CBSE Class 6 Mathematics Revision Assignment 3

Question 1. Place commas at the correct places and also write the numbers in words : a) 86391 ; b) 903512 ; c) 2854709
Answer: Using the Indian numbering system:
a) 86,391 = Eighty-six thousand three hundred ninety-one
b) 9,03,512 = Nine lakh three thousand five hundred twelve
c) 28,54,709 = Twenty-eight lakh fifty-four thousand seven hundred nine
In simple words: Commas help us read large numbers easily by grouping digits into periods like thousands and lakhs.

Exam Tip: In the Indian system, the first comma is after three digits from the right, and thereafter after every two digits.

Question 2. Rewrite the following in descending order : a) 97,654 ; 98,000 ; 96,999 ; b) 10,56,783 ; 9,99,999 ; 12,00,101
Answer: Descending order means arranging numbers from the largest to the smallest.
a) Comparing 97,654, 98,000, and 96,999:
98,000 > 97,654 > 96,999
b) Comparing 10,56,783, 9,99,999, and 12,00,101:
12,00,101 > 10,56,783 > 9,99,999
In simple words: Descending order is stepping down from the biggest number to the smallest number.

Exam Tip: Always compare digits from the leftmost place value to determine which number is larger.

Question 3. Write the place value and face value of the underlined digits : a) 37\(\underline{2}\)814 ; b) 5\(\underline{6}\)082431
Answer: a) In 372,814, the underlined digit is 2.
Face value of 2 = 2
Place value of 2 = \(2 \times 1000 = 2000\)
b) In 56,082,431, the underlined digit is 6.
Face value of 6 = 6
Place value of 6 = \(6 \times 10,00,000 = 60,00,000\) (or Sixty Lakh)
In simple words: Face value is the digit itself, and place value is the digit multiplied by its position value.

Exam Tip: Face value never changes regardless of the digit's position, but place value depends entirely on its position.

Question 4. Use all the digits only once to make the smallest and largest 6 digit number : 3, 7, 2, 0, 6, 4
Answer: Given digits: 3, 7, 2, 0, 6, 4
To form the largest 6-digit number, arrange the digits in descending order: 7, 6, 4, 3, 2, 0
Largest number = 7,64,320
To form the smallest 6-digit number, arrange the digits in ascending order without putting 0 at the extreme left: 2, 0, 3, 4, 6, 7
Smallest number = 2,03,467
In simple words: Big numbers start with the largest digits, and small numbers start with the smallest non-zero digit.

Exam Tip: Never place 0 as the first digit on the far left when forming the smallest number, as it makes the number a 5-digit number instead of 6.

Question 5. Find the sum of the smallest 4 digit number and the largest 6 digit number.
Answer: Smallest 4-digit number = 1,000
Largest 6-digit number = 9,99,999
Sum = \(9,99,999 + 1,000 = 10,00,999\)
In simple words: Add one thousand to nine lakh ninety-nine thousand nine hundred ninety-nine.

Exam Tip: Smallest n-digit number is 1 followed by \(n-1\) zeros; largest n-digit number is n nines.

Question 6. Write the following numerals in Indian place value chart : a) 246813579 ; b) 24136857099
Answer: a) 246,813,579 in Indian system with commas: 24,68,13,579 (Twenty-four crore sixty-eight lakh thirteen thousand five hundred seventy-nine)
b) 24136857099 in Indian system with commas: 24,13,68,57,099 (Two thousand four hundred thirteen crore sixty-eight lakh fifty-seven thousand ninety-nine)
In simple words: Group digits into periods of 3 (hundreds) and then groups of 2 from right to left.

Exam Tip: Indian periods are Ones (3 digits), Thousands (2 digits), Lakhs (2 digits), Crores (2 digits or more).

Question 7. Write the following numerals in International place value chart : a) 2453190876 ; b) 5400958203
Answer: a) 2,453,190,876 = Two billion four hundred fifty-three million one hundred ninety thousand eight hundred seventy-six
b) 5,400,958,203 = Five billion four hundred million nine hundred fifty-eight thousand two hundred three
In simple words: In the International system, we put commas after every three digits from the right.

Exam Tip: International periods consist of groups of three: Ones, Thousands, Millions, Billions.

Question 8. Write the numerals for the following : a) Four crore twenty-eight lakh three hundred and six ; b) Seventy-nine lakh and eight
Answer: a) Four crore twenty-eight lakh three hundred and six = 4,28,00,306
b) Seventy-nine lakh and eight = 79,00,008
In simple words: Fill missing place values with zeros carefully according to the Indian periods.

Exam Tip: Check every period (Crore, Lakh, Thousand, Ones) to ensure no place value is missed.

Question 9. Round off the following numbers to the nearest thousand : a) 7654 ; b) 98177 ; c) 120841
Answer: a) 7,654: The digit in the hundreds place is 6 (which is \(\geq 5\)), so round up.
Rounded value = 8,000
b) 98,177: The digit in the hundreds place is 1 (which is \(< 5\)), so round down.
Rounded value = 98,000
c) 120,841: The digit in the hundreds place is 8 (which is \(\geq 5\)), so round up.
Rounded value = 121,000
In simple words: Look at the hundreds digit. If it is 5 or more, increase the thousand digit by 1. If less, keep it the same.

Exam Tip: Always examine the digit one place to the right of the place you are rounding to.

Question 10. Estimate the difference between 74582 and 65382 by rounding them off to the nearest thousand, and compare it with the actual difference.
Answer: Rounding 74,582 to the nearest thousand = 75,000 (since hundreds digit 5 is \(\geq 5\))
Rounding 65,382 to the nearest thousand = 65,000 (since hundreds digit 3 is \(< 5\))
Estimated difference = \(75,000 - 65,000 = 10,000\)
Actual difference = \(74,582 - 65,382 = 9,200\)
Comparison: Estimated difference (10,000) is close to the actual difference (9,200).
In simple words: Round both numbers first, subtract them to get the estimate, then subtract the original numbers for the actual value.

Exam Tip: Always perform rounding before doing operations for estimation problems.

Question 11. Estimate 56 X 77 by rounding off the first number up and the second number down.
Answer: First number = 56. Rounding up to the nearest ten = 60
Second number = 77. Rounding down to the nearest ten = 70
Estimated product = \(60 \times 70 = 4,200\)
In simple words: Change 56 to 60 and 77 to 70 as instructed, then multiply.

Exam Tip: Carefully follow specific instructions like "rounding up" or "rounding down" instead of standard rounding rules.

Question 12. Estimate the quotient in the following : a) 58 \(\div\) 26 ; b) 4800 \(\div\) 97
Answer: a) \(58 \div 26\):
Rounding 58 to nearest ten = 60
Rounding 26 to nearest ten = 30
Estimated quotient = \(60 \div 30 = 2\)
b) \(4800 \div 97\):
Rounding 4800 to nearest hundred = 4800
Rounding 97 to nearest hundred = 100
Estimated quotient = \(4800 \div 100 = 48\)
In simple words: Round numbers to compatible values that divide easily.

Exam Tip: Choose numbers that form simple division facts when estimating quotients.

Question 13. In a town there are 4352 women, 5821 men, and 3670 children. Estimate the town's population by rounding off the numbers to the nearest hundred.
Answer: Number of women = 4,352. Rounded to nearest hundred = 4,400
Number of men = 5,821. Rounded to nearest hundred = 5,800
Number of children = 3,670. Rounded to nearest hundred = 3,700
Estimated total population = \(4,400 + 5,800 + 3,700 = 13,900\)
In simple words: Round each group to the nearest hundred first, then add them up.

Exam Tip: Look at the tens digit when rounding to the nearest hundred.

Question 14. Simplify : a) 538 X 76 ; b) 149270 \(\div\) 46
Answer: a) \(538 \times 76\):
\(538 \times 6 = 3,228\)
\(538 \times 70 = 37,660\)
\(3,228 + 37,660 = 40,888\)
b) \(149270 \div 46\):
\(149270 \div 46 = 3,245\) (Since \(46 \times 3245 = 149270\))
In simple words: Multiply step by step using standard multiplication, or perform long division carefully.

Exam Tip: Double-check multiplication and division steps to avoid calculation slips.

Question 15. Find the L.C.M. of the following : a) 12, 14, 16 ; b) 3, 12, 15
Answer: a) L.C.M. of 12, 14, and 16 by prime factorization:
\(12 = 2^2 \times 3\)
\(14 = 2 \times 7\)
\(16 = 2^4\)
L.C.M. = highest powers of all prime factors = \(2^4 \times 3 \times 7 = 16 \times 3 \times 7 = 48 \times 7 = 336\)
b) L.C.M. of 3, 12, and 15:
\(3 = 3\)
\(12 = 2^2 \times 3\)
\(15 = 3 \times 5\)
L.C.M. = \(2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60\)
In simple words: LCM is the smallest common multiple found by taking the highest exponent of each prime factor.

Exam Tip: Use prime factorization method to find LCM quickly and accurately.

Question 16. Find the H.C.F. of the following by finding the factors : a) 18, 27 ; b) 12, 36, 64
Answer: a) Factors of 18 = 1, 2, 3, 6, 9, 18
Factors of 27 = 1, 3, 9, 27
Common factors = 1, 3, 9
H.C.F. of 18 and 27 = 9
b) Factors of 12 = 1, 2, 3, 4, 6, 12
Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36
Factors of 64 = 1, 2, 4, 8, 16, 32, 64
Common factors = 1, 2, 4
H.C.F. of 12, 36, and 64 = 4
In simple words: List all factors of each number and pick the largest number that appears in all lists.

Exam Tip: Ensure all factor pairs are listed correctly so no common factor is missed.

Question 17. Find the H.C.F. of the following by long division method : a) 24, 36, 60 ; b) 12, 16, 20
Answer: a) For 24, 36, and 60:
First, find H.C.F. of 24 and 36:
\(36 \div 24\) gives quotient 1, remainder 12.
\(24 \div 12\) gives quotient 2, remainder 0.
So H.C.F. of 24 and 36 is 12.
Now find H.C.F. of 12 and 60:
\(60 \div 12\) gives quotient 5, remainder 0.
H.C.F. of 24, 36, and 60 = 12.
b) For 12, 16, and 20:
First, find H.C.F. of 12 and 16:
\(16 \div 12\) gives quotient 1, remainder 4.
\(12 \div 4\) gives quotient 3, remainder 0.
So H.C.F. of 12 and 16 is 4.
Now find H.C.F. of 4 and 20:
\(20 \div 4\) gives quotient 5, remainder 0.
H.C.F. of 12, 16, and 20 = 4.
In simple words: Divide the larger number by the smaller, then divide the divisor by the remainder until the remainder is 0.

Exam Tip: For three numbers, find the HCF of any two first, then find the HCF of that result and the third number.

Question 18. The H.C.F. and L.C.M. of two numbers are 16 and 672. If one of the numbers is 112, find the other.
Answer: We know that for any two numbers:
\(\text{First Number} \times \text{Second Number} = \text{H.C.F.} \times \text{L.C.M.}\)
\(112 \times \text{Second Number} = 16 \times 672\)
\(\text{Second Number} = \frac{16 \times 672}{112}\)
Simplify \(\frac{672}{112} = 6\)
Second Number = \(16 \times 6 = 96\)
In simple words: The product of two numbers always equals the product of their HCF and LCM.

Exam Tip: Memorize the relation Product of numbers = HCF \(\times\) LCM.

Question 19. In a hall 594 seats are arranged in 33 rows. How many seats are there in each row ?
Answer: Total number of seats = 594
Total number of rows = 33
Number of seats in each row = \(594 \div 33 = 18\)
In simple words: Divide total seats equally among all rows to find how many fit in one row.

Exam Tip: Division is used when equal grouping or sharing is required.

Question 20. Vikram sold 22 books every day, for 18 days. Tanya sold 19 books every day for 19 days. Who sold more books ?
Answer: Books sold by Vikram = \(22 \times 18 = 396\) books
Books sold by Tanya = \(19 \times 19 = 361\) books
Comparing the totals: 396 > 361
Therefore, Vikram sold more books.
In simple words: Multiply daily sales by number of days for each person, then compare the totals.

Exam Tip: Compute both totals completely before making a comparison conclusion.