ML Aggarwal Class 6 Maths Solutions Chapter 01 Knowing Our Numbers

 

Question 1. Write the smallest natural number. Can you write the largest natural number?
Answer: Numbers used for counting like 1, 2, 3, 4, and so on are called natural numbers. The smallest natural number is 1. You cannot write the largest natural number because natural numbers go on forever - for each natural number, you can always add 1 to get a bigger one. So there is no end to natural numbers.
In simple words: The smallest natural number is 1. There is no largest natural number because you can keep adding 1 forever.

Exam Tip: Remember that natural numbers start from 1 and have no upper limit. This is a key definition in number theory.

 

Question 2. Insert commas suitably and write each of the following numbers in words in the Indian system and the International system of numeration:
(i) 506723
(ii) 180018018
Answer:
(i) 506723
In the Indian system, the first comma comes after 3 digits from the right, and then another comma comes after every 2 digits.

506723 = 5,06,723

In words (Indian system): Five lakh six thousand seven hundred twenty three.

In the International system, commas come after every 3 digits from the right.

506723 = 506,723

In words (International system): Five hundred six thousand seven hundred twenty three.

(ii) 180018018
In the Indian system: 18,00,18,018

In words (Indian system): Eighteen crore eighteen thousand eighteen.

In the International system: 180,018,018

In words (International system): One hundred eighty million eighteen thousand eighteen.
In simple words: Place commas in the right spots, then read the number out loud. The Indian way groups digits differently than the world way.

Exam Tip: Always mark where the commas go first, then read carefully from left to right. Practice both Indian and International systems equally.

 

Question 3. Write the following numbers in expanded form:
(i) 750687
(ii) 5032109
Answer:
(i) Expanding 750687:

750687 = 7 × 1,00,000 + 5 × 10,000 + 0 × 1,000 + 6 × 100 + 8 × 10 + 7 × 1

= 7,00,000 + 50,000 + 600 + 80 + 7

(ii) Expanding 5032109:

5032109 = 5 × 10,00,000 + 0 × 1,00,000 + 3 × 10,000 + 2 × 1,000 + 1 × 100 + 0 × 10 + 9 × 1

= 50,00,000 + 30,000 + 2,000 + 100 + 9
In simple words: Break the number into parts. Each digit times its place value. Add them all up to check your work.

Exam Tip: Show every place value step, even the ones that are zero. This shows clear understanding.

 

Question 4. Write the following numbers in figures:
(i) Seven lakh three thousand four hundred twenty
(ii) Eighty crore twenty three thousand ninety three

Also write the above numbers in the place value chart.
Answer:
(i) Seven lakh three thousand four hundred twenty

This number has 7 lakhs, 0 ten thousands, 3 thousands, 4 hundreds, 2 tens and 0 ones.

In figures: 7,03,420

In the Indian place value chart:

(ii) Eighty crore twenty three thousand ninety three

This number has 8 ten crores, 0 crores, 0 ten lakhs, 0 lakhs, 2 ten thousands, 3 thousands, 0 hundreds, 9 tens and 3 ones.

In figures: 80,00,23,093

In the Indian place value chart:


In simple words: Look at what each place says and write down the digit. Match the word name to the right columns in the table.

Exam Tip: Count the word groups carefully - each group (lakh, crore, thousand) tells you where to place the digits.

 

Question 5. Write each of the following numbers in numeral form and place commas correctly:
(i) Seventy three lakh seventy thousand four hundred seven.
(ii) Nine crore five lakh forty one.
(iii) Fifty eight million four hundred twenty three thousand two hundred two.
Answer:
(i) Seventy three lakh seventy thousand four hundred seven.

This number has 73 lakhs, 70 thousands, 4 hundreds, 0 tens and 7 ones.

In figures (Indian system): 73,70,407

(ii) Nine crore five lakh forty one.

This number has 9 crores, 5 lakhs, 0 thousands, 0 hundreds, 4 tens and 1 one.

In figures (Indian system): 9,05,00,041

(iii) Fifty eight million four hundred twenty three thousand two hundred two.

This number has 58 millions, 423 thousands and 202 ones.

In figures (International system): 58,423,202
In simple words: Read each word group. Write down those digits in the right order. Put commas in their correct spots.

Exam Tip: When you see "lakh" or "crore," use Indian system commas. When you see "million," use International system commas.

 

Question 6. Write the face value and place value of the digit 6 in the number 756032.
Answer: In the Indian system, this number is written as 7,56,032.

The face value of a digit in a number is simply that digit itself. So the face value of 6 is 6.

The place value of 6 is 6 × 1,000 = 6,000.

Therefore, the face value is 6 and the place value is 6,000.
In simple words: Face value is just the digit itself. Place value is that digit times how much its spot is worth.

Exam Tip: Face value never changes - it is always the digit. Place value depends on where the digit sits in the number.

 

Question 7. Write the face value and the place value of the odd digits in the number 36510692.
Answer: In the Indian system, this number is written as 3,65,10,692.

The odd digits in this number are 3, 5, 1 and 9.

For digit 3: Face value = 3, Place value = 3 × 1,00,00,000 = 3,00,00,000

For digit 5: Face value = 5, Place value = 5 × 1,00,000 = 5,00,000

For digit 1: Face value = 1, Place value = 1 × 10,000 = 10,000

For digit 9: Face value = 9, Place value = 9 × 10 = 90

Therefore, the face values of the odd digits 3, 5, 1 and 9 are 3, 5, 1 and 9 respectively, and their place values are 3,00,00,000; 5,00,000; 10,000 and 90 respectively.
In simple words: Find all odd digits (1, 3, 5, 7, 9). For each one, the face value is the digit itself and the place value is that digit times its position's worth.

Exam Tip: First identify which digits are odd. Then find face and place value for each one separately. List them in the order they appear.

 

Question 8. Find the difference between the place value and the face value of the digit 9 in the number 229301.
Answer: In the Indian system, this number is written as 2,29,301.

The face value of 9 is 9.

The place value of 9 is 9 × 1,000 = 9,000.

Difference = Place value - Face value = 9,000 - 9 = 8,991

Therefore, the difference between the place value and the face value of the digit 9 in 2,29,301 is 8,991.
In simple words: Find what the 9 is worth in its spot (9,000). Subtract what the digit itself is (9). The answer is 8,991.

Exam Tip: Always subtract face value FROM place value, not the other way around. The difference will be large when the digit is in a big place.

 

Question 9. Determine the product of place value and the face value of the digit 4 in the number 5437.
Answer:

The face value of 4 is 4.

The place value of 4 is 4 × 100 = 400.

Product = Place value × Face value = 400 × 4 = 1,600

Therefore, the product of the place value and the face value of the digit 4 in 5437 is 1,600.
In simple words: Find the place value (400). Find the face value (4). Multiply them together: 400 × 4 = 1,600.

Exam Tip: Multiply place value by face value, not the other way around. Show your steps clearly to earn full marks.

 

Question 10. Find the difference between the number 895 and that obtained on reversing its digits.
Answer: The given number is 895.

When you reverse the digits of 895, the new number becomes 598.

Difference = 895 - 598 = 297

Therefore, the required difference is 297.
In simple words: Write the digits backwards. Subtract the smaller from the bigger. That is your answer.

Exam Tip: Always subtract the smaller number from the larger one. Double-check your digit reversal before doing the subtraction.

 

Question 11. Determine the difference of the place value of two 7's in 37014472 and write it in words in International system.
Answer: In the International system, this number is written as 37,014,472.

The place value of 7 at the millions place is 7 × 1,000,000 = 7,000,000.

The place value of 7 at the tens place is 7 × 10 = 70.

Difference = 7,000,000 - 70 = 6,999,930

In words (International system): Six million nine hundred ninety nine thousand nine hundred thirty.

Therefore, the required difference is 6,999,930, which in words is six million nine hundred ninety nine thousand nine hundred thirty.
In simple words: Find both 7's and their place values. Subtract the smaller from the larger. Then say it in words using the International system.

Exam Tip: Mark both occurrences of the digit you are looking for. Be careful to use the correct system (Indian or International) for writing the answer in words.

 

Exercise 1.2

 

Question 1. Use the appropriate symbol < or > to fill in the blanks:
(i) 173 ... 189
(ii) 1058 ... 1074
(iii) 8315 ... 8037
Answer:
(i) 173 ... 189

Both numbers have 3 digits. Looking at digits from left to right, the hundreds digit is 1 in both. The tens digit is 7 in 173 and 8 in 189. Since 7 is smaller than 8, 173 is smaller than 189.

Therefore, 173 < 189.

(ii) 1058 ... 1074

Both numbers have 4 digits. Looking from left to right, the thousands digit is 1 in both. The hundreds digit is 0 in both. The tens digit is 5 in 1058 and 7 in 1074. Since 5 is smaller than 7, 1058 is smaller than 1074.

Therefore, 1058 < 1074.

(iii) 8315 ... 8037

Both numbers have 4 digits. Looking from left to right, the thousands digit is 8 in both. The hundreds digit is 3 in 8315 and 0 in 8037. Since 3 is bigger than 0, 8315 is bigger than 8037.

Therefore, 8315 > 8037.
In simple words: Line up the numbers. Start from the left. Find where they first differ. Decide which digit is bigger. Use that to pick your symbol.

Exam Tip: Always compare from left to right, digit by digit. The first place where numbers differ tells you the answer.

 

Question 2. In each of the following pairs of numbers, state which number is smaller:
(i) 553, 503
(ii) 41338, 1139
(iii) 25431, 24531
Answer:
(i) 553, 503

Both numbers have 3 digits. Looking from left to right, the hundreds digit is 5 in both. The tens digit is 5 in 553 and 0 in 503. Since 0 is smaller than 5, 503 is smaller than 553.

Therefore, 503 is the smaller number.

(ii) 41338, 1139

41338 has 5 digits and 1139 has 4 digits. A number with fewer digits is always smaller than a number with more digits.

Therefore, 1139 is the smaller number.

(iii) 25431, 24531

Both numbers have 5 digits. Looking from left to right, the ten thousands digit is 2 in both. The thousands digit is 5 in 25431 and 4 in 24531. Since 4 is smaller than 5, 24531 is smaller than 25431.

Therefore, 24531 is the smaller number.
In simple words: First, count the digits. Fewer digits means a smaller number. If they have the same number of digits, compare left to right.

Exam Tip: If digit counts differ, the answer is quick. If they match, compare position by position from left to right.

 

Question 9. Rearrange the digits of the number 5701024 to get the largest number and the smallest number of 7 digits.
Answer: The digits in 5701024 are 5, 7, 0, 1, 0, 2, and 4. To form the largest 7-digit number, place all digits in descending order: 7, 5, 4, 2, 1, 0, 0. This gives 75,42,100. For the smallest 7-digit number, the leftmost position must have the smallest non-zero digit, which is 1 (since 0 cannot be the leading digit). After placing 1, arrange the remaining digits 0, 0, 2, 4, 5, 7 in ascending order: 0, 0, 2, 4, 5, 7. This gives 10,02,457.
In simple words: To make the biggest number, put the largest digits first. To make the smallest number, put the smallest non-zero digit first, then arrange the rest from smallest to largest.

Exam Tip: When finding the smallest n-digit number, always place the smallest non-zero digit in the highest place value - never start with 0.

 

Question 10. Keeping the place value of digit 3 in the number 730265 same, rearrange the digits of the given number to get the largest number and smallest number of 6 digits.
Answer: In the number 7,30,265, digit 3 is in the ten thousand's place. The place value of 3 = 3 × 10,000 = 30,000. To keep this place value unchanged, digit 3 must stay at the ten thousand's place. The remaining digits are 7, 0, 2, 6 and 5.

For the largest 6-digit number, arrange the remaining digits in descending order while keeping 3 fixed at the ten thousand's place:
Lakhs place: 7 (largest)
Ten thousands place: 3 (fixed)
Thousands place: 6
Hundreds place: 5
Tens place: 2
Ones place: 0
The largest 6-digit number = 7,36,520.

For the smallest 6-digit number, the lakh's place cannot be 0. So, place the smallest non-zero digit (excluding 3) at the lakh's place, which is 2. Then arrange the remaining digits in ascending order:
Lakhs place: 2
Ten thousands place: 3 (fixed)
Thousands place: 0
Hundreds place: 5
Tens place: 6
Ones place: 7
The smallest 6-digit number = 2,30,567.

Hence, the largest 6-digit number is 7,36,520 and the smallest 6-digit number is 2,30,567.
In simple words: Keep digit 3 where it belongs. Then put the bigger digits first for the largest number, and use the smaller non-zero digit first for the smallest number.

Exam Tip: Always identify which digit must stay fixed first, then arrange the remaining digits based on whether you need the largest or smallest number.

 

Question 11. Form the smallest and greatest 4-digit numbers by using any one digit twice from the digits:
(i) 5, 2, 3, 9
(ii) 6, 0, 1, 4
(iii) 4, 6, 1, 5, 8
Answer:
(i) The given digits are 5, 2, 3 and 9. For the greatest 4-digit number, use the largest digit twice. So, use 9 twice. The next largest digits are 5 and 3. Arrange in descending order: 9, 9, 5, 3. The greatest 4-digit number = 9,953. For the smallest 4-digit number, use the smallest digit twice. So, use 2 twice. The other two smaller digits are 3 and 5. Arrange in ascending order: 2, 2, 3, 5. The smallest 4-digit number = 2,235. Hence, the greatest 4-digit number is 9,953 and the smallest 4-digit number is 2,235.

(ii) The given digits are 6, 0, 1 and 4. For the greatest 4-digit number, use the largest digit twice. So, use 6 twice. The next largest digits are 4 and 1. Arrange in descending order: 6, 6, 4, 1. The greatest 4-digit number = 6,641. For the smallest 4-digit number, use the smallest digit twice. So, use 0 twice. The other two smaller digits are 1 and 4. Since 0 cannot be at the thousand's place, the next smallest non-zero digit (1) is placed at the thousand's place, then 0, 0, 4 in order. The smallest 4-digit number = 1,004. Hence, the greatest 4-digit number is 6,641 and the smallest 4-digit number is 1,004.

(iii) The given digits are 4, 6, 1, 5 and 8. For the greatest 4-digit number, use the largest digit twice. So, use 8 twice. The next largest digits from the remaining are 6 and 5. Arrange in descending order: 8, 8, 6, 5. The greatest 4-digit number = 8,865. For the smallest 4-digit number, use the smallest digit twice. So, use 1 twice. The other two smaller digits from the remaining are 4 and 5. Arrange in ascending order: 1, 1, 4, 5. The smallest 4-digit number = 1,145. Hence, the greatest 4-digit number is 8,865 and the smallest 4-digit number is 1,145.
In simple words: To get the biggest number, repeat the largest digit and fill other places with the next biggest digits. To get the smallest number, repeat the smallest digit and fill with next smallest digits, but never put 0 first.

Exam Tip: Remember that 0 can never occupy the leftmost (thousands) place in a 4-digit number - always use the smallest non-zero digit there when finding the smallest number.

 

Question 12. Write (i) the greatest number of 6 digits (ii) the smallest number of 7 digits. Also find their difference.
Answer: (i) The greatest number of 6 digits is formed by placing the largest digit (9) in all six positions. The greatest 6-digit number = 9,99,999.

(ii) The smallest number of 7 digits is formed by placing the smallest non-zero digit (1) at the highest place and 0s in all the remaining positions. The smallest 7-digit number = 10,00,000.

Difference = 10,00,000 - 9,99,999 = 1.

Hence, the greatest 6-digit number is 9,99,999, the smallest 7-digit number is 10,00,000, and their difference is 1.
In simple words: Fill all places with 9 to get the biggest 6-digit number. Put 1 first and fill the rest with 0 to get the smallest 7-digit number. When you subtract one from the other, you get just 1.

Exam Tip: The difference between the largest n-digit number and the smallest (n+1)-digit number is always 1 - a useful shortcut.

 

Question 13. Write the greatest 4-digit number having distinct digits.
Answer: The digits available are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. For the greatest 4-digit number with distinct digits, arrange the largest digits in descending order:
Thousands place: 9 (largest)
Hundreds place: 8 (next largest)
Tens place: 7 (next largest)
Ones place: 6 (next largest)
Hence, the greatest 4-digit number having distinct digits is 9,876.
In simple words: Take the four biggest different digits and arrange them from largest to smallest.

Exam Tip: When all digits must be distinct, always start with the highest available digit in the leftmost position and work downward for the maximum value.

 

Question 14. Write the smallest 4-digit number having distinct digits.
Answer: The digits available are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. For the smallest 4-digit number with distinct digits, the thousands place cannot be 0. So, the smallest non-zero digit (1) is placed at the thousand's place. Then arrange the remaining smallest digits in ascending order:
Thousands place: 1
Hundreds place: 0
Tens place: 2
Ones place: 3
Hence, the smallest 4-digit number having distinct digits is 1,023.
In simple words: Start with 1 (the smallest non-zero digit). Then fill the remaining spots with 0, 2, and 3 in order.

Exam Tip: The leftmost position can never be 0 in any multi-digit number - always place the smallest non-zero digit there when finding the minimum value.

 

Question 15. Write the greatest 6-digit number using three different digits.
Answer: For the greatest 6-digit number using exactly three different digits, the digits at the higher places should be as large as possible. Use the digit 9 as many times as possible (without exceeding three different digits in total). Using 9 four times makes only 3 different digits (9, 8, 7) when combined with 8 and 7. Let us check: in the number 9,99,987, the digits are 9, 9, 9, 9, 8, 7. The different digits are 9, 8 and 7, which gives exactly 3 different digits. Arranging in descending order: 9, 9, 9, 9, 8, 7. The greatest 6-digit number = 9,99,987. Hence, the greatest 6-digit number using three different digits is 9,99,987.
In simple words: Use the largest digit (9) as many times as you can, then add the next two largest digits (8 and 7) to make exactly three different digits total.

Exam Tip: When forming numbers with a fixed count of distinct digits, maximize repetition of the largest digit and place smaller distinct digits at the end.

 

Question 16. Write the smallest 7-digit number using four different digits.
Answer: For the smallest 7-digit number using exactly four different digits, the digit at the highest place (ten lakhs) cannot be 0. So, the smallest non-zero digit (1) is placed at the highest place. Then use 0 in as many remaining positions as possible. To make exactly four different digits in total, we need two more digits apart from 1 and 0. The smallest such digits are 2 and 3. The digits to be used: 1, 0, 0, 0, 0, 2, 3. Arranging the digits with 1 at the highest place and the rest in ascending order:
Ten lakhs place: 1
Lakhs place: 0
Ten thousands place: 0
Thousands place: 0
Hundreds place: 0
Tens place: 2
Ones place: 3
Hence, the smallest 7-digit number using four different digits is 10,00,023.
In simple words: Put 1 first, then fill as many middle spots as you can with 0, and save the other two small digits (2 and 3) for the end.

Exam Tip: To minimize a number with a set count of distinct digits, use 1 in the first position, fill the middle with 0s, and place the remaining smallest digits at the end.

 

Question 17. Write the greatest and the smallest 4-digit numbers using four different digits with the conditions as given:
(i) Digit 7 is always at units place.
(ii) Digit 4 is always at tens place.
(iii) Digit 9 is always at hundreds place.
(iv) Digit 2 is always at thousands place.
Answer:
(i) Digit 7 is always at units place. The number is of the form _ _ _ 7 (with all four digits different). For the greatest 4-digit number, the remaining three digits should be as large as possible (different from 7 and from each other). The largest such digits are 9, 8 and 6. Greatest 4-digit number = 9867. For the smallest 4-digit number, the thousands place cannot be 0. The smallest non-zero digit (other than 7) is 1. Then the remaining digits are 0 and 2 (smallest available, different from 1 and 7). Smallest 4-digit number = 1027. Hence, the greatest 4-digit number is 9,867 and the smallest 4-digit number is 1,027.

(ii) Digit 4 is always at tens place. The number is of the form _ _ 4 _ (with all four digits different). For the greatest 4-digit number, fill the remaining places with the largest distinct digits other than 4. Use 9 at thousands, 8 at hundreds and 7 at units. Greatest 4-digit number = 9847. For the smallest 4-digit number, the thousands place cannot be 0. Place 1 at thousands, 0 at hundreds, 4 at tens and 2 at units. Smallest 4-digit number = 1042. Hence, the greatest 4-digit number is 9,847 and the smallest 4-digit number is 1,042.

(iii) Digit 9 is always at hundreds place. The number is of the form _ 9 _ _ (with all four digits different). For the greatest 4-digit number, fill the remaining places with the largest distinct digits other than 9. Use 8 at thousands, 7 at tens and 6 at units. Greatest 4-digit number = 8976. For the smallest 4-digit number, the thousands place cannot be 0. Place 1 at thousands, 9 at hundreds, 0 at tens and 2 at units. Smallest 4-digit number = 1902. Hence, the greatest 4-digit number is 8,976 and the smallest 4-digit number is 1,902.

(iv) Digit 2 is always at thousands place. The number is of the form 2 _ _ _ (with all four digits different). For the greatest 4-digit number, fill the remaining places with the largest distinct digits other than 2. Use 9 at hundreds, 8 at tens and 7 at units. Greatest 4-digit number = 2987. For the smallest 4-digit number, fill the remaining places with the smallest distinct digits other than 2 (since 2 is already at thousands). Use 0 at hundreds, 1 at tens and 3 at units. Smallest 4-digit number = 2013. Hence, the greatest 4-digit number is 2,987 and the smallest 4-digit number is 2,013.
In simple words: When a digit is fixed at one place, fill the other three places with the largest digits you can find to get the biggest number, and with the smallest digits to get the smallest number. Remember 0 can never be first.

Exam Tip: Always identify which digit is fixed and at which place first, then fill remaining places strategically based on whether you need the maximum or minimum number.

 

Exercise 1.4

 

Question 1. In a particular year, a company manufactured 8570435 bicycles and next year it manufactured 8756430 bicycles. In which year more bicycles were manufactured and by how many?
Answer: Number of bicycles manufactured in the first year = 85,70,435. Number of bicycles manufactured in the next year = 87,56,430. Both numbers have 7 digits. Comparing from the leftmost digit: The ten lakhs digit is 8 in both, the lakhs digit is 5 in 85,70,435 and 7 in 87,56,430. Since 7 > 5, 87,56,430 > 85,70,435. So, more bicycles were manufactured in the next year. Difference = 87,56,430 - 85,70,435 = 1,85,995. Hence, more bicycles were manufactured in the next year, by 1,85,995.
In simple words: When you compare the two numbers from left to right, the second year's number is bigger because the lakhs digit (7) is larger than the first year's (5). The extra bicycles made are 1,85,995.

Exam Tip: Always compare large numbers digit by digit from the leftmost position - the first difference you find tells you which number is greater.

 

Question 2. What number must be subtracted from 1,02,59,756 to get 77,63,835?
Answer: Let the required number to be subtracted be x. Then, 1,02,59,756 - x = 77,63,835

\[ \Rightarrow x = 1,02,59,756 - 77,63,835 \]

\[ \Rightarrow x = 24,95,921 \]

Hence, the required number to be subtracted is 24,95,921.
In simple words: If you take away something from the first number and get the second number, then that "something" is the difference between the two numbers.

Exam Tip: When asked "what must be subtracted," set up an equation and solve for the unknown by rearranging - subtracting the second number from the first gives you the answer.

 

Question 3. The sale receipt of a company during a year was Rs 30587850. Next year it increased by Rs 6375490. What was the total sale receipt of the company during these two years?
Answer: Sale receipt of the company in the first year = Rs 3,05,87,850. Increase in sale receipt in the next year = Rs 63,75,490. Sale receipt of the company in the next year = Rs 3,05,87,850 + Rs 63,75,490 = Rs 3,69,63,340. Total sale receipts during these two years = Sale in first year + Sale in next year = Rs 3,05,87,850 + Rs 3,69,63,340 = Rs 6,75,51,190. Hence, the total sale receipt of the company during these two years was Rs 6,75,51,190.
In simple words: First add the increase to the first year's amount to find the second year's total. Then add both years together to get the grand total.

Exam Tip: Break the problem into steps: first find the second year's receipt by adding the increase, then add both years to find the total.

 

Question 4. A machine manufactures 23875 screws per day. How many screws did it produce in the year 2016? Assume that the machine worked on all the days of the year.
Answer: Number of screws manufactured per day = 23,875. The year 2016 is a leap year (since 2016 is divisible by 4 and, being divisible by 4 but not by 100, it is a leap year). Number of days in 2016 = 366. Total number of screws produced in 2016 = 23,875 × 366 = 87,397,500. Hence, the machine produced 87,397,500 screws in the year 2016.
In simple words: Check if the year is a leap year (has 366 days instead of 365). Then multiply the daily production by the total number of days to get the yearly total.

Exam Tip: Always check whether the given year is a leap year - a year is a leap year if it is divisible by 4 (with a special rule for century years: they must be divisible by 400).

 

Question 5. A merchant had Rs.78,592 with him. He placed an order for purchasing 30 bicycles at Rs.2450 each. How much money will remain with him after the purchase?
Answer: The merchant started with Rs.78,592. Each bicycle costs Rs.2,450, and he wanted to buy 30 of them. The total cost is 30 times Rs.2,450, which equals Rs.73,500. After paying this amount, the remaining money is Rs.78,592 minus Rs.73,500, giving Rs.5,092.
In simple words: The merchant had Rs.78,592. He spent Rs.73,500 on bicycles. He was left with Rs.5,092.

Exam Tip: Always identify what you know (starting amount, cost per item, number of items), calculate the total cost (multiply), then subtract from the starting amount to find what remains.

 

Question 6. Amitabh is 1 m 82 cm tall and his wife is 35 cm shorter than him. What is his wife's height?
Answer: Amitabh's height is 1 m 82 cm. Converting to centimetres, this is (1 × 100 + 82) cm = 182 cm. His wife is 35 cm shorter, so her height is 182 cm - 35 cm = 147 cm. Converting back to metres and centimetres: 147 cm = 1 m 47 cm.
In simple words: Amitabh is 182 cm tall. His wife is 35 cm shorter, so she is 147 cm tall, or 1 m 47 cm.

Exam Tip: When comparing heights or lengths, convert all measurements to the same unit first, do the subtraction, then convert back if needed.

 

Question 7. The mass of each gas cylinder is 21 kg 270 g. What is total mass of 28 such cylinders?
Answer: Each cylinder has a mass of 21 kg 270 g. Converting to grams: (21 × 1,000 + 270) g = 21,270 g. For 28 cylinders, the total mass is 28 × 21,270 g = 5,95,560 g. Converting back to kilograms and grams: (595 × 1,000 + 560) g = 595 kg 560 g.
In simple words: One cylinder weighs 21,270 grams. Twenty-eight cylinders weigh 28 times this amount, which is 5,95,560 grams or 595 kg 560 g.

Exam Tip: For mass or weight problems involving multiple identical items, convert to a single unit, multiply, then convert back to the required form.

 

Question 8. In order to make a shirt, 2 m 25 cm cloth is needed. What length of cloth is required to make 18 such shirts?
Answer: One shirt requires 2 m 25 cm of cloth. Converting to centimetres: (2 × 100 + 25) cm = 225 cm. For 18 shirts, the total cloth needed is 18 × 225 cm = 4,050 cm. Converting back: (40 × 100 + 50) cm = 40 m 50 cm.
In simple words: Each shirt needs 225 cm of cloth. Eighteen shirts need 18 times that, which is 4,050 cm or 40 m 50 cm.

Exam Tip: For problems with quantities per unit, convert to the smaller unit, multiply by the number of units, then convert back to mixed units if required.

 

Question 9. The total mass of 12 packets of sweets, each of the same size, is 15 kg 600 g. What is the mass of each such packet?
Answer: The total mass is 15 kg 600 g. Converting to grams: (15 × 1,000 + 600) g = 15,600 g. Since all 12 packets are identical, the mass of one packet is 15,600 g ÷ 12 = 1,300 g. Converting back: (1 × 1,000 + 300) g = 1 kg 300 g.
In simple words: Twelve packets together weigh 15,600 grams. One packet weighs 15,600 ÷ 12, which is 1,300 grams or 1 kg 300 g.

Exam Tip: When finding the mass of one item from a total, divide the total mass by the number of items. Remember to convert units when needed.

 

Question 10. A vessel has 4 litres 500 millilitres of orange juice. In how many glasses, each of 25 mL capacity, can it be filled?
Answer: The total amount of juice is 4 L 500 mL. Converting to millilitres: (4 × 1,000 + 500) mL = 4,500 mL. Each glass holds 25 mL. The number of glasses that can be filled is 4,500 mL ÷ 25 mL = 180 glasses.
In simple words: There are 4,500 mL of juice. Each glass holds 25 mL. Dividing 4,500 by 25 gives 180 glasses.

Exam Tip: For division problems with capacity, ensure both quantities are in the same unit before dividing.

 

Question 11. To stitch a trouser, 1 m 30 cm cloth is needed. Out of 25 m cloth, how many trousers can be stitched and how much cloth will remain?
Answer: The available cloth is 25 m. Converting to centimetres: (25 × 100) cm = 2,500 cm. Each trouser needs 1 m 30 cm = (1 × 100 + 30) cm = 130 cm. Dividing 2,500 by 130: the quotient is 19 and the remainder is 30. This means 19 trousers can be stitched and 30 cm of cloth will be left over.
In simple words: There is 2,500 cm of cloth available. Each trouser uses 130 cm. Dividing gives 19 trousers with 30 cm leftover.

Exam Tip: When a problem asks how many items can be made and how much is left, use division with remainder. The quotient is the number of items, and the remainder is what is left.

 

Exercise 1.5

 

Question 1. Round off each of the following numbers to their nearest tens:
(i) 77
(ii) 903
(iii) 1205
(iv) 999
Answer:
(i) 77 - The ones digit is 7. Since 7 is greater than 5, increase the tens digit by 1 and replace the ones digit with 0. Result: 80.
(ii) 903 - The ones digit is 3. Since 3 is less than 5, replace the ones digit with 0 and keep all other digits the same. Result: 900.
(iii) 1205 - The ones digit is 5. Increase the tens digit by 1 and replace the ones digit with 0. Result: 1,210.
(iv) 999 - The ones digit is 9. Since 9 is greater than 5, increase the tens digit by 1 and replace the ones digit with 0. The tens digit is 9, so adding 1 carries over to the hundreds place. Result: 1,000.
In simple words: Look at the ones digit. If it is 5 or more, round up. If it is less than 5, round down by changing the ones digit to 0.

Exam Tip: The key to rounding is checking the digit in the place below the one you are rounding to - if it is 5 or greater, round up; otherwise, round down.

 

Question 2. Estimate each of the following numbers to their nearest hundreds:
(i) 1246
(ii) 32057
(iii) 53961
(iv) 555555
Answer:
(i) 1246 - The tens digit is 4. Since 4 is less than 5, replace both tens and ones digits with 0. Result: 1,200.
(ii) 32057 - The tens digit is 5. Increase the hundreds digit by 1 and replace both tens and ones digits with 0. Result: 32,100.
(iii) 53961 - The tens digit is 6. Since 6 is greater than 5, increase the hundreds digit by 1 and replace both tens and ones digits with 0. The hundreds digit is 9, so adding 1 carries over. Result: 54,000.
(iv) 555555 - The tens digit is 5. Increase the hundreds digit by 1 and replace both tens and ones digits with 0. Result: 5,55,600.
In simple words: Look at the tens digit. If it is 5 or more, round the hundreds digit up. If it is less than 5, keep the hundreds digit the same. Replace tens and ones with zeros.

Exam Tip: When rounding to the nearest hundred, examine the tens place - this determines whether you round up or down.

 

Question 3. Estimate each of the following numbers to their nearest thousands:
(i) 5706
(ii) 378
(iii) 47,599
(iv) 1,09,736
Answer:
(i) 5706 - The hundreds digit is 7. Since 7 is greater than 5, increase the thousands digit by 1 and replace the hundreds, tens, and ones digits with 0. Result: 6,000.
(ii) 378 - The hundreds digit is 3. Since 3 is less than 5, replace the hundreds, tens, and ones digits with 0 and keep the thousands place as is. Result: 0.
(iii) 47,599 - The hundreds digit is 5. Increase the thousands digit by 1 and replace the hundreds, tens, and ones digits with 0. Result: 48,000.
(iv) 1,09,736 - The hundreds digit is 7. Since 7 is greater than 5, increase the thousands digit by 1 and replace the hundreds, tens, and ones digits with 0. Result: 1,10,000.
In simple words: Look at the hundreds digit. If it is 5 or more, round the thousands digit up. If it is less than 5, keep it the same. Replace all smaller digits with zeros.

Exam Tip: The hundreds place is the key digit when rounding to the nearest thousand. Always check whether it is at least 5 before rounding.

 

Question 4. Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):
(i) 439 + 334 + 4317
(ii) 8325 - 491
(iii) 1,08,734 - 47,599
(iv) 4,89,348 - 48,365
Answer:
(i) 439 + 334 + 4317
Rough estimate (to nearest hundreds): 439 becomes 400 (tens digit 3 < 5); 334 becomes 300 (tens digit 3 < 5); 4317 becomes 4300 (tens digit 1 < 5). Sum = 400 + 300 + 4300 = 5,000.
Closer estimate (to nearest tens): 439 becomes 440 (ones digit 9 > 5); 334 becomes 330 (ones digit 4 < 5); 4317 becomes 4320 (ones digit 7 > 5). Sum = 440 + 330 + 4320 = 5,090.

(ii) 8325 - 491
Rough estimate (to nearest hundreds): 8325 becomes 8300 (tens digit 2 < 5); 491 becomes 500 (tens digit 9 > 5). Difference = 8300 - 500 = 7,800.
Closer estimate (to nearest tens): 8325 becomes 8330 (ones digit 5); 491 becomes 490 (ones digit 1 < 5). Difference = 8330 - 490 = 7,840.

(iii) 1,08,734 - 47,599
Rough estimate (to nearest hundreds): 1,08,734 becomes 1,08,700 (tens digit 3 < 5); 47,599 becomes 47,600 (tens digit 9 > 5). Difference = 1,08,700 - 47,600 = 61,100.
Closer estimate (to nearest tens): 1,08,734 becomes 1,08,730 (ones digit 4 < 5); 47,599 becomes 47,600 (ones digit 9 > 5). Difference = 1,08,730 - 47,600 = 61,130.

(iv) 4,89,348 - 48,365
Rough estimate (to nearest hundreds): 4,89,348 becomes 4,89,300 (tens digit 4 < 5); 48,365 becomes 48,400 (tens digit 6 > 5). Difference = 4,89,300 - 48,400 = 4,40,900.
Closer estimate (to nearest tens): 4,89,348 becomes 4,89,350 (ones digit 8 > 5); 48,365 becomes 48,370 (ones digit 5). Difference = 4,89,350 - 48,370 = 4,40,980.
In simple words: A rough estimate uses larger rounding (hundreds), while a closer estimate uses smaller rounding (tens). Both help check if your exact answer makes sense.

Exam Tip: Always show both rough and closer estimates clearly. This demonstrates your understanding of rounding at different place values and helps validate your final answer.

 

Question 5. Estimate each of the following by rounding off each number nearest to its greatest place:
(i) 730 + 998
(ii) 5,290 + 17,986
(iii) 796 - 314
(iv) 28,292 - 21,496
Answer:
(i) 730 + 998
For 730, the greatest place is hundreds. The tens digit is 3 (< 5), so 730 rounds to 700.
For 998, the greatest place is hundreds. The tens digit is 9 (≥ 5), so 998 rounds to 1000.
Estimated sum = 700 + 1000 = 1,700.

(ii) 5,290 + 17,986
For 5,290, the greatest place is thousands. The hundreds digit is 2 (< 5), so 5,290 rounds to 5,000.
For 17,986, the greatest place is ten thousands. The thousands digit is 7 (≥ 5), so 17,986 rounds to 20,000.
Estimated sum = 5,000 + 20,000 = 25,000.

(iii) 796 - 314
For 796, the greatest place is hundreds. The tens digit is 9 (≥ 5), so 796 rounds to 800.
For 314, the greatest place is hundreds. The tens digit is 1 (< 5), so 314 rounds to 300.
Estimated difference = 800 - 300 = 500.

(iv) 28,292 - 21,496
For 28,292, the greatest place is ten thousands. The thousands digit is 8 (≥ 5), so 28,292 rounds to 30,000.
For 21,496, the greatest place is ten thousands. The thousands digit is 1 (< 5), so 21,496 rounds to 20,000.
Estimated difference = 30,000 - 20,000 = 10,000.
In simple words: Find the greatest place value (leftmost non-zero digit) in each number and round there. Then add or subtract the rounded numbers.

Exam Tip: "Greatest place" means the highest place value in each number - for 730 it is hundreds, for 17,986 it is ten thousands. Rounding to this place makes the numbers easier to work with while keeping estimates reasonably accurate.

 

Question 6. Estimate the following products by rounding off each of its factors nearest to its greatest place:
(i) 578 × 161
(ii) 9650 × 27
Answer:
(i) 578 × 161
Round 578 to its greatest place (hundreds): the tens digit is 7 (≥ 5), so 578 becomes 600.
Round 161 to its greatest place (hundreds): the tens digit is 6 (≥ 5), so 161 becomes 200.
Estimated product = 600 × 200 = 1,20,000.

(ii) 9650 × 27
Round 9650 to its greatest place (thousands): the hundreds digit is 6 (≥ 5), so 9650 becomes 10,000.
Round 27 to its greatest place (tens): the ones digit is 7 (≥ 5), so 27 becomes 30.
Estimated product = 10,000 × 30 = 3,00,000.
In simple words: Look at each number and round it up or down based on its biggest place. Then multiply the rounded numbers to get your estimate.

Exam Tip: Identify the highest place value in each number first, then check the next digit to decide whether to round up or down. Always round to the same place value for all numbers in the problem.

 

Question 7. Estimate the following products by rounding off each of its factors nearest to its hundreds:
(i) 5281 × 3491
(ii) 1387 × 888
Answer:
(i) 5281 × 3491
Round 5281 to the nearest hundreds: the tens digit is 8 (≥ 5), so 5281 becomes 5,300.
Round 3491 to the nearest hundreds: the tens digit is 9 (≥ 5), so 3491 becomes 3,500.
Estimated product = 5,300 × 3,500 = 1,85,50,000.

(ii) 1387 × 888
Round 1387 to the nearest hundreds: the tens digit is 8 (≥ 5), so 1387 becomes 1,400.
Round 888 to the nearest hundreds: the tens digit is 8 (≥ 5), so 888 becomes 900.
Estimated product = 1,400 × 900 = 12,60,000.
In simple words: Round each number to the nearest hundreds place, then multiply the rounded values.

Exam Tip: When rounding to hundreds, look at the tens digit. If it is 5 or more, round up; otherwise, round down. Apply this rule to both factors before multiplying.

 

Question 8. Estimate the following quotients by rounding off each number to its nearest tens:
(i) 423 ÷ 29
(ii) 777 ÷ 27
Answer:
(i) 423 ÷ 29
Round 423 to the nearest tens: the ones digit is 3 (< 5), so 423 becomes 420.
Round 29 to the nearest tens: the ones digit is 9 (≥ 5), so 29 becomes 30.
Estimated quotient = 420 ÷ 30 = 14.

(ii) 777 ÷ 27
Round 777 to the nearest tens: the ones digit is 7 (≥ 5), so 777 becomes 780.
Round 27 to the nearest tens: the ones digit is 7 (≥ 5), so 27 becomes 30.
Estimated quotient = 780 ÷ 30 = 26.
In simple words: Round each number to the nearest tens, then divide the rounded numbers.

Exam Tip: For division problems, round both the dividend and divisor first, then perform the division. Check that your estimate is reasonable by comparing it to rough mental calculations.

 

Question 9. Estimate the following quotients by rounding off each number to its nearest hundreds:
(i) 2472 ÷ 493
(ii) 7459 ÷ 286
Answer:
(i) 2472 ÷ 493
Round 2472 to the nearest hundreds: the tens digit is 7 (≥ 5), so 2472 becomes 2,500.
Round 493 to the nearest hundreds: the tens digit is 9 (≥ 5), so 493 becomes 500.
Estimated quotient = 2,500 ÷ 500 = 5.

(ii) 7459 ÷ 286
Round 7459 to the nearest hundreds: the tens digit is 5, so 7459 becomes 7,500.
Round 286 to the nearest hundreds: the tens digit is 8 (≥ 5), so 286 becomes 300.
Estimated quotient = 7,500 ÷ 300 = 25.
In simple words: Round both numbers to the nearest hundreds, then divide to find your estimate.

Exam Tip: When the tens digit is exactly 5, round up to the next hundred. Always work with the rounded figures, not the original numbers.

 

Mental Maths

 

Question 1. Fill in the blanks:
(i) The digit ... has the highest place value in the number 2309
(ii) The digit ... has the highest face value in the number 2039
(iii) The digit ... has the lowest place value in the number 2039
(iv) Both Indian and International systems of numeration have ... period in common.
(v) In the International system of numeration, commas are placed from ... after every ... digits.
(vi) The bigger number from the numbers 57,631 and 57,361 is ....
(vii) 1 crore = ... million
(viii) The smallest 4-digit number with 3 different digits is ...
(ix) The greatest 4-digit number with 3 different digits is ...
(x) 15 km 300 m = .... m
(xi) 7850 cm = ... m ... cm
(xii) The number 5079 when estimated to the nearest hundreds is ...
Answer:
(i) In 2309, the digit 2 sits in the thousands place with a place value of 2000, which is the greatest among all digits. So the answer is 2.

(ii) In 2039, the digits are 2, 0, 3, and 9. Face value means the digit's own value regardless of position. The largest digit is 9.

(iii) In 2039, the digit 0 is in the hundreds place. Since 0 × 100 = 0, this has the lowest place value.

(iv) Both the Indian and International numeration systems share the ones period in common.

(v) In the International system, commas are inserted starting from the right, with one comma placed after every 3 digits.

(vi) Comparing 57,631 and 57,361: both have 5 and 7 in the same places, but in the hundreds position, 6 > 3. Therefore, 57,631 is bigger.

(vii) 1 crore equals 10 million.

(viii) To build the smallest 4-digit number using 3 different digits, place 1 (the smallest non-zero digit) in the thousands place, then use 0, 0, and 2. This gives 1,002.

(ix) To build the greatest 4-digit number using only 3 different digits, use 9, 9, 8, and 7, which gives 9,987.

(x) 15 km 300 m = (15 × 1,000 + 300) m = 15,300 m.

(xi) 7,850 cm = 7,850 ÷ 100 m = 78 m 50 cm.

(xii) In 5079, look at the tens digit, which is 7 (≥ 5). Round up the hundreds digit from 0 to 1, and replace the tens and ones digits with 0. The result is 5,100.
In simple words: Place value tells you what a digit is worth based on where it sits. Face value is just the digit itself. Round numbers by checking the digit to the right of your target place.

Exam Tip: Memorize key conversions like 1 crore = 10 million, 1 km = 1,000 m, 1 m = 100 cm. For fill-in-the-blank questions, identify which concept is being tested (place value, face value, rounding, numeration systems) before answering.

 

Question 2. State whether the following statements are true (T) or false (F):
(i) The difference between the place value and the face of the digit 7 in the number 2701 is 693.
(ii) The smallest 4-digit number - 1 = the greatest 3-digit number.
(iii) The place of a digit is independent of whether the number is written in the Indian system or International system of numeration.
(iv) In the International system, a number having less number of digits is always smaller than the number having more number of digits.
(v) The estimated value of 9999 to the nearest tens is 10000
Answer:
(i) In 2701, the digit 7 is in the hundreds place. Place value of 7 = 7 × 100 = 700. Face value of 7 = 7. The difference is 700 - 7 = 693. Statement is True.

(ii) The smallest 4-digit number is 1,000. The greatest 3-digit number is 999. Since 1,000 - 1 = 999, the statement is True.

(iii) A digit's position in a number remains the same whether written in the Indian or International system; only the names given to higher place values change. Statement is True.

(iv) When comparing natural numbers, the one with more digits is always greater, and the one with fewer digits is always smaller. This rule applies in both systems. Statement is True.

(v) In 9999, the ones digit is 9 (≥ 5). Rounding to the nearest tens: increase the tens digit by 1 and replace the ones digit with 0. Since the tens digit is also 9, this carry continues through hundreds and thousands, yielding 10,000. Statement is True.
In simple words: Place value is what a digit is worth in its position. Face value is the digit itself. When rounding, check the digit to the right of where you want to round.

Exam Tip: For true-false questions, verify each statement by working through the calculation step-by-step rather than relying on intuition. Pay special attention to rounding rules, especially when a carry occurs across multiple places.

 

Multiple Choice Questions

 

Question 3. The face value of the digit 5 in the number 36,503 is
(a) 5
(b) 503
(c) 500
(d) None of the options
Answer: (a) 5
Face value refers to the digit itself, independent of its position in the number. The digit 5 always has a face value of 5, no matter where it appears.
In simple words: Face value is simply the digit's own value. Ignore where it sits in the number.

Exam Tip: Never confuse face value (the digit itself) with place value (what the digit is worth based on its position). This is a common source of error.

 

Question 4. The difference between the place values of 6 and 3 in 76834 is
(a) 3
(b) 5700
(c) 5930
(d) 5970
Answer: (d) 5970
In 76,834:
The digit 6 occupies the thousands place: place value = 6 × 1,000 = 6,000.
The digit 3 occupies the tens place: place value = 3 × 10 = 30.
Difference = 6,000 - 30 = 5,970.
In simple words: Find where each digit sits, calculate its place value, then subtract.

Exam Tip: Always identify the position of each digit clearly before calculating its place value. Write out the place value calculation (digit × positional value) to avoid arithmetic errors.

 

Question 5. The sum of the place values of all the digits in 5003 is
(a) 8
(b) 53
(c) 5003
(d) 8000
Answer: (c) 5003
In 5003:
Place value of 5 (thousands place) = 5 × 1,000 = 5,000.
Place value of 0 (hundreds place) = 0 × 100 = 0.
Place value of 0 (tens place) = 0 × 10 = 0.
Place value of 3 (ones place) = 3 × 1 = 3.
Sum = 5,000 + 0 + 0 + 3 = 5,003.
In simple words: Calculate the place value of each digit, then add all of them together. You get the original number itself.

Exam Tip: When finding the sum of place values of all digits, you always get back the original number. Use this as a check on your work.

 

Question 6. The total number of 4-digit numbers is
(a) 9000
(b) 9999
(c) 10000
(d) None of the options
Answer: (a) 9000
The greatest 4-digit number is 9,999.
The smallest 4-digit number is 1,000.
Total count of 4-digit numbers = 9,999 - 1,000 + 1 = 9,000.
In simple words: Count from the smallest 4-digit number to the largest, including both endpoints.

Exam Tip: When counting numbers in a range, use the formula: (largest - smallest + 1). The "+1" is crucial because you include both the starting and ending numbers.

 

Question 7. The product of the place values of two-threes in 73532 is
(a) 9000
(b) 90000
(c) 99000
(d) 1000
Answer: (b) 90000
In 73,532, there are two threes:
The first 3 (from the left) is at the thousands place: place value = 3 × 1,000 = 3,000.
The second 3 is at the tens place: place value = 3 × 10 = 30.
Product = 3,000 × 30 = 90,000.
In simple words: Find the place values of both threes, then multiply them.

Exam Tip: When a digit appears multiple times in a number, identify each occurrence separately and calculate its place value based on its specific position.

 

Question 8. The smallest 4-digit number having distinct digits is
(a) 1234
(b) 1023
(c) 1002
(d) 3210
Answer: (b) 1023
To form the smallest 4-digit number with all different digits: place the smallest non-zero digit (1) in the thousands position. Then arrange the remaining smallest available digits in ascending order: 0, 2, 3. This gives 1,023.
In simple words: Start with 1 in the thousands place (smallest non-zero), then add the next smallest unused digits: 0, 2, 3.

Exam Tip: The thousands place cannot be 0 in a 4-digit number, so start with 1. Then arrange remaining digits in ascending order to keep the number as small as possible.

 

Question 9. The largest 4-digit number having distinct digits is
(a) 9999
(b) 9867
(c) 9786
(d) 9876
Answer: (d) 9876
In simple words: To make the biggest 4-digit number where each digit is different, arrange the largest four digits (9, 8, 7, 6) from left to right in order from largest to smallest.

Exam Tip: Always arrange digits in descending order to form the largest number with distinct digits.

 

Question 10. The largest 4-digit number is
(a) 9999
(b) 9876
(c) 9990
(d) None of these
Answer: (a) 9999
In simple words: The biggest 4-digit number is made by putting the digit 9 in all four places.

Exam Tip: When there is no restriction like "distinct digits," repeat the largest digit to maximize the number's value.

 

Question 11. The difference between the largest number of 3-digit and the largest number of 3-digit with distinct digits is
(a) 0
(b) 10
(c) 12
(d) 14
Answer: (c) 12
In simple words: Subtract the biggest 3-digit number with all different digits (987) from the biggest 3-digit number overall (999). The answer is 999 - 987 = 12.

Exam Tip: Identify which number applies the distinct-digits rule and which does not before calculating the difference.

 

Question 12. If we write numbers from 1 to 100, the number of times the digit 5 has been written is
(a) 11
(b) 15
(c) 19
(d) 20
Answer: (d) 20
In simple words: Count how many times 5 shows up in the ones place (5, 15, 25, ..., 95 = 10 times) and in the tens place (50, 51, 52, ..., 59 = 10 times). Total = 20 times.

Exam Tip: Always count each position separately and watch for numbers like 55 where the digit appears twice.

 

Question 13. The number 28,549 when rounded off to the nearest hundreds is
(a) 28,000
(b) 28,500
(c) 28,600
(d) 29,000
Answer: (b) 28,500
In simple words: Look at the tens digit (4). Since 4 is less than 5, keep the hundreds digit the same (5) and put 0 in the tens and ones places. Result: 28,500.

Exam Tip: When rounding to the nearest hundred, check the tens digit - if it's less than 5, round down; if it's 5 or more, round up.

 

Question 14. The smallest natural number which when rounded off to the nearest hundreds as 500 is
(a) 499
(b) 501
(c) 450
(d) 549
Answer: (c) 450
In simple words: Any number from 450 to 549 will round to 500. The smallest one in this range is 450.

Exam Tip: To find the range of numbers that round to a specific value, work backwards - what numbers are closest to that value?

 

Question 15. The greatest natural number which when rounded off to the nearest hundreds as 500 is
(a) 549
(b) 599
(c) 450
(d) None of these
Answer: (a) 549
In simple words: Numbers from 450 to 549 all round to 500. The largest of these is 549.

Exam Tip: The upper boundary number that still rounds to 500 is 549; anything at 550 or above rounds to 600.

 

Question 16. The greatest 5-digit number formed by the digits 3, 0, 7 is
(a) 33077
(b) 77730
(c) 77330
(d) None of these
Answer: (b) 77730
In simple words: To make the biggest 5-digit number using only 3, 0, and 7 (and repeating them as needed), use the largest digit (7) as many times as possible first. Place 7 three times, then 3, then 0: 77,730.

Exam Tip: When forming the largest number with a limited set of digits and repetition allowed, prioritize larger digits in the leftmost positions.

 

Question 17. In the International place value system, we write 1 billion for
(a) 10 lakh
(b) 1 crore
(c) 10 crore
(d) 100 crore
Answer: (d) 100 crore
In simple words: In the world system (International), 1 billion = 1,000,000,000. In the Indian system, 1 crore = 1,00,00,000 and 100 crore = 1,00,00,00,000. So 1 billion equals 100 crore.

Exam Tip: Learn the conversions between International and Indian place value systems - 1 billion = 100 crore = 10 million is a key relationship.

 

Question 18. Statement I: The place value of 6 in the numbers 126 and 621 is different. Statement II: The place value of a non-zero digit depends upon the place it occupies in the given number.
(a) Statement I is true but statement II is false.
(b) Statement I is false but statement II is true.
(c) Both Statement I and statement II are true.
(d) Both Statement I and statement II are false.
Answer: (c) Both Statement I and statement II are true.
In simple words: In 126, the digit 6 is in the ones place and has a value of 6. In 621, the same digit 6 is in the hundreds place and has a value of 600. These are different, so Statement I is true. Statement II is also true because where a digit sits determines how much it is worth.

Exam Tip: Always verify both statements separately before selecting your answer - do not assume one is false just because the other is true.

 

Question 19. Statement I: Both the face value and the place value of 1 in 531 is 1. Statement II: The face value of a digit in a number is the digit itself.
(a) Statement I is true but statement II is false.
(b) Statement I is false but statement II is true.
(c) Both Statement I and statement II are true.
(d) Both Statement I and statement II are false.
Answer: (c) Both Statement I and statement II are true.
In simple words: The digit 1 in 531 is at the ones place. Its face value is 1, and its place value is also 1 (1 × 1 = 1), so Statement I is correct. The face value of any digit is always the digit itself, no matter where it appears in a number, so Statement II is also correct.

Exam Tip: Remember: face value never changes, but place value depends on position - the ones place always has a place value equal to the digit itself.

 

Question 20. Statement I: 4560 < 1234 Statement II: Given two numbers, the number having more digits is lesser.
(a) Statement I is true but statement II is false.
(b) Statement I is false but statement II is true.
(c) Both Statement I and statement II are true.
(d) Both Statement I and statement II are false.
Answer: (d) Both Statement I and statement II are false.
In simple words: Both 4560 and 1234 have 4 digits. Comparing their leftmost digits: 4 is bigger than 1, so 4560 is greater than 1234, not smaller. Statement I is false. Statement II is also false because when comparing two numbers, the one with more digits is actually greater, not lesser.

Exam Tip: To compare numbers with the same digit count, start from the leftmost digit and work right - the first difference tells you which is larger.

 

Question 21. Statement I: The largest five-digit number is 99999. Statement II: Ten million = one lakh
(a) Statement I is true but statement II is false.
(b) Statement I is false but statement II is true.
(c) Both Statement I and statement II are true.
(d) Both Statement I and statement II are false.
Answer: (a) Statement I is true but statement II is false.
In simple words: The biggest 5-digit number is formed by putting 9 in all five places: 99,999. So Statement I is correct. For Statement II, ten million in the International system is 10,000,000. In the Indian system, this equals 1 crore (1,00,00,000), not 1 lakh (1,00,000). So Statement II is wrong.

Exam Tip: Keep the International and Indian numbering systems separate - ten million is not the same as one lakh; it equals one crore.

 

Question 22. Statement I: The estimated value of 2132 + 1234 by estimating the numbers to the nearest hundred = 3300. Statement II: Sum of two natural numbers is always a natural number.
(a) Statement I is true but statement II is false.
(b) Statement I is false but statement II is true.
(c) Both Statement I and statement II are true.
(d) Both Statement I and statement II are false.
Answer: (c) Both Statement I and statement II are true.
In simple words: Round 2132 to 2,100 (tens digit 3 is less than 5) and 1234 to 1,200 (tens digit 3 is less than 5). Their sum is 2,100 + 1,200 = 3,300, so Statement I is true. When you add any two natural numbers, you always get another natural number, so Statement II is also true.

Exam Tip: When estimating, always round to the specified place value and then perform the operation - do not operate first and then round.

 

Question 1. Write the numeral for each of the following numbers and insert commas correctly:
(i) Six crore nine lakh forty seven.
(ii) One hundred four million seven hundred twenty two thousand three hundred ninety four.
Answer:
(i) Six crore nine lakh forty seven consists of 6 crores, 9 lakhs, 0 thousands, 0 hundreds, 4 tens and 7 ones.
In figures (Indian system): 6,09,00,047

(ii) One hundred four million seven hundred twenty two thousand three hundred ninety four consists of 104 millions, 722 thousands and 394 ones.
In figures (International system): 104,722,394
In simple words: Break the number into its parts - crores/millions, lakhs/thousands, and ones - then write each part in the correct position with commas following the system you are using.

Exam Tip: Always identify which system (Indian or International) you are working with first, then place commas accordingly - the Indian system uses commas at 2-digit intervals after the initial period, while the International system uses them every 3 digits from the right.

 

Question 2. Insert commas suitably and write the number 30189301 in words in Indian and International system of numeration.
Answer:
In the Indian system: 3,01,89,301
In words (Indian system): Three crore one lakh eighty nine thousand three hundred one.

In the International system: 30,189,301
In words (International system): Thirty million one hundred eighty nine thousand three hundred one.
In simple words: Take the same number and break it into groups using the rules of each system - Indian uses 2-2-2-3 digit grouping from right to left, while International uses 3-3-3 digit grouping.

Exam Tip: Write out the place value names separately for each system to avoid mixing them - "crore" and "lakh" are Indian system terms, while "million" is International system terminology.

 

Question 3. Find the difference between the place value and the face value of the digit 6 in the number 72601
Answer: In the number 72601, the digit 6 occupies the hundreds place. Its place value is 6 × 100 = 600. The face value of 6 is simply 6. The difference between place value and face value is 600 - 6 = 594.
In simple words: Find where the digit sits (hundreds place), multiply it by that place's value (6 × 100 = 600), then subtract the digit itself (600 - 6 = 594).

Exam Tip: Always identify the correct place value of the digit by counting from right to left - the hundreds place is the third position from the right.

 

Question 4. Write all possible two-digit numbers using the digits 4 and 0. Repetition of digits is allowed.
Answer: To form two-digit numbers using 4 and 0 where digits may repeat, we must remember that 0 cannot occupy the tens place, as that would give us a one-digit number instead. If we place 4 in the tens position, we can fill the ones place with either 4 or 0, making our numbers 44 and 40. Therefore, all possible two-digit numbers are 40 and 44.
In simple words: Zero cannot be in the tens place of a two-digit number. So we can only use 4 in the tens place, giving us 40 and 44.

Exam Tip: Remember that zero in the leftmost position makes a number smaller than it should be — always check this constraint when forming multi-digit numbers with zero.

 

Question 5. Write all possible natural numbers using the digits 7, 0, 6. Repetition of digits is not allowed.
Answer: Since we cannot repeat digits, we must use each of 7, 0, and 6 at most once. Start with one-digit natural numbers: 6 and 7 qualify (0 is not considered a natural number). For two-digit numbers, 0 cannot be in the tens spot. Placing 6 in the tens spot gives 60 and 67. Placing 7 in the tens spot gives 70 and 76. For three-digit numbers, again 0 cannot be in the hundreds position. With 6 in the hundreds place: 607 and 670. With 7 in the hundreds place: 706 and 760. Therefore, the complete list is 6, 7, 60, 67, 70, 76, 607, 670, 706, and 760.
In simple words: List one-digit numbers first (6, 7), then two-digit numbers avoiding 0 in the front (60, 67, 70, 76), then three-digit numbers avoiding 0 in front (607, 670, 706, 760).

Exam Tip: When listing numbers without repeating digits, organize by number of digits and ensure the leading position never contains 0.

 

Question 6. Arrange the following numbers in ascending order: 3706, 58019, 3760, 59801, 560023
Answer: First, group the numbers by their digit count. The numbers 3706 and 3760 each have 4 digits. The numbers 58019 and 59801 each have 5 digits. The number 560023 has 6 digits. Since 4-digit numbers are smaller than 5-digit numbers, which are smaller than 6-digit numbers, we begin with the 4-digit numbers. Comparing 3706 and 3760: the thousands and hundreds digits match (3 and 7), but the tens digit differs (0 vs 6), so 3706 comes first. Next, compare the 5-digit numbers: 58019 and 59801 both start with 5, but the thousands digit is smaller in 58019 (8 vs 9), so 58019 comes before 59801. Thus the order is 3706, 3760, 58019, 59801, 560023.
In simple words: Count the digits first. Fewer digits means a smaller number. If digit counts match, compare digit by digit from left to right.

Exam Tip: Always count digits as the primary sorting criterion — it is the quickest way to organize numbers, especially when dealing with large numbers.

 

Question 7. Write the greatest six-digit number using four different digits.
Answer: To maximize a six-digit number with only four distinct digits, place the largest possible digits in the highest place values. The digit 9 should appear as often as allowed. We need exactly four different digits across six positions, so use 9 three times. The remaining three positions should hold the next-largest available digits, which are 8, 7, and 6 — each used once. Our complete digit set is 9, 9, 9, 8, 7, 6. Arrange these in descending order to get the largest number: 999876. Therefore, the greatest six-digit number using four different digits is 999,876.
In simple words: Use the biggest digit (9) as many times as you can, then fill remaining spots with the next biggest available digits in descending order.

Exam Tip: For "greatest" numbers, always arrange digits in descending order; for "smallest," descending order applies after placing the smallest non-zero digit first.

 

Question 8. Write the smallest eight-digit number using four different digits.
Answer: For the smallest eight-digit number using exactly four different digits, the leftmost position (crores) cannot be 0 — it must be the smallest non-zero digit, which is 1. Fill as many remaining positions as possible with 0. We need exactly four distinct digits, so after using 1 and 0, we need two more. The next-smallest available digits are 2 and 3. Our complete digit set is 1, 0, 0, 0, 0, 0, 2, 3. Arrange these with 1 at the front and the rest in ascending order: 10000023. Therefore, the smallest eight-digit number using four different digits is 1,00,00,023.
In simple words: Place 1 first (cannot use 0). Fill middle positions with as many 0s as needed. Add the next-smallest digits at the end.

Exam Tip: For "smallest" numbers, use 0 repeatedly after placing the smallest non-zero digit first, ensuring you meet the constraint on distinct digits.

 

Question 9. Find the difference between the greatest and the smallest 4-digit numbers formed by the digits 0, 3, 6, 9
Answer: Each digit (0, 3, 6, 9) must be used exactly once. For the greatest 4-digit number, arrange in descending order: 9630. For the smallest 4-digit number, the thousands place cannot be 0, so place the smallest non-zero digit (3) there, then arrange the remaining digits (0, 6, 9) in ascending order to get 3069. The difference is 9630 - 3069 = 6561. Therefore, the difference between the greatest and smallest 4-digit numbers is 6,561.
In simple words: Make the largest number by putting digits in descending order. Make the smallest by putting the smallest non-zero digit first, then the rest in ascending order. Subtract.

Exam Tip: Always verify that your "smallest" number doesn't start with 0 — rearrange to place the smallest non-zero digit first.

 

Question 10. Find the sum of the greatest and the smallest 6-digit numbers formed by the digits 2, 0, 4, 7, 6, 5 using each digit only once.
Answer: Each digit (2, 0, 4, 7, 6, 5) must be used once. For the greatest 6-digit number, arrange in descending order: 765420. For the smallest 6-digit number, the lakh's place (leftmost) cannot be 0, so place the smallest non-zero digit (2) first, then arrange the remaining digits (0, 4, 5, 6, 7) in ascending order: 204567. The sum is 765420 + 204567 = 969987. Therefore, the sum of the greatest and smallest 6-digit numbers is 9,69,987.
In simple words: For the largest: arrange all digits from biggest to smallest. For the smallest: put the smallest non-zero digit first, then arrange the rest from smallest to biggest. Add them.

Exam Tip: Double-check your addition when the sum involves rearrangement of many digits — a single arithmetic error changes your final answer.

 

Question 11. Find the sum of the four-digit greatest number and the five-digit smallest number, each number having three different digits.
Answer: For the greatest 4-digit number with three different digits, use 9 twice (the largest digit allowed to repeat), then add the next two largest digits, 8 and 7, each once: 9987. For the smallest 5-digit number with three different digits, place 1 (smallest non-zero) in the ten-thousands position. Fill the remaining four positions with 0 as much as possible. To have exactly three different digits total, add one more digit — the smallest available is 2. This gives 10002. The sum is 9987 + 10002 = 19989. Therefore, the required sum is 19,989.
In simple words: For the 4-digit greatest: use 9 twice, then 8 and 7 once each. For the 5-digit smallest: start with 1, fill with 0s, then add 2 at the end. Add them together.

Exam Tip: When a problem specifies "different digits" (meaning distinct values), count carefully to ensure you meet the requirement without over-using or under-using any digit.

 

Question 12. Write the greatest and the smallest four-digit numbers using four different digits with the conditions:
(i) Digit 3 always at hundred's place.
(ii) Digit 0 always at ten's place.
Answer: (i) Digit 3 always at hundred's place. The number has the form _3__. For the greatest 4-digit number, fill the remaining positions with the largest available digits: 9 in the thousands place, 8 in the tens place, and 7 in the units place, giving 9387. For the smallest 4-digit number, place 1 (smallest non-zero, different from 3) in the thousands position, 0 in the tens, and 2 in the units, giving 1302. Thus the greatest is 9,387 and the smallest is 1,302.
(ii) Digit 0 always at ten's place. The number has the form __0_. For the greatest 4-digit number, place 9 in the thousands, 8 in the hundreds, and 7 in the units: 9807. For the smallest 4-digit number, place 1 (smallest non-zero) in the thousands, 2 in the hundreds, and 3 in the units: 1203. Thus the greatest is 9,807 and the smallest is 1,203.
In simple words: (i) With 3 fixed in the hundreds place, put 9, 8, 7 in other spots for greatest; put 1, 0, 2 for smallest. (ii) With 0 fixed in the tens place, put 9, 8, 7 in other spots for greatest; put 1, 2, 3 for smallest.

Exam Tip: When digits are fixed in certain positions, treat those positions as "locked" and arrange only the free positions to maximize or minimize the number.

 

Question 13. Find the greatest and the smallest 5-digit numbers using different digits in which 5 appears at ten's place.
Answer: With 5 fixed in the tens position, the number has the form ___5_. All five digits must be different. For the greatest 5-digit number, fill the remaining positions with the largest available digits: 9 in the ten-thousands, 8 in the thousands, 7 in the hundreds, and 6 in the units, giving 98756. For the smallest 5-digit number, place 1 (smallest non-zero, different from 5) in the ten-thousands, 0 in the thousands, 2 in the hundreds, 5 in the tens (fixed), and 3 in the units: 10253. Therefore, the greatest is 98,756 and the smallest is 10,253.
In simple words: With 5 locked in the tens place, use 9, 8, 7, 6 for the greatest. For the smallest, use 1 first (not 0), then 0, 2, 5, 3.

Exam Tip: Always check the leftmost position for the number 0 — if it appears there, rearrange by placing the smallest non-zero digit first.

 

Question 14. A mobile number consists of ten digits. First four digits are 9, 9, 7 and 9. Make the smallest mobile number by using only one digit twice from the digits 8, 3, 5, 0, 6
Answer: The first four positions are fixed as 9, 9, 7, 9. The remaining six positions must be filled using the digits 8, 3, 5, 0, 6, with exactly one digit appearing twice. To create the smallest mobile number, use the smallest digit twice: 0 appears twice. The remaining digits are 3, 5, 6, 8. The six digits to arrange are 0, 0, 3, 5, 6, 8. Arrange these in ascending order: 0, 0, 3, 5, 6, 8. The complete mobile number is 9979003568. Therefore, the smallest mobile number is 9979003568.
In simple words: The first four digits are fixed. For the remaining six spots, use 0 twice and 3, 5, 6, 8 once each. Arrange all six in order from smallest to largest.

Exam Tip: When making the "smallest" arrangement, always use the smallest digit available for repetition and arrange all remaining digits in ascending order.

 

Question 15. To stitch a uniform, 1 m 75 cm cloth is needed. Out of 153 m cloth, how many uniforms can be stitched and how much cloth will remain?
Answer: First, convert all measurements to the same unit. The available cloth is 153 m = 15,300 cm. Each uniform requires 1 m 75 cm = 175 cm. Divide the total cloth by the cloth per uniform: 15,300 ÷ 175. This division gives a quotient of 87 and a remainder of 75. Therefore, 87 uniforms can be stitched from the available cloth, and 75 cm of cloth will remain unused.
In simple words: Change meters to centimeters so all numbers are the same. Divide total cloth by cloth needed for one uniform. The quotient is how many uniforms; the remainder is leftover cloth.

Exam Tip: Always convert to a common unit before dividing — mixing units leads to incorrect answers. Clearly state both the quotient (number of items) and remainder (leftover material).

 

Question 16. Medicine is packed in boxes, each weighing 4 kg 500 g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Answer: Convert all masses to the same unit. Each box weighs 4 kg 500 g = 4,500 g. The van's maximum capacity is 800 kg = 8,00,000 g. Divide the maximum capacity by the mass per box: 8,00,000 ÷ 4,500. This division yields a quotient of 177 and a remainder of 3,500. Since 178 boxes would weigh 178 × 4,500 = 8,01,000 g = 801 kg, which exceeds the van's 800 kg limit, the maximum safe number of boxes is 177. Therefore, 177 boxes can be loaded in the van.
In simple words: Convert to the same unit (grams or kilograms). Divide the van's capacity by the weight of one box. The quotient tells you how many boxes fit safely.

Exam Tip: When the remainder is significant, check whether the next whole number would exceed the limit — this determines if you round down.

 

Question 17. Estimate: 6554 - 677 by estimating the numbers to their nearest
(i) thousands
(ii) hundreds
(iii) greatest places
Also point out the most reasonable estimate.
Answer: (i) Estimating to nearest thousands: For 6554, the hundreds digit is 5, so round up to 7,000. For 677, the hundreds digit is 6 (which is ≥ 5), so round up to 1,000. Estimated difference: 7,000 - 1,000 = 6,000.
(ii) Estimating to nearest hundreds: For 6554, the tens digit is 5, so round up to 6,600. For 677, the tens digit is 7 (which is ≥ 5), so round up to 700. Estimated difference: 6,600 - 700 = 5,900.
(iii) Estimating to greatest places: The greatest place of 6554 is thousands, so round 6554 to 7,000 (using its hundreds digit 5). The greatest place of 677 is hundreds, so round 677 to 700 (using its tens digit 7). Estimated difference: 7,000 - 700 = 6,300. The actual difference is 6,554 - 677 = 5,877. Comparing estimates: 5,900 is closest to the actual value 5,877, making estimation to nearest hundreds the most reasonable.
In simple words: Round each number separately, then subtract. The "hundreds" estimate (5,900) comes closest to the true answer (5,877).

Exam Tip: After calculating estimates, always compare them with the actual result to identify which rounding method gives the most accurate approximation — this teaches you which level of rounding works best for different situations.

 

Question 18. Write all 4-digit numbers that can be formed with the digits 2 and 5, using both digits equal number of time. Also find their sum.
Answer: To create 4-digit numbers using only the digits 2 and 5 where each appears the same number of times, we must use each digit twice. The digits we need to arrange are 2, 2, 5 and 5.

The possible 4-digit numbers are:
2255, 2525, 2552, 5225, 5252, 5522

To find the sum:
2255 + 2525 + 2552 + 5225 + 5252 + 5522 = 23,331

Therefore, the required 4-digit numbers are 2255, 2525, 2552, 5225, 5252 and 5522, with their sum being 23,331.
In simple words: Take the digits 2 and 2 and 5 and 5. Mix them to make different 4-digit numbers. Then add all those numbers together to get 23,331.

Exam Tip: List all arrangements systematically by fixing one digit in the first position and varying the rest. Always verify your final sum by adding carefully.

 

Question 19. What is the difference between the smallest 6-digit number with five different digits and the greatest 5-digit number with four different digits?
Answer: Finding the smallest 6-digit number with five different digits:

The lakh's place cannot be 0, so we place 1 (the smallest non-zero digit) there. To get exactly five different digits, we need three more digits besides 1 and 0. The smallest choices are 2, 3 and 4.

Our six digits are: 1, 0, 0, 2, 3, 4 (which gives us five different digits: 1, 0, 2, 3, 4).

Arranging with 1 at the lakh's place and the remaining digits in ascending order:
Smallest 6-digit number = 1,00,234

Finding the greatest 5-digit number with four different digits:

To make the largest number, we use 9 as many times as we can. Using 9 twice, the next three largest different digits are 8, 7 and 6.

Our five digits are: 9, 9, 8, 7, 6 (which gives us four different digits: 9, 8, 7, 6).

Arranging in descending order:
Greatest 5-digit number = 99,876

The difference is:
1,00,234 - 99,876 = 358

Therefore, the required difference is 358.
In simple words: The smallest 6-digit number using 5 different digits is 1,00,234. The biggest 5-digit number using 4 different digits is 99,876. When we take away one from the other, we get 358.

Exam Tip: For smallest numbers, always put the smallest non-zero digit at the leftmost place, then fill remaining positions with 0. For largest numbers, fill leftmost positions with 9.

 

Question 20. Write the smallest 7-digit number using all the even digits.
Answer: The even digits are 0, 2, 4, 6 and 8 (five different digits in total).

To form a 7-digit number that uses all five even digits (each at least once), we need two extra positions. These can be filled by repeating any even digit.

The digit at the ten lakhs place (highest place) cannot be 0. Therefore, we place the smallest non-zero even digit, which is 2, at the highest place.

To make the number as small as possible, we fill as many remaining positions as we can with 0 (the smallest digit).

Our seven digits are: 2, 0, 0, 0, 4, 6, 8 (which uses all five even digits: 0, 2, 4, 6, 8).

Arranging with 2 at the highest place and the rest in ascending order:
Smallest 7-digit number = 20,00,468

Therefore, the smallest 7-digit number using all the even digits is 20,00,468.
In simple words: The even digits are 0, 2, 4, 6, and 8. Put 2 first, then fill the spaces with three 0s, then put 4, 6, and 8 to get the smallest 7-digit number: 20,00,468.

Exam Tip: Remember that 0 cannot be the first digit of a number. Always place the smallest non-zero digit first when forming the smallest number with a given set of digits.

 

Question 21. How many times does the digit 3 occur at ten's place in natural numbers from 100 to 1000?
Answer: We need to count how many natural numbers between 100 and 1000 have 3 in the ten's place.

Such numbers have the form _3_ (where the first blank is the hundred's place and the second blank is the unit's place). Note that 1000 itself has 0 at its ten's place, so it is not included.

For the hundred's place, the digit can be any of: 1, 2, 3, 4, 5, 6, 7, 8, or 9. This gives us 9 choices.

For the ten's place, the digit is fixed as 3. This gives us 1 choice.

For the unit's place, the digit can be any of: 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. This gives us 10 choices.

Using the counting principle:
Total such numbers = 9 × 1 × 10 = 90

Therefore, the digit 3 occurs at the ten's place 90 times in natural numbers from 100 to 1000.
In simple words: We make 3-digit numbers where 3 must be in the middle (ten's place). The first digit can be 1-9, the middle is always 3, and the last can be 0-9. That gives us 9 × 1 × 10 = 90 numbers.

Exam Tip: Use the counting principle by finding how many choices exist for each position independently, then multiply them together. This method is faster than listing all numbers.