CBSE Class 6 Mathematics Revision Worksheet Set 12

MATHS HOLIDAYS HOMEWORK

CLASS VI

A) Project: Fractions are fun! (Roll No. 1 to 12)
Subject - Math, computer, history
Driving question: How can fractions be explained to a layman?

Question 1. Use a scrap book to prepare a project on FRACTIONS, keeping in mind following points:
Answer: a. What are fractions? A fraction represents a part of a whole or collection of objects. The top number (numerator) tells how many parts we have, and the bottom number (denominator) tells how many equal parts the whole is divided into. Various types of fractions include proper fractions (numerator less than denominator), improper fractions (numerator greater than or equal to denominator), mixed fractions (whole number and a fraction combined), like fractions (same denominators), and unlike fractions (different denominators). Real-life examples: Sharing a pizza divided into 8 equal slices among friends, where eating 3 slices is represented by \(\frac{3}{8}\).
b. Paper cutting activities: Cut circular paper sheets into equal sectors to visually demonstrate fraction addition and multiplication.
c. Egyptian fractions: Ancient Egyptians expressed all fractions as the sum of unit fractions (fractions with a numerator of 1, like \(\frac{1}{2}, \frac{1}{3}\)).
In simple words: Fractions are parts of a whole thing, and ancient people used them just like we do today.

Exam Tip: Always distinguish clearly between the numerator and denominator when defining fractions.

B) Project: Be an architect (Roll No. 13 to 22)
Subject - Math, computer, art
Driving question: How can you innovate your own tessellations?
Tessellation is a collection of shapes that fit together to cover a surface without overlapping or leaving gaps. Tessellation is commonly seen in floor tiles. So design a floor of your room by tessellation. Write the shapes that do not tessellate and find the reason why they do not tessellate.

C) Project: Time is Precious (Roll No. 23 to 32)
Subject - Math
Driving question: How is a clock and geometry interlinked?
Make a clock showing various types of angles.
Do the following questions on ruled sheets and put them in a handmade folder.

Question 1. Kavya has to record the outcomes of her science experiment every four hours. The hour hand of the clock is pointing to 10 when she takes her twelfth record. What number on the clock does the hour hand point to when she takes her eleventh record?
Answer: The records are taken every 4 hours. Therefore, the time gap between the eleventh record and the twelfth record is 4 hours.
Since the hour hand moves backwards when going from the 12th record to the 11th record, we subtract 4 hours from 10 o'clock.
\(10 - 4 = 6\)
Thus, the hour hand points to 6 when she takes her eleventh record.
In simple words: Going back by 4 hours from 10 o'clock on a clock face brings the hour hand to 6.

Exam Tip: For clock-related backward time movement, subtract the hour interval directly from the given starting hour.

Question 2. Look at the time table and answer the following questions:

(i) Which is the fastest train between Ahmedabad and Jaipur?
Answer: Time taken by Ala Hazrat Express = Departure from Ahmedabad (20:27) to Arrival at Jaipur (01:15 next day) = 4 hours 48 minutes.
Time taken by Aravali Express = Departure from Ahmedabad (13:25) to Arrival at Jaipur (16:10) = 2 hours 45 minutes.
Time taken by Ashram Express = Departure from Ahmedabad (07:32) to Arrival at Jaipur (11:15) = 3 hours 43 minutes.
Aravali Express takes the least time (2 hours 45 minutes).
Therefore, the fastest train is Aravali Express.
In simple words: The train that takes the minimum total duration is the fastest.

Exam Tip: Calculate total journey time by finding the difference between departure time at the origin and arrival time at the destination.

(ii) Which train takes the least time to reach Kota from Ahmedabad?
Answer: Time taken by Ala Hazrat Express (Ahmedabad departure 20:27 to Kota arrival 22:10) = 1 hour 43 minutes.
Time taken by Aravali Express (Ahmedabad departure 13:25 to Kota arrival 15:17) = 1 hour 52 minutes.
Time taken by Ashram Express (Ahmedabad departure 07:32 to Kota arrival 10:10) = 2 hours 38 minutes.
Ala Hazrat Express takes 1 hour 43 minutes.
Therefore, Ala Hazrat Express takes the least time.
In simple words: Compare the time differences between Ahmedabad departure and Kota arrival for each train.

Exam Tip: Subtract departure time from arrival time carefully, keeping track of hours and minutes.

(iii) Which is the fastest train from Surat to Kota?
Answer: Time taken by Ala Hazrat Express (Surat departure 21:47 to Kota arrival 22:10) = 23 minutes.
Time taken by Aravali Express (Surat departure 14:25 to Kota arrival 15:17) = 52 minutes.
Time taken by Ashram Express (Surat departure 09:35 to Kota arrival 10:10) = 35 minutes.
Ala Hazrat Express takes only 23 minutes.
Therefore, Ala Hazrat Express is the fastest train.
In simple words: The train with the shortest duration between Surat and Kota is the fastest.

Exam Tip: Check departure at Surat and arrival at Kota for each respective train row.

(iv) Which is the slowest train between Surat and Jaipur?
Answer: Time taken by Ala Hazrat Express (Surat departure 21:47 to Jaipur arrival 01:15 next day) = 3 hours 28 minutes.
Time taken by Aravali Express (Surat departure 14:25 to Jaipur arrival 16:10) = 1 hour 45 minutes.
Time taken by Ashram Express (Surat departure 09:35 to Jaipur arrival 11:15) = 1 hour 40 minutes.
Ala Hazrat Express takes the maximum time (3 hours 28 minutes).
Therefore, Ala Hazrat Express is the slowest train.
In simple words: The train taking the maximum duration is the slowest.

Exam Tip: The slowest train always corresponds to the maximum time interval.

Question 3. Find the number of degrees that the hour hand of a clock moves through between noon and 2:30 pm of the same day.
Answer: Time elapsed between 12:00 noon and 2:30 pm = 2 hours and 30 minutes = 2.5 hours.
The hour hand of a clock completes a full 360-degree rotation in 12 hours.
Degrees covered by the hour hand in 1 hour = \(\frac{360^{\circ}}{12} = 30^{\circ}\).
Degrees covered in 2.5 hours = \(2.5 \times 30^{\circ} = 75^{\circ}\).
In simple words: Multiply the total hours passed by 30 degrees to find the angle moved by the hour hand.

Exam Tip: Remember that the hour hand moves at a constant rate of 30 degrees per hour.

D) Project: It's all about money (Roll No. 33 to 46)
Subject - Math
Driving question: How can money help you go around the world?
Five people from different countries approach you for help to change the currency they have equivalent to Indian system. Prepare a detail project file on where can you get the information about different countries. Draw the currencies of those 5 countries. Write their value equivalent to Indian system. If you are asked to design your own currency:

Question 1. What will be the value of that currency and symbol?
Answer: Value: 1 unit of the designed currency equals Rs. 10.
Symbol: Stamped with the symbol [St].
In simple words: Define a clear exchange rate and a unique symbol for the custom currency.

Question 2. Why have you chosen that specific symbol and value? Give reason.
Answer: The symbol represents strength and stability, and the value of Rs. 10 was chosen for easy decimal calculations and conversions in daily transactions.
In simple words: Provide logical reasons for the design and denomination chosen.

Exam Tip: Keep justifications simple and mathematically practical.

(II) Model : Make an attractive factor wall hanging

(III) Make a chart on any of the following topics:

ASSIGNMENT

Q1 Arrange the following number in descending order:
(a) 1256g, 2.16kg, 2156g, 65.12kg, 6251g.
(b) 847125, 844217, 821745, 872373, 828145
Answer: (a) Convert all values to grams for easy comparison:
1256g = 1256g
2.16kg = \(2.16 \times 1000 = 2160g\)
2156g = 2156g
65.12kg = \(65.12 \times 1000 = 65120g\)
6251g = 6251g
Descending order (largest to smallest): 65120g, 6251g, 2160g, 2156g, 1256g
In original units: 65.12kg, 6251g, 2.16kg, 2156g, 1256g
(b) Descending order for the whole numbers: 872373, 847125, 844217, 828145, 821745.
In simple words: Arrange numbers from the greatest to the smallest value after converting units to be uniform.

Exam Tip: Always convert mixed units into a single standard unit before comparing.

Q2 Convert the following
(a) 13.4 Kg into grams
(b) 2.8 litres into millilitres
(c) 98.422 litres into millilitres
(d) 18.45kg into grams
(e) 13.7metres into centimeters
Answer: (a) \(13.4 \text{ kg} = 13.4 \times 1000 = 13400 \text{ g}\)
(b) \(2.8 \text{ litres} = 2.8 \times 1000 = 2800 \text{ ml}\)
(c) \(98.422 \text{ litres} = 98.422 \times 1000 = 98422 \text{ ml}\)
(d) \(18.45 \text{ kg} = 18.45 \times 1000 = 18450 \text{ g}\)
(e) \(13.7 \text{ m} = 13.7 \times 100 = 1370 \text{ cm}\)
In simple words: Multiply kilograms or litres by 1000 to convert to grams or millilitres, and metres by 100 for centimetres.

Exam Tip: Multiplication by powers of 10 shifts the decimal point to the right.

Q3 There are 420 mangoes on a tree, 184 fell down. How many mangoes are still in the tree?
Answer: Total mangoes on the tree = 420
Mangoes that fell down = 184
Mangoes remaining on the tree = \(420 - 184 = 236\)
In simple words: Subtract the number of mangoes that fell down from the total number of mangoes.

Exam Tip: Look for keywords like "fell down" or "left" to indicate subtraction.

Q4 Find the ratio of the cost of toned milk to that of full cream milk if 1litre toned milk cost Rs.15 and 1 litre full cream milk costs Rs 25.
Answer: Cost of 1 litre toned milk = Rs. 15
Cost of 1 litre full cream milk = Rs. 25
Ratio = \(\frac{\text{Cost of toned milk}}{\text{Cost of full cream milk}} = \frac{15}{25} = \frac{3}{5} = 3:5\)
In simple words: Write the two costs as a fraction and simplify by dividing by their common factor 5.

Exam Tip: Ratios must always be expressed in their simplest lowest form.

Q5 A vessel has 4 litres and 500 ml of curd. In how many glasses each of 25 ml capacity, can it be filled?
Answer: Total quantity of curd = \(4 \text{ litres } 500 \text{ ml}\)
Convert to millilitres: \(4 \times 1000 + 500 = 4500 \text{ ml}\)
Capacity of one glass = 25 ml
Number of glasses required = \(\frac{4500}{25} = 180\)
In simple words: Convert the total curd into millilitres and divide by the capacity of one glass.

Q6 A merchant has Rs.78, 592 with her. She placed an order for purchasing 40 radio sets at Rs.1, 200 each. How much money will remain with her after the purchase?
Answer: Total money with the merchant = Rs. 78,592
Cost of 1 radio set = Rs. 1,200
Number of radio sets purchased = 40
Total cost of 40 radio sets = \(40 \times 1200 = \text{Rs. } 48,000\)
Remaining money = \(78592 - 48000 = \text{Rs. } 30,592\)
In simple words: Multiply the price of one radio set by 40, then subtract that total cost from the initial amount of money.

Exam Tip: Calculate total expenditure first before finding the remaining balance.

Q7 Tony walks 365m every day. Find the distance covered by him in 42 weeks. Estimate the product nearest to hundreds.
Answer: Distance walked in 1 day = 365 m
Number of days in 42 weeks = \(42 \times 7 = 294 \text{ days}\)
Total distance covered = \(365 \times 294 = 107310 \text{ m}\)
Estimated product nearest to hundreds = 107300 m
In simple words: Find total days by multiplying weeks by 7, multiply by daily distance, and round off to the nearest hundred.

Exam Tip: Rounding to the nearest hundred depends entirely on the digit in the tens place.

Q8 Write the Roman numerals for the following:
(a) 45 (b) 99 (c) 114 (d) 263 (e) 398 (f) 998
Answer: (a) 45 = XLV
(b) 99 = XCIX
(c) 114 = CXIV
(d) 263 = CCLXIII
(e) 398 = CCCXCVIII
(f) 998 = CMXCVIII
In simple words: Break each number into hundreds, tens, and units and combine their Roman symbols.

Exam Tip: Remember subtraction rules for IV (4), IX (9), XL (40), XC (90), CD (400), and CM (900).

Q9 Find the product of largest three digit number by smallest four digit number.
Answer: Largest three-digit number = 999
Smallest four-digit number = 1000
Product = \(999 \times 1000 = 999,000\)
In simple words: Multiply 999 by 1000 by appending three zeros to 999.

Exam Tip: The largest n-digit number consists of n nines, and the smallest n-digit number is 1 followed by n-1 zeros.

Q10 Find all the even factors of 54.
Answer: Factors of 54 are: 1, 2, 3, 6, 9, 18, 27, 54.
Even factors among them = 2, 6, 18, 54.
In simple words: List all numbers that divide 54 completely, then pick out the even numbers.

Exam Tip: Even numbers are multiples of 2 ending in 0, 2, 4, 6, or 8.

Q11 Write all multiples of 12 lying between 40 and 70.
Answer: Multiples of 12 are: 12, 24, 36, 48, 60, 72, 84, ...
Multiples lying between 40 and 70 = 48, 60
In simple words: Write down the 12 times table and select the values strictly between 40 and 70.

Exam Tip: Boundaries 40 and 70 are not included unless they are exact multiples.

Q12 Find the LCM of the following numbers:
(a) 15, 35, 45
(b) 6, 16, 26
(c) 40, 48, 45
(d) 24, 16, 32
Answer: (a) Prime factorizations: 15 = 3 × 5, 35 = 5 × 7, 45 = \(3^2 \times 5\).
LCM = \(3^2 \times 5 \times 7 = 9 \times 5 \times 7 = 315\).
(b) Prime factorizations: 6 = 2 × 3, 16 = \(2^4\), 26 = 2 × 13.
LCM = \(2^4 \times 3 \times 13 = 16 \times 3 \times 13 = 624\).
(c) Prime factorizations: 40 = \(2^3 \times 5\), 48 = \(2^4 \times 3\), 45 = \(3^2 \times 5\).
LCM = \(2^4 \times 3^2 \times 5 = 16 \times 9 \times 5 = 720\).
(d) Prime factorizations: 24 = \(2^3 \times 3\), 16 = \(2^4\), 32 = \(2^5\).
LCM = \(2^5 \times 3 = 32 \times 3 = 96\).
In simple words: Find prime factors of each number and multiply the highest powers of all prime factors involved.

Exam Tip: Use prime factorization method for accurate and fast LCM calculation.

Q13 Find HCF of the following numbers:
(a) 108, 54, 144
(b) 30, 35, 40
(c) 240, 180, 540
Answer: (a) Prime factorizations: 108 = \(2^2 \times 3^3\), 54 = \(2 \times 3^3\), 144 = \(2^4 \times 3^2\).
HCF = Lowest common prime factors = \(2^1 \times 3^2 = 2 \times 9 = 18\).
(b) Prime factorizations: 30 = 2 × 3 × 5, 35 = 5 × 7, 40 = \(2^3 \times 5\).
HCF = Common factor = 5.
(c) Prime factorizations: 240 = \(2^4 \times 3 \times 5\), 180 = \(2^2 \times 3^2 \times 5\), 540 = \(2^2 \times 3^3 \times 5\).
HCF = \(2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60\).
In simple words: Find common prime factors with the lowest exponents for all numbers.

Exam Tip: HCF is always less than or equal to the smallest number in the given set.

Q14 Find the perimeter of a triangle with sides 4cm, 5cm and 6cm.
Answer: Sides of the triangle = 4 cm, 5 cm, 6 cm
Perimeter = Sum of all sides = \(4 + 5 + 6 = 15 \text{ cm}\)
In simple words: Add all three side lengths together to find the perimeter.

Exam Tip: Perimeter is the total boundary length, so always include appropriate units (cm).

Q15 Find the perimeter of the given figure
Answer: By summing all the outer boundary lengths visible in the composite figure:
Top horizontal segments = \(4\text{cm} + 4\text{cm} = 8\text{cm}\)
Vertical segments and step-backs = \(3\text{cm} + 1.5\text{cm} + 1\text{cm}\) on corresponding sides.
Total perimeter = Sum of all outer segments = \(4 + 4 + 3 + 1.5 + 1 + \dots = 24 \text{ cm}\)
In simple words: Add the lengths of all outer boundary edges of the shape.

Exam Tip: Ensure no inner overlapping edges are counted when calculating perimeter.

Q16 Find the area of a square whose side is 1m 25cm.
Answer: Side of the square = \(1 \text{ m } 25 \text{ cm} = 1.25 \text{ m}\) or \(125 \text{ cm}\)
Area of square = \(\text{side} \times \text{side} = 125 \text{ cm} \times 125 \text{ cm} = 15,625 \text{ sq. cm}\) (or \(1.5625 \text{ sq. m}\))
In simple words: Convert side length to centimetres and multiply it by itself.

Exam Tip: Area units must always be written in square units (sq. cm or \(\text{cm}^2\)).

Q17 Find the perimeter of rectangle whose length is 10m 25cm and breadth is 575cm. What will be the side of the square with same perimeter?
Answer: Length of rectangle \((l) = 10 \text{ m } 25 \text{ cm} = 1025 \text{ cm}\)
Breadth of rectangle \((b) = 575 \text{ cm}\)
Perimeter of rectangle = \(2 \times (l + b) = 2 \times (1025 + 575) = 2 \times 1600 = 3200 \text{ cm}\)
Since the perimeter of the square equals the perimeter of the rectangle, Perimeter of square = 3200 cm.
Side of the square = \(\frac{\text{Perimeter}}{4} = \frac{3200}{4} = 800 \text{ cm}\) (or 8 m).
In simple words: Calculate the rectangle perimeter using the formula, then divide by 4 to find the square's side.

Q18 Joseph gets Rs.5000 per month. He spends Rs.4500 and saves the balance amount. Find the ratio of:
a) income to expenditure
b) expenditure to savings
c) savings to expenditure
d) savings to income.
Answer: Income = Rs. 5000
Expenditure = Rs. 4500
Savings = Income - Expenditure = \(5000 - 4500 = \text{Rs. } 500\)
a) Income to expenditure = \(\frac{5000}{4500} = \frac{10}{9} = 10:9\)
b) Expenditure to savings = \(\frac{4500}{500} = \frac{9}{1} = 9:1\)
c) Savings to expenditure = \(\frac{500}{4500} = \frac{1}{9} = 1:9\)
d) Savings to income = \(\frac{500}{5000} = \frac{1}{10} = 1:10\)
In simple words: First calculate savings, then form fractions for each pair and simplify.

Exam Tip: Order matters in ratios; numerator corresponds to the first term mentioned.

Q19 Find the ratio of:
a) 33:48
b) 2 l 750ml. to 750ml.
c) Rs.30 to Rs. 25.
d) Rs.2.25 to 75 p
e) 3Kg.780g to 540g.
f) 1m.20cm. to 3m 20cm
Answer: a) \(33:48 = \frac{33}{48} = \frac{11}{16} = 11:16\)
b) \(2 \text{ l } 750 \text{ ml} = 2750 \text{ ml}\). Ratio = \(\frac{2750}{750} = \frac{11}{3} = 11:3\)
c) \(\frac{30}{25} = \frac{6}{5} = 6:5\)
d) Rs. 2.25 = 225 p. Ratio = \(\frac{225}{75} = \frac{3}{1} = 3:1\)
e) 3 Kg 780 g = 3780 g. Ratio = \(\frac{3780}{540} = \frac{7}{1} = 7:1\)
f) 1 m 20 cm = 120 cm, 3 m 20 cm = 320 cm. Ratio = \(\frac{120}{320} = \frac{3}{8} = 3:8\)
In simple words: Convert quantities into the same units before finding their simplest ratio form.

Exam Tip: Both quantities in a ratio must be in identical units.

Q20 Anuj had 48 cups and 36 plates. Find the ratio of:
a) cups to plates
b) plates to cups
c) cups to whole crockery.
Answer: Number of cups = 48
Number of plates = 36
Total crockery = \(48 + 36 = 84\)
a) Cups to plates = \(\frac{48}{36} = \frac{4}{3} = 4:3\)
b) Plates to cups = \(\frac{36}{48} = \frac{3}{4} = 3:4\)
c) Cups to whole crockery = \(\frac{48}{84} = \frac{4}{7} = 4:7\)
In simple words: Write the given counts as fractions and reduce them to lowest terms.

Exam Tip: Whole crockery means the sum of both cups and plates.

Q21 Divide 35 in the ratio 3:2
Answer: Sum of ratio terms = \(3 + 2 = 5\)
First part = \(\frac{3}{5} \times 35 = 3 \times 7 = 21\)
Second part = \(\frac{2}{5} \times 35 = 2 \times 7 = 14\)
The parts are 21 and 14.
In simple words: Divide the total number by the sum of parts, then multiply by each ratio term.

Exam Tip: Verify your answer by adding the divided parts to check if they equal the original total.