Rd Sharma 2018 Solutions for Class 6 Math Chapter 1 Knowing Our Numbers are provided here with simple step-by-step explanations. These solutions for Knowing Our Numbers are extremely popular among Class 6 students for Math Knowing Our Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2018 Book of Class 6 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2018 Solutions. All Rd Sharma 2018 Solutions for class Class 6 Math are prepared by experts and are 100% accurate.

Page No 1.13:

Question 1:

Write each of the following numbers in digits by using international place value chart. Also, write them in expanded form
(i) Seven million three hundred three thousand two hundred six.
(ii) Fifty five million twenty nine thousand seven.
(iii) Six billion one hundred ten million three thousand seven.

Answer:

(i) 7,303,206

Expanded form = 7 × 1000000 + 3 × 100000 + 0 × 10000 + 3 × 1000 + 2 × 100 + 0 × 10 + 6 × 1

(ii) 55,029,007

Expanded form = 5 × 10000000 + 5 × 1000000 + 0 × 100000 + 2 × 10000 + 9 × 1000 + 0 × 100 + 0 × 10 + 7 × 1

(iii) 6,110,003,007

Expanded form = 6 × 1000000000 + 1 × 100000000 + 1 × 10000000 + 0 × 1000000 + 0 × 100000 + 0 × 10000 + 3 × 1000 + 0 × 100 + 0 × 10 + 7 × 1


 

Page No 1.13:

Question 2:

Rewrite each of the following numerals with proper commas in the international system of numeration. Also, write the number name of each in the international system of numeration
(i) 513625
(ii) 4035672
(iii) 65954923
(iv) 70902005

Answer:

(i) 513,625 or Five hundred thirteen thousand six hundred twenty five.

(ii) 4,035,672 or Four million thirty five thousand six hundred seveny two.

(iii) 65,954,923 or Sixty five million nine hundred fifty four thousand nine hundred twenty three

(iv) 70,902,005 or Seventy million nine hundred two thousand five

Page No 1.13:

Question 3:

Write each of the following numbers in the International system of numeration:
(i) Forty three lakh four thousand eighty four.
(ii) Six crore thirty four lakh four thousand forty four.
(iii) Seven lakh thirty five thousand eight hundred ninety nine only.

Answer:

(i) 4,304,084 or Four million three hundred four thousand eighty four.

(ii) 63,404,044 or Sixty three million four hundred four thousand forty four.

(iii) 735,899 or Seven hundred thirty five thousand eight hundred ninety nine.

Page No 1.13:

Question 4:

Write the following numbers in the Indian system of numeration:
(i) Six million five hundred forty three thousand two hundred ten.
(ii) Seventy six million eighty five thousand nine hundred eighty seven.
(iii) Three hundred twenty five million four hundred seventy nine thousand eight hundred thirty eight.

Answer:

(i) 65,43,210 or Sixty five lakh forty three thousand two hundred ten.

(ii) 7,60,85,987 or Seven crore sixty lakh eighty five thousand nine hundred eighty seven.

(iii) 32,54,79,838 or Thirty two crore fifty four lakh seventy nine thousand eight hundred thirty eight.

Page No 1.13:

Question 5:

A certain nine digit number has only ones in ones period, only twos in the thousands period and only threes in millions period. Write this number in words in the Indian system.

Answer:

The number is 333, 222,111.
In Indian system, the number is written as 33,32,22,111 ⇒ Thirty-three crore thirty-two lakh twenty-two thousand one hundred and eleven.

Page No 1.13:

Question 6:

How many thousands make a million?

Answer:

1,000 thousands make a million.



Page No 1.14:

Question 7:

How many millions make a billion?

Answer:

1,000 millions make a billion.

Page No 1.14:

Question 8:

(i) How many lakhs make a million?
(ii) How many lakhs make a billion?

Answer:

(i) Ten lakhs make a million.
(ii) Ten thousand lakhs make a billion.

Page No 1.14:

Question 9:

Write each of the following in numeral form:

(i) Eight million seven hundred eight thousand four.
(ii) Six hundred seven million twelve thousand eighty four.
(iii) Four billion twenty five million forty five thousand.

Answer:

(i) 8,708,004
(ii) 607,012,084
(iii) 4,025,045,000

Page No 1.14:

Question 10:

Write the number names of each of the following in international system of numeration:

(i) 435,002
(ii) 1,047,509
(iii) 59,064,523
(iv) 25,201,905

Answer:

(i) Four hundred thirty-five thousand and two
(ii) One million, forty-seven thousand, five hundred and nine
(iii) Fifty-nine million, sixty-four thousand, five hundred and twenty-three

(iv) Twenty-five million, two hundred one thousand, nine hundred and five



Page No 1.16:

Question 1:

How many four-digit numbers are there in all?

Answer:

There are 10 digits i.e., 0, 1, 2, 3, 4, 5, 6 ,7, 8, 9.
We can not use '0' at thousand's place.
So, we can use only 9 digits at thousand's place.
Also, we can use 10 digits at hundrend's, 10 digits at ten's and 10 digits at unit's place.
So, total numbers of four-digit numbers = 9 × 10 × 10 × 10 = 9000

Page No 1.16:

Question 2:

Write the smallest and the largest six digit numbers. Howmany numbers are between these two.

Answer:

The smallest digit is 0. But we cannot use 0 at the place having the highest place value in six digit numbers. So, we will use the second smallest digit i.e., 1. All other places are filled by 9.
Hence, the required number = 100000
smallest six digit number will be 100000.
 
The largest digit is 9. We can use 9 at any place. Infact, we can use 9 in all places in six digit numbers.
Hence, the required number = 999999
largest six digit number will be 999999
Required difference = 999999 − 100000 = 899999

So, the total numbers between 999999 and 100000 will be 899998.

 

Page No 1.16:

Question 3:

How many 8-digit numbers are there in all?

Answer:

There are 10 digits i.e., 0, 1, 2, 3, 4, 5, 6 ,7, 8, 9.
We can not use '0' at the place having the highest place value in 8 digit numbers
So, we can use only 9 digits at the place having the highest place value in 8 digit numbers
Also, we can use 10 digits at the remaining places in 8 digit numbers
So, total numbers of 8-digit numbers = 9 × 10 × 10 × 10 × 10 × 10 × 10 × 10 = 90000000

Page No 1.16:

Question 4:

Write 10075302 in words and rearrange the digits to get the smallest and the largest numbers.

Answer:

One crore seventy five thousand three hundred two.

In order to write the smallest 8-digit number using digits 0, 1, 2, 3, 5 and 7, we put the smallest digit 1 (Except 0) at the place having the highest place value. The largest digit 7 is put at the right most place i.e. at unit's place, the digit 5 is put at the ten's place, the digit 3 is put at the hundred's place and the digit 2 is put at the thousand's place. All other places are filled by 0.
Hence, the required largest number is 10002357.
In order to write the largest 8-digit number using digits 0, 1, 2, 3, 5 and 7, we put the largest digit 7 at the place having the highest place value. The smallest digit 5 is put at the place after the highest place value. We put the next smallest digit(i.e., 3) after the previous one. After it we place the next smallest digit(i.e., 2) and after that we put the digit 1. All other places are filled by 0.
Hence, the required largest number is 75321000.
 

Page No 1.16:

Question 5:

What is the smallest 3-digit number with unique digits?

Answer:

The smallest three-digit number with unique digits is 102.

Page No 1.16:

Question 6:

What is the largest 5-digit number with unique digits?

Answer:

The largest five-digit number with unique digits is 98,765.

Page No 1.16:

Question 7:

Write the smallest 3-digit number which does not change if the digits are written in reverse order.

Answer:

The smallest three-digit number that does not change if the digits are written in reverse order is 101.

Page No 1.16:

Question 8:

Find the difference between the number 279 and that obtained on reversing its digits.

Answer:

The number obtained on reversing 279 = 972

Difference = 972 − 279 = 693
Thus, the difference between 279 and that obtained on reversing its digits is 693.

Page No 1.16:

Question 9:

Form the largest and smallest 4-digit numbers using each of digits 7,1,0,5 only once.

Answer:

The largest and smallest four-digit numbers formed using 7, 1, 0 and 5 are 7,510 and 1,057.

 



Page No 1.18:

Question 1:

Put the appropriate symbols (<,>) in each of the following boxes:

(i) 102394      99887
(ii) 2507324      2517324
(iii) 3572014      10253104
(iv) 47983505      47894012

Answer:

(i) 1,02,394 > 99,887
(ii) 25,07,324 
< 25,17,324
(iii) 35,72,014 < 1,02,53,104
(iv) 4,79,83,505 > 4,78,94,012

Page No 1.18:

Question 2:

Arrange the following numbers in ascending order:

(i) 102345694, 8354208, 6539542, 63547201, 12345678
(ii) 1808090, 1808088, 181888, 190909, 16060666

Answer:

(i) 65,39,542, 83,54,208, 1,23,45,678, 6,35,47,201, 10,23,45,694

(ii) 1,81,888, 1,90,909, 18,08,088, 18,08,090, 1,60,60,666 

Page No 1.18:

Question 3:

Arrange the following numbers in descending order:

(i) 56943300, 56943201, 5695440, 56944000, 5694437
(ii) 1020216, 1020308, 1021430, 893245,893425

Answer:

(i) 5,69,44,000, 5,69,43,300, 5,69,43,201, 56,95,440, 56,94,437
(ii) 10,21,430, 10,20,308, 10,20,216, 8,93,425, 8,93,245



Page No 1.21:

Question 1:

How many milligrams make on kilogram?

Answer:

Ten lakh or one million (10,00,000) milligrams make one kilogram.

Page No 1.21:

Question 2:

A box of medicine tablets contains 2,00,000 tablets each weighing 20 mg. What is the total weight of all the tablets in the box in grams? In kilograms?

Answer:

∵ Each tablet weighs = 20 mg

∴ The weight of 2,00,000 tablets = 2,00,000 × 20 = 40,00,000 mg
​ ∴ The total weight of all the tablets in the box = 40,00,000 mg
 
We know 1 g = 1,000 mg
∴ Weight of the box having all tablets = 40,00,000 ÷ 1,000 = 4,000 g

And, as 1 kg = 1,000 g
∴ Weight of the box having all tablets = 4,000 ÷ 1,000 = 4 kg

Page No 1.21:

Question 3:

Population of Sundar nagar was 2,35,471 in the year 1991. In the year 2001 it was found to have increased by 72,958. What was the population of the city n 2001?

Answer:

The population of Sundar Nagar in 2001 = Sum of the population of city in 1991 + Increase in population over the given time period
∵ The population of Sundar Nagar in 1991 = 2,35,471
∵ Increase in population over the given time period = 72,958
∴ 
The population of Sundar Nagar in 2001 = 2,35,471 + 72,958 = 3,08,429

Page No 1.21:

Question 4:

A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final days wre respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.

Answer:

Total number of tickets sold on all four days is the sum of the tickets sold on the first, second, third and final days.
∴ Total number of tickets sold on all four days = 1,094 + 1,812 + 2,050 + 2,751 = 7,707

Page No 1.21:

Question 5:

The town newspaper is published everyday. One copy has 12 pages. Everyday 11,980 copies are printed. How many pages are in all printed everyday? Every months?

Answer:

∵ Number of pages in 1 copy of newspaper = 12

∴ Number of pages in 11,980 copies of newspaper = 11,980 × 12 = 1,43,760

 Thus, 1,43,760 pages are printed every day.

Now, number of pages printed every day = 1,43,760
∴ Number of pages printed in a month = 1,43,760 × 30 = 43,12,800

Thus, 43,12,800 pages are printed in a month.

Page No 1.21:

Question 6:

A machine, on an average, manufactures 2825 screws a day.  How many screws did it produce in the month of January 2006?

Answer:

∵ Number of screws produced by a machine in a day = 2,825
∴ Number of screws produced by the same machine in the month of January 2006 = 2,825 × 31 = 87,575
    Thus, machine produced 87,575 screws in the month of January 2006.

Page No 1.21:

Question 7:

A famous cricket player has so far scored 6978 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?

Answer:

Runs scored by cricket player in test matches = 6,978

∴ Remaining runs required to complete 10,000 runs = 10,000 − 6,978 = 3,022
 
Thus, the player needs to score 3,022 more runs to complete 10,000 runs.

Page No 1.21:

Question 8:

Ravish has Rs 78,592 with him. He placed an order for purchasing 39 radio sets at Rs 1234 each. How much money will remain with him after the purchase?

Answer:

Ravish's initial money = ₹78,592
He purchased 39 radio sets at ₹1,234 each.
∴ Money spent by him on purchasing 39 radio sets = ₹1,234 × 39 = ₹48,126

∴ Remaining money with Ravish after the purchase = Initial money − Money spent on purchasing 39 radio sets = ₹78,592 − ₹48,126 = ₹30,466

Thus, ₹30,466 is left with him after the purchase.
 

Page No 1.21:

Question 9:

In an election, the successful candidate registered 5,77,570 votes and his nearest rival secured 3,48,685 votes. By what margin did the successful candidate win the election?

Answer:

Margin of victory in the election for the successful candidate = Number of votes registered by the winner − Number of votes secured by nearest rival candidate
 Votes registered by the winner = 5,77,570
 Votes secured by the rival = 3,48,685
 Margin of victory for the successful candidate = 5,77,570 − 3,48,685 = 2,28,885

Page No 1.21:

Question 10:

To stitch a shirt 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stiched and how much cloth will ramain?

Answer:

∵ Total length of available cloth = 40 m = 4,000 cm (1 m = 100 cm)

∵ Length of cloth required to stitch a shirt = 215 cm =  200 + 15 = 215 cm 
∴ The number of shirts that can be stitched from the 40-metre cloth = 4,000/215 = 18.60
As the number of shirts has to be a whole number, we consider the whole part only. That is, 18 such shirts can be stitched.

∴ Cloth required for stitching 18 shirts = 215 × 18 = 3870 cm
∴ Remaining cloth = 4,000 − 3870 = 130 cm = 1.3 m

Page No 1.21:

Question 11:

A vessel has 4 litre and 650 ml of curd. In how many glasses, each of 25 ml capacity, can it be distributed?

Answer:

Number of glasses in which curd can be distributed = Total amount of curd/Capacity of each glass.

Total amount of curd in the vessel =  4,650 mL =  4,000 + 650 = 4,650 mL (1 L = 1,000 mL)
Capacity of each glass = 25 mL
∴ Number of glasses in which curd can be distributed = 4,650/25 = 186

Page No 1.21:

Question 12:

Medicine is packed in boxes, each such box weighing 4 kg 500 g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg.

Answer:

∵ Total capacity of a van carrying boxes of medicines = 800 kg = 8,00,000 g (1 kg = 1,000 g)

∵ Weight of each packed box  = 4,500 g  = 4,000 + 500 = 4,500 g
∴ Total number of boxes that can be loaded in the van = 8,00,000/4,500 = 177.77

The obtained number of boxes is not a whole number.
∴ Weight of 177 boxes = 177 ´ 4,500 = 7,96,500 g (under permissible limit)
∴ Weight of 178 boxes = 178 ´ 4,500 = 8,01,000 g (beyond permissible limit)
Therefore, we can't load 178 boxes; hence, we can say that 177 boxes can be loaded in the van.

Page No 1.21:

Question 13:

The distance between the school and the house of a student is 1 km 875 m. Everyday she walks both ways between her school and home. Find the total distance covered by her in a week.

Answer:

∵ Distance between the school and the house of a student = 1,875 m = 1,000 + 875 = 1,875 m (1 km = 1,000 m)
∵ Distance covered by a student in a day = 2 × 1,875 = 3,750 m
 ∴ Total distance covered by her in a week = 7 × 3,750 = 26,250 m = 26.25 km



Page No 1.26:

Question 1:

Round off each of the following numbers to nearest tens:

(i) 84
(ii) 98
(iii) 984
(iv) 808
(v) 998
(vi) 12,096
(vii) 10,908
(viii) 28,925

Answer:

(i) 80
(ii) 100
(iii) 980
(iv) 810
(v) 1,000
(vi) 12,100
(vii) 10,910
(viii) 28,930

Page No 1.26:

Question 2:

Round off each of the following numbers to nearest hundreds:

(i) 3,985
(ii) 7,289
(iii) 8,074
(iv) 14,627
(v) 28,826
(vi) 4,20,387
(vii) 43,68,973
(viii) 7,42,898

Answer:

(i) 4,000
(ii) 7,300
(iii) 8,100
(iv) 14,600
(v)​ 28,800
(vi) 4,20,400
(vii) 43,69,000

(viii) 7,42,900

Page No 1.26:

Question 3:

Round off each of the following numbers of the nearest thousands:

(i) 2,401
(ii) 9,600
(iii) 4,278
(iv) 7,832
(v) 9,567
(vi) 26,019
(vii) 20,963
(viii) 4,36,952

Answer:

(i) 2,000
(ii) 10,000
(iii) 4,000
(iv) 8,000
(v) 10,000
(vi) 26,000
(vii) 21,000
(viii) 4,37,000

Page No 1.26:

Question 4:

Round off each of the following to the nearest tens, hundreds and thousands:

(i) 964
(ii) 1,049
(iii) 45,634
(iv) 79,085

Answer:

     Tens               Hundreds                  Thousands 
(i) 970                    1,000                     1,000
(ii)1,050                 1,000                     1,000
(iii) 45,630              45,600                   46,000
(iv) 79,090              79,100                   79,000

Page No 1.26:

Question 5:

Round off the following measures to the nearest hundreds:

(i) Rs 666
(ii) Rs 850
(iii) Rs 3,428
(iv) Rs 9,080
(v) 1,265 km
(vi) 417 m
(vii) 550 cm
(viii) 2,486 m
(ix) 360 gm
(x) 940 kg
(xi) 273 l
(xii) 820 mg

Answer:

(i) ₹700
(ii) ₹900
(iii) ₹3,400
(iv) ₹9,100
(v) 1,300 km
(vi) 400 m
(vii) 600 cm
(viii) 2,500 m
(ix) 400 gm
(x) 900 kg
(xi) 300 L
(xii) 800 mg

Page No 1.26:

Question 6:

List all numbers which are rounded off to nearest ten as 370.

Answer:

365, 366, 367, 368, 369, 370, 371, 372, 373, 374

Page No 1.26:

Question 7:

Find the smallest and greatest numbers which are rounded off to the nearest hundreds as 900.

Answer:

Smallest number = 850
Greatest number = 949

Page No 1.26:

Question 8:

Find the smallest and greatest numbers which are rounded off to the nearest thousands as 9000.

Answer:

Smallest number = 8,500

Greatest number = 9,499



Page No 1.29:

Question 1:

Estimate the following by rounding off each factor to nearest hundreds:

(i) 730 + 998
(ii) 796 − 314
(iii) 875 − 384

Answer:

(i) 700 + 1,000 = 1,700
(ii) 800 − 300 = 500
(iii) 900 − 400 = 500

Page No 1.29:

Question 2:

Estimate the following by rounding off each factor to nearest thousands:

(i) 12,904 + 2,888
(ii) 28,292 − 21,496

Answer:

(i) 13,000 + 3,000 = 16,000
(ii) 28,000 − 21,000 = 7,000

Page No 1.29:

Question 3:

Estimate the following by rounding off each number to its greatest place:

(i) 439 + 334 + 4,317
(ii) 8,325 − 491
(iii) 1,08,734 − 47,599
(iv) 898 × 785
(v) 9 × 795
(vi) 87 × 317

Answer:

(i) 400 + 300 + 4,000 = 4,700
(ii) 8,000 − 500 = 7,500
(iii) 1,00,000 − 50,000 = 50,000
(iv) 900 ×
800​ = 7,20,000
(v) 10 × 800​ = 8,000
(vi) 90 × 300​ = 27,000​

Page No 1.29:

Question 4:

Find the estimated quotient for each of the following by rounding off each number of its greatest place:

(i) 878÷28
(ii) 745÷24
(iii) 4489÷394

Answer:

(i) 900 ÷ 30 = 30
(ii) 700 ÷ 20 = 35
(iii) 4,000 ÷ 400 = 10



Page No 1.30:

Question 1:

Write the expression for each of the following statements using brackets:
(i) Four multiplied by the sum of 13 and 7.
(ii) Eight multiplied by the difference of four from nine.
(iii) Divide the difference of twenty eight and seven by 3.
(iv) The sum of 3 and 7 in multiplied by the difference of twelve and eight.

Answer:

(i) 4 × (13 + 7)
(ii) 8 × (9 − 4)
(iii) (28 − 7) ÷ 7
(iv) (3 + 7) × (12 − 8)

Page No 1.30:

Question 2:

Simplify each of the following:
(i) 124 − (12 − 2) × 9
(ii) (13 + 7) × (9 − 4) − 18
(iii) 210 − (14 − 4) × (18 + 2) − 10

Answer:

(i) 124 − (12 − 2) × 9
= 124 − 10 × 9
= 124 − 90
= 34
(ii) (13 + 7) × (9 − 4) − 18
= 20 × 5 − 18
= 100 − 18
= 82
(iii) 210 − (14 − 4) × (18 + 2) − 10
= 210 − 10 × 20 − 10
= 210 − 200 − 10
= 210 − 210
= 0



Page No 1.31:

Question 1:

Simplify each of the following:
7 × 109

Answer:

7 × 109 = 7 × (100 + 9)
= 7 × 100 + 7 × 9
= 700 + 63
= 763

Page No 1.31:

Question 2:

Simplify the following:
6 × 112

Answer:

6 × 112 = 6 × (100 + 12)
= 6 × 100 + 6 × 12
= 600 + 6 × (10 + 2)
= 600 + 6 × 10 + 6 × 2
= 600 + 60 + 12
= 672

Page No 1.31:

Question 3:

Simplify the following:

9 × 105
 

Answer:

9 × 105 = 9 × (100 + 5)
= 9 × 100 + 9 × 5
= 900 + 45
= 945

Page No 1.31:

Question 4:

Simplify the following:
17 × 109

Answer:

17 × 109 = (10 + 7) × 109
= 10 × 109 + 7 × 109
= 10 × (100 + 9) + 7 × (100 + 9)
= 1000 + 90 + 700 + 63
= 1853

Page No 1.31:

Question 5:

Simplify the following:
16 × 108

Answer:

16 × 108 = (10 + 6) × 108
= 10 × 108 + 6 × 108
= 10 × (100 + 8) + 6 × (100 + 8)
= 1000 + 80 + 600 + 48
= 1728

Page No 1.31:

Question 6:

Simplify the following:
12 × 105

Answer:

12 × 105 = (10 + 2) × 105
= 10 × 105 + 2 × 105
= 10 × (100 + 5) + 2 × (100 + 5)
= 1000 + 50 + 200 + 10
= 1260

Page No 1.31:

Question 7:

Simplify the following:
102 × 103

Answer:

102 × 103 = (100 + 2) × 103
= 100 × 103 + 2 × 103
= 100 × (100 + 3) + 2 × (100 + 3)
= 10000 + 300 + 200 + 6
= 10506

Page No 1.31:

Question 8:

Simplify the following:

101 × 105
 
 

Answer:

101 × 105 = (100 + 1) × 105
= 100 × 105 + 1 × 105
= 100 × (100 + 5) + 105
= 10000 + 500 +105
= 10605

Page No 1.31:

Question 9:

Simplify the following:
109 × 107

Answer:

109 × 107 = (100 + 9) × 107
= 100 × 107 + 9 × 107
= 100 × (100 + 7) + 9 × (100 + 7)
= 10000 + 700 + 900 + 63
= 11663



Page No 1.35:

Question 1:

Write the roman-numerals for each of the following:

(i) 33
(ii) 48
(iii) 76
(iv) 95

Answer:

(i) XXXIII
(ii) XLVIII
(iii) LXXVI
(iv) XCV

Page No 1.35:

Question 2:

Write the following in Roman numerals:

(i) 154
(ii) 173
(iii) 248
(iv) 319

Answer:

(i) CLIV
(ii) CLXXIII
(iii) CCXLVIII
(iv) CCCXIX

Page No 1.35:

Question 3:

Write the following in Roman numerals:

(i) 1008
(ii) 2718
(iii) 3906
(iv) 3794

Answer:

(i) MVIII
(ii) MMDCCXVIII
(iii) MMMCMVI
(iv) MMMDCCXCIV



Page No 1.36:

Question 4:

Write the following in Roman numerals:

(i) 4201
(ii) 10009
(iii) 44000
(iv) 25819

Answer:

(i) IVCCI
(ii) XIX
(iii) XLIV
(iv) XXVDCCCXIX

Page No 1.36:

Question 5:

Write the following in Hindu-Arabic numerals:

(i) XXVI
(ii) XXIX
(iii) LXXII
(iv) XCI

Answer:

(i) 10 + 10 + 6 = 26
(ii) 10 + 10 + 9 = 29
(iii) 50 + 10 + 10 + 2 = 72
(iv) (100 − 10) + 1 = 91

Page No 1.36:

Question 6:

Write the corresponding Hindu-Arabic numerals for each of the following:

(i) CIX
(ii) CLXXII
(iii) CCLIV
(iv) CCCXXIX

Answer:

(i) 100 + 9 = 109
(ii) 100 + 50 + 10 + 10 + 2 = 172
(iii) 100 + 100 + 50 + 4 = 254
(iv) 100 +100 +100 + 10 +10 + 9 = 329

Page No 1.36:

Question 7:

Write the corresponding Hindu-Arabic numerals for each of the following:

(i) KXIX
(ii) KDLXV
(iii) KKCXXIII
(iv) KKKDCXL

Answer:

(i) 1,000 +10 + 9 = 1,019
(ii) 1,000 + 500 + 50 +10 + 5 = 1,565

(iii) 1,000 + 1,000 + 100 + 10 + 10 + 3 = 2,123
(iv) 1,000 + 1,000 + 1,000 + 500 + 100 + (50 − 10) = 3,640

Page No 1.36:

Question 8:

Write the following in Hindu-Arabic numerals:

(i) IVCDXLIC
(ii)VICKXLIX
(iii) IXCCCXCI
(iv) LXXIX

Answer:

(i) 4,000 + (500 − 100) + (50 − 10) + 4 = 4,444
(ii) 6,000 + (1,000 − 100)  + (50 − 10) + 9 = 6,949
(iii) 9,000 + 100 + 100 + 100 + (100 − 10) + 1 = 9,391
(iv) 70,000 + 9 = 70,009

Page No 1.36:

Question 9:

Which of the following are meaningless?

(i) III¯CC
(ii) KKKCCXI
(iii) XD
(iv) VC

Answer:

The following numbers are meaningless:
(i) Since bar is not placed above any symbol
(iii) Since X can't be subtracted from D
(iv) Since V is never written to the left of a symbol of greater value

Page No 1.36:

Question 1:

The difference between the place value and face value of 8 in 658742 is
(a) 0
(b) 42
(c) 735
(d) 693

Answer:

The options given are incorrect.
The place value of 8 in 6,58,742 = 8 thousands = 8,000
The face value of 8 = 8
∴ Difference = 8,000 − 8 = 7,992

Page No 1.36:

Question 2:

The difference between the place values of 6 and 3 in 256839 is
(a) 3
(b) 9
(c) 6800
(d) 5930

Answer:

The options given are incorrect.
The place value of 6 in 2,56,839 = 6 thousands = 6,000
The place value of 3 in 2,56,839 = 3 tens = 30
 Difference = 6,000 − 30 = 5,970

Page No 1.36:

Question 3:

The difference of the smallest three digit number and the largest two digit number is
(a) 100
(b) 1
(c) 10
(d) 99

Answer:

(b) 1.

The smallest three-digit number is 100 and the largest two-digit number is 99.
∴ Difference = 100 − 99 = 1

Page No 1.36:

Question 4:

The largest three digit number formed by the digits 8, 5, 9 is
(a) 859
(b) 985
(c) 958
(d) 589

Answer:

(b) 985
The largest number is formed by writing the digits in descending order.

Page No 1.36:

Question 5:

The smallest three digit number having three distinct digits is
(a) 123
(b) 101
(c) 102
(d) 201

Answer:

(c) 102

The three distinct smallest digits are 0, 1 and 2. To form a smallest number using these digits, we need to arrange them in ascending order, but by this we get 012, which results in a two-digit number. So, the smallest three-digit number having distinct digits is 102.

Page No 1.36:

Question 6:

The largest three digit number having distinct digits is
(a) 987
(b) 789
(c) 999
(d) 900

Answer:

(a) 987

The largest three distinct digits are 9, 8 and 7. So, the largest number using these digits can be obtained by arranging the digits in descending order.

Page No 1.36:

Question 7:

The difference between the largest three digit number and the largest three digit number with distinct digits is
(a) 10
(b) 0
(c) 12
(d) 13

Answer:

(c) 12
The largest three-digit number = 999

The largest three-digit number with distinct digits = 987
∴ Difference = 999 − 987 = 12

Page No 1.36:

Question 8:

The product of the place values of two three in 53432 is
(a) 9000
(b) 90000
(c) 10000
(d) 99000

Answer:

(b) 90,000
The 3 in the second place from right is at tens place.
Therefore, the place value of 3 at tens place = 30.
The 3 in the fourth place from right is at thousands place.
Therefore, the place value of 3 at thousands place  = 3,000.
The product of the place values of two 3's = 3,000 × 30 = 90,000.



Page No 1.37:

Question 9:

The smallest counting number is
(a) 0
(b) 1
(c) 10
(d) None of these

Answer:

(b) 1

 The smallest digit is 0, but the smallest counting number is 1.

Page No 1.37:

Question 10:

The total number of 4 digit numbers is
(a) 8999
(b) 9000
(c) 8000
(d) 9999

Answer:

(b) 9,000

The smallest four-digit number is 1,000 and the largest four-digit number is 9,999.
∴ Total number of four-digit numbers = (9,999 − 1,000 ) + 1 = 9,000

Page No 1.37:

Question 11:

The number of 3 digit numbers formed by using 3, 5, 9 taking each digit exactly once, is
(a) 3
(b) 4
(c) 5
(d) 6

Answer:

(d) 6

The numbers are 359, 395, 539, 593, 935, 953.

Page No 1.37:

Question 12:

Total number of numbers which when rounded off to nearest ten give us 200 is
(a) 9
(b) 10
(c) 8
(d) 7

Answer:

(b) 10
The numbers that when rounded off to nearest tens give us 200 are 195, 196, 197, 198, 199, 200, 201, 202, 203, 204.

Page No 1.37:

Question 13:

The smallest number which when rounded off the nearest hundred as 600, is
(a) 550
(b) 595
(c) 604
(d) 599

Answer:

(a) 550
All numbers from 550 to 649 are rounded off to the nearest hundred as 600. Therefore, the smallest number is 550.

Page No 1.37:

Question 14:

The greatest number which when rounded off to the nearest thousand as 7000, is
(a) 6500
(b) 6549
(c) 7499
(d) 6499

Answer:

(c) 7,499

All numbers from 6,500 to 7,499 are rounded off to the nearest thousand as 7,000. Therefore, the greatest number is 7,499.

Page No 1.37:

Question 15:

The difference between the greatest and smallest numbers which when rounded off a number to the nearest tens as 540, is
(a) 10
(b) 9
(c) 8
(d) 10

Answer:

(b) 9
544 is the greatest number that when rounded off to the nearest tens will become 540.
535 is the least number that when rounded off to the nearest tens will become 540.
∴ Difference: 544 − 535 = 9

Page No 1.37:

Question 16:

The difference between the greatest and smallest numbers which when rounded off a number to the nearest hundred as 6700, is
(a) 100
(b) 99
(c) 98
(d) 101

Answer:

(b) 99
6,749 is the greatest number that when rounded off to the nearest hundred will become 6,700.
6,650 is the least number that when rounded off to the nearest hundred will become 6,700.
∴ Difference = 6,749 − 6,650 = 99

Page No 1.37:

Question 17:

The difference between the greatest and the smallest numbers which when rounded off to the nearest thousand as 9000, is
(a) 1000
(b) 990
(c) 999
(d) 900

Answer:

(c) 999
9,499 is the greatest number that when rounded off to the nearest thousand will become 9,000.
8,500 is the smallest number that when rounded off to the nearest thousand will become 9,000.
∴ Difference = 9,499 − 8,500 = 999

Page No 1.37:

Question 18:

Which of the following numbers is equal to 1 billion?
(a) 10 lakh
(b) 1 crore
(c) 100 lakh
(d) 100 crore

Answer:

(d) 100 crore

Page No 1.37:

Question 19:

In the international place vlaue system, we write one million for
(a) 1 lakh
(b) 10 lakh
(c) 100 lakh
(d) 1 crore

Answer:

(b) 10 lakh

Page No 1.37:

Question 20:

Which of the following is not meaningful?
(a) XIV
(b) XXXV
(c) XXV
(d) VX

Answer:

We know that 'V' stands for '5' and 'X' stands for '10' in Roman
So, we can not write 5 before 10.
Hence, the correct answer is option (d)

Page No 1.37:

Question 21:

Which of the following is not meaningful?
(a) XXIII
(b) XII
(c) XVV
(d) XIV

Answer:

We know that 'V' stands for '5' and 'X' stands for '10' in Roman
So, we can not write 5 two times
Hence, the correct answer is option (c)

Page No 1.37:

Question 1:

How many three digit numbers are there in all?
(a) 100
(b) 101
(c) 99
(d) 102

Answer:

There are 10 digits i.e., 0, 1, 2, 3, 4, 5, 6 ,7, 8, 9.
We can not use '0' at the hundred's place.
So, we can use only 9 digits at the hundred's place.
Also, we can use 10 digits at the remaining two places.
So, total numbers of 3-digit numbers = 9 × 10 × 10 = 900
The total number of 3 digit numbers =  900
Disclaimer: none of the given options is correct.

Page No 1.37:

Question 2:

The difference between the successor and predecessor of 99999 is
(a) 1
(b) 0
(c) 2
(d) − 1

Answer:

Successor of 99999 = 99999 + 1 = 100000
Predecessor of 99999 = 99999 − 1 = 99998
Required difference = 100000 − 99998 = 2
Hence, the correct answer is option (c).



Page No 1.38:

Question 3:

How many lakhs are there in one million?
(a) 100
(b) 10
(c) 1000
(d) None of these

Answer:

We know 1 million = 10 lakhs
Hence, the correct answer is option (b).

Page No 1.38:

Question 4:

The difference between the place value and face value of 8 in 357864 is
(a) 808
(b) 800
(c) 792
(d) None of these

Answer:

Place value of 8 = 8 × 100 = 800
Face value of 8 = 8
Required difference = 800 − 8 = 792
Hence, the correct answer is option (c).

Page No 1.38:

Question 5:

Ten millions equal to
(a) 10lakh
(b) 1crore
(c) 10crore
(d) 11akh

Answer:

We know 1 million = 10 lakhs
∴ 10 millions = 10 × 10 lakhs = 1 crore
Hence, the correct answer is option (b).

Page No 1.38:

Question 6:

One trillion equals
(a) 100 billion
(b) 10 billion
(c) 1000 billion
(d) None of these

Answer:

We know 1 trillion = 1000 billions
Hence, the correct answer is option (c).

Page No 1.38:

Question 7:

One billion equals
(a) 1 crore
(b) 10 crore
(c) 100 crore
(d) 1000 crore

Answer:

We know 1 billion = 100 crores
Hence, the correct answer is option (c).

Page No 1.38:

Question 8:

How many times does the digit 9 occur between 1 and 100?
(a) 11
(b) 15
(c) 18
(d) 20

Answer:

In units place, i.e., 9, 19, 29, 39, 49, 59, 69, 79, 89, 99
9 occur 10 times
In ten's place i.e., starting from 90 to 99,
9 occur 10 times
∴Total = 20 times
Hence, the correct answer is option (d).

Page No 1.38:

Question 9:

1827 when rounded off to the nearest hundred is
(a) 1900
(b) 1820
(c) 1850
(d) 1800

Answer:

In 1827 the last two digits i.e., 27 is less than 50.
Hence, the 3rd last digit will remain same and the last two digits become zero
Hence, 1827 when rounded off to the nearest hundred is 1800
Hence, the correct answer is option (d).

Page No 1.38:

Question 10:

2985 is rounded off to the nearest hundred and the nearest tens. The difference between the two values is
(a) 5
(b) 100
(c) 10
(d) 150

Answer:

In 2985 the last two digits i.e., 85 is greater than 50.
Hence, 1 will be added to the 3rd last digit and the last two digits will become zero.
After rounding to nearest hundred we will get 3000.
Again, In 2985 the last digit i.e., 5 is equal to 5.
Hence, 1 will be added to the 2nd last digit and the last digit will become zero.
After rounding to nearest ten we will get 2990.
Required difference = 3000 − 2990 = 10
Hence, the correct answer is option (c).

Page No 1.38:

Question 11:

Write each of the following numerals in Indian system of numeration:
(i) 27943
(ii) 823514
(iii) 7421932
(iv) 53428769

Answer:

(i) 27,943 or Twenty seven thousand nine hundred forty three.
(ii) 8,23,514 or Eight lakh twenty three thousand five hunderd fourteen
(iii) 74,21,932 or Seventy four lakh twenty one thousand nine hundred thirty two
(iv) 5,34,28,769 or Five crore thirty four lakh twenty eight thousand seven hunderd sixty nine

Page No 1.38:

Question 12:

Write each of the following numerals in International system of numeration:
(i) 83254
(ii) 457923
(iii) 2543876
(iv) 700430084

Answer:

(i) Eighty three thousand two hundred fifty four
(ii) Four hundred fifty seven thousand nine hundred twenty three
(iii) Two million five hundred forty three thousand eight hundred seventy six
(iv) Seven hundred million four hundred thirty thousand eighty four

Page No 1.38:

Question 13:

Estimate the sum (4505 + 27807 + 21397) to the nearest thousand.

Answer:

4505 + 27807 + 21397 = 53709
In 53709 the last three digits i.e., 709 is greater than 500.
Hence, 1 will be added to the 4th last digit and the last three digits will become zero.
After rounding to nearest thousand we will get 54000.

Page No 1.38:

Question 14:

Estimate the product 475 × 225 rounding off each number to the nearest hundred.

Answer:

In 475 the last two digits i.e., 75 is greater than 50.
Hence, 1 will be added to the 3rd last digit and the last two digits will become zero.
After rounding to nearest hundred we will get 500.
In 225 the last two digit i.e., 25 is less than 50.
Hence, the 3rd last will remain same and the last digit will become zero.
After rounding to nearest ten we will get 200.
Required product 500 × 200 = 1,00,000

Page No 1.38:

Question 15:

Write each of the following in Hindu-Arabic numeral:
(i) CCXXIV
(ii) CCCLXV
(iii) DCCLXVI

Answer:

(i) 224
(ii) 365
(iii) 766

Page No 1.38:

Question 16:

Write the greatest and the smallest numbers of 4 digits that can be formed using the digits 0, 8, 7, 5 ; using each digit only once.

Answer:

In order to write the largest 4-digit number using digits 0, 5, 7 and 8, we put the largest digit 8 at the place having the highest place value. The smallest digit 0 is put at the right most place i.e. at unit's place, the digit 7 is put at the hundred's place and the digit 5 is put at the ten's place. Hence, the required largest number is 8750.
In order to write the smalest 4-digit number using digits 0, 5, 7 and 8, we put the smallest digit 5 (Except 0) at the place having the highest place value. The largest digit 8 is put at the right most place i.e. at unit's place, the digit 7 is put at the ten's place and the digit 0 is put at the hundred's place. Hence, the required largest number is 5078.

Page No 1.38:

Question 17:

Write all natural numbers between 500 and 600 which do not change if the digits are written in the reverse order.

Answer:

To write the natural numbers between 500 and 600 which do not change if the digits are written in the reverse order we must have the same digit at the hundred's place and unit's place.
Hence, the required numbers are
505, 515, 525, 535, 545. 555, 565, 575, 585, 595.

Page No 1.38:

Question 18:

Write (i) the smallest (ii) the largest 6 digit numbers having four different digits.

Answer:

(i) Four smallest digits are 0, 1,2 and 3. In order to generate the smallest 6-digit number using digits 0, 1, 2 and 3, we write the smallest non-zero digit at the place having the highest place value and the largest digit at the place having least place value. Thus, we put 1 in the left-most place and 3 in the right-most place. Digit 2 is put at ten's place and at all other places we write zero.
Hence, the required number = 100023.

(ii) To write the greatest 6-digit number having four different digits, we will have to use four largest digits. Clearly 9, 8, 7 and 6 are four largest digits. In order to write the largest 6-digit number using digits 6, 7, 8 and 9, we put the largest digit 9 at the place having the highest place value. The smallest digit 8 is put at the hundred's place, the smallest digit 6 is put at the right most place i.e. at unit's place and the digit 7 is put at the ten's place. All other places are filled by 9.
Hence, the required number = 999876.

Page No 1.38:

Question 19:

In two-digit numbers, how many times does the digit 5 occur in
(i) the ten's place (ii) the unit's place?

Answer:

(i) In ten's place i.e., starting from 50 to 59,
5 occur 10 times

(ii) In units place, i.e., 15, 25, 35, 45, 55, 65, 75, 85, 95
5 occur 9 times

Page No 1.38:

Question 20:

Write the greatest 6-digit number formed by three different digits.

Answer:

To write the greatest 6-digit number having three different digits, we will
have to use three largest digits. Clearly 9, 8 and 7 are three largest digits. In
order to write the largest 6-digit number using digits 7, 8 and 9, we put the
largest digit 9 at the place having the highest place value. The smallest digit
7 is put at the right most place i.e. at unit's place and the digit 8 is put at the
ten's place. All other places are filled by 9.
Hence, the required number = 999987.

Page No 1.38:

Question 21:

1 billion = __________million.

Answer:

1 billion = 1000000000
= 1000 × 1000000
= 1000 millions               [∵ 1 million = 1000000]
∴ 1 billion = 1000 millions.

Page No 1.38:

Question 22:

1 billion = ..........crore

Answer:

1 billion = 1000000000
= 100 × 10000000
= 100 crores               [∵ 1 crore = 10000000]
∴ 1 billion = 100 crores.

Page No 1.38:

Question 23:

The smallest four digit number with four different digits is ______________

Answer:

Four smallest digits are 0, 1,2 and 3. In order to write the smalest 4-digit number using digits 0, 1, 2 and 3, we put the smallest digit 1 (Except 0) at the place having the highest place value. The largest digit 3 is put at the right most place i.e. at unit's place, the digit 2 is put at the ten's place and the digit 0 is put at the hundred's place. Hence, the required largest number is 1023.

The smallest four digit number with four different digits is  1023 .

Page No 1.38:

Question 24:

1 crore =              millions.

Answer:

1 crore = 10000000
= 10 × 1000000
= 10 millions               [∵ 1 millions = 1000000]
∴ 1 crore = 10  millions.

Page No 1.38:

Question 25:

The total number of 5 digit numbers =  ..................

Answer:

There are 10 digits i.e., 0, 1, 2, 3, 4, 5, 6 ,7, 8, 9.
We can not use '0' at the place having the highest place value.
So, we can use only 9 digits at the place having the highest place value.
Also, we can use 10 digits at the remaining places.
So, total numbers of four-digit numbers = 9 × 10 × 10 × 10 × 10 = 90000
The total number of 5 digit numbers =  90000



Page No 1.7:

Question 1:

Write each of the following in numeral form:
(i) Eight thousand twelve.
(ii) Seventy thousand fifty three.
(iii) Five lakh seven thousand four hundred six.
(iv) Six lakh two thousand nine.
(v) Thirty lakh eleven thousand one.
(vi) Eight crore four lakh twenty five.
(vii) Three crore three lakh three thousand three hundred three.
(viii) Seventeen crore sixty lakh thirty thousand fifty seven.

Answer:

(i) 8,012
(ii) 70,053
(iii) 5,07,406
(iv) 6,02,009
(v) 30,11,001
(vi) 8,04,00,025
(vii) 3,03,03,303
(viii) 17,60,30,057

Page No 1.7:

Question 2:

Write the following numbers in words in the Indian system of numeration:
(i) 42,007
(ii) 4,05,045
(iii) 35,42,012
(iv) 7,06,04,014
(v) 25,05,05,500
(vi) 5,50,50,050
(vii) 5,03,04,012

Answer:

(i) Forty two thousand seven.
(ii) Four lakh five thousand forty five.
(iii) Thirty five lakh forty two thousand twelve.
(iv) Seven crore six lakh four thousand fourteen.
(v) Twenty five crore five lakh five thousand five hundred.
(vi) Five crore fifty lakh fifty thousand fifty.
(vii) Five crore three lakh four thousand twelve.



Page No 1.8:

Question 3:

Insert commas in the correct positions to separate periods and write the following numvers in words:
(i) 4375
(ii) 24798
(iii) 857367
(iv) 9050784
(v) 10105607
(vi) 10000007
(vii) 910107104

Answer:

(i) 4,357
(ii) 24,798
(iii) 8,57,367
(iv) 90,50,784
(v) 1,01,05,607
(vi) 1,00,00,007
(vii) 91,01,07,104

Page No 1.8:

Question 4:

Write each of the following expanded notation:

(i) 3057
(ii) 12345
(iii) 10205
(iv) 235060
 

Answer:

(i) 3000 + 50 + 7

(ii) 10000 + 2000 + 300 + 40 + 5

(iii) 10000 + 200 + 5


(iv) 200000 + 30000 + 5000 + 60

 

Page No 1.8:

Question 5:

Write the corresponding numeral for each of the following:
(i) 7 × 10000 + 2 × 1000 + 5 × 100 + 9 × 10 + 6 × 1
(ii) 4 × 100000 + 5 × 1000 + 1 × 100 + 7 × 1
(iii) 8 × 1000000 + 3 × 1000 + 6 × 1
(iv) 5 × 10000000 + 7 × 1000000 + 8 × 1000 + 9 × 10 + 4

Answer:

(i) 70000 + 2000 + 500 + 90 + 6 = 72,596
(ii) 400000 + 5000 + 100 + 7 = 4,05,107
(iii) 8000000 + 3000 + 6 = 80,03,006
(iv) 50000000 + 7000000 + 8000 + 90 + 4 = 5,70,08,094

Page No 1.8:

Question 6:

Find the place value of the digit 4 in each of the following:
(i) 74983160
(ii) 8745836

Answer:

(i) Place value of 4 = 4 × 10,00,000 = 40,00,000

(ii) Place value of 4 = 4 × 10,000 = 40,000

Page No 1.8:

Question 7:

Determine the product of the place values of two fives in 450758.

Answer:

Place value of first 5 = 5 × 10 = 50
Place value of second 5 = 5 × 10,000 = 50,000
Required product = 50 × 50,000 = 25,00,000

Page No 1.8:

Question 8:

Determine the difference of the place values of two 7's in 257839705.

Answer:

Place value of first 7 = 7 × 100 = 700
Place value of second 7 = 7 × 10,00,000 = 70,00,000
Required difference = 70,00,000 − 700 = 69,99,300

Page No 1.8:

Question 9:

Determine the difference between the place value and the face value of 5 in 78654321.

Answer:

The number = 7,86,54,321
The place value of 5 = 5 ten thousands = 50,000
The face value of 5 = 5

∴ Difference = 50,000 − 5 = 49,995

Page No 1.8:

Question 10:

Which digits have the same face value and place value in 92078634?

Answer:

The place value of a digit depends on the place where it occurs, while the face value is the value of the digit itself.
In a number, the digit that have same face value and place value are the ones digit and all the zeroes of the number.

Therefore, in 9,20,78,634,  4 (the ones digit) and 0 (the lakhs digit) have the same face value and place value.

Page No 1.8:

Question 11:

How many different 3-digit numbers can be formed by using the digits 0,2,5 without repeating any digit in the number?

Answer:

The three-digit numbers formed using the digits 0, 2 and 5 (without repeating any digit in the number) are 250, 205, 502 and 520.
Therefore, four such numbers can be formed.

Page No 1.8:

Question 12:

Write all possible 3-digit numbers using the digits 6,0,4 when

(i) repetition of digits is not allowed
(ii) repetition of digits is allowed

Answer:

(i) 604, 640, 460, 406
(ii) 666, 664, 646, 660, 606, 600, 644, 640, 604, 444, 466, 440, 446,464, 400, 404, 406, 460

Page No 1.8:

Question 13:

Fill in the blank:
(i) 1 lakh = ... ten thousand
(ii) 1 lakh = ... thousand
(iii) 1 lakh = ... hundred
(iv) 1 lakh = ... ten
(v) 1 crore = ... ten lakh
(vi) 1 crore = ... lakh
(vii) 1 crore = ... ten thousand
(viii) 1 crore = ... thousand
(ix) 1 crore = ... hundred
(x) 1 crore = ... ten

Answer:

(i) 1 lakh = 10 ten thousand
(ii) 1 lakh = 100 thousand
(iii) 1 lakh = 1000 hundred
(iv) 1 lakh = 10000 ten
(v) 1 crore = 10 ten lakh
(vi) 1 crore = 100 lakh
(vii) 1 crore = 1000 ten thousand
(viii) 1 crore = 10000 thousand
(ix) 1 crore = 100000 hundred
(x) 1 crore = 1000000 ten



View NCERT Solutions for all chapters of Class 6