CBSE Class 6 Mathematics Revision Worksheet Set 02

Revision papers for class VI

Question 1. Fill in the blanks;
a. 1 lakh = -------- ten thousand
b. 1 million = ------- hundred thousand
c. 1 Crore = -------------- million
d. 1 million = ------------ 1 lakh
Answer: a. 1 lakh = 10 ten thousand
b. 1 million = 10 hundred thousand
c. 1 Crore = 10 million
d. 1 million = 10 lakh
In simple words: Each higher place value in the number system is 10 times the preceding place value.

Exam Tip: Memorize the conversion factors between Indian and International numbering systems to solve rapidly.

Question 2. Add the difference of the smallest 4 digit number and the smallest 3 digit number to the largest 2 digit number.
Answer: Smallest 4-digit number = 1000
Smallest 3-digit number = 100
Difference = \(1000 - 100 = 900\)
Largest 2-digit number = 99
Sum = \(900 + 99 = 999\)
In simple words: First find the difference between 1000 and 100, then add 99 to that result.

Exam Tip: Carefully identify place values for smallest and largest numbers before performing operations.

Question 3. Write the following in Roman Numerals
a. 1990 b. 687 c. 1089
Answer: a. 1990 = 1000 + (1000 - 100) + 90 = M + CM + XC = MCMXC
b. 687 = 500 + 100 + 100 + 50 + 10 + 10 + 5 + 2 = DCLXXXVII
c. 1089 = 1000 + 50 + 10 + 10 + 10 + (10 - 1) = MLXXXIX
In simple words: Break each number into thousands, hundreds, tens, and units, then convert using Roman symbols.

Exam Tip: Use subtraction rules for numbers like 90 (XC) and 9 (IX) to keep Roman numerals concise.

Question 4. Write the following in Indo Arabic numerals.
a. CDXXXVII b. MCM c. MCMIX
Answer: a. CDXXXVII = 400 + 30 + 7 = 437
b. MCM = 1000 + (1000 - 100) = 1900
c. MCMIX = 1000 + (1000 - 100) + (10 - 1) = 1909
In simple words: Add the values of symbols, remembering that a smaller symbol before a larger one means subtraction.

Exam Tip: Group Roman letters into subtractive and additive pairs before calculating the final value.

Question 5. Place the commas correctly and write the numerals
a. Seventy three lakh seventy five thousand three hundred and seven.
b. Nine crore five lakh forty one.
c. Fifty eight million four hundred twenty three thousand two hundred and two.
Answer: a. 73,75,307
b. 9,05,00,041
c. 58,423,202
In simple words: Insert commas according to Indian (Lakhs, Crores) or International (Millions) period rules.

Exam Tip: Fill missing place values with zeros carefully, especially for crore and lakh periods.

Question 6. In the sums given below, round off the numbers to the nearest hundred and find the estimated answer. Also find the actual answer.
a. 7923 + 3021
b. 8865 - 2360
c. 207 \times 111
d. 4775 \div 25
e. 2350 \times 23
Answer: a. 7923 rounds to 7900, 3021 rounds to 3000.
Estimated sum = \(7900 + 3000 = 10900\)
Actual sum = \(7923 + 3021 = 10944\)

b. 8865 rounds to 8900, 2360 rounds to 2400.
Estimated difference = \(8900 - 2400 = 6500\)
Actual difference = \(8865 - 2360 = 6505\)

c. 207 rounds to 200, 111 rounds to 100.
Estimated product = \(200 \times 100 = 20000\)
Actual product = \(207 \times 111 = 22977\)

d. 4775 rounds to 4800, 25 rounds to 0 (or use compatible numbers: 4775 / 25). Let us round divisor to nearest ten = 30 or keep 25, 4800 / 25 = 192.
Estimated quotient = \(4800 \div 25 = 192\)
Actual quotient = \(4775 \div 25 = 191\)

e. 2350 rounds to 2400, 23 rounds to 20.
Estimated product = \(2400 \times 20 = 48000\)
Actual product = \(2350 \times 23 = 54050\)
In simple words: Round numbers to the nearest hundred by checking the tens digit, then perform the required operation.

Exam Tip: If the tens digit is 5 or more, round up; if less than 5, round down.

Question 7. Give the greatest 5 digits number, only three digits being the same.
Answer: 99987
In simple words: To make the greatest 5-digit number with only three identical digits, place the largest possible digits (9) at the highest place values, leaving the remaining lower places for 8 and 7.

Exam Tip: Start filling from the leftmost digit with the highest possible number (9).

Question 8. The short form of the number \(4 \times 1000000 + 2 \times 1000 + 3 \times 100 + 2\) is ------.
Answer: 4,002,302 (or 4002302)
In simple words: Combine the expanded terms by putting each digit in its correct place value position.

Exam Tip: Watch out for missing place values like ten-thousands and thousands; they must be filled with zeros.

Question 9. Write the number name of 36700007 in the Indian and International systems with the commas in the right place.
Answer: Indian System: 3,67,00,007 - Three crore sixty-seven lakh seven.
International System: 36,700,007 - Thirty-six million seven hundred thousand seven.
In simple words: Place commas according to the chosen system and read the number period by period.

Exam Tip: Double check zero placeholders when spelling out number names.

Question 10. Give the predecessor of 9968 and the successor of 4563.
Answer: Predecessor of 9968 = \(9968 - 1 = 9967\)
Successor of 4563 = \(4563 + 1 = 4564\)
In simple words: Subtract 1 to find the predecessor, and add 1 to find the successor.

Exam Tip: Predecessor means one less, successor means one more.

2. Playing with Numbers.

Question 11. State True or false and if false, mention the true statement.
a. Twin prime numbers are pairs of numbers which have only prime numbers between them.
b. If b is a multiple of a then, b \ge a.
c. Co primes are pairs of numbers that share a common divisor.
d. 1 is a factor of every number.
Answer: a. False. True statement: Twin prime numbers are pairs of prime numbers whose difference is 2.
b. True.
c. False. True statement: Co-primes are pairs of numbers that have only 1 as their common factor (HCF is 1).
d. True.
In simple words: Verify each mathematical definition against standard number theory rules.

Exam Tip: Remember that co-prime numbers do not need to be prime numbers individually; they only need an HCF of 1.

Question 12. Write all the factors of 36.
Answer: 1, 2, 3, 4, 6, 9, 12, 18, 36
In simple words: List all whole numbers that divide 36 without leaving a remainder.

Exam Tip: Write factors in pairs (1 × 36, 2 × 18, 3 × 12, 4 × 9, 6 × 6) to ensure none are missed.

Question 13. Write three multiples of 23, 34.
Answer: Multiples of 23: 23, 46, 69
Multiples of 34: 34, 68, 102
In simple words: Multiply the given number by 1, 2, and 3 to get its first three multiples.

Exam Tip: Multiples are infinite, whereas factors are finite.

Question 14. Express 44 as a sum of two odd prime numbers.
Answer: \(3 + 41 = 44\) (Both 3 and 41 are odd prime numbers). Other valid pairs include \(7 + 37\) and \(13 + 31\).
In simple words: Find two prime numbers other than 2 that add up to 44.

Exam Tip: Since 44 is even and the sum of two odd numbers is even, any valid pair of odd primes will work.

Question 15. Check the divisibility of 476918 by 11.
Answer: Sum of digits at odd places (from right): \(8 + 9 + 7 = 24\)
Sum of digits at even places (from right): \(1 + 6 + 4 = 11\)
Difference = \(24 - 11 = 13\)
Since 13 is not divisible by 11, the number 476918 is not divisible by 11.
In simple words: Subtract the sum of even-position digits from the sum of odd-position digits; if the result is 0 or a multiple of 11, it is divisible.

Exam Tip: Always count digit positions starting from the rightmost digit as position 1 (odd).

Question 16. The number that should be added to 5819 to be divisible by 7?
Answer: Divide 5819 by 7:
\(5819 \div 7 = 831\) with a remainder of 2.
To make it divisible, we need to add the difference between the divisor and the remainder:
\(7 - 2 = 5\)
Check: \(5819 + 5 = 5824\), and \(5824 \div 7 = 832\).
In simple words: Find the remainder when divided by 7, then calculate how much more is needed to reach the next multiple of 7.

Question 17. Find the HCF by prime factorization method.
a. 658, 940, 1125
b. 762, 1270
Answer: a. Prime factorization of 658: \(2 \times 7 \times 47\)
Prime factorization of 940: \(2^2 \times 5 \times 47\)
Prime factorization of 1125: \(3^2 \times 5^3\)
Common prime factors: None (other than 1).
HCF = 1.

b. Prime factorization of 762: \(2 \times 3 \times 127\)
Prime factorization of 1270: \(2 \times 5 \times 127\)
Common prime factors: \(2 \times 127 = 254\)
HCF = 254.
In simple words: Find the prime factors of each number and multiply the common prime factors to get the HCF.

Exam Tip: If no common prime factor exists among all numbers, their HCF is 1.

Question 18. Find the HCF by division method. 806, 663, 377.
Answer: Step 1: Find HCF of 806 and 663.
\(806 \div 663 = 1\) with remainder 143.
\(663 \div 143 = 4\) with remainder 91.
\(143 \div 91 = 1\) with remainder 52.
\(91 \div 52 = 1\) with remainder 39.
\(52 \div 39 = 1\) with remainder 13.
\(39 \div 13 = 3\) with remainder 0.
So HCF of 806 and 663 is 13.
Step 2: Find HCF of 13 and the third number, 377.
\(377 \div 13 = 29\) with remainder 0.
Thus, the HCF of 806, 663, and 377 is 13.
In simple words: Apply long division successively until the remainder becomes 0, then use that divisor with the next number.

Exam Tip: The last non-zero divisor is always the HCF.

Question 19. Reduce 527/1147 to the lowest terms.
Answer: Find the HCF of 527 and 1147 using long division:
\(1147 \div 527 = 2\) with remainder \(1147 - 1054 = 93\)
\(527 \div 93 = 5\) with remainder \(527 - 465 = 62\)
\(93 \div 62 = 1\) with remainder \(93 - 62 = 31\)
\(62 \div 31 = 2\) with remainder 0.
HCF = 31.
Divide numerator and denominator by 31:
\(\frac{527 \div 31}{1147 \div 31} = \frac{17}{37}\)
In simple words: Divide both the numerator and denominator by their HCF to simplify the fraction.

Exam Tip: Reducing a fraction to lowest terms means the numerator and denominator become co-prime.

Question 20. Find the greatest number by which one can divide 11296 and 13528, leaving remainders 11 and 23 respectively.
Answer: Subtract the respective remainders from each number:
\(11296 - 11 = 11285\)
\(13528 - 23 = 13505\)
Now, find the HCF of 11285 and 13505.
\(13505 \div 11285 = 1\) with remainder 2220.
\(11285 \div 2220 = 5\) with remainder 185.
\(2220 \div 185 = 12\) with remainder 0.
HCF = 185.
In simple words: Subtract the remainders first, then find the HCF of the resulting numbers.

Exam Tip: Subtracting remainders first ensures the new numbers are exactly divisible by the required number.

Question 21. Find the smallest number which when divided by 15, 20, 48 will in each case leave 9 as the remainder.
Answer: First, find the LCM of 15, 20, and 48.
Prime factorizations:
\(15 = 3 \times 5\)
\(20 = 2^2 \times 5\)
\(48 = 2^4 \times 3\)
\(\text{LCM} = 2^4 \times 3 \times 5 = 16 \times 15 = 240\)
Required number = \(\text{LCM} + \text{Remainder} = 240 + 9 = 249\)
In simple words: Find the LCM of the divisors and add the common remainder to it.

Exam Tip: Whenever a problem asks for the "smallest number divisible by", calculate the LCM.

Question 22. If the HCF of two numbers is 26 and their LCM is 5148. If one number is 572, find the other number.
Answer: We know that \(\text{Product of two numbers} = \text{HCF} \times \text{LCM}\)
Let the other number be \(x\).
\(572 \times x = 26 \times 5148\)
\(x = \frac{26 \times 5148}{572}\)
\(x = \frac{133848}{572} = 234\)
In simple words: Multiply the HCF and LCM, then divide by the given number to find the missing number.

Exam Tip: This formula only works for pairs of numbers.

Question 23. Can two numbers have 16 and 204 as their HCF and LCM respectively? Why?
Answer: No, two numbers cannot have 16 as HCF and 204 as LCM.
Reason: The HCF of two numbers must always exactly divide their LCM. Here, \(204 \div 16 = 12.75\), which is not a whole number (16 does not divide 204 evenly).
In simple words: An LCM must always be a multiple of the HCF.

Exam Tip: Always check if LCM is divisible by HCF when testing such conditions.

Question 24. Write all the prime numbers between 40 to 50
Answer: 41, 43, 47
In simple words: List the numbers between 40 and 50 that have no factors other than 1 and themselves.

Exam Tip: Check odd numbers in the range against divisibility by 3 and 7.

Question 25. Check whether 72912 is divisible by 2, 3, 4, 5, 6, 7, 8, 9 explaining the rules.
Answer: - Divisible by 2: Yes, because the last digit (2) is even.
- Divisible by 3: Yes, because the sum of digits \(7 + 2 + 9 + 1 + 2 = 21\), which is divisible by 3.
- Divisible by 4: Yes, because the number formed by the last two digits (12) is divisible by 4.
- Divisible by 5: No, because the last digit is 2 (not 0 or 5).
- Divisible by 6: Yes, because it is divisible by both 2 and 3.
- Divisible by 7: Yes, \(72912 \div 7 = 10416\) (remainder is 0).
- Divisible by 8: Yes, because the number formed by the last three digits (912) is divisible by 8 (\(912 \div 8 = 114\)).
- Divisible by 9: No, because the sum of digits (21) is not divisible by 9.
In simple words: Test each divisibility rule step-by-step for the given number.

Exam Tip: Memorize divisibility rules for 2 through 10 to solve quickly.

Question 26. To test for divisibility by 12, the number 20028 should be divisible by
Answer: 3 and 4 (since 3 and 4 are co-prime factors of 12, and \(3 \times 4 = 12\)).
In simple words: A number is divisible by 12 if it is divisible by both 3 and 4.

Exam Tip: Always use co-prime factors when breaking down composite divisors like 12, 15, or 18.

Question 27. Find the prime factors of 625 and 3900.
Answer: Prime factorization of 625: \(5 \times 5 \times 5 \times 5 = 5^4\)
Prime factorization of 3900: \(2 \times 2 \times 3 \times 5 \times 5 \times 13 = 2^2 \times 3 \times 5^2 \times 13\)
In simple words: Repeatedly divide by prime numbers until only primes remain.

Exam Tip: Use divisibility tests to quickly determine prime factors like 2, 3, and 5.

Question 28. Find the LCM of 39, 65, 130, 156.
Answer: Prime factorizations:
\(39 = 3 \times 13\)
\(65 = 5 \times 13\)
\(130 = 2 \times 5 \times 13\)
\(156 = 2^2 \times 3 \times 13\)
Take the highest power of each prime factor: \(2^2 \times 3 \times 5 \times 13 = 4 \times 3 \times 5 \times 13 = 780\)
In simple words: Find the prime factors of all numbers and multiply the highest power of each prime factor.

Exam Tip: Common factor 13 appears in all numbers; ensure it is included with its highest exponent.

Question 29. Four bells toll at intervals of 10 minutes, 15 minutes, 30 minutes and 45 minutes. If all of them tolled together at 8 a.m., when will they next toll together?
Answer: Find the LCM of 10, 15, 30, and 45.
\(10 = 2 \times 5\)
\(15 = 3 \times 5\)
\(30 = 2 \times 3 \times 5\)
\(45 = 3^2 \times 5\)
\(\text{LCM} = 2 \times 3^2 \times 5 = 90 \text{ minutes}\)
90 minutes = 1 hour 30 minutes.
Next toll time = 8:00 a.m. + 1 hour 30 minutes = 9:30 a.m.
In simple words: Calculate the LCM of the time intervals to find when the bells will ring together again.

Exam Tip: Convert minutes into hours when adding to the starting clock time.

Question 30. Find the least number of five digits exactly divisible by 12, 15, 18, 24.
Answer: Find the LCM of 12, 15, 18, and 24.
\(12 = 2^2 \times 3\)
\(15 = 3 \times 5\)
\(18 = 2 \times 3^2\)
\(24 = 2^3 \times 3\)
\(\text{LCM} = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360\)
The smallest 5-digit number is 10000.
Divide 10000 by 360: \(10000 \div 360 = 27\) with a remainder of 280.
Required number = \(10000 + (360 - 280) = 10000 + 80 = 10080\)
In simple words: Find the LCM, divide the smallest 5-digit number by this LCM, and add the difference between the LCM and the remainder.

Exam Tip: Always add the difference between divisor and remainder to the base number to get the exact multiple.

Operations on whole numbers.

Question 31. Find the sum of the following by suitable arrangement: 222 + 333 + 667 + 778.
Answer: Group numbers to make tens/hundreds:
\(= (222 + 778) + (333 + 667)\)
\(= 1000 + 1000\)
\(= 2000\)
In simple words: Rearrange the numbers so that units digits add up to multiples of 10 for easier calculation.

Exam Tip: Use commutative and associative properties of addition to group numbers strategically.

Question 32. Find the product of the following by suitable arrangement: 818 \times 2 \times 50; 212 \times 8 \times 50 \times 10
Answer: Part 1: \(818 \times 2 \times 50\)
\(= 818 \times (2 \times 50)\)
\(= 818 \times 100\)
\(= 81800\)

Part 2: \(212 \times 8 \times 50 \times 10\)
\(= 212 \times (8 \times 50) \times 10\)
\(= 212 \times 400 \times 10\)
Wait, rearrange as: \(212 \times (8 \times 50 \times 10) = 212 \times 4000 = 848000\)
In simple words: Multiply numbers like 2 and 50 first to get powers of 10, making the rest of the multiplication much simpler.

Exam Tip: Look for pairs like \(2 \times 50\) or \(4 \times 25\) or \(8 \times 125\) that yield powers of 10.

Question 33. Using the property of distribution of multiplication over addition, solve the following: 859 \times 102; 112 \times 1004
Answer: a. \(859 \times 102 = 859 \times (100 + 2)\)
\(= (859 \times 100) + (859 \times 2)\)
\(= 85900 + 1718\)
\(= 87618\)

b. \(112 \times 1004 = 112 \times (1000 + 4)\)
\(= (112 \times 1000) + (112 \times 4)\)
\(= 112000 + 448\)
\(= 112448\)
In simple words: Split the larger number into a round hundred or thousand plus a small number, then distribute multiplication.

Exam Tip: Distributive property is \(a \times (b + c) = (a \times b) + (a \times c)\).

Question 34. Do the subtraction sums and check your answer by suitable addition: 23605 - 19878; 70546 - 34789.
Answer: a. \(23605 - 19878 = 3727\)
Check: \(3727 + 19878 = 23605\) (Correct)

b. \(70546 - 34789 = 35757\)
Check: \(35757 + 34789 = 70546\) (Correct)
In simple words: Subtract to find the difference, then add the result to the subtrahend to get the original minuend.

Exam Tip: Always verify subtraction by adding the answer back to the number subtracted.

Question 35. State the property used in each case:
a. \(56 \times (19 + 91) = 56 \times 19 + 56 \times 91\)
b. \(1 \times 6892 = 6892\)
c. \((212 \times 27) + (212 \times 23) = 212 \times (27 + 23) = 212 \times 50\)
d. \(386 \times (9 - 6) = (386 \times 9) - (386 \times 6)\)
e. \(1 \times 0 = 0\)
Answer: a. Distributive property of multiplication over addition.
b. Multiplicative identity property (Identity element for multiplication).
c. Distributive property of multiplication over addition (used in reverse / factoring out).
d. Distributive property of multiplication over subtraction.
e. Multiplication property of zero.
In simple words: Identify the algebraic rule being demonstrated in each equation.

Exam Tip: Know the names of fundamental whole number properties thoroughly.

Question 36. Divide and check your answer: 3779520 \div 32; 514228 \div 52.
Answer: a. \(3779520 \div 32 = 118110\)
Check: \(118110 \times 32 = 3779520\) (Correct)

b. \(514228 \div 52 = 9889\) with remainder 0 (Exact division: \(9888.99...\) wait, let us check: \(514228 \div 52 = 9888.99\) - let us re-verify: \(9889 \times 52 = 514228\)).
Check: \(9889 \times 52 = 514228\) (Correct)
In simple words: Perform long division and verify by multiplying the quotient by the divisor.

Exam Tip: Double-check each subtraction step during long division.

Question 37. A school has 5746 students. If 1298 students leave the school and 2478 new students join in, find the total number of students.
Answer: Initial number of students = 5746
Students who left = 1298
Students remaining = \(5746 - 1298 = 4448\)
New students who joined = 2478
Total number of students = \(4448 + 2478 = 6926\)
In simple words: Subtract the students who left from the total, then add the new students who joined.

Exam Tip: Perform addition and subtraction step-by-step in the order of occurrence.

Question 38. Find the number which when divided by 27 gives 125 as quotient and 25 as remainder.
Answer: Using the division algorithm: \(\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}\)
Number = \((27 \times 125) + 25\)
\(27 \times 125 = 3375\)
Number = \(3375 + 25 = 3400\)
In simple words: Multiply the divisor by the quotient and add the remainder to find the original number.

Exam Tip: Formula: \(\text{Dividend} = (d \times q) + r\).

Question 39. Which property states that the sum of any two whole numbers is always a whole number.
Answer: Closure property of addition for whole numbers.
In simple words: It means combining two whole numbers always results in another whole number.

Exam Tip: Closure property applies to addition and multiplication of whole numbers, but not always to subtraction or division.

Question 40. How much less than the sum of 4236, 1281 and 45 is the product of 42 and 126.
Answer: Sum = \(4236 + 1281 + 45 = 5562\)
Product = \(42 \times 126 = 5292\)
Difference = \(5562 - 5292 = 270\)
In simple words: First find the sum of the three numbers and the product of the two numbers, then subtract the product from the sum.

Exam Tip: Read word problems carefully to determine the correct order of operations.

Question 41. If a and b are two whole numbers, then the commutative property of multiplication states-----
Answer: \(a \times b = b \times a\)
In simple words: Changing the order of numbers in multiplication does not change the product.

Exam Tip: Commutative property works for addition and multiplication, but not for subtraction and division.

Basic geometrical ideas.

Question 42. Fill in the blanks:
A line has ---- end points.
A ray has ----- end points.
A line segment has ---- end points.
A ----- is a three sided polygon.
A ----- is a four sided polygon.
The length of a diameter is ----- that of a radius.
The ----- is the longest chord of a circle.
Concentric circles have ----- center.
Distance around the circle is -------.
Points that fall together on the same line are ----------. The line segment joining the opposite vertices of a quadrilateral is called -------------.
Answer: - A line has zero (0) end points.
- A ray has one (1) end point.
- A line segment has two (2) end points.
- A triangle is a three sided polygon.
- A quadrilateral is a four sided polygon.
- The length of a diameter is twice (2 times) that of a radius.
- The diameter is the longest chord of a circle.
- Concentric circles have the same (common) center.
- Distance around the circle is circumference.
- Points that fall together on the same line are collinear. The line segment joining the opposite vertices of a quadrilateral is called a diagonal.
In simple words: Basic geometric definitions for lines, polygons, and circles.

Exam Tip: Memorize fundamental geometric terms and their properties.

Question 43. Name the following in a triangle:
a. The three sides
b. The three vertices
c. The side opposite to the vertex R.
Answer: a. Side PQ, QR, and RP
b. Vertex P, Q, and R
c. Side PQ
In simple words: Identify the sides, vertices, and opposite relationships in triangle PQR.

Exam Tip: The side opposite to any vertex is the side that does not touch that vertex.

Question 44. Write the three letter names of all the angles for the above figure.
Answer: Based on the intersecting lines figure with points B, C, D and ray/line intersections:
\(\angle ABD\), \(\angle DBC\), \(\angle ABC\) (or linear pair angles formed around vertex B).
In simple words: Name each angle using three letters with the vertex letter in the middle.

Exam Tip: The middle letter in a three-letter angle notation always represents the vertex.

Question 45. What is a median and altitude. In a triangle how many medians and altitudes are there?
Answer: - Median: A line segment joining a vertex of a triangle to the mid-point of the opposite side.
- Altitude: A perpendicular line segment from a vertex of a triangle to the opposite side.
- Number of medians in a triangle: 3
- Number of altitudes in a triangle: 3
In simple words: Medians connect vertices to midpoints, while altitudes show perpendicular heights from vertices.

Exam Tip: Every triangle always has exactly 3 medians and 3 altitudes.

Question 46. The point where all the lines meet is called -------
Answer: Point of intersection (or concurrent point).
In simple words: When three or more lines pass through the same single point, that point is called the point of intersection.

Exam Tip: Lines meeting at a single point are termed concurrent lines.

Question 47. What are parallel lines? What are perpendicular lines? Give some examples.
Answer: - Parallel lines: Lines in a plane that never intersect and always remain at the same distance apart (e.g., opposite edges of a ruler, railway tracks).
- Perpendicular lines: Lines that intersect at a right angle (90 degrees) (e.g., adjacent edges of a blackboard, the letter L).
In simple words: Parallel lines run side-by-side without meeting, while perpendicular lines cross at a square 90-degree angle.

Exam Tip: Use the symbol \(\parallel\) for parallel and \(\perp\) for perpendicular.

Question 48. Draw a triangle ABC and mark an altitude and median from vertex A.
Answer: Construction steps:
1. Draw triangle ABC with vertices A, B, and C.
2. From vertex A, draw a perpendicular line down to side BC and mark the foot of the perpendicular as D. Line AD is the altitude.
3. Find the exact midpoint of side BC and label it M. Join vertex A to M. Line AM is the median.

In simple words: The altitude goes straight down at a right angle, while the median connects to the exact center of the opposite side.

Exam Tip: Clearly label the altitude foot (D) and median midpoint (M) to avoid losing marks.

Question 49. Draw a quadrilateral PQRS and name all the vertices, all the sides, all the angles, all the triangles, adjacent and opposite sides, adjacent and opposite angles.
Answer: - Vertices: P, Q, R, S
- Sides: PQ, QR, RS, SP
- Angles: \(\angle P\), \(\angle Q\), \(\angle R\), \(\angle S\)
- Triangles formed by diagonals: \(\triangle PQR\), \(\triangle QRS\), \(\triangle RSP\), \(\triangle SPQ\) (and triangles formed by intersecting diagonals inside).
- Adjacent sides: (PQ & QR), (QR & RS), (RS & SP), (SP & PQ)
- Opposite sides: PQ & RS, QR & SP
- Adjacent angles: (\(\angle P\) & \(\angle Q\)), (\(\angle Q\) & \(\angle R\)), (\(\angle R\) & \(\angle S\)), (\(\angle S\) & \(\angle P\))
- Opposite angles: \(\angle P\) & \(\angle R\), \(\angle Q\) & \(\angle S\)
In simple words: List all geometric components of quadrilateral PQRS systematically.

Exam Tip: Adjacent sides share a common vertex, while opposite sides do not share any common endpoint.

Question 50. 'The surface of a book is a plane'. Is this statement correct? Explain.
Answer: Yes, the statement is correct (practically/mathematically modeled as a flat surface).
Explanation: A plane is a flat, two-dimensional surface that extends infinitely in all directions. The cover or surface of a book represents a finite portion of a plane.
In simple words: A flat book cover models a mathematical plane because it has length and breadth without thickness.

Exam Tip: In geometry, physical objects like paper sheets and tabletops are standard real-world examples of planes.

Question 51. Draw a circle of radius 4 cm. Draw a chord AB of the circle. Mark points M, N to show minor arc and major arc, minor segment and major segment. In which segment does the center of the circle lie?
Answer: 1. Draw a circle with center O and radius 4 cm.
2. Draw a chord AB dividing the circle into two parts.
3. Mark point M on the smaller curved boundary (minor arc) and point N on the larger curved boundary (major arc).
4. The region between chord AB and the minor arc is the minor segment. The region between chord AB and the major arc is the major segment.
5. The center of the circle always lies in the major segment (unless AB is a diameter, which divides the circle into two equal semi-circles).
In simple words: A chord splits a circle into two segments; the larger portion is the major segment containing the center.

Exam Tip: Remember that a diameter divides a circle into two equal semi-circular segments.

Integers

Question 52. Fill in the blanks with the symbol \(<\), or \(>\):
a. 4 \(\square\) -2
b. -1 \(\square\) 5
c. -1 \(\square\) 0
d. 0 \(\square\) 5
e. 3 \(\square\) -3
f. 0 \(\square\) -7
g. -6 \(\square\) -8
h. -2 \(\square\) -5
Answer: a. \(4 > -2\)
b. \(-1 < 5\)
c. \(-1 < 0\)
d. \(0 < 5\)
e. \(3 > -3\)
f. \(0 > -7\)
g. \(-6 > -8\)
h. \(-2 > -5\)
In simple words: On the number line, numbers to the right are always greater than numbers to the left.

Exam Tip: Every positive integer is greater than any negative integer, and 0 is greater than any negative integer.

Question 53. Use the number line and write 3 more than -2, 2 less than -1, 6 more than -6, 3 less than -5, 10 less than -10.
Answer: - 3 more than -2: \(-2 + 3 = 1\)
- 2 less than -1: \(-1 - 2 = -3\)
- 6 more than -6: \(-6 + 6 = 0\)
- 3 less than -5: \(-5 - 3 = -8\)
- 10 less than -10: \(-10 - 10 = -20\)
In simple words: Move right on the number line for "more" and left for "less".

Exam Tip: "More than" means add (+) and "less than" means subtract (-).

Question 54. Arrange the following in ascending order... -1, -2, -3, 0, 1, 2, 3.
Answer: -3, -2, -1, 0, 1, 2, 3
In simple words: Arrange numbers from smallest to largest, keeping negative numbers ordered from most negative to least negative.

Exam Tip: Ascending order means smallest to largest.

Question 55. Write all the integers between -3 and 3.
Answer: -2, -1, 0, 1, 2
In simple words: List the whole numbers strictly between -3 and 3.

Exam Tip: The endpoints -3 and 3 are excluded when writing integers "between" two numbers.

Question 56. Write the absolute value of -10 and 10, 0 and -200.
Answer: - Absolute value of -10 = 10
- Absolute value of 10 = 10
- Absolute value of 0 = 0
- Absolute value of -200 = 200
In simple words: Absolute value is the numerical value without regard to its sign (always positive or zero).

Exam Tip: Absolute value is denoted by vertical bars, e.g., \(|-10| = 10\).

Question 57. Add using the number line: (-3) + 2 + (-5)
Answer: 1. Start at 0 on the number line.
2. Move 3 units to the left to reach -3.
3. Move 2 units to the right to reach -1.
4. Move 5 units to the left to reach -6.
Result = -6.
In simple words: Combine movements on the number line: start at 0, go left 3, right 2, and left 5.

Exam Tip: Addition of positive numbers moves right; addition of negative numbers moves left.

Question 58. Use the associative law to add these integers:
a. 200 + (-33) + (-67)
b. 0 + (-1) + (1) + 932 + (-332)
c. (-2) + (-35) + 90 + 985
d. (-345) + 908 + (-76)
Answer: a. \(200 + [(-33) + (-67)] = 200 + (-100) = 100\)
b. \([(-1) + 1] + [932 + (-332)] + 0 = 0 + 600 + 0 = 600\)
c. \([(-2) + (-35)] + [90 + 985] = -37 + 1075 = 1038\)
d. \([(-345) + (-76)] + 908 = -421 + 908 = 487\)
In simple words: Group numbers with convenient sums or opposite signs first to simplify addition.

Exam Tip: Associative law states that grouping does not affect the sum: \((a + b) + c = a + (b + c)\).

Question 59. What is the sum of the largest 4-digit positive integer and the smallest 3-digit negative integer.
Answer: Largest 4-digit positive integer = 9999
Smallest 3-digit negative integer = -999
Sum = \(9999 + (-999) = 9999 - 999 = 9000\)
In simple words: Add 9999 and -999 by finding their difference.

Exam Tip: The smallest 3-digit negative integer is -999, not -100.

Question 60. Subtract:
- 0 from -32
- -84 from 0
- -256 from 256
Answer: - Subtracting 0 from -32: \(-32 - 0 = -32\)
- Subtracting -84 from 0: \(0 - (-84) = 0 + 84 = 84\)
- Subtracting -256 from 256: \(256 - (-256) = 256 + 256 = 512\)
In simple words: Remember that subtracting a negative number is the same as adding its positive counterpart.

Exam Tip: Rule: \(a - (-b) = a + b\).

Question 61. Simplify:
a. 256 - 78 + 620
b. 89 - 90 - 120
c. 64 + [(-20) - (-30)] - (-6)
Answer: a. \(256 - 78 + 620 = 178 + 620 = 798\)
b. \(89 - 90 - 120 = -1 - 120 = -121\)
c. \(64 + [(-20) + 30] + 6 = 64 + 10 + 6 = 80\)
In simple words: Follow order of operations (BODMAS), simplifying brackets and performing addition/subtraction from left to right.

Exam Tip: Be careful with negative signs when opening brackets.

Topic : Number System

Objective type questions:

1. How many six digit numbers are there in all ?
(a) 900000 (b) 9000 (c) 90000 (d) 900

2. How many hundreds make a lakh ?
(a) One hundred hundreds (b) One thousand hundreds (c) Ten thousand hundreds

3. Find the difference between the number 946 and that obtained on reversing its digits.
(a) 72 (b) 297 (c) 279 (d) 27

4. What comes just after 3674999.
(a) 3675000 (b) 3674998 (c) 4674999 (d) 3675001

5. What comes just before 9000000.
(a) 899999 (b) 8999999 (c) 999999 (d) 9000001

6. Write all three digit number using 1, 4, 5 taking each digit only once.
(option not required)

7. Write the smallest number of different digit formed by 7, 5, 0, 4, 6.
(option not required)

8. Write the place value of six in 56943821 in international system.
(a) 6 million (b) 6 thousand (c) 60 million (d) 60 thousand

9. A number exceed 48359610 by 2163405
(a) 4602615 (b) 46186205 (c) 50523015 (d) 51523115

10. The total population of city is 2783645. If the number males 1640329.
(a) 1413136 (b) 4249347 (c) 4423974 (d) 1143316

11. By how much is 5643879 smaller than one crore.
(a) 1564387 (b) 4356121 (c) 4531621 (d) 4356111

12. The sum of two number is 19705308 if one of them is 5492094.
(a) 32114221 (b) 14213214 (c) 2419402 (d) 4212204

13. A man had Rs 20000000 with him. He spent Rs 13607085 on buying a school building. How much money is left with him?
(a) 6319915 (b) 6391911 (c) 6392911 (d) 6392915

14. How much money was collected from 1786 students of a school for a charity show if each student contributed Rs.725.
(a) 169670 (b) 696271 (c) 217966 (d) 1769760

15. Mr. Bhaskar saves Rs. 3567 every month how much money did he saves in 15 years?
(a) 206640 (b) 426600 (c) 642060 (d) 240606

Please refer to attached file for CBSE Class 6 Mathematics Revision Assignment 3