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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 5 Geometry Ex 5.1

Question 1.
Name the pairs of adjacent angles.
Solution:
(i) ∠ABG and ∠GBC are adjacent angles.
(ii) ∠BCF and ∠FCD are adjacent angles.
(iii) ∠BCF and ∠FCE are adjacent angles.
(iv) ∠FCE and ∠ECD are adjacent angles.

Question 2.
Find the angle ∠JIL from the given figure.
Solution:
∠LIK and ∠KIJ are adjacent angles.
∴ ∠JIL = ∠LIK + ∠KIJ
= 38° + 27°
= 65°
∴ ∠JIL = 65°

Question 3.
Find the angles ∠GEH from the given figure.
Solution:

∠HEF = ∠HEG + ∠GEF
120° = ∠HEG + 34°
120°- 34° = ∠GEH + 34° – 34°
∠GEH = 86°

Question 4.
Given that AB is a straight line. Calculate the value of x° in the following cases.

Solution:
(i) Since the angles are linear pair ∠AOC + ∠BOC = 180°
72° + x° = 180°
72° + x° – 12° = 180°- 72°
x° = 108°

(ii) Since the angles are linear pair
∠AOC + ∠BOC = 180°
3x + 42° = 180°

(iii) Since the angles are linear pair
∠AOC + ∠BOC = 180°
4x° + 2x° = 180°
6x° = 180°
x° = 180°

Question 5.
One angle of a linear pair is a right angle. What can you say about the other angle?
Solution:
If the angle are linear pair, then their sum is 180°.
Given one angle is right angle ie 90°.
∴ The other angle = 180° -90° = 90°
∴ The other angle also a right angle

Question 6.
If the three angles at a point are in the ratio 1 : 4 : 7, find the value of each angle?
Solution:
We know that the sum of angles at a point is 360°.
Given the three angles are in the ratio 1:4:7.
Let the three angles be 1x, 4x, 7x.

∴ The three angles are 30°, 120° and 210°.

Question 7.
Three are six angles at a point. One of them is 45° and the other five angles are all equal. What is the measure of all the five angles.
Solution:
We know that the sum of angles at a point is 360°.
One angle = 45°
Let the equal angles be x° each
∴ x° + x° + x° + x° + x° + 45° = 360°
5x° + 45° – 45° = 360° – 45°

5x° = 315°

∴ Measure of all 5 equal angles = 63°.

Question 8.
In the given figure, identify
(i) Any two pairs of adjacent angles.
(ii) Two pairs of vertically opposite angles.
Solution:
(i) (a) ∠PQT and ∠TOS are adjacent angles.
(b) ∠PQU and ∠UQR are adjacent angles.
(ii) (a) ∠PQT and ∠RQU are vertically opposite angles.
(b) ∠TQR and ∠PQU are vertically opposite angles.

Question 9.
The angles at a point are x°, 2x°, 3x°, 4x° and 5x°. Find the value of the largest angle?
Sum of angles at a point = 360°
∴ x° + 2x° + 3x° + 4x° + 5x° = 360°
15x° = 360°
x° = \(\frac{360^{\circ}}{15}\)
x° = 24°.
∴ The largest angle = 5x°
= 5 × 24° = 120°
The largest angle is 120°

Question 10.
From the given figure, find the missing angle.
Solution:
Lines \(\overleftrightarrow { RP }\) and \(\overleftrightarrow { SQ }\) are interesting at ‘O’

∴ Vertically opposite angles are equal.
∴ x = 105°
∴ Missing angle = 105°

Question 11.
Find the angles x° and y° in the figure shown.

Solution:
Consider the line m.
x° and 3x° are linear pair of angles
∴ x° + 3x° = 180°


x° = 45°
Also lines l and m intersects.
Vertically opposite angles are equal.
ie 3x° = y°
3 × 45° = y°
y = 135°
x° = 15° and
y° = 135°

Question 12.
Using the figure, answer the following questions.

(i) What is the measure of angle x°?
(ii) What is the measure of angle y°?
Solution:
From the figure x° and 125° are vertically opposite angles. So they are equal ie
ie x° = 125°
Also y° and 125° are linear pair of angles.
∴ y° + 125° = 180°
y° + 125° – 125° = 180°- 125°
y° = 55°
x° = 125°,
y° = 55°

Question 13.
Adjective angles have
(i) No common interior, no common arm, no common vertex.
(ii) One common vertex, one common arm, common interior
(iii) One common arm, one common vertex, no common interior.
(iv) One common arm, no common vertex, no common interior.
Solution:
(iii) one common arm, one common vertex, no common interior

Question 14.
In the given figure the angles ∠1 and ∠2 are
(i) Opposite angles
(ii) Adjacent angles
(iii) Linear angles
(iv) Supplementary angles

Solution:
(iii) Linear pair

Question 15.
Vertically opposite angles are
(i) not equal in measure
(ii) Complementary
(iii) supplementary
(iv) equal in measure
Solution:
(iv) equal in measure

Question 16.
The sum of all angles at a point is
(i) 360°
(ii) 180°
(iii) 90°
(iv) 0°
Solution:
(i) 360°

Question 17.
The measure of ∠BOC is
(i) 90°
(ii) 180°
(iii) 80°
(iv) 100°

Solution:
(iii) 80°

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Intext Questions

Recap (Textbook Page No. 108)

Question 1.
Count the objects in the following figure and complete the table that follows :

From Fig 5.1 and the table, answer the following question.
(i) The total number of objects in the above picture is ______
(ii) The difference between the number of squares and the number of bats is ______
(iii) The ratio of the number of balls to the number of bats is _______
(iv) What are the objects equal in number?
(v) How many more balls are there than the number of bats?
Solution:
(i) 24
(ii) 0
(iii) \(\frac{8}{6}=\frac{4}{3}\)
(iv) Bat and Square
(v) 2 balls more

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 6 Information Processing Ex 6.1

Question 1.
A tetromino is a shape obtained by …… squares together.
Solution:
4

Question 2.
Draw a tetromino which passes symmetry ……..
Solution:

Question 3.
Complete the table.

Solution:

Question 4.
Shade the figure completely, by using five tetrominoes shapes only once.

Solution:

Question 5.

Solution:

More possible ways are the

Question 6.
Match the tetrominoes of same type.

Solution:

Question 7.
Using the given tetrominoes with numbers, compute the 4 × 4 magic square.

Solution:

(more possible ways are these)

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 6 Information Processing Additional Questions

Question 1.
Form a 5 × 4 square shade the square completely by tetromino shapes.

Solution:

Question 2.
Shade the following figure using the five tetrominoes exactly once.

Solution:

Question 3.
Find the Shortest distance to reach the play ground from school.

Also find out all possible routes.
Solution:
Route 1: School ➝ Swimming pool ➝ canteen ➝ playground
Distance 9km + 2km + 10km = 21 km.
Route 2 : School ➝ canteen ➝ playground
Distance : 10 km + 10 km = 20 km.
Route 3 : School ➝ garden ➝ playground
Distance : 2 km + 20 km = 22 km.
Route 4 : School ➝ garden ➝ teashop ➝ playground
Distance : 2km + 17km + 2km = 21 km.
Route 5 : School ➝ Park ➝ teashop ➝ playground
Distance : 12 km + 12 km + 2 km = 24 km.
∴ Route 2 is the shortest distance.

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions

Additional Questions And Answers

Exercise 3.1

Very Short Answers [2 Marks]

Question 1.
Find the product of the following.
(i) (x, y)
(ii) (10x, 5y)
(iii) (2x², 5y²)
Solution:
(i) x × y = xy
(ii) 10x × 5y = (10 × 5) x × xy
= 50 xy
(iii) 2x² x 5y² = (2 x 5) x (x² + y²)
= 10x²y²

Short Answers [3 Marks]

Question 1.
Find the product of the following.
(i) 3ab² c³ by 5a³b²c
(ii) 4x²yz by \(\frac{3}{2}\) x²yz²
Solution:
(i) (3ab³c³) × (5a³b²c)
= (3 × 5)(a × a³ × b² × b² × c² × c)
= 15a¹⁺³.b²⁺².c³⁺¹ = 15a⁴b⁴c⁴

(ii) 4x²yz by \(\frac{3}{2}\) x²yz²
= (4 × \(\frac{3}{2}\)) × (x² × x² × y × y × z × z²)
= -6x²⁺² y¹⁺¹ x¹⁺² = -6x⁴y²z³

Long Answers [5 Marks]

Question 1.
Simplify (3x – 2) (x – 1) (3x + 5).
Solution:
(3x – 2) (x – 1) (3x + 5)
= {(3x – 2) (x – 1)} × (3x + 5) [∴Multiplication in associative]
= {3x (x – 1) – 2 x – 1)} × (3x + 5)
= (3x² – 3x – 2x + 2) × (3x + 5)
= (3x² – 5x + 2) (3x + 5)
= 3x² × (3x + 5) – 5x (3x + 5) + 2 (3x + 5)
= (9x³ + 15x²) + (-15x² – 25x) + (6x + 10)
= 9x³ + 15x² – 15x² – 25x + 6x + 10
= 9x³ – 19x + 10

Question 2.
Simplify (5 – x) (3 – 2x) (4 – 3x).
Solution:
(5 – x) (3 – 2x) (4 – 3x)
= {(5 – x)(3 – 2x)} × (4 – 3x) [∴ Multiplication in association]
= {5 (3 – 2x) -x (3 – 2x)} × (4 – 3x)
= (15 – 10x – 3x + 2x²) × (4 – 3x)
= (2x² – 13x + 15) (4 – 3x)
= 2x² × (4 – 3x) – 13x (4 – 3x) + 15 (4 – 3x)
= 8x³ – 6³ – 52x + 39x² + 60 – 45x
= -6x³ + 47x² – 97x + 60

Exercise 3.2

Very Short Answers [2 Marks]

Question 1.
Divide.
(i) 12x³y³ by 3x²y
(ii) -15a² bc³ by 3ab
(iii) 25x³y² by – 15x²y
Solution:

Short Answers [3 Marks]

Question 1.
Divide
(i) 15m²n³ by 5m²n²
(ii) 24a³b³ by -8ab
(iii) -21 abc² by 7 abc
Solution:

Question 2.
Divide
(i) 16m³y² by 4m²y
(ii) 32m² n³p² by 4mnp
Solution:

Long Answers [5 Marks]

Question 1.
Divide.
(i) 9m⁵ + 12m⁴ – 6m² by 3m²
(ii) 24x³y + 20x²y² – 4xy by 2xy
Solution:

Exercise 3.3

Question 1.
Evaluate:
(i) (2x + 3y)²
(ii) (2x – 3y)²
Solution:
(i) (2x + 3y)²
= (2x)² + 2 × (2x) × (3y) + (3y)²
[using (a + b)² = a² + 2ab + b²]
= 4x² + 12xy + 9y²

(ii) (2x – 3y)²
= (2x)² – 2(2x) (3y) + (3y)²
[∵ using (a – b)² = a² – 2ab + b²]
= 4x² – 12xy + 9y²

Short Answers [3 Marks]

Question 1.
Evaluate the following
(i) (2x – 3) (2x + 5)
(ii) (y – 7) (y + 3)
(iii) 107 × 103
Solution:
(i) (2x – 3) (2x + 5)
= (2x)² + (-3 + 5) (2x) + (-3) (5)
[∵ (x + a) (x + b) = x² + (a + b)x + ab]
= 2²x² + 2 × 2x + (-15)
= 4x² + 4x – 15

(ii) (y – 7) (y + 3)
= y² + (-7 + 3)y + (-7) (3)
[∵ (x + a)(x + b) = x² + (a + b)x + ab]
= y² – 4y + (-21) = y² – 4y – 21

(iii) 107 × 103
= (100 + 7) × (100+ 3)
= 100² + (7 + 3) × 100 +(7 × 3)
= 10000 + 10 × 100 + 21 = 10000 + 1000 + 21 = 11021

Long Answers [5 Marks]

Question 1.
If x + y = 12 and xy = 14 find x² + y².
Solution:
(x + y)² = x² + y² + 2xy
12² = x² + y² + 2 × 14
144 = x² + y² + 28
x² + y² = 144 – 28
x² + y² = 116

Question 2.
If 3x + 2y = 12 and xy = 6 find the value of 9x² + 4y².
Solution:
(3x + 2y)² = (3x)² + (2y)² + 2 (3x) (2y)
= 9x² + 4y² + 12xy
12² = 9x² + 4y² + 12 × 6
144 = 9x² + 4y² + 72
144 – 72 = 9x² + 4y²
∴ 9x² + 4y² = 72

Exercise 3.4

Question 1.
Factorize:
(i) 7(2x + 5) + 3 (2x + 5)
(ii) 12x³y⁴ + 16x²y⁵ – 4x⁵y²
Solution:
(i) 7(2x + 5) + 3 (2x + 5)
= (2x + 5) (7 + 3)
(ii) 12x³y⁴ + 16x²y⁵ – 4x⁵y²
= 4x²y² (3xy² + 4y³ – x³)

Short Answers [3 Marks]

1. Factorize
(i) 81a² – 121b²
(ii) x² + 8x + 16
Solution:
(i) 81a² – 121b²
= (9a)² – (11b)²
[∵ using a² – b² = (a + b)²]
= (9a + 11b) (9a – 11b)

(ii) x² + 8x + 16 = x² + 2 × x × 4 + 4²
[∵ using a² + 2ab + b² = (a + b)²]
= (x + 4)² = (x + 4)(x + 4)

Long Answers [5 Marks]

Question 1.
Factorize
(i) x² + 2xy + y² – a² + 2ab – b²
(ii) 9 – a⁶ + 2a³ – b⁶
Solution:
(i) x² + 2xy + y² – a² + 2ab – b².
= (x² + 2xy + y²) – (a² – 2ab + b²)
= (x + y)² – (a – b)²
= {(x + y) + (a – b)} {(x + y) – (a – b)}
= (x + y + a – b) (x + y – a + b)

(ii) a – a⁶ + 2a³b³ – b⁶
= 9 – (a⁶ – 2a³b³ + b⁶)
= 3² -{(a³)² – 2 × a³ × b³ + (b³)²}
= 3² – (a³ – 6³)²
= {3 + (a³ – b³)} {3 – (a³ – b³)}
= (3 + a³ – b³) (3 – a³ + b³)
= (a³ – b³ + 1){-a³ + b³ + 3)

Question 2.
Factorize
(i) 100 (x + y )² – 81 (a + b)²
(ii)(x + 1)² – (x – 2)²
Solution:
(i) 100 (x + y)² – 81 (a + b)²
= {10 (x + y)}² – {(a (a + b)}²
= {10 (x + y) + 9 (a + b)}
{10 (x + y) – 9(a + b)}
= (10x + 10y + 9a – 9b)}
(10x + 10y – 9a – 9b)

(ii) (x – 1)² – (x – 2)²
= {(x – 1 +(x – 2)}
{(x – 1) – (x – 2)}
= (2x – 3) – (x – 1 – x + 2)
= (2x – 3) × 1 = 2x – 3

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.5

Miscellaneous Practice Problems

Question 1.
Subtract: -2(xy)² (y³ + 7x²y + 5) from 5y² (x²y³ – 2x⁴y + 10x²)
Solution:
5y² (x²y³ – 2x⁴y + 10x²) – [(-2)(xy)² (y³ + 7x²y + 5)]
= [5y² (x²y³) – 5y² (2x⁴y) + 5y² (10x²)] – [(-2)x²y² (y³ + 7x²y + 5)]
= (5y⁵5x² – 10x⁴y³ + 50x²y²)
= 5x²y⁵ – 10x⁴y³ + 50x²y² – [(-2x²y⁵) – 14x⁴y³ – 10x²y²]
= 5x²y⁵ – 10x⁴y³ + 50x²)y² + 2x² y⁵ + 14x⁴)y³ + 10x²)y²
= (5 + 2)x²y⁵ + (-10 + 14)x⁴y³ + (+50 + 10)x²y²
= 7x²y⁵ + 4x⁴y³ + 60x²y²

Question 2.
Multiply (4x² + 9) and (3x – 2).
Solution:
(4x² + 9) (3x – 2) = 4x²(3x – 2) + 9(3x – 2)
= (4x²)(3x) – (4x²)(2) + 9(3x) – 9(2) = (4 × 3 × x × x²) – (4 × 2 × x²) + (9 × 3 × x) – 18
= 12x³ – 8x² + 27x – 18 (4x³ + 9) (3x – 2) = 12x³ – 8x² + 27x – 18

Question 3.
Find the simple interest on Rs. 5a²b² for 4ab years at 7b% per annum.
Solution:

Question 4.
The cost of a note book is Rs. 10ab. If Babu has Rs. (5a²b + 20ab² + 40ab). Then how many note hooks can he buy?
Solution:
For ₹ 10 ab the number of note books can buy = 1.

Question 5.
Factorise : (7y² – 19y – 6)
Solution:
7y² – 19y – 6 is of the form ax² + bx + c where a = 7; b = -19; c = – 6

The product a × c = 7 × -6 = -42
sum b = – 19

The middle term – 19y can be written as – 21y + 2y
7y² – 19y – 6 = 7y² – 21y + 2y – 6
= 7y(y – 3) + 2(y – 3) = (y – 3)(7y + 2)
7y² – 19y – 6 = (y – 3)(7y + 2)

Challenging Problems

Question 6.
A contractor uses the expression 4x² + 11x + 6 to determine the amount of wire to order when wiring a house. If the expression comes from multiplying the number of rooms times the number of outlets and he knows the number of rooms to be (x + 2), find the number of outlets in terms of ‘x’ [Hint: factorise 4x² + 11x + 6]
Solution:
Given Number of rooms = x + 2
Number of rooms × Number of outlets = amount of wire.

Now factorising 4x² + 11x + 6 which is of the form ax² + bx + c with a = 4 b = 11 c = 6.
The product a × c = 4 × 6 = 24
sum b = 11


The middle term 11x can be written as 8x + 3x
∴ 4x² + 11x + 6 = 4x² + 8x + 3x + 6 = 4x (x + 2) + 3 (x + 2)
4x² + 11x + 6 = (x + 2) (4x + 3)
Now from (1) the number of outlets

∴ Number of outlets = 4x + 3

Question 7.
A mason uses the expression x² + 6x + 8 to represent the area of the floor of a room. If the decides that the length of the room will be represented by (x + 4), what will the width of the room be in terms of x ?
Solution:
Given length of the room = x + 4
Area of the room = x² + 6x + 8
Length × breadth = x² + 6x + 8

Question 8.
Find the missing term: y² + (-) x + 56 = (y + 7)(y + -)
Solution:
We have (x + a) (x + b) = x² + (a + b)x + ab
56 = 7 × 8 .
∴ y² + (7 + 8)x + 56 = (y + 7)(y + 8)

Question 9.
Factorise: 16p⁴ – 1
Solution:
16p⁴ – 1 = 2⁴p⁴ – 1 = (2²)²(p²)² – 1²
Comparing with a² – b² = (a + b)(a – b) where a = 2²p² and b = 1
∴ (2²p²)² – 1² = (2²p² + 1) (2²p² – 1) = (4p² + 1) (4p² – 1)
∴ 16p⁴ – 1 = (4p² + 1)(4p² – 1)(4p² + 1)(2²p² – 1²)
= (4p² + 1) [(2p)²– 1²] = (4p² + 1) (2p + 1)(2p – 1)
[∵ using a² – b² = (a + b) (a – b)]
∴ 16p⁴ – 1 = (4psup>2 + 1)(2p + 1)(2p – 1)

Question 10.
Factorise : x⁶ – 64y³
Solution:
x⁶ – 64y³ = (x²)³ – 4³y³ = (x²)³ – (4y)³
This is of the form a³ – b³ with a = x², b = 4y
a³ – b³ = (a – b)(a² + ab + b²)
(x²)³ – (4y)³ = (x² – 4y) [(x²)² + (x²)(4y) + (4y)²]
= (x² – 4y) [x⁴ + 4x²y + 16y²]
∴ x⁶ – 64y³ = (x² – 4y) [x⁴ + 4x²y + 16y²]

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 1 Number System Ex 1.3

Question 1.
Fill in the blanks.
(i) -80 × ____ = -80
(ii) (-10) × ____ = 20
(iii) 100 × ___ = -500
(iv) ____ × (-9) = -45
(v) ___ × 75 = 0
Solution:
(i) 1
(ii) -2
(iii) -5
(iv) 5
(v) 0

Question 2.
Say True or False:
(i) (-15) × 5 = 75
(ii) (-100) × 0 × 20 = 0
(iii) 8 × (-4) = 32
Solution:
(i) False
(ii) True
(iii) False

Question 3.
What will be the sign of the product of the following:
(i) 16 times of negative integers.
(ii) 29 times of negative integers.
Solution:
(i) 16 is an even interger.
If negative integers are multiplied even number of times, the product is a positive integer.
∴ 16 times a negative integer is a positive integer.

(ii) 29 times negative integer.
If negative integers are multiplied odd number of times, the product is a negative integer. 29 is odd.
∴ 29 times negative integers is a negative integer.

Question 4.
Find the product of
(i) (-35) × 22
(ii) (-10) × 12 × (-9)
(iii) (-9) × (-8) × (-7) × (-6)
(iv) (-25) × 0 × 45 × 90
(v) (-2) × (+50) × (-25) × 4
Solution:
(i) 35 × 22 = -770
(ii) (-10) × 12 × (-9) = (-120) × (-9) = +1080
(iii) (-9) × (-8) × (-7) × (-6) = (+72) × (-7) × (-6) = (-504) × (-6) = +3024
(iv) (-25) × 0 × 45 × 90 = 0 × 45 × 90 = 0 × 90 = 0
(v) (-2) × (+50) × (-25) × 4 = (-100) × -25 × 4 = 2500 × 4 = 10,000

Question 5.
Check the following for equality and if they are equal, mention the property.
(i) (8 – 13) × 7 and 8 – (13 × 7)
Solution:
Consider (8 – 13) × 7 = (-5) × 7 = -35
Now 8 – (13 × 7) = 8 – 91 = -83
∴ (8 – 13) × 7 ≠ 8 – (13 × 7)

(ii) [(-6) – (+8)] × (-4) and (-6) – [8 × (-4)]
Solution:
[(-6) – (+8)] × (-4) = [(-6) + (-8)] × (-4) = (-14) × (-4) = +56
Now (-6) – [8 × (-4)] = (-6) – (-32)
= (-6) + (+32) = +26
∴ [(-6) – (+8)] × (-4) ≠ (-6) – [8 × (-4)]

(iii) 3 × [(-4) + (-10)] and [3 × (-4) + 3 × (-10)]
Solution:
Consider 3 × [(-4) + (-10)] = 3 × -14 = -42
Now [3 × (-4) + 3 × (-10)] = (-12) + (-30) = -42
Here 3 × [(-4) + (-10)] = [3 × (-4) + 3 × (-10)]
It is the distributive property of multiplication over addition.

Question 6.
During summer, the level of the water in a pond decreases by 2 inches every week due to evaporation. What is the change in the level of the water over a period of 6 weeks?
Solution:
Level of water decreases a week = 2 inches.
Level of water decreases in 6 weeks = 6 × 2 = 12 inches

Question 7.
Find all possible pairs of integers that give a product of -50.
Solution:
Factor of 50 are 1, 2, 5, 10, 25, 50.
Possible pairs of integers that gives product -50:
(-1 × 50), (1 × (-50)), (-2 × 25), (2 × (-25)), (-5 × 10), (5 × (-10))

Objective Type Questions

Question 8.
Which of the following expressions is equal to -30.
(i) -20 – (-5 × 2)
(ii) (6 × 10) – (6× 5)
(iii) (2 × 5)+ (4 × 5)
(iv) (-6) × (+5)
Solution:
(iv) (-6) × (+5)
Hint:
(i) -20 + (10) = -10
(ii) 60 – 30 = 30
(iii) 10 + 20 = 30
(iv) (-6) × (+5) = – 30

Question 9.
Which property is illustrated by the equation: (5 × 2) + (5 × 5) = 5 × (2 + 5)
(i) commutative
(ii) closure
(iii) distributive
(iv) associative
Solution:
(iii) distributive

Question 10.
11 × (-1) = _____
(i) -1
(ii) 0
(iii) +1
(iv) -11
Solution:
(iv) -11

Question 11.
(-12) × (-9) =
(i) 108
(ii) -108
(iii) +1
(iv) -1
Solution:
(i) 108

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Ex 5.4

Miscellaneous Practice Problems

Question 1.
The heights (in centimetres) of 40 children are

Prepare a tally mark table.
Solution:

Question 2.
There are 1000 students in a school. Data regarding the mode of transport of the students as given below. Draw a pictograph to represent the data.

Solution:
The pictograph for the mode of transport of students.

Question 3.
The following pictograph shows the total savings of a group of friends in a year. Each picture represents a saving of Rs. 100. Answer the following questions.

(i) What is the ratio of Ruby’s saving to that of Thasnim’s?
(ii) What is the ratio of Kuzhali’s savings to that of others?
(iii) How much is Iniya’s savings?
(iv) Find the total amount of savings of all your friends?
(v) Ruby and Kuzhali save the same amount. Say True or False.
Solution:
(i) Ratio of Ruby’s saving to that of Thasnim’s
\(=\frac{\text { Ruby’s saving }}{\text { Thasnim’s saving }}=\frac{5 \times 100}{4 \times 100}=\frac{5}{4}=5: 4\)
Ratio of Ruby’s saving to that of Thasnims = 5 : 4
(ii) Ratio of Kuzhali’s savings to that of others
\(=\frac{\text { kuzhali’s saving }}{\text { others saving }}=\frac{5 \times 100}{19 \times 100}=\frac{5}{19}=5: 19\)
Ratio of Kuzhali’s saving to that of others = 5 : 19
(iii) Iniya’s saving = 3 × 100 = ₹ 300
(iv) Saving of all the friends = (5 + 7 + 4 + 5 + 3) × 100 = 24 × 100 = ₹ 2400.
Total savings = ₹ 2400
(v) True.

Challenging Problems

Question 4.
The table shows the number of moons that orbit each of the planets in our solar system.

Make a Bar graph for the above data.
Solution:

Question 5.
The predictions of Weather in the month of September is given below:

(i) Make a frequency table of the types of weather by reading the calender.
(ii) How many days are either cloudy or partly cloudy
(iii) How many days do not have rain? Give two ways to find the answer?
(iv) Find the ratio of the number of sunny days to Rainy days.
Solution:
Frequency Table for the Type of weather for the month of September

(ii) 14 days.
(iii) For 24 days there is no rain.
(a) From the total 30 days, we subtract the rainy days i.e 30 – 6 = 24 days (from the frequency table)
(b) From the picture, we can count the non-rainy days.
(iv) Ratio of number of Sunny day to Rainy days
\(=\frac{\text { Number of Sunny days }}{\text { Number of Rainy days }}=\frac{10}{6}=\frac{5}{3}=5: 3\)
The ratio of a number of Sunny days to Rainy days = 5 : 3.

Question 6.
26 students were interviewed to find out what they want to become in future. Their responses are given in the following table.

Represent this data using the pictograph
Solution:

Question 7.
Yasmin of class VI was given a task to count the number of books which are biographies, in her school library. The information collected by her is represented as follows.

Observe the pictograph and answer the following questions.
(i) Which title has the maximum number of biographies?
(ii) Which title has the minimum number of biographies?
(iii) Which title has exactly half the number of biographies as Novelists?
(iv) How many biographies are there on the title of Sportspersons?
(v) What is the total number of biographies in the library?
Solution:
(i) ‘The title Novelists’ have the maximum number of biographies
(ii) ‘The title Scientists’ have the minimum number of biographies.
(iii) ‘Sportspersons’ title has exactly half the number of biographies as Novelist.
(iv) \((1 \times 20)+\frac{20}{4}=20+5=25\) biographies are there in the title sportsperson.
(v) 8 × 20 = 160 biographies are there in the library.

Question 8.
The bar graph illustrates the results of a survey conducted on vehicles crossing over a Toll Plaza in on hour.

Observe the bar graph carefully and fill up the following table.

Solution:

Question 9.
The lengths (in the nearest centimetre) of 30 drumsticks are given as follows.

Draw the bar graph showing the same information.
Solution:
The bar graph showing the same information is given below:

Question 10.
Given two angles are supplementary i.e. their sum = 180°.
Solution:
Let the angle be x.
Then anothcf angle = x + 20 (given)

The two angles are 80° and 100°

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 1 Number System Ex 1.2

Question 1.
Fill in the blanks
(i) -44 + ____ = -88
(ii) ___ – 75 = -45
(iii) ___ – (+50) = -80
Solution:
(i) -44
(ii) 30
(iii) -30

Question 2.
Say True or False.
(i) (-675) – (-400) = -1075
(ii) 15 – (-18) is the same as 15 + 18
(iii) (-45) – (-8) = (-8) – (-45)
Solution:
(i) False
(ii) True
(iii) False

Question 3.
Find the value of the following.
(i) -3 – (-4) using number line.
Solution:
We start at zero facing positive direction. Move 3 units backward to represent (-3). Then turn towards the negative side and move 4 units backwards. We reach+1.

∴ (-3) – (-4) = +1

(ii) 7 – (-10) using number line
Solution:

We start at zero facing positive direction. Move 7 units forward to represent (+7). Then turn towards the negative side and move 10 units backwards.
We reach +17
∴ 1 – (-10) = +17

(iii) 35 – (-64)
Solution:
35 – (-64) = 35+ (Additive inverse of-64) = 35 + (+64) = 99
∴ 35 – (-64) = 99

(iv) -200 – (+100)
Solution:
-200 – (+100) = -200 + (Additive inverse of+100) = -200 + (-100) = -300
-200 – (+100) = -300

Question 4.
Kabilan was having 10 pencils with him. He gave 2 pencils to senthil and 3 to Karthick. Next day his father gave him 6 more pencils, from that he gave 8 to his sister. How many pencils are left with him?
Solution:
Total pencils Kabilan had = 10
No. of pencils given to Senthil = 2
No. of pencils given to Karthick = 3.
Now number of pencils left with Kabilan = 10 – 2 – 3 = 8 – 3 = 5
Number of pencils got from his father = 6
No. total pencils Kabilan had = 5 + 6 = 11
Number of pencils given to his sister = 8
Number of pencils left with Kabilan = 11 – 8 = 3

Question 5.
A lift is on the ground floor. If it goes 5 floors down and then moves up to 10 floors from there, then in which floor will the lift be?
Solution:
Initially the lift will be in the ground floor representing ‘0’
It goes to 5 floors down ⇒ -5
Then it moves 10 floors up +10.
Now the lift will be = 0 – 5 + 10 = -5 + 10
= 5^th floor (above the ground floor)

Question 6.
When Kala woke up, her body temperature was 102°F. She took medicine for fever. After 2 hours it was 2°F lower. What was her temperature then?
Solution:
Kala’s temperature initially = 102°F
After two hours the temperature decreased = -2°F
Now the final temperature = 102°F – 2°F = 100°F

Question 7.
What number should be added to (-17) to get -19?
Solution:
According to the problem = -17 + A number = -19
The number = -19 + 17 = -2
∴ -2 should be added to -17 to get -19

Question 8.
A student was asked to subtract (-12) from -47. He got -30. Is he correct? Justify.
Solution:
Subtracting -12 from -47, we get
-47 – (-12) = -47 + (Additive inverse of-12)
= -47 + (+12) = -35
But the students answer is -30.
So he is not correct.

Objective Type Questions

Question 9.
(-5) – (-18)
(i) 23
(ii) -13
(iii) 13
(iv) -23
Solution:
(iii) 131

Question 10.
(-100) – 0 + 100 =
(i) 200
(ii) 0
(iii) 100
(iv)-200
Solution:
(ii) 0

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 1 Number System Ex 1.4

Question 1.
Fill in the blanks.
(i) (-40) ÷ ___ =40
(ii) 25 ÷ ____ = -5
(iii) ____ ÷ (-4) = 9
(iv) (-62) ÷ (-62) = ____
Solution:
(i) -1
(ii) -5
(iii) -36
(iv) 1

Question 2.
Say True or False:
(i) (-30) ÷ (-6) = -6
(ii) (-64) ÷ (-64) is 0
Solution:
(i) False
(ii) False

Question 3.
Find the values of the following.
(i) (-75) ÷ 5
(ii) (-100) ÷ (-20)
(iii) 45 ÷ (-9)
(iv) (-82) ÷ 82
Solution:
(i) \(\frac{-75}{5}\) = -15
(ii) \(\frac{-100}{-20}\) = 5
(iii) \(\frac{45}{-9}\) = -5
(iv) \(\frac{-82}{82}\) = -1

Question 4.
The product of two integers is -135. If one number is -15. Find the other integer.
Solution:
Given the product of two integers = -135
One of them = -15
∴ -15 × Another number = -135
Other number = \(\frac{-135}{-15}\) = 9
∴ The other number = 9.

Question 5.
In 8 hours duration, with uniform decrease in temperature, the temperature dropped 24°C. How many degrees did the temperature drop each hour?
Solution:
In 8 hours the drop in temperature = 24
In 1 hour the drop in temperature = \(\frac{24}{8}\) = 3°
The temperature dropped 3°C every hour.

Question 6.
An elevator descends into a mine shaft at the rate of 5 m/min. If the descent starts from 15 m above the ground level, how long will it take to reach -250 m?
Solution:
The elevator’s position = 15 m above ground level = +15 m
It should reach = -250 m
The distance to be travelled = 15 – (-250) m = 15 + (+250) m 265 m
Time taken to descend 5 m = 1 min
∴ Time required to descend 265 m = \(\frac{24}{8}\) = 53 min

Question 7.
A person lost 4800 calories in 30 days. If the calory loss is uniform, calculate the loss of calory per day.
Solution:
Loss of calory in 30 days = 4800
∴ Loss of calory in 1 day = \(\frac{4800}{30}\) = 160 calories
∴ 160 calories lost per day.

Question 8.
Given 168 × 32 = 5376 then fined (-5376) ÷ (-32).
Solution:
Given 168 × 32 = 5376
∴ \(\frac{5376}{32}\) = 168
Also, \(\frac{-5376}{-32}\) = 168

Question 9.
How many -4’s are there is (-20)?
Solution:
Number of -4’s in (-20) = \(\frac{-20}{-4}\) = 5

Question 10.
(-400) divided into 10 equal parts gives ____
Solution:
\(\frac{-400}{10}\) = -4

Objective Type Questions

Question 11.
Which of the following does not represent an integer?
(i) 0 ÷ (-7)
(ii) 20 ÷ (-4)
(iii) (-9) ÷ 3
(iv) 12 ÷ 5
Solution:
(iv) 12 ÷ 5

Question 12.
(-16) ÷ 4 is the same as
(i) -(-16 ÷ -4)
(ii) -(16) ÷ (-4)
(iii) 16 ÷ (-4)
(iv) -4 ÷ -16
Solution:
(iii) 16 ÷ (-4)

Question 13.
(-200) ÷ 10 is
(i) 20
(ii) -20
(iii) -190
(iv) 210
Solution:
(ii) 20

Question 14.
The set of integers is not closed under
(i) Addition
(ii) Subtraction
(iii) Multiplication
(iv) Division
Solution:
(iv) Division