Category: Class 7

Students can Download Maths Chapter 3 Algebra Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Additional Questions

Additional Questions and Answers

Exercise 3.1

Question 1.
If 4x² + y² = 40 and xy = b find the value of 2x + y.
Solution:
We have (a + b)² = a² + 2ab + b²
(2x + y)² = (2x)² + (2 × 2x × y) + y² = (4x² + y²) + 4xy = 40 + 4 × 6 = 40 + 24
(2x + y)² = 64
(2x + y)² = 82
2x + y = 8

Question 2.
If x² + \(\frac{1}{x^{2}}\) =23 find x + \(\frac { 1 }{ x } \)
Solution:
We have (a + b)² = a² + 2ab + b²
So (x + \(\frac { 1 }{ x } \))² = x² + 2 × x × \(\frac { 1 }{ x } \) + \(\frac{1}{x^{2}}\)
= x² + 2 + \(\frac{1}{x^{2}}\) = x² + \(\frac{1}{x^{2}}\) + 2 = 23 + 2
∵ x² + \(\frac{1}{x^{2}}\) = 23
(x + \(\frac { 1 }{ x } \))² = 25
(x + \(\frac { 1 }{ x } \))² = 52
x + \(\frac { 1 }{ x } \) = 5

Question 3.
Find the product of (\(\frac { 2 }{ 3 } \) x² + 5y²) (\(\frac { 2 }{ 3 } \) x² + 5y²)
Solution:
(\(\frac { 2 }{ 3 } \) x² + 5y²) (\(\frac { 2 }{ 3 } \) x² + 5y²) = (\(\frac { 2 }{ 3 } \) x² + 5y²)²
We have (a + b)² = a² + 2ab + b²
Here a = \(\frac { 2 }{ 3 } \) x² b = 5y²
(\(\frac { 2 }{ 3 } \) x² + 5y²)² = (\(\frac { 2 }{ 3 } \) x²)² + 2 × \(\frac { 2 }{ 3 } \) x² × 5y² + (5y²)²
= (\(\frac { 2 }{ 3 } \))² (x²)² + \(\frac{20 x^{2} y^{2}}{3}\) + 52 (y²)²
(\(\frac { 2 }{ 3 } \) x² + 5y²)² = \(\frac { 4 }{ 9 } \) x⁴ + \(\frac{20 x^{2} y^{2}}{3}\) + 25y⁴

Exercise 3.2

Question 1.
Solve 2x + 5 < 15 where x is a natural number and represent the solution in a number line.
Solution:
2x + 5 < 15
Subtracting 5 on both sides 2x + 5 – 5 < 15 – 5
2x < 10
Dividing by 2 on both the sides
\(\frac { 2x }{ 2 } \) < \(\frac { 10 }{ 2 } \)
Since x is a natural number and it is less than 5, the solution is 4, 3, 2 and 1. It is shown in the number line as below.

Question 2.
Solve 2c + 4 < 14, where c is a whole number.
Solution:
2c + 4 < 14
Subtracting 4 on both sides 2c + 4 – 4 < 14 – 4
2c < 10
Dividing by 2 on both the sides
\(\frac { 2c }{ 2 } \) < \(\frac { 10 }{ 2 } \)
c < 5
Since the solutions are whole numbers which are less than r equal to 5, the solution set is 0, 1, 2, 3, 4 and 5.

Question 3.
Solve -8 < -2n + 4, n is a natural number.
Solution:
-8 < – 2n + 4
Subtracting 4 on both sides
-8 -4 < -2n + 4 – 4
– 12 < – 2 n
÷ by -2, we have
\(\frac { -2n }{ -2 } \) < \(\frac { -12 }{ -2 } \) [ ∵ Dividing by negative number, the inequation get reversed]
n < 6
Since the solutions are natural numbers which are less then 6, we have the solution as 1, 2, 3, 4 and 5.

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 4 Geometry Ex 4.3

Miscellaneous Practice problems

Question 1.
The bishop, in given picture of chess board, can move diagonally along dark squares. Describe the translations of the bishop after two moves as shown in the figure.

Solution:
For first: 2 →, 2↓; For second move: 5 ←, 5↓

Question 2.
Write a possible translation for each of chess piece for a single move.

Solution:
Pawn – 1 ↑ or 2↑
Rook – 1 to 8 ↑
Knight – 2 →,1 ↑or 2 ←,1 ↑ or 1 →,2 ↑or 1 ←,2↑
Bishop – 1 →,1 ↑ or 2 →,2↑or 3 →,3↑ or 4 →,4↑ or 5 → 5 ↑1 ←,1↑ or 2 ←,2↑or 3 ←,3↑or 4 ←,4↑ or 5 ← 5↑
Queen – 1 to 8 ,1 →, 1 ↑ or 2 →,2↑ or 3 → ,3↑or 4 → ,4↑or 5 →,5↑or 1 ←,↑1 or 2 ←,2 ↑or
3 ←,3 ↑ or 4 ←,4 ↑ or 5 ← 5↑
King – 1 → or ← or ↑

Question 3.
Referring the graphic given, answer the following questions. Each bar of the category is made up of boy-girl-boy unit, (i) Which categories show a boy- girl-boy unit that is translation within the bar? (ii) Which categories show a boy-girl-boy unit that is reflected within the bar?

Solution:
(i) Essay Writing category shows translation
(ii) Essay Writing and Mono Acting categories shows reflection

Question 4.
Given figure is a floor design in which the length of the small red equilateral triangle is 30 cm. All the triangles and hexagons are regular. Describe the translations in cm, represented by the (i) yellow line (ii) black line (iii) blue line.

Solution:
(i) 120cm →, 210cm ↓
(ii) 270cm ← ,330cm ↑
(iii) 150 cm →

Question 5.
Describe the transformation involved in the following pair of figures (letters). Write translation, reflection or rotation.

Solution:
(i) rotation
(ii) reflection
(iii) translation
(iv) reflection
(v) rotation
(vi) reflection
(vii) rotation
(viii) translation

Challenge problems

Question 1.
In chess, a knight can move only in an L-shaped pattern:

• two vertical squares, then one horizontal square;
• two horizontal squares, then one vertical square;
• one vertical square, then two horizontal squares; or
• one horizontal square, then two vertical squares.

Write a series of translations to move the knight from g8 to g5 (at most two moves)
Solution:.
2 ←,1↓ and then 1 ←, 2↓ (or) 2 ←, 1↓and then 1 ← ,2↓

Question 2.
The pink shape is congruent to blue shape. Describe a sequence of transformations in which the blue shape is the image of pink shape.

Solution:
(i) Translation 3 ←, 5↑ and 90° counter clockwise rotation about the green point and translates 5 ←, 2↓,
(ii) Translation 2 ← 90° counter clockwise rotation about the green point and translates 2 ←, 2↓.

Question 3.

Solution:

Question 4.
Draw concentric circles given that radius of inner circle is 4.5 cm and width of circular ring is 2.5 cm.
Solution:
Give radius of inner circle = 4.5 cm
Width of circular ring is 2.5 cm
Radius of outer circle = 4.5 + 2.5 = 7 cm

Step 1 : Drawn a rough diagram and marked the given measurements
Step 2 : Taken any point O and marked it as the center.
Step 3 : With O as center and drawn a circle of radius OA = 4.5 cm.
Step 4 : With O as center drawn a circle of radius OB = 4.5 + 2.5 = 7 cm. Thus the concentric circles C₁ and C₂ are drawn.
Width of the circular ring = OB – OA
= 7 – 4.5 = 2.5 cm

Question 5.
Draw concentric circles given that radius of outer circle is 5.3 cm and width of circular ring is 1.8 cm.
Solution:
Give radius of outer circle = 5.3 cm
Width of circular ring = 1.8 cm
Radius of the inner circle = 5.3 – 1.8
= 3.5 cm

Step 1: Drawn a rough diagram and marked the given measurements
Step 2 : Taken any point O and marked it as the center.
Step 3 : With O as center drawn a circle of radius OA =3.5 cm.
Step 4 : With O as center, drawn a circle of radius OB = 5.3 cm. Thus concentric circles
C₁ and C₂ are drawn.
Width of the circular ring = OB – OA
= 5.3 – 3.5 = 1.8 cm

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 4 Geometry Ex 4.2

Question 1.
Draw circles for the following measurements of radius (r)/ diameters(d).
(i) r = 4 cm
(ii) d = 12 cm
(iii) r = 3.5 cm
(iv) r = 6.5 cm.
(v) d = 6 cm
Solution:
(i) r = 4 cm

Step 1 : Market a point ‘O’ on the paper.
Step 2 : Extended the compass distance equal to radius 4 cm.
Step 3 : At center ‘O’, helded the compass firmly and placed the pointed end of the compass.
Step 4 : Slowly rotated the compass around to get the circle.

(ii) d = 12 cm
given d= 12 cm
∴ radius r = \(\frac { d }{ 2 } \) = \(\frac { 12 }{ 2 } \) = 6 cm

Step 1: Marked a point ‘O’ on the paper.
Step 2: Extended the compass distance equal to radius 6 cm.
Step 3: At center ‘O’, held the compass firmly and placed the pointed end of the compass.
Step 4: Slowly rotated the compass around to get the circle.

(iii) r = 3.5 cm

Step 1: Market a point ‘O’ on the paper.
Step 2: Extended the compass distance equal to radius 3.5 cm.
Step 3: At center ‘O’, held the compass firmly and placed the pointed end of the compass.
Step 4: Slowly rotated the compass around to get the circle.

(iv) r = 6.5 cm

Step 1: Market a point ‘O’ on the paper.
Step 2: Extended the compass distance equal to radius 6.5 cm.
Step 3: At center ‘O’, held the compass firmly and placed the pointed end of the compass.
Step 4: Slowly rotated the compass around to get the circle.

(v) d = 6 cm
∴ radius r = \(\frac { d }{ 2 } \) = \(\frac { 6 }{ 2 } \) = 3 cm

Step 1: Market a point ‘O’ on the paper.
Step 2: Extended the compass distance equal to radius 3 cm.
Step 3: At center ‘O’, held the compass firmly and placed the pointed end of the compass.
Step 4: Slowly rotated the compass around to get the circle.

Question 2.
Draw concentric circles for the following measurements of radii / diameters. Find out the width of each circular ring.
(i) r = 3 cm and r = 5 cm.
(ii) r = 3.5 cm and r = 6.5 cm.
(iii) d = 6.4 cm and d = 11.6 cm.
(iv) r = 5 cm and r = 7.5 cm.
(vi) r = 7.1 cm and d = 12 cm.
Solution:
(i) r = 3 cm and r = 5 cm.

Width of the circular ring = OB – OA = 5 – 3 = 2 cm
Step 1: Drawn a rough diagram and market the given measurements
Step2: Taken any point O and marked it as center.
Step 3: With O as center drawn a circle of radius OA = 3 cm.
Step 4: With O as center drawn a circle of radius OB = 5 cm. Thus concentric circles C₁ and C₂ are drawn.
Width of the circular ring = OB – OA = 5 – 3 = 2 cm

(ii) r = 3.5 cm and r = 6.5 cm

Step 1: Drawn a rough diagram and market the given measurements
Step 2: Taken any point O and marked it as the center.
Step 3: With O as center drawn a circle of radius OA = 3.5 cm.
Step 4: With O as center drawn a circle of radius OB = 6.5 cm. Thus the concentric circles C₁ and C₂ are drawn.
Width of the circular ring = OB – OA = 6.5 – 3.5 = 3 cm

(iii) d = 6.4 cm and d = 11.6 cm
r = \(\frac { d }{ 2 } \)
r = \(\frac { 6.4 }{ 2 } \) = 3.2 cm; r = \(\frac { 11.6 }{ 2 } \) = 5.8 cm

Step 1: Drawn a rough diagram and market the given measurements
Step 2: Taken any point O and marked it as the center.
Step 3: With O as center drawn a circle of radius OA =3.2 cm.
Step 4: With O as center and drawn a circle of radius OB = 5.8 cm. Thus the concentric circles C₁ and C₂ are drawn.

Width of the circular ring = OB – OA = 5.8 – 3.2 = 2.6 cm

(iv) r = 5 cm and r = 7.5 cm

Step 1 : Drawn a rough diagram and market the given measurements
Step 2 : Taken any point O and marked it as the center.
Step 3 : With O as center drawn a circle of radius OA = 5 cm.
Step 4 : With O as center drawn a circle of radius OB = 7.5 cm. Thus the concentric circles C₁ and C₂ are drawn.
Width of the circular ring = OB – OA = 7.5 – 5 = 2.5 cm

(v) d = 6.2 cm and r = 6.2 cm.
r = \(\frac { d }{ 2 } \)
∴ r = \(\frac { 6.2 }{ 2 } \) = 3.1 cm and r = 6.2 cm

Step 1: Drawn a rough diagram and market the given measurements
Step 2: Taken any point O and marked it as the center.
Step 3: With O as center, drawn a circle of radius OA = 3.1 cm.
Step 4: With O as center, drawn a circle of radius OB = 6.2 cm.
Thus the concentric circles C₁ and C₂ are drawn.
Width of the circular ring = OB – OA = 6.2 – 3.1 = 3.1 cm

(vi) r = 7.1 cm and d = 12 cm.
r = \(\frac { d }{ 2 } \)
∴ r = \(\frac { 12 }{ 2 } \) = 6 cm and r = 7.1 cm

Step 1: Drawn a rough diagram and marked the given measurements
Step 2 : Taken any point O and marked it as center.
Step 3 : With O as center, drawn a circle of radius OA = 6 cm.
Step 4 : with O as center, drawn a circle of radius OB = 7.1 cm. Thus concentric circles
C₁ and C₂ are drawn.
Width of the circular ring = OB – OA = 7.1 – 6 = 1.1 cm

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 4 Geometry Additional Questions

Additional Questions and Answers

Exercise 4.1

Question 1.
Give some examples of shapes with no line of symmetry.
Solution:
1. A scalene Triangle
2. The letter F

Question 2.
What is the order of rotational symmetry of the letter N?
Solution:
2

Exercise 4.2

Question 1.
Construct a circle of radius 5.2 cm with center ‘O’.

Solution:
Step 1: Market a point‘O’on the paper.
Step 2: Extended the compass distance equal to radius 5.2 cm.
Step 3: At center ‘O’, held the compass firmly and placed the pointed end of the compass.
Step 4: Slowly rotated the compass around to get the circle.

Question 2.
Draw concentric circles with radii 5.2cm and 6.6cm. Find the width of the circular ring.
Solution:

Step 1: Drawn a rough diagram and marked the given measurements Taken any
Step 2: Take any point O and marked it as the center.
Step 3: With O as center, drawn a circle of radius OA = 5.2 cm.
Step 4:With O as center, drawn a circle of radius OB = 6.6 cm. Thus the concentric circles C₁ and C₂ are drawn.
Width of the circular ring = OB – OA = 6.6 – 5.2 = 1.4 cm

Students can Download Maths Chapter 4 Geometry Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 4 Geometry Intext Questions

Exercise 4.1
Try These (Text book Page No. 72)

Question 1.
Can you draw a shape which has no line of symmetry?
Solution:
Yes

Question 2.
Draw all possible line of symmetry for the following shapes.
Solution:

Think (Text book Page No. 73)

Question 1.
What can you say about the number of lines of symmetry of a circle?
Solution:
A circle has infinite number of lines of symmetry.

Try These (Text book Page No. 73)

Question 1.
Reflect the words CHEEK, BIKE, BOX with horizontal line.
Solution:

Question 2.
Reflect the following words with vertical line.

Solution:

Think (Textbook Page no.73)

Question 1.
Will the figure be symmetric about both the diagonals?
Solution:
Yes, it is symmetric about both the diagonals.

Try These (Text book Page No. 74)

Question 1.
Find the order of rotational symmetry of the following figures.

Solution:
Order of symmetry : 6
Order of symmetry : 3

Question 2.
Find the order of rotational symmetry for an equilateral triangle.
Solution:

For an equilateral triangle order of rotational symmetry is 3.

Think (Text book Page No. 74)

Question 1.
Can a parallelogram have a rotational symmetry?
Solution:

Yes, order of rotational symmetry is 2.

Try These (Text book Page No. 75)

Question 1.
Using translational symmetry make new pattern with the given figure.

Solution:

Try These (Text book Page No. 77)

Question 1.
Translate this figure to 4 → 3↑

Solution:

Question 2.
Translate this figure to 2 ↓ 1 ←

Solution:

Question 3.
How is the pre-image A translated to image A’ in each of the following figures?
(i)

(ii)

Solution:
(i) 8 →

(ii) 5 → 3 ↑

Think (Text book Page No. 78)

Question 1.
The pre-image and the image after a translation coincide. What can you say about the translation
Solution:
There is no right, left, up or down movement took place.

Try These (Text book Page No. 80)

Question 1.
Draw the line reflection in the following pictures.

Solution:

Question 2.
Reflect the shape with given line of reflection.

Solution:

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 5 Statistics Intext Questions

Exercise 5.1
Try These (Text book Page No. 96)

Question 1.
Collect the height of students of your class. Organise the data in ascending order.
Solution:
Height of 15 students in our class.
130cm, 150 cm, 155 cm, 142 cm, 138 cm, 145 cm, 148 cm, 147 cm, 148cm, 143 cm, 141cm, 152 cm, 147 cm, 139 cm, 155 cm.

Ascending order:
130cm, 138cm, 139cm, 141cm, 142cm, 143cm, 145cm, 147cm, 147cm, 148cm, 148 cm, 150cm, 152 cm, 155cm, 155cm.

Try These (Text book Page No. 97)

Find the Arithmetic Mean or average of the following data.

Question 1.
The study time spent by Kathir in a week is 3 hrs, 4 hrs, 5 hrs, 3 hrs, 4 hrs, 3:45 hrs; 4:15 hrs.
Solution:


mean = 3 : 52 hrs

Question 2.
The marks scored by Muhil in five subjects are 75, 91, 48, 63, 51.
Solution:

Arithmetic Mean = 65.6

Question 3.
Money spent on vegetables for five days is ₹ 120, ₹ 80, ₹ 75, ₹ 95 and ₹ 86.
Solution:

Arithmetic Mean = 91.2

Think (Text book Page No. 99)

Check the properties of arithmetic mean for the example given below:

Question 1.
If the mean is increased by 2, then what happens to the individual observations.
Solution:
Given number are 3, 6, 9, 12, 15

If mean is increased by 2 then,

Sum of observations = 5 × 11 = 55
Difference in sum = 55 – 45 = 10
∴ Each number is increased by 2 if the mean is increased by 2.

Question 2.
If first two items are increased by 3 and last two items are reduced by 3, then what will be the new mean?
Solution:
If the first two items is increased by 3, then the numbers will be 3 + 3, 6 + 3 ⇒ 6, 9.
If last two numbers are decreased by 3, then the numbers will be 12 – 3, 15 – 3 ⇒ 9,12.

= \(\frac { 45 }{ 5 } \) = 9
There is no change in the mean.

Exercise 5.2
Try These (Text book Page No. 101)

Question 1.
Find the mode of the following data. 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2
Solution:
Arranging the numbers in ascending order we get 0, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6
Since 2 and 3 occurs the maximum of 3 times. So mode of this data is 2 and 3.

Question 2.
Find the mode of the following data set. 3, 12, 15, 3, 4, 12, 11, 3, 12, 9, 19.
Solution:
Arranging the given data in ascending order : 3, 3, 3, 4, 9,11, 12, 12,12, 15, 19.
The data 3 and 12 occurs the maximum of 3 times.
So mode of this data is 3 and 12.

Question 3.
Find the mode of even numbers within 20.
Solution:
Even numbers within 20 are 2, 4, 6, 8, 10, 12, 14, 16, 18.
There is no mode for this data.

Think (Text book Page No. 102)

Question 1.
A toy factory making variety of toys for kids, wants to know the most popular toy liked by all the kids. Which average will be the most appropriate for it?
Solution:
Mode.

Question 2.
Is there a mode exists between the odd numbers from 20 to 40? Discuss.
Solution:
Odd number between 20 to 40 are 21, 23, 25, 27, 29, 31, 33, 35, 37, 39.
As all numbers occurs only once there is no mode for this data.

Think (Text book Page No. 103)

Question 3.
Which average will be most aprropriate for the companies producing the following goods? why?
(i) Diaries and notebooks
(ii) School bags
(iii) Jeans and T-shirts.
Solution:
for all the above data mode will be more appropriate.

Exercise 5.3
Try These (Text book Page No. 106)

Question 1.
Find the median of 3, 8, 7, 8,4, 5, 6.
Solution:
Arranging in ascending order: 3, 4, 5, 6, 7, 8, 8.
Here n = 7, which is odd.

Hence the median is 6.

Question 2.
Find the median: 11, 14, 10, 9, 14, 11, 12, 6, 7, 7.
Solution:
Arranging in ascending order: 6, 7, 7, 9, 10, 11, 11, 12, 14, 14
Here n = 10, which is even.

Think (Text book Page No. 108)

Complete the table given below and observe it to answer the following questions.

Solution:

Question 1.
Which are all the series having common mean and median?
Solution:
Solution:
A, B and C.

Question 2.
Why median is same for all the 4 series?
Solution:
Since the middle value is 100.

Question 3.
How mean is unchanged in the series A, B and C ?
Solution:
The difference between the given numbers are equal.

Question 4.
What change is to be made in the data, so that mean and median of ‘D’ series is equal to other series?
Solution:
If 99 becomes 0 or 200 becomes 101 then mean becomes 100.

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 5 Statistics Ex 5.4

Miscellaneous Practice problems

Question 1.
Arithmetic mean of 15 observations was calculated as 85. In doing so an observation was wrongly taken as 73 for 28. What would be correct mean?
Solution:

85 × 15 = sum of 15 observations
1275 = sum of 15 observations
Wrong observation = 73
Correct observation = 28

Correct mean = 82

Question 2.
The median of 25,16,15,10, 8, 30.
Solution:
Arranging is ascending order : 8, 10, 15, 16, 25, 30
Here n = 6, even

∴ Median = 15.5

Question 3.
Find the mode of 2, 5, 5, 1, 3, 2, 2, 1, 3, 5, 3.
Solution:
Arranging the data in ascending order: 1, 1, 2, 2, 2, 3, 3, 3, 5, 5, 5
Here 2, 3 and 5 occurs 3 times each.
Which is the maximum number of times.
∴ Mode is 2, 3 and 5.

Question 4.
The marks scored by the students in social test out of 20 marks are as follows.
12, 10, 8, 18, 14, 16. Find the mean and the median?
Solution:
Arranging the given data in ascending order: 8, 10, 12, 14, 16, 18.

There are n = 6 observations, which is even

Question 5.
The number of goals scored by a football team is given below.
Find the mode and median for the data of 2 ,3, 2, 4, 6, 1, 3, 2, 4, 1, 6.
Solution:
Arranging the given data in ascending order: 1, 1, 2, 2, 2, 3, 3, 4, 4, 6, 6
Clearly 2 occurs at the maximum of 3 times and so mode = 2
Here number of data of data n = 11, odd.
∴ Median = (\(\frac { n+1 }{ 2 } \))^th term
= (\(\frac { 11+1 }{ 2 } \))^th
term = (\(\frac { 12 }{ 2 } \))^th term
= 6^th term
Median = 3

Question 6.
Find the mean and mode of 6, 11, 13, 12, 4, 2.
Answer:
Arranging is ascending order : 2, 4, 6, 11, 12, 13

Mean = \(\frac { 48 }{ 6 } \) = 8
All observation occurs only once and so there is no mode for this date.

Challenge Problems

Question 1.
The average marks of six students is 8. One more student mark is added and the mean is still 8. Find the student mark that has been added.
Solution:

Sum of observation = 6 × 8 = 48
If one more mark is added then number of observations = 6 + 1 = 7
Let the number be x
Still average = 8
∴ 8 = \(\frac { 48+x }{ 7 } \)
48 + x = 7 × 8
48 + x = 56
48 + x = 56 – 48
x = 8
∴ The number that is added = 8

Question 2.
Calculate the mean, mode and median for the following data: 22, 15, 10, 10, 24, 21.
Solution:
Arranging in ascending order: 10, 10, 15, 21, 22, 24

Here n = 6, even

∴ Median = 18
Clearly the data 10 occurs maximum number of times and so 10 is the mode.
∴ Mode = 10

Question 3.
Find the median of the given data: 14, -3, 0, -2, -8, 13, -1, 7.
Solution:
Arranging the data is ascending order: -8, -3, -2, -1, 0, 7, 13, 14
Here number of data n = 8, even
∴ Median

∴ Median = – 0.5

Question 4.
Find the mean of first 10 prime numbers and first 10 composite numbers.
Solution:
First 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Mean = 12.9
Mean of first 10 prime numbers = 12.9
First 10 composite numbers are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18

Mean of first 10 composite numbers = 11.2

Students can Download Maths Chapter 5 Statistics Ex 5.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 5 Statistics Ex 5.3

Question 1.
Fill in the blanks.

(i) The median of the data 12, 14, 23, 25, 34, 11, 42, 45, 32, 22, 44 is ________
(ii) The median of first ten even natural numbers is ________
Answers:
(i) 25
(ii) 11

Question 2.
Find the median of the given data: 35, 25, 34, 36, 45, 18, 28.
Solution:
Arranging the given data in ascending order 18, 25, 28, 34, 35, 36, 45.
Here the number of observations n = 7, which is odd.

Hence Median = 34

Question 3.
The weekly sale of motor bikes in a showroom for the past 14 weeks given below.
10, 6, 8, 3, 5, 6, 4, 7, 12, 13, 16, 10, 4, 7.
Find the median of the data.
Solution:
Arranging the given data in ascending order 3, 4, 4, 5, 6, 6, 7, 7, 8, 10, 10, 12, 13, 16.
Here number of data n = 14, which is even


∴ Median = 7

Question 4.
Find the median of the 10 observations 36, 33, 45, 28, 39, 45, 54, 23, 56, 25.
If another observation 35 is added to the above data, what would be the new median?
Solution:
Arranging the given 10 observations in ascending order 23, 25, 28, 33, 36, 39, 45, 45, 54, 56.
Here number of data n = 10, which is even

∴ Median = 37.5
If 35 is added to the above data then it will be the 5th term then number of data n = 11, odd
∴ Median = (\(\frac { n+1 }{ 2 } \))^th term = (\(\frac { 11+1 }{ 2 } \))^th term
= (\(\frac { 12 }{ 2 } \))^th term = 6th term
New median = 36

Objective Type Questions

Question 1.
If the median of a, 2a, 4a, 6a, 9a is 8, then find the value of a is
(i) 8
(ii) 6
(iii) 2
(iv) 10
Answer:
(iii) 2
Hint:

Median = 4a = 8
a = 2

Question 2.
The median of the data 24, 29, 34, 38, 35 and 30, is ________
(i) 29
(ii) 30
(iii) 34
(iv) 32
Answer:
(iv) 32
Hint:

Median = \(\frac { 30+34 }{ 2 } \)
= \(\frac { 64 }{ 2 } \)
= 32

Question 3.
The median first 6 odd natural numbers is
(i) 6
(ii) 7
(iii) 8
(iv) 14
Answer:
(i) 6
Hint:

Median = \(\frac { 5+7 }{ 2 } \)
= \(\frac { 12 }{ 2 } \)
= 6

Students can Download Maths Chapter 5 Statistics Ex 5.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 5 Statistics Ex 5.2

Question 1.
Find the mode of the following data. 2, 4, 5, 2, 6, 7, 2, 7, 5,4, 8, 6, 10, 3, 2, 4, 2.
Solution:
Arranging the given data in ascending order 2, 2, 2, 2, 2, 3, 4, 4, 4, 5, 5, 6, 6, 7, 7, 8, 10.
Here the number 2 occurs 5 times which is the maximum
∴ Mode of this data is 2.

Question 2.
The number of points scored by a Kabaddi team in 20 matches are
36, 35, 27, 28, 29, 31, 32, 31, 35, 38, 38, 31, 28, 31, 34, 33, 34, 31, 30, 29.
Find the mode of the points scored by the team.
Solution:
Arranging the given data in ascending order
27, 28, 28, 29, 29, 30, 31, 31, 31, 31, 31, 32, 33, 34, 34, 35, 35, 36, 38, 38.
Here the number 31 occurs 5 times which is the maximum.
∴ Mode of this data is 31.

Question 3.
The ages (in years) of 11 cricket players are given below.
25, 36, 39,38 40, 36, 25, 25, 38, o 26,36. Find the mode of the ages.
Solution:
Arranging the ages is ascending order: 25, 25, 25, 26, 36, 36, 36, 36, 38, 38, 39, 40.
The ages (in years) of 11 cricket players are given below. 25, 36, 39,38 40, 36, 25, 25, 38, o 26,36. Find the mode of the ages.
Arranging the ages is ascending order: 25, 25, 25, 26, 36, 36, 36, 36, 38, 38, 39, 40.
25 and 36 occurs maximum number of times.
∴ Mode is 25 and 36.

Question 4.
Find the mode of the following data.
12, 14, 12, 16, 15, 13, 14, 18, 19, 12, 14, 15, 16, 15, 16, 16, 15, 17, 13, 16, 16, 15, 15, 13, 15, 17, 15, 14, 15, 13, 15, 14.
Solution:
Tabulating the given data

The highest frequency is 10 which corresponds to the value 15.
Hence mode of this data is 15.

Objective Type Questions

Question 1.
The colors used by the six students for drawing is blue, orange, yellow, white, green and blue then the mode is ________
(i) blue
(ii) green
(iii) white
(iv) yellow
Answer:
(i) blue

Question 2.
Find the mode of data 3, 6, 9, 12, 15.
(i) 1
(ii) 2
(iii) 3
(iv) No mode
Answer:
(iv) No mode

Question 3.
Find the modes of the data 2, 1, 1, 3, 4, 5, 2.
(1) 1 and 5
(2) 2 and 3
(3) 2 and 1
(4) 1 and 4
Answer:
(3) 2 and 1

Students can Download Maths Chapter 6 Information Processing Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 6 Information Processing Additional Questions

Exercise 6.1

Question 1.
Draw a flowchart to calculate simple interest for your bank deposit.
Solution:
Flow Chart:

Algorithm:
(i) Read principal
(ii) Read years
(iii) Read rate of interest per year
(iv) Calculate the interest with formula Interest = Principal × Years × Rate/100
(v) Print Interest

Question 2.
Draw a flowchart to convert the given Fahrenheit temperature to Celsius. Formula is C = \(\frac { 5 }{ 9 } \) × (F – 32)
Solution:
Algorithm:
(i) Read temperature in Fahrenheit
(ii) Calculate temperature in Celsius using formula C = \(\frac { 5 }{ 9 } \) × (F – 32)
(iii) Print C