CBSE Class 6 Mathematics Revision Worksheet Set 19

CH - KNOWING OUR NUMBERS

I Fill In the blanks

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

Exam Tip: Remember the conversion chain: kilometer to meter is 1,000, meter to millimeter is 1,000.

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

Exam Tip: Prefix 'milli' always denotes one-thousandth part, so 1 g = 1,000 mg.

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

Exam Tip: Memorize basic Roman numerals: I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1000.

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

Exam Tip: For rounding to nearest hundred, if the tens digit is 5, 6, 7, 8, or 9, increase the hundreds digit by 1.

Question 5. 79 can be written in Roman numeral as ___________
Answer: LXXIX
In simple words: 79 is split as 70 + 9, where 70 is LXX and 9 is IX.

Exam Tip: Break down numbers into tens and units before converting to 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 and then add them up.

Exam Tip: Clearly show the rounding of each individual term before performing the final addition.

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 off both numbers first and then subtract.

Exam Tip: Be careful with subtraction when rounding; ensure you subtract the rounded subtrahend from the rounded minuend.

III Write the roman numeral for

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

Exam Tip: Remember that XC represents 90 and V represents 5.

Question b) 49
Answer: XLIX
In simple words: 49 is 50 - 10 + 9, written as XLIX.

Exam Tip: XL represents 40 and IX represents 9.

Question c) 57
Answer: LVII
In simple words: 50 is L and 7 is VII, so 57 is LVII.

Question d) 35
Answer: XXXV
In simple words: 30 is XXX and 5 is V.

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

Question e) 16
Answer: XVI
In simple words: 10 is X and 6 is VI.

Exam Tip: Larger values precede smaller values in addition.

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 to the original population to find the new population.

Exam Tip: Align digits properly by place value during column addition.

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 = 30 m = 3000 cm
Cloth needed for 1 shirt = 2 m 25 cm = 225 cm
Number of shirts = \(3000 \div 225\)
\(3000 \div 225 = 13\) with a remainder of 75
Number of shirts that can be stitched = 13
Remaining cloth = 75 cm (or 0.75 m)
In simple words: Convert both measurements to centimeters and divide the total cloth by the cloth needed per shirt.

Exam Tip: Always convert all units into the smaller unit (centimeters) 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 formed by arranging digits in descending order = 96531
Least number formed by arranging digits in ascending order = 13569
Difference = 96531 - 13569
\(= 82962\)
In simple words: Make the largest possible number by putting biggest digits first, smallest number by putting smallest digits first, and subtract them.

Exam Tip: Ensure digits are not repeated and the least number does not start with 0 (though 0 is not present here).

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: Associative property applies to addition and multiplication, but not subtraction or division.

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: This demonstrates the commutative property of multiplication.

Exam Tip: Order does not matter in multiplication.

Question 4. 57 x 103 = (57 x ________) + (57 x ________)
Answer: 100, 3
In simple words: 103 is split into 100 + 3 using the distributive property.

Exam Tip: Distributive property links multiplication over addition.

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

Exam Tip: Natural numbers start from 1, whereas whole numbers start from 0.

II Find the product by suitable rearrangement.

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

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, then attach zeros.

Exam Tip: Always look for pairs like (2, 50), (4, 25), or (8, 125).

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: Rearrangement saves time and reduces calculation errors.

Question d) 25 x 125 x 40 x 8
Answer:
\(= (25 \times 40) \times (125 \times 8)\)
\(= 1000 \times 1000\)
\(= 1000000\)
In simple words: Pair 25 with 40 and 125 with 8.

Exam Tip: Grouping powers of 10 simplifies large multiplications instantly.

III Find the sum by suitable re-arrangement

Question a) 425 + 1326 + 575 + 674
Answer:
\(= (425 + 575) + (1326 + 674)\)
\(= 1000 + 2000\)
\(= 3000\)
In simple words: Group numbers whose unit digits add up to 10 (425 with 575, and 1326 with 674).

Exam Tip: Look for combinations that result in round hundreds or thousands.

Question b) 685 + 840 + 315
Answer:
\(= (685 + 315) + 840\)
\(= 1000 + 840\)
\(= 1840\)
In simple words: Add 685 and 315 first to make 1000.

Exam Tip: Rearrangement makes mental math fast and reliable.

IV Find the value using suitable property

Question a) 348 x 75 + 348 x 25
Answer:
Using distributive property: \(a \times b + a \times c = a \times (b + c)\)
\(= 348 \times (75 + 25)\)
\(= 348 \times 100\)
\(= 34800\)
In simple words: Take 348 common and add 75 and 25 to get 100.

Exam Tip: Distributive property in reverse (taking common terms) simplifies calculations.

Question b) 63475 x 145 - 63475 x 45
Answer:
Using distributive property:
\(= 63475 \times (145 - 45)\)
\(= 63475 \times 100\)
\(= 6347500\)
In simple words: Factor out 63475 and subtract 45 from 145.

Exam Tip: Look for common factors across addition or subtraction signs.

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

Exam Tip: Expanding numbers near 100, 1000 makes multiplication much easier.

Question d) 159 x 1006
Answer:
\(= 159 \times (1000 + 6)\)
\(= (159 \times 1000) + (159 \times 6)\)
\(= 159000 + 954\)
\(= 159954\)
In simple words: Break 1006 into 1000 + 6 and multiply separately.

Exam Tip: Always split the larger number into a base of 10 plus/minus a small digit.

V Find using distributive property

Question a) 8425 x 37
Answer:
\(= 8425 \times (30 + 7)\)
\(= (8425 \times 30) + (8425 \times 7)\)
\(= 252750 + 58975\)
\(= 311725\)
In simple words: Express 37 as 30 + 7 and multiply.

Exam Tip: Splitting into tens and units simplifies multiplication.

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: Break 125 into 100 + 20 + 5 and distribute.

Exam Tip: Alternatively, \(348 \times \frac{1000}{8}\) can also be used for 125.

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 apply distribution.

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 = Rs. 25 + Rs. 8 = Rs. 33
Total money spent in 6 days = \(33 \times 6\)
\(= 198\) Rs.
Alternatively using distribution: \((25 + 8) \times 6 = 25 \times 6 + 8 \times 6 = 150 + 48 = 198\)
In simple words: Find daily expense by adding lunch and tea cost, then multiply by 6 days.

Exam Tip: Word problems can often be solved in multiple ways; showing distributive property earns extra credit.

PLAYING WITH NUMBERS

I Fill in the blanks

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

Exam Tip: Prime numbers have exactly two factors, composite numbers have more than two.

Question 2. A number for which sum of all its factors is equal to twice the number is called ________
Answer: perfect number
In simple words: Example: 6 has factors 1, 2, 3, 6 whose sum is 12 (twice of 6).

Exam Tip: 6 and 28 are classic examples of 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; 13 is the greatest.

Exam Tip: Check odd numbers when looking for primes greater than 2.

Question 4. The smallest multiple of 123 is _______
Answer: 123
In simple words: Every number is a multiple of itself (number multiplied by 1).

Exam Tip: Smallest multiple is the number itself; smallest factor is always 1.

Question 5. Two prime numbers, whose difference is 2 are called _________
Answer: twin primes
In simple words: Example: (3, 5) or (11, 13).

Question 6. The smallest digit in the blank space of 421__4, so that the number so formed is divisible by 4.
Answer: 1
In simple words: For divisibility by 4, the number formed by the last two digits must be divisible by 4. With blank as 1, the number is 14, which is not divisible by 4. Wait, let us check: last two digits form __4. If blank is 0, 04 is divisible by 4. Let us re-verify: 04 is 4, which is divisible by 4. So the smallest digit is 0.

Exam Tip: Check digits from 0 to 9; 04 is divisible by 4.

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

Exam Tip: Factors are finite and bounded by the number.

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

Exam Tip: Product of co-prime numbers equals their LCM.

Question 9. The smallest 3 digit prime number is ________
Answer: 101
In simple words: 100 is composite, so the next number 101 is prime.

Exam Tip: Remember 101 is the first 3-digit prime number.

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: Divide successively by prime numbers starting from 2.

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

Exam Tip: Even x Even = Even.

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

Exam Tip: 2 is the only even prime number.

Question 13. The number of different factors in the factorization of 21 is ________
Answer: 2 (prime factors are 3 and 7, total factors are 4: 1, 3, 7, 21)
In simple words: 21 has 4 factors: 1, 3, 7, and 21.

Question 14. The greatest two digit multiple of 9 is _____
Answer: 99
In simple words: 9 x 11 = 99.

Exam Tip: 99 is the largest 2-digit number and is divisible by 9.

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 share no common factors other than 1.

Exam Tip: Consecutive numbers are always co-prime.

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.
Product = \(3 \times 5 \times 7\)
\(= 105\)
In simple words: Multiply the three smallest odd prime numbers together.

Exam Tip: Smallest prime is 2, but since the question specifies odd prime factors, start from 3.

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

Exam Tip: Express final prime factorization in exponential form if required.

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

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:
a) 1586:
- By 2: Yes (ends in 6)
- By 3: No (sum of digits = 20, not divisible by 3)
- By 4: No (86 not divisible by 4)
- By 5: No (does not end in 0 or 5)
- By 6: No (not divisible by 3)
- By 8: No (586 not divisible by 8)
- By 9: No (sum is 20)
- By 10: No (does not end in 0)
- By 11: \((1+8) - (5+6) = 9 - 11 = -2\) (No)

b) 990:
- By 2: Yes, By 3: Yes, By 4: No, By 5: Yes, By 6: Yes, By 8: No, By 9: Yes, By 10: Yes, By 11: Yes

c) 275:
- By 2: No, By 3: No, By 4: No, By 5: Yes, By 6: No, By 8: No, By 9: No, By 10: No, By 11: Yes

d) 6686:
- By 2: Yes, By 3: No, By 4: No, By 5: No, By 6: No, By 8: No, By 9: No, By 10: No, By 11: No

e) 639210:
- By 2: Yes, By 3: Yes, By 4: No, By 5: Yes, By 6: Yes, By 8: No, By 9: Yes, By 10: Yes, By 11: Yes
In simple words: Apply standard divisibility rules for each number.

Exam Tip: Memorize divisibility rules for 2, 3, 4, 5, 6, 8, 9, 10, and 11 to solve quickly.

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

Exam Tip: Whenever a common remainder is asked, find LCM first and add the remainder.

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

Exam Tip: Finding the LCM of divisors is the first crucial step for such problems.

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 will be the HCF of 850 and 680.
Prime factorization of 850 = \(2 \times 5^2 \times 17\)
Prime factorization of 680 = \(2^3 \times 5 \times 17\)
\(\text{HCF} = 2 \times 5 \times 17 = 170 \text{ litres}\)
In simple words: Find the HCF of the two quantities to find the maximum measuring container.

Exam Tip: Keywords like 'maximum capacity' indicate finding the 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 will be the LCM of 80 cm, 85 cm, and 90 cm.
\(80 = 2^4 \times 5\)
\(85 = 5 \times 17\)
\(90 = 2 \times 3^2 \times 5\)
\(\text{LCM} = 2^4 \times 3^2 \times 5 \times 17 = 16 \times 9 \times 5 \times 17 = 12240 \text{ cm} = 122 \text{ m } 40 \text{ cm}\)
In simple words: Find the LCM of the step lengths to determine the minimum distance covered in complete steps.

Exam Tip: Keywords like 'minimum distance' or 'together again' indicate finding the LCM.

GEOMETRY

I FILL IN THE BLANKS

Question 1. The polygon with least number of sides is _________
Answer: triangle
In simple words: A polygon must have at least 3 sides.

Exam Tip: A polygon with 3 sides is a triangle.

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

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

Exam Tip: Do not confuse segment (chord + arc) with sector (radii + arc).

Question 4. __________ is the longest chord.
Answer: diameter
In simple words: The chord passing through the center is the diameter.

Exam Tip: Diameter = 2 x 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: Region bounded by two radii and an arc is a sector.

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

Question 7. _________ is the chord passing through the centre of the circle.
Answer: diameter
In simple words: Diameter is the special chord that passes through the center.

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

II Do as directed:

Question 1. Identify the triangles in the figure (Rectangle ABCD with diagonals intersecting at P)
Answer:
Triangles formed in the figure are:
\(\triangle DAB\), \(\triangle DCB\), \(\triangle ABC\), \(\triangle ADC\), \(\triangle PAB\), \(\triangle PBC\), \(\triangle PCD\), \(\triangle PDA\)
In simple words: List all combinations of 3 non-collinear vertices forming triangles in the given rectangle.

Exam Tip: Systematically list triangles by looking at the halves and quarters divided by diagonals.

Question 2. Name the points which are a) in the interior of \(\angle PQR\) b) in the exterior of \(\angle PQR\) c) on \(\angle PQR\)
Answer:
a) Interior: G, D
b) Exterior: P, A, B, F
c) On the angle: Q, C, R
In simple words: Points inside the angle are interior, outside are exterior, and on the rays are on the angle.

Exam Tip: Carefully observe the position of each point relative to the arms of the angle.

Question 3. Draw a rough 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:
a) Vertices: A, B, C, D
b) Angles: \(\angle A\), \(\angle B\), \(\angle C\), \(\angle D\)
c) Diagonals: AC, 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), (AD, BC)
g) Opposite angles: (\(\angle A, \angle C\)), (\(\angle B, \angle D\))
In simple words: Identify parts of quadrilateral ABCD using standard geometric definitions.

Exam Tip: Draw a clear, labeled figure before answering each sub-part.

Question 4. a) Identify the triangles in the figure b) Write the name of angles c) Write the name of line segments
Answer:
a) Triangles: \(\triangle ABC\), \(\triangle ADE\), \(\triangle ABF\), etc. (depending on inner partitions)
b) Angles: \(\angle ABC\), \(\angle BCA\), \(\angle CAB\), etc.
c) Line segments: AB, BC, CA, AD, DE, EA, BF, etc.
In simple words: List all geometric components visible in the given polygon diagram.

Exam Tip: Use three letters to name angles unambiguously.

Question 5. 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:
Draw a circle with center O. Mark center O, radii OA and OB, diameter AB, sector AOB, arc AB, interior points P and Q, exterior point R, segment bounded by chord CD, and points E and F on the circle.
In simple words: Follow each instruction step-by-step to label features on a drawn circle.

Exam Tip: Use a compass for a neat circle and a ruler for straight lines like chords and radii.

FRACTIONS

1. Fill in the blanks

Question 1) A part of a whole is called a ___________
Answer: fraction
In simple words: A fraction represents a part of a whole.

Question 2) A fraction having its numerator less than the denominator is called a _________ fraction.
Answer: proper
In simple words: Proper fractions have a smaller numerator than denominator.

Question 3) The combination of a whole and a proper fraction is called a _________ fraction.
Answer: mixed
In simple words: Example: \(2\frac{1}{3}\) is a mixed fraction.

Question 4) A fraction having the numerator equal to or greater than the denominator is called a / an _________ fraction.
Answer: improper
In simple words: Improper fractions have numerators greater than or equal to denominators.

Question 5) A proper fraction has 1 as its numerator, it is a called a _________ fraction.
Answer: unit
In simple words: Fractions with 1 as numerator are unit fractions.

Question 6) Fractions with equal value are called ____________ fractions.
Answer: equivalent
In simple words: Fractions that represent the same value are equivalent.

Question 7) Fractions with same denominators are called __________ fractions.
Answer: like
In simple words: Like fractions share the same denominator.

Question 8) Fractions with different denominators are called __________ fractions.
Answer: unlike
In simple words: Unlike fractions have different denominators.

Question 9) Proper fraction is less than _____________
Answer: 1
In simple words: Proper fractions are always strictly less than 1.

Question 10) Improper fraction is greater than _________ or equal to 1.
Answer: 1
In simple words: Improper fractions are always greater than or equal to 1.

2. Express the following as mixed fractions

Question a) 18/7
Answer:
\(18 \div 7 = 2\) quotient with remainder 4
\(= 2\frac{4}{7}\)
In simple words: Divide numerator by denominator to get the whole number and remainder.

Question b) 40/11
Answer:
\(40 \div 11 = 3\) quotient with remainder 7
\(= 3\frac{7}{11}\)
In simple words: 40 divided by 11 gives 3 with remainder 7.

Question c) 79/9
Answer:
\(79 \div 9 = 8\) quotient with remainder 7
\(= 8\frac{7}{9}\)
In simple words: 79 divided by 9 gives 8 with remainder 7.

3. Express the following as improper fractions

Question a) \(3\frac{7}{12}\)
Answer:
\(= \frac{(3 \times 12) + 7}{12}\)
\(= \frac{36 + 7}{12} = \frac{43}{12}\)
In simple words: Multiply whole number by denominator and add numerator.

Question b) \(9\frac{3}{5}\)
Answer:
\(= \frac{(9 \times 5) + 3}{5}\)
\(= \frac{45 + 3}{5} = \frac{48}{5}\)
In simple words: Multiply 9 by 5 and add 3.

Question c) \(4\frac{13}{19}\)
Answer:
\(= \frac{(4 \times 19) + 13}{19}\)
\(= \frac{76 + 13}{19} = \frac{89}{19}\)
In simple words: Multiply 4 by 19 and add 13.

4. Write the next five fractions equivalent to each of the following fractions.

Question a) 3/4
Answer:
Multiply numerator and denominator by 2, 3, 4, 5, 6:
\(\frac{6}{8}, \frac{9}{12}, \frac{12}{16}, \frac{15}{20}, \frac{18}{24}\)
In simple words: Multiply both numerator and denominator by the same counting numbers.

Question b) 7/9
Answer:
\(\frac{14}{18}, \frac{21}{27}, \frac{28}{36}, \frac{35}{45}, \frac{42}{54}\)
In simple words: Multiply numerator and denominator by 2, 3, 4, 5, 6.

Question c) 13/17
Answer:
\(\frac{26}{34}, \frac{39}{51}, \frac{52}{68}, \frac{65}{85}, \frac{78}{102}\)
In simple words: Multiply numerator and denominator by 2, 3, 4, 5, 6.

5. Write the equivalent fraction of 3/5 having

Question a) denominator 20
Answer:
Since denominator becomes 20 (\(5 \times 4 = 20\)), multiply numerator by 4:
\(\frac{3 \times 4}{5 \times 4} = \frac{12}{20}\)
In simple words: Scale numerator and denominator by the same factor.

Question b) numerator 9
Answer:
Since numerator becomes 9 (\(3 \times 3 = 9\)), multiply denominator by 3:
\(\frac{3 \times 3}{5 \times 3} = \frac{9}{15}\)
In simple words: Scale numerator and denominator by 3.

Question c) numerator 27
Answer:
Numerator becomes 27 (\(3 \times 9 = 27\)), multiply denominator by 9:
\(\frac{3 \times 9}{5 \times 9} = \frac{27}{45}\)
In simple words: Scale numerator and denominator by 9.

Question d) denominator 30
Answer:
Denominator becomes 30 (\(5 \times 6 = 30\)), multiply numerator by 6:
\(\frac{3 \times 6}{5 \times 6} = \frac{18}{30}\)
In simple words: Scale numerator and denominator by 6.

6. Write the equivalent fraction of 36/48 having

Question a) numerator 6
Answer:
Numerator goes from 36 to 6 (divided by 6), so divide denominator by 6:
\(\frac{36 \div 6}{48 \div 6} = \frac{6}{8}\)
In simple words: Divide numerator and denominator by 6.

Question b) denominator 12
Answer:
Denominator goes from 48 to 12 (divided by 4), so divide numerator by 4:
\(\frac{36 \div 4}{48 \div 4} = \frac{9}{12}\)
In simple words: Divide numerator and denominator by 4.

7. Reduce the following fractions to their simplest form

Question A) 45/75
Answer:
HCF of 45 and 75 is 15.
\(\frac{45 \div 15}{75 \div 15} = \frac{3}{5}\)
In simple words: Divide both numerator and denominator by their HCF.

Question B) 65/169
Answer:
HCF of 65 and 169 is 13.
\(\frac{65 \div 13}{169 \div 13} = \frac{5}{13}\)
In simple words: Divide both by 13.

Question C) 48/72
Answer:
HCF of 48 and 72 is 24.
\(\frac{48 \div 24}{72 \div 24} = \frac{2}{3}\)
In simple words: Divide both by 24.

Question 8. Jaya had 54 toffees and Reshma had 48 toffees, Jaya used 36 toffees on her birthday and Reshma used 32 toffees on her birthday. What fraction of toffees did each of them used? Did they use equal fraction of their toffees?
Answer:
Fraction of toffees used by Jaya = \(\frac{36}{54} = \frac{2}{3}\)
Fraction of toffees used by Reshma = \(\frac{32}{48} = \frac{2}{3}\)
Yes, both used equal fractions (\(\frac{2}{3}\)) of their respective toffees.
In simple words: Simplify the fraction of used toffees for both girls and compare them.

9. Fill appropriate sign (<, =, >)

Question a) \(\frac{3}{5} \square \frac{4}{5}\)
Answer: \(\frac{3}{5} < \frac{4}{5}\)
In simple words: With same denominators, the fraction with the smaller numerator is smaller.

Question b) \(\frac{3}{5} \square \frac{2}{3}\)
Answer:
Cross multiplication: \(3 \times 3 = 9\) and \(5 \times 2 = 10\)
Since 9 < 10, \(\frac{3}{5} < \frac{2}{3}\)
In simple words: Compare using cross multiplication.

Question c) \(\frac{1}{2} \square \frac{2}{8}\)
Answer:
\(\frac{2}{8}\) simplifies to \(\frac{1}{4}\)
Since \(\frac{1}{2} > \frac{1}{4}\), \(\frac{1}{2} > \frac{2}{8}\)
In simple words: Simplify fractions or find common denominator to compare.

Question d) \(\frac{3}{5} \square \frac{3}{4}\)
Answer:
Cross multiplication: \(3 \times 4 = 12\) and \(5 \times 3 = 15\)
Since 12 < 15, \(\frac{3}{5} < \frac{3}{4}\)
In simple words: With same numerators, the fraction with the larger denominator is smaller.

Question 10. Ahmed read 25 pages of a book contains 100 pages. Ravi read \(\frac{2}{5}\) pages of the same book. Who read more?
Answer:
Pages read by Ahmed = 25 out of 100 = \(\frac{25}{100} = \frac{1}{4}\) of the book.
Pages read by Ravi = \(\frac{2}{5}\) of the book = \(\frac{2}{5} \times 100 = 40\) pages.
Comparing \(\frac{1}{4}\) and \(\frac{2}{5}\):
Cross multiplication: \(1 \times 5 = 5\) and \(4 \times 2 = 8\). Since 8 > 5, Ravi read more.
In simple words: Convert both into fractions of the whole book or total pages read and compare.

11. Simplify

Question a) \(2\frac{3}{8} + \frac{1}{8}\)
Answer:
\(= \frac{19}{8} + \frac{1}{8} = \frac{20}{8} = \frac{5}{2} = 2\frac{1}{2}\)
In simple words: Convert mixed fraction to improper fraction and add.

Question b) \(\frac{2}{7} + \frac{3}{7}\)
Answer:
\(= \frac{2 + 3}{7} = \frac{5}{7}\)
In simple words: Add numerators directly since denominators are the same.

Question c) \(\frac{15}{8} + \frac{3}{8} + \frac{2}{8}\)
Answer:
\(= \frac{15 + 3 + 2}{8} = \frac{20}{8} = \frac{5}{2} = 2\frac{1}{2}\)
In simple words: Combine all numerators over the common denominator 8.

Question d) \(\frac{2}{5} - \frac{1}{5}\)
Answer:
\(= \frac{2 - 1}{5} = \frac{1}{5}\)
In simple words: Subtract numerators with the common denominator.

Question e) \(4\frac{3}{10} + 2\frac{4}{15} - 3\frac{2}{5}\)
Answer:
Convert to improper fractions:
\(\frac{43}{10} + \frac{34}{15} - \frac{17}{5}\)
LCM of 10, 15, 5 is 30.
\(= \frac{43 \times 3}{30} + \frac{34 \times 2}{30} - \frac{17 \times 6}{30}\)
\(= \frac{129 + 68 - 102}{30}\)
\(= \frac{95}{30} = \frac{19}{6} = 3\frac{1}{6}\)
In simple words: Take LCM of denominators, convert all fractions, and compute.

Question f) \(\frac{2}{7} \text{ of } 21\)
Answer:
\(= \frac{2}{7} \times 21 = 2 \times 3 = 6\)
In simple words: Multiply 21 by 2 and divide by 7.

12. Subtract

Question a) \(\frac{7}{10} \text{ from } 4\frac{1}{15}\)
Answer:
\(4\frac{1}{15} = \frac{61}{15}\)
\(\frac{61}{15} - \frac{7}{10}\)
LCM of 15 and 10 is 30.
\(= \frac{61 \times 2}{30} - \frac{7 \times 3}{30} = \frac{122 - 21}{30} = \frac{101}{30} = 3\frac{11}{30}\)
In simple words: Convert mixed fraction to improper fraction, find common denominator, and subtract.

Question b) \(1\frac{3}{20} \text{ from } 2\frac{3}{16}\)
Answer:
\(1\frac{3}{20} = \frac{23}{20}\)
\(2\frac{3}{16} = \frac{35}{16}\)
LCM of 20 and 16 is 80.
\(= \frac{35 \times 5}{80} - \frac{23 \times 4}{80}\)
\(= \frac{175 - 92}{80} = \frac{83}{80} = 1\frac{3}{80}\)
In simple words: Convert to improper fractions, use LCM 80, and subtract.

Question 13. Subtract the sum of \(1\frac{7}{10}\) and \(2\frac{1}{15}\) from 5
Answer:
Sum = \(1\frac{7}{10} + 2\frac{1}{15} = \frac{17}{10} + \frac{31}{15}\)
LCM of 10 and 15 is 30.
\(\text{Sum} = \frac{17 \times 3}{30} + \frac{31 \times 2}{30} = \frac{51 + 62}{30} = \frac{113}{30}\)
Now subtract from 5:
\(5 - \frac{113}{30} = \frac{150 - 113}{30} = \frac{37}{30} = 1\frac{7}{30}\)
In simple words: First find the sum of the two mixed fractions, then subtract it from 5.

Question 14. A vessel had \(4\frac{3}{4}\) lit of milk. Suman drank \(\frac{3}{5}\) lit. How much milk is left in the vessel?
Answer:
Initial milk = \(4\frac{3}{4} = \frac{19}{4}\) liters
Milk drunk = \(\frac{3}{5}\) liters
Milk left = \(\frac{19}{4} - \frac{3}{5}\)
LCM of 4 and 5 is 20.
\(= \frac{19 \times 5}{20} - \frac{3 \times 4}{20} = \frac{95 - 12}{20} = \frac{83}{20} = 4\frac{3}{20} \text{ liters}\)
In simple words: Subtract the consumed milk from the total initial milk.

Question 15. There are two packets. One weight is \(1\frac{4}{5}\) kg and other weight is \(1\frac{1}{2}\) kg. Find the total weight of the two packets.
Answer:
Weight 1 = \(1\frac{4}{5} = \frac{9}{5}\) kg
Weight 2 = \(1\frac{1}{2} = \frac{3}{2}\) kg
Total weight = \(\frac{9}{5} + \frac{3}{2}\)
LCM of 5 and 2 is 10.
\(= \frac{9 \times 2}{10} + \frac{3 \times 5}{10} = \frac{18 + 15}{10} = \frac{33}{10} = 3\frac{3}{10} \text{ kg}\)
In simple words: Add the weights of both packets after converting to improper fractions.

Question 16. Javed was given \(\frac{3}{5}\) of a basket of oranges. What fraction of oranges was left in the basket?
Answer:
Total basket = 1 whole
Fraction given to Javed = \(\frac{3}{5}\)
Fraction left = \(1 - \frac{3}{5} = \frac{5}{5} - \frac{3}{5} = \frac{2}{5}\)
In simple words: Subtract the given fraction from 1.

Exam Tip: Whole is always represented as 1 in fraction word problems.