Class 6 Mathematics

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.

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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

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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

Read And Learn More WBBSE Solutions For Class 6 Maths

 

∴ 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

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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

Arithmetic Chapter 11 Square Root

Question 1. 

1. The square of 7

Solution:

Given Number 7

The square of 7 = 7²

= 7 × 7

= 49.

∴ The square of 7 = 49.

2. The square root of 121

Solution:

Given number 121

The square root of 121 = √I21

= (11)²

=11.

And 121 = 11 x 11 = (11)²

∴ The square root of 121 = 11.

3. √144

Solution:

Given Number √144

√144 = √(12)²

= 12

∴ √144 = 12.

Read and Learn More  WBBSE Solutions For Class 6 Maths

∴ 144 = 2 x 2 x 2 x 2 x 3 x 3

= 2² x 2² x 3²

= (2 x 2 x 3)²

= (12)²

√144 = (12)²

WBBSE Class 6 Square Root Notes

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4. √(3² x 2²)

Solution:

= √3²  x √2²

= 3 x 2

= 6.

∴ √3² x 2² = 6.

5. √(5 x 7 x 5 x 7)

Solution:

= √(5 x 5 x 7 x 7)

= √(5² x 7²)

= √5² x √7²

= 5 x 7

= 35.

∴ √5 x 7 x 5 x 7

= √(14)²

= 14.

∴ √14 x 14

= 14.

√(5 x 7 x 5 x 7) = 14.

Understanding Square Roots

Question 2. Using factorization find the square root of the following numbers:

1. 144

Solution:

 

 

∴ 144 = 2 x 2 x 2 x2 x 3 x 3

∴ √144 = 2 x 2 x 3

= 12.

∴ The required square root = 12.

2. 169

Solution:

 

 

∴ 169 = 13 x 13

∴ √169 = 13 (one number is taken from one pair)

∴ The required square = 13.

Short Questions on Square Roots

3. 225

Solution:

 

 

 

∴ 225 = 3 x 3 x 5 x 5

∴ √225 = 3 x 5

= 15.

The required square = 15.

 

4. 900

Solution:

 

∴ 900 = 3 x 3 x 5 x 5

∴ √225 = 3 x 5

= 15.

Common Questions About Finding Square Roots

5. 15² + 20².

Solution:

 

 

∴ 625 = 5 x 5 x 5 x 5

∴ √625 = √5 x 5 x 5 x 5

= 5 x 5

= 25.

∴ The required square root = 25.

 

Question 3. Write the following numbers correctly in the room (space) given below:

 

 

1. 20

Solution:

 

 

∴ 20 = 2 x 2 x 5

= 2² x 5

Practice Problems on Square Roots

2. 27

Solution:

 

 

∴ 27 = 3 x 3 x 3

= 3² x 3.

 

3. 50

Solution:

 

 

∴ 50 = 20 x 5 x 5

= 2 x 5²

 

4. 100

 

 

∴ 100 = 2 x 2 x 5 x 5

= 2² x 5²

 

5. 108

 

 

∴ 108 = 2 x 2 x 3 x 3 x 3

= 2² x 3² x 3.

Important Definitions Related to Square Roots

6. 169

 

 

∴ 169 = 13 x 13

= (13)².

The table is as follows:

 

 

Question 4. Arrange the following number in ascending order of magnitude:

1. √9 + √49

Solution:

√9 + √49

= √(3)²  +√(7)²

= 3 + 7

= 10.

√9 + √49 = 10.

2. √25 + √36

Solution:

√25 + √36

= √(5)² + √(6)²

= 5 + 6

= 11.

√25 + √36 = 11.

Examples of Real-Life Applications of Square Roots

3. √4 + √16

Solution:

√4 + √16

= √(2)² + √(4)²

= 2 + 4

= 6.

√4 + √16 = 6.

∵ 6 < 10< 11< 15,

∴ √4 + √16 < √9 + √49 < √25 + √36.

∴ Arranging the given numbers in ascending order of magnitudes,

We get,

√4 + √16 < √9 + √49 < √25 + √36.

 

Question 5. Without finding the actual square root determine the units place digit of the square root of the following numbers and also determine the number of digits in the square root of the given numbers :

1. 784

Solution:

1. The unit’s place digit of 784 is 4.

∴ The unit’s place digit of the square root of 784 is either 2 or 8 and the number of digits in the square root of 784 is 2.

2. 1225

Solution:

The unit’s place digit of 1225 is 5.

∴ The units’

The units’ place digit of the square root of 1225 is 5 and the number of digits in the square root of 1225 is 2.

3. 10201

Solution:

The unit’s place digit of 10201 is 1

The units’, place digit of the square root of 10201 is either 1 or 9 and the number of digits in the square root of 10201 is 3.

4. 160000. 

Solution:

The unit’s place digit of 160000 is 0.

The units’ place digit of the square root of 160000 is 0 and the number of digits in the square root of 160000 is 3.

Conceptual Questions on Methods to Find Square Roots

Question 6. What is the value of N in the number 202N, so that the number becomes a perfect square number?

Solution:

 

 

∴ The required value of N is 5.

 

Question 7. Determine the square root of the following numbers by the division method:

1. 529 

Solution:

 

 

∴ The required square root is 23.

2. 1764

Solution:

 

 

∴ The required root is 42

 

Question 8. Find the perfect square number nearer to 1000.

Solution:

 

 

∴ The remainder is 39,

∴ (1000 – 39) = 961 which is a perfect square number.

Again the quotient is 31 and the next number to it is 32.

The square of 32 is (32)²

= 32 x 32

= 1024.

Now the preceding perfect square number to 1000 is 961 and the succeeding perfect square number to 1000 is 1024.

Between 961 and 1024, the number 1024 is nearer to 1000 as 1024-1000 = 24 < 1000 – 961 = 39.

∴ The required perfect square number = 1024.

 

Question 9. Find the least perfect square number of 4 digits.

Solution: The least number of 4 digits = 1000.

The required least perfect square number of 4 digits is the least perfect square number greater than 1000.

 

 

∴ It is seen that the least number of 4 digits i.e., 1000 is not a perfect square number.

So the least perfect square number of 4 digits will be the square of (31+1)

i.e., the square of 32.

∴ The required least perfect square number of 4 digits = (32)²

= 32 x 32

= 1024.

 

Question 10. Find the greatest perfect square number of 4 digits.

Solution: The greatest number of 4 digits = 9999.

The required greatest perfect square number of 4 digits is the greatest perfect square number not greater than 9999.

 

 

∴ It is seen that the greatest number of 4 digits i.e., 9999 is not a perfect square. It is 198 more than the square of 99. So the greatest perfect square number of 4 digits is the square of 999.

∴ The required greatest perfect square number of 4 digits = (99)² = 9801.

This can also be 9999 – 198 = 9801 which is a perfect square number.

It is to be noted that the quotient in the square root is 99. Its next integer is 99 + 1 = 100 whose square is 100² = 100 x 100 = 10000, which is a perfect square number but not a number of 4 digits.

∴ The required greatest perfect square number of 4 digits = 9801.

 

Question 11. What least number is to be subtracted from 9585 so that the result of subtraction is a perfect square?

Solution:

 

 

∴ The required number = is 176.

In the determination of the square root of 9585, we see that there is a remainder of 176 at the end i.e., the given number is 176 more than the square of 97. So if we subtract 176 from the given number, then the remainder will be a perfect square.

176 is the required least number that will be subtracted from 9585 so that the result of subtraction is a perfect square.

 

Question 12. What least number is to be added to 5320 so that the sum is a perfect square number?

Solution:

 

 

∴ The required least number  = 9.

 

Question 13. What is the least perfect square number (other than zero) which is exactly divisible by 15, 25, 35, and 45?

Solution: Since the required number is a perfect square number and divisible by 15, 25, 35, and 45, the required number will be the L. C. M. of the given numbers 15, 25, 35, and 45 or a multiple of L. C. M. of them.

 

 

∴ L.C. M. of 15, 25, 35, 45 = 3 x 5 x 5 x 7 x 3 = 3² x 5² x7 = 1575.

But 1575 is not a perfect square number. (This is because factor 7 is once only).

If the L. C. M. i.e., the number 1575 is multiplied by 7, Ifren the product is a perfect square number, and also if is divisible by the given numbers.

The required least perfect square number = 1575 x 7 = 11025.

 

Question 14. What is the least perfect square number which has a factor of 17?

Solution : The multiples of 17 are 17 x 1, 17 x 2, 17 x 3, 17 x 4,……………..  17 x 16, 17 x 17, 17 x 18…………

These are the numbers each of which has a factor of 17.

But the least perfect square number having 17 as a factor is 17 x 17 = 289.

∴ The required least perfect square number = 289.

 

Question 15. The product of two positive numbers is 1575 and their quotient is 9/7. Find the numbers.

Solution: Let the greater number be x and the smaller number be y. (Here both x, and y are positive).

∴ By the given condition, we get,

xy = 1575  …………….(1)

x/y = 9/7  ……………..(2)

Multiplying (1) & (2)

 

 

or, x² = 2025

or, x = 2025

= (45)²

=45           ( ∵ > 0)

From (1), we get, 45y = 1575

or, y = 1575 / 45 = 35.

∴ The numbers are 35 and 45.

Real-Life Scenarios Involving Area Calculations

Question 16. All the students of a school on republic day can be arranged in 12, 15, or 20 rows and also they can be arranged in solid squares. What is the least number of students in that school?

Solution:

Since the students of a school are arranged in 12,15 or20 rows and they can also be arranged in solid squares, the least number of students in the school will be the L. C. M. (if it is a perfect square) or the multiple of L.C.M. which is a perfect square.

 

 

∴ L.C.M. of 12, 15, 20

= 2 x 2 x 3 x 5

= 2² x 3 x 5

= 60.

But 60 is not a perfect square number (Because it contains only one 3 and one 5 as factors).

The least multiple of L. C. M. which is a perfect square number = 60 x 3 x 5 = 900.

∴ The required number of students in the school = 900.

 

Question 17. Shanti Devi has plucked 441 oranges from her fruit garden. She has kept the oranges in baskets such that the number of oranges in each basket is equal to the number of baskets. Find the number of baskets.

Solution: Here the number of baskets and the number of oranges in each basket are equal.

The product of the number of baskets and the number of oranges in each basket the total number of oranges plucked = 441.

The product of two equal numbers = 441.

 

 

So each number = √441

= 21.

 

Question 18. There are 140 students in your class. They are arranged in rows in solid squares. While arranging them in this way it is observed that there are 4 students. What is the number of rows in the solid square?

Solution: If we include these 4 fewer students in the total number of students in the class, then the total number of students would be

140 + 4 = 144.

These 144 students can be arranged in rows of solid squares.

∴ Number of rows = √144

 

 

∴ 144 = 2 x 2 x 2 x 2 x 3 x 3

= 2² x 2² x 3²

= (2 x 2 x 3)²

= (12)²

∴ √144 = √(12)²

= 12.

∴ The number of rows in the arrangements

= √44

=12.

The number of rows in the solid square =12.

 

Question 19. Each member of Sukanta Smriti Pathagar has been given a subscription in rupees as the total number of members in Pathagar. If the total amount of subscriptions is Rs. 2601, then what is the number of members in the Pathagar?

Solution: Each member of the Pathagar has been given a subscription in rupees = the total number of members in the Pathagar.

The total amount of subscriptions = is Rs. 2601.

.’. Product, of two equal numbers = 2601.

∴ Each number = Number of members = √2601 =51

 

 

Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple

Question 1. Find the H.C.F. of the following numbers by factorization 24, 36, 54

Solution:

Given:

24, 36, 54

∴ 24 = 2 x 2 x 2 x 3

∴ 36 = 2 x 2 x 3 x 3

Read and Learn More WBBSE Solutions For Class 6 Maths

∴ 54 = 2 x 3 x 3 x 3

∴ The common prime factors of given 3 numbers are 2, 3.

∴ The required H.C.F. = 2 x 3

= 6.

WBBSE Class 6 HCF and LCM Notes

Question 2. Find the H.C.F. of the following numbers by division method: 160, 165, 305

Solution:

Given: 160, 165, 305

The required H.C.F. = 5.

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Question 3. Find the H.C.F. of the following numbers by short division method 165, 264, 286

Solution :

Given: 165, 264, 286

 

∴ The required H.C.F = 11.

Understanding HCF and LCM

Question 4. Find the L.C.M. of the following numbers by factorization: 36, 60, 72

Solution:

Given: 36, 60, 72

∴ 36 = 2 x 2 x 3 x 3.

∴ 72 = 2 x 2 x 2 x 3 x 3

∴ 60 = 2 x 2 x 3 x 5

∴ The required L.C.M = 2 x 2 x 3 x 3 x 2 x 5

= 360

Short Questions on HCF and LCM

Question 5. Find the L.C.M. of the following numbers by the short division method :

1. 24, 36, 45, 60

Solution:

Given: 24, 36, 45, 60

2. 105, 119, 289

Solution:

Given: 105, 119, 289

Common Questions About Finding HCF

Question 6. Find the H.C.F. and L.C.M. of the following quantities :

1. 2 m 28 cm; 3 m 42 cm; 4 m 56 cm

Solution:

Given: 2 m 28 cm; 3 m 42 cm; 4 m 56 cm

2 m 28 cm = (2 x 100 + 28) cm = 228 cm

3 m 42 cm = (3 x 100 + 42) cm =. 342 cm

4 m 56 cm = (4 x 100 + 56) cm = 456 cm

 

∴  H.C.F. of 228, 342, 456 = 114.

∴ The required H.C.F. = 114 cm = 1 m 14 cm

Again,

 

 

L.C.M. of 228, 342, 456 = 2x2x3x 19 x3x2 = 1368.

The required L.C.M. = 1368 cm = 13 m 68 cm.

Practice Problems on HCF and LCM

2. 6 paisa 50; Rs. 5 Paisa 20; Rs. 7 Paisa 80.

Solution:

Given: 6 paisa 50; Rs. 5 Paisa 20; Rs. 7 Paisa 80.

Rs. 6 paisa 50 = 650 paisa;

Rs. 5 paisa 20 = 520 paisa;

Rs. 7 paisa 80 = 780 paisa.

Now,

 

∴ The H.C.F of 650, 520, and 780 = 130.

∴ The required H.C.F. = 130 paise = Re 1 Paisa 30.

Again,

∴ L.C.M. of 650, 520, 780 = 10 x 13 x 2 x 5 x 2 x 3 = 7800

∴ The required L.C.M = 7800 Paisa = Rs. 78.

Question 7. Prove of the following numbers that the product of the numbers is equal to the product of their H.C.F. and L.C.M. 87, 145

Solution:

Given: 87, 145

∴ The H.C.F of 87 and 145 = 29.

Again,

∴ L.C.M. of 87 and 145 = 29 x 3 x 5 = 435.

∴ Product of the H.C.F. and L.C.M. of the numbers 87, 145 = 29 x 435 = 12615. Product of the numbers = 87 x 145 = 12615.

The product of the numbers = Product of their H.C.F. and L.C.M. (Proved).

Conceptual Questions on Prime Factorization Method for HCF and LCM

Question 8. Which least number is exactly divisible by 15, 20, 24, and 32?

Solution:

Given: 15, 20, 24, And 32

∴ L.C.M. of 15, 20, 24, 32 = 2 x 2 x 2 x 3 x 5 x 4 = 480.

The required least number will be the L.C.M. of 15, 20, 24, and 32, and this last number will be exactly divisible by the given numbers.

∴ The required number = 480.

Question 9. By which greatest number 306, 810, and 2214 will be exactly divisible?

Solution:

Given: 306, 810, And 2214

The required greatest number will be the H.C.F. of 306, 810, and 2214 because H.C.F. will be the greatest number by which 306, 810, and 2214 will be divisible.

 

∴ The H.C.F. of 306, 810, and 2214 = 18.

∴ The required greatest number = is 18.

Real-Life Scenarios Involving Equal Distribution

Question 10.

1. Find the L.C.M. of 145 and 232 with the help of their H.C.F.

Solution:

Given: 145 And 232

 

∴ H.C.F. of 145 and 232 = 29.

We know that, the product of two numbers = Their H.C.F x L.C.M.

∴ L.C.M of two given numbers = Product of the numbers / H.C.F

= 145 x 232 / 29

= 1160.

∴ The required L.C.M. = 1160.

2. Find the H.C.F. of 144 and 384 with the help of their L.C.M.

Given: 144 and 384

∴ L.C.M. of 144 and 384 = 2 x 2 x 2 x 2 x 3 x 3 x 8 = 1152.

We know that the product of two given numbers = Their H.C.F x L.C.M.

∴ H.C.F of two numbers = Product of two given numbers

= 144 x 384 / 1152

= 48.

∴ The required H.C.F. = 48.

Examples of Real-Life Applications of HCF and LCM

Question 11. Find two pairs of numbers whose H.C.F. is 16 between 50 and 100.

Solution:

Given:

50 and 100

Since the H.C.F. of two numbers is 16, let the numbers be 16m and 16n where m and n are two co-prime natural numbers.

Now, 2 and 3 are co-prime natural numbers, and 2 x 16 = 32, and 3 x 16 = 48. But they do not lie between 50 and 100.

3 and 4 are co-prime natural numbers and taking m = 3, n = 4, we get 16 x 3 = 48, 16 x 4 = 64.

But 48 does not lie between 50 and 100.

Now, if we take m = 4 and n = 5 as 4, 5 are two co-prime natural numbers, and 16 x 4 = 64, 16 x 5 = 80.

Here both 64 and 80 lie between 50 and 100.

Again if we take m = 5 and n = 6 as 5,-6 are also two co-prime natural numbers, and 5 x 16 = 80, 6 x 16 = 96.

Here both 80 and 96 lie between 50 and 100.

∴ The required two pairs of numbers are 64, 80, and 80, 96.

Question 12. The H.C.F. and L.C.M. of the two numbers are 145 and 2175 respectively. If one number is 725, then what is the other number?

Solution:

We know that, Product of two numbers = Their H.C.F. x L.C.M.

Here H.C.F. = 145 and L.C.M. = 2175

∴ Product of two numbers = H.C.F. x L.C.M. = 145 x 2175

Since one number = 725 (given),

∴ The other number = 145 x 2175 / 725

= 435.

Question 13. Which least number is subtracted from 5834 so that the result of subtraction is divisible by 20, 28, 32, and 35?

Solution:

∴ L.C.M. of 20, 28, 32, 35 = 2 x 2 x 5 x 7 x 8 = 1120. So 1120 is the least number that is divisible by each of the numbers 20, 28, 32, and 35.

∴ Any multiple of 1120 is also divisible by each of the given numbers.

Now the multiple of 1120 are

1120 x 1 = 1120,

1120 x 2 = 2240,

1120 x 3 = 3360,

1120 x 4 = 4480,

1120 x 5 = 5600,

1120 x 6 = 6720

But the number 5834 lies between 5600 and 6720 and is very near to 5600 and 5834 > 5600.

Now, if we subtract (5834 – 5600) = 234 from 5834 the remainder is 5600, which is a multiple of 1120 and so 5600 is divisible by each of the numbers 20, 28, 32, and 35.

∴ The required least number = is 234.

Question 14. Find the greatest number that will divide 2300 and 3500 leaving the remainder 32 and 56 respectively.

Solution:

We have to find the greatest number that will divide 2300 and 3500 leaving the remainder 32 and 56 respectively.

2300 – 32 = 2268;

3500 – 56 = 3444.

Therefore, the numbers 2268 and 3444 must be divisible by the required greatest number and so the H.C.F. of 2268 and 3444 will be the required greatest number.

 

∴ The H.C.F. of 2268 and 3444 = 84.

∴ The required number = is 84.

Question 15. Find the greatest number which will divide 650, 775, and 1250 so as to leave the same remainder in each case.

Solution:

Since the required number when divides 650, 775, and 1250 leaves the

same remainder in each case, therefore (775 – 650) or 125 and (1250 – 775) or 475 must be divisible by the required greatest number. So the required greatest number will be the H.C.F. of 125 and 475.

∴ The H.C.F. of 125 and 475 = 25.

∴ The required greatest number is 25.

Question 16. The sum of two numbers is 384 and the H.C.F. of two numbers is 48. What are the two possible numbers?

Solution:

Given:

The sum of two numbers is 384 and the H.C.F. of two numbers is 48

The H.C.F. of two numbers is 48.

Let the numbers be 48x and 48y, where x and y are two co-prime natural numbers.

Again, by the questions,

48x + 48y = 384

or, 48 (x + y) = 384,

or, x + y = 384/48

= 8

or, x + y = 8

or, y = 8 – x ………………….(1)

8 = 1+7 = 2 + 6 = 3 + 5= 4 + 4

Among them, (1, 7), (3, 5), (5, 3), and (7, 1) are co-prime natural numbers

and they are values of (x, y)

∴ When x = 1, y = 7, the numbers are 48 x 1 = 48, 48 x 7 = 336.

When x = 3, y = 5, the numbers are 48 x 3 = 144, 48 x 5 = 240.

When x = 5, y = 3, the numbers are 48 x 5 = 240, 48 x 3 = 144.

When x = 7, y = 1, the numbers are 48 x 7 = 336, 48 x 1 = 48.

∴ The two possible numbers are (48, 336)

or, (336, 48) and (144, 240)

or, (240, 144).

The two possible numbers (48, 336)

Question 17. From which least number 4000 is subtracted so that the remainder is divisible by 7, 11, and 13?

Solution:

Here 7, 11, and 13 are prime to each other.

∴ L.C.M. of 7, 11, and 13 = 7 x 11 x 13 = 1001.

So the least number which is divisible by each of 7, 11, and 13 is 1001.

The required number is (1001 + 4000) = 5001.

Important Definitions Related to HCF and LCM

Example 18. The H.C.F. and L.C.M. of two numbers are 12 and 720 respectively. What are the possible numbers?

Solution:

Since the HCF of two numbers is 12, let the numbers be 12jc and 12y where x and y are two co-prime natural numbers.

Now the L.C.M. of 12jc and 12y = 12xy (x, y are co-prime to each other.)

By the given question, 12xy = 720

or, xy = .60

∴ 60

= 1 x 60

= 2 x 30

= 3 x 20

= 4 x 15

= 5 x 12

= 6 x 10

∴ Among these the pairs (1, 60), (3, 20), (4, 15), and (5, 12) are co-prime and they are the values of (;c, y). _

When x = 1, y = 60;

when x = 3, y = 20;

when x = 4, y = 15;

when x = 5, y = 12.

The numbers are

12 x 1 = 12, 12 x 60 = 720;

12 x 3 = 36; 12 x 20 = 240;

12 x 4 = 48; 12 x 15 = 180;

12 x 5 = 60; 12 x 12 = 144.

∴ The possible required numbers are 12, 720; 36, and 240; 48, 180; 60, and 144.

Question 19. Find the number which is divisible by 28, 33, 42, and 77 and which is nearer to 98765.

Solution:

Given:

28, 33, 42, and 77 and which is nearer to 98765

∴ 924 is the least number that is divisible by the given numbers.

Now,

 

98765 – 821 = 97944 is divisible by 924 and so 97944 is divisible by each of the given numbers 28, 33, 42, 77.

Again, 924 x 107 = 98868 is also divisible by 924 and so 98868 is divisible by each of the numbers 28, 33, 42, 77.

Since 98765 – 97944 = 821 and 98868 – 98765 = 103, the number 98868 is very near to 98765.

Hence the required number is 98868.

Question 20. Find the least number which is divisible by 13 and which when divided by 8, 12, 16, and 20 will leave a remainder of 1 in each case.

Solution:

Given:

8, 12, 16, And 20

The L.C.M. of 8, 12, 16, 20 = 2 x 2 x 2 x 3 x 2 x 5 = 240.

This implies that 240 is the least number that is divisible by each of the numbers 8, 12, 16, and 20.

But we have found the least number which is divisible by 13 and the number which when divided by each of the numbers 8, 12, 16, 20 will leave the remainder of 1 in each case.

Let the number (240 k + 1) be divisible by 13 where k is a positive integer.

Putting k = 1, 2, 3, 4, etc, we get,

240 k + 1 = 240 x 1 + 1 = 241 which is not divisible by 13.

240 k + 1 = 240 x 2 + 1 = 481 which is divisible by 13.

The required least number is 481.

Question 21. The circumference of the front wheel of a carriage is 1 m 4 dm and the circumference of the rear wheel is 2 1/2 times that of the front wheel. Find the least distance to be covered so that both wheels complete their full rotation simultaneously.

Solution :

Given:

The circumference of the front wheel of a carriage is 1 m 4 dm and the circumference of the rear wheel is 2 1/2 times that of the front wheel.

1 m 4 dm = (1 x 10 + 4) dm = 14 dm.

∴ The circumference of the front wheel = 14 dm.

So the circumference of the rear wheel = (14 x 2 1/2)dm

= 14 x 5/2 dm

= 35 dm.

Here the least required distance will be the L.C.M. of 14 dm and 35 dm.

L.C.M. of 14 dm and 35 dm = (7x2x5) dm

= 70 dm

= 7 m.

The least distance that the carriage will go so that both the front wheel and the rear wheel will rotate complete full rotation is 7 meters.

Question 22. There are 3 traffic signals at the crossing of 3 different ways. The lights of 3 traffic signals are changed at an interval every 16 sec, 28 sec, and 40 sec respectively. If all the lights in the traffic signals changed their lights simultaneously at 8 a.m. in the morning, when will they again change their lights simultaneously?

Solution:

Given:

There are 3 traffic signals at the crossing of 3 different ways. The lights of 3 traffic signals are changed at an interval every 16 sec, 28 sec, and 40 sec respectively. If all the lights in the traffic signals changed their lights simultaneously at 8 a.m. in the morning

The required time will be the L.C.M. of 16 sec, 28 sec, and 40 sec.

L.C.M. of 16 sec, 28 sec, 40 sec = (2 x 2 x 2 x 2 x 7 x 5) sec = 560 sec.

= 9 m sec.

Since the first time, the traffic signals simultaneously change their light again they will change their lights at (8 am + 9 m 20 sec) or 8 hr 9 m 20 sec in the morning.

Arithmetic Chapter 9 Recurring Decimal Number

Conversion of Recurring Decimal Into Vulgar Fraction:

There are two types of recurring decimals; pure recurring decimals and mixed recurring decimals. First, we shall discuss the conversion of pure recurring into vulgar fractions.

Conversion of pure recurring decimal into a vulgar fraction :

Question 1: Convert 0.2 into vulgar fractions.

Solution:

Given:

0.2

0.2 x 10 = (2222 ) x 10 = 2.2222 …………  (1)        (∵ Multiplying 0.2 by 10)

0.2x 1 = (0.2222 ) x 1 = 0.2222 ……… (2)              (∵ Multiplying 0.2 by 1)

Subtracting (2) from (1), we get,

0.2 (10 – 1) = (2.2222 ) – (2222……) = 2

or, 0.2 x 9 = 2

or 0.2 = 2/9

0.2 into vulgar fractions is 2/9

WBBSE Class 6 Recurring Decimals Notes

Question 2: Convert 0.35 into a vulgar fraction.

Solution :

Given:

0.35

0.35 = 0.353535

Multiplying both sides by 100 and 1 respectively, we get,

0.35 x 100 = (0.353535 ) x 100 = 35.353535………(1)

0.35 x 1 = (0.353535 ) x 1 = 0.353535……(2)

Subtracting (2) from (1), we get,

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0.35 (100 – 1) = 35

or, 0.35 x 99 = 35

or, 0.35 = 35/99

∴ 0.35 = 35/99

0.35 into a vulgar fraction is  35/99

From the above examples, we get the following rule of conversion of pure recurring decimal into a vulgar fraction

For Example 0.54632 = 54632/99999;

0.025 = 205/999;

0.51 = 51/99

= 17/13.

Short Questions on Recurring Decimals

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2. Conversion of mixed recurring decimal into a vulgar fraction:

For this observe the following examples :

Question 1: Convert 0.1275 into a vulgar fraction.

Solution:

Given: 0.1275

0.1275 = 0.12757575 ……..(1)

Multiplying both sides by 10000, we get,

0.1275 x 10000 = (0.12757575 ) x 10000 = 1275.757575 …………(2)

Multiplying both sides of (1) by 100, we get,

0.1275 x 100 = (0.12757575 ) x 100 = 12.757575.

Subtracting (3) from (2), we get,

(10000 – 100) x 0.1275 = 1275 – 12

or, 9900 x 0.1275 = 1275 – 12

or, 0.1275 = (1275-12) / 9900

= 1263/9900

= 421/3300

0.1275 into a vulgar fraction is 421/3300

Question 2. Convert 0.26321 into a vulgar fraction.

Solution:

Given: 0.26321

0.26321 = 0.26321321321…………………..(1)

Multiplying both sides of (1) by 100000, we get,

0-2632i x 100000 = (0-26321321321 ) x 100000 = 26321.321321……………………….(2)

Again multiplying both sides of (1) by 100, we get,

0.26321 x 100 = (0.26321321321……….) x 100 = 26-321321321…………………..(3)

Subtracting (3) from (2) we get,

0.26321 (100000 – 100) = 26321 – 26

or, 0.26321 x 99900 = 26321 – 26

or, 0.26321 = (26321 – 26) / 99900

= 26295/99900

= 8765 / 33300

0.26321 into a vulgar fraction is 8765 / 33300

Common Questions About Converting Recurring Decimals

Question 3: Convert 3.128 into a vulgar fraction.

Solution :

Given: 3.128

3.128 = 3.1282828……………………..(1)

Multiplying both sides of (1) by 1000, we get,

3.128 x 1000 = (3.1282828 ) x 1000 = 3128.282828……………………….(2)

Again multiplying both sides of (1) by 10, we get,

3.128 x 10 = (3.1282828 ) x 10 = 31.282828……………………….(3)

Subtracting (3) from (2), we get,

(1000 – 10) x 3.128 = 3128 – 31

or, 990 x 3.128 = 3128 – 31

or, 3.128 = (3128-31) / 990

= 3097 / 990

= 3 127 / 990.

∴ 3.128 = 3.127 / 990.

3.128 into a vulgar fraction is 3.127 / 990

Some Examples Are given Below:

1. 0.02028 

0.02028 = (2028 – 2) / 99900

= 2026 / 99900

= 1013 / 49950.

2. 10.293 

10.293 = (10293 – 102) / 990

= 10191 / 990

= 3397 / 330

= 10 97 / 330.

3. 5.2476 

5.2476 = (52476 – 52) / 9990

= 52424 / 9990

= 5 2474 / 9990

= 5 1237 / 4995

Practice Problems on Recurring Decimals

This can also be done in the following way:

If the recurring part contains only 9 (one or more), then omitting the recurring part, 1 is added to the number just before the recurring part.

For Example:

0.9 = 9/9

= 1;

0.19 =(19 – 1) / 90

= 18 / 90

= 1/5

= 0.2;

2.349 = (2349 – 234) / 900

= 2115 / 900

= 423 / 180

= 141 / 60

= 2.35;

1.099 = (1099 – 10) / 990

= 1089 / 990

= 121 / 110

= 11 / 10

= 1.1

Arithmetic Chapter 8 Percentage

Question 1: Convert the following fractions the percentages:

1. 9/10;

Solution:

9/10 = (9/10 x 100)%

= 90%.

9/10 = 90%.

2. 43/50;

Solution:

43/50 = (43/50 x 100)%

= 86%

43/50 = 86%

3.1 2/5;

Solution:

1 2/5 = 7/5

= 7/5 x 100 %

= 140

1 2/5 = 140

WBBSE Class 6 Percentage Notes

4. 4 3/8.

Solution:

4 3/8 = 35/8

= (35/8 x 100)%

= 437.5%

4 3/8 = 437.5%

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Question 2. Convert the following percentage into proper fractions:

1. 10%;

Solution:

10% = 10/100

= 1/10;

10% = 1/10

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2. 70%;

Solution:

70% = 70/100

7/10;

70% = 7/10

3. 257%;

Solution:

257% = 257/100

= 2 57/100;

257% = 2 57/100

Short Questions on Percentage Calculations

4. 33 1/3%;

Solution:

33 1/3% = 100/3 %

= 100/3 x 1/100

= 1/3.

33 1 = 1/3.

 

Question 3. Convert the following decimal fractions into percentages:

1. 0.6;

Solution:

0.6 = 6/10

= (6/10 x 100)%

= 60%

0.6 = 60%

2. 0.02

Solution:

0.02 = 2/100

= (2/100 x 100)%

= 2%

0.02 = 2%

Common Questions About Finding Percentages

3. 1.21

Solution:

1.21 = 121/100

=(121/100 x 100)%

= 121%

1.21 = 121%

4. 0.003

Solution:

0.003 = 3/1000

=(3/1000 x 100)%

= 3/10%

0.003 = 3/10%

 

Question 4. Convert the following percentages into decimal fractions:

1. 61%

Solution:

61% = 61/100

= 0.61;

61% = 0.61;

Practice Problems on Percentage

2. 105%

Solution:

105 = 105/100

= 1.05

105 = 1.05

3. 1.26% 

Solution:

1.26% = 1.26/100

= 0.0126

1.26% = 0.0126

4. 0.07%

Solution:

0.07% = 0.07/100

3= 0.0007

0.07% 0.0007

 

Question 5. Express the following vulgar fractions and decimal fractions in percentages and arrange them in ascending order:

1. 2/5 , 13/25 , 7/10

Solution:

2/5,

2/5 = (2/5 x 100)%

= 40%;

2/5 = 40%

13/25,

13/25 = (13/25 x 100)%

= 52%;

13/25 = 52%

Important Definitions Related to Percentages

7/10

7/10 = (7/10 x 100)%

=70%

7/10 =70%

∴ Arranging in ascending order of magnitudes, we get,

2/5 ,13/25,7/10

2. 1 2/5, 1 1/2, 1 9/10

Solution:

1 2/5,

1 2/5 = 7/5

= (7/5 x 100)%

= 140%;

1 2/5 = 140%;

1 1/2,

1 1/2 = 3/2

= (3/2 x 100)%

= 150%

1 1/2 = 150%

1 9/10,

1 9/10= 19/10

= (19/10 x 100)%

190%

1 9/10 = 190%

∴ Arranging in ascending order, we get,

1 2/5, 1 1/2, 1 9/10

Examples of Real-Life Applications of Percentages

3. 0.02, 0.15, 0.6

Solution:

0.02,

= 02/100

= (2/100 x 100)%

= 2%

0.02 = 2%

0.15

0.15 = 15/100

= (15/100 x 100)%

= 15%

0.15 = 15%

0.6

0.6 = 6/10

=(6/10 x 100)%

= 60%

0.6 = 60%

∴ Arranging in ascending order of magnitude, we get,

0.02, 0.15, 0.6.

 

Question 6. Today out of 35 students 7 are absent from class of Riya. What is the percentage of the total students present today in that class?

Solution: Total number of students in Riya’s class = 35

Number of students absent today = 7

∴ Total number of students present today = 35 – 7

= 28

∴ Percentage of students present today = (28/35 x 100)

= 80

So today 80% of the total students are present.

Alternative method :

The number of students present today = 35 – 7 = 28

∴ Out of 35 students, the present number of students = 28

∴ Out of 1 student part of the students present = 28/35

∴ Out of 100 students, the present number of students = 28/35 x 100 = 80

∴ Today present student = 80%.

Conceptual Questions on Percentage Problems

Question 7. There are 2100 storybooks in the library of Palpara. If 30% of it, more storybooks are purchased, then what will be the total number of storybooks in the library now, and what number of additional storybooks are purchased?

Solution: Total number of storybooks in the library = 2100. 30% of it is purchased.

30% of 2100 = 2100 x 30/100 = 630.

∴ An additional number of story books that are purchased = 630.

The total present number of storybooks in the library after purchasing = is 2100 + 630 = 2730.

∴ There will be 2730 story books in the library after purchasing.

 

Question 8. Abanibabu pays 22% of his monthly salary as house rent. If he pays Rs. 1870 as house rent every month, then what is his monthly salary?

Solution: Abanibabu pays as house rent = 22% of his monthly salary and which is ₹ 1870.

∴ 22% of monthly salary = ₹ 1870

∴ 1% of monthly salary = ₹ 1870/22

∴ 100% of monthly salary

 

 

= ₹ 8500

∴ Abanibabu’s monthly salary = ₹ 8500.

Real-Life Scenarios Involving Percentage Increase and Decrease

Question 9: The present population in a village is 26250. If the population increases at the rate of 4% every year, then what will be the total population in the next year? What will be the total population after two years?

Solution: Total number of present population in the village = 26250.

It increases every year by 4%.

∴ 4% of 26250 = 4/100 x 26250

= 1050

∴ The total population will be in the next year = (26250 + 1050) = 27300.

Again 4% of 27300

 

 

= 1092

∴ After 2 years, the total population village will be

(27300 + 1092) or 28392.

 

Question 10. Out of the total monthly expenses of the family of Reba, ₹ 4750 is spent on food and for all other expenses ₹ 5900 is spent. If the expenses on food are increased by 10% and other expenses are decreased by 16%, then calculate whether the total monthly expenses will increase or decrease and by how much.

Solution: The expenses on food is increased by 10%.

∴ 10% of ₹ 4750 = 10/100 x ₹ 4750 = ₹ 475.

The expenses for food will be increased by ₹ 475.

So the expenses for food will be after increment = ₹ (4750 + 475) = ₹ 5225.

Again the expenses for all other items are decreased by 16%.

∴ 16% of ₹ 5900 = 16/100 x Rs. 5900

= ₹ 944

∴ Other expenses will be decreased by = ₹ 944.

For all other expenses total amount to be spent = is ₹ (5900 – 944)

= ₹ 4956.

Total previous expenditure = ₹ (4750 + 5900)

= ₹ 10650.

Total present expenditure = ₹ (5225 + 4956)

= ₹ 10181.

As Rs. 10650 > ₹ 10181, the total expenditure will decrease by

₹ (10650 – 10181)

= ₹ 469.

 

Question 11. During harvesting, the price of rice was ₹ 1080 per quintal. In monsoon, the price of rice is increased by 15%. How much more money will be earned by a farmer who sold 12 acquittals of rice earlier if he will sell the same amount of rice during monsoon? Find also the total money he will receive by selling the rice during monsoon.

Solution: During monsoon, the price of rice is increased by 15%, whose earlier price was ₹ 1080 per quintal.

∴ 15% of ₹ 1080 = 15/100 x ₹ 1080 = ₹ 162.

∴ The price of rice per quintal during monsoon = ₹ (1080 + 162)

= ₹ 1242.

∴ The farmer will earn ₹ 162 more by selling per quintal of rice during monsoon.

∴ The farmer will earn more by selling 12 quintals of rice in monsoon than in earlier = ₹ (162 x 12)

= ₹ 1944.

The total amount of money will be received by the farmer by selling 12 quintals of rice in monsoon = ₹ (1242 x 12)

= ₹ 14904.

 

Question 12. 80 students have appeared for the Madhyamika examination this year from the school of Geeta. If 65% of them have passed, how many students have failed this year?

Solution: Total number of students who appeared for the Madhyamika examination = 80.

65% of 80 = 65/100 x 80 = 52.

∴ 52 students have passed the examination this year.

∴ (80 – 52) or 28 students have failed in Madhyamika examination this year.

 

Question 13. Due to the rise in the price of sugar, a family decides that the consumption of sugar will be reduced by 4%. If the family consumed 625 gm of sugar every day previously, how much gm of sugar consumption will be reduced, and also how much gm of sugar will be consumed each day now?

Solution: Previous consumption of sugar each day = 625 gm.

It is reduced by 4%.

∴ 4% of 625 gm

= 4/100 x 625

= 25 gm.

∴ The family will reduce sugar consumption each day = to 25 gm.

Again (625 – 25) = 600 gm

∴ The family will consumes sugar each day now = 600 gm.

 

Question 14. In a special kind of brass, copper is 70% and the rest is zinc. How many kg of copper and how many kg of zinc will be required to prepare 20 kg of this type of brass?

Solution: In the special kind of brass, copper is 70% and the rest is zinc.

∴ 70% of 20 kg = (70/100 x20) kg = 14 kg.

∴ Copper = 14 kg. So zinc = (20 – 14) = 6 kg.

∴ To prepare 20 kg of brass 14 kg of copper and 6 kg of zinc will be required.