The subject experts present the concepts in a clear and precise manner based on the IQ level of students. In previous years, students have learned about basic addition, subtraction, multiplication and division. By solving this Chapter, the students will understand the sequence of numbers and interesting problems based on it.
Here, the students will learn two methods of expressing numbers in words and digits, i.e., the Indian System and International System. The solutions PDF helps students to solve exercise-wise problems on a daily basis. RD Sharma Solutions for Class 6 Chapter 1 Knowing Our Numbers are provided here.
RD Sharma Solutions for Class 6 Maths Chapter 1: Knowing Our Numbers
Access answers to Maths RD Sharma Solutions for Class 6 Chapter 1: Knowing Our Numbers
Exercise 1.1 page: 1.7
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.
Solution:
(i) The numeral form of eight thousand twelve is 8,012.
(ii) The numeral form of seventy thousand fifty three is 70,053.
(iii) The numeral form of five lakh seven thousand four hundred six is 5, 07, 406.
(iv) The numeral form of six lakh two thousand nine is 6, 02, 009.
(v) The numeral form of thirty lakh eleven thousand one is 30, 11, 001.
(vi) The numeral form of eight crore four lakh twenty five is 8, 04, 00, 025.
(vii) The numeral form of three crore three lakh three thousand three hundred three is 3, 03, 03, 303.
(viii) The numeral form of seventeen crore sixty lakh thirty thousand fifty seven is 17, 60, 30, 057.
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
Solution:
(i) 42,007 is written as forty two thousand seven in the Indian system of numeration.
(ii) 4,05,045 is written as four lakh five thousand forty five in the Indian system of numeration.
(iii) 35, 42, 012 is written as thirty five lakh forty two thousand twelve in the Indian system of numeration.
(iv) 7, 06, 04, 014 is written as seven crore six lakh four thousand fourteen in the Indian system of numeration.
(v) 25, 05, 05,500 is written as twenty five crore five lakh five thousand five hundred in the Indian system of numeration.
(vi) 5, 50, 50, 050 is written as five crore fifty lakh fifty thousand fifty in the Indian system of numeration.
(vii) 5, 03, 04, 012 is written as five crore three lakh four thousand twelve in the Indian system of numeration.
3. Insert commas in the correct positions to separate periods and write the following numbers in words:
(i) 4375
(ii) 24798
(iii) 857367
(iv) 9050784
(v) 10105607
(vi) 10000007
(vii) 910107104
Solution:
(i) 4375 by inserting commas is written as 4, 375.
(ii) 24798 by inserting commas is written as 24, 798.
(iii) 857367 by inserting commas is written as 8,57,367.
(iv) 9050784 by inserting commas is written as 90,50,784.
(v) 10105607 by inserting commas is written as 1,01,05,607.
(vi) 10000007 by inserting commas is written as 1,00,00,007.
(vii) 910107104 by inserting commas is written as 91,01,07,104.
4. Write each of the following in expanded notation:
(i) 3057
(ii) 12345
(iii) 10205
(iv) 235060
Solution:
(i) 3057 = 3 × 1000 + 0 × 100 + 5 × 10 + 7 × 1
(ii) 12345 = 1 × 10000 + 2 × 1000 + 3 × 100 + 4 × 10 + 5 × 1
(iii) 10205 = 1 × 10000 + 0 × 1000 + 2 × 100 + 0 × 10 + 5 × 1
(iv) 235060 = 2 × 100000 + 3 × 10000 + 5 × 1000 + 0 × 100 + 6 × 10 + 0 × 1
5. Write the corresponding numeral for each of the following:
(i) 7 × 1000 + 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
Solution:
(i) 7 × 1000 + 2 × 1000 + 5 × 100 + 9 × 10 + 6 × 1 = 72, 596.
(ii) 4 × 100000 + 5 × 1000 + 1 × 100 + 7 × 1 = 4, 05, 107.
(iii) 8 × 1000000 + 3 × 1000 + 6 × 1 = 80, 03, 006.
(iv) 5 × 10000000 + 7 × 1000000 + 8 × 1000 + 9 × 10 + 4 = 5, 70, 08, 094.
6. Find the place value of the digit 4 in each of the following:
(i) 74983160
(ii) 8745836
Solution:
(i) The place value of digit 4 in 74983160 is 4 × 10, 00, 000 = 40, 00, 000.
(ii) The place value of digit 4 in 8745836 is 4 × 10, 000 = 40, 000.
7. Determine the product of the place values of two fives in 450758.
Solution:
The place value of first five is 5 × 10 = 50
The place value of second five is 5 × 10, 000 = 50, 000
So the required product = 50 × 50, 000 = 25, 00, 000.
Therefore, the product of the place values of two fives in 450758 is 25, 00, 000.
8. Determine the difference of the place values of two 7’s in 257839705.
Solution:
The place value of first seven is 7 × 100 = 700
The place value of second seven is 7 × 10, 00, 000 = 70, 00, 000
So the required difference = 70, 00, 000 – 700 = 69, 99, 300
Therefore, the difference of the place values of two 7’s in 257839705 is 69, 99, 300.
9. Determine the difference between the place value and the face value of 5 in 78654321.
Solution:
The given number is 78654321
The place value of five = 5 × 10, 000 = 50, 000
The face value of five = 5
So the required difference = 50, 000 – 5 = 49, 995
Therefore, the difference between the place value and the face value of 5 in 78654321 is 49, 995.
10. Which digits have the same face value and place value in 92078634?
Solution:
We know that the face value depends on the value of the digit and the place value depends on its place of occurrence.
So the digits which have same place and face value in a number are the ones digit and zeros of the number.
The given number is 9, 20, 78, 634
We know that 4 which is the ones digit and 0 which is the lakhs digit have same face and place value.
11. How many different 3-digit numbers can be formed by using the digits 0, 2, 5 without repeating any digit in the number?
Solution:
The different 3-digit numbers which can be formed by using the digits 0, 2, 5 without repeating any digit in the number are 205, 250, 502 and 520.
Therefore, four 3 digit numbers can be formed by using the digits 0, 2, 5.
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.
Solution:
(i) The possible 3 digit numbers using the digits 6, 0, 4 when repetition of digits is not allowed are 604, 640, 406 and 460.
(ii) The possible 3 digit numbers using the digits 6, 0, 4 when repetition of digits is allowed are 400, 406, 460, 466, 444, 404, 440, 446, 464, 600, 604, 640, 644, 646, 664, 606, 660 and 666.
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
Solution:
(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
Exercise 1.2 page: 1.13
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.
Solution:
(i) Seven million three hundred three thousand two hundred six is written as 7, 303, 206 using international place value chart.
It can be written in expanded form as
7, 303, 206 = 7 × 1000000 + 3 × 100000 + 0 × 10000 + 3 × 1000 + 2 × 100 + 0 × 10 + 6 × 1
(ii) Fifty five million twenty nine thousand seven is written as 55, 029, 007 using international place value chart.
It can be written in expanded form as
55, 029, 007 = 5 × 10000000 + 5 × 1000000 + 0 × 100000 + 2 × 10000 + 9 × 1000 + 0 × 100 + 0 × 10 + 7 × 1
(iii) Six billion one hundred ten million three thousand seven is written as 6, 110, 003, 007 using international place value chart.
It can be written in expanded form as
6, 110, 003, 007 = 6 × 1000000000 + 1 × 100000000 + 1 × 10000000 + 0 × 1000000 + 0 × 100000 + 0 × 10000 + 3 × 1000 + 0 × 100 + 0 × 10 + 7 × 1
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
Solution:
(i) 513625 is written as 513, 625 which is five hundred thirteen thousand six hundred twenty five in the international system of numeration.
(ii) 4035672 is written as 4, 035, 672 which is four million thirty five thousand six hundred seventy two in the international system of numeration.
(iii) 65954923 is written as 65, 954, 923 which is sixty five million nine hundred fifty four thousand nine hundred twenty three in the international system of numeration.
(iv) 70902005 is written as 70, 902, 005 which is seventy million nine hundred two thousand five in the international system of numeration.
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.
Solution:
(i) Forty three lakh four thousand eighty four can be written as 4, 304, 084 which is four million three hundred four thousand eighty four using International system of numeration.
(ii) Six crore thirty four lakh four thousand forty four can be written as 63, 404, 044 which is sixty three million four hundred four thousand forty four using International system of numeration.
(iii) Seven lakh thirty five thousand eight hundred ninety nine only can be written as 735, 899 which is seven hundred thirty five thousand eight hundred ninety nine only using International system of numeration.
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.
Solution:
(i) Six million five hundred forty three thousand two hundred ten can be written as 65, 43, 210 which is sixty five lakh forty three thousand two hundred ten using the Indian system of numeration.
(ii) Seventy six million eighty five thousand nine hundred eighty seven can be written as 7, 60, 85, 987 which is seven crore sixty lakh eight five thousand nine hundred eighty seven using the Indian system of numeration.
(iii) Three hundred twenty five million four hundred seventy nine thousand eight hundred thirty eight can be written as 32, 54, 79, 838 which is thirty two crore fifty four lakh seventy nine thousand eight hundred thirty eight using the Indian system of numeration.
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.
Solution:
It is given that a nine digit number has only ones in ones period, only twos in the thousands period and only threes in millions period.
So we get the number as 333, 222, 111
We can write it as 33, 32, 22, 111 which is thirty three crore thirty two lakh twenty two thousand one hundred and eleven using the Indian system.
6. How many thousands make a million?
Solution:
We know that
One thousand = 1000 and One million = 1, 000, 000
Number of thousands which make a million = One million/ One thousand
By substituting the values
Number of thousands which make a million = 1, 000, 000/ 1000
So we get
Number of thousands which make a million = 1000
Therefore, 1000 thousands make a million.
7. How many millions make a billion?
Solution:
We know that
One million = 1, 000, 000 and One billion = 1, 000, 000, 000
Number of millions which make a billion = One billion/ One million
By substituting the values
Number of millions which make a billion = 1, 000, 000, 000/ 1, 000, 000
So we get
Number of millions which make a billion = 1000
Therefore, 1000 millions make a billion.
8. (i) How many lakhs make a million?
(ii) How many lakhs make a billion?
Solution:
(i) We know that
One lakh = 1, 00, 000 and One million = 1, 000, 000
Number of lakhs which make a million = One million/ One lakh
By substituting the values
Number of lakhs which make a million = 1, 000, 000/ 1, 00, 000
So we get
Number of lakhs which make a million = 10
Therefore, 10 lakhs make a million.
(ii) We know that
One lakh = 1, 00, 000 and One billion = 1, 000, 000, 000
Number of lakhs which make a billion = One billion/ One lakh
By substituting the values
Number of lakhs which make a billion = 1, 000, 000, 000/ 1, 00, 000
So we get
Number of lakhs which make a billion = 10, 000
Therefore, 10, 000 lakhs make a billion.
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.
Solution:
(i) Eight million seven hundred eight thousand four can be written as
8 × 1, 000, 000 + 7 × 100, 000 + 0 × 10, 000 + 8 × 1000 + 0 × 100 + 0 × 10 + 4 × 1 = 8, 708, 004
(ii) Six hundred seven million twelve thousand eighty four can be written as
6 × 100, 000, 000 + 0 × 10, 000, 000 + 7 × 1, 000, 000 + 0 × 100, 000 + 1 × 10, 000 + 2 × 1000 + 0 × 100 + 8 × 10 + 4 × 1 = 607, 012, 084
(iii) Four billion twenty five million forty five thousand can be written as
4 × 1, 000, 000, 000 + 0 × 100, 000, 000 + 2 × 10, 000, 000 + 5 × 1, 000, 000 + 0 × 100, 000 + 4 × 10, 000 + 5 × 1000 + 0 × 100 + 0 × 10 + 0 × 1 = 4, 025, 045, 000
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
Solution:
(i) 435, 002 can be written as four hundred thirty five thousand and two in international system of numeration.
(ii) 1, 047, 509 can be written as one million forty seven thousand five hundred and nine in international system of numeration.
(iii) 59, 064, 523 can be written as fifty nine million sixty four thousand five hundred and twenty three in international system of numeration.
(iv) 25, 201, 905 can be written as twenty five million two hundred one thousand nine hundred and five in international system of numeration.
Exercise 1.3 page: 1.16
1. How many four digit numbers are there in all?
Solution:
We know that the 10 digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9
0 cannot be used in thousands place so only nine digits can be used.
10 digits can be used in hundreds, tens and units place
The number of four digit numbers = 9 × 10 × 10 × 10 = 9000
Therefore, 9000 four digit numbers are there in all.
2. Write the smallest and the largest six digit numbers. How many numbers are between these two.
Solution:
We know that the smallest digit is 0 which cannot be used in the highest place value.
So 1 which is the second smallest digit can be used in the highest place value
The required smallest six digit number is 100000
We know that the largest digit is 9 which can be used in any place
The required largest six digit number is 999999
So we get the difference = 999999 – 100000 = 899999
Therefore, the smallest six digit number is 100000, the largest six digit number is 999999 and 899999 numbers are between these two numbers.
3. How many 8-digit numbers are there in all?
Solution:
We know that the 10 digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9
0 cannot be used in the highest place value and 9 can be used in the highest place value
So the 10 digits can be used in the remaining places of 8 digit numbers
The total number of 8 digit numbers = 9 × 10 × 10 × 10 × 10 × 10 × 10 × 10 = 90000000
Therefore, 90000000 eight digit numbers are there in all.
4. Write 10075302 in words and rearrange the digits to get the smallest and the largest numbers.
Solution:
The given number 10075302 can be written as one crore seventy five thousand three hundred and two.
To get smallest 8 digit number using 0, 1, 2, 3, 5 and 7
We use 1 which is the smallest digit in the highest place and largest digit 7 at the units place
Further we put 5 in the tens place, 3 in the hundreds place and 2 in thousands place
So the required smallest number is 10002357
To get largest 8 digit number using 0, 1, 2, 3, 5 and 7
We use 7 which is the largest digit in the highest place value, 5 in a place after highest place, 3 as the next one, 2 as the smallest digit and then 1.
So the required largest number is 75321000.
5. What is the smallest 3-digit number with unique digits?
Solution:
102 is the smallest 3-digit number with unique digits.
6. What is the largest 5-digit number with unique digits?
Solution:
98765 is the largest 5-digit number with unique digits.
7. Write the smallest 3-digit number which does not change if the digits are in reverse order.
Solution:
101 is the smallest 3-digit number which does not change if the digits are in reverse order.
8. Find the difference between the number 279 and that obtained on reversing its digits.
Solution:
The reverse of 279 is 972
Difference between both the numbers = 972 – 279 = 693
Therefore, the difference between the number 279 and that obtained on reversing its digits is 693.
9. Form the largest and smallest 4-digit numbers using each of digits 7, 1, 0, 5 only once.
Solution:
The largest 4 digit number = 7510
Smallest 4 digit number = 1057
Therefore, the largest and smallest 4-digit numbers using each of digits 7, 1, 0, 5 only once is 7510 and 1057.
Exercise 1.4 page: 1.18
1. Put the appropriate symbol (<, >) in each of the following boxes:
(i) 102394 ☐ 99887
(ii) 2507324 ☐ 2517324
(iii) 3572014 ☐ 10253104
(iv) 47983505 ☐ 47894012
Solution:
(i) 102394 > 99887
(ii) 2507324 < 2517324
(iii) 3572014 < 10253104
(iv) 47983505 > 47894012
2. Arrange the following numbers in ascending order:
(i) 102345694, 8354208, 6539542, 63547201, 12345678
(ii) 1808090, 1808088, 181888, 190909, 16060666.
Solution:
(i) It can be written as
6539542 < 8354208 < 12345678 < 63547201 < 102345694
(ii) It can be written as
181888 < 190909 < 1808088 < 1808090 < 16060666.
3. Arrange the following numbers in descending order:
(i) 56943300, 56943201, 5695440, 56944000, 5694437
(ii) 1020216, 1020308, 1021430, 893245, 893425.
Solution:
(i) It can be written as
56944000 > 56943300 > 56943201 > 5695440 > 5694437
(ii) It can be written as
1021430 > 1020308 > 1020216 > 893425 > 893245.
Exercise 1.5 page: 1.21
1. How many milligrams make one kilogram?
Solution:
One million or ten lakh milligrams make one kilogram.
2. A box of medicine tablets contains 2, 00, 000 tablets each weighing 20mg. What is the total weight of all the tablets in the box in grams? In kilograms?
Solution:
It is given that
Weight of each tablet = 20mg
So we get
Weight of 2, 00, 000 tablets = 2, 00, 000 × 20
We get the weight of 2, 00, 000 tablets = 40, 00, 000mg
We know that
Total weight of tablets in the box = 40, 00, 000mg
It can be written as
1g = 1000mg
So the weight of box having tablets = 40, 00, 000/ 1000 = 4000g
We know that 1kg = 1000g
So the weight of box having tablets = 4000/ 1000 = 4kg
Therefore, the total weight of all the tablets in the box is 4000g or 4kg.
3. Population of Sundarnagar 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 in 2001?
Solution:
It is given that
Population of Sundarnagar in 1991 = 2, 35, 471
Increase in population in 2001 = 72, 958
We know that
Population of the city in 2001 = Population of Sundarnagar in 1991 + Increase in population in 2001
By substituting the values
Population of the city in 2001 = 2, 35, 471 + 72, 958
By addition
Population of the city in 2001 = 3, 08, 429
Therefore, the population of the city in 2001 is 3, 08, 429.
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 were respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.
Solution:
It is given that
Number of tickets sold at the counter on the first = 1094
Number of tickets sold at the counter on the second = 1812
Number of tickets sold at the counter on the third = 2050
Number of tickets sold at the counter on the final day = 2751
We know that
Total number of tickets sold = 1094 + 1812 + 2050 + 2751 = 7707
Therefore, the total number of tickets sold on all the four days is 7, 707.
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 month?
Solution:
It is given that
1 copy of newspaper contains = 12 pages
So the number of pages in 11, 980 copies = 12 × 11, 980 = 1, 43, 760
Hence, 1, 43, 760 pages are printed everyday
We know that
Number of pages printed in a month = 1, 43, 760 × 30 = 43, 12, 800
Therefore, 1, 43, 760 pages are printed every day and 43, 12, 800 pages are printed every month.
6. A machine, on an average, manufactures 2825 screws a day. How many screws did it produce in the month of January 2006?
Solution:
It is given that
Number of screws produced per day = 2, 825
So the number of screws produced in the month of January = 2, 825 × 31 = 87, 575
Therefore, the machine produced 87, 575 screws in the month of January 2006.
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?
Solution:
It is given that
Runs scored by a famous cricket player = 6978
So the runs required to complete 10, 000 runs = 10, 000 – 6978 = 3022
Therefore, 3, 022 runs are needed by the cricket player to complete 10, 000 runs.
8. Ravish has ₹ 78, 592 with him. He placed an order for purchasing 39 radio sets at ₹ 1234 each. How much money will remain with him after the purchase?
Solution:
It is given that
Amount with Ravish = ₹ 78, 592
Cost of 1 radio set = ₹ 1234
Number of radio sets purchased = 39
We know that
Amount spent to purchase 39 radio sets = ₹ (1234 × 39) = ₹ 48126
Amount remaining with Ravish = Amount with Ravish – Amount spent to purchase 39 radio sets
By substituting the values
Amount remaining with Ravish = ₹ 78, 592 – ₹ 48126 = ₹ 30466
Therefore, ₹ 30466 money will remain with Ravish after the purchase.
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?
Solution:
It is given that
Number of votes registered by successful candidate = 5, 77, 570
Number of votes secured by nearest rival = 3, 48, 685
We know that
Margin of victory the successful candidate obtain to win the election = Number of votes registered by successful candidate – Number of votes secured by nearest rival
By substituting the values
Margin of victory the successful candidate obtain to win the election = 5, 77, 570 – 3, 48, 685 = 2, 28, 885
Therefore, the margin of victory for the successful candidate is 2, 28, 885.
10. To stitch a shirt 2m 15cm cloth is needed. Out of 40m cloth, how many shirts can be stitched and how much cloth will remain?
Solution:
It is given that
Length of cloth = 40m
Length of cloth required to stitch a shirt = 2m 15cm = 200 + 15 = 215cm
Number of shirts that can be stitched using 40m cloth = 4000/ 215 = 18.60
We know that the number of shirts should be a whole number, 18 shirts can be stitched.
Length of cloth required to stitch 18 shirts = 215 × 18 = 3870cm
So the remaining cloth = 4000 – 3870 = 130cm = 1.3m = 1m 30cm
Therefore, 18 shirts can be stitched and 1m 30cm cloth will remain.
11. A vessel has 4 litre and 650ml of curd. In how many glasses, each of 25ml capacity, can it be distributed?
Solution:
It is given that
Total amount of curd = 4 litre 650ml
We know that 1 litre = 1000ml
Total amount of curd = 4650ml
Capacity of each glass = 25ml
So we get
Number of glasses of curd which can be distributed = Total amount of curd/ Capacity of each glass
By substituting the values
Number of glasses of curd which can be distributed = 4650/ 25 = 186
Therefore, 186 glasses each of 25ml capacity of curd can be distributed.
12. Medicine is packed in boxes, each such box weighing 4kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800kg?
Solution:
It is given that
Capacity of van carrying medicine boxes = 800kg
We know that 1kg = 1000g
Capacity of van carrying medicine boxes = 8, 00, 000g
Weight of each box = 4kg 500g = 4500g
We know that
Total number of boxes that can be loaded in a van = Capacity of van carrying medicine boxes/ Weight of each box
By substituting the values
Total number of boxes that can be loaded in a van = 8, 00, 000/ 4500 = 177.77
Number of boxes should be a whole number
We get
Weight of 177 boxes = 177 × 4500 = 7, 96, 500g which is under the permissible limit
Weight of 178 boxes = 178 × 4500 = 8, 01, 000g which is beyond the permissible limit
Therefore, 177 boxes can be loaded in a van which cannot carry beyond 800kg.
13. The distance between the school and the house of a student is 1km 875m. Everyday she walks both ways between her school and home. Find the total distance covered by her in a week.
Solution:
It is given that
Distance between school and house = 1km 875m = 1875m
We know that
Distance covered by a student per day = 2 × 1875 = 3750m
So we get
Total distance covered by the student in a week = 7 × 3750 = 26250m = 26km 250m
Therefore, the total distance covered by her in a week is 26km 250m.
Exercise 1.6 page: 1.26
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
Solution:
(i) 84 is rounded off to the nearest tens as 80.
(ii) 98 is rounded off to the nearest tens as 100.
(iii) 984 is rounded off to the nearest tens as 980.
(iv) 808 is rounded off to the nearest tens as 810.
(v) 998 is rounded off to the nearest tens as 1000.
(vi) 12, 096 is rounded off to the nearest tens as 12, 100.
(vii) 10, 908 is rounded off to the nearest tens as 10, 910.
(viii) 28, 925 is rounded off to the nearest tens as 28, 930.
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
Solution:
(i) 3, 985 is rounded off to the nearest hundreds as 4, 000.
(ii) 7, 289 is rounded off to the nearest hundreds as 7, 300.
(iii) 8, 074 is rounded off to the nearest hundreds as 8, 100.
(iv) 14, 627 is rounded off to the nearest hundreds as 14, 600.
(v) 28, 826 is rounded off to the nearest hundreds as 28, 800.
(vi) 4, 20, 387 is rounded off to the nearest hundreds as 4, 20, 400.
(vii) 43, 68, 973 is rounded off to the nearest hundreds as 43, 69, 000.
(viii) 7, 42, 898 is rounded off to the nearest hundreds as 7, 42, 900.
3. Round off each of the following numbers to 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
Solution:
(i) 2, 401 is rounded off to the nearest thousands as 2, 000.
(ii) 9, 600 is rounded off to the nearest thousands as 10, 000.
(iii) 4, 278 is rounded off to the nearest thousands as 4, 000.
(iv) 7, 832 is rounded off to the nearest thousands as 8, 000.
(v) 9, 567 is rounded off to the nearest thousands as 10, 000.
(vi) 26, 019 is rounded off to the nearest thousands as 26, 000.
(vii) 20, 963 is rounded off to the nearest thousands as 21, 000.
(viii) 4, 36, 952 is rounded off to the nearest thousands as 4, 37, 000.
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
Solution:
(i) 964 is rounded off to the nearest tens as 960.
964 is rounded off to the nearest hundreds as 1, 000.
964 is rounded off to the nearest thousands as 1, 000.
(ii) 1, 049 is rounded off to the nearest tens as 1, 050.
1, 049 is rounded off to the nearest hundreds as 1, 000.
1, 049 is rounded off to the nearest thousands as 1, 000.
(iii) 45, 634 is rounded off to the nearest tens as 45, 630.
45, 634 is rounded off to the nearest hundreds as 45, 600.
45, 634 is rounded off to the nearest thousands as 46, 000.
(iv) 79, 085 is rounded off to the nearest tens as 79, 090.
79, 085 is rounded off to the nearest hundreds as 79, 100.
79, 085 is rounded off to the nearest thousands as 79, 000.
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) 417m
(vii) 550cm
(viii) 2, 486m
(ix) 360gm
(x) 940kg
(xi) 273l
(xii) 820mg
Solution:
(i) Rs 666 is rounded off to the nearest hundreds as Rs 700.
(ii) Rs 850 is rounded off to the nearest hundreds as Rs 900.
(iii) Rs 3, 428 is rounded off to the nearest hundreds as Rs 3, 400.
(iv) Rs 9, 080 is rounded off to the nearest hundreds as Rs 9, 100.
(v) 1, 265 km is rounded off to the nearest hundreds as 1, 300km.
(vi) 417m is rounded off to the nearest hundreds as 400m.
(vii) 550cm is rounded off to the nearest hundreds as 600cm.
(viii) 2, 486m is rounded off to the nearest hundreds as 2, 500m.
(ix) 360gm is rounded off to the nearest hundreds as 400gm.
(x) 940kg is rounded off to the nearest hundreds as 900kg.
(xi) 273l is rounded off to the nearest hundreds as 300l.
(xii) 820mg is rounded off to the nearest hundreds as 800mg.
6. List all numbers which are rounded off to nearest ten as 370.
Solution:
The list of numbers which are rounded off to nearest ten as 370 are
365, 366, 367, 368, 369, 370, 371, 372, 373, 374.
7. Find the smallest and greatest numbers which are rounded off to the nearest hundreds as 900.
Solution:
The smallest number which is rounded off to the nearest hundreds as 900 is 850.
The greatest number which is rounded off to the nearest hundreds as 900 is 949.
8. Find the smallest and greatest numbers which are rounded off to the nearest thousands as 9000.
Solution:
The smallest number which is rounded off to the nearest thousands as 9000 is 8, 500.
The greatest number which is rounded off to the nearest thousands as 9000 is 9, 499.
Exercise 1.7 page: 1.29
1. Estimate the following by rounding off each factor to nearest hundreds:
(i) 730 + 998
(ii) 796 – 314
(iii) 875 – 384
Solution:
(i) It can be rounded off to the nearest hundreds as 700 + 1000 = 1, 700.
(ii) It can be rounded off to the nearest hundreds as 800 – 300 = 500.
(iii) It can be rounded off to the nearest hundreds as 900 – 400 = 500.
2. Estimate the following by rounding off each factor to nearest thousands:
(i) 12, 094 + 2, 888
(ii) 28, 292 – 21, 496
Solution:
(i) It can be rounded off to the nearest thousands as 13, 000 + 3, 000 = 16, 000.
(ii) It can be rounded off to the nearest thousands as 28, 000 – 21, 000 = 7, 000.
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
Solution:
(i) It can be rounded off to its greatest place as 400 + 300 + 4, 000 = 4, 700.
(ii) It can be rounded off to its greatest place as 8, 000 – 500 = 7, 500.
(iii) It can be rounded off to its greatest place as 1, 00, 000 – 50, 000 = 50, 000.
(iv) It can be rounded off to its greatest place as 900 × 800 = 7, 20, 000.
(v) It can be rounded off to its greatest place as 10 × 800 = 8, 000.
(vi) It can be rounded off to its greatest place as 90 × 300 = 27, 000.
4. Find the estimated quotient for each of the following by rounding off each number to its greatest place:
(i) 878 ÷ 28
(ii) 745 ÷ 24
(iii) 4489 ÷ 394
Solution:
(i) It can be rounded off to its greatest place as 900 ÷ 30 = 30.
(ii) It can be rounded off to its greatest place as 700 ÷ 20 = 35.
(iii) It can be rounded off to its greatest place as 4000 ÷ 400 = 10.
Exercise 1.8 page: 1.30
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 is multiplied by the difference of twelve and eight.
Solution:
(i) The expression for Four multiplied by the sum of 13 and 7 is 4 × (13 + 7).
(ii) Eight multiplied by the difference of four from nine is 8 × (9 – 4).
(iii) Divide the difference of twenty eight and seven by 3 is (28 – 7) ÷ 3.
(iv) The sum of 3 and 7 is multiplied by the difference of twelve and eight is (3 + 7) × (12 – 8).
2. Simplify each of the following:
(i) 124 – (12 – 2) × 9
(ii) (13 + 7) × (9 – 4) – 18
(iii) 210 – (14 – 4) × (18 + 2) – 10
Solution:
(i) 124 – (12 – 2) × 9
On further simplification we obtain
124 – (12 – 2) × 9
= 124 – 10 × 9
= 124 – 90
= 34
(ii) (13 + 7) × (9 – 4) – 18
On further simplification we obtain
(13 + 7) × (9 – 4) – 18
= 20 × 5 – 18
= 100 – 18
= 82
(iii) 210 – (14 – 4) × (18 + 2) – 10
On further simplification we obtain
210 – (14 – 4) × (18 + 2) – 10
= 210 – 10 × 20 – 10
= 210 – 200 – 10
= 0
Exercise 1.9 page: 1.31
Simplify each of the following:
1. 7 × 109
2. 6 × 112
3. 9 × 105
4. 17 × 109
5. 16 × 108
6. 12 × 105
7. 102 × 103
8. 101 × 105
9. 109 × 107
Solution:
1. On simplification of 7 × 109 we get
7 × 109 = 763.
2. On simplification of 6 × 112 get
6 × 112 = 672.
3. On simplification of 9 × 105 we get
9 × 105 = 945.
4. On simplification of 17 × 109 we get
17 × 109 = 1853.
5. On simplification of 16 × 108 we get
16 × 108 = 1728.
6. On simplification of 12 × 105 we get
12 × 105 = 1260.
7. On simplification of 102 × 103 we get
102 × 103 = 10506.
8. On simplification of 101 × 105 we get
101 × 105 = 10605.
9. On simplification of 109 × 107 we get
109 × 107 = 11663.
Exercise 1.10 page: 1.35
1. Write the roman-numerals for each of the following:
(i) 33
(ii) 48
(iii) 76
(iv) 95
Solution:
(i) The roman-numeral of 33 is XXXIII.
(ii) The roman-numeral of 48 is XLVIII.
(iii) The roman-numeral of 76 is LXXVI.
(iv) The roman-numeral of 95 is XCV.
2. Write the following in Roman numerals:
(i) 154
(ii) 173
(iii) 248
(iv) 319
Solution:
(i) The roman-numeral of 154 is CLIV.
(ii) The roman-numeral of 173 is CLXXIII.
(iii) The roman-numeral of 248 is CCXLVIII.
(iv) The roman-numeral of 319 is CCCXIX.
3. Write the following in Roman numerals:
(i) 1008
(ii) 2718
(iii) 3906
(iv) 3794
Solution:
(i) The roman-numeral of 1008 is MVIII.
(ii) The roman-numeral of 2718 is MMDCCXVIII.
(iii) The roman-numeral of 3906 is MMMCMVI.
(iv) The roman-numeral of 3794 is MMMDCCXCIV.
4. Write the following in Roman numerals:
(i) 4201
(ii) 10009
(iii) 44000
(iv) 25819
Solution:
(i) The roman-numeral of 4201 is MVCCI.
(ii) The roman-numeral of 10009 is XIX.
(iii) The roman-numeral of 44000 is
.
(iv) The roman-numeral of 25819 is
DCCCXIX.
5. Write the following in Hindu-Arabic numerals:
(i) XXVI
(ii) XXIX
(iii) LXXII
(iv) XCI
Solution:
(i) The Hindu-Arabic numerals of XXVI is 26.
(ii) The Hindu-Arabic numerals of XXIX is 29.
(iii) The Hindu-Arabic numerals of LXXII is 72.
(iv) The Hindu-Arabic numerals of XCI is 91.
6. Write the corresponding Hindu-Arabic numerals for each of the following:
(i) CIX
(ii) CLXXII
(iii) CCLIV
(iv) CCCXXIX
Solution:
(i) The Hindu-Arabic numerals of CIX is 109.
(ii) The Hindu-Arabic numerals of CLXXII is 172.
(iii) The Hindu-Arabic numerals of CCLIV is 254.
(iv) The Hindu-Arabic numerals of CCCXXIX is 329.
7. Write the corresponding Hindu-Arabic numerals for each of the following:
(i) KXIX
(ii) KDLXV
(iii) KKCXXIII
(iv) KKKDCXL
Solution:
(i) The Hindu-Arabic numerals of KXIX is 1019.
(ii) The Hindu-Arabic numerals of KDLXV is 1565.
(iii) The Hindu-Arabic numerals of KKCXXIII is 2123.
(iv) The Hindu-Arabic numerals of KKKDCXL is 3640.
8. Write the following in Hindu-Arabic numerals:
Solution:
(i) The Hindu-Arabic numerals of
is 4444.
(ii) The Hindu-Arabic numerals of
is 6949.
(iii) The Hindu-Arabic numerals of
is 9391.
(iv) The Hindu-Arabic numerals of
is 70009.
9. Which of the following are meaningless?
(i) 
CC
(ii) KKKCCXI
(iii) XD
(iv) VC
Solution:
In the given question we know that (i), (iii) and (iv) are meaningless.
Objective Type Questions PAGE: 1.36
Mark the correct alternative in each of the following:
1. The difference between the place value and face value of 8 in 658742 is
(a) 0 (b) 42 (c) 735 (d) 693
Solution:
The given options are incorrect.
Place value of 8 in 6,58,742 = 8,000
Face value of 8 = 8
Difference between them = 8,000 − 8 = 7,992
2. The difference between the place values of 6 and 3 in 256839 is
(a) 3 (b) 9 (c) 6800 (d) 5930
Solution:
The given options are incorrect.
Place value of 6 in 2,56,839 = 6,000
The place value of 3 in 2,56,839 = 30
Difference between them = 6,000 − 30 = 5,970
3. The difference of the smallest three digit number and the largest two digit number is
(a) 100 (b) 1 (c) 10 (d) 99
Solution:
The option (b) is the correct answer.
We know that
The smallest three-digit number is 100 and the largest two-digit number is 99.
So the difference between them= 100 − 99 = 1
4. The largest three digit number formed by the digits 8, 5, 9 is
(a) 859 (b) 985 (c) 958 (d) 589
Solution:
The option (b) is the correct answer.
We know that the largest number is formed by writing the digits in descending order.
5. The smallest three digit number having three distinct digits is
(a) 123 (b) 101 (c) 102 (d) 201
Solution:
The option (c) is the correct answer.
We know that the smallest three digit numbers are 0, 1 and 2
By arranging them in ascending order we get 012 which is a two digit number.
Hence, 102 is the smallest three digit number.
6. The largest three digit number having distinct digits is
(a) 987 (b) 789 (c) 999 (d) 900
Solution:
The option (a) is the correct answer.
The largest three digit numbers are 7, 8 and 9. The largest number is obtained by arranging them in descending order.
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
Solution:
The option (c) is the correct answer.
We know that Largest three digit number = 999
Largest three-digit number with distinct digits = 987
Difference between them = 999 − 987 = 12
8. The product of the place values of two threes in 53432 is
(a) 9000 (b) 90000 (c) 10000 (d) 99000
Solution:
The option (b) is the correct answer.
We know that 3 in the second place from right is at tens place.
So the place value of 3 at tens place = 30
We know that 3 in the fourth place from right is at thousands place.
So the place value of 3 at thousands place = 3,000
Product of the place values = 3,000 × 30 = 90,000
9. The smallest counting number is
(a) 0 (b) 1 (c) 10 (d) None of these
Solution:
The option (b) is the correct answer.
We know that the smallest digit is zero and the smallest counting number is 1.
10. The total number of 4 digit numbers is
(a) 8999 (b) 9000 (c) 8000 (d) 9999
Solution:
The option (b) is the correct answer.
We know that the smallest four digit number = 1,000
Largest four digit number = 9,999.
So the total number of four-digit numbers = [9,999 − 1,000 ] + 1 = 9,000
11. The number of 3 digit numbers formed by using digits 3, 5, 9, taking each digit exactly once, is
(a) 3 (b) 4 (c) 5 (d) 6
Solution:
The option (d) is the correct answer.
359, 395, 539, 593, 935 and 953 are the numbers.
12. Total number of numbers which when rounded off to nearest ten give us 200 is
(a) 9 (b) 10 (c) 8 (d) 7
Solution:
The option (b) is the correct answer.
195, 196, 197, 198, 199, 200, 201, 202, 203, 204 are the numbers that when rounded off to nearest ten give us 200.
13. The smallest number which when rounded off to the nearest hundred as 600, is
(a) 550 (b) 595 (c) 604 (d) 599
Solution:
The option (a) is the correct answer.
The number from 550 to 649 are rounded off to the nearest hundred as 600. Hence, the smallest number is 550.
14. The greatest number which when rounded off to the nearest thousand 7000, is
(a) 6500 (b) 6549 (c) 7499 (d) 6499
Solution:
The option (c) is the correct answer.
Numbers from 6500 to 7499 are rounded off to the nearest thousand as 7000. Hence, the greatest number is 7499.
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
Solution:
The option (b) is the correct answer.
We know that the greatest number that when rounded off to the nearest tens will become 540 = 544
Least number that when rounded off to the nearest tens will become 540 = 535
So, the difference between them = 544 − 535 = 9
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
Solution:
The option (b) is the correct answer.
We know that the greatest number which when rounded off to the nearest hundred will become 6,700 = 6749
Least number which when rounded off to the nearest hundred will become 6,700 = 6650
So, the difference between them = 6,749 − 6,650 = 99
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
Solution:
The option (c) is the correct answer.
We know that the greatest number which when rounded off to the nearest thousand becomes 9,000 = 9499
Smallest number which when rounded off to the nearest thousand becomes 9,000 = 8500
So, the difference between them = 9,499 − 8,500 = 999
18. Which of the following numbers is equal to 1 billion?
(a) 10 lakh (b) 1 crore (c) 10 crore (d) 100 crore
Solution:
The option (d) is the correct answer.
1 billion = 100 crore
19. In the international place value system, we write one million for
(a) 1 lakh (b) 10 lakh (c) 100 lakh (d) 1 crore
Solution:
The option (b) is the correct answer.
One million = 10 lakh
20. Which of the following is not meaningful?
(a) XIV (b) XXXV (c) XXV (d) VX
Solution:
The option (d) is the correct answer.
In Roman numerals V is 5 and X is 10
5 cannot be written before 10
21. Which of the following is not meaningful?
(a) XXII (b) XII (c) XVV (d) XIV
Solution:
The option (c) is the correct answer.
In Roman numerals V is 5 and X is 10
5 cannot be written two times.
RD Sharma Solutions for Class 6 Maths Chapter 1: Knowing Our Numbers
Chapter 1 Knowing Our Numbers has 10 exercises with problems based on methods to represent numbers in figures or in words. The concepts covered in RD Sharma Solutions Chapter 1 are as follows:
- Introduction
- Natural numbers and whole numbers
- Methods of numeration
- Indian System of Numeration
- The place value and the face value of a digit in a numeral
- The international system of numeration
- Knowing more about numbers
- Comparison of numbers
- Large numbers in practice
- Estimation
- Estimating sum, difference, product and quotient
- Using brackets
- Roman Numerals
Chapter Brief of RD Sharma Solutions for Class 6 Maths Chapter 1 – Knowing Our Numbers
The students can download PDFs of RD Sharma Solutions to solve exercise-wise problems effortlessly and improve their understanding of the topics. These solutions are prepared by experienced faculty after vast research conducted on each concept. The exercise-wise solutions are explained in simple language to make it easy for the students to understand them.
Numbers play a very important role in determining the quantity or amount of things around us. From calling a friend to counting the number of eggs in the fridge, numbers are very important in day-to-day life. By using PDFs of the solutions, students can solve the problems in a shorter duration to boost their exam preparation.
Comments