Class 7

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 4 Geometry Ex 4.1

Question 1.

Solution:

Question 2.
How is the pre-image translated to the image?

Solution:
(i) 3 →, 4↑
(ii) 3 ←, 3↑
(iii) 4 ←, 4↓
(iv) 2 ←, 2↓

Question 3.
Find the image of the given triangle with given translation.

Solution:

Question 4.
Reflect the shape with given line of reflection

Solution:

Question 5.
Reflect the shape in each of the following pictures with given line of reflection.

Solution:

Question 6.
Rotate the preimages in each case as directed about the red point.


Solution:

Identify the transformation:

Question 7.

Solution:
Reflection

Question 8.

Solution:
Rotation

Question 9.

Solution:
Translation

Question 10.
A pool of fish translates from point F to point D.
a. Describe the translation of the pool of fish.
b. Can the fishing boat make the same translation? Explain.
c. Describe a translation the fishing boat could make to get to point D.

Solution:
(a) Translation of pool of fish is 7 →, 2↓
(b) No, the fishing boat will be landed on the island if translated.
(c) To get point D, the translation will be 5 →, 3↓

Question 11.
Name the transformation that will map footprint A onto the indicated footprint.

(i) Footprint B
(ii) Footprint
(iii) Footprint D
(iv) Footprint E
Solution:
(i) It is translation
(ii) Reflection about horizontal line.
(iii) Reflection about vertical line.
(iv) Rotation about the heel.

Question 12.
In given diagram, the blue figure is an image of the pink figure.

(i) Choose an angle or a vertex from the preimage and name its image.
(ii) List all pairs of corresponding sides.
Solution:
(i) Image of ∠L is ∠L’, Image of ∠M is ∠M’,
Image of ∠N is ∠N’, Image of ∠O is ∠O’
Image of vertex L is L’, Image of vertex M is ∠M’
Image of vertex N is ∠N’, Image of vertex O is O’

(ii) Corresponding sides are LM and L’M’, MN and M’N’, NO and N’O’ and OL and O’L’

Question 13.
In the diagram at the right, the green figure is a translation image of the pink figure. Write a coordinate rule that describes the translation.

Solution:
The rule bind here in 3→, 1↓

Objective Type Questions

Question 1.
A _______ is a turn about a point.
(i) Translation
(ii) Rotation
(iii) Reflection
(iv) Glide Reflection
Answer:
(ii) Rotation

Question 2.
A _______ is a flip over a line.
(i) Translation
(ii) Rotation
(iii) Reflection
(iv) Glide Reflection
Answer:
(iii) Reflection

Question 3.
A _______ is a slide; move without turning or flipping the shape.
(i) Translation
(ii) Rotation
(iii) Reflection
(iv) Glide Reflection
Answer:
(i) Translation

Question 4.
The transformation used in the picture is

(i) Translation
(ii) Rotation
(iii) Reflection
(iv) Glide Reflection
Answer:
(ii) Rotation

Question 5.
The transformation used in the picture is

(i) Translation
(ii) Rotation
(iii) Reflection
(iv) Glide Reflection
Answer:
(i) Translation

Question 6.
You must rotate the puzzle piece 270° clockwise about point P to fit it into a puzzle. Which piece fits in the puzzle as shown?

Solution:

Students can Download Maths Chapter 6 Information Processing Ex 6.1 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 Ex 6.1

Question 1.
Match the following:

Solution:

Question 2.
The steps of withdrawing cash from your saving bank account using ATM card are explained in the figures given below. Construct an appropriated flow chart.

Solution:
Algorithm:
(i) Swipe your bank Debit card / credit card.

(ii) Select banking
(iii) Select language
(iv) Select transaction
(v) Enter your PIN
(vi) Enter the required amount
(vii) Collect your money

Question 3.
Using given step by step process to recharge mobile phone, draw a sequence flowchart.

Step by Step process:

• Login the mobile recharge web browser
• Select prepaid or postpaid
• Enter mobile number
• Select operator and browse plans to choose your recharge plan
• Enter amount to recharge
• Proceed to recharge

Solution:

Question 4.
Complete the direction of the flowchart using arrows for the flow chart explaining the traffic rule given below.

Solution:

Question 5.
Complete the given flowchart, input names of things and check whether it is living or non – living.

Solution:
Input name Horse
Horse = a Yes
Print Living thing

Question 6.
Complete the given flowchart.
(i) Fill in the flow chart to print the average mark by giving your I term or II term marks as inputs.

(ii) Construct the flow chart to print teachers comment as “very good” if your average mark is above 75 out of 100 or else, as “still try more” can be inserted in the flow chart with earlier one.

Solution:

Question 7.
A merchant calculates the cost price (CP) and the selling price (SP) of the product bought by him. Construct the flow chart to print ‘PROFIT’ if the selling price (SP) is more than the cost price (CP) or else ‘LOSS’.
Solution:
First we have to input the cost price and selling price. Then we check whether cost price less than selling price. If S.P > CP, then print “PROFIT” otherwise print ‘Loss’.

Algorithm:

1. Enter cost price
2. Enter selling price
3. Checking whether CP < SP
4. If Yes print ‘PROFIT’
5. If No print ‘Loss’

Students can Download Maths Chapter 4 Geometry 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 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

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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:

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 3 Algebra Ex 3.1

Question 1.
Fill in the blanks.
1. (p – q)² = _______
2. The product of (x + 5) and (x – 5) is _______
3. The factors of x² – 4x + 4 are _______
4. Express 24ab²c² as product of its factors is _______
Answers:
1. p² – 2pq + q²
2. x² – 25
3. (x – 2) and (x – 2)
4. 2 × 2 × 2 × 3 × a × b × b × c × c

Question 2.
Say whether the following statements are True or False.
(i) (7x + 3) (7x – 4) = 49 x² – 7x – 12
(ii) (a – 1)² = a² – 1.
(iii) (x² + y²)(y² + x²) = (x² + y²)²
(iv) 2p is the factor of 8pq.
Answers:
(i) True
(ii) False
(iii) True
(iv) True

Question 3.
Express the following as the product of its factors.
(i) 24ab²c²
(ii) 36 x³y²z
(iii) 56 mn²p²
Solution:
(i) 24ab²c² = 2 × 2 × 2 × 3 × a × b × b × c × c
(ii) 36 x³y²z = 2 × 2 × 3 × 3 × x × x × x × y × y × z
(iii) 56 mn²p² = 2 × 2 × 2 × 7 × m × n × n × p × p

Question 4.
Using the identity (x + a)(x + b) – x² + x(a + b) + ab, find the following product.
(i) (x + 3) (x + 7)
(ii) (6a + 9) (6a – 5)
(iii) (4x + 3y) (4x + 5y)
(iv) (8 + pq) (pq + 7)
Solution:
(i) (x + 3) (x + 7)
Let a = 3; b = 7, then
(x + 3) (x + 7) is of the form x² + x (a + b) + ab
(x + 3) (x + 7) = x² + x (3 + 7) + (3 × 7) = x² + 10x + 21

(ii) (6a + 9) (6a – 5)
Substituting x = 6a ; a = 9 and b = -5
In (x + a) (x + b) = x² + x (a + b) + ab, we get
(6a + 9)(6a – 5) = (6a)² + 6a (9 + (-5)) + (9 × (-5))
6² a² + 6a (4) + (-45) = 36a² + 24a – 45
(6a + 9) (6a – 5) = 36a² + 24a – 45

(iii) (4x + 3y) (4x + 5y)
Substituting x = 4x ; a = 3y and b = 5y in
(x + a) (x + b) = x² + x (a + b) + ab, we get
(4x + 3y) (4x – 5y) = (4x)² + 4x (3y + 5y) + (3y) (5y)
= 4² x² + 4x (8y) + 15y² = 16x² + 32xy + 15y²
(4x + 3y) (4x + 5y) = 16x² + 32xy + 15y²

(iv) (8 + pq) (pq + 7)
Substituting x = pq ; a = 8 and b = 7 in
(x + a) (x + b) = x² + x (a + b) + ab, we get
(pq + 8) (pq + 7) = (pq)² + pq (8 + 7) + (8) (7)
= p² q² + pq (15) + 56
(8 + pq) (pq + 7) = p² q² + 15pq + 56

Question 5.
Expand the following squares, using suitable identities.
(i) (2x + 5)²
(ii) (b – 7)²
(iii) (mn + 3p)²
(iv) (xyz – 1)²
Solution:
(i) (2x + 5)²
Comparing (2x + 5)² with (a + b)² we have a = 2x and b = 5
a = 2x and b = 5,
(a + b)² = a² + 2ab + b²
(2x + 5)² = (2x)² + 2(2x) (5) + 5² = 2² x² + 20x + 25
= 2² x² + 20x + 25
(2x + 5)² = 4x² + 20x + 25

(ii) (b – 7)²
Comparing (b – 7)² with (a – b)² we have a = b and b = 7
(a – b)² = a² – 2ab + b²
(b – 7)² = b² – 2(b) (7) + 7²
(b – 7)² = b² – 14b + 49

(iii) (mn + 3p)²
Comparing (mn + 3p)² with (a + b)² we have
(a + b)² = a² + 2ab + b²
(mn + 3p)² = (mn)² + 2(mn) (3p) + (3p)²
(mn + 3p)² = m² n² + 6mnp + 9p²

(iv) (xyz – 1)²
Comparing (xyz – 1)² with (a – b)² we have = a + xyz and b = 1
a = xyz and b = 1
(a – b)² = a² – 2ab + b²
(xyz – 1)² = (xyz)² – 2 (xyz) (1) + 1²
(xyz -1)² = x² y² z² – 2 xyz + 1

Question 6.
Using the identity (a + b)(a – b) = a² – b², find the following product.
(i) (p + 2) (p – 2)
(ii) (1 + 3b) (3b – 1)
(iii) (4 – mn) (mn + 4)
(iv) (6x + 7y) (6x – 7y)
Solution:
(i) (p + 2) (p – 2)
Substituting a = p ; b = 2 in the identity (a + b) (a – b) = a² – b², we get
(p + 2) (p – 2) = p² – 2²

(ii) (1 + 3b)(3b – 1)
(1 + 3b) (3b -1) can be written as (3b + 1) (3b – 1)
Substituting a = 36 and b = 1 in the identity
(a + b) (a – b) = a² – b², we get
(3b + 1)(3b – 1) = (3b)² – 1² = 3² × b² – 1²
(3b + 1) (3b – 1) = 9b² – 1²

(iii) (4 – mn) (mn + 4)
(4 – mn) (mn + 4) can be written as (4 – mn) (4 + mn) = (4 + mn) (4 – mn)
Substituting a = 4 and b = mn is
(a + b) (a – b) = a² – b², we get
(4 + mn) (4 – mn) = 4² – (mn)² = 16 – m² n²

(iv) (6x + 7y) (6x – 7y)
Substituting a = 6x and b = 7y in
(a + b) (a – b) = a² – b², We get
(6x + 7y) (6x – 7y) = (6x)² – (7y)² = 6²x² – 7²y²
(6x + 7y) (6x – 7y) = (6x)² – (7y)² = 6²x² – 7²y²
(6x + 7y) (6x – 7y) = 36x² – 49y²

Question 7.
Evaluate the following, using suitable identity.
(i) 51²
(ii) 103²
(iii) 998²
(iv) 47²
(v) 297 × 303
(vi) 990 × 1010
(vii) 51 × 52
Solution:
51²
= (50 + 1)²
Taking a = 50 and b = 1 we get
(a + b)² = a² + 2ab + b²
(50 + 1)² = 50² + 2 (50) (1) + 1² = 2500 + 100 + 1
51² = 2601

(ii) 103²
103² = (100 + 3)²
Taking a = 100 and b = 3
(a + b)² = a² + 2ab + b² becomes
(100 + 3)² = 100² + 2 (100) (3) + 3² = 10000 + 600 + 9
103² = 10609

(iii) 998²
998² = (1000 – 2)²
Taking a = 1000 and b = 2
(a – b)² = a² + 2ab + b² becomes
(1000 – 2)² = 1000² – 2 (1000) (2) + 2²
= 1000000 – 4000 + 4
998² = 10,04,004

(iv) 47²
47² = (50 – 3)²
Taking a = 50 and b = 3
(a – b)² = a² – 2ab + b² becomes
(50 – 3)² = 50² – 2 (50) (3) + 3²
= 2500 – 300 + 9 = 2200 + 9
47² = 2209

(v) 297 × 303
297 × 303 = (300 – 3) (300 + 3)
Taking a = 300 and b = 3, then
(a + b) (a – b) = a² – b² becomes
(300 + 3) (300 – 3) = 300² – 3²
303 × 297 = 90000 – 9
297 × 303 = 89,991

(vi) 990 × 1010
990 × 1010 = (1000 – 10) (1000 + 10)
Taking a = 1000 and b = 10, then
(a – b) (a + b) = a² – b² becomes
(1000 – 10) (1000 + 10) = 1000² – 10²
990 × 1010 = 1000000 – 100
990 × 1010 = 999900

(vii) 51 × 52
= (50 + 1) (50 + 1)
Taking x = 50, a = 1 and b = 2
then (x + a) (x + b) = x² + (a + b) x + ab becomes
(50 + 1) (50 + 2) = 50² + (1 + 2) 50 + (1 × 2)
2500 + (3) 50 + 2 = 2500 + 150 + 2
51 × 52 = 2652

Question 8.
Simplify: (a + b)² – 4ab
Solution:
(a + b)² – 4ab = a² + b² + 2ab – 4ab = a² + b² – 2ab = (a – b)²

Question 9.
Show that (m – n)² + (m + n)² = 2(m² + n²)
Solution:
Taking the LHS = (m – n)² + (m + n)²

Question 10.
If a + b = 10 , and ab = 18, find the value of a² + b².
Solution:
We have (a + b)² = a² + 2ab + b²
(a + b)² = a² + b² + 2ab
given a + b = 0 and ab = 18
10² = = a² + b² + 2(18)
100 = = a² + b² + 36
100 – 36 = a² + b²
a² + b² = 64

Question 11.
Factorise the following algebraic expressions by using the identity a² – b² = (a + b)(a – b).
(i) z² – 16
(ii) 9 – 4y²
(iii) 25a² – 49b²
(iv) x⁴ – y⁴
Solution:
(i) z² – 16
z² – 16 = z² – 4²
We have a² – b² = (a + b) (a – b)
let a = z and b = 4,
z² – 4² = (z + 4) (z – 4)

(ii) 9 – 4y²
9 – 4y² = 3² – 2² y² = 3² – (2y)²
let a = 3 and b = 2y, then
a² – b² = (a + b) (a – b)
∴ 3² – (2y)² = (3 + 2y) (3 – 2y)
9 – 4y² = (3 + 2y) (3 – 2y)

(iii) 25a² – 49b²
25a2 – 49b2 = 52 – a2 – 72 = (5a)2 – (7b)2
let A = 5a and B = 7b
A²  B²
(5a)² – (7b)² = (5a + 7b) (5a – 7b)

(iv) x⁴ – y⁴
Let x⁴ – y⁴ = (x²)² – (y²)²
We have a² – b² = (a + b) (a – b)
(x²)² – (y²)² = (x² + y²) (x² – y²)
x⁴ – y⁴ = (x² + y²) (x² – y²)
Again we have x² – y² = (x + y) (x – y)
∴ x⁴ – y⁴ = (x² + y²) (x + y) (x – y)

Question 12.
Factorise the following using suitable identity.
(i) x² – 8x + 16
(ii) y² + 20y + 100
(iii) 36m² + 60m + 25
(iv) 64x² – 112xy + 49y²
(v) a² + 6ab + 9b² – c²
Solution:
(i) x² – 8x + 16
x² – 8x + 16 = x² – (2 × 4 × x) + 4²
This expression is in the form of identity
a² – 2ab + b² = (a – b)²
x² – 2 × 4 × x + 4² = (x – 4)²
∴ x² – 8x + 16 = (x – 4) (x – 4)

(ii) y² + 20y + 100
y² + 20y + 100 = y² + (2 × (10)) y + (10 × 10)
= y² + (2 × 10 × y) + 10²
This is of the form of identity
a² + 2 ab + b² = (a + b)²
y² + (2 × 10 × y) + 10² = (y + 10)²
y² + 20y + 100 = (y + 10)²
y² + 20y + 100 = (y + 10) (y + 10)

(iii) 36m² + 60m + 25
36m² + 60m + 25 = 62 m² + 2 × 6m × 5 + 5²
This expression is of the form of identity
a² + 2ab + b² = {a + b)²
(6m)² + (2 × 6m × 5) + 5²
= (6m + 5)²
36m² + 60m + 25 = (6m + 5) (6m + 5)

(iv) 64x² – 112xy + 49y²
64x² – 112xy + 49y² = 82 x² – (2 × 8x × 7y) + 7²y²
This expression is of the form of identity
a² – 2ab + b² = (a- b)²
(8x)² – (2 × 8x × 7y) + (7y)² = (8x – 7y)²
64x² – 112xy + 49y² = (8x – 7y) (8x – 7y)

(v) a² + 6ab + 9b² – c²
a² + 6ab + 9b² – c² = a² + 2 × a × 3b + 3² b² – c²
= a² + (2 × a × 3b) + (3b)² – c²
This expression is of the form of identity
[a² + 2ab + b²] – c² = (a + b)² – c²
a² + (2 × a × 36) + (3b)² – c² = (a + 3b)² – c²
Again this RHS is of the form of identity
a² – b² = (a + b) (a – b)
(a + 3b)² – c² = [(a + 3b) + c] [(a + 3b) – c]
a² + 6ab + 9b² – c² = (a + 3b + c) (a + 3b – c)

Objective Type Questions

Question 1.
If a + b = 5 and a² + b² = 13, then ab = ?
(i) 12
(ii) 6
(iii) 5
(iv) 13
Answer:
(ii) 6
Hint: (a + b)² = 25
13 + 2ab = 25
2ab = 12
ab = 6

Question 2.
(5 + 20)(-20 – 5) = ?
(i) -425
(ii) 375
(iii) -625
(iv) 0
Answer:
(iii) -625
Hint: (50 + 20) (-20 – 5) = -(5 + 20)² = – (25)² = – 625

Question 3.
The factors of x² – 6x + 9 are
(i) (x – 3)(x – 3)
(ii) (x – 3)(x + 3)
(iii) (x + 3)(x + 3)
(iv) (x – 6)(x + 9)
Answer:
(i) (x – 3)(x – 3)
Hint: x² – 6x + 9 = x² – 2(x) (3) + 3²
a² – 2ab + b² – (a- b)² = (x – 3)² = (x – 3) (x – 3)

Question 4.
The common factors of the algebraic expression ax²y, bxy² and cxyz is
(i) x²y
(ii) xy²
(iii) xyz
(iv) x
Ans :
(iv) xy
Hint: ax²y = a × x × x × y
bxy² = b × x × y × y
cxyz = C × x × y × z
Common factor = xy

Students can Download Maths Chapter 3 Algebra Ex 3.3 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 3 Algebra Ex 3.3

Miscellaneous Practice problems

Question 1.
Using identity, find the value of
(i) (4.9)²
(ii) (100.1)²
(iii) (1.9) × (2.1)
Solution:
(i) (4.9)²
(4.9)² = (5 – 0.1)²
Substituting a = 5 and b = 0.1 in
(a – b)² = a² – 2ab + b², we have
(5 – 0.1)² = 5² – 2(5) (0.1) + (0.1)²
(4.9)² = 25 – 1 + 0.01 = 24 + 0.01
(4.9)² = 24.01

(ii) (100.1)²
(100.1)² = (100 + 0.1)²
Substituting a = 100 and b = 0.1 in
(a + b)² = a² + 2ab + b², we have
(100 + 0.1)² = (100)² + 2(100) (0.1) + (0.1)²
(100.1)² = 10000 + 20 + 0.01
(100.1)² = 10020.01

(iii) (1.9) × (2.1)
(1.9) × (2.1) = (2 – 0.1) × (2 + 0.1)
Substituting a = 100 and b = 0.1 in
(a – b) (a + b) = a² – b² we have
(2 – 0.1) (2 + 0.1) = 2² – (0.1)²
(1.9) × (2.1) = 4 – 0.01
(9.9) (2.1) = 3.99

Question 2.
Factorise: 4x² – 9y²
Solution:
4x² – 9y² = 22 x² – 3² y² = (2x)² – (3y)²
Substituting a = 2x and b = 3y in
(a² – b²) = (a + b) (a – b), we have
(2x)² – (3y)² = (2x + 3y) (2x – 3y)
∴ Factors of 4x² – 9y² are (2x + 3y) and (2x – 3y)

Question 3.
Simplify using identities
(i) (3p + q) (3p + r)
(ii) (3p + q) (3p – q)
Solution:
(i) (3p + q) (3p + r)
Substitute x = 3p,a = q and b = r in
(x + a) (x + b) = x² + x(a + b) + ab
(3p + q)(3p + r) = (3p)² + 3p (q + r) + (q × r)
= 3² p² + 3p (q + r) + qr
(3p + q)(3p + r) = 9p² + 3p(q + r) + qr

(ii) (3p + q) (3p – q)
Substitute a = 3p and b = q in
(a + b) (a – b) = a² – b², we have
(3p + q) (3p – q) = (3p)² – q² = 32 p² – q²
(3P + q) (3p – q) = 9p² – q²

Question 4.
Show that (x + 2y)² – (x – 2y)² = 8xy.
Solution:
LHS = (x + 2y)² – (x – 2y)²
= x² + (2 × x × 2y) + (2y)² – [x² – (2 × x × 2y) + (2y)²]
= x² + 4xy + 4y² – [x² – 4xy + 2²y²]
= x² + 4xy + 4y² – x² + 4xy – 4y²
= x² – x² + 4xy + 4xy + 4y² – 4y²
= x² (1 – 1) + xy (4 + 4) + y² (4 – 4)
= 0x² + 8xy + 0y² = 8xy = RHS
∴ (x + 2y)² – (x – 2y)² = 8xy
[∵ (a + b)² = a² + 2ab + b² (a – b)² = a² – 2ab + b²]

Question 5
The pathway of a square paddy field has 5 m width and length of its side is 40 m. Find the total area of its pathway. (Note: Use suitable identity)
Solution:
Given side of the square = 40 m
Also width of the pathway = 5 m
∴ Side of the larger square = 40m + 2(5)m = 40m + 10m = 50m
Area of the path way = area of large square – area of smaller square
= 50² – 40²

Substituting a = 50 and b = 40 in
a² – b² = (a + b) (a – b) we have
50² – 40² = (50 + 40) (50 – 40)
Area of pathway = 90 × 10
Area of the pathway = 900 m²

Challenge Problems

Question 1.
If X = a² – 1 and Y = 1 – b², then find X + Y and factorize the same.
Solution:
Given X = a² – 1
Y = I – b²
X + Y = (a² – 1) + (1 – b²)
= a² – 1 + 1 – b²
We know the identity that a² – b² = (a + b) (a – b)
∴ X + Y = (a + b) (a – b)

Question 2.
Find the value of (x – y) (x + y) (x² + y²).
Solution:
We know that (a – b) (a + b) = a² – b²
Put a = x and b = y in the identity (1) then
(x – y) (x + y) = x² – y²
Now (x – y) (x + y)(x² + y²) = (x² – y²) (x² + y²)
Again put a = x² and b = y² in (1)
We have (x² – y²) (x² + y²) = (x²)² – (y²)² = x⁴ – y⁴
So (x – y) (x + y) (x² + y²) = x⁴ – y⁴

Question 3.
Simplify (5x – 3y)² – (5x + 3y)².
Solution:
We have the identities
(a + b)² = a² + 2ab + b²
(a – b)² = a² – 2ab + b²
So (5x – 3y)² – (5x + 3y)² = (5x)² – (2 × 5x × 3y) + (3y)²
= 5²x² – 30xy + 3² y² – [5²x² – 30xy + 3² y²]
= 25x² – 30xy + 9y² – [25x² + 30xy + 9y²]
= 25x² – 30xy + 9y² – 25x² – 30xy – 9y²
= x² (25 – 25) – xy (30 + 30) + y² (9 – 9)
= 0x² – 60xy + 0y² = – 60 xy
∴ (5x – 3y)² – (5x + 3y)² = -60xy

Question 4.
Simplify : (i) (a + b)² – (a – b)²
(ii) (a + b)² + (a – b)²
Solution:
Applying the identities
(a + b)² = a² + 2ab + b²
(a – b)² = a² – 2ab + b²

(i) (a + b)² – (a – b)² = a² + 2ab + b² – [a² – 2ab + b²]
= a² + 2ab + b² – a² + 2ab – b²
= a² (1 – 1) + ab (2 + 2) + b² (1 – 1)
= 0a² + 4 ab + 0b² = 4ab
(a + b)² – (a – b)² = 4ab

(ii) (a + b)² + (a – b)² = a² + 2ab + b² + (a² – 2ab + b²)
= a² + 2ab + b² + a² – 2ab + b²
= a² (1 + 1) + ab (2 – 2) + b² (1 + 1)
= 2a² + 0 ab + 2b² = 2a² + 2b² = 2 (a² + b²)
∴ (a + b)² – (a – b)² = 2 (a² + b²)

Question 5.
A square lawn has a 2 m wide path surrounding it. If the area of the path is 136 m², find the area of lawn.
Solution:
Let the side of the lawn = a m
then side Of big square = (a + 2(2)) m
= (a + 4)m

Area of the path – Area Of large square – Area of smaller square
136 = (a + 4)2 – a2
136 = a² + (2 × a × 4) + 4² – a²
136 = a² + 8a + 16 – a²
136 = 8a + 16
136 = 8 (a + 2)
Dividing by 8
17 = a + 2
Subtracting 2 on both sides
17 – 3 = a + 2 – 2
15 = a
∴ side of small square = 15 m
Area of square = (side × side) Sq. units
∴ Area of the lawn = (15 × 15)m² = 225 m²
∴ Area of the lawn = 225 m²

Question 6.
Solve the following inequalities.
(i) 4n + 7 > 3n + 10, n is an integer
(ii) 6(x + 6) > 5 (x – 3), x is a whole number.
(iii) -13 < 5x + 2 < 32, x is an integer.
Solution:
(i) 4n + 7 > 3n + 10, n is an integer.
4n + 7 – 3n > 3n + 10 – 3n
n(4 – 3) + 7 > 3n + 10 – 3n
n (4 – 3) + 7 > n (3 – 3) + 10
n + 7 > 10
Subtracting 7 on both sides
n + 7 – 7 > 10 – 7
n > 3
Since the solution is an integer and is greater than or equal to 3, the solution will be 3,
4, 5, 6, 7, …..
n = 3, 4, 5, 6,7, ….

(ii) 6 (x + 6) > 5 (x – 3), x is a whole number.
6x + 36 > 5x – 15
Subtracting 5x on both sides
6x + 36 – 5x > 5x – 15 – 5x
x (6 – 5) + 36 > x(5 – 5) – 15
x + 36 > -15
Subtracting 36 on both sides
x + 36 – 36 > -15 -36
x > -51
The solution is a whole number and which is greater than or equal to -51
∴ The solution is 0, 1, 2, 3, 4,…
x = 0,1,2, 3,4,…

(iii) -13 < 5x + 2 < 32, x is an integer.
Subtracting throughout by 2
-13 – 2 < 5x + 2 – 2 < 32 – 2
-15 < 5x < 30
Dividing throughout by 5
\(\frac { -15 }{ 5 } \) < \(\frac { 5x }{ 5 } \) < \(\frac { 30 }{ 5 } \)
– 3 < x < 6
∴ Since the solution is an integer between -3 and 6 both inclusive, we have the solution
as -3, -2, -1,0, 1,2, 3, 4, 5, 6.
i.e. x = -3, -2, 0, 1, 2, 3,4, 5 and 6.

Students can Download English Lesson 1 Journey by Train Questions and Answers, Summary, Notes Pdf, Activity, Samacheer Kalvi 7th English 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 English Solutions Term 3 Prose Chapter 1 Journey by Train

Read And Understand

A. Fill in the table
Answer:

Problems Faced By Mr. Fogg  And His Team	Solution
The train stopped in the middle of the forest.	The passengers must provide themselves a means of transport from Kholby to Allahabad.
They couldn’t hire the elephant.	Mr. Fogg purchased it for 2000 pounds.
They were in need Of an elephant driver.	A young man offered his service as a guide.

B. Answer the questions briefly

Question 1.
Who inhabited the jungles that the train passed through?
Answer:
The elephants inhabited the jungles that the train passed through.

Question 2.
What was the reaction of the inhabitants?
Answer:
Snakes and tigers fled at the noise Of the train. The elephants stood gazing with sad eyes at the train, as it passed.

Question 3.
What did Mr. Fogg mean by, ‘it was foreseen’?
Answer:
Mr. Fogg knew that some obstacle would arise on his route. So he said that the difficulty was foreseen.

Question 4.
Describe the elephant driver in your own words.
Answer:
The elephant driver was intelligent and a skilled young man. He covered the elephant’s back with saddle-cloth. He attached seats on each of its sides. He sat on the neck of the elephant and set out from the village with the three passengers.

C. Think and Answer.

Question 1.
What qualities of Mr. Phileas Fogg are highlighted in this extract? Support your answer with suitable examples.
Answer:
Mr. Phileas Fogg was a man of adventure. He was ready to take a risk. Otherwise, he would not have accepted a wager. 20,000 dollars set by his friends at the Reform Club. Mr. Phileas Fogg always maintained his calm and cool temperament. He did not get angry even when the train stopped suddenly.

Mr. Fogg was a man of farsightedness. He had foreseen some obstacles on his route sooner or later to arise. Mr. Fogg was a man of good planning. He had planned to catch the steamer that would leave Calcutta for Hongkong on 25th. Mr. Fogg was a man of generosity. He promised a generous reward for the guide.

When the elephant owner refused to hire, at the excessive sum of two thousand pounds, for the role of the animal, Mr. Fogg was not in the least flurried. These incidents show the determination of Mr. Fogg in executing his decisions.

Vocabulary

D. Fill in the blanks with correct travel words.

(schedule,reach,book,railway,pack,board)
Answer:
To make travel convenient, we must book tickets well in advance. Then we have to Pack our things and schedule our trip. We have to reach the railway station in time and board the train in order to reach our destination.

E. Match the phrasal verbs with their meanings

Answer:

1. b
2. f
3. a
4. e
5. c
6. d

F. Dictionary Task

Refer to a dictionary. Find the meaning of the following words and write them down.
Answer:

1. journey – The act of travelling from one place to another.
2. picnic – a trip or excursion to the country, seaside, etc, on which people bring food to be eaten in the open air
3. pilgrimage – a religious journey.
4. tour – a journey for pleasure in which several places are visited
5. vacation – holiday.
6. excursion – a short journey or trip.

Listening

G. Listen to the teacher reading the Weather forecast and complete the report.
Answer:
The name of the Cyclone is Gaja it may affect the places North Tamil Nadu and Puducherry Heavy rains are expected on November 14th and 15th It is 880 km away from Nagapattinam.

Speaking

H. Your family has decided to go on a tour during the vacation. You are calling a travel agency and seeking information regarding package, places of visit, cost, etc. Work in pairs and role play as a receptionist and a customer.
Answer:
Receptionist: Good morning. This is Sai Dwaraka Mai Travel Agency. How can I help you?
Customer: We would like to go on a tour of the Shirdi temple. Can you give us the information regarding the package, places of visit, and the cost for it?
Receptionist: Yes, Madam. We offer flight packages from Chennai to Shirdi at affordable prices.
Customer: Okay Sir, what are the places of visit over there?
Receptionist: We take you to places like Dwarkamai, Chavadi, Gurusthan, Sai Museum, Maruthi temple, and Nandadeep.
Customer: Okay fine. We would like to book tickets for the tour, as we would like to go any day between 23rd November and 27th November. What is the cost for four people?
Receptionist: It costs Rs. 9,9991- per person. So the total amount for four people would be Rs. 39,996/-
Customer: Thank you, Sir, I will get back to you immediately, after discussing with my family members.
Receptionist: Thank you for calling, Madam. Have a nice day.

Grammar

I. Rewrite the story in the past tense.
Answer:

The boy chased a cat. The cat climbed up the tree and purred from the branch of the tree. The cat jumped to another tree. The boy who was chasing the cat noticed a snake under the tree. He left his attempt to catch the cat and he ran home screaming for help

J. Look at the picture and complete the following

Answer:
Nila told Miruthula that that hotel was famous for masala idli Miruthula said that the idly was soft and spongy Nila said that it was delicious too. She also said that they had added Kashmiri chili. Miruthula said very much.

Writing

K. Your friend is coming to your city/town to spend a week with you. He /she wants to visit some tourist places and enjoy the special food items in the place. Prepare a two-day itinerary for the visit.
Answer:

Creative Writing

L. You are waiting to board a train at a railway station. The train is delayed by an hour. Write a paragraph about the crowded scene in the railway station based on your observation.

Answer:
I was waiting to board a train at a railway station. The train was delayed by an hour. A lot of families were waiting eagerly for the announcement of the arrival of the train. Some of the porters were carrying the luggage of the passengers. Piles of luggage were seen all over. There were fruit vendors, tea vendors, and other shops selling water, milk, and snacks. By the ticket counter, there was a queue, waiting to buy tickets. The scroll board was displaying the necessary information for the passengers. The porters were also using trolleys to carry the luggage.

Journey by Train Additional Questions

I. Choose the correct Synonyms from the options below.

Question 1.
proceeded
(a) stopped
(b) halted
(c) moved
(d) passed
Answer:
(c) moved

Question 2.
fertile
(a) vast
(b) productive
(c) barren
(d) incapable
Answer:
(b) productive

Question 3.
territory
(a) barrier
(b) border
(c) Limit
(d) region
Answer:
(d) region

Question 4.
gazing
(a) look steadily
(b) looking away
(c) scanning
(d) blinking
Answer:
(a) look steadily

Question 5.
hasty
(a) slow
(b) wise
(c) quickly
(d) lazy
Answer:
(c) quick

Question 6.
curled
(a) straightened
(b) erected
(c) evened
(d) twined
Answer:
(d) twined

Question 7.
obstacle
(a) assist
(b) aid
(c) support
(d) difficulty
Answer:
(d) difficulty

Question 8.
halt
(a) start
(b) stop
(c) go
(d) continue
Answer:
(b) stop

Question 9.
foreseen
(a) predicted
(b) neglected
(c) disregarded
(d) failed
Answer:
(a) predicted

Question 10.
rapidly
(a) slowly
(b) suddenly
(c) quickly
(d) easily
Answer:
(c) quickly

II. Choose the correct Antonyms from the options below.

Question 1.
different
(a) same
(b) unlike
(c) dissimilar
(d) contrast
Answer:
(a) same

Question 2.
punctually
(a) promptly
(b) timely
(c) regularly
(d) early
Answer:
(d) early

Question 3.
straggling
(a) few
(b) rare
(c) abundant
(d) irregular
Answer:
(c) abundant

Question 4.
waking
(a) sleeping
(b) arousing
(c) raising
(d) getting up
Answer:
(a) sleeping

Question 5.
abandoned
(a) deserted
(b) forsakened
(c) casted
(d) inhabited
Answer:
(d) inhabited

Question 6.
skilled
(a) trained
(b) qualified
(c) inexperienced
(d) practised
Answer:
(c) inexperienced

Question 7.
announced
(a) reported
(b) declared
(c) notified
(d) suppressed
Answer:
(d) suppressed

Question 8.
snapped
(a) became happy
(b) broke
(c) crackled
(d) fractured
Answer:
(a) became happy

Question 9.
disadvantage
(a) effectiveness
(b) unfavourable
(c) drawback
(d) advantage
Answer:
(d) advantage

Question 10.
hesitation
(a) fluctuation
(b) unwillingness
(c) willingness
(d) stutter
Answer:
(c) willingness

III. Choose the Correct Answer (MCQ).

Question 1.
The train had started _________
(a) delayed
(b) punctually
(c) late
(d) after an hour
Answer:
(b) punctually

Question 2.
An hour after leaving _________ the train had passed the bridges and the island of salcette.
(a) Allahabad
(b) Calcutta
(c) Lucknow
(d) Bombay
Answer:
(d) Bombay

Question 3.
At half-past _________ the train stopped at Burhampoor.
(a) ten
(b) twelve
(c) six (d) eight
Answer:
(b) twelve

Question 4.
The _________ at once stepped out.
(a) General
(b) attendant
(c) club member
(d) driver
Answer:
(a) General

Question 5.
Sir Francis was _________
(a) happy
(b) furious
(c) cool
(d) cheerful
Answer:
(b) furious

Question 6.
A steamer leaves Calcutta for _________ at noon on the 25th.
(a) Hongkong
(b) Persia
(c) Norway
(d) Burhampoor
Answer:
(a) Hongkong

Question 7.
‘Good Heavens’, what a price for an _________
(a) eagle
(b) ox
(c) elephant
(d) ostrich
Answer:
(c) elephant

Question 8.
The driver _________ himself on the elephant’s neck.
(a) lost
(b) perched
(c) unsettled
(d) lowered
Answer:
(b) perched

Question 9.
A young man, with an _________ face, offered his services as a guide.
(a) averse
(b) ugly
(c) intelligent
(d) inapt
Answer:
(c) intelligent

Question 10.
The travellers made a hasty _________
(a) lunch
(b) dinner
(c) moment
(d) breakfast
Answer:
(d) breakfast

7th Standard Journey By Train  Short Questions with Answers.

Question 1.
Who is the author of this novel ‘Around the world in Eighty Days’?
Answer:
The author of this novel is Jules Verna.

Question 2.
Who was there among the passengers?
Answer:
There were a number of officers, Government officials, and merchants.

Question 3.
Who occupied a seat opposite Mr. Fogg?
Answer:
Sir Francis Cromarty occupied a seat opposite to him.

Question 4.
Who was Sir Francis?
Answer:
Sir Francis was one of the friends of Mr. Fogg.

Question 5.
Where did he meet him?
Answer:
He met him on the ship Mongolia that brought him to Bombay.

Question 6.
When will the steamer leave Calcutta for Hong Kong?
Answer:
The steamer would leave Calcutta for Hong Kong at noon on the 25th.

Question 7.
What was Mr. Fogg resolved to hire?
Answer:
He was resolved to hire an Indian elephant for his journey to Allahabad.

Question 8.
How much did Mr. Fogg offer to hire the Elephant at first?
Answer:
He offered ten pounds per hour to hire the elephant.

Question 9.
Where did the train stop at half-past twelve?
Answer:
It stopped at Burhampoor at half-past twelve.

Question 10.
Which mountains separated the Khandesh from Bundelcund?
Answer:
The Sutpour mountains separated the Khandesh from Bundelcund.

V. Paragraph Questions with Answers.

Question 1.
What did Passepartout see, when he was crossing India in a railway train?
Answer:
Passepartout on waking looked out. He could not believe that he was actually crossing India. The locomotive, guided by an English engineer and fed with coal, threw out its smoke upon cotton, coffee, nutmeg, clove, and pepper plantations. The steam curled in spirals around groups of palm-trees. In the midst of these trees were attractive bungalows, viharas, and marvellous temples, decorated by the rich work of Indian architecture. There were vast areas extending to the horizon with jungles and forests. There he saw snakes, tigers, and elephants.

Question 2.
Who served as a guide to Mr. Fogg and others? How did he manage to take all three to Allahabad?
Answer:
A young intelligent man offered his services as a guide. The elephant was led out and equipped. The skilled driver covered the elephant’s back with a sort of saddle-cloth and attached to each of its side some uncomfortable howdahs. While Sir Francis and Mr. Fogg took the howdahs on either side, Passepartout got on to the saddle-cloth between them. The driver positioned himself on the elephant’s neck and they set out from the village at nine o’clock by short cuts through the dense forests of palms.

Journey by Train Grammar Additional

Reported Speech

The actual words spoken by a person are Direct speech. They are enclosed within quotation marks.
When we later report this, making changes to the words the speaker originally said, it is Reported Speech. (Indirect Speech)

Sentence	Direct Speech	Indirect Speech
I want an ice cream	Ram said to Rakesh, “I want an ice cream”	Ram told Rakesh that he wanted ice cream.
I am coming to Chennai tomorrow.	My uncle said to me, “I am coming to Chennai tomorrow.”	My uncle told me that he was coming to Chennai the next day
I want to become a doctor.	Monica said, “I want to become a doctor.”	Monica said that she wanted to become a doctor.
The comic books are kept on the second shelf.	The librarian said, “The comic books are kept on the second shelf”	The librarian said that the comic books were kept on the second shelf.
The monument is beautiful.	Sidharthan said, “The monument is beautiful.”	Sidharthan said that the monument was beautiful.

I. Change into Indirect Speech.

Question 1.
He said, “I live in the city centre”.
Answer:
He said that he lived in the city centre.

Question 2.
Radha said, “I am going out”.
Answer:
Radha said that she was going out.

Question 3.
Ravi said, “I can swim”
Answer:
Ravi said that he could swim.

Question 4.
He said, “I arrived before you”.
Answer:
He said that he had arrived before him.

Question 5.
My father said, “I will be in Pairs on Monday”.
Answer:
My father said that he would be in Paris on Monday.

Warm-Up

Work in pairs. Choose six items which are essential for a camp. Rank their priority and justify your answer.

Answer:

1. Torch Light
2. Tent
3. Cell Phone
4. Rope
5. Hammer
6. Stove

These things are essential for a camp because, without these things, we cannot enjoy our camp. Camping is an amazing experience. So we should prepare ourselves properly with the essential items taken along with us. This will avoid frustration and stress.

Journey by Train Summary

Section – I

This story tells us of an amazing journey by an Englishman Mr. Phileas Fogg. He is a ruthless perfectionist who cares more about the bet than the native places, he sees on his travels. He will do anything, even lie and cheat, to get what he wants.
Mr. Phileas Fogg along with his French attendant, Passepartout, attempts to go round the world in eighty days by taking a bet for $ 20,000. They travel through some parts of India. That was the time when the railways were being built in the country. The train started at the scheduled time from Bombay. There were a number of passengers like officers, government officials, and merchants on the train. Sir Francis Cromarty, one of Mr. Fogg’s friends, occupied a seat opposite to them.

The train passed the bridges, the Island of Salcette, mountains, jungles, forests, and the fertile territory of Khandesh.
Passepartout couldn’t believe that he was actually crossing India on a railway train. Tigers and snakes fled at the noise of the train. Elephants stood gazing at the train with sad eyes. At half-past twelve, the train stopped at Burhampoor. The travellers had a hasty breakfast. The train entered the valleys of Sutpour Mountains at three in the morning.

Question 1.
Give a picturesque view through which the train travels.
Answer:

Section – II

The train stopped at 8 o’clock some fifteen miles beyond Rothal. Phileas Fogg and others were annoyed. They were wondering why there was a halt of the train in the midst of the forest. When they enquired, the conductor informed them that the railroad lines ends at Kholby, 50 miles short of Allahabad. The line begins again from there in Allahabad. Sir Francis got angry because they sold the tickets from Bombay to Calcutta, without confirming about the railroad lines.

The passengers had to provide means of transportation for themselves from Kholby to Allahabad. So, Mr. Fogg, his attendant, and Sir. Francis planned to reach Allahabad in time by some means of transport, as a steamer was ready to leave Calcutta for Hongkong on the 25th. They had two days left to reach Calcutta.

Discuss and answer – Intext Questions.

Question 1.
Why did the train stop in the middle of the forest?
Answer:
The train stopped in the middle of the forest as there were no railroad lines beyond that place.

Question 2.
Why was Sir Francis angry?
Answer:
There were no railroad lines from Kholby to Allahabad. But the tickets were sold from Bombay to Calcutta. So Sir Francis was angry.

Section – III

After searching the village from end to end, Mr. Fogg and Sir. Francis came back without finding any means of transport. But Passepartout suggested that they can travel on an Indian elephant. As it could travel rapidly for a long time, Mr. Fogg determined to hire it. Even though Mr. Fogg offered an excessive amount to hire the elephant, the elephant’s owner refused to hire it. So Mr. Fogg finally purchased the elephant for two thousand pounds. A young man offered his services as a guide to them.

While Mr. Fogg and Sir Francis took the howdahs on either side, Passepartout got on to the saddle-cloth between them. The driver sat on the elephant’s neck and at 9 o’clock set out from the village to Allahabad.

Discuss and answer.

Question 1.
Which mode of transport did Fogg choose?
Answer:
At first, Fogg chose to go on foot to Allahabad. Then he purchased an elephant to take them to Allahabad.

Question 2.
Did he hire Kiouni? Why?
Answer:
No, he did not hire Kiouni, but purchased it for 2000 pounds, because the owner of the elephant intended to make a big bargain and so refused to hire him.

Question 3.
Why was the elephant owner happy with the deal?
Answer:
The elephant owner was happy with the deal because Mr. Fogg offered him 2000 pounds to purchase the elephant.

Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.4 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 2 Percentage and Simple Interest Ex 2.4

Question 1.
Find the simple interest on ₹ 35,000 at 9% per annum for 2 years?
Solution:

Principal P = ₹ 35,000
Rate of interest r = 9 % Per annum
Time (n) = 2 years
Simple Interest I = \(\frac { Pnr }{ 100 } \) = \(\frac{35000 \times 2 \times 9}{100}\) = ₹ 6300
Simple intrest I = ₹ 6300

Question 2.
Aravind borrowed a sum of ₹ 8,000 from Akash at 7% per annum. Find the interest and amount to be paid at the end of two years.
Solution:
Here Principal P = ₹ 8,000
Rate of interest r = 7% Per annum
Time (n) = 2 Years
Simple Interest (I) = \(\frac { Pnr }{ 100 } \) = \(\frac{8000 \times 2 \times 7}{100}\)
I = ₹ 1120
Amount = P + I
I = ₹ 8000 + 1120 = 9120
Interest to be paid = ₹ 1,120
Amount to be paid = ₹ 9,120

Question 3.
Sheela has paid simple interest on a certain sum for 4 years at 9.5% per annum is ₹ 21,280. Find the sum.
Solution:
Let the Principal be ₹ P
Rate of interest r = 9.5% per annum
Time (n) = 4 years
Simple Interest I = \(\frac { Pnr }{ 100 } \)
Given I = ₹ 21,280

∴ Sum of money Sheela bought = ₹ 56,000

Question 4.
Basha borrowed ₹ 8,500 from a bank at a particular rate of simple interest. After 3 years, he paid ₹ 11,050 to settle his debt. At what rate of interest he borrowed the money?
Solution:
Let the rate of interest be r% per annum
Here Principal P = ₹ 8,500
Time n = 3 years
Total amount paid = ₹ 11,050
A = P + 1 = ₹ 11,050
i.e. 8,500 + 1 = ₹ 11,050
I = ₹ 11,050 – ₹ 8,500 = ₹ 2,550

Question 5.
In What time will ₹ 16,500 amount to ₹ 22,935 at 13% per annum?
Solution:
Rate of interest r = 13% per annum
Here Amount A = ₹ 22,935
Principal P = ₹ 16,500
A = P + I
22935 = 16,500 + I
∴ Interest I = 22935 – 16,500 = ₹ 6,435
Simple Interest I = \(\frac { pnr }{ 100 } \)

6435 = \(\frac{16500 \times n \times 13}{100}\)
n =\(\frac{6435 \times 100}{16500 \times 13}\)
n = 3 years
Required time n = 3 years

Question 6.
In what time will ₹ 17800 amount to ₹ 19936 at 6% per annum?
Solution:
Let the require time be n years
Here Principal P = ₹ 17,800
Rate of interest r = 6% per annum
Amount A = ₹ 19,936
A = P + I
19936 = 17800 + 1
19936 – 17800 = I
2136 = I
Simple Interest (I) = \(\frac { pnr }{ 100 } \)
2136 = \(\frac{17800 \times n \times 6}{100}\)
n = \(\frac{2136 \times 100}{17800 \times 6}\)

n = 2 Years
Required time = 2 years

Question 7.
A sum of ₹ 48,000 was lent out at simple interest and at the end of 2 years and 3 months the total amount was ₹ 55,560. Find the rate of interest per year.
Solution:
Given Principal P = ₹ 48,000
Time n = 2 years 3 months
= 2 + \(\frac { 3 }{ 12 } \) years = 2 + \(\frac { 1 }{ 4 } \) years
= \(\frac { 8 }{ 4 } \) + \(\frac { 1 }{ 4 } \) years = \(\frac { 9 }{ 4 } \) years
Amount A = ₹ 55,660
A = p + 1
55660 = 48000 + I
I = 55660 – 48000 = ₹ 7660
∴ Interest for \(\frac { 9 }{ 4 } \) years = ₹ 7660
Simple intrest = \(\frac { pnr }{ 100 } \)
7660 = 48000 × \(\frac { 9 }{ 4 } \) × \(\frac { r }{ 100 } \)
r = \(\frac{7660 \times 4 \times 100}{9 \times 48000}\) = 7.09 % = 7 %
Rate of interest = 7 % Per annum

Question 8.
A principal becomes ₹ 17,000 at the rate of 12% in 3 years. Find the principal.
Solution:
Given the Principal becomes ₹ 17,000
Let the principle initially be P
Rate of Interest r Time = 12 % Per annum
Time n = 3 years
According to the problem given I = 17000 – P = \(\frac{P \times 3 \times 12}{100}\)
17000 = \(\frac { 36 }{ 100 } \) p + p
17000 = p(\(\frac { 36 }{ 100 } \) + 1)
17000 = p(\(\frac { 136 }{ 100 } \))
p = \(\frac{17000 \times 100}{136}\) = 12,500
∴ Principal P = ₹ 12,500

Objective Type Questions

Question 9.
The interest for a principle of? 4,500 which gives an amount of? 5,000 at end of certain period is
(i) ₹ 500
(ii) ₹ 200
(iii) 20%
(iv) 15%
Hint: Interest = Amount – Principle = ₹ 5000 – ₹ 4500 = ₹ 500
Answer:
(i) ₹ 500

Question 10.
Which among the following is the simple interest for the principle of ₹ 1,000 for one year at the rate of 10% interest per annum?
(i) ₹ 200
(ii) ₹ 10
(iii) ₹ 100
(iv) ₹ 1,000
Hint: Intrest = \(\frac { pnr }{ 100 } \) = \(\frac{1000 \times 1 \times 10}{100}\) = ₹ 100
Answer:
(iii) ₹ 100

Question 11.
Which among the following rate of interest yields an interest of ₹ 200 for the principle of ₹ 2,000 for one year.
(i) 10%
(ii) 20%
(iii) 5%
(iv) 15%
Hint: r = \(\frac{I \times 100}{P \times n}\) = \(\frac{200 \times 100}{2000 \times 1}\) = 10 %
Answer:
(i) 10%

Students can Download Maths Chapter 2 Percentage and Simple Interest 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 2 Percentage and Simple Interest Additional Questions

Additional Questions and Answers

Exercise 2.1

Question 1.
72% of 25 students are good at science. How many are not good at science?
Solution:
Number of students who are good at science
= 72% of 25 = \(\frac { 72 }{ 100 } \) × 25 = 18 students
∴ Number of students who are not good at science
= 25 – 18 = 7 students

Question 2.
A flower garden has 1000 plants. 5% of the plants are roses and 1% are daisy plants. What is the total number of other plants.
Solution:
Total plants = 1000
Number of rose plants = 5 % of 1000 = \(\frac { 5 }{ 100 } \) × 1000 = 50
Number of Daisy plants = 1 % of 1000 = \(\frac { 1 }{ 100 } \) × 1000 = 10
Total of rose and daisy = 50 + 10 = 60
Number of other plants = 1000 – 60 = 940

Question 3.
Find 135 % of 80 ₹.
Solution:
135 % of 80 = \(\frac { 135 }{ 100 } \) × 80 = ₹ 108

Exercise 2.2

Question 1.
Neka bought 72.3m of cloth from a role of 100m. Express the cloth bought in terms of percentage.
Solution:
Total length of the cloth = 100 m
Length of cloth bought = 72.3 m
Percentage of cloth bought = \(\frac { 72.3 }{ 100 } \) = 72.3 %

Question 2.
Convert (i) 88 % (ii) 1.86 % into decimals.
Solution:
(i) 88 % = \(\frac { 88 }{ 100 } \) = 0.88
(ii) 1.86 % = \(\frac { 1.86 }{ 100 } \) = 0.0186

Question 3.
Convert (i) 3.35 (ii) 0.5 into percentage.
Solution:
(i) 3.35 = \(\frac { 335 }{ 100 } \) × 100 % = 335 %
(ii) 0.5 = \(\frac { 5 }{ 10 } \) × 100 % = 50 %

Exercise 2.3

Question 1.
If Gayathri had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?
Solution:
Suppose Gayathri had ₹ X in the beginning.
Then money Spend = 75 % of X = \(\frac { 75 }{ 100 } \) X = \(\frac { 3X }{ 4 } \)
Money left with her = X – \(\frac { 3X }{ 4 } \) = \(\frac { 4X-3X }{ 4 } \) = \(\frac { X }{ 4 } \)
But it is given that money left = ₹ 600
i.e. \(\frac { X }{ 4 } \) = 600
X = 600 × 4 = 2400
∴ Gayathri had ₹ 2,400

Question 2.
Mohan gets 98 marks in her exams. This amounts to 56% of the total marks, What are the maximum marks?
Solution:
Let the maximum marks be X. 56 % of X = 98
\(\frac { 56 }{ 100 } \) × (X) = 98
⇒ X = 98 × \(\frac { 100 }{ 56 } \)
X = 175
∴ Maximum marks = 175

Exercise 2.4

Question 1.
On what sum of money lent out at 9% per annum for 6 years does the simple interest amount to ₹ 810?
Solution:
Given Simple Interest I = ₹ 810
Let the sum of money (Principal) be P
Rate of interest r = 9 % Per annum.
Time n = 6 years
I = \(\frac { pnr }{ 100 } \)
810 = \(\frac{P \times 6 \times 9}{100}\)
P = \(\frac{810 \times 100}{6 \times 9}\)
P = ₹ 1500
Sum of money required = ₹ 1500

Question 2.
Find the simple interest on ₹ 1120 for 2 \(\frac { 2 }{ 5 } \) years at the rate of 5% per annum.
Solution:
Simple Interest I = \(\frac { pnr }{ 100 } \)
Principal P = ₹ 1120
Time n = 2 \(\frac { 2 }{ 5 } \) years
= \(\frac { 12 }{ 5 } \) years
Rate of Interest r = 5 %

∴ I = 1120 × \(\frac { 12 }{ 5 } \) × \(\frac { 5 }{ 100 } \)
= \(\frac { 672 }{ 5 } \)
= ₹ 134.4
Simple interest = = ₹ 134.4