Class 6 Mathematics

Arithmetic Chapter 7 Metric System

Arithmetic Chapter 7 Unit

Definition: How many times or a part of a body or a physical quantity is measured in comparison to that of a definite and convenient part of the body or a physical quantity taken as standard measurement then this definite and convenient part is called the unit.

Unit is of two types namely:

1. Fundamental unit and

2. Derived unit.

Fundamental Unit :

Definition:

1. The units of those physical quantities which do not depend on one another and from whose units, the units of all other quantities can be formed are called Fundamental Units.

2. For example, the unit of length, the unit of mass, the unit of time, etc. are the fundamental units.

Important Definitions Related to the Metric System

Derived Unit :

Definition:

1. The units formed from one or more fundamental units are called Derived Units.

2. For example, the unit of area, the unit of volume, etc. are the derived units.

 

Arithmetic Chapter 7 International Fundamental unit

At present, there are 7 internationally recognized fundamental units.

These are:

 

WBBSE Class 6 Metric System Notes

Arithmetic Chapter 7 Metric system or Decimal system

Definition:

1. The system of measurements in which the interrelation among the units can be determined by taking 10 as the base is called the metric system or decimal system.
2. In other words, in this system, any unit in comparison to its preceding or succeeding unit is 10 times or parts.

The list of measurements in the metric system is given below:

1. Units in the measurement of length:

1. 10 millimeters (mm) = 1 centimeter (cm)
2. 10 centimeters = 1 decimeter (dcm)
3. 10 decimeters = 1 meter (m)
4. 10 meters = 1 decameter (Dm)
5. 10 decameters = 1 hectometer (Hm)
6. 10 hectometers = 1 kilometer (Km)

Understanding the Metric System

From the above list, we get the following relations:

1 kilometer

= 1 x 10 hectometers = 10 hectometers

=10 x 10 decameters = 100 decameters

= 100 x 10 meters = 1000 meters

= 1000 x 10 decimeters = 10000 decimeters

= 10000 x 10 centimeters = 100000 centimeeters

= 100000 x 10 millimeters = 1000000 millimeters

Again, 1 meter

= 1 x 10 decimeters = 10 diameters

= 10 x 10 centimeters = 100 centimters

= 100 x 10 millimeters = 1000 millimeters

Also 10 kilometers = 1 myriametre.

From the following, we can keep in mind these units easily

 

Short Questions on Metric Units

2. Unit of the measurement of mass:

 

 

From the above list, we can write,

 1 kilogram 

(1 x 10) hectograms =  10 hectograms

 (10 × 10) decagrams  = 100 decagrams

 (100 x 10) grams = 1000 grams

 (1000 x 10) decigrams = 10000 decigrams

 (10000 × 10) centigrams = 100000 centigrams

(100000 × 10) milligrams = 1000000 milligrams

Again,

10 kilograms = 1 myriagram

100 kilograms = 1 Quintal

10 Quintals = 1 metric ton

1 Metric ton = (10 x 100) kilograms

= 1000 kilograms

Examples of Real-Life Applications of the Metric System

3. Unit of measurement of volume

1. Unit of measurement of the volume of liquid

The unit of measurement of the volume of liquid matter is Litre.

10 millilitres = 1 centilitre

10 centilitres = 1 decilitre

10 decilitres = 1 litre

10 litres = 1 decalitre

10 decalitres = 1 hectolitre 

10 hectolitres = 1 kilolitre

So from the above list, we can write,

1 kilolitre 

(1 × 10) hectolitres = 10 hectolitres

(10x 10) decalitres = 100 decalitres 

(100 x 10) litres = 1000 litres

(1000 x 10) decilitres = 10000 decilitres 

(10000 × 10) centilitres = 100000 centilitres

= (100000 x 10 millilitres = 1000000 millilitres

Again,

1 Litre = 10 decilitre

= 100 centilitres

= 1000 millilitres

1 kilolitre = 1000 Litres

 

2.  Unit of measurement of the volume of solid body

The unit measurement of the volume of a solid body is cubic meters.

 

Now, 

1 cubic metre = 1 metre x,1 metre x 1 metre

= 10 decimetres x 10 decimetres x 10 decimetres

= 1000 cubic decimetres

1 cubic metre = 1 metre x 1 metre x 1 metre

= 100 cm x 100 cm x 100 cm

= \(10^6\) cubic cm

1 cubic kilometre = 1 km x 1 km x 1 km

= 1000 m x 1000 m x 1000 m

= \(10^9\) cubic meters

= \(10^9 \times 10^6\) cubic centimeters

= \(10^15\) cubic cm

 

4. Unit of measurement of Area

The unit of measurement of the area of any surface is Are.

 

 

∴ 1 sq. kilometre = 1 km x 1 km

= 1000 m x 1000 m

= \(10^6\) sq. meters

1 sq. meter 1m x 1m

= 100 cm x 100 cm

= 10000 sq. cm

= \(10^4\) sq. cm

1 Are = 1 sq decametre

= 100 sq. meters

1 hectare = 100 ares

= 100 x 100 sq. metres

= 10000 sq. metres

1 centesimal = 40.4678 sq. metres

= 1000 sq. links [ 1 link = 0-2011 metres, 100 links 1 chain]

1 Acre = 100 centesimal

= 4046.78 sq. metres 

= 60 katas

1 Katha = 1.66 centesimal

1 Bigha = 20 Kathas 

= 33.2 centesimals

 

5. Unit of measurement of Time

1 minute = 60 seconds

1 hour = 60 minutes

= 60 x 60 seconds

= 3600 seconds

1 day = 24 hours

1 week = 7 days

= (7 x 24) hours

= 168 hours

1 Paksha = 15 days

1 month = 30 days

1 year = 12 months

= 52 weeks

= 365 days

1 century = 100 years.

Leap-Year: If the number of any year is divisible by 4, then the year is called Leap-Year.

The number of days of a Leap Year = 365+ 1

= 366 days

and the number of days in the month of February in a leap-year = 28+ 1

= 29 days.

Real-Life Scenarios Involving Distance and Weight Measurements

Arithmetic Chapter 7 Unit Of Different Measurements

1. Unit of Length

Unit of Length:

The smallest unit of length is Farmy and the greatest unit is Parsec. 

1 Farmy (F) 10-15 metre  

1 Angstram (Å) = 10-10 metre 

1 micrometer or micron (um) = 10-6 meter 

1 nanometer (nm) = 10-9 metre 

1 light year = 9.46 x 1015 metres

[The distance traveled by light in 1 year in zero medium is called 1 light year.]

1 parsec = 3.08 x 1016 metres = 3.26 light years.

2. The unit of force is Newton 

3. The unit of work or energy is Joule 

4. The unit of power is Watt

5. The unit of electric charge is Coulomb

6. The unit of electrical potential-Difference is Volt

7. The unit of electrical resistance is the Ohm 

8. The unit of pressure is Paskel

9. The unit of the Plane cone is Radian

10. The unit of frequency is Hertz.

Arithmetic Chapter 6 Multiplication And Division Of A Decimal Fraction By Whole Numbers And Decimal Fraction

 

1. Multiplication of Decimal Numbers by 10, 100, 1000……

1. The effective rule of multiplication of any decimal number by 10, 100, 1000,
2. Move the decimal point according to the following rule
3. Move the decimal point 1 digit towards the right when a decimal number is multiplied by 10.
4. Move the decimal point 2 digits towards the right when a decimal number is multiplied by 100.
5. Move the decimal point 3 digits towards the right when a decimal number is multiplied by 1000.
6. Move the decimal point 4 digits towards the right when a decimal number is multiplied by 10000.
7. ∴  Move the decimal point n digits towards the right when a decimal number is multiplied by 10n. [n = 1, 2, 3, ]
8. These imply that when a decimal number is multiplied by the multiple of 10, move the decimal point as many digits towards the right as many zeroes (0) after 1 are there in the multiple of 10.
9. If the given decimal number does not contain digits after the decimal point as many zeroes are there in the multiple of 10, put the necessary number of zeroes to the right side of
10. the given decimal number then put the decimal point and then put one or two more zeroes.
11. For example, observe the following two examples: 2-54 x 10 = 25-4
12. Here 10 contains one zero after 1, so the decimal point is moved one digit towards the right.
    4.75 x 100000 = 475000.00
13. Here 100000 contains 5 zeroes after 1.
14. ∴ The decimal point should be moved 5 digits towards the right but in the given decimal numbers 4.75,
15. There are only two digits after decimal points 7 and 5.
16. That’s, why 3 zeroes are put to the right side of the given number as 4.75000 and then the decimal point.
17. Is moved 5 digits towards the right and two zeroes are put after the decimal point in the final answer and we get 475000.00.

WBBSE Class 6 Decimal Fractions Notes

2. Multiplication of Decimal Number by Integer :

Effective Rule:

1. First step: First find the number of digits after the decimal point;

2. Second step: Removing the decimal point, the multiplication process is done according to the general rule.

3. Third step:

1. In this step, the decimal point is to be put in the product obtained (which is obtained in the 2nd step).
2. Put the decimal point as many digits (towards left) before the number of digits from the end digit of the product obtained as many digits lie after the decimal point in the first step.
3. If the product contains digits less than the number of digits after the decimal point in the first step.
4. Then put the necessary number of zeroes to the left of the product and then put the decimal point.
5. For this, observe the following example.
6. 0-0002 x 12 = 0-0024.
7. Here, in the given decimal number, we have 4 digits after the decimal point and after removing the decimal point, the product is 2 x 12 = 24 which contains only 2 digits.
8. For this, we insert 2 zeroes to the left side of 24, and then the decimal point is put as shown above and there will be 4 digits after the decimal point in the product.

Important Definitions Related to Decimal Operations

3. Multiplication of a Decimal number by a Decimal number :

Effective Rule:

Step 1:

1. First, determine the number of digits after the decimal point of the multiplicand decimal number, and then determine the number of digits after the decimal point of the multiplier decimal number.
2. Now add these two numbers of digits after the decimal point in both the multiplier and multiplicand.

Step 2:

1. Determine the product of the multiplicand and multiplier after removing the decimal point of both.
2. The multiplicand and multiplier are according to the general rule.

Understanding Multiplication of Decimal Fractions

Step 3:

1. Now put the decimal point in the product obtained in step 2 according to the following rule Place the decimal point as many digits (towards left) before the number of digits from the end digit of the product obtained in step 2 as many digits obtained in step 1.
2. If the product contains digits less than the number of digits as obtained in step 1, then put the necessary number of zeroes to the left of the product obtained in step 2 and then put the decimal point.
3. Observe the following examples so that you will have a clear concept of the multiplication of a decimal number by a decimal number.

 

Arithmetic Chapter 8 Percentage

Arithmetic Chapter 8 What is meant by percentage

1. In our daily life, we often say that the price of an article increases by 4%; the student has got 60% marks, etc.

2. Now, naturally, the question arises what is the meaning of 4% or 60%?

3. In simple language, the meaning of an increase in price by 4% of an article is that the previous price of that article was Rs. 100 and the present price of the article is ₹ (100 + 4)

= ₹ 104.

4. This implies that the increase in the price of the article is ₹ 4 over ₹ 100.

5. Similarly “getting 60% marks” means that the student has got 60 marks in an examination over 100 marks.

6. Again, if a student has obtained 70 marks in an examination of 100 marks and we say that the student has obtained 70% marks.

7. So the process in which a number or a quantity is expressed as a part of 100 is called percentage.

WBBSE Class 6 Percentage Notes

Arithmetic Chapter 8 Conversion Of Percentages Into Fractions

1. When x% is converted into a fraction, we get x/100, etc.

2. So x% = x/100

∴ 10% = 10/100

= 1/10

25% = 1/4

a% = a/100 etc.

3. Therefore the increase in the price of an article by 4% means that the price of the article increases by 4/100 = 1/25 part i.e., the price of the article increases by 1/25 part or

the previous price of the article.

Understanding Percentage Calculation

Arithmetic Chapter 8 Conversion Of Fraction Into Percentage

1. If n is a natural number, then 1/n is a fraction.

2. If it is expressed in percentage, then we get \(\frac{1}{n8} \times 100 \%\)

3. ∴ 1/2 = (\(\frac{1}{2} \times 100 \))%

= 50%

1/2 = (\(\frac{1}{5} \times 100\))%

= 20%

3/4 = \(\frac{3}{4} \times 100\)

= 75%

 

Arithmetic Chapter 8 Conversion Of Decimal Fraction Into Percentage

1. Suppose that 0-4 is a decimal fraction. We want to convert it into a percentage.

2. At first, we convert 0-4 into a vulgar fraction and we get 0.4

= 4/10

= 2/5.

3. Then according to the previous article, we get, 2/5 = ( \(\frac{2}{5} \times 100\) )%

= 40%

∴ 0.4 = 40%

Short Questions on Percentage Problems

1. Similarly, 1.5 

1.5 = 15/10

( \(\frac{3}{4} \times 100\))%

= 150%

1.5 = 150%

2. Similarly 2.4 

2.4 = 24/10

= 12/5

= (\(\frac{3}{4} \times 100\))%

= 240% etc.

2.4 = 240%

 

Arithmetic Chapter 8 Conversion Of Percentage Into Decimal Fraction

1. We know that x% = x/100

= (0.01)x.

∴ x% = (0.01)x

1. Similarly, 2%

2% = 2/100

= 0.02

2% = 0.02

Common Questions About Finding Percentage

2. Similarly 10%

10% = 10/100

= 1/10

= 0.1

10% = 0.1

3. Similarly 25%

= 25/100

= 0.25.

25% = 0.25.

∴ The rule is that the percentage → Vulgar fraction → Decimal fraction.

 

 

 

Arithmetic Chapter 5 Multiplication And Division Of A Number And By Fraction

Arithmetic Chapter 5 the Rule Of Multiplication Of A Fraction By Whole Number

1. Multiplication of a proper fraction by the whole number

1. We know that 2/7 is a proper fraction (v the numerator < the denomination).
2. Suppose we have to multiply this proper fraction by any whole number say 7.
3. So the mathematical problem is  2/7 x 7 = What is the value?
4. Here the rule is:

A proper fraction x the whole number = \(\frac{The number of the proper fraction x whole number}{The denominator of the proper fraction}\)

According to the above rule, \(\frac{2}{7} \times 7\)

= 2

1. Similarly \(\frac{5}{6} \times 8\)

= \(\frac{5 \times 8}{6}\)

= \(\frac{5 \times 4}{3}\)

= 20/3

= \(6 \frac{2}{3}\)

\(\frac{5}{6} \times 8\) = \(6 \frac{2}{3}\)

WBBSE Class 6 Multiplication and Division Notes

2. similarly  \(\frac{11}{14} \times 21\)

= \(\frac{11}{14} \times 21\)

= \(\frac{11 \times 21}{14}\)

= \(\frac{11 \times 3}{2}\)

= 33/2

= \(16 \frac{1}{2}\)

\(\frac{11}{14} \times 21\) = \(16 \frac{1}{2}\)

 

3. similarly \(\frac{17}{25} \times 35\)

= \(\frac{17}{25} \times 35\)

= \(\frac{17 \times 35}{25}\)

= \(\frac{17 \times 7}{5}\)

= 119/5

= \(23 \frac{4}{5}\)

\(\frac{17}{25} \times 35\) = \(23 \frac{4}{5}\)

Understanding Multiplication of Fractions

2.  Multiplication of an improper fraction by the whole number :

1. We know that “ is an improper fraction (v the numerator > the denominator).
2. Suppose we have to multiply this improper fraction by any whole number say 15.
3. So the mathematical problem is:

\(\frac{7}{18} \times 15\)

Here also the rule is: The fraction x whole number = \(\frac{The numerator of the fraction x whole number}{The denominator of the fraction}\)

According to the above rule,

= \(\frac{8}{5} \times 15\)

= \(\frac{8 \times 15}{5}\)

= 8 x 3

= 24

Short Questions on Division of Fractions

1. Similarly \(\frac{11}{7} \times 4\)

= \(\frac{11}{7} \times 4\)

= \(\frac{11 \times 4}{7}\)

= 44/7

= \(4 \frac{2}{7}\)

\(\frac{11}{7} \times 4\) = \(4 \frac{2}{7}\)

 

2. Similarly \(\frac{15}{4} \times 9\)

= \(\frac{15}{4} \times 9\)

= \(\frac{15 \times 9}{4}\)

= 135/2

= \(33 \frac{3}{4}\)

\(\frac{15}{4} \times 9\) = \(33 \frac{3}{4}\)

 

3. Similarly \(\frac{24}{8} \times 14\)

= \(\frac{24}{8} \times 14\)

= \(\frac{21 \times 14}{8}\)

= 147/4

= \(36 \frac{3}{4}\)

\(\frac{24}{8} \times 14\) = \(36 \frac{3}{4}\)

Common Questions About Multiplying Fractions

3. Multiplication of a mixed fraction or a complex fraction by a whole number :

1. To multiply a mixed fraction or a complex fraction by a whole number, first, we have to convert the given fraction to the improper fraction and then according to the rule of 5.1.B.
2. The multiplication is to be completed.

1. For Example \(5 \frac{1}{3} \times 7\)

= \(5 \frac{1}{3} \times 7\)

= \(\frac{16}{3} \times 7\)

= \(\frac{16 \times 7}{3}\)

= 112/3

= \(37 \frac{1}{3}\)

\(5 \frac{1}{3} \times 7\) = \(37 \frac{1}{3}\)

 

2. For Example \(7 \frac{1}{2} \times 12\)

= \(7 \frac{1}{2} \times 12\)

= \(\frac{15}{2} \times 12\)

= \(\frac{15 \times 12}{2}\)

= 15 x 6

= 90

\(7 \frac{1}{2} \times 12\) = 90

Practice Problems on Fraction Multiplication and Division

3. For Example \(\frac{\frac{3}{4}}{\frac{5}{7}} \times 14\)

= \(\frac{\frac{3}{4}}{\frac{5}{7}} \times 14\)

= \(\left(\frac{3}{4} \times \frac{7}{5}\right) \times 14\)

= \(\frac{21}{20} \times 14\)

= \(\frac{21 \times 14}{20}\)

= \(\frac{21 \times 7}{10}\)

= 147/10

= \(14 \frac{7}{10}\)

\(\frac{\frac{3}{4}}{\frac{5}{7}} \times 14\) = \(14 \frac{7}{10}\)

 

 

 

Class 6 Maths Roman Numerals

Arithmetic Chapter 4 Roman Numbers Up To One Hundred

Arithmetic Chapter 4 Introduction To Roman Numbers

1. Among all the human civilizations which have been developed in different terminals of the world, Roman civilization is one of the distinguishable civilizations.

2. From ancient times the Romans are prospering in knowledge science.

3. They are also very experts in mathematical accounts.

4. The signs and symbols which the Romans used to express the numbers are called the Romans’ Number System.

5. The learned people thought that the Roman Numbers originated from the fingers of the hands of the people and from the formation of the hands of the people.

6. For example, the numbers I, II, III, etc. are invented from the formation of the fingers of hands, and the numbers V, X, etc. are also invented from the formation of hands.

7. These are very ancient, and still we use these Roman Numbers abundantly today.

8. We use Roman Numbers in the dial of a clock, for writing the classes of an educational institution, for writing the chapters of a printed book, etc.

9. There are some remarkable number systems other than the Roman Number System.

For example:

1. Hindu Arabic Numbers System.

2. Binary numbers system.

WBBSE Class 6 Roman Numbers Notes

1. Hindu Arabic Numbers System.

1. We are also well-known for this system excessively.

2. Generally, we use this system in our mathematical calculations.

3. At present this system is accepted universally and used internationally system.

4. The number of principal signs or symbols in this system is ten such as 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.

5. Any number (however large or small) can be expressed using these ten symbols one or more times.

6. These signs or symbols are called digits.

2. Binary number system.

1. In this system, any number can be expressed by only two numbers 0 and 1.

Understanding Roman Numerals

Class 6 Maths Roman Numerals 

Arithmetic Chapter 4 The Principal Sings or Symbols Of Roman Numbers

1. The number of main alphabets of signs which are used to represent the numbers in Roman numerals is seven such as V, X, L, C, D, M.

2. Using these 7 symbols any number can be expressed in this system.

3. In comparison to the Hindu Arabic Number system, the values of 7 symbols are given in the following table

 

 

Arithmetic Chapter 4 Rules For Writing Roman Numbers

We can write according to the values of the alphabetical symbols of Roman Numbers:

1. I<V<X<L<C<D<M

2. So the value of any symbol is always less than that of any symbol on its right side and is always greater than that of any symbol on its left side.

3. If we take two symbols V and C, between these two V is less than C and C is greater than V.

4. This is because C lies on the right side of V.

5. For the same reason between X and D, X is smaller than D and D is greater than X.

Important Definitions Related to Roman Numerals

There are some rules for writing Roman Numbers. Now we shall discuss these rules:

Rule 1:

1. In Roman Numbers only I, X, C, and M signs are repeated consecutively not more than 3 times.

2. They can be repeated a maximum of 3 times only, V, L, and D can not be used in any number consecutively more than once.

3. We can not use 1 four times to write 4 as IIII.

4. To write 40, X cannot be used 4 times as XXXX.

5. To write 600, C can not be used 6 times as CCCCCC.

1. We can use I as II, III.
2. X as XX, XXX.
3. C as CC, CCC.
4. M as MM, MMM.
5. II = 1 + 1 = 2.
6. III= 1 + 1 + 1 = 3.
7. XX = 10 + 10 = 20.
8. XXX =10 + 10 + 10 = 30.
9. CC = 100 + 100 = 200.
10. CCC = 100 + 100 + 100 = 300.
11. MM = 1000 + 1000 = 2000.
12. MMM = 1000 + 1000 + 1000 = 3000.

Rule 2:

1. In the Roman Number System, if a sign is repeated in any number, it implies addition.

1. II = 1 + 1 = 2.
2. III = 1 + 1 + 1 + 3.
3. XX = 10 + 10 = 20.
4. XXX = 10 + 10 + 10 = 30.
5. CC = 100 + 100 = 200.
6. CCC = 100 + 100 + 100 = 300.
7. MM = 1000 + 1000 = 2000.
8. MMM = 1000 + 1000 + 1000 = 3000.

Roman Numerals Worksheet for Class 6 

Rule 3:

1. In the Roman Number System, if a number is expressed when a sign of a smaller value is inserted after a sign of greater value then the value of the number will be the sum of the values of the signs.

2. For example, suppose a number is expressed as VI, then the value of the sign (or symbol) V is greater than the value of the sign (or symbol) I i.e., here a sign of smaller value is placed after a sign of greater value.

1. The value of the number which is expressed as VI = 5 + 1 = 6
2. Similarly, XII = 10 + 1 + 1 = 12,
3. LIII = 50 + 1 + 1 + 1 = 53, .
4. LXXV = 50 + 10 + 10 + 5 = 75,
5. LXXXI = 50 + 10 + 10 + 10 + 1 = 81

Rule 4:

1. In the Roman Number System, if a number is expressed when a sign of smaller value is inserted before a sign of greater value, then the value of the number will be the difference between the values of the Signs (here the smaller value of the sign is subtracted from the greater value of the sign).

2. For example, suppose a number is expressed by IV. In this number, the value of the sign V is greater than the value of the sign I, and the sign I is placed on the left side of V.

1. So the value of the number which is expressed as IV = 5 – 1 = 4.
2. Similarly, IX = 10 – 1 = 9.
3. XL = 50 – 10 = 40.
4. CD = 500 – 100 = 400 etc.

Real-Life Applications of Roman Numerals

Rule 5:

1.  If a sign (or symbol) of smaller value is placed between the two signs of greater values, then the value of the sign of smaller value is subtracted from the greater value of the next sign to be placed and never be added to the greater value of sign which is placed before the sign of smaller value.

2. For example, suppose a number is expressed by Roman Number System XIV.

3. Here the sign I of smaller value is placed between two signs X and V of greater values.

4. So according to the rule, the value of I is subtracted from the value of V (v V is the sign of greater value which is placed next to I) and the value of the number will be (XIV)

= 10 + (5 – 1)

= 14

1. Similarly, X3X = 10 + (10 – 1) = 10 + 9 = 19.
2. LXL = 50 + (50 – 10) = 50 + 40 = 90.
3. CXC = 100 + (100 – 10) = 100 + 90 = 190.
4. DCD = 500 + (500 – 100) = 500 + 400 = 900.
5. DCM = 500 + (1000 – 100) = 500 + 900 = 1400.
6. MCD = 1000 + (500 – 100) = 1000 + 400 = 1400.
7. MCM = 1000 + (1000 – 100) = 1000 + 900 = 1900.
8. MCDLX = 1000 + (500 – 100) + 50 + 10 = 1000 + 400 + 50 + 10 = 1460.
9. MCMLX = 1000 + (1000 – 100) + 50 + 10 = I960.

Roman Numbers up to 100 

Rule 6:

1. To mean one thousand times any number or of the value of any sign, a small line above the sign is drawn.

2. For example, 1000 times of X = X = X x 1000 = 10 x 1000 = 10000.

1. In the same way, we have,
2. C = C x 1000 = 100 x 1000 = 100000.
3. M = M x 1000 = 100 x 1000 = 1000000.
4. XXTV = XXIV x 1000 = [10 + 10 + (5 – 1)] x 100 = 24 x 1000 = 24000.

Examples of Roman Numerals in Real Life

WBBSE Class 6 Maths Chapter 4 

Arithmetic Chapter 4 Characteristics Of The Roman Number System

1. There is no sign to express zero in the Roman Number System.

2. The same number can be arranged in different ways.

3. For Example 1400 = DCM ; 1400 = MCD

4. In the Roman Number System, the difference in the values of any two consecutive signs of the main 7 alphabetic signs is not fixed.

5. For example,

1. The difference between the values of I and V = 5 – 1 = 4.
2. The difference between the values of V and X = 10 – 5 = 5.
3. The difference between the values of X and L = 50 – 10 = 40.
4. The difference between the values of L and C = 100 – 50 = 50,
5. The difference between the values of C and D = 500 – 100 = 400,
6. The difference of the values of D and M = 1000 – 500 = 500.

6. On the other hand, in Hindu Arabic Number System, the difference between 0 and 1 = 1— 0=1

1. The difference of 1 and 2 = 2 — 1 = 1
2. The difference of 2 and 3 = 3 – 2 = 1
3. The difference of 8 and 9 = 9 — 8= 1

7. ∴ The difference between two consecutive numbers is always the same 1.

 

 

 

Arithmetic Chapter 2 Concept Of Seven And Eight Digit Numbers

Arithmetic Chapter 2 Seven And Eight Digit Numbers

1. You have already learned that there are ten digits namely : 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.

2. Among these 10 digits, taking any seven or eight digits perfect numbers can be formed.

3. These numbers are called seven or eight-digit numbers.

4. For example, taking seven digits such as 0, 1, 2, 3, 4, 5, 6 a perfect number is formed which is 2041365.

5. This number is a seven-digit number.

6. It is to be noted that, with these seven digits actually

7 x 6 x 5 x 4 x 3 x 2 x 1 – 6 x 5 x 4 x 3 x 2 x 1

= 5040 – 720

= 4320

numbers can be formed (using one digit once in each number only).

7. Similarly taking eight digits such as 0, 1, 2, 3, 4, 5, 6, and 7 a perfect number 40326751 is formed.

8. This number is an eight-digit number.

9. Here actually (using one digit once in each number only)

8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 – 7 x 6 x 5 x 4 x 3 x 2 x 1

= 40320 – 5040

= 35280 numbers can be formed.

10. So you can form seven and eight-digit perfect numbers by taking any seven and eight digits from ten digits.

WBBSE Class 6 Seven and Eight Digit Numbers Notes

Arithmetic Chapter 2 Value Or Principal Value Or Absolute Value Of A Digit

1. The actual value of a digit in a number is the own value of the digit, which is always the same.

2. The actual value or principal value of a digit is the absolute value of the digit. For example, let us consider the number 25032189.

3. In this number (from left),

1. The principal value of 2 is 2
2. The principal value of 5 is 5
3. The principal value of 0 is 0
4. The principal value of 3 is 3
5. The principal value of 2 is 2
6. The principal value of 1 is 1
7. The principal value of 8 is 8
8. The principal value of 9 is 9.

4. So in any number, the actual value of 2 is always 2; the actual value of 4 is 4; the actual value of 5 is 5, etc.

5. Anywhere a digit in a number is placed, the actual value of the digit is always the same.

Understanding Seven Digit Numbers for Kids

Arithmetic Chapter 2 The Place Value Of A Digit

1. You know that, in a number, the digit in the extreme right is called the unit’s place digit; the digit next to its left place is called the ten’s place digit.

2. The digit next to its left place is called the hundred’s place digit, …… etc.

3. Thus the list of the places of the

Digits are arranged from extreme right to left as:

 

In short, they can be written as:

 

 

4. According to this rule, if we proceed towards left then the place value of any place will be multiplied by 10 and if we proceed towards right then the place value of any place will be divided by 10.

5. For example, if we proceed from the lac place towards the left which is in the ten lac place, then the place value will be multiplied by 10.

6. Again if we proceed from the lac place towards the right that is in the ten thousand places, then the place value will be divided by 10.

7. In this way we get,

1. The value in the Unit place = 1
2. The value in the Ten place = 1 x 10 = 10
3. The value in the Hundred place = 1 x 100 or 10 x 10 = 100
4. The value in the Thousand place = 100 x 10 = 1000 ,
5. The value in the Ten Thousand place = 1000 x 10 = 10000
6. The value in the Lac place = 10000 x 10 = 100000
7. The value in the Ten Lac place = 100000 x 10 = 1000000
8. The value in the Crore place = 1000000 x 10 = 10000000.

8. So the place value of any digit in a number is the product of the actual value of the digit and the place value of the digit where it lies.

9. Let us consider the number 1583246.

10. We want to determine the place value of 8. Then the actual value of 8 = 8.

11. 8 is placed in ten thousand places and the place value of ten thousand is 10000.

12. The place value of the digit 8 = 8 x 10000 = 80000

13. Similarly, the actual value of 2 is 2 and the digit 2 is placed in the hundred places whose place value = 100.

14. The place value of 2 = 2 x 100 = 200 In this way we can determine the place value of any digit in any number.

Important Definitions Related to Large Numbers

 

Arithmetic Chapter 2 Expansion of a number according to the place value

 

1. To expand a given number according to the place value write the place value of each digit (keeping the order of the digits correctly) with a “+” sign between any two.

2. Let us take the example that we have, and expand the number 65432019 according to the place value.

3. Then first write the list of the places of the digits, and next find the place values of the digits as follows.

 

1. Place value of 6 = 6 x 10000000 = 60000000
2. Place value of 5 = 5 x 1000000 = 5000000
3. Place value of 4 = 4 x 100000 = 400000
4. Place value of 3 = 3 x 10000 = 30000
5. Place value of 2 = 2 x 1000 = 2000
6. Place value of 0 = 0 x 100 = 0
7. Place value of 1 = 1 x 10 = 10
8. Place value of 9 = 9 x 1 = 9

∴ The expanded form of the number 65432019 is

65432019 = 60000000 + 5000000 + 400000 + 30000 + 2000 + 0 + 10 + 9

= 60000000 + 5000000 + 400000 + 30000 + 2000 + 10 + 9

 

Arithmetic Chapter 2 Writing a number given in numeral in words and a number given in words in numeral

Writing a number in words (the number is given in numerals):

1. Here first write the digits (given in numerals) in the list of the place values.

2. Then express the place values of the digits in words.

3. We consider the number 7540351 which is to be expressed in words.

4. First, write the digits of the given number according to the list of place values as follows.

 

5. Now expressing the place values in words, we get, Seven ten lac five lac four ten thousand three hundred five ten one unit.

6. This can be written as Seventy-five lac forty thousand three hundred fifty-one.

7. So the given number when expressed in words we get.

8. Seventy-five lac forty thousand three hundred fifty one.

9. According to the place value, when expressed in words, we get,

10. Seven ten lac five lac four ten thousand three hundred five ten one unit.

Examples of Real-Life Applications of Large Numbers

Expressing a number in numerals (when the number is given in words) :

1. When a number is given in words, we have to express it in numerals, then first the digits in the number can be listed properly according to the list of place values.

2. Here special care is to be taken that if there is no digit mentioned in the place value, it should be written “0” in that place.

3. For example, if the number is given in words say “Six crore Three thousand two”.

4. We have to express this number in numerals.

5. Then at first the digits (written in words in the given number) are listed in the list of place values and we get,

 

 

6. Here it is observed that there is no digit (in words) mentioned in ten lac, lac, ten thousand, hundred, and tens places so we write 0 in each of these places in the list of place values.

∴ “Six crore three thousand two” = 60003002

 

 

 

Arithmetic Chapter 3 Logical Approximation Of Number

Arithmetic Chapter 3 To express any number to the nearest multiple of 10, 100, 1000,…. etc. in integers :

1. In our daily life, we often face different types of problems such as “How many populations are there in our village ?” or “How many populations are there in India ?” etc.

2. In answering this type of question, we all generally do not answer the exact correct number of the population. But we say a convenient and approximate nearest integer.

3. For example, in the case of the first question, if the population of the village is 2145, then we say in its answer 2000.

Understanding Approximation in Maths for Kids

4. Again, in the case of the second question, if the correct population in India is one hundred twenty-five crore seventy-five lac twenty-seven thousand seven hundred twenty-two, then we say in answer that the population in India is about one hundred twenty-six crore.

5. The tendency to say this type of answer in integers is remaining with almost all of us.

6. Now the question arises that whether there are any rules to answer in integers or not.

7. We shall illuminate this in the following discussions.

WBBSE Class 6 Logical Approximation Notes

Arithmetic Chapter 3 The Rules To Express Any Number To The Nearest Multiple Of 10, 100, 1000 …… Etc

1. To express a number to the nearest multiple of 10:

1. To express a number to the nearest multiple of 10, if the digit in the units place is 5 or greater than 5, then the digit in the tens place increases by 1, and the digit in the units place will be 0.
2. To express a number of nearest multiple of 10, if the digit in the units place is less than 5, then the digit in the tens place will be the same and the digit in the units place will be 0

2. To express a number to the nearest multiple of 100.

1. To express a number to the nearest multiple of 100, if the digit in the tens place of the number is less than 5, then the hundreds place digit of the number will be the same as before, and the digits in the tens place and the units place both will be zero.
2. To express a number to the nearest multiple of 100, if the digit in the tens place of the number is 5 or greater than 5 then the digit in the hundreds place increases by 1 and both the digits.
3. In the tens place and the units, the place should be 0.

Important Definitions Related to Approximation

 

3. To express a number to the nearest multiple of 1000.

1. To express a number to the nearest multiple of 1000, if the digit in the hundred places of the number is less than 5, then the thousand place digit will be the same as before and the digits in the hundreds place, tens place, and the units place all will be 0.
2. To express a number to the nearest multiple of 1000, if the digit in the hundred places of the number is 5 or greater than 5, then the digit in the thousands place increases by 1, and all the digits in the hundreds, tens, and units place is 0.

4. General rule To express a number to the nearest multiple of 10ⁿ (n = 1, 2, 3, ), if the digit in nth place (from right) is

1. less than 5, then (n + 1)th digit will be the same as before and all the next digits will be zero.
2. 5 or greater than 5 then (n + 1)th digit increases by 1 and next all digits should be zero.

Examples of Real-Life Applications of Approximation

Example: Suppose that we want to express a number to the nearest multiple of 10⁷ = 10000000 in an integer.   Then if the 7th place digit (from right) i.e., the digit in the ten lacs place is

1. less than 5, then the digit in the 8th place i.e., the digit in the crores place will be the same as before and all the next digits i.e., ten lac, lac, ten thousand, thousand, hundred, tens, units places will be 0.
2. 5 or greater than 5, then the digit in the 8th place i.e., the digit in the crores place increases by 1, and all the next place digits should be zero.

 

You observe the following examples so that you will have a clear concept about the above discussions and also about the chapter.

 

 

 

Geometry Chapter 2 Points Lines Line Segment Ray And Their Concepts

Question 1. Answer the following questions:

1. How many line segments can be drawn through a fixed point?

Solution:

An infinite number of line segments can be drawn through a fixed point.

Simplification Maths Class 6

2. What is the greatest number of points of intersection at which three noncoincident line segments can be intersected?

Solution:

A maximum number of three non-coincident line segments that can be intersected is three.

Read And Learn More WBBSE Solutions For Class 6 Maths

3. How many line segments can be drawn through three non-collinear points?

Solution:

Three line segments can be drawn through three non-collinear fixed points.

4. What is the maximum number of points of intersection at which two noncoincident line segments can be intersected?

Solution:

The maximum number of points of intersection that the given two non-coincident line segments can be intersected is one only.

WBBSE Class 6 Points and Lines Notes

5. How many endpoints are there in line segment \(\overline{\mathbf{A B}}\)? What are they?

Solution:

There are two endpoints of a line segment \(\overline{\mathbf{A B}}\) and they are A and B.

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6. How many endpoints are there in the ray \(\overrightarrow{\mathrm{AB}}\)?

Solution:

The ray \(\overrightarrow{\mathrm{AB}}\) has only one endpoint and this is A.

7. Are the rays \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{BA}}\) the same?

Solution:

No, the rays \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{BA}}\) are not the same, they are in opposite directions to each other.

8. Are the line segments \(\overline{\mathbf{A B}}\) and \(\overline{\mathbf{B A}}\) the same? From which point of view are they the same?

Class 6 Math Solutions WBBSE English Medium

Solution:

Yes, the line segments \(\overline{\mathbf{A B}}\) and \(\overline{\mathbf{B A}}\) are the same.

They are the same in respect of length.

9. Which of the three straight lines, line segments, and a ray has a definite length?

Solution:

The line segment has a definite length

10. How many line segments can be drawn through two given fixed points?

Solution:

Only one line segment can be drawn through two given fixed points.

Question 2. Write the name of the points of intersection of the following figures

1.

 

Solution:

The points of intersection are A, B, C, D, and O.

Short Questions on Points and Lines

2.

 

 

Solution:

The points of intersection are P, Q, R, S, and T.

Question 3. Write the names of line segments and rays of each of the following.

1.

 

 

Solution:

The line segments are \(\overline{\mathbf{B C}}\) and \(\overline{\mathbf{C D}}\).

The rays are \(\overrightarrow{\mathrm{BA}}\) and \(\overrightarrow{\mathrm{DE}}\).

2.

 

Solution:

The only one-line segment is \(\overline{\mathbf{A B}}\)

The rays are \(\overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{BD}}\)

Question 4. From the following figure, answer the following questions.

1. What are the points X, Y, and Z called?

Solution :

The points X, Y, and Z are called collinear points.

2. How many line segments can be drawn through X, Y, and Z ? Name the line segments.

Common Questions About Line Segments

Solution:

Three line segments can be drawn through X, Y, and Z.

The names of the line segments are XY, YZ, and XZ.

Question 5. Give the answer from the figure given alongside.

1. Write the names of the points of intersection in the figure.

Solution :

The names of the points of intersection are A, B, C, D, E, F, and O.

2. Write the names of the collinear points.

The collinear points are

A, O, D.

B, O, E.

C, O, F.

A, E, C.

A, F, B.

B, D, C.

3. Write the names of the line segments.

The line segments are \(\overline{\mathbf{A E}}\), \(\overline{\mathbf{C E}}\), \(\overline{\mathbf{A F}}\), \(\overline{\mathbf{B F}}\), \(\overline{\mathbf{A O}}\), \(\overline{\mathbf{O D}}\), \(\overline{\mathbf{B O}}\), \(\overline{\mathbf{O E}}\), \(\overline{\mathbf{O C}}\), \(\overline{\mathbf{F O}}\), \(\overline{\mathbf{A D}}\), \(\overline{\mathbf{B E}}\), \(\overline{\mathbf{C F}}\), \(\overline{\mathbf{A B}}\), \(\overline{\mathbf{B C}}\), \(\overline{\mathbf{A C}}\), \(\overline{\mathbf{B D}}\), \(\overline{\mathbf{D C}}\).

4. Write the names of the line segments which are either concurrent or coincident.

The concurrent line segments are \(\overline{\mathbf{A D}}\), \(\overline{\mathbf{B C}}\), \(\overline{\mathbf{C F}}\)

Simplification Questions For Class 6

Question 6. From the given figure answer the following questions :

1. Name six line segments:

Solution :

The names of the six line segments are \(\overline{\mathbf{P Q}}\), \(\overline{\mathbf{R S}}\), \(\overline{\mathbf{U V}}\),\(\overline{\mathbf{M N}}\), \(\overline{\mathbf{G H}}\), and \(\overline{\mathbf{A B}}\).

2. Write the names of 3 pairs of intersecting straight lines 

3. Write the names of 3 pairs of parallel straight lines.

3 pairs of parallel straight lines are

\(\stackrel{\leftrightarrow}{\mathrm{PQ}}\) || \(\stackrel{\leftrightarrow}{\mathrm{RS}}\)

\(\stackrel{\leftrightarrow}{\mathrm{RS}}\) || \(\stackrel{\leftrightarrow}{\mathrm{UV}}\)

and \(\stackrel{\leftrightarrow}{\mathrm{PQ}}\) || \(\stackrel{\leftrightarrow}{\mathrm{UV}}\)

 

Question 7. From the given figure answer the following questions.

 

 

1. PQ + QR = ?

Solution :

PQ + QR

= 2 cm + 3 cm

= 5 cm

PQ + QR = 5 cm

Examples of Real-Life Applications of Lines and Segments

2. PR + QS = ?

Solution :

PR + QS

= (PQ + QR) + (QR + RS)

= (2 cm + 3 cm) + (3 cm + 4 cm)

= 5 cm + 7 cm

= 12 cm.

PR + QS = 12 cm.

3. PS – QS =?

Solution :

PS-QS = (PQ + QR + RS)-(QR + RS)

= (2 cm + 3 cm + 4 cm) – (3 cm + 4 cm)

= 9 cm – 7 cm

= 2 cm.

PS – QS = = 2 cm.

4. PS =?

Solution :

PS = PQ + QR +RS

= 2 cm + 3 cm + 4 cm

= 9 cm.

PS =9 cm.

WBBSE Solutions For Class 6 Maths Algebra Chapter 3 Statistical Data Its Handling And Analysis

Question 1. In class VI of Mitra Institution, classes have been held for 22 days in the month of June. The number of students attending classes on these 22 days are given below :

30, 28, 34, 29, 25, 30, 28, 26, 29, 30, 22, 25, 26, 29, 30, 31, 21, 27, 25, 13, 32, 28.

Using the above raw data, prepare a frequency distribution table with tally marks.

Solution:

Arranging the given raw data in ascending order of magnitudes, we get: 

13, 21, 22, 25, 25, 25, 26, 26, 27, 28, 28, 28, 29, 29, 29, 30, 30, 30, 30, 31, 32, 34.

Read And Learn More WBBSE Solutions For Class 6 Maths

Frequency Distribution Table of the number of students attending the classes in class VI of Mitra Institution in the month of June:

WBBSE Class 6 Statistical Data Notes

Question 2. The weight (in kg) of 30 students of class VI of Saraswati Vidyamandir is given below :

32, 32, 37, 34, 37, 35, 35, 36, 37, 39, 40, 36, 37, 36, 32, 33, 31, 36, 37, 38, 40, 34, 36, 34, 35, 33, 34, 35, 32, 35.

Arranging these raw data, prepare a frequency distribution table.

Class 6 Math Solution WBBSE

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Solution: Arranging the given raw data, in ascending order of magnitudes, we get:

31, 32, 32, 32, 32, 33, 33, 34, 34, 34, 34, 35, 35, 35, 35, 35, 36, 36, 36, 36, 36, 37, 37, 37, 37, 37, 38, 39, 40, 40.

Frequency Distribution Table of the weights (kg) of 30 students of class VI of Sar&swati Vidyamandir:

Understanding Data Handling 

Question 3. A survey has been conducted of over 150 students of Habra High School. The content of the survey is how many students like to study what subject. The raw data obtained from the survey is given below:

 

 

Taking 1 unit = 5 students, prepare a Bar diagram.

Solution:  1 unit = 5 students

25 students = 25/5 = 5 units

30 students = 30/5 = 6 units

40 students = 40/5 = 8 units.

Frequency Distribution Table:

Short Questions on Statistical Analysis

 

 

Scale:

Along X-axis:

1 unit = 1 subject

(1 small square division)

Along Y-axis:

1 unit = 1 small square division

= 5 students

Common Questions About Data Representation

Question 4. The following is the raw data obtained from 20 famous mathematicians coming from different countries of the world who like to eat what types of fruits.

 

 

Taking 1 unit = 2 Mathematicians, prepare a Horizontal Bar Diagram based on this data. 

Solution:

∵ 2 Mathematicians = 1 unit

∴ 4 Mathematicians = \(\frac{1}{2} \times 4\)

6 Mathematicians = \(\frac{1}{2} \times 6\)

Frequency Distribution Table:

Class 6 Math Solutions WBBSE English Medium

 

Practice Problems on Data Handling

Question 5. The following is the bar diagram showing the number of students appearing at the Madhyamik Examination each year during the last 5 years of a school:

 

 

From the above bar diagram answer the following questions:

1. In which year maximum number of students have appeared for the examination?

Solution:

The length of the bar diagram in the year 2012 has the greatest of all. So the maximum number of students appeared at the examination in the year 2012. This year 120 students have appeared for the examination.

Important Definitions Related to Statistics

2. In which year least number of students have appeared at the examination?

Solution:

The length of the bar diagram in the year 2010 has the least of all.

So the least number of students appeared at the examination in the year 2010.

In this year 80 students appeared at the examination in the year 2010.

3. How many more students appeared at the examination in the year 2011 than in the year 2010?

Solution:

From the bar diagram, we see that 80 students appeared at the examination in the year 2010 and 110 students appeared at the examination in the year 2011.

110 – 80 = 30 more students have appeared at the examination in the year. 2011 than the year 2010.

4. How many fewer students appeared for the examination in the year 2010 than the year 2009?

Solution:

From the bar diagram, we see that 90 students appeared in the examination in the year 2009 and 80 students appeared at the examination.

90 – 80 = 10 fewer students appeared at the examination in the year 2009 than in the year 2010.

Examples of Real-Life Applications of Statistics

5. How many total students appeared at the examination between 2008 and 2010 both years included?

From the diagram, we set that, 100 students appeared in the examination in the year 2008.

90 students appeared in the examination in the year 2009.

80 students appeared at the examination in the year 2010.

From the year 2008 to the year 2010, a total number of students have appeared at the examination

= 100 + 90 + 80

= 270 students.

WBBSE Solutions For Class 6 Maths Algebra Chapter 2 Concept Of Directed numbers And Numbers Line

Question 1. What do the following quantities mean:

1. Profit of Rs. (- 7)

Solution:

A profit of ₹ (- 7) means a Loss of Rs. 7

2. – 5m. above

Solution:

– 5 metres above means 5 metres below or down

3. – 26 gm less

Solution:

– 26 gm less means 26 gm more

Read And Learn More WBBSE Solutions For Class 6 Maths

4. – 18 meters towards the east

Solution:

– 18 metres towards east means 18 metres towards West

5. savings of ₹ (- 23).

Solution:

Savings of ₹ (- 23) means the expenditure of ₹ 23

WBBSE Class 6 Directed Numbers Notes

Question 2. Write the opposite of the following quantities:

1. Expenditure of Rs. 15

Solution:

The opposite quantity of “Expenditure of ₹ 15” is “Income of  ₹ 15” or “Expenditure of ₹ (- 15)”.

Class 6 Math Solution WBBSE

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2. Climbing – 12 metres up

Solution:

The opposite quantity of “climbing (- 12) metres up” is “Descing (- 12) metres down” or “climbing 12 metres up”.

3. Profit of ₹ 80 

Solution:

The opposite of “Profit of ₹ 80” is “Loss of ? 80” or “Profit of ₹ (- 80)”.

4. Descending – 35 m down 

Solution:

The opposite of “Descending (- 35) metres down” is “climbing (- 35) metres up” or “Descending 35 metres down”.

5. – 24 kg increase in weight 

Solution:

The opposite quantity of “- 24 kg increase in weight” is 24 kg decrease in weight” or “24 kg increase in weight”.

6. 28 metres towards the right 

Solution:

The opposite of “28 metres towards the right” is “28 metres towards left” or “- 28 metres towards the right”.

Class 6 Math Solution WBBSE

7. 9 kg decrease in weight 

Solution:

The opposite of “9 kg decrease of weight” is “9 kg increase in weight” or “(- 9) kg decrease of weight”.

8. Loss of ₹ (- 5).

Solution:

The opposite of “Loss of ₹ (- 5)” is “Profit of ₹ (- 5)” or “Loss of ₹ 5”.

Question 3. Write the synonyms of the following quantities:

1. 10 km towards the north 

Solution:

The synonym of “10 km towards the north” is “-10 km towards the south”.

Understanding Number Line

2. – 3 kg decrease in weight 

Solution:

The synonym of “-3 kg decrease of weight” is “3 kg increase in weight”.

3. climbing 11 metres up 

Solution:

The synonym of “climbing 11 metres up” is “descending – 11 metres down”.

4. Profit of ₹ (- 18).

Solution:

The synonym of “Profit of ₹ (- 18)” is “Loss of ₹ 18”.

Class 6 Math Solutions WBBSE English Medium

Question 4. Write the opposite numbers of the following

1. – 17,  

Solution :

The opposite number of – 17 is + 17.

2. 0

Solution :

The opposite number of 0 is 0.

3. 1

Solution :

The opposite number of 1 is – 1.

4. 00

Solution :

The opposite number of 100 is – 100.

Short Questions on Directed Numbers

Question 5. Write the absolute values of the following numbers :

1. – 12, 

Solution:

|- 12| = – (- 12) = + 12 = 12 (v – 12 < 0)

2. + 13.

Solution:

|+ 13| = + (+ 13) = + 13 = 13 (v + 13 > 0)

Question 6.

1. Write 4 negative integers less than (- 8).

Solution:

4 negative integers less than – 8, are – 9, – 10, – 11, – 12.

Common Questions About Number Line Operations

2. Write 4 negative integers greater than (- 12).

Solution:

4 negative integers greater than (- 12) are – 11, – 10, – 9, – 8.

Question 7. Using the concept of opposite numbers, subtract the following :

1. (+ 14) – (+ 16)

Solution :

(+ 14) – (+ 16) = (+ 14) + (- 16)

[∵ The opposite number of (+ 16) is – 16]

= [- (16 – 14)]

= (- 2)

= – 2.

(+ 14) – (+ 16) = – 2.

2. (+ 25) – (+ 21)

Solution :

(+ 25) – (+ 21) = (+ 25) + (- 21)

[∵ the opposite number of (- 21) is + 21]

= [+ (25 – 21)]

= (+ 4)

= 4.

(+ 25) – (+ 21) = 4.

3. (+ 34) – (- 19)

Solution :

(+ 34) – (- 19) = (+ 34) + (+ 19)

[∵ the opposite number of (- 19) is (+ 19)]

= [+ (34 + 19)] (+ 53)

= 53.

(+ 34) – (- 19) = 53.

4. (- 15) – (- 27).

Solution :

(- 15) – (- 27) = (- 15) +- (+ 27)

[∵ the opposite number of (- 27)) is (+27)]

= [+ (27 – 15)]

= (+ 12)

= 12.

(- 15) – (- 27) = 12.

Practice Problems on Directed Numbers

5. (- 25) – (+ 13)

Solution :

(- 25) – (+ 13) = (- 25) + (- 13)

[∵ the opposite number of (+13) is (- 13)]

= [- (25 + 13)]

= (- 38)

= – 38.

(- 25) – (+ 13) = – 38.

Question 8. Put the numbers in blank spaces:

1. (- 3) + □ = 0

Solution:

We know that the sum of two opposite numbers is always zero.

Now, the opposite number of (- 3) is (+ 3).

∴ (- 3) + (+3) = 0

2. (+ 16) + □ = 0

Since the sum of two opposite numbers is always zero

Now, the opposite number of (+ 16) is (- 16), we have

∴ (+ 16) + (- 16) = 0

3. (- 7) + □ = (- 10).

Let (- 7) + m= (- 10)

∴ x = (- 10) -(-7)

= (- 10) + (+ 7)

[∵ The opposite number of (- 1) is (+ 7)]

= h (10 – 7) = [(- 3)]

∴ x = (- 3)

So, (-7) +03= (- 10)

Important Definitions Related to Directed Numbers

Question 9. Simplify :

1. (+ 12) – (- 3) + [opposite number of (+ 6)]

Solution: (i) (+ 12) – (- 3) + (opposite number of + 6)

= (+ 12) + (+ 3) + (- 6)

= [+ (12 + 3)] + (- 6)

= (+ 15) + (- 6)

= [+ (15 – 6)]

= (+ 9)

= 9.

(+ 12) – (- 3) + [opposite number of (+ 6)] = 9.

2. (Opposite number of + 20) – (opposite number of – 7) – (- 8).

Solution:

(opposite number of + 20) – (opposite number of – 7) – (- 8)

= (- 20) – (+ 7) + (+ 8)

= (- 20) + (- 7) + (+ 8)

= [- (20 + 7)] + (+ 8)

= (- 27) + (+ 8)

= [- (27 – 8)]

= (- 19)

= – 19.

(Opposite number of + 20) – (opposite number of – 7) – (- 8) = – 19.

3. 15 – (+ 14) + (opposite number of + 9)

Solution:

15 – (+ 4) + (opposite number of + 9)

= 15 + (- 4) + (- 9)

= 15 + [- (4 + 9)]

= 15 + (- 13)

= (15 – 13)

= 2.

15 – (+ 14) + (opposite number of + 9) = 2.

4. (- 5) + (opposite number of – 7) – (- 5).

Solution:

(- 5) + (opposite number of – 7) – (- 5)

= (- 5) + (+ 7) + (+ 5)

= (- 5) + [+ (7 + 5)] ‘ ’

= (- 5) + ( + 12)

= [+ (12 – 5)]

= (+7)

= 7.

(- 5) + (opposite number of – 7) – (- 5) = 7.

Examples of Real-Life Applications of Directed Numbers

Question 10. What must be added to the first to get the second in the following:

1. – 7, – 12

Solution:

The required number = – 12 – (- 7)

= – 12 + (+ 7)

= – (12 – 7) = – 5

2. 24, – 32

Solution:

The required number = – 32 – (+ 24)

= – 32 + (- 24)

= – (32 + 24)

= – 56.

24, – 32 = – 56.

3. – 12, 17

Solution:

The required number = 17 – (- 12)

= 17 + (+ 12)

= 17 + 12

= 29

– 12, 17 = 29

4. 25, – 42.

Solution:

The required number = – 42 – ( + 25)

= – 42 + ( – 25)

= – (42 + 25)

= – 67

25, – 42 = – 67

Question 11. Put <, > or = sign properly in the appropriate blank spaces of the following :

1. (+ 13) + (- 8) □ (+ 3) – (- 2)

Solution : (i) (+ 13) + (- 8) = [+ (13 – 8)] = (+ 5) = 5

(+ 3) – (- 2) = (+ 3) + (+ 2) = [+ (3 + 2)] = (+ 5) = 5

But 5 = 5

(+ 13) + (- 8) C=] (+ 13) – (- 2)

2. (- 18) – (+ 6) □ (- 18) – (- 6)

Solution:

(- 18) – (+ 6) = (- 18) + (- 6) = [- (18 + 6)] = (- 24) = – 24

(- 18) – (- 6) = (- 18) + (+ 6) = [- (18 – 6)] = (- 12) = – 12

But – 24 < – 12

∴ (18) – (+ 6)  < (- 18) – (- 6)

Verify the commutative property of addition for the following :

(+ 5) + (- 3) ; 

Solution : (i) (+ 5) + (- 3)

= [+ (5 – 3)] = (+ 2)

= 2 (- 3) + (+ 5)

= [+ (- 3 + 5)]

= (+ 2)

= 2

∴ (+5) +(-3) = (-3) +(+5)

So commutative law of addition is verified.

Conceptual Questions on Positive and Negative Numbers

2. (- 5) + (+ 3).

Solution:

(- 5) + (+ 3) = [- (5 – 3)]

= (- 2)

= – 2

(- 5) + (+ 3) = – 2

(+ 3) + (- 5) = [- (5 – 3)]

= (- 2)

= – 2

∴ (-5) + (+ 3) = (+ 3) + (- 5)

So commutative law of addition is verified.