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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.

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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

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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.

 

 

 

Geometry Chapter 3 Geometrical Box Its Instruments And Their Uses

Question 1. Draw a line segment of length 2.8 cm. using a ruler or a scale.

Solution:

Given:

length 2.8 cm.

Construction: Here we have to draw a line segment of length equal to AB = 2.8 cm.

1. Place the ruler or scale graduated with centimeters on the plane of the paper where the line segment (of length 2.8 cm) is to be drawn.

Hold the ruler with the left hand and take a pencil with the right hand.

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2. Put a point A on the paper at the 0 mark of the scale and another point B at the 2-8 mark of the scale (or ruler). ,

3. Join points A and B with the help of the pencil and scale.

Then the line segment AB— = 2-8 cm is constructed.

WBBSE Class 6 Geometry Box Instruments

Question 2. Draw a line segment of length 2.6 cm using a pair of dividers.

Solution:

Given: 2.6 cm

Construction: We have to draw a line segment of length 2.6 cm.

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1. At first place one of the needle points of the dividers on the 0 markings on the ruler and the other needlepoint of the dividers is drawn apart in such a way that it reaches up to the 6th small marking beyond the 2 bold markings.

 

Short Questions on Geometry Instruments

2. lift the dividers without disturbing the distance between the needles mark two points by pressing the dividers on the paper.

3. Join these two points with the ruler and pencil to draw the required line segment of length 2-6 cm.

4. Give the name of the line segment PQ.

Thus PQ— = 2.6 cm.

 

Question 3. Draw a circle of radius 2 cm. Using a pencil compass and scale.

Solution:

Given:

2 cm

Construction: We have to draw a circle of radius 2 cm. We follow the following steps

First, we draw a line segment AB of length 2 cm using a ruler (or a scale) and pencil.

 

 

2. Mark a point O on the plane of the paper. Place the needlepoint of the pencil compass at A and keep the tip of the pencil lying on the other leg of the compass such that it falls on B. Thus at this stage, the distance between the ends of the legs is equal to the length AB = 2 cm.

3. Now remove the compass from the line segment AB. Without disturbing the distance between the ends of the legs, place the needle point of the compass at O and keep it fixed at O.

4. Turn the compasses i.e., rotate the .tip of the pencil on the plane of the paper between the fourth finger and thumbs so as to draw a closed curve line.

5. The region enclosed by the curve obtained is the circle and the curve itself is the circumference of the circle. Thus the circle of radius 2 cm. is drawn.

Common Geometry Box Instruments and Their Uses

Question 4. Measure the following angle ZAOB with the help of a protractor.

 

 

Solution:

Here we have to measure the given angle ZAOB. We follow the following steps:

1. First, we extend the arms of the angle ZAOB, OA to C, and OB to D such that if we place the protractor on the angle, the ends of the arms OC (along OA) and OD (along OB) lie outside the protractor.

2. Now the protractor is placed on the angle ZAOB such that the center of the protractor falls on O and the 0°—180° line of the protractor coincides with the line OB.

 

 

We see that the arm OC (along OA) falls along the mark of 60 ° on the circumference of the protractor.

From the lilac figure, we see that ∠AOB measures the angle of 60°

∴ ∠AOB = 60°.

∠AOB measures the angle of 60°

Practice Problems Involving Geometry Tools

Question 5. Draw the angle 82° with the help of a protractor.

Solution:

Given: 82°

Construction: Here we have to draw the angle 82° with the help of the Motractor.

We perform the following steps:

First, draw a straight line \(\stackrel{\leftrightarrow}{\mathrm{AB}}\) on the plane of the paper.

Mark a point O on \(\stackrel{\leftrightarrow}{\mathrm{AB}}\)

 

2. Now place the protractor on the straight line AB such that the center C of the protractor falls on O and the baseline

i.e., the 0°- 180° line coincides with \(\stackrel{\leftrightarrow}{\mathrm{AB}}\).

The semi-circular portion of the protractor lies above \(\stackrel{\leftrightarrow}{\mathrm{AB}}\).

3. Mark a point P on the paper against the mark 82° (80° and 2 small markings Tier it) on the protractor starting from 0° lies on AB and proceeding in the anticlockwise direction.

4. Now remove the protractor and join the point P with O by scale and pencil,

5. The angle ∠BOP = 82° is drawn.

Algebra Chapter 1 Concept Of Algebraic Variables Or Quantities Or Symbols

Question 1. Write in words the following quantities:

1. 10a

Solution:

10a means that the variable a is multiplied by 10.

2. a + c

Solution:

a + c means that a is added to c or the addition of a and c.

3. x – y 

Solution:

x – y means that the value of y is subtracted from the value of x.

Class 6 Math Solutions WBBSE English Medium

4. 9x + 2

Solution:

9x + 2 means that 2 is added to the value of 9 times of x.

WBBSE Class 6 Algebraic Variables Notes

5. 3x – 7

Solution:

3x – 7 means that 7 is subtracted from the value of 3 times of x.

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6. (p/3) – 4

Solution:

(p/3) – 4 means that 4 is subtracted from the quotient p/3

i.e., 4 is subtracted from the value of p divided by 3.

7. x > y

Solution:

x > 6 means that x is greater than 6.

8. p >≠ 9.

Solution :

p >≠ 9 means that p is not greater than 9.

Understanding Algebraic Symbols

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Question 2. Express the following statements in algebraical quantities by sign and symbols :

1. The addition of x and 7.

Solution:

x + 7.

2. Subtraction of 9 from y 

Solution:

y – 9.

3. Multiplication of a by 3.

Solution:

3a.

4. x is greater than 13.

Solution:

x > 13.

5. Division of x by 8.

Solution:

x/8.

6. y is not equal to 5.

Solution:

y ≠ 5.

7. The addition of 7 to 10 times p.

Solution:

10p + 7.

8. Subtraction of 1 from the multiplication of x by 3.

Solution:

3x – 1.

Short Questions on Algebraic Variables

9. x is less than y.

Solution:

x < y.

10. b is not less than 8.

Solution:

b <≠ 8.

 

Question 3.

1. The present age of Suvadra is x years, after 4 years her age will be years.

Solution:

Given: 

The present age of Suvadra is x years, after 4 years her age will be years

x + 4

2. The present age of Avisekh is y years, 7 years before his age was years.

Solution:

Given:

The present age of Avisekh is y years, 7 years before his age was years.

y – 7

3. There are x rose flower plants in the garden of Kamala. There are rose flower plants in the garden of Kamalini 1/4 the that in the garden of Kamala. So the number of rose flower plants in the garden of Kamalini is.

Common Questions About Algebraic Expressions

Solution:

Given:

There are x rose flower plants in the garden of Kamala. There are rose flower plants in the garden of Kamalini 1/4 the that in the garden of Kamala. So the number of rose flower plants in the garden of Kamalini is.

x/4

4. Shibu has given a subscription of Rs x for Durgapuja at Muktipara of Bamangachi. Manish has given a subscription of Rs 10 more than twice what Shibu has given. So Manish has given Rs. as a subscription for Durgapuja.

Solution:

Given:

Shibu has given a subscription of Rs x for Durgapuja at Muktipara of Bamangachi. Manish has given a subscription of Rs 10 more than twice what Shibu has given. So Manish has given Rs. as a subscription for Durgapuja.

2x + 10

5. The height of Himangshu is 3 less than that of Hindal. If the height of Hindal is x cm, then the height of Himangshu is cm.

Solution:

Given:

The height of Himangshu is 3 less than that of Hindal. If the height of Hindal is x cm, then the height of Himangshu is cm.

x – 3.

Practice Problems on Algebraic Quantities

Question 4. Express the following statements algebraically using symbols and operation signs:

1. The height of Kankana is y The height of Kumudini is 21 cm less than that of Kankana. What is the height of Kumudini?

Solution:

Given:

The height of Kankana is y The height of Kumudini is 21 cm less than that of Kankana.

The height of Kumudini = (y – 21) cm.

2. The present age of the father is 5 times that of his son. The present age of the son is x What is the present age of the father?

Solution:

Given:

The present age of the father is 5 times that of his son. The present age of the son is x

The present age of the father = 5x years

Conceptual Questions on Properties of Algebraic Operations

3. Anindita has pnote books. Sunanda has notebooks 1/4th that of Anindita. How many notebooks have Sunanda?

Solution:

Given:

Anindita has pnote books. Sunanda has notebooks 1/4th that of Anindita.

Sunanda has p/4 notebooks.

4. Mihir takes x hours to go from his house to school. Mitali takes 5 hours less than 3 times that Mihir takes to go from house to school. How many hours does Mitali take to go from her house to school?

Solution:

Given:

Mihir takes x hours to go from his house to school. Mitali takes 5 hours less than 3 times that Mihir takes to go from house to school.

Mitali takes (3x – 5) hours to go from her house to school.

Real-Life Scenarios Involving Problem Solving with Variables

Question 5.

1. Your present age is 4 years less than that of your elder brother. If the present age of your elder brother is x years, then what is your present age?

Solution:

Given:

Your present age is 4 years less than that of your elder brother. If the present age of your elder brother is x years,

Your present age is (x – 4) years.

2. Thirtha has caught x Partha has caught 5 fishes more than that Thirtha. How many fishes have been caught by Partha?

Solution :

Given:

Thirtha has caught x Partha has caught 5 fishes more than that Thirtha.

Partha has caught (x+ 5) fish.

 

Question 6.

1. My father has brought x packets of sweets today. Each packet contains 5 sweets. How many sweets have been brought my father?

Solution:

Given:

My father has brought x packets of sweets today. Each packet contains 5 sweets.

Father has brought x packets of sweets. Each packet contains 5 sweets.

So x packets of sweets contain 5x sweets.

My father brought today 5x sweets.

2. Today Pradip has worked out sums 2 less than 4 times that Prabir has worked out. Prabir has worked out x sums today. How many sums Pradip has worked out today?

Solution :

Given:

Today Pradip has worked out sums 2 less than 4 times that Prabir has worked out. Prabir has worked out x sums today.

Prabir has worked out sums today = x.

Pradip has worked out today = (4x – 2) sums.

 

Question 7.

1. The length of each side of a square is “a” meter, then what is the perimeter of the square?

Solution:

Given:

The length of each side of a square is “a” meter,

We know that the length of each side of the square is the same and the number of sides of the square = 4.

The length of each side = meters (given)

The perimeter of the square = 4 x a meters

= 4a meters.

2. The length and breadth of a rectangle are x cm, and y cm respectively. What is the perimeter of the rectangle?

Solution :

Given:

The length and breadth of a rectangle are x cm, and y cm respectively.

The length of the given rectangle = x cm

and the breadth of the rectangle = y cm

∴ The perimeter of the rectangle = 2 (length + Breadth)

= 2 (x + y) cm.

Examples of Real-Life Applications of Algebraic Variables

Question 8. Moumita and Madhumita have purchased some birds of different designs and different colors from a fair in their locality. Moumita has purchased n birds and Madhumita has purchased birds 8 more than 1/4th part that Moumita has purchased. How many birds have been purchased by Madhumita ?

Solution:

Given:

Moumita and Madhumita have purchased some birds of different designs and different colors from a fair in their locality. Moumita has purchased n birds and Madhumita has purchased birds 8 more than 1/4th part that Moumita has purchased.

Moumita has purchased = n birds

Madhumita has purchased = 8 birds more than 1/4th part that Moumita has

purchased = \(\left(8+\frac{1}{4} \times n\right)\) birds.

= \(\left(8+\frac{n}{4}\right)\) birds.

 

 

 

Geometry Chapter 4 Geometrical Concept of Circle

Question 1. From the given following.

 

 

1 . O is of the circle.

Solution: Centre.

2. OA is an off-the-circle.

Solution: Radius

WBBSE Class 6 Circle Concepts Notes

3. AB is an off-the-circle.

Solution: Diameter

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4. PR is an off-the-circle.

Solution: Chord

5. MN is an off-the-circle.

Solution: arc

6. The shaded region COD is a circle.

Solution:  Sector

Class 6 Math Solutions WBBSE English Medium

7. \(\overparen{P Q R}\) is a of the circle.

Solution: minor arc

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8. PACDMNBR is an off-the-circle.

Solution: major arc

9. APQRB is of the circle.

Solution: semi-circle

10. AB = 2 x ? or 2 x ?

Solution: OA or OB.

Short Questions on Circle Geometry

Question 2. The radius of a circle is 2*5 cm, what is the diameter of the circle?

Solution:

Given:

The radius of a circle is 2*5 cm

The diameter of the circle = 2 x the radius of the circle

= (2 x 2-5) cm

= 5 cm.

The diameter of the circle is 5 cm

Common Questions About Parts of a Circle

Question 3. The diameter of a circle is 21 cm, What is the radius of the circle?

Solution:

Given:

The diameter of a circle is 21 cm

The radius of the circle = 1/2 x the diameter of the circle

= \(\left(\frac{1}{2} \times 21\right)\) cm

= 10.5 cm

The radius of the circle is 10.5 cm

Practice Problems on Circles

Question 4. The radius of a circle is 14 cm, what is the circumference of the circle?

Solution:

Given:

The radius of a circle is 14 cm

The circumference of the circle

= 2πr, (r = radius)

=\(\left(2 \times \frac{22}{7} \times 14\right)\) cm

= 88 cm.

The circumference of the circle is 88 cm

Examples of Real-Life Applications of Circles

Question 5. The radius of a semi-circle is 7 cm, what is the circumference of the semi-circle?

Solution:

Given:

The radius of a semi-circle is 7 cm

The circumference of the semi-circle

= (πr + 2r), r = radius of semi-circle.

= \(\left(\frac{22}{7} \times 7+2 \times 7\right)\) cm

= (22 + 14) cm

= 36 cm.

The circumference of the semi-circle is 36 cm

Short Questions on Fraction and Decimal Equivalence

Arithmetic Chapter 14 Equivalence Of Fraction Decimal Fraction Percentage And Ratio

Question 1. Express the following vulgar fractions in Decimal fractions, Percentages, and Ratios:

1. 7/18

Solution:

Given:

7/18

∴ 7/18 = 0.388……

= 0.38

Now 7/18 can be converted into percentages as

= \(\frac{7}{18} \times 100 \%\)

= \(\frac{7}{9} \times 50 \%\)

= \(\frac{350}{9} \%\)

= \(38 \frac{8}{9} \%\)

Also 7/18 = 7: 18.

∴ The required decimal fraction = 0.38

The required percentage = \(38 \frac{8}{9} \%\)

The required ratio = 7: 18.

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2. 25/64

Solution:

Given:

25/64

 

∴ 25/64 = 0.390625.

Now,

= \(\frac{25}{64} \times 100 \%\)

= \(\frac{25}{16} \times 25 \%\)

= \(\frac{625}{16} \%\)

= \(39 \frac{1}{6} \%\)

Again, 25/64

= 25: 64.

WBBSE Class 6 Equivalence of Fractions Notes

∴ The required decimal fraction = 0.390625

The required percentage = \(39 \frac{1}{6} \%\)

The required ratio = 25: 64.

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Question 2. Express 0.9375 in vulgar fraction, percentage, and ratio.

Solution:

Given:

0.9375

0.9375 = 9375 / 10000

= 15/16

(Dividing both numerator and denominator by 625)

Now, 0.9375 = \(\frac{15}{16} \times 100 \%\)

= \(\frac{15}{4} \times 25 \%\)

= \(\frac{375}{4} \%\)

= \(93 \frac{3}{4} \%\)

Again, 0.9375 = 15/16

= 15: 16.

∴ The required vulgar fraction = 15/16;

The required percentages = \(93 \frac{3}{4} \%\);

The required ratio = 15: 16.

Common Questions About Converting Fractions to Percentages

Question 3. Express 40% in vulgar fractions, decimal fractions, and ratios.

Solution:

Given:

40%

40% = 40/100

= 2/5;

Now,

 

∴ 2/5 = 0.4

Again, 2/5 = 2: 5.

∴ The required vulgar fraction = 2/5;

The required decimal fraction = 0.4;

The required ratio = 2: 5.

Conceptual Questions on Finding Equivalent Forms

Question 4. Convert the ratio 5: 8 into vulgar fractions, decimal fractions, and percentages.

Solution:

Given:

5: 8

5: 8 = 5/8

Now,

 

 

∴ 5/8 = 0.625

Again,

= \(\frac{5}{8} \times 100 \%\)

= \(\frac{5}{2} \times 25 \%\)

= \(\frac{125}{2} \%\)

= \(62 \frac{1}{2} \%\)

∴ The required vulgar fraction = 5/8;

The required decimal fraction = 0.625;

The required percentage = \(62 \frac{1}{2} \%\)

Arithmetic Chapter 13 fundamental Concept of Ration And Proportion

Determine in which of the following cases a ratio can be obtained :

1. The weight of Ram and the height of Shyam.

Solution:

The weight and the height are not the same kinds of quantities. So a ratio can not be obtained in this case.

2. The amount of money that Raghab had and the amount of money that Raghab had spent.

Solution:

Here the two quantities of the same kind are comparable. So a ratio can be determined in this case.

3. The amount of water in litres that your bottle contains and the temperature of the water.

Solution:

Since the amount of water and the temperature of the water is not the same kind of quantities, any ratio can not be determined in this case.

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4. The time that you have read the book and the time that your sister have read the book.

Solution:

Here the two quantities of the same kind (i.e. the time) are comparable. Hence a ratio can be determined in this case.

WBBSE Class 6 Ratio and Proportion Solutions

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Question 2. Determine the ratio in each of the following quantities and write whether the ratio is a ratio of greater inequality or ratio of lesser inequality or a ratio of equality 

1. ₹ 30 and ₹ 22.5

Solution:

Given:

₹ 30 and ₹ 22.5

The required ratio = ₹ 30 : ₹ 22.5

= ₹ 30 / ₹ 22.5

= 30 / 22.5

 

= 4/3

= 4:3.

∵ 4 < 3 i.e., the antecedent > the radio is greater inequality.

₹ 30 and ₹ 22.5 = 4:3.

Short Questions on Ratio and Proportion

2. 4.9 liters and 8.4 liters

Solution:

Given:

4.9 liters and 8.4 liters

The required ratio = 4-9 litres: 8.4 litres

= 4.9 / 8.4

 

 

= 7/12

= 7:12

∵ 7 < 12 i.e, the antecedent < the Consequent, the ratio is the ratio of lesser inequality.

4.9 liters and 8.4 liters = 7:12

3. l hour 24 minutes and 6 hours 18 minutes.

Solution:

Given:

l hour 24 minutes and 6 hours 18 minutes

1 hour 24 minutes = (1 x 60 + 24) minutes = 84 minutes.

6 hours 18 minutes = (6 x 60 + 18) minutes = 378 minutes.

∴ The required ratio = 84 minutes: 378 minutes

 

 

∵ 2 < 9 i.e., the antecedent < the consequent, the ratio is a ratio of lesser inequality.

l hour 24 minutes and 6 hours 18 minutes = 2:9

4. 52 m 25 cm and 522.5 cm.

Solution:

Given:

52 m 25 cm and 522.5 cm

52 m. 25 cm. = (52 x 100 + 25). cm

= 522.5 cm 522.5 dcm

= (522.5 x 10) cm = 522.5 cm.

The required ratio

= 522.5 cm: 522.5 cm

= \(\frac{5225 \mathrm{~cm}}{5225 \mathrm{~cm}}\)

= 1/1

= 1: 1

Here Antecedent = Consquent.

So the ratio is the ratio of equality.

52 m 25 cm and 522.5 cm = 1: 1

5. x²yand xy² (x > y).

Solution:

Given:

x²yand xy² (x > y)

The required ratio = x²y : xy²

= \(\frac{x^2 y}{x y^2}\)

= x/y

= x: y

As x > y i.e., the antecedent > the consequent, the ratio is a ratio of equality.

x²yand xy² (x > y) = x: y

Common Questions About Ratios in Real Life

6. a²bc and ab²c (where a = b).

Solution:

Given:

a²bc and ab²c (where a = b)

The required ratio = a²bc : ab²c

= \(\frac{a^2 b c}{a b^2 c}\)

= a / b

= a:b.

As a = b i.e., the antecedent = the consequent, the ratio is a ratio of equality.

a²bc and ab²c (where a = b) = a:b

 

Question 3. What is the ratio of the 3 angles of an equilateral triangle?

Solution:

Given:

3 angles of an equilateral triangle

In an equilateral triangle, the sum of the angles = 180° and the 3 angles of the triangle are the same.

∴ The measure of each angle of the triangle = 180° / 3

= 60°

∴ The required ratio = 60° : 60° : 60°

= 1: 1: 1.

The ratio of the 3 angles of an equilateral triangle = 1: 1: 1.

Practice Problems on Ratios and Proportions

Question 4. What is the ratio of the 3 angles of an isosceles right-angled triangle?

Solution:

Given:

3 angles of an isosceles right-angled triangle

One angle of an isosceles right-angled triangle = 90° and the sum of the 3 angles of the triangle = 180°. So the sum of the angles of the remaining two angles

= 180° – 90°

= 90°

and they are equal. Hence the measure of each of the equal angles

= 180° /  90°

= 45°

The angles of the isosceles right-angled triangle

= 45°, 45°, 90°.

The required ratio of the measure of the 3 angles of the isosceles right-angled triangle

= 45°: 45°: 90° = 1:1:2.

The ratio of the 3 angles of an isosceles right-angled triangle = 45°: 45°: 90° = 1:1:2.

Important Definitions Related to Ratio and Proportion

Question 5. How much money Bidhu and Bivu will get if we divide ₹ 210 between them in a ratio of 3:4?

Solution:

Sum of the components of the ratio = 3 + 4 = 7

So Bidhy will get 3/7 of the part and Bivu will get 4/7 part of the total amount.

∴ Bidhu will et = ₹ \(\left(210 \times \frac{3}{7}\right)\)

= ₹ 90.

Bivu will get = ₹ \(\left(210 \times \frac{4}{7}\right)\)

= ₹ 120.

₹ 120 money Bidhu and Bivu will get if we divide ₹ 210 between them in a ratio of 3:4

 

Question 7. The ratio of your reading books and story books is 4 :1; if the number of your reading books is 16, then what is the number of your story books? What is the total number of your books?

Solution:

Given:

The ratio of your reading books and story books is 4 :1; if the number of your reading books is 16

It is given that the ratio of your reading books: story books = 4:1.

∴ If the number of reading books is 4, then the number of storybooks = is 1

So If the number of reading books is 1, then the number of story books = 1/4

∴ If the number of reading books is 16, then the number of story books = 1/4 x 15

= 4.

∴ Number of story books = 4

Total number of books = 16 + 4

= 20.

Total number of books = 20.

Examples of Real-Life Applications of Ratios

Question 7. In one kind of ornament, the ratio of the weight of gold to that of silver = 4: 7. In such a type of ornament, how many milligrams of gold be mixed with 357 milligrams of silver?

Solution :

Let the required weight of gold = x milligram

∵ \(\frac{the weight of gold}{the weight of silver}\) = 4/7

∴ \(\frac{x}{357}\) = 4/7

or, x x  7 = 357 x 4

or, 7x = 357 x 4

 

= 204

∴ the required weight of gold = 204 milligrams.

 

Question 8.  The length of bamboo is 2 meters. 75 cm. of its length is painted with orange color and the remaining part of its length is painted with white color.

Solution:

1. What is the ratio of the total length of the bamboo to that of the portion painted with the orange color?

Solution :

Given:

The length of bamboo is 2 meters. 75 cm. of its length is painted with orange color and the remaining part of its length is painted with white color.

The length of the bamboo = 2 metres = (2 x 100) cm. = 200 cm.

The length of the bamboo in which portion is painted with the orange color

= 75 cm.

The required ratio = 200 : 75 = 200 / 75 = 8/3 =8:3.

2. What is the ratio of the total length of the bamboo to that of the portion painted with the white color?

Solution:

Given:

The length of bamboo is 2 meters. 75 cm. of its length is painted with orange color and the remaining part of its length is painted with white color.

The total length of the bamboo = 200 cm

The length of the bamboo in which portion is painted with the white color

= (200 – 75) cm = 125 cm.

The required ratio = 200 : 125 = 200 / 125 = 8/5 = 8 : 5

3. What is the ratio of the length of the bamboo painted with orange color to that of the portion painted with white color?

Solution:

Given:

The length of bamboo is 2 meters. 75 cm. of its length is painted with orange color and the remaining part of its length is painted with white color.

The required ratio = 75 cm ÷ 125 cm

 

= 3/5

= 3:5.

3:5 ratio of the length of the bamboo painted with orange color to that of the portion painted with white color

Conceptual Questions on Equivalent Ratios

Question 9. Kamala Devi bought 6 bananas for ₹ 18. But Sarala Devi bought 2 dozen bananas for ₹ 72. Determine who gave more price for purchasing banana by expressing in ratio.

Solution:

Given:

Kamala Devi bought 6 bananas for ₹ 18. But Sarala Devi bought 2 dozen bananas for ₹ 72.

Kamala Devi bought 6 bananas in ₹ 18.

Cost of 6 bananas = ₹ 18

So the cost of 1 banana = ₹ 18/6

= ₹

Sarala Devi bought 2 dozen bananas in ₹ 72

2 dozens = 2 x 12 = 24 bananas.

Cost of 24 bananas =v₹ 72.

So the cost of 1 banana = ₹ 72/24 = ₹ 3.

The ratio of the cost of 1 banana paid by Kamala Devi to that by Sarala Devi

= ₹ 3: ₹ 3

= 1: 1.

It is a ratio of equality.

So both of them paid the same rate for bananas.

Real-Life Scenarios Involving Scale Models

Question 10. Determine which of the following ratios are equal :

1.  20: 24 and 25: 30

Solution:

20: 24

= 20 / 24

= 5/6

= 5: 6

20: 24 = 5: 6

25: 30

= 25/30

= 5/6

= 5: 6.

25: 30 = 5: 6.

2. 1.4: 0.6 and 6.3: 2.7

Solution:

1.4: 0.6 

= 1.4 / 0.6

= 14/6

= 7/3

= 7: 3

1.4: 0.6 = 7: 3

6.3: 2.7

= 6.3/2.7

= 63/27

= 7/3

= 7:3

6.3: 2.7 = 7:3

∴ The two given ratios are equal.

 

Question 11. Determine whether the following numbers are according to the given order in proportion.

1. 9, 7, 36, 28

Solution:

Here first number = 9 ; second number = 7;

Third number = 36 ; fourth number = 28.

∴ first number x fourth number = 9 x 28

= 252

second number x third number = 7 x 36

= 525.

So first number x third number = second number x third number.

∴ The given ordered numbers are in proportion.

∴ 9: 7: : 36: 28.

2. 1/2, 1, 3/5, \(1 \frac{1}{5}\)

Solution:

Here first number = 1/2 ; second number = 1 ; third number = 3/5 ;

fourth number = \(1 \frac{1}{5}\)

= 6/5

∴ The first number x fourth number =  \(\frac{1}{2} \times \frac{6}{5}\)

= 3/5

The third number x second number = \(\frac{3}{5} \times{1}\)

= 3/5

So first number x fourth number = second number x third number

∴ The given order numbers are in proportion

∴ 1/2 : 1 : : 3/5 : 6/5

or, 1/2 : 1 : : 3/5 : \(1 \frac{1}{5}\)

Class 6 Math Solutions WBBSE English Medium

Question 12. Examine whether any proportion can be formed with the quantities 12 km, 16 km, 21 kg, 28 kg. If it is possible, then how many proportions can be formed?

Solution: Here we see that 12 x 28 = 336 and 16 x 21 = 336.

∴ 12 x 28 = 16 x 21

so the given quantities form a proportion.

The proportions are :

12 km : 16 km : : 21 kg : 28 kg

 and 16 km : 12 km : : 28 kg : 21 kg

Only 2 proportions can be formed.

 

Question 13. Your height is 160 cm and your mother’s height is 170 cm. Your weight is 40 kg and your mother’s weight is 42*5 kg. Examine whether there exists a proportion between height and weight or not.

Solution:

Given:

My height is 160 cm and my mother’s height is 170 cm. MY weight is 40 kg and my mother’s weight is 42*5 kg.

The heights of you and your mother are 160 cm and 170 cm.

The ratio of the height = 160 cm : 170 cm

= 160 / 170

= 16: 17.

Again the weight of you and your mother is 40 kg and 42.5 kg.

∴ The ratio of the weights = 40 kg : 42.5 kg = 40 / 42.5

= 400 / 425

= 16 / 17

= 16: 17

∴ the height and weights of you and your mother are in proportion.

Arithmetic Chapter 12 Measurement Of Time

Question 1. Add the following :

1. (8 hours 32 minutes 41 seconds) + (18 hours 42 minutes 25 seconds)

Solution:

Given:

(8 hours 32 minutes 41 seconds) + (18 hours 42 minutes 25 seconds)

Here for Sec:

41 + 25 = 66 seconds = 1 min. 6 sec.

(60 sec. = 1 min.)

For Min:

32 + 42 + 1 = 75 min.

= 1 hr. 15 min (60 Min. = 1 hr.)

For Hours:

8 + 18 + 1 = 27 hours.

Class 6 Math Solution WBBSE

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∴ The required sum = 27 hours 15 minutes 6 seconds

(8 hours 32 minutes 41 seconds) + (18 hours 42 minutes 25 seconds) = 27 hours 15 minutes 6 seconds

WBBSE Class 6 Measurement of Time Notes

WBBSE Solutions For Class 6 Geography	WBBSE Solutions For Class 6 History	WBBSE Solutions For Class 6 Maths
WBBSE Class 6 Geography Notes	WBBSE Class 6 History Notes	WBBSE Solutions For Class 6 School Science
WBBSE Class 6 Geography Multiple Choice Questions	WBBSE Class 6 History MCQs	WBBSE Notes For Class 6 School Science

 

2. (12 days 10 hours 42 minutes) + (25 days 14 hours 28 minutes) + (17 days 11 hours 32 minutes)

Solution:

Given:

(12 days 10 hours 42 minutes) + (25 days 14 hours 28 minutes) + (17 days 11 hours 32 minutes)

42 + 28 + 32 = 102 min.

= 1 hr. 42 min. (60 min. = 1 hr.)

For Hour:

1 + 10 + 14 + 11 = 36 hr.

= 1 day 12 hr.

(24 hr. = 1 day)

For Days:

(1 + 12 + 25 + 17) = 55 days.

Class 6 Math Solution WBBSE

 

 

∴ The required sum = 55 days 12 hours 42 minutes.

(12 days 10 hours 42 minutes) + (25 days 14 hours 28 minutes) + (17 days 11 hours 32 minutes) = 55 days 12 hours 42 minutes.

Understanding Time Measurement

3. (4 years 7 months 17 days) + (9 years 11 months 7 days) + (8 years 5 months 27 days)

Solution:

Given:

(4 years 7 months 17 days) + (9 years 11 months 7 days) + (8 years 5 months 27 days)

For Days:

17 + 7 + 27 = 51 days

= 1 month 21 days (30 days = 1 month)

For Mths:

1 + 7 + 11 + 5 = 24 mths.

= 2 yrs.

(12 Months = 1 year)

For Year:

(2+ 4 + 9 + 8 = 23 years.

 

 

∴ The required sum = 23 years 21 days.

(4 years 7 months 17 days) + (9 years 11 months 7 days) + (8 years 5 months 27 days) = 23 years 21 days.

Short Questions on Measurement of Time

Question 2. Subtract :

1. (6 hours 2 minutes 2 seconds) – (2 hours 55 minutes 42 seconds)

Solution:

Given:

(6 hours 2 minutes 2 seconds) – (2 hours 55 minutes 42 seconds)

1 min. = 60 sec.

∴ 60 + 2  = 62 sec.

62 – 42 = 20 secs

1 hr. = 60 min.

101 – 55 = 46 min.

5-2 = 3 hrs.

 

(6 hours 2 minutes 2 seconds) – (2 hours 55 minutes 42 seconds)= 3 hrs 46 min 20 secs

 

2. (11 years 2 months 16 days) – (5 years 10 months 21 days)

Solution:

Given:

(11 years 2 months 16 days) – (5 years 10 months 21 days)

1 month = 30 days

46 – 21 = 25 days

1 yr. = 12 months

13 – 10 = 3 months

10 – 5 = 5 yrs.

 

 

∴ The required value of subtraction = 5 years. 3months 25 days

(11 years 2 months 16 days) – (5 years 10 months 21 days) =  5 years. 3months 25 days

Practice Problems on Time Measurement

Question3. Multiply:

1. (7 hrs. 32 mins. 41 secs.) x 3

Solution:

Given:

(7 hrs. 32 mins. 41 secs.) And 3

1 min. = 60 secs.

 

 

1 hr = 60 min.

 

 

(7 hrs. 32 mins. 41 secs.) x 3

= 21 hrs. 96 mins. 123 secs.

= 21 hrs. (96 + 2) mins. (123 – 120 secs.) = 21 hrs. 98 mins. 3 secs.

= (21 + 1) hrs. (98 – 60) mins. 3 secs.

= 22 hrs. 38 mins. 3 secs.

∴ The required value of the product

= 22 hours 38 minutes 3 secs.

(7 hrs. 32 mins. 41 secs.) x 3 = 22 hours 38 minutes 3 secs.

 

2. (7 months 12 days 8 hrs.) x 12

Solution :

Given:

(7 months 12 days 8 hrs.) And 12

1 day = 24 hrs.

 

 

1 month = 30 days

 

 

1 year = 12 months

 

 

= (7 months 12 days 8 hours.) x 12

= 84 months 144 days 96 hours.

= 84 months (144 + 4) days 0 hours.

= 84 months 148 days = (84 + 4) months 28 days = 88 months 28 days = 7 yrs. 4 months 28 days.

∴ The required value of the product

= 7 years 4 months. 28 days.

(7 months 12 days 8 hrs.) x 12 = 7 years 4 months. 28 days.

Important Definitions Related to Time Measurement

Question 4. Divide:

1. (15 hrs.) ÷12

Solution:

Given:

(15 hrs.) And 12

1 hour = 60 minutes.

(15 hours) ÷ 12

= 15/12 hours.

= 5/4 hours.

= \(1 \frac{1}{4}\) hours

= 1 hour (\(\frac{1}{4}\) x 60 ) minutes.

= 1 hour. 15 minutes.

∴ The required quotient = 1 hour 15 minutes.

(15 hrs.) ÷12 = 1 hour 15 minutes.

2. (3 hrs. 27 mins.) ÷ 9

Solution:

Given:

(3 hrs. 27 mins.) And 9

∵ 1 hour. = 60 minutes.

Class 6 Math Solution WBBSE In English

= (3 hrs. 27 minutes) ÷ 9

= (3 x 60 + 27) minutes ÷ 9

= 207 minutes ÷ 9

= 23 minutes.

∴ The required quotient = 23 minutes.

 

Question 5. Your present age is 11 years 7 months 11 days. Any person can have the right to give a vote when the person is 18 years old. After how many years will you have to fight to give a vote?

Solution:

Given:

My present age is 11 years 7 months 11 days. Any person can have the right to give a vote when the person is 18 years old.

When you will be 18 years old, then will have the right to give a vote.

But your present age is 11 years 7 months 11 days.

1 month = 30 days

30 – 11 = 19 days

1 years = 12 months

11 – 7 = 4 months

17 – 11 = 6 years.

 

∴ After 6 years 4 months 19 days you will attain the age of 18 years, then only you will have the right to give a vote.

Examples of Real-Life Applications of Time Measurement

Question 6. Your father’s present age is 52 years 8 months 21 days. Your elder uncle’s age is 3 years 10 months 25 days more than your father’s. What is the present age of your elder uncle?

Solution:

Given:

My father’s present age is 52 years 8 months 21 days. My elder uncle’s age is 3 years 10 months 25 days more than My father’s.

1 month = 30 days

21 + 25 = 46 days

 

 

10 + 8 + 1 = 19 months

1 yr. = 12 months

 

 

 

 

∴ The present age of your elder uncle is 56 years and 7 months.

 

Question 7. Bablu’s date of birth is 19-11-1975. What was his age on 11–10-2000?

Solution :

Given:

Bablu’s date of birth is 19-11-1975.

1 month = 30 days

41 – 19 = 22 days

1 years = 12 months

21 – 11 = 10 months

1999 – 1975 = 24 yrs.

 

Bablu’s age on 11-10-2000 is 24 yrs 24 yrs 22 days

 

Question 8.

1. Write down the leap-years from 1895 year to the 1915 year

Solution:

The required leap years from 1895 year to the 1915 year are

1896, 1904, 1908, and 1912.

2. Write down the leap years from 2010 year to 2030 year.

Solution :

The required leap years from 2010 year to the 2030 year are

2012, 2016, 2020, 2024, 2028.

Conceptual Questions on Calculating Duration

Question 9. Write down all the leap years up to the year 2015 after the independence of India.

Solution: India was independent in 1947. The year 1947 is not a leap year.

The first leap year after the independence of India is 1948.

The required leap years are :

1948, 1952, 1956, 1960, 1964, 1968, 1972, 1976, 1980, 1984, 1988, 1992, 1996, 2000, 2004, 2008, 2012.

There are 17 leap years up to the year 1915 after the independence of India.

 

Question 10.  If the 1st February of 2010 was Monday, what days were the 1st March and 1st April of 2010?

Solution: Total number of days after 1st February 2010 to 1st March 2010 (only one day of 1st February and 1st March 2010 is included) = 28 days.

(∵ The year 2010 is not a leap year)

 

 

Since the remainder is zero, the 1st of March 2010 was Monday.

Total number of days after 1st February 2010 to 1st April 2010 only one day of 1st February 2010 and 1st April 2010 is included

= 27 + 31 + 1

= 59 days.

 

 

Since the remainder is 3, the 3rd-day counting after Monday will be the required day and this is Thursday.

∴ 1st April 2010 was Thursday.

Real-Life Scenarios Involving Schedules and Timetables

Question 11. The present age of Jhilmil is 1/4 part of the present age of her father. The present age of her father is 53 years. What is the present age of Jhilmil?

Solution:

Given:

The present age of Jhilmil is 1/4 part of the present age of her father. The present age of her father is 53 years.

The present age of Jhilmil’s father = is 53 years.

∴ The present age of Jhilmil = (53 years) x 1/4 = (53 years +  4)

= (53 x 12 months) + 4

= (636 months) + 4

= 159 months

= 13 years 3 months.

∴ The present age of Jhilmil is 13 years and 3 months

1 year = 12 months

 

The present age of Jhilmil is 13 years and 3 months

 

Question 12. Shekhar Pal of our village can make 4 earthen images in 11 hours and 36 minutes. What will be the time taken by him to make 1 image if it is assumed that each image can be made in equal time?

Solution:

Given:

Shekhar Pal of our village can make 4 earthen images in 11 hours and 36 minutes.

The time taken by Shekhar Pal to make 1 image

= (11 hours 36 minutes ) + 4

= (11 x 60 + 36) minutes + 4

= 696 minutes + 4

= 174 minutes

= 2 hours 54 minutes.

∴ The required time = 2 hours 54 minutes.

1 hour = 60 minutes.

 

The required time = 2 hours 54 minutes