RS Aggarwal Solutions for Class 6 Chapter 1 Number System

Exercise 1.1

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 tow thousand nine
(v) Thirty lakh eleven thousand one
(vi) Eight crore four lakh twenty-five
(vii) Three crore three thousand three hundred three
(viii) Seventeen crores 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
In simple words: Convert word form to numeral form by placing each digit in its correct position based on the Indian place value system where lakh and crore are used for grouping.

Exam Tip: Remember the Indian system groups: ones, tens, hundreds, thousands, ten thousands, lakhs, ten lakhs, crores. Write commas in the correct spots to separate periods accurately.

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
In simple words: Break the number into periods using commas, then read each period from left to right, naming it with its place value group such as crore, lakh, or thousand.

Exam Tip: Always separate the number into groups of two digits from the right in Indian system, then name each group by its value.

Question 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
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
In simple words: Place commas starting from the right: first after every three digits for thousands, then after every two digits for lakhs and crores following Indian system rules.

Exam Tip: In the Indian system, commas go after the ones place (3 digits), then every 2 digits to the left. Practice this pattern carefully to avoid mistakes.

Question 4. Write each of the following in expanded form:
(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
In simple words: Break down the number into the sum of its place values, writing each digit multiplied by its place value and adding them together.

Exam Tip: Expanded form shows the place value of each digit. Skip any place with a zero, as it adds nothing to the sum.

Question 5. Write the corresponding numeral for each of the following:
(i) 7 x 10000 + 2 x 1000 + 5 x 100 + 9 x 10 + 6 x 1
(ii) 4 x 1000000 + 5 x 1000 + 1 x 100 + 7 x 1
(iii) 8 x 1000000 + 3 x 1000 + 6 x 1
(iv) 5 x 10000000 + 7 x 1000000 + 8 x 1000 + 9 x 10 + 4
Answer:
(i) 70000 + 2000 + 500 + 90 + 6 = 72,596
(ii) 4000000 + 5000 + 100 + 7 = 4,05,107
(iii) 8000000 + 3000 + 6 = 80,03,006
(iv) 50000000 + 7000000 + 8000 + 90 + 4 = 5,70,08,094
In simple words: Multiply each digit by its place value, then add all the products together to get the final number.

Exam Tip: Carefully work out each multiplication step, then add all parts in order. A small error in multiplication or addition changes the whole answer.

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 x 10,00,00 = 40,00,00
(ii) Place value of 4 = 4 x 10,000 = 40,000
In simple words: Identify where digit 4 appears in the number, then multiply it by the value of that position to find its place value.

Exam Tip: The place value is the digit multiplied by the value of its position. A digit in the millions place has a different place value than the same digit in the thousands place.

Question 7. Determine the product of the place values of two fives in 450758.
Answer: Place value of first 5 = 5 x 10 = 50
Place value of second 5 = 5 x 10,000 = 50,000
Required product = 50 x 50,000 = 25, 00,000
In simple words: Find the place value of each 5, then multiply those values together to get the result.

Exam Tip: Always find the place value of each digit separately, then perform the operation asked (in this case, multiply them).

Question 8. Determine the difference of the place values of 7's in 257839705.
Answer: Place value of first 7 = 7 x 10 = 700
Place value of second 7 = 7 x 10,000 = 70, 00,000
Required difference = 70, 00,000 - 700 = 69, 99,300
In simple words: Find the place value of each 7, then subtract the smaller value from the larger value.

Exam Tip: When subtracting place values, identify which 7 has the larger place value and subtract in the correct order.

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
Therefore, the difference = 50,000 - 5 = 49,995
In simple words: Place value tells you what the digit is worth based on its position. Face value is just the digit itself. Subtract face value from place value to find the difference.

Exam Tip: Remember: face value is always just the digit, while place value depends on where the digit sits in the number.

Question 10. Which digits have the same face value and place value in 92078634?
Answer: The place value of a digit relies on the place where it occurs, while the face value is the value of the digit itself.
In a number, the digits 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
In simple words: Only the digit in the ones place and any zeros in the number have place values equal to their face values.

Exam Tip: The digit 1 in the ones place and any zeros anywhere in the number are the only digits where place value equals face value.

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.
In simple words: Since 0 cannot be in the first position of a three-digit number, try placing 2 and 5 there, then arrange the remaining digits.

Exam Tip: Remember that a three-digit number cannot start with 0. Count all valid arrangements where 0 is not in the first position.

Question 12. Write all possible 3- digit numbers using 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
In simple words: When repetition is not allowed, use each digit once per number. When repetition is allowed, each digit can be used multiple times, creating many more combinations.

Exam Tip: Always remember that three-digit numbers cannot start with 0. Count all valid combinations carefully, noting the difference between allowing and not allowing repetition.

Question 13. Fill in the blanks:
(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
In simple words: To convert from one place value unit to another, divide the larger value by the smaller value. For example, since 1 lakh equals 100,000, it equals 100 groups of thousand.

Exam Tip: Memorize the relationships: 1 lakh = 100,000 and 1 crore = 10,000,000. Use these base facts to figure out how many of any smaller unit fit into them.

Exercise 1.2

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 x 1000000 + 3 x 100000 + 0 x 10000 + 3 x 1000 + 2 x 100 + 0 x 10 + 6 x 1
(ii) 55,029,007
Expanded form = 5 x 10000000 + 5 x 1000000 + 0 x 100000 + 2 x 10000 + 9 x 1000 + 0 x 100 + 0 x 10 + 7 x 1
(iii) 6,110,003,007
Expanded form = 6 x 10000000000 + 1 x 1000000000 + 1 x 100000000 + 0 x 10000000 + 0 x 1000000 + 0 x 100000 + 0 x 10000 + 3 x 1000 + 0 x 100 + 0 x 10 + 7 x 1
In simple words: In the international system, group digits in threes from the right: ones, thousands, millions, and billions. Write each digit times its place value and add them all together for expanded form.

Exam Tip: The international system uses millions and billions, while the Indian system uses lakhs and crores. Make sure you know which system the question asks for.

Question 2. Rewrite each of the following numerals with proper commas 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 seventy 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
In simple words: In the international system, insert the first comma after the third digit from the right, then every three digits moving left. Name each group by its value: thousand, million, or billion.

Exam Tip: In international system, commas come after every 3 digits from the right (unlike Indian system which uses 2-3 pattern). Practice grouping: ones, thousands, millions, billions.

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
In simple words: First convert the Indian system number to digits, then regroup using international commas (after every 3 digits from the right) and read it using millions and thousands.

Exam Tip: To change from Indian to international system, first write the digits, then reposition the commas. One lakh = 100,000; one crore = 10,000,000.

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
In simple words: Convert from international numbers to Indian numbers by first writing all digits, then regrouping with commas in the Indian pattern: after the first 3 from right, then every 2 digits.

Exam Tip: Remember the Indian grouping pattern starting from right: ones-tens-hundreds (3 digits), then ten thousands-lakhs (2 digits), then ten lakhs-crores (2 digits).

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 thousand one hundred and eleven
In simple words: Write each digit group in its place: 3's in millions (3,3,3), 2's in thousands (2,2,2), and 1's in ones (1,1,1). Then add commas following Indian rules and read it aloud.

Exam Tip: Understand what "period" means: ones period is the last 3 digits, thousands period is the next 2, and millions period is the next 2. Arrange digits correctly in each period.

Question 6. How many thousands make a million?
Answer: 1,000 thousands makes a million
In simple words: A million is 1,000,000. A thousand is 1,000. Dividing 1,000,000 by 1,000 gives 1,000.

Exam Tip: Know the key facts: 1 thousand = 1,000; 1 million = 1,000,000; 1 billion = 1,000,000,000. Use division to find how many of one unit make another.

Question 7. How many millions make a billion?
Answer: 1,000 millions make a billion
In simple words: A billion is 1,000,000,000. A million is 1,000,000. Dividing 1 billion by 1 million gives 1,000.

Exam Tip: Remember that billion is the next major place value after million in the international system. 1 billion = 1,000 millions.

Question 8. (i) How many lakhs make a million? (ii) How many lakhs make billion?
Answer:
(i) Ten lakhs make a million
(ii) Ten thousand lakhs make a billion
In simple words: One lakh = 100,000. One million = 1,000,000. So 1 million divided by 1 lakh equals 10. One billion = 1,000,000,000. So 1 billion divided by 1 lakh equals 10,000.

Exam Tip: Build conversion bridges between Indian and international systems: 10 lakhs = 1 million; 100 lakhs = 10 million; 10,000 lakhs = 1 billion.

Question 9. Write each of the following in numerical form:
(i) Nighty-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) 98,708,004
(ii) 607,012,084
(iii) 4,025,045,000
In simple words: Break the words into groups (billions, millions, thousands, ones), then write each group as a three-digit block with commas between them in international system.

Exam Tip: Use place value knowledge to convert word form to numbers. Write the groups one by one, left to right, using zeros to fill any missing places.

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
In simple words: Read the commas from left to right, naming each group by its value: millions, thousands, ones. Connect groups with "and" or "comma" as needed.

Exam Tip: Practice reading large numbers by groups. Always identify which group (millions, thousands, or ones) each comma separates, then name the digits in that group correctly.

Exercise 1.3

Question 1. How many four – digit numbers are there in all?
Answer: There are 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. We cannot use 0 at the thousand's place, so we can use only 9 digits at thousand's place. Also, we can use 10 digits at hundred'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.
In simple words: The first digit cannot be 0, so we have 9 choices. The remaining three positions can be any digit from 0 to 9, giving us 10 choices each. Multiply all together: 9 × 10 × 10 × 10 = 9000.

Exam Tip: Remember that the leftmost digit can never be 0 in a multi-digit number. Always account for this restriction in counting problems.

Question 2. Write the smallest and the largest six digit numbers. How many 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 0. Hence, the required number = 100000. Smallest six digit number will be 100000. The largest digit is 9. We can use 9 at any place. In fact, 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.
In simple words: The smallest six-digit number is 100000 (start with 1, rest are 0s). The largest six-digit number is 999999 (all 9s). The count of numbers between them is their difference minus 1, which equals 899998.

Exam Tip: When finding numbers "between" two values, always subtract 1 from the difference to exclude the boundary values themselves.

Question 3. How many 8 – digit numbers are there in all?
Answer: There are 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. We cannot 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.
In simple words: The first position has 9 choices (1-9). Each of the remaining 7 positions has 10 choices (0-9). Multiply them: 9 × 10^7 = 90000000.

Exam Tip: For n-digit numbers, the first digit always has one fewer choice than other positions because 0 cannot be the leading digit.

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 rightmost 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 75321200.
In simple words: For the smallest number, arrange non-zero digits in ascending order starting with the smallest non-zero digit, then fill other positions with 0. For the largest number, arrange all digits in descending order from left to right.

Exam Tip: To form the smallest n-digit number, put the smallest non-zero digit first, then arrange remaining digits in ascending order. To form the largest, arrange all digits in descending order.

Question 5. What is smallest 3-digit number with unique digits?
Answer: The smallest three-digit number with unique digits is 102.
In simple words: Start with the smallest non-zero digit (1), then use the next smallest available digit (0), then 2. All three digits are different, so it has unique digits.

Exam Tip: For the smallest number with unique (distinct) digits, start with the smallest non-zero digit in the highest place, then add the smallest available digits in order.

Question 6. What is the largest 5- digits number with unique digits?
Answer: The largest five-digit number with unique digits is 98,765.
In simple words: Use the five largest digits (9, 8, 7, 6, 5) arranged in descending order from left to right. No digit repeats.

Exam Tip: For the largest number with unique digits, arrange the largest available digits in descending order from the leftmost position.

Question 7. Write is 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.
In simple words: A palindromic number reads the same forwards and backwards. For a 3-digit palindrome, the first and last digits must match. The smallest is 101.

Exam Tip: A palindromic number has matching digits at equal distances from each end. For three digits, the form is aba where a and b are digits and a ≠ 0.

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.
In simple words: Write the digits in reverse order to get 972. Subtract the original from this new number: 972 - 279 = 693.

Exam Tip: Always subtract the smaller number from the larger one to get a positive difference. Here, 972 > 279, so subtract 279 from 972.

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.
In simple words: For the largest number, arrange digits in descending order: 7, 5, 1, 0 to get 7510. For the smallest, put the smallest non-zero digit first (1), then arrange the rest in ascending order: 1, 0, 5, 7 to get 1057.

Exam Tip: When forming the smallest n-digit number, always place the smallest non-zero digit first to ensure it remains an n-digit number.

Exercise 1.4

Question 1. Put the appropriate symbol ( < > ) in each of the following boxes:
(i) 102394 ___ 99887
(ii) 2507324 ___ 2517324
(iii) 3572014 ___ 10253104
(iv) 47983505 ___ 47894012
Answer:
(i) 102394 > 99887
(ii) 2507324 < 2517324
(iii) 3572014 < 10253104
(iv) 47983505 > 47894012
In simple words: Compare numbers by looking at the number of digits first, then at digits from left to right in the same place values. The number with more digits is greater, or if digits match, the next digit decides.

Exam Tip: Always count digits first - a 6-digit number is always less than a 7-digit number. For same-digit numbers, compare from the leftmost digit position.

Question 2. Arrange the following numbers in ascending order:
(i) 6,35,47,201, 10,23,45,694 , 65,39,542 , 83,54,208 , 1,23,45,678
(ii) 18,08,088, 1,81,888, 1,90,909, 18,08,090, 1,60,60,666
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
In simple words: Sort by the number of digits first (fewer digits come first). For numbers with the same digit count, compare digit by digit from left to right.

Exam Tip: Group numbers by their digit count before comparing within each group. This speeds up the ordering process.

Question 3. Arrange the following numbers in descending order:
(i) 05,69,44,000, 5,69,43,201 , , 56,95,440, 5,69,43,300, 56,94,437
(ii) 10,20,216, 10,20,308 , 10,21,430, 8,93,425, 8,93,245
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
In simple words: Arrange numbers from largest to smallest by first comparing digit counts (more digits means larger). Then for same-digit-count numbers, compare leftmost digits first.

Exam Tip: For descending order, start with the largest numbers (those with more digits or larger leftmost digits) and work down to smaller ones.

Exercise 1.5

Question 1. How many milligrams make one kilogram?
Answer: Ten lakh or one million (10,00,000) milligrams make one kilogram.
In simple words: 1 kilogram = 1000 grams, and 1 gram = 1000 milligrams. So 1 kg = 1000 × 1000 = 10,00,000 milligrams.

Exam Tip: Memorize key conversions: 1 kg = 1000 g, 1 g = 1000 mg. Use these to convert between units quickly.

Question 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?
Answer: Given data: Each tablet weighs = 20 mg. Therefore, The weight of 2,00,000 tablets = 2,00,000 × 20 = 40,00,000 mg. Therefore, 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 = 4000g. And, as 1 kg = 1,000 g. Therefore, Weight of the box having all tablets = 4,000 ÷ 1,000 = 4000g = 4 kg.
In simple words: Multiply the number of tablets by the weight of each tablet to get total weight in milligrams. Then divide by 1000 to convert to grams, and again by 1000 to convert to kilograms.

Exam Tip: Convert all quantities to the same unit before performing calculations. Always show conversion steps clearly in your answer.

Question 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?
Answer: The population of Sundar Nagar in 2001 = Sum of the population of city in 1991 + Increase in population over the given time period. As given in the question, The population of Sundar Nagar in 1991 = 2,35,471. As given in the question, Increase in population over the given time period = 72,958. Therefore, The population of Sundar Nagar in 2001 = 2,35,471 + 72,958 = 3,08,429.
In simple words: Add the initial population to the increase to find the new population: 2,35,471 + 72,958 = 3,08,429.

Exam Tip: For problems involving increase or decrease, clearly identify the initial value and the change, then apply addition or subtraction accordingly.

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 were 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. Therefore, total number of tickets sold on all four days is given by: = 1094 + 1812 + 2050 + 2751 = 7707.
In simple words: Add all four daily ticket sales together: 1094 + 1812 + 2050 + 2751 = 7707.

Exam Tip: When adding multiple numbers, group them strategically (e.g., pair numbers that sum to round values) to reduce arithmetic errors.

Question 5. The town newspaper is published every day. One copy has 12 pages. Everyday 11,980 copies are printed. How many pages are in all printed every day? Every month?
Answer: As given in the question, Number of pages in 1 copy of newspaper = 12. Therefore, 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. Therefore, Number of pages printed in a month = 1,43,760 × 30 = 43,12,800. Thus, 43,12,800 pages are printed in a month.
In simple words: Multiply copies by pages per copy to get daily total, then multiply the daily total by 30 days to get the monthly total.

Exam Tip: Break multi-step problems into stages: first find the daily amount, then scale up for the month by multiplying by 30.

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: As given in the question, Number of screws produced by a machine in a day = 2,825. Therefore, 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.
In simple words: Multiply the daily production by the number of days in January (31 days) to get the total monthly production.

Exam Tip: Remember the number of days in each month: January has 31 days. Check the calendar for the specific month mentioned in the problem.

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. Therefore, 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.
In simple words: Subtract the runs already scored from the target of 10,000 to find how many more runs are needed.

Exam Tip: For "how many more" problems, subtract the current amount from the target amount.

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 = Rs.78,592. He purchased 39 radio sets at Rs.1,234 each. Therefore, Money spent by him on purchasing 39 radio sets = 1,234 × 39 = Rs.48,126. Therefore, Remaining money with Ravish after the purchase = Initial money - Money spent on purchasing 39 radio sets = Rs.78,592 - Rs.48,126 = Rs.30,466. Thus, 230,466 are left with him after the purchase.
In simple words: Multiply the unit price by the quantity to find total spending, then subtract from the starting amount.

Exam Tip: Always calculate the total cost before subtracting from the available amount. Show all steps to avoid careless errors.

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. Therefore, Margin of victory for the successful candidate = 5,77,570 - 3,48,685 = 2,28,885.
In simple words: To find the winning margin, subtract the rival's votes from the winner's votes.

Exam Tip: Always subtract the smaller number from the larger number. The margin shows how decisively the winner defeated their rival.

Question 10. To stitch a shirt 2m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?
Answer: As given in the question, Total length of available cloth = 40 m = 4,000 cm (1 m = 100 cm). As given in the question, Length of cloth required to stitch a shirt = 215 cm = 200 + 15 = 215 cm. Therefore, 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. Therefore, Cloth required for stitching 18 shirts = 215 × 18 = 3870 cm. Therefore, Remaining cloth = 4,000 - 3870 = 130 cm = 1.3 m.
In simple words: Convert all measurements to the same unit (cm), divide total cloth by cloth per shirt, take the whole number result, then find leftover cloth.

Exam Tip: When division gives a decimal, use only the whole number part for counting items. Calculate leftover by subtracting used amount from total.

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: The 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. Therefore, Number of glasses in which curd can be distributed = 4,650/25 = 186.
In simple words: Convert total curd to millilitres, then divide by the capacity of each glass to find how many glasses can be filled.

Exam Tip: Always convert mixed units (litres and millilitres) to a single unit before dividing or multiplying.

Question 12. Medicine in packed in boxes, each such boxes weighing 4kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800 Kg?
Answer: Sol:
As given in the question, Total capacity of a van carrying boxes of medicines = 800 kg = 8,00,000 g (1 kg = 1,000 g). As given in the question, Weight of each packed box = 4,500 g = 4,000 + 500 = 4,500 g. Therefore, 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. Therefore, Weight of 177 boxes = 177 × 4,500 = 7,96,500 g (under permissible limit). Therefore, 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.
In simple words: Convert all weights to grams, divide van capacity by box weight, and take the whole number result. Verify that this many boxes do not exceed the van's limit.

Exam Tip: When the quotient is not a whole number, always check that using the whole number part does not exceed the stated capacity or limit.

Question 13. The Distance between the school and the house of a student is 1 Km 875 m. Every day she walks both ways between her school and home. Find the total distance covered by her in a week?
Answer: Therefore, Distance between the school and the house of a student = 1,875 m = 1,000 + 875 = 1,875 m (1 km = 1,000 m). As given in the question, 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.
In simple words: Multiply the one-way distance by 2 for daily travel (to and from school). Then multiply by 7 for one week of such daily travel.

Exam Tip: Always read carefully - if the problem mentions "both ways," multiply the one-way distance by 2 before scaling up by days or weeks.

Exercise 1.6

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
In simple words: Look at the ones digit. If it is 5 or more, round up. If it is less than 5, round down to the nearest ten.

Exam Tip: Always check the digit in the ones place — that digit tells you whether to round up or down to the nearest ten.

Question 2. Round off each of the following numbers to nearest hundreds:
(i) 3,985
(ii) 7289
(iii) 8074
(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
In simple words: Look at the tens digit. If it is 5 or more, round up. If it is less than 5, round down to the nearest hundred.

Exam Tip: Focus on the tens digit to decide whether to round up or down to the nearest hundred.

Question 3. Round off each of the numbers to nearest thousands:
(i) 2401
(ii) 9600
(iii) 4278
(iv) 7832
(v) 9567
(vi) 26,019
(vii) 20,963
(viii) 4,36,952
Answer:
(i) 2000
(ii) 10000
(iii) 4000
(iv) 8000
(v) 10000
(vi) 26000
(vii) 21000
(viii) 4,37,000
In simple words: Look at the hundreds digit. If it is 5 or more, round up. If it is less than 5, round down to the nearest thousand.

Exam Tip: Check the hundreds digit carefully — it decides whether you round up or down to the nearest thousand.

Question 4. Round off each of the following numbers to nearest tens, hundreds and thousands.
(i) 964
(ii) 1049
(iii) 45,634
(iv) 79,085
Answer:
Tens:
(i) 970
(ii) 1050
(iii) 45,630
(iv) 79,090

Hundreds:
(i) 1000
(ii) 1000
(iii) 45,600
(iv) 79,100

Thousands:
(i) 1000
(ii) 1000
(iii) 46000
(iv) 79000
In simple words: To round to tens, look at the ones digit. To round to hundreds, look at the tens digit. To round to thousands, look at the hundreds digit. In each case, if that digit is 5 or more, round up; otherwise, round down.

Exam Tip: Write your answer in three separate sections - tens, hundreds, and thousands - to show all three roundings clearly.

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) 1265 km
(vi) 417 m
(vii) 550 cm
(viii) 2486 m
(ix) 360 gm
(x) 940 kg
(xi) 273 l
(xii) 820 mg
Answer:
(i) Rs. 700
(ii) Rs. 900
(iii) Rs. 3,500
(iv) Rs.9100
(v) 1300 km
(vi) 400 m
(vii) 600 cm
(viii) 2500 m
(ix) 400 gm
(x) 900 kg
(xi) 300 l
(xii) 800 mg
In simple words: For any measurement - money, distance, weight, or volume - look at the tens digit. If it is 5 or more, round up; if less than 5, round down to the nearest hundred.

Exam Tip: The rounding method stays the same whether you round money, distance, weight, or volume - always check the tens digit.

Question 6. List all numbers which are rounded off to the nearest ten as 370.
Answer: 365, 366, 367, 368, 369, 370, 371, 372, 373, 374
In simple words: All numbers from 365 to 374 round to 370 when rounding to the nearest ten, because 370 is the midpoint, and these are the closest whole numbers that round to it.

Exam Tip: For any rounded number, the range of numbers that round to it goes from (rounded number - 5) to (rounded number + 4).

Question 7. Find the smallest and the greatest numbers which are rounded off to the nearest hundreds as 900.
Answer: Smallest number: 850
Greatest number: 949
In simple words: When rounding to the nearest hundred, any number from 850 to 949 will round to 900. The smallest of these is 850 and the greatest is 949.

Exam Tip: For rounding to hundreds, the range is always (rounded number - 50) to (rounded number + 49).

Question 8. Find the smallest and the greatest numbers which are rounded off to the nearest thousands as 9000.
Answer: Smallest number: 8,500
Greatest number: 9,499
In simple words: When rounding to the nearest thousand, any number from 8,500 to 9,499 will round to 9,000. The smallest of these is 8,500 and the greatest is 9,499.

Exam Tip: For rounding to thousands, the range is always (rounded number - 500) to (rounded number + 499).

Exercise 1.7

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 + 1000 = 1700
(ii) 800 - 300 = 500
(iii) 900 - 400 = 500
In simple words: First round each number to the nearest hundred, then do the addition or subtraction using the rounded numbers. This gives you a quick estimate of the answer.

Exam Tip: Always round both numbers before doing the operation - this makes the mental maths easier and faster.

Question 2. Estimate the following by rounding off each factor to nearest thousands:
(i) 12904 + 2888
(ii) 28292 - 21496
Answer:
(i) 13000 + 3000 = 16000
(ii) 28000 - 21000 = 7000
In simple words: Round each number to the nearest thousand first, then add or subtract the rounded values. This makes large numbers easier to work with in your head.

Exam Tip: Rounding to thousands helps with large numbers - always do the rounding step before the operation.

Question 3. Estimate the following by rounding off each number to its greatest place:
(i) 439 + 334 + 4317
(ii) 8325 - 491
(iii) 108734 - 47599
(iv) 898 × 785
(v) 9 × 795
(vi) 87 × 317
Answer:
(i) 400 + 300 + 4000 = 4700
(ii) 8000 - 500 = 7500
(iii) 100000 - 50000 = 50000
(iv) 900 × 800 = 720000
(v) 10 × 800 = 8000
(vi) 90 × 300 = 27000
In simple words: Round each number to its greatest place value (the place of its leftmost digit), then do the operation using these rounded values. This gives you a quick rough estimate.

Exam Tip: Rounding to the greatest place makes calculations faster - it's useful when you just need a rough idea, not an exact answer.

Question 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
Answer:
(i) 900 ÷ 30 = 30
(ii) 700 ÷ 20 = 35
(iii) 4000 ÷ 400 = 10
In simple words: Round both the dividend and divisor to their greatest place values, then do the division using these rounded numbers to get an estimate of the quotient.

Exam Tip: For division problems, round both numbers to their greatest place before dividing - this makes the calculation much simpler.

Question 5. 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) ÷ 3
(iv) (3 + 7) × (12 - 8)
In simple words: Use brackets to group the operations that must be done first. The phrases like "sum of" mean addition, "difference of" means subtraction, "multiplied by" means multiply, and "divide" means division.

Exam Tip: Read each statement slowly and translate each operation phrase into its matching symbol before writing the brackets.

Question 6. 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) 34
(ii) 82
(iii) 0
In simple words: Follow the order of operations: first solve what is in brackets, then multiply or divide from left to right, and finally add or subtract from left to right.

Exam Tip: Always complete brackets first, then multiplication or division, then addition or subtraction - this order is called BODMAS.

Question 7. Simplify each of the following:
(i) 7 × 109
(ii) 6 × 112
(iii) 9 × 105
(iv) 17 × 109
(v) 16 × 108
(vi) 12 × 105
(vii) 102 × 103
(viii) 101 × 105
(ix) 109 × 107
Answer:
(i) 763
(ii) 672
(iii) 945
(iv) 1853
(v) 1728
(vi) 1260
(vii) 10506
(viii) 10605
(ix) 11663
In simple words: Break down each number into its parts. For example, 109 = 100 + 9, so 7 × 109 = 7 × 100 + 7 × 9 = 700 + 63 = 763. This method makes it easier to multiply in your head.

Exam Tip: Use the distributive property to break numbers into tens and ones - this trick makes mental multiplication much faster.

Question 8. 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
In simple words: Remember: I = 1, V = 5, X = 10, L = 50, C = 100. A smaller numeral before a larger one means subtraction (like IV = 4). Repeat symbols only up to three times in a row.

Exam Tip: Always write the largest values first, then smaller ones - this makes roman numerals easier to read and write correctly.

Question 9. Write the following in roman numerals:
(i) 154
(ii) 173
(iii) 248
(iv) 319
Answer:
(i) CLIV
(ii) CLXXIII
(iii) CCXLVIII
(iv) CCCXIX
In simple words: Break the number into hundreds, tens, and ones, then convert each part to roman numerals separately. For example, 154 = 100 + 50 + 4 = C + L + IV = CLIV.

Exam Tip: Split larger numbers into smaller parts - convert hundreds first, then tens, then ones to avoid mistakes.

Question 10. Write the following in roman numerals:
(i) 1008
(ii) 2718
(iii) 3906
(iv) 3794
Answer:
(i) KVIII
(ii) KKDCCXVIII
(iii) KKKCKVI
(iv) KKKDCCXCIV
In simple words: For numbers greater than 1000, use an overline or special notation. A bar or underline above a roman numeral means multiply that value by 1000. So K with a line means 1000 + something.

Exam Tip: For very large numbers in roman numerals, remember that a line above a symbol multiplies it by 1000 - this is less common but important to know.

Question 11. Write the following in roman numerals:
(i) 4201
(ii) 10009
(iii) 44000
(iv) 25819
Answer: (Answers not provided in source. Generation based on context: for numbers above 3999, special notation with vinculum or overline above the roman numeral is used to indicate multiplication by 1000. Standard notation: K̅ for 1000, M̅ for 1000, etc.)
In simple words: Very large numbers in roman numerals use a bar or line above the numeral to show that value is multiplied by 1000. This allows representation of numbers much bigger than 3999.

Exam Tip: For modern large roman numerals above 3999, always use vinculum (overline) notation - this is the standard method taught in schools today.

Question 12. Write the following in Hindu - Arabic numeral:
(i) IVCCI
(ii) XIX
(iii) XLIV
(iv) XXVDCCCXIX
Answer:
(i) 26
(ii) 29
(iii) 72
(iv) 91
In simple words: Roman numerals use letters like I, V, X, L, C, D, and M to show numbers. Each letter has a value, and you combine them to make different numbers. When a smaller value comes before a larger one, you subtract it.

Exam Tip: Learn the basic Roman numeral values (I=1, V=5, X=10, L=50, C=100, D=500, M=1000) and remember the subtraction rule for letters placed before larger values.

Question 13. Write the corresponding Hindu - Arabic numeral for each of the following:
(i) CIX
(ii) CLXXII
(iii) CCLIV
(iv) CCCXXIX
Answer:
(i) 109
(ii) 172
(iii) 254
(iv) 329
In simple words: Break each Roman numeral into smaller parts. Add up the value of each letter. If a smaller letter appears before a bigger one, subtract it from the bigger value instead of adding.

Exam Tip: Work left to right and identify subtraction cases (like IV, IX, XL, XC, CD, CM) first to avoid errors in conversion.

Question 14. Write the corresponding Hindu - Arabic numeral for each of the following:
(i) KXIX
(ii) KDLXV
(iii) KKCXXIII
(iv) KKKDCXL
Answer:
(i) 1019
(ii) 1565
(iii) 2123
(iv) 3640
In simple words: The letter K represents 1000 in Roman numerals. So KK means 2000, and KKK means 3000. Then add the other letters according to their values.

Exam Tip: When you see K (1000), note how many times it appears and multiply accordingly, then add the remaining Roman numeral values.

Question 15. Write the following in Hindu - Arabic numeral:
(i) IVCDXLIV
(ii) VICXLIX
(iii) IXCCCXCI
(iv) LXXIX
Answer:
(i) 4444
(ii) 6949
(iii) 9391
(iv) 70009
In simple words: These are complex Roman numerals with many letters combined together. Break them down piece by piece, starting from the left. Add each value, and subtract when a smaller letter comes before a larger one.

Exam Tip: For complex numerals, underline or separate the subtraction pairs (like IV, IX, XL) first to make the conversion clearer and less error-prone.

Question 16. Which of the following is meaningless?
(i) IIICC
(ii) KKKCCXI
(iii) XD
(iv) VC
Answer: (i) and (iii) are meaningless.
In simple words: In Roman numerals, you cannot write more than three of the same letter in a row. Also, you can only subtract a smaller value from a larger one if they follow certain rules - not any smaller from any larger.

Exam Tip: Remember the rules: no more than three identical letters in a row, and only specific subtraction pairs are allowed (I before V or X, X before L or C, C before D or M).