CBSE Class 6 Mathematics Revision Worksheet Set 09

I Fill In the blanks

Question 1. 1km = ___________ mm
Answer: \(10,00,000\)
In simple words: Since 1 km = 1,000 m and 1 m = 1,000 mm, multiplying them gives 10,00,000 mm.

Exam Tip: Remember metric conversions: \(1\text{ km} = 1000\text{ m}\), \(1\text{ m} = 100\text{ cm}\), and \(1\text{ cm} = 10\text{ mm}\).

Question 2. 1 gram = ____________ milligrams
Answer: 1,000
In simple words: 1 gram is equal to 1,000 milligrams.

Exam Tip: Always check the prefix 'milli-', which means one-thousandth (\(\frac{1}{1000}\)).

Question 3. The roman numeral M stands for the number _________
Answer: 1000
In simple words: In Roman numerals, M represents 1,000.

Exam Tip: Basic Roman numerals to remember are I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, and M = 1000.

Question 4. 1453 when rounded off to the nearest hundreds, we get __________
Answer: 1500
In simple words: Look at the digit in the tens place, which is 5. Since it is 5 or more, round up the hundreds digit by 1.

Exam Tip: For rounding to the nearest hundred, examine the tens digit. If it is \(\ge 5\), increase the hundred's digit by 1 and put zeros behind.

Question 5. 79 can be written in Roman numeral as ___________
Answer: LXXIX
In simple words: 79 is broken down as \(50 + 20 + 9\), which is written as LXX + IX = LXXIX.

Exam Tip: Break numbers into standard values: tens and units before writing Roman numerals.

II Give a rough estimate ( by rounding off to nearest hundreds ) and also a closer estimate ( by rounding off to nearest tens ).

Question a) 468 + 243 + 5416
Answer:
Rough estimate (nearest hundreds):
\(468\) rounds to \(500\)
\(243\) rounds to \(200\)
\(5416\) rounds to \(5400\)
Sum = \(500 + 200 + 5400 = 6100\)
Closer estimate (nearest tens):
\(468\) rounds to \(470\)
\(243\) rounds to \(240\)
\(5416\) rounds to \(5420\)
Sum = \(470 + 240 + 5420 = 6130\)
In simple words: Round each number to the requested place value first, then add them up.

Exam Tip: Clearly label your rough and closer estimates to avoid losing marks.

Question b) 9471 - 596
Answer:
Rough estimate (nearest hundreds):
\(9471\) rounds to \(9500\)
\(596\) rounds to \(600\)
Difference = \(9500 - 600 = 8900\)
Closer estimate (nearest tens):
\(9471\) rounds to \(9470\)
\(596\) rounds to \(600\)
Difference = \(9470 - 600 = 8870\)
In simple words: Round each number first, then subtract.

Question a) 95
Answer: XCV
In simple words: 95 is \(100 - 10 + 5\), which is XCV.

Exam Tip: Remember that 90 is written as XC (10 subtracted from 100).

Question b) 49
Answer: XLIX
In simple words: 49 is \(50 - 10 + 9\), which is XLIX.

Question c) 57
Answer: LVII
In simple words: 57 is \(50 + 5 + 2\), which is LVII.

Exam Tip: L represents 50.

Question d) 35
Answer: XXXV
In simple words: 35 is \(30 + 5\), which is XXXV.

Exam Tip: Symbols can be repeated up to three times.

Question e) 16
Answer: XVI
In simple words: 16 is \(10 + 6\), which is XVI.

Exam Tip: Combine X and VI for 16.

Question IV. Population of a city was 2,46,972 in the year 2010. In the year 2012, it was found to be increased by 25,650. What was the population of the city in 2012?
Answer:
Population in 2010 = 2,46,972
Increase in population = 25,650
Population in 2012 = \(2,46,972 + 25,650 = 2,72,622\)
In simple words: Add the increase in population to the initial population of 2010.

Exam Tip: Align digits properly according to place value columns when adding large numbers.

Question V. To stitch a shirt 2m 25 cm cloth is needed. Out of 30m cloth, how many shirts can be stitched and how much cloth will remain?
Answer:
Total cloth available = \(30\text{ m} = 3000\text{ cm}\)
Cloth needed for one shirt = \(2\text{ m } 25\text{ cm} = 225\text{ cm}\)
Number of shirts = \(3000 \div 225\)
\(3000 \div 225 = 13\) with a remainder of \(75\)
Number of shirts that can be stitched = 13
Cloth remaining = \(75\text{ cm}\) (or \(0.75\text{ m}\))
In simple words: Convert the total length and individual length into the same unit (cm), then divide.

Exam Tip: Always convert all units to the smaller unit (centimetres in this case) before performing division.

Question VI. Find the difference between the greatest and the least number that can be formed using the digits 9, 3, 1, 5, 6 each only once.
Answer:
Digits given: 9, 3, 1, 5, 6
Greatest number = 96531
Least number = 13569
Difference = \(96531 - 13569 = 82962\)
In simple words: Arrange digits in descending order for the greatest number, and ascending order for the least number, then subtract.

Exam Tip: Make sure not to start the least number with 0 if 0 is present, though here 0 is not in the list.

CH - WHOLE NUMBERS

I Fill in the blanks

Question 1. (3 + 4) + 6= 3 + (4+6). This statement shows that addition of whole numbers is _____________
Answer: associative
In simple words: Changing the grouping of numbers in addition does not change the result, which is the associative property.

Exam Tip: Remember: Associative property deals with grouping numbers using parentheses.

Question 2. __________ is the multiplicative identity for the whole numbers.
Answer: 1
In simple words: Multiplying any whole number by 1 gives the same number back.

Exam Tip: Do not confuse multiplicative identity (1) with additive identity (0).

Question 3. 425 x 36 = 36 x _______
Answer: 425
In simple words: Changing the order of factors does not change the product (commutative property of multiplication).

Exam Tip: This illustrates the commutative property of multiplication: \(a \times b = b \times a\).

Question 4. 57 x 103 = (57 x ________) + (57 x ________)
Answer: 100, 3
In simple words: Split 103 into \(100 + 3\) and distribute 57 to both numbers.

Exam Tip: This applies the distributive property of multiplication over addition: \(a \times (b + c) = (a \times b) + (a \times c)\).

Question 5. The smallest whole number is ________
Answer: 0
In simple words: Whole numbers start from 0 and go on infinitely.

Question a) 3 x 125 x 9 x 8
Answer:
\(= (125 \times 8) \times (3 \times 9)\)
\(= 1000 \times 27\)
\(= 27000\)
In simple words: Group numbers that multiply to give powers of 10 (like 125 and 8).

Exam Tip: Always look for pairs like \(2 \times 5 = 10\), \(4 \times 25 = 100\), or \(8 \times 125 = 1000\).

Question b) 2 x 3465 x 50
Answer:
\(= (2 \times 50) \times 3465\)
\(= 100 \times 3465\)
\(= 346500\)
In simple words: Multiply 2 and 50 first to get 100, making multiplication easy.

Exam Tip: Multiplying by 100 just requires adding two zeros at the end.

Question c) 4 x 272 x 25
Answer:
\(= (4 \times 25) \times 272\)
\(= 100 \times 272\)
\(= 27200\)
In simple words: Group 4 and 25 to make 100.

Exam Tip: Rearranging terms saves calculation time.

Question d) 25 x 125 x 40 x 8
Answer:
\(= (25 \times 40) \times (125 \times 8)\)
\(= 1000 \times 1000\)
\(= 10,00,000\)
In simple words: Pair 25 with 40 and 125 with 8.

Question a) 425 + 1326 + 575 + 674
Answer:
\(= (425 + 575) + (1326 + 674)\)
\(= 1000 + 2000\)
\(= 3000\)
In simple words: Group numbers whose units/tens digits add up to multiples of 10 or 100 (\(25 + 75 = 100\), \(26 + 74 = 100\)).

Exam Tip: Look for number combinations that end in zero.

Question b) 685 + 840 + 315
Answer:
\(= (685 + 315) + 840\)
\(= 1000 + 840\)
\(= 1840\)
In simple words: Add 685 and 315 first to get 1,000, then add 840.

Question a) 348 x 75 + 348 x 25
Answer:
\(= 348 \times (75 + 25)\)
\(= 348 \times 100\)
\(= 34800\)
In simple words: Take out the common factor 348 using the distributive property.

Exam Tip: This is the reverse of the distributive property (taking common terms outside).

Question b) 63475 X 145 - 63475 X 45
Answer:
\(= 63475 \times (145 - 45)\)
\(= 63475 \times 100\)
\(= 63,47,500\)
In simple words: Factor out 63475 and subtract 45 from 145.

Exam Tip: Distributive property works for both addition and subtraction.

Question c) 327 x 108
Answer:
\(= 327 \times (100 + 8)\)
\(= (327 \times 100) + (327 \times 8)\)
\(= 32700 + 2616\)
\(= 35316\)
In simple words: Expand 108 as \(100 + 8\) and distribute.

Question d) 159 x 1006
Answer:
\(= 159 \times (1000 + 6)\)
\(= (159 \times 1000) + (159 \times 6)\)
\(= 159000 + 954\)
\(= 159954\)
In simple words: Split 1006 into \(1000 + 6\) and apply distributive property.

Exam Tip: Always expand the number closest to 10, 100, or 1000.

V Find using distributive property

Question a) 8425 x 37
Answer:
\(= 8425 \times (30 + 7)\)
\(= (8425 \times 30) + (8425 \times 7)\)
\(= 2,52,750 + 58,975\)
\(= 3,11,725\)
In simple words: Split 37 into \(30 + 7\) and multiply separately.

Exam Tip: Show all expansion steps clearly.

Question b) 348 x 125
Answer:
\(= 348 \times (100 + 20 + 5)\)
\(= (348 \times 100) + (348 \times 20) + (348 \times 5)\)
\(= 34800 + 6960 + 1740\)
\(= 43500\)
In simple words: Expand 125 into \(100 + 20 + 5\). Alternatively, \(348 \times \frac{1000}{8}\).

Question c) 742 x 102
Answer:
\(= 742 \times (100 + 2)\)
\(= (742 \times 100) + (742 \times 2)\)
\(= 74200 + 1484\)
\(= 75684\)
In simple words: Write 102 as \(100 + 2\) and distribute.

Exam Tip: This is a classic application of the distributive property.

Question VI. The school canteen charges Rs. 25 for lunch and Rs. 8 for tea each day. How much money do you spend in 6 days on these things?
Answer:
Cost of lunch per day = Rs. 25
Cost of tea per day = Rs. 8
Total cost per day = \(25 + 8 = \text{Rs. } 33\)
Total amount spent in 6 days = \(33 \times 6 = \text{Rs. } 198\)
Alternatively, using distributive property:
\((25 \times 6) + (8 \times 6) = 150 + 48 = \text{Rs. } 198\)
In simple words: Calculate the daily total and multiply by 6, or calculate lunch and tea costs separately for 6 days and add them.

Exam Tip: Word problems can often be solved in multiple ways; choose the method that is clearest.

Ch. 9 : DATA HANDLING

Fill in the blanks

Question 1. The graph represents data in the form of pictures is known as _____________.
Answer: pictograph
In simple words: A pictograph uses symbols or pictures to represent data.

Exam Tip: Remember: 'Picto' refers to pictures.

Question 2. A graph drawn using bars of uniform width is called _______________.
Answer: bar graph
In simple words: Bar graphs use rectangular bars to compare data categories.

Exam Tip: Ensure bars have equal spacing and uniform width when drawing manually.

Question 3. A collection of numbers gathered to give some information is called _______________.
Answer: data
In simple words: Raw information in numerical form is called data.

Exam Tip: Data is plural for datum.

Question II The height (in cm) of 20 students of class VI are given below
125, 130, 135, 142, 125, 139, 150, 147,
142, 139, 125, 130, 125, 150, 147, 139,
152, 130, 152, 142.
Prepare a table using tally marks.
Now, answer the following :
Answer:
Frequency Distribution Table:

In simple words: Count how many times each height appears and represent it using tally marks (groups of 5 with a slash).

Exam Tip: Always cross out numbers as you tally them to ensure you do not miss or double-count any data point.

Question a. Find how many students are of the height 142 cm ?
Answer: 3 students
In simple words: Look at the table row for 142 cm to find the frequency.

Exam Tip: Double-check your table count before answering sub-questions.

Question b. How many students are there of height less than or equal to 135 cm?
Answer:
Heights less than or equal to 135 cm are 125 cm, 130 cm, and 135 cm.
Number of students = \(4 \text{ (for 125)} + 3 \text{ (for 130)} + 1 \text{ (for 135)} = 8\) students
In simple words: Add the frequencies of all heights up to 135 cm.

Exam Tip: Read carefully whether 'less than' includes 'equal to'.

Question c. How many students are of the height more than 147 cm?
Answer:
Heights strictly greater than 147 cm are 150 cm and 152 cm.
Number of students = \(2 \text{ (for 150)} + 2 \text{ (for 152)} = 4\) students
In simple words: Sum the frequencies of heights above 147 cm.

Exam Tip: 'More than 147' means 147 itself is excluded.

Question III Observe the following bar graph and answer the following questions :
*(Note: Based on the standard bar graph provided in the worksheet representing shirt sales across days: MON = 40, TUE = 30, WED = 40, THU = 15, FRI = 45, SAT = 25)*
Answer: (See sub-questions below)

Exam Tip: Read axis labels and scale carefully before answering bar graph questions.

Question a. What information does the above bar graph give ?
Answer: The bar graph shows the number of shirts sold on different days of the week.
In simple words: It tells us shirt sales day by day.

Exam Tip: The title of the graph or axis labels usually gives the exact answer.

Question b. How many shirts were sold on Friday?
Answer: 45 shirts
In simple words: Look at Friday's bar height, which reaches 45.

Question c. On which day the minimum no. of shirts sold?
Answer: Thursday (15 shirts)
In simple words: Thursday has the shortest bar.

Exam Tip: Minimum means the smallest value.

Question d. What is the scale chosen on the vertical line representing no. of shirts ?
Answer: 1 unit length = 5 shirts (or increments of 5 units: 0, 5, 10, 15, 20...)
In simple words: Each grid line on the vertical axis goes up by 5.

Exam Tip: Always check the difference between consecutive tick marks on the axis.

Question e. How many more shirts sold on Wednesday than Tuesday ?
Answer:
Shirts sold on Wednesday = 40
Shirts sold on Tuesday = 30
Difference = \(40 - 30 = 10\) shirts
In simple words: Subtract Tuesday's sales from Wednesday's sales.

Exam Tip: 'How many more' always indicates finding the difference.

Question IV Following table shows the monthly expenditure of a family on various items. Draw a bar graph.
Answer:
To draw the bar graph:
1. Draw horizontal X-axis for Items and vertical Y-axis for Expenditure (in Rs.).
2. Choose a suitable scale on the vertical axis, such as 1 unit = 100 Rs. (0, 100, 200, 300, 400, 500, 600, 700, 800, 900).
3. Draw bars of uniform width for each item with equal gaps:
- House Rent: bar height up to 900
- Food: bar height up to 500
- Education: bar height up to 700
- Electricity: bar height up to 200
- Transport: bar height up to 400
- Clothes: bar height up to 800
In simple words: Represent each item as a bar whose height matches its expenditure.

Exam Tip: Always label both axes and give your bar graph a clear title.

Question V The following are the number of bulbs purchased for a house during the first months of a year. Represent the details by a pictograph using one symbol to represent 10 bulbs :
Months: January (30), February (35), March (20), April (26)
Answer:
Let one symbol (e.g., \(\bigodot\)) represent 10 bulbs, and half symbol (\(\bigodot\) cut or smaller) represent 5 bulbs.
- January (30 bulbs) = 3 symbols (\(\bigodot \bigodot \bigodot\))
- February (35 bulbs) = 3 full symbols and 1 half symbol (\(\bigodot \bigodot \bigodot\) + half symbol)
- March (20 bulbs) = 2 symbols (\(\bigodot \bigodot\))
- April (26 bulbs) = 2 full symbols and 1 fraction representing 6 bulbs.
In simple words: Divide the number of bulbs by 10 to find how many symbols to draw for each month.

Exam Tip: Always include a key stating what each symbol represents (e.g., Key: \(\bigodot = 10\) bulbs).

PLAYING WITH NUMBERS

I Fill in the blanks

Question 1. ________ is neither a prime nor a composite.
Answer: 1
In simple words: 1 has only one factor (itself), so it is neither prime nor composite.

Exam Tip: Prime numbers start from 2.

Question 2. A number for which sum of all its factors is equal to twice the number is called ________
Answer: a perfect number
In simple words: Examples include 6 (factors: 1, 2, 3, 6; sum = 12 = \(2 \times 6\)).

Exam Tip: 6 and 28 are the first two perfect numbers.

Question 3. The greatest prime number between 10 and 15 is _________
Answer: 13
In simple words: Primes between 10 and 15 are 11 and 13, of which 13 is the greatest.

Exam Tip: Check odd numbers for primality in a given range.

Question 4. The smallest multiple of 123 is _______
Answer: 123
In simple words: The first multiple of any number is the number itself (\(123 \times 1\)).

Question 5. Two prime numbers, whose difference is 2 are called _________
Answer: twin primes
In simple words: Prime numbers that differ by 2 (like 3 and 5, or 11 and 13) are twin primes.

Exam Tip: Twin primes must both be prime numbers.

Question 6. The smallest digit in the blank space of 421__4, so that the number so formed is divisible by 4.
Answer: 0
In simple words: A number is divisible by 4 if the number formed by its last two digits is divisible by 4. Here, if the blank is 0, the last two digits are 04, which is divisible by 4.

Exam Tip: Test digits from 0 upwards: 04, 14, 24, etc.

Question 7. The greatest factor of 144 is __________
Answer: 144
In simple words: Every number is the greatest factor of itself.

Exam Tip: The factors of a number are always less than or equal to the number itself.

Question 8. The LCM of two prime numbers is ________
Answer: their product
In simple words: Since prime numbers share no common factors other than 1, their LCM is simply their product.

Exam Tip: For example, the LCM of 3 and 5 is \(3 \times 5 = 15\).

Question 9. The smallest 3 digit prime number is ________
Answer: 101
In simple words: 100 is composite, but 101 cannot be divided by any number other than 1 and 101.

Exam Tip: Test 101 for divisibility by primes up to 10 (2, 3, 5, 7).

Question 10. 180 = 2 x 2 x ___ x 3 __ is the prime factorization of 180
Answer: 3, 5
In simple words: Prime factorization of 180 is \(2 \times 2 \times 3 \times 3 \times 5\).

Exam Tip: Use prime factorization tree or repeated division by prime numbers.

Question 11. The product of any two even numbers is _______
Answer: even
In simple words: Multiplying any two even numbers always results in an even number.

Exam Tip: Test with examples: \(2 \times 4 = 8\) (even).

Question 12. The smallest odd prime number is _____
Answer: 3
In simple words: 2 is the only even prime number, so 3 is the smallest odd prime number.

Exam Tip: 1 is not prime, and 2 is even.

Question 13. The number of different factors in the factorization of 21 is ________
Answer: 2 (or 3 and 7)
In simple words: Prime factorization of 21 is \(3 \times 1 = 3 \times 7\), which has two distinct prime factors: 3 and 7.

Exam Tip: 'Different factors' refers to distinct prime factors.

Question 14. The greatest two digit multiple of 9 is _____
Answer: 99
In simple words: \(9 \times 11 = 99\), which is the largest two-digit number divisible by 9.

Exam Tip: 100 is a three-digit number.

Question 15. _______ and _________ are the factors of every number.
Answer: 1, the number itself
In simple words: 1 divides every number, and every number divides itself.

Exam Tip: Universal rule for factors.

Question 16. HCF of any two consecutive numbers is _______
Answer: 1
In simple words: Consecutive numbers (like 4 and 5) share no common factor other than 1.

Exam Tip: Coprime numbers always have an HCF of 1.

II Do the following

Question 1. I am the smallest number having 3 different odd prime factors. Can you find me?
Answer:
The smallest odd prime numbers are 3, 5, and 7.
The smallest number having these 3 different odd prime factors is their product:
\(3 \times 5 \times 7 = 105\)
In simple words: Multiply the three smallest odd prime numbers (3, 5, 7).

Exam Tip: The first few odd primes are 3, 5, 7, 11, 13...

Question 2. Express the smallest 3 digit number in the form of prime factors.
Answer:
Smallest 3-digit number = 100
Prime factorization of 100:
\(100 = 2 \times 50 = 2 \times 2 \times 25 = 2 \times 2 \times 5 \times 5\)
In simple words: Break down 100 into its prime factors using division.

Exam Tip: Express answers in exponential form if needed: \(2^2 \times 5^2\).

Question 3. Write all prime numbers between 10 and 30.
Answer: 11, 13, 17, 19, 23, 29
In simple words: List all odd numbers between 10 and 30 that have no divisors other than 1 and themselves.

Exam Tip: Memorize prime numbers up to 50 for quick problem-solving.

Question 4. Using divisibility test check whether the following are divisible by 2,3,4,5,6,8,9,10 and 11.
a) 1586 b) 990 c) 275 d) 6686 e) 639210
Answer:
Let us test each number:
a) 1586:
- 2: Yes (ends in 6)
- 3: No (sum of digits = \(1+5+8+6 = 20\), not divisible by 3)
- 4: No (last two digits 86 not divisible by 4)
- 5: No (ends in 6)
- 6: No (not divisible by both 2 and 3)
- 8: No (\(1586 \div 8 = 198.25\), not divisible)
- 9: No (sum 20 not divisible by 9)
- 10: No (does not end in 0)
- 11: Check: \((6+5) - (8+1) = 11 - 9 = 2\) (No)

b) 990:
- 2: Yes (ends in 0)
- 3: Yes (sum = \(9+9+0 = 18\))
- 4: No (90 not divisible by 4)
- 5: Yes (ends in 0)
- 6: Yes (divisible by both 2 and 3)
- 8: No (\(990 \div 8 = 123.75\))
- 9: Yes (sum 18 is divisible by 9)
- 10: Yes (ends in 0)
- 11: Check: \((0+9) - (9+0) = 0\) (Yes)

c) 275:
- 2: No (ends in 5)
- 3: No (sum = \(2+7+5 = 14\))
- 4: No (ends in 75)
- 5: Yes (ends in 5)
- 6: No
- 8: No
- 9: No
- 10: No
- 11: Check: \((5+2) - 7 = 0\) (Yes)

d) 6686:
- 2: Yes (ends in 6)
- 3: No (sum = \(6+6+8+6 = 26\))
- 4: No (86 not divisible by 4)
- 5: No
- 6: No
- 8: No
- 9: No
- 10: No
- 11: Check: \((6+6) - (8+6) = 12 - 14 = -2\) (No)

e) 639210:
- 2: Yes (ends in 0)
- 3: Yes (sum = \(6+3+9+2+1+0 = 21\))
- 4: No (10 not divisible by 4)
- 5: Yes (ends in 0)
- 6: Yes (divisible by both 2 and 3)
- 8: No
- 9: No (sum 21 not divisible by 9)
- 10: Yes (ends in 0)
- 11: Check: \((0+2+3) - (1+9+6) = 5 - 16 = -11\) (Yes)
In simple words: Apply standard divisibility rules for each number from 2 to 11.

Exam Tip: Memorize divisibility rules for 2, 3, 4, 5, 9, 10, and 11 to save time.

Question 5. Find the smallest number when divided by 24 and 81 leave a remainder 5 in each case.
Answer:
First, find the LCM of 24 and 81.
Prime factorization:
\(24 = 2^3 \times 3\)
\(81 = 3^4\)
\(\text{LCM}(24, 81) = 2^3 \times 3^4 = 8 \times 81 = 648\)
The smallest number that leaves a remainder of 5 is:
\(648 + 5 = 653\)
In simple words: Find the LCM of the divisors and add the remainder to it.

Exam Tip: 'Smallest number when divided by...' always indicates finding the LCM.

Question 6. Find the greatest 3 digit number which is exactly divisible by 6, 10, 12 and 4.
Answer:
First, find the LCM of 6, 10, 12, and 4.
\(6 = 2 \times 3\)
\(10 = 2 \times 5\)
\(12 = 2^2 \times 3\)
\(4 = 2^2\)
\(\text{LCM} = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60\)
Now, find the greatest 3-digit number, which is 999.
Divide 999 by 60:
\(999 \div 60 = 16\) with a remainder of 39.
Subtract the remainder from 999:
\(999 - 39 = 960\)
In simple words: Find the LCM, then divide the largest 3-digit number (999) by the LCM and subtract the remainder.

Exam Tip: Always subtract the remainder from the target number to get the greatest exact multiple.

Question 7. Two tankers contain 850 litres and 680 litres of petrol respectively. Find the maximum capacity of the container which can measure the petrol of either tanker in exact number of times.
Answer:
Maximum capacity = HCF of 850 and 680.
Using prime factorization or division method for HCF:
\(850 = 2 \times 5^2 \times 17 = 2 \times 5 \times 5 \times 17\)
\(680 = 2^3 \times 5 \times 17 = 2 \times 2 \times 2 \times 5 \times 17\)
Common prime factors = \(2 \times 5 \times 17 = 170\)
Maximum capacity = 170 litres.
In simple words: 'Maximum capacity' means find the HCF of the two quantities.

Exam Tip: Keywords like 'maximum', 'greatest', or 'highest' indicate HCF.

Question 8. In a morning walk, three persons step off together. Their steps measure 80cm, 85cm, and 90cm respectively. What is the minimum distance each should walk so that all can cover the distance in complete steps?
Answer:
Minimum distance = LCM of 80, 85, and 90.
Prime factorization:
\(80 = 2^4 \times 5\)
\(85 = 5 \times 17\)
\(90 = 2 \times 3^2 \times 5\)
\(\text{LCM} = 2^4 \times 3^2 \times 5 \times 17 = 16 \times 9 \times 5 \times 17\)
\(16 \times 5 = 80\)
\(80 \times 9 = 720\)
\(720 \times 17 = 12,240\text{ cm} = 122.4\text{ m}\)
In simple words: 'Minimum distance' means find the LCM of the step lengths.

Exam Tip: Keywords like 'minimum' or 'least' indicate LCM.

I ) FILL IN THE BLANKS

Question 1. The polygon with least number of sides is _________
Answer: triangle
In simple words: A polygon needs at least 3 straight sides, which forms a triangle.

Exam Tip: A 2-sided shape cannot enclose an area, so polygons start with 3 sides.

Question 2. The distance around a circle is called _____________
Answer: circumference
In simple words: Perimeter of a circle is called its circumference.

Exam Tip: Formula for circumference is \(2\pi r\).

Question 3. A region in the interior of a circle enclosed by a chord and an arc is called ___________
Answer: segment
In simple words: The region bounded by a chord and an arc is a segment of a circle.

Exam Tip: Do not confuse segment with sector.

Question 4. __________ is the longest chord.
Answer: diameter
In simple words: A chord passing through the center is the diameter, which is the longest possible chord in a circle.

Exam Tip: Diameter = \(2 \times \text{radius}\).

Question 5. The region in the interior of a circle enclosed by an arc and a pair of radii is called ____________
Answer: sector
In simple words: Like a slice of pizza, a sector is enclosed by two radii and an arc.

Exam Tip: Radii is plural for radius.

Question 6. The line segment joining any two points on the circle is _____
Answer: chord
In simple words: Any line connecting two points on a circle's boundary is a chord.

Exam Tip: Diameter is a special chord that passes through the center.

Question 7. _________ is the chord passing through the centre of the circle.
Answer: diameter
In simple words: A chord through the center is the diameter.

Exam Tip: It is the longest chord of the circle.

Question 8. The joint of any two non adjacent vertices of a polygon is __________
Answer: diagonal
In simple words: Connecting non-adjacent vertices creates a diagonal.

Exam Tip: Triangles have no diagonals since all their vertices are adjacent.

II Do as directed:

Question 1. Identify the triangles in the figure
*(Note: Based on the rectangle ABCD with diagonals intersecting at P)*
Answer:
The triangles formed in the figure are:
- \(\triangle PAB\)
- \(\triangle PBC\)
- \(\triangle PCD\)
- \(\triangle PDA\)
- \(\triangle DAB\)
- \(\triangle DCB\)
- \(\triangle ABC\)
- \(\triangle ADC\)
In simple words: List all 3-sided closed figures visible in the diagram.

Exam Tip: Count both small triangles inside and larger triangles formed by diagonals and sides.

Question 2. Name the points which are
a) in the interior of \(\angle PQR\)
b) in the exterior of \(\angle PQR\)
c) on \(\angle PQR\)br /> Answer:
*(Based on standard reference angle diagram)*
a) Interior points: Points inside the angle arms (e.g., D)
b) Exterior points: Points outside the angle arms (e.g., B, F)
c) Points on \(\angle PQR\): Points lying directly on the rays forming the angle (e.g., P, Q, R, G, E)
In simple words: Interior means inside, exterior means outside, and 'on' means directly on the lines.

Exam Tip: Vertex and arms are considered 'on' the angle.

Question 3. Draw a sketch of a quadrilateral ABCD and draw its diagonals .Name the
a) vertices b) angles c) diagonals d) adjacent sides e) adjacent angles f) opposite sides g) opposite angles of the quadrilateral.
Answer:
Let ABCD be a quadrilateral.
a) Vertices: A, B, C, D
b) Angles: \(\angle A, \angle B, \angle C, \angle D\) (or \(\angle DAB, \angle ABC, \angle BCD, \angle CDA\))
c) Diagonals: AC and BD
d) Adjacent sides: (AB, BC), (BC, CD), (CD, DA), (DA, AB)
e) Adjacent angles: (\(\angle A, \angle B\)), (\(\angle B, \angle C\)), (\(\angle C, \angle D\)), (\(\angle D, \angle A\))
f) Opposite sides: (AB, DC) and (AD, BC)
g) Opposite angles: (\(\angle A, \angle C\)) and (\(\angle B, \angle D\))
In simple words: Adjacent means next to each other, opposite means facing each other across the shape.

Exam Tip: Always sketch a clear diagram and label vertices in order (clockwise or counter-clockwise).

Question 5. Identify the triangles in the figure:
a) Write the name of angles
b) Write the name of line segments.
Answer:
*(Based on the given pentagon/polygon figure with internal triangles like ABC, ACD, ADE)*
a) Angles: \(\angle A, \angle B, \angle C, \angle D, \angle E\) and internal angles like \(\angle BAC, \angle CAD, \angle DAE\), etc.
b) Line segments: AB, BC, CD, DE, EA, AC, AD, AF, etc.
In simple words: Identify all intersecting lines and corners in the given geometric figure.

Exam Tip: Use three letters to name an angle precisely when multiple angles share the same vertex.

Question 6. Draw any circle and mark
a) its centre
b) two radii
c) a diameter
d) a sector
e) an arc
f) two points in its interior
g) a point in its exterior
h) a segment
i) two points on the circle
Answer:
Construction / Labelling guide for a circle with centre O:
a) Centre: O
b) Two radii: OA, OB
c) Diameter: PQ (passing through O)
d) Sector: Region bounded by radii OA, OB and arc AB
e) Arc: Curved portion of the boundary (e.g., arc AB)
f) Interior points: X, Y (inside the circle)
g) Exterior point: Z (outside the circle)
h) Segment: Region bounded by chord RS and arc RS
i) Two points on the circle: A and B
In simple words: Draw a circle with a compass and label each requested part clearly.

Exam Tip: Use a sharp pencil and compass for neat geometrical drawings.