Sunday, December 17, 2023

Class – 6 CH-12 RATIO AND PROPORTION MATHS NCERT SOLUTIONS

Class – 6 CH-12 RATIO AND PROPORTION

MATHS NCERT SOLUTIONS

Exercise 12.1

Question 1:

There are 20 girls and 15 boys in a class.

(a) What is the ratio of number of girls to the number of boys?

(b) What is the ratio of girls to the total number of students in the class? Solution 1:

20 4

(a) The ratio of girls to that of boys =  = 4:3

15 3

20 20 4

(b) The ratio of girls to total students =   = 4:7

20 15 35 7

Question 2:

Out of 30 students in a class, like football, 12 like cricket and remaining like tennis. Find the ratio of:

(a) Number of students liking football to number of students liking tennis. (b) Number of students liking cricket to total number of students.

Solution 2:

Total number of students = 30

Number of students like football = 6

Number of students like cricket = 12

Thus number of students like tennis = 30 – 6 – 12 = 12

6 1

(a) The ratio of students like football that of tennis =  = 1 : 2

12 2

(b) The ratio of students like cricket to that of total students =  = 2 : 5

30 5

Question 3:

See the figure and find the ratio of:

(a) Number of triangles to the number of circles inside the rectangle.

(b) Number of squares to all the figures inside the rectangle.

Solution 3:

(a) Ratio of number of triangle to that of circles = = 3 : 2

(b) Ratio of number of squares to all figures = = 2 : 7

Question 4:

Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of speed of Hamid to the speed of Akhtar.

Solution 4:

Distance

We know that, Speed =

Time

9 m 12 m

Speed of Hamid = = 9 km/h and Speed of Akhtar = = 12 km/h

1 h 1 h

9 3

Ratio of speed of Hamid to that of speed of Akhtar =  = 3 : 4

12 4

Question 5:

Fill in the following blanks:

  18 630

[Are these equivalent ratios?] Solution 5:

15 5 10 25

  

18 6 12 30

Yes, these are equivalent ratios.

Question 6:

Find the ratio of the following:

(a) 81 to 108 (b) 98 to 63

(a) Ratio of 81 to 108 =  = 3 : 4

108

(b) Ratio of 98 to 63 =  = 14 : 9

121

(d) Ratio of 30 minutes to 45 minutes =  = 2 : 3 45

Question 7:

Find the ratio of the following:

(a) 30 minutes to 1 hour (b) 40 cm to 1.5 m (c) 55 paise to ₹ 1 (d) 500 ml to 2 litres Solution 7:

(a) 30 minutes to 1.5 hour

1.5 hours = 1.5 x 60 = 90 minutes [ 1 hour = 60 minutes]

Now, ratio of 30 minutes to 1.5 hour = 30 minutes : 1.5 hour

 30 minutes : 90 minutes =  = 1 : 3

(b) 40 cm to 1.5 m

1.5 m = 1.5 x 100 cm = 150 cm [ 1 m = 100 cm]

Now, ratio of 40 cm to 1.5 m = 40 cm : 1.5 m

 40 cm : 150 cm =  = 4 : 15

150

₹ 1 = 100 paise

Now, ratio of 55 paise to ₹1 = 55 paise : 100 paise

  = 11 : 20

100

(d) 500 ml to 2 litters

2 litres = 2 x 1000 ml = 2000 ml [ 1 litre = 1000 ml]

Now, ratio of 500 ml to 2 litres = 500 ml : 2 litres

 500 ml : 2000 ml =  = 1 : 4

Question 8:

In a year, Seema earns ₹1,50,000 and saves ₹50,000. Find the ratio of:

(a) Money that Seema earns to the money she saves. (b) Money that she saves to the money she spends. Solution 8:

Total earning = ₹1,50,000 and Saving = ₹50,000

 Money spent = ₹1,50,000 - ₹50,000 = ₹1,00,000

(a) Ratio of money earned to money saved =  = 3 : 1

(b) Ratio of money saved to money spend =  = 1 : 2

Question 9:

There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students.

Solution 9:

Ratio of number of teachers to that of students =  = 17 : 550

Question 10:

In a college out of 4320 students, 2300 are girls. Find the ratio of:

(a) Number of girls to the total number of students.

(b) Number of boys to the number of girls.

Solution 10:

Total number of students in school = 4320

Number of girls = 2300

Therefore, number of boys = 4320 – 2300 = 2020

(a) Ratio of girls to total number of students =  = 115 : 216

(b) Ratio of boys to that of girls =  = 101 : 115

Question 11:

Out of 1800 students in a school, 750 opted basketball, 800 opted cricket and remaining opted table tennis. If a student can opt only one game, find the ratio of:

(a) Number of students who opted basketball to the number of students who opted table tennis.

(b) Number of students who opted cricket to the number of students opting basketball. (c) Number of students who opted basketball to the total number of students.

Solution 11:

Total number of students = 1800

Number of students opted basketball = 750

Number of students opted cricket = 800

Therefore, number of students opted tennis = 1800 – (750 + 800) = 250

(a) Ratio of students opted basketball to that of opted table tennis =  =3:1

(b) Ratio of students opted cricket to students opted basketball = =16:15

Question 12:

Cost of a dozen pens is ₹180 and cost of 8 ball pens is ₹56. Find the ratio of the cost of a pen to the cost of a ball pen.

Solution 12:

Cost of a dozen pens (12 pens) = ₹180

 Cost of 1 pen = = ₹15

Cost of 8 ball pens = ₹56

 Cost of 1 ball pen = = ₹7

Ratio of cost of one pen to that of one ball pen = = 15 : 7

Question 13:

Consider the statement: Ratio of breadth and length of a ball is 2 : 5. Complete the following table that shows some possible breadths and lengths of the hall.

Breadth of the hall (in meters) 10 40

Length of the hall (in meters) 25 50

Solution 13:

Ratio of breadth to length = 2 : 5 =

2 10 20 2 20 40

 Other equivalent ratios are =   ,  

5 10 50 5 20 100

Thus,

Breadth of the hall (in meters) 10 20 40

Length of the hall (in meters) 25 50 100

Question 14:

Divide 20 pens between Sheela and Sangeeta in the ratio 3 : 2.

Solution 14:

Ratio between Sheela and Sangeeta = 3 : 2 Total these terms = 3 + 2 = 5

Therefore, the part of Sheela = of the total pens and the part of Sangeeta = of total pens Thus, Sheela gets =  20 = 12 pens and Sangeeta gets =  20 = 8 pens

Question 15:

Mother wants to divide ₹36 between her daughters Shreya and Bhoomika in the ratio of their ages. If the age of Shreya is 15 years and age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.

Solution 15:

Ratio of the age of Shreya to that of Bhoomika =  = 5 : 4

Thus, ₹36 divide between Shreya and Bhoomika in the ratio of 5 : 4.

Shreya gets = of ₹36 =  36 = ₹20

Bhoomika gets = of ₹36 =  36 = ₹16

Question 16:

Present age of father is 42 years and that of his son is 14 years. Find the ratio of:

(a) Present age of father to the present age of son.

(b) Age of the father to the age of the son, when son was 12 years old.

(d) Age of father to the age of son when father was 30 years old.

Solution 16:

(a) Ratio of father’s present age to that of son =  = 3 : 1

(b) When son was 12 years, i.e., 2 years ago, then father was (42 – 2) = 40 years

Therefore, the ratio of their ages =  = 10 : 3

Therefore, ratio of their ages =  = 13 : 6

(d) When father was 30 years old,

i.e., 12 years ago, then son was (14 – 12) = 2 years old

Therefore, the ratio of their ages =  = 15 : 1 2

Exercise 12.2

Question 1:

Determine the following are in proportion:

(a) 15, 45, 40, 120

(b) 33, 121, 9, 96

(d) 32, 48, 70, 210 (e) 4, 6, 8, 12 (f) 33, 44, 75, 100 Solution 1:

15 (a) 15 : 45 =  = 1 : 3

40 40 : 120 =  = 1 : 3

120

Since 15 : 45 = 40 : 120

Therefore, 15, 45, 40, 120 are in proportion.

33 (b) 33 : 121 =  = 3 : 11

121

9 9 : 96 =  = 3 : 32

Since 33 : 121  9 : 96

Therefore, 33, 121, 9, 96 are not in proportion.

24 (c) 24 : 28 =  = 6 : 7

36 36 : 48 =  = 3 : 4

Since 24 : 28  36 : 48

Therefore, 24, 28, 36, 48 are not in proportion.

32 (d) 32 : 48 =  = 2 : 3

70 70 : 210 =  = 1 : 3

210

Since 32 : 48  70 : 210

Therefore, 32, 48, 70, 210 are not in proportion.

4 (e) 4 : 6 =  = 2 : 3

8 8 : 12 =  = 2 : 3

Since 4 : 6 = 8 : 12

Therefore, 4, 6, 8, 12 are in proportion.

33 (f) 33 : 44 =  = 3 : 4

75 75 : 100 =  = 3 : 4

100

Since 33 : 44 = 75 : 100

Therefore, 33, 44, 75, 100 are in ratio.

Question 2:

Write True (T) or False (F) against each of the following statements:

(a) 16 : 24 : : 20 : 30 (b) 21 : 6 : : 35 : 10

(d) 8 : 9 : : 24 : 27

(e) 5.2 : 3.9 : : 3 : 4

(f) 0.9 : 0.36 : : 10 : 4

Solution 2:

(a) 16 : 25 : : 20 : 30

16 20

 

24 30

2 2  

3 3

Hence, it is True.

(b) 21 : 6 : : 35 : 10

21 35

 

6 10

7 7

 

2 2

Hence, it is True.

12 28

 

18 12

2 7

 

3 3

Hence, it is False.

(d) 8 : 9 : : 24 : 27

8 24

 

9 27

8 8  

9 9

Hence, it is True.

(e) 5.2 : 3.9 : : 3 : 4

5.2 3

 

3.9 4

4 3

 

3 4

Hence, it is False.

(f) 0.9 : 0.36 : : 10 : 4

0.9 10

 

0.36 4

5 5

 

2 2

Hence, it is True.

Question 3:

Are the following statements true:

(a) 40 persons : 200 persons = ₹15 : ₹75

(b) 7.5 litres : 15 litres = 5 kg : 10 kg

(e) 45 km : 60 km = 12 hours : 15 hours

Solution 3:

40 (a) 40 persons : 200 persons =  = 1 : 5

200

15 ₹15 : ₹75 =  = 1 : 5

Since, 40 persons : 200 persons = ₹15 : ₹75 Hence, the statement is true.

7.5 75

(b) 7.5 litres : 15 litres =   = 1 : 2

15 150

5 5 kg : 10 kg =  = 1 : 2

Since, 7.5 litres : 15 litres = 5 kg : 10 kg Hence, the statement is true.

99 (c) 99 kg : 45 kg =  = 11 : 5

44 ₹44 : ₹20 =  = 11 : 5

Since, 99 kg : 45 kg = ₹44 : ₹20 Hence, the statement is true.

32 (d) 32 m : 64 m =  = 1 : 2

6 6 sec : 12 sec =  = 1 : 2

Since, 32 m : 64 m = 6 sec : 12 sec Hence, the statement is true.

45 (e) 45 km : 60 km =  = 3 : 4

12 12 hours : 15 hours =  = 4 : 5

Since, 45 km : 60 km  12 hours : 15 hours Hence, the statement is not true.

Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion:

(a) 25 cm : 1 m and ₹40 : ₹160

(b) 39 litres : 65 litres and 6 bottles : 10 bottles

(a) 25 cm : 1 m = 25 cm : (1 x 100) cm = 25 cm : 100 cm =  = 1 : 4

100

₹40 : ₹160 =  = 1 : 4

160

Since the ratios are equal, therefore these are in proportion. Middle terms = 1 m, ₹40 and Extreme terms = 25 cm, ₹160

39 (b) 39 litres : 65 litres = 

6 bottles : 10 bottles =  = 3 : 5

Since the ratios are equal, therefore these are in proportion.

Middle terms = 65 litres, 6 bottles and Extreme terms = 39 litres, 10 bottles

25 g : 625 g =  = 1 : 25

625

Since the ratios are not equal, therefore these are not in proportion.

(d) 200 ml :2.5 litres = 200 ml:(25000) litres = 200 ml : 2500 ml =  = 2:25

₹4 : ₹50 =  = 2 : 25

Since the ratios are equal, therefore these are in proportion.

Middle terms = 2.5 litres, ₹4 and Extreme terms = 200 ml, ₹50

Exercise 12.3

Question 1:

If the cost of 7 m of cloth is ₹294, find the cost of 5 m of cloth.

Solution 1:

Cost of 7 m of cloth = ₹294

294  Cost of 1 m of cloth = = ₹42

 Cost of 5 m of cloth = 42 x 5 = ₹210

Thus, the cost of 5 m of cloth is ₹210.

Question 2:

Ekta earns ₹1500 in 10 days. How much will she earn in 30 days?

Solution 2:

Earning of 10 days = ₹1500

1500  Earning of 1 day = = ₹150

 Earning of 30 days = 150 x 30 = ₹4500

Thus, the earning of 30 days is ₹4,500.

Question 3:

If it has rained 276 mm in the last 3 days, how many cm of rain will fall in one full week (7 days)? Assume that the rain continues to fall at the same rate.

Solution 3:

Rain in 3 days = 276 mm

276  Rain in 1 day = = 92 mm

 Rain in 7 days = 92 x 7 = 644 mm

Thus, the rain in 7 days is 644 mm.

Cost of 5 kg of wheat is ₹30.50.

(a) What will be the cost of 8 kg of wheat?

(b) What quantity of wheat can be purchased in ₹61?

Solution 4:

(a) Cost of 5 kg of wheat = ₹30.50

30.50 3050 Cost of 1 kg of wheat =  = ₹6.10

5 500

 Cost of 8 kg of wheat = 6.10 x 8 = ₹48.80

(b) From ₹30.50, quantity of wheat can be purchased = 5 kg

From ₹1, quantity of wheat can be purchased =

5 5

 From ₹61, quantity of wheat can be purchased =  61  6100

30.50 3050

= 10 kg

Question 5:

The temperature dropped 15 degree Celsius in the last 30 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next ten days?

Solution 5:

Degree of temperature dropped in last 30 days = 15 degrees

 Degree of temperature dropped in last 30 days =  degree

 Degree of temperature dropped in last 10 days =  10 = 5 degree

Thus, 5 degree Celsius temperature dropped in 10 days.

Question 6:

Shaina pays ₹7500 as rent for 3 months. How much does she has to pay for a whole year, if the rent per month remains same?

Solution 6:

Rent paid for 3 months = ₹7500

Rent paid for 1 months = = ₹2500

Rent paid for 12 months = 2500 x 12 = ₹30,000 Thus, the total rent of one year is ₹30,000.

Question 7:

Cost of 4 dozens bananas is ₹60. How many bananas can be purchased for ₹12.50?

Solution 7:

Cost of 4 dozen bananas = ₹60

Cost of 48 bananas = ₹60 [4 dozen = 4 x 12 = 48]

From ₹60, number of bananas can be purchased = 48

48  From ₹1, number of bananas can be purchased = 

From ₹12.50, number of bananas can be purchased = 12.50  

= 10 bananas

Thus, 10 bananas can be purchased for ₹12.50.

Question 8:

The weight of 72 books is 9 kg what is the weight of 40 such books?

Solution 8:

The weight of 72 books = 9 kg

 The weight of 1 book =

 The weight of 40 books =  40 = 5 kg

Thus, the weight of 40 books is 5 kg.

A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?

Solution 9:

For covering 594 km, a truck will be required diesel = 108 litres

108  For covering 1 km, a truck will be required diesel = 

594

 For covering 1650 km, a truck will be required diesel =  1650 = 300 litres

Thus, 300 litres diesel required by the truck to cover a distance of 1650 km.

Question 10:

Raju purchases 10 pens for ₹150 and Manish buys 7 pens for ₹84. Can you say who got the pen cheaper?

Solution 10:

Raju purchase 10 pens for = ₹150

150  Raju purchases 1 pen for = = ₹15

Manish purchases 7 pens for = ₹84

84 Manish purchases 1 pen for = = ₹12

Thus, Manish got the pens cheaper.

Question 11:

Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?

Solution 11:

Anish made in 6 overs = 42 runs

 Anish made in 1 overs = = 7 runs

Anup made in 7 overs = 63 runs

63  Anup made in 1 overs = = 9 runs

Thus, Anup made more runs per over.

Class – 6 CH-13 SYMMETRY MATHS NCERT SOLUTIONS

Class – 6 CH-13 SYMMETRY

MATHS NCERT SOLUTIONS

Exercise 13.1

Question 1:

List any four symmetrical from your home or school.

Solution 1:

Notebook, Blackboard, Glass, Inkpot.

Question 2:

For the given figure, which one is the mirror line, l1 or l2 ?

Solution 2:

l2 is the mirror line as both sides of the lines are symmetric.

Question 3:

Identify the shapes given below. Check whether they are symmetric or not. Draw the line of symmetry as well.

Solution 3:

(a) Symmetric

(b) Symmetric

(d) Symmetric

(e) Symmetric

(f) Symmetric

Question 4:

Copy the following on a square paper. A square paper is what you would have used in your arithmetic notebook in earlier classes. Then complete them such that the dotted line is the line of symmetry.

Question 5:

In the figure, l is the line of symmetry. Complete the diagram to make it symmetric.

Solution 5:

Question 6:

In the figure, l is the line of symmetry. Draw the image of the triangle and complete the diagram, so that it becomes symmetric.

Solution 6:

Exercise 13.2

Question 1:

Find the number of lines of symmetry for each of the following shapes:

Solution 1:

(a) 4

(b) 4

(d) 1

(e) 6

(f) 4

(g) 0

(h) 0

(i) 3

Question 2:

Copy the triangle in each of the following figures on squared paper. In each case, draw the line(s) of symmetry, if any and identify the type of triangle. (Some of you may like to trace the figures and try paper-folding first!)

(a) l1 is the line of symmetry.

(b) l1 is the line of symmetry.

(d) No line of symmetry.

Question 3:

Complete the following table:

Shape Rough figure No. of lines of symmetry

Equilateral triangle 3

Square

Rectangle

Isosceles triangle

Rhombus

Circle

Solution 3:

Shape Rough figure No. of lines of symmetry

Equilateral triangle 3

Square 4

Rectangle 2

Isosceles triangle 1

Rhombus 2

Circle Infinite

Question 4:

Can you draw a triangle which has:

(a) exactly one line of symmetry?

(b) exactly two lines of symmetry?

(d) no lines of symmetry? Sketch a rough figure in each case. Solution 4:

(a) Yes, Isosceles triangle

(b) No such triangle cannot be formed.

(d) Yes, Scalene triangle

Question 5:

On a squared paper, sketch the following:

(a) A triangle with a horizontal line of symmetry but no vertical line of symmetry.

(b) A quadrilateral with both horizontal and vertical lines of symmetry.

symmetry.

(d) A hexagon with exactly with two lines of symmetry.

(e) A hexagon with six lines of symmetry.

(Hint: It will be helpful if you first draw the lines of symmetry and then complete the figures) Solution 5:

(a)

(b)

(c)

(d)

(e)

Question 6:

Trace each figure and draw the lines of symmetry, if any:

Solution 6:

(a) No line

(b) Two lines

(d) Two lines

(e) One line

(f) Four lines

Question 7:

Consider the letters of English alphabets A to Z. List among them the letters which have:

(a) vertical lines of symmetry (like A)

(b) horizontal lines of symmetry (like B) (c) no lines of symmetry (like Q) Solution 7:

Vertical lines: A, H, I, M, O, T, U, V, W, X, Y

Horizontal lines: B, C, D, E, H, I, K, O, X

No line of symmetry: F, G, J, N, P, Q, R, S, Z

Question 8:

Given here are figures of a few folded sheets and designs drawn about the fold. In each case, draw a rough diagram of the complete figure that would be seen when the design is cut off.

Solution 8:

Exercise 13.3

Question 1:

Find the number of lines of symmetry in each of the following shapes. How will you check your Solution?

(a) (b) (c)

Question 2:

Copy the following drawing on squared paper. Complete each one of them such that the resulting figure has two dotted lines as two lines of symmetry.

How did you go about completing the picture?

Question 3:

In each figure below, a letter of alphabet is shown along with a vertical line. Take the mirror image of the letter in the given line. Find which letters look the same after reflection (i.e., which letters look the same in the image) and which do not. Can you guess why?

Try for

Different after reflection

Same after reflection

Different after reflection

Same after reflection

Different after reflection

Same after reflection

Class – 6 CH-14 PRACTICAL GEOMETRY MATHS NCERT SOLUTIONS

Class – 6 CH-14 PRACTICAL GEOMETRY

MATHS NCERT SOLUTIONS

Exercise 14.1

Question 1:

Draw a circle of radius 3.2 cm.

Solution 1:

Steps of construction:

(a) Open the compass for the required radius of 3.2 cm.

(b) Make a point with a sharp pencil where we want the centre of circle to be.

(d) Place the pointer of compasses on O.

(e) Turn the compasses slowly to draw the circle.

Hence, it is the required circle.

Question 2:

With the same centre O, draw two circles of radii 4 cm and 2.5 cm.

Solution 2:

Steps of construction:

(a) Marks a point ‘O’ with a sharp pencil where we want the centre of the circle.

(b) Open the compasses 4 cm.

(d) Turn the compasses slowly to draw the circle.

(e) Again open the compasses 2.5 cm and place the pointer of the compasses on D.

(f) Turn the compasses slowly to draw the second circle.

Hence, it is the required figure.

Question 3:

Draw a circle and any two of its diameters. If you join the ends of these diameters, what is the figure obtained if the diameters are perpendicular to each other? How do you check your Solution?

Solution 3:

(i) By joining the ends of two diameters, we get a rectangle. By measuring, we find AB = CD = 3 cm, BC = AD = 2 cm, i.e., pairs of opposite sides are equal and also  A =  B =  C =  D = 90, , i.e. each angle is of 90 .

Hence, it is a rectangle.

(ii) If the diameters are perpendicular to each other, then by joining the ends of two diameters, we get a square.

By measuring, we find that AB = BC = CD = DA = 2.5 cm, i.e., all four sides are equal.

Also  A =  B =  C =  D = 90, , i.e. each angle is of 90 .

Hence, it is a square.

Question 4:

Draw any circle and mark points A, B and C such that: (a) A is on the circle.

(b) B is in the interior of the circle. (c) C is in the exterior of the circle.

Solution 4:

(i) Mark a point ‘O’ with sharp pencil where we want centre of the circle.

(ii) Place the pointer of the compasses at ‘O’. Then move the compasses slowly to draw a circle.

(a) Point A is on the circle.

(b) Point B is in interior of the circle.

Question 5:

Let A, B be the centres of two circles of equal radii; draw them so that each one of them passes through the centre of the other. Let them intersect at C and D. Examine whether AB and CD are at right angles.

Solution 5:

Draw two circles of equal radii taking A and B as their centre such that one of them passes through the centre of the other. They intersect at C and D. Join AB and CD. Yes, AB and CD intersect at right angle as  COB is 90 .

Exercise 14.2

Question 1:

Draw a line segment of length 7.3 cm, using a ruler.

Solution 1:

Steps of construction:

(i) Place the zero mark of the ruler at a point A.

(ii) Mark a point B at a distance of 7.3 cm from A. (iii) Join AB.

Hence, AB is the required line segment of length 7.3 cm.

Question 2:

Construct a line segment of length 5.6 cm using ruler and compasses.

Solution 2:

Steps of construction:

(i) Draw a line ' '.l Mark a point A on this line.

(ii) Place the compasses pointer on zero mark of the ruler. Open it to place the pencil point up to 5.6 cm mark.

(iii) Without changing the opening of the compasses. Place the pointer on A and cut an arc ' 'l at B.

AB is the required line segment of length 5.6 cm.

Question 3:

Construct AB of length 7.8 cm. From this cut off AC of length 4.7 cm. Measure BC.

Solution 3:

Steps of construction:

(i) Place the zero mark of the ruler at A.

(ii) Mark a point B at a distance 7.8 cm from A.

(iii) Again, mark a point C at a distance 4.7 from A.

Hence, by measuring BC , we find that BC = 3.1 cm

Question 4:

Solution 4:

Steps of construction:

(i) Draw a line ' '.l

(ii) Construct PX such that length of PX = length of AB

(iii) Then cut of XQ such that XQ also has the length of AB.

(iv) Thus the length of PX and the length of XQ added together make twice the

length of AB.

Verification:

Hence, by measurement we find that PQ = 7.8 cm

= 3.9 cm + 3.9 cm

= AB + AB = 2 x AB

Question 5:

Given AB of length 7.3 cm and CD of length 3.4 cm, construct a line segment XY such

that the length of XY is equal to the difference between the lengths of AB and CD. Verify by measurement. Solution 5:

Steps of construction:

(i) Draw a line ' 'l and take a point X on it.

(ii) Construct XZ such that length XZ = length of AB = 7.3 cm

(iii) Then cut off ZY = length of CD = 3.4 cm

(iv) Thus the length of XY = length of AB – length of CD

Verification:

Hence, by measurement we find that length of XY = 3.9 cm

= 73. Cm – 3.4 cm

= ABCD

EXERCISE 14.3

Question 1:

Draw any line segment PQ. Without measuring PQ, construct a copy of PQ.

Solution 1:

Steps of construction:

(i) Given PQ whose length is not known.

(ii) Fix the compasses pointer on P and the pencil end on Q. The opening of the

instrument now gives the length of PQ .

(iii) Draw any line ' '.l Choose a point A on ' '.l Without changing the compasses setting, place the pointer on A.

(iv) Draw an arc that cuts ' 'l at a point, say B.

Hence, AB is the copy of PQ .

Question 2:

Given some line segment AB, whose length you do not know, construct PQ such that the

length of PQ is twice that of AB.

Solution 2:

Steps of construction:

(i) Given AB whose length is not known.

(ii) Fix the compasses pointer on A and the pencil end on B. The opening of the

instrument now gives the length of AB .

(iii) Draw any line ' '.l Choose a point P on ' '.l Without changing the compasses setting, place the pointer on Q.

(iv) Draw an arc that cuts ' 'l at a point R.

(v) Now place the pointer on R and without changing the compasses setting, draw another arc that cuts ' 'l at a point Q.

Hence, PQ is the required line segment whose length is twice that of AB.

EXERCISE 14.4

Question 1:

Draw any line segment AB. Mark any point M on it. Through M, draw a perpendicular to

AB. (Use ruler and compasses) Solution 1:

Steps of construction:

(i) With M as centre and a convenient radius, draw an arc intersecting the line AB at two points C and B.

(ii) With C and D as centres and a radius greater than MC, draw two arcs, which cut each other at P.

(iii) Join PM. Then PM is perpendicular to AB through the point M.

Question 2:

Draw any line segment PQ. Take any point R not on it. Through R, draw a perpendicular to PQ. (Use ruler and set-square) Solution 2:

Steps of construction:

(i) Place a set-square on PQ such that one arm of its right angle aligns along PQ .

(ii) Place a ruler along the edge opposite to the right angle of the set-square.

(iii) Hold the ruler fixed. Slide the set square along the ruler till the point R touches the other arm of the set square.

(iv) Join RM along the edge through R meeting PQ at M. Then RM  PQ.

Draw a line l and a point X on it. Through X, draw a line segment XY perpendicular to l.

Now draw a perpendicular to XY to Y. (use ruler and compasses) Solution 3:

Steps of construction:

(i) Draw a line ' 'l and take point X on it.

(ii) With X as centre and a convenient radius, draw an arc intersecting the line ' 'l at two points A and B.

(iii) With A and B as centres and a radius greater than XA, draw two arcs, which cut each other at C.

(iv) Join AC and produce it to Y. Then XY is perpendicular to ' 'l .

(v) With D as centre and a convenient radius, draw an arc intersecting XY at two points C and D.

(vi) With C and D as centres and radius greater than YD, draw two arcs which cut each other at F.

(vii) Join YF, then YF is perpendicular to XY at Y.

EXERCISE 14.5

Question 1:

Draw AB of length 7.3 cm and find its axis of symmetry. Solution 1:

Axis of symmetry of line segment AB will be the perpendicular bisector of AB. So, draw the perpendicular bisector of AB. Steps of construction:

(i) Draw a line segment AB = 7.3 cm

(ii) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.

(iii) Join CD. Then CD is the axis of symmetry of the line segment AB.

Question 2:

Draw a line segment of length 9.5 cm and construct its perpendicular bisector.

Solution 2:

Steps of construction:

(i) Draw a line segment AB = 9.5 cm

(ii) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.

(iii) Join CD. Then CD is the perpendicular bisector of AB .

Draw the perpendicular bisector of XY whose length is 10.3 cm.

(a) Take any point P on the bisector drawn. Examine whether PX = PY.

(b) If M is the mid-point of XY , what can you say about the lengths MX and XY?

Solution 3:

Steps of construction:

(i) Draw a line segment XY = 10.3 cm

(ii) Taking X and Y as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.

(iii) Join CD. Then CD is the required perpendicular bisector of XY .

Now:

(a) Take any point P on the bisector drawn. With the help of divider we can check

that PX PY .

(b) If M is the mid-point of XY, then MX XY.

Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.

(i)

(ii) Draw the perpendicular bisector of AB which cuts it at C. Thus, C is the mid-

point of AB.

(iii) Draw the perpendicular bisector of AC which cuts it at D. Thus D is the midpoint of .

(iv) Again, draw the perpendicular bisector of CB which cuts it at E. Thus, E is

the mid-point of CB.

(v) Now, point C, D and E divide the line segment AB in the four equal parts.

(vi) By actual measurement, we find that

AD  DC  CE  EB  3.2 cm

With PQ of length 6.1 cm as diameter, draw a circle.

Solution 5:

Steps of construction:

(i) Draw a line segment PQ = 6.1 cm.

(ii) Draw the perpendicular bisector of PQ which

cuts, it at O. Thus O is the mid-point of PQ .

(iii) Taking O as centre and OP or OQ as radius draw a circle where diameter is the line segment PQ .

Question 6:

Draw a circle with centre C and radius 3.4 cm. Draw any chord AB. Construct the

perpendicular bisector AB and examine if it passes through C.

Solution 6:

Steps of construction:

(i) Draw a circle with centre C and radius 3.4 cm.

(ii) Draw any chord AB.

(iii) Taking A and B as centres and radius more than half of AB, draw two arcs which cut each other at P and Q.

(iv) Join PQ. Then PQ is the perpendicular bisector of AB.

(v) This perpendicular bisector of AB passes through the centre C of the circle.

Repeat Question 6, if AB happens to be a diameter.

Solution 7:

Steps of construction:

(i) Draw a circle with centre C and radius 3.4 cm.

(ii) Draw its diameter AB .

(iii) Taking A and B as centres and radius more than half of it, draw two arcs which intersect each other at P and Q.

(iv) Join PQ. Then PQ is the perpendicular bisector of AB

(v) We observe that this perpendicular bisector of AB passes through the centre C of the circle.

Question 8:

Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?

Solution 8:

Steps of construction:

(i) Draw the circle with O and radius 4 cm.

(ii) Draw any two chords AB and

CD in this circle.

(iii) Taking A and B as centres and radius more than half AB, draw two arcs which intersect each other at E and F.

(iv) Join EF. Thus EF is the perpendicular bisector of chord

CD.

(v) Similarly draw GH the

perpendicular bisector of chord CD.

(vi) These two perpendicular bisectors meet at O, the centre of the circle.

Draw any angle with vertex O. Take a point A on one of its arms and B on another such

that OA = OB. Draw the perpendicular bisectors of OA and OB. Let them meet at P. Is PA = PB?

Solution 9:

Steps of construction:

(i) Draw any angle with vertex O.

(ii) Take a point A on one of its arms and B on another such that OA = OB.

(iii) Draw perpendicular bisector of OA and

OB.

(iv) Let them meet at P. Join PA and PB.

(v) With the help of divider, we check that

PA  PB.

Exercise 14.6

Question 1:

Draw  POQ of measure 75 and find its line of symmetry.

Solution 1:

Steps of construction:

(a) Draw a line l and mark a point O on it.

(b) Place the pointer of the compasses at O and draw an arc of any radius which intersects the line l at A.

(d) Join OB, then  BOA = 60 .

(e) Taking same radius, with centre B, cut the previous arc at C.

(f) Draw bisector of  BOC. The angle is of

90 . Mark it at D. Thus,  DOA = 90

(g) Draw OP as bisector of  DOB. Thus,  POA = 75

Question 2: Draw an angle of measure 147 and construct its bisector.

Solution 2:

Steps of construction:

(a) Draw a ray OA.

(b) With the help of protractor, construct 

AOB = 147 .

arms OA and OB at P and Q respectively.

(d) Taking P as centre and radius more than half of PQ, draw an arc.

(e) Taking Q as centre and with the same radius, draw another arc which intersects the previous at R.

(f) Join OR and produce it.

(g) Thus, OR is the required bisector of  AOB.

Draw a right angle and construct its bisector.

Solution 3:

Steps of construction:

(a) Draw a line PQ and take a point O on it.

(b) Taking O as centre and convenient radius, draw an arc which intersects PQ at A and B.

(d) Join OC. Thus,  COQ is the required right angle.

(e) Taking B and E as centre and radius more than half of BE, draw two arcs which intersect each other at the point D.

(f) Join OD. Thus, OD is the required bisector of  COQ.

Question 4: Draw an angle of measure 153 and divide it into four equal parts.

Solution 4:

Steps of construction:

(a) Draw a ray OA.

(b) At O, with the help of a protractor, construct  AOB = 153 . (c) Draw OC as the bisector of  AOB.

(d) Again, draw OD as bisector of  AOC.

Question 5:

Construct with ruler and compasses, angles of following measures:

(a) 60 (b) 30 (c) 90 (d) 120 (e) 45 (f) 135

Solution 5:

Steps of construction:

(a) 60

(i) Draw a ray OA.

(ii) Taking O as centre and convenient radius, mark an arc, which intersects OA at P.

(iii) Taking P as centre and same radius, cut previous arc at Q.

(iv) Join OQ.

Thus,  BOA is required angle of 60 .

(b) 30

(i) Draw a ray OA.

(ii) Taking O as centre and convenient radius, mark an arc, which intersects OA at P.

(iii) Taking P as centre and same radius, cut previous arc at Q.

(iv) Join OQ. Thus,  BOA is required angle of

60 .

(v) Put the pointer on P and mark an arc.

(vi) Put the pointer on Q and with same radius, cut the previous arc at C. Thus,  COA is required angle of 30 .

(i) Draw a ray OA.

(ii) Taking O as centre and convenient radius, mark an arc, which intersects OA at X.

(iii) Taking X as centre and same radius, cut previous arc at Y.

(iv) Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.

(v) Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.

(vi) Join OS and produce it to form a ray OB.

Thus,  BOA is required angle of 90 .

(d) 120

(i) Draw a ray OA.

(ii) Taking O as centre and convenient radius, mark an arc, which intersects OA at P.

(iii) Taking P as centre and same radius, cut previous arc at Q.

(iv) Taking Q as centre and same radius cut the arc at S.

(v) Join OS.

Thus,  AOD is required angle of 120 .

(e) 45

(i) Draw a ray OA.

(ii) Taking O as centre and convenient radius, mark an arc, which intersects OA at X.

(iii) Taking X as centre and same radius, cut previous arc at Y.

(iv) Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.

(v) Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.

(vi) Join OS and produce it to form a ray OB. Thus,  BOA is required angle of

90 .

(vii) Draw the bisector of  BOA.

Thus,  MOA is required angle of 45 .

(f) 135

(i) Draw a line PQ and take a point O on it.

(ii) Taking O as centre and convenient radius, mark an arc, which intersects PQ at A and B.

(iii) Taking A and B as centres and radius more than half of AB, draw two arcs intersecting each other at R.

(iv) Join OR. Thus,  QOR =  POQ = 90 .

(v) Draw OD the bisector of  POR.

Thus,  QOD is required angle of 135 .

Question 6:

Draw an angle of measure 45 and bisect it.

Solution 6:

Steps of construction:

(a) Draw a line PQ and take a point O on it.

(b) Taking O as centre and a convenient radius, draw an arc which intersects PQ at two points A and B.

(d) Join OC. Then  COQ is an angle of 90

(e) Draw OE as the bisector of  COE. Thus,  QOE = 45 (f) Again draw OG as the bisector of  QOE.

Thus,  QOG =  EOG = 22 .

Question 7:

Draw an angle of measure 135 and bisect it.

Solution 7:

Steps of construction:

(a) Draw a line PQ and take a point O on it.

(b) Taking O as centre and convenient radius, mark an arc, which intersects PQ at A and B.

(d) Join OR. Thus,  QOR =  POQ = 90 .

(e) Draw OD the bisector of  POR. Thus,  QOD is required angle of 135 .

(f) Now, draw OE as the bisector of  QOD.

Thus,  QOE =  DOE = 67 

Question 8: Draw an angle of 70 . Make a copy of it using only a straight edge and compasses.

Solution 8:

Steps of construction:

(a) Draw an angle 70 with protractor, i.e.,  POQ = 70

(b) Draw a ray AB.

(d) Use the same compasses, setting to draw an arc with A as centre, cutting AB at X. (e) Set your compasses setting to the length LM with the same radius.

(f) Place the compasses pointer at X and draw the arc to cut the arc drawn earlier at Y.

(g) Join AY.

Thus,  YAX = 70

Question 9: Draw an angle of 40 . Copy its supplementary angle.

Solution 9:

Steps of construction:

(a) Draw an angle of 40 with the help of protractor, naming  AOB.

(b) Draw a line PQ.

(d) Place the compasses at O and draw an arc to cut the rays of  AOB at L and N.

(e) Use the same compasses setting to draw an arc O as centre, cutting MQ at X.

(f) Set your compasses to length LN with the same radius.

(g) Place the compasses at X and draw the arc to cut the arc drawn earlier Y.

(h) Join MY.

Thus,  QMY = 40 and  PMY is supplementary of it.

Class – 6 CH-9 DATA HANDLING MATHS NCERT SOLUTIONS

Class – 6 CH-9 DATA HANDLING

MATHS NCERT SOLUTIONS

Exercise 9.1

Question 1:

In a mathematics test the following marks were obtained by 40 students. Arrange these marks in a table using tally marks.

8 1 3 7 6 5 5 4 4 2

4 9 5 3 7 1 6 5 2 7

7 3 8 4 2 8 9 5 8 6

7 4 5 6 9 6 4 4 6 6

(a) Find how many students obtained marks equal to or more than 7?

(b) How many students obtained marks below 4?

Solution 1:

Marks Tally Marks No. of students

9 2

(a) Twelve students

(b) Eight students

Question 2:

Following is the choice of sweets of 30 students of Class VI.

Ladoo, Barfi, Ladoo, jalebi, Ladoo, Rashulla, Jalebi, Ladoo, Barfi, Rasgulla, Ladoo, Jalebi, Jalebi, Rashulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Ladoo, Ladoo, Barfi, Rasgulla, Rasgulla, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo (a) Arrange the names of sweets in a table using tally marks.

(b) Which sweet is preferred by most of the students?

Solution 2:

(a) Table using tally marks:

Sweets Tally Marks No. of students

Ladoo

Barfi

Jalebi

Rasgulla 11

(b) Ladoo. Because 11 students prefer eat.

Question 3:

Catherine threw a dice 40 times and noted the number appearing each time as shown below:

1 3 5 6 6 3 5 4 1 6

2 5 3 4 6 1 5 5 6 1

1 2 2 3 5 2 4 5 5 6

5 1 6 2 3 5 2 4 1 5

Make a table and enter the data using tally marks. Find the number that appeared.

(a) The minimum number of times.

(b) The maximum number of times. (c) Find those numbers that appear an equal number of times.

Solution 3:

Numbers Tally Marks How many times?

6 7

(a) The minimum number of times = 4

(b) The maximum number of times = 5

Following pictograph shows the number of tractors in five villages:

Villages Number of tractors

Observe the pictograph and Solution the following questions:

(i) Which village has the minimum number of tractors?

(ii) Which village has the maximum number of tractors?

(iii) How many more tractors village C has as compared to village B. (iv) What is the total number of tractors in all the five villages? Solution 4:

(i) Village D (ii) Village C (iii) 3 (iv) 28

Question 5:

The number of girl students in each class of a co-educational middle school is depicted by the pictograph.

Classes Number of girl students

Observe this pictograph and Solution the following questions:

(a) Which class has the minimum number of girl students?

(b) Is the number of girls in class VI less than the number of girls in class V?

Solution 5:

(a) Class VIII (b) No (c) 3 × 4 = 12 girls

Question 6:

The sale of electric bulbs on different days of a week is shown below:

Days Number of electric bulbs

Observe the pictograph and Solution the following questions:

(a) How many bulbs were sold on Friday?

(b) On which day were the maximum number of bulbs sold?

(d) On which of the days minimum number of bulbs were sold?

(e) If one big carton can hold 9 bulbs. How many cartons were needed in the given week?

Solution 6:

(a) Number of bulbs sold on Friday are 14.

(b) Maximum number of bulbs (18) were sold on Sunday.

(d) Then minimum number of bulbs (8) were sold on Wednesday and Saturday.

(e) The total number of bulbs sold in the given week were 86.

So the number of carton required to hold 86 bulb = 86/9 10. Hence, 10 cartons required to hold the bulbs.

In a village six fruit merchants sold the following number of fruit baskets in a particular season:

Name of fruit merchants Number of fruit baskets

Observe this pictograph and Solution the following questions:

(a) Which merchant sold the maximum number of baskets?

(b) How many fruit baskets were sold by Anwar?

(c) The merchants who have sold 600 or more number of baskets are planning to buy a godown for the next season. Can you name them?

Solution 7:

(a) Martin

(b) 7 × 100 = 700 fruit basket

Exercise 9.2

Question 1:

Total number of animals in five villages are as follows:

Village A : 80 Village B : 120

Village C : 90 Village D : 40

Village E : 60

Prepare a pictograph of these animals using one symbol to represent 10 animals and Solution the following questions:

(a) How many symbols represent animals of village E?

(b) Which village has the maximum number of animals? (c) Which village has more animals: village A or village C?

Solution 1:

10 animals =

Village A

Village B

Village C

Village D

Village E

120

(a) 6

(b) Village B

Total number of students of a school in different years is shown in the following table:

Years Number of students

1996 400

1998 535

2000 472

2002 600

2004 623

A. Prepare a pictograph of students using one symbol to represent 100 students and Solution the following questions:

(a) How many symbols represent total number of students in the year 2002?

(b) How many symbols represent total number of students for the year 1998?

B. Prepare another pictograph of students using any other symbol each representing 50 students. Which pictograph do you find more informative? Solution 2:

Years

100 students =

1996

1998

2000

2002

2004

(a) 6

(b) Five completed and one incomplete.

Years 50 students =

1996

1998

2000

2002

2004

Pictograph B is more informative than A.

Exercise 9.3

Question 1:

The bar graph given below shows the amount of wheat purchased by government during the year 1998 – 2002.

Read the bar graph and write down your observations.

(a) In which year was the wheat production maximum?

(b) In which year was the wheat production minimum?

Solution 1:

(a) In 2002, production of wheat was maximum.

(b) In 1998, production of wheat was minimum.

Question 2:

Observe this bar graph which is showing the sale of shirts in a readymade shop from Monday to Saturday.

Now Solution the following questions:

(a) What information does the above bar graph give?

(b) What is the scale chosen on the horizontal line representing number of shirts?

(d) On which day were the minimum number of shirts sold? (e) How many shirts were sold on Thursday?

Solution 2:

(a) The bar graph shows the sale of shirt in a readymade shop from Monday to Saturday.

(b) 1 unit = 5 shirts

(d) On Tuesday, minimum number of shirts were sold. (e) On Tuesday, 35 shirts were sold.

Observe this bar graph which shows the marks obtained by Aziz in half yearly examination in different subjects:

Solution the given questions:

(a) What information is does the bar graph give?

(b) Name the subject in which Aziz scored maximum marks.

(c) Name the subject in which he has scored minimum marks. (d) State the name of the subjects and marks obtained in each of them.

Solution 3:

(a) The bar graph shows the marks obtained by Aziz in half yearly examination in different subjects.

(b) Hindi.

(d) Hindi 80, English 60, Mathematics 70, Science 50, Social Studies 40.

Exercise 9.4

Question 1:

A survey of 120 school students was done to find which activity they prefer to do in their free time:

Preferred activity Number of students

Playing

Reading story books

Watching TV

Listening to music

Painting 45

Draw a bar graph to illustrate the above data taking scale of 1 unit length = 5 students.

Which activity is preferred by most of the students other than playing?

Solution 1:

Reading story books is preferred by most of the students other than playing.

The number of mathematics books sold by a shopkeeper on six consecutive days is shown below:

Days Sunday Monday Tuesday Wednesday Thursday Friday

No. of books sold 65 40 30 50 20 70

Draw a bar graph to represent the above information choosing the scale of your choice.

Solution 2:

Following shows the number of bicycles manufactured in a factory during the year 1998 to 2002. Illustrate this data using a bar graph. Choose a scale your choice.

Years Number of bicycles manufactured

1998

1999

2000

2001

2002 800

600

900

1100

1200

(a) In which year were the maximum number of bicycles manufactures? (b) In which year were the minimum number of bicycles manufactured?

(a) The maximum number of bicycles manufactures in 2002.

(b) The minimum number of bicycles manufactured in 1999.

Number of persons in various age groups in a town is given in the following table:

Age Group Number of persons

1 – 14

15 – 29

30 – 44

45 – 59

60 – 74

75 and above 2 Lakhs

1 lakh 60 thousands

1 lakh 20 thousands

80 thousands

40 thousands

Draw a bar graph to represent the above information and Solution the following questions. (Take 1 unit length = 20 thousands)

(a) Which two age groups have same population?

(b) All persons in the age group of 60 and above are called senior citizens. How many senior citizens are there in the town?

(a) Group 30 – 44 and group 45 – 59 have same population.

(b) 80,000 + 40,000 = 1,20,000 senior citizens are there in the town.

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Sunday, December 17, 2023