(d) Following table shows the money received from selling items:

The entries of amount of money received in descending order are as follows:
Rs. 253670 >Rs. 240012 >Rs. 160040 > Rs. 98280 > Rs. 90120 >Rs. 68280 > Rs. 38350
The item, which brought him the highest amount is tooth brushes. This amount is Rs. 253670.

Given, van can carry a weight of 800 kg
= 800 \(\times\) 1000 g = 800000 g
According to the question,
Weight of one packet = 4 kg 500 g
= 4 \(\times\) 1000 + 500

= 4000 + 500
= 4500 g
∴ Number of packets that can be loaded in the van\(=\frac { Total\quad weight\quad strength\quad of\quad the\quad van }{ Weight\quad of\quad one\quad packet } \)
\(=\frac { 800000 }{ 4500 } =\frac { 8000 }{ 45 } \)
Hence, only 177 boxes can be loaded in the van.

Rs.297600

31 km 640 m

In roman numerals, there are seven basic symbols to write any numeral. These symbols are given as below:

Now, we will break the given numbers into small numbers in such a way, that these numbers can be written in place of those numbers.
(i) Here, 73 can be written as sum of 50, 10, 10, 1, 1 and 1. Also, we can replace these numbers by Roman numerals easily.
∴  73 = 70 + 3 = 50 + 10 + 10 + 1 +  1+ 1
Hence, roman numerals = LXXIII
(ii) Similarly, 92= 90 + 2= 90 + 1+1 = (100 - 10) + 1+ 1
Hence, roman numerals = XCII
(If we subtract the smaller value from greater value then, we write symbol for smaller value to the left of symbol of greater value).

To find, the minimum number of square slabs to cover the floor, we have to find the greatest size of each such slab.
For this purpose, we have to find the HCF of 450 and 300.
[since, 4.5 m = 450cm and 3 m = 300 cm)
Now, HCF of 450 and 300 = 150
∴ Required size of the slab =150 cm \(\times\) 150 cm
∴ Number of required slabs
\(=\frac { Area\quad of\quad the\quad floor }{ Area\quad of\quad one\quad slab } \)
\(=\frac { 450\times 300 }{ 150\times 150 } =6\)
Hence, the number of slabs is required 6.

We have,
Total stock of wheat = 17380200 quintal
Quantity of wheat sent to Delhi = 2756744 quintal
Quantity of wheat sent to UP = 4863108 quintal
Total quantity of wheat taken out of the godown
= 2756744 + 4863108 = 7619852 quintal
Balance stock of wheat in godown
= (17380200 - 7619852)
= 9760348 quintal
Hence, 9760348 quintal balanced stock wheat in godown.

Given digits are 2, 0, 4, 7, 6, 3.
Using each digit only once,
The greatest six-digit number is 764320.
The smallest six-digit number is 203467.
Now, sum of these numbers
= 764320 + 203467 = 967787

Given, total number of packets = 50
Weight of each packet = 120 g
Weight of a box = 50 \(\times\) 120 g = 6000 g= 6 kg [∵ 1000 g = 1 kg]
∴ Required number of boxes = \(\frac{900}{6}=150\) 

(a) Distance between her house to the school =\(\frac{8}{10}\)km
Distance covered by bus =\(\frac{1}{2}\)km
She walks distance =\(\left( \frac { 8 }{ 10 } -\frac { 1 }{ 2 } \right) =\frac { 8-5 }{ 10 } =\frac { 3 }{ 10 } km\)
\(=\frac { 3\times 1000 }{ 10 } =300m\)
(b) She walk some distance daily, because
(i) its strengthen her heart.
(ii) its lower disease risk.
(iii) it keeps her weight balance.

Given, population of India in 1961 = 439 millions
= 439 \(\times\)1000000 = 439000000 [∵ 1million = 1000000]
and population of India in 2001 = 1028 millions
= 1028\(\times\) 1000000 = 1028000000 [∵ 1 million = 1000000]
∴ Total increase in population from 1961 to 2001
= Population in 2001 - Population in 1961
= 1028000000 - 439000000 = 589000000
= 589 \(\times\) 1000000 = 589 millions
So, the increase population in Indian system of numeration = 58,90,00,000

Given, total money = Rs.10,00,000
Money spent on a colour TV = Rs. 16,580
Money spent on a motorcycle = Rs. 45,890
and money spent on a flat = Rs. 8,70,000
Total amount spent = 16,580 + 45,890 + 8,70,000
= Rs. 9,32,470
∴ Money left with him = 10,00,000 - 9,32,470
= Rs.67,530
Hence, Rs. 67,530 was left with him.

Given, Chinmay's total money =Rs.6,10,000
Money given to Jyoti by Chinmay = Rs. 87,500
Money given to Javed by Chinmay =Rs. 1,26,380
Money given to John by Chinmay = Rs. 3,50,000
∴ Money left with Chinmay = Total money - Distributed money
=6,10,000 -(87,500 + 1,26,380 + 3,50,000)
= 6,10,000 - 5,63,880 =Rs. 46,120
Hence, Rs. 46,120 was left with him.
The value depicted by Chinmay is that he is helpful nature towards his friends.

According to the question,
Digit at ten's place = 4
Digit at unit's place = \(\frac{1}{4}\) of ten's place digit =\(\frac{1}{4}\times4=1\)
Digit at hundred's place = 0
Digit at thousand's place = 5 x Digit of unit's place
=5\(\times\)1=5
Digit at ten thousand's place = 2 x Digit of ten's place
=2\(\times\)4=8
∴ Required number = 85041

According to the question,
A garment factory produced shirts = 216315
Produced trousers = 182736
Produced jackets = 58704
∴Total production of all the three items in that year
= Sum of all items
= 216315 + 182736 + 58704 = 457755
Hence, the total production of all the three items in that year is 457755.

m 5 cm

Rs.1499568

2 m 35 cm

Rs.1692279

356666004

32812 km

We know that,
1 lakh = 100000 (in digits) = 10 \(\times\) 10000
= ten \(\times\) ten thousand (in number name)
1 lakh = ten ten thousand

We know that,
1 million = 1000000 (in digits) = 10 \(\times\) 100000
= ten\(\times\) hundred thousands
(in number name)
∴ 1 million = ten hundred thousands

We know that,
1 crore = 10000000 (in digits) = 10 \(\times\)1000000
= ten \(\times\)ten lakhs (in number name)
∴1 crore = ten ten lakhs

We know that,
1 crore = 10000000 (in digits) = 10 \(\times\) 1000000
= ten \(\times\) one million (in number name)
∴ 1 crore = ten millions

We know that,
1 million = 1000000 (in digits)
= 10 \(\times\)100000 = ten \(\times\) one lakhs (in number name)
∴1 million = ten lakhs

According to Indian system of numeration, 6,17,11,682.

1023 is required smallest four-digit number.

Here, given number is 60 i.e. LX.

Given population is 2538473.
Rounded off to nearest thousands = 2538000