Class 7

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.4

Students can Download Maths Chapter 2 Measurements 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 2 Chapter 2 Measurements Ex 2.4

Miscellaneous Practice Problems

Question 1.
A wheel of a car covers a distance of 3520 cm in 20 rotations. Find the radius of the wheel?
Solutions:
Distance covered by circular wheel in 20 rotation = 3520 cm
∴ Distance covered ini rotation = \(\frac { 3520 }{ 20 } \) cm = 176 cm
∴ Circumference of the wheel = 176 cm
∴ 2πr = 176
2 × \(\frac { -2 }{ 6 } \) × r = 176
r = \(\frac{176 \times 7}{2 \times 22}\)
r = 28 cm
Radius of the wheel = 28 cm

Question 2.
The cost of fencing a circular race course at the rate of ₹ 8 per metre is ₹2112. Find the diameter of the race course.
Solution:
Cost of fencing the circumference = ₹ 2112
Cost of fencing one meter = ₹ 8
∴ Circumference of the circle = \(\frac { 2112 }{ 8 } \) = 264 m
πd = 264 m
\(\frac { 22 }{ 7 } \) × d = 264
d = \(\frac{264 \times 7}{22}\) = 12 × 7 m = 84 m
∴ Diameter of the race cource = 84 m

Question 3.
A path 2 m long and 1 m broad is constructed around a rectangular ground of dimensions 120 m and 90 m respectively. Find the area of the path.
Solution:
Length of the rectangular ground l = 120 m
Breadth b = 90 m
Length of the path W1 = 2m
Length of the path W2 = 1m
Length of the ground with path L = 1 + 2 (W2) = 120 + 2(1) m
= 120 + 2 = 122 m
Breadth of the ground with path B = l + 2(W1) units
= 90 + 2(2) m = 90 + 4 m = 94 m
∴ Area of the path = (L × B) – (1 × b) sq. units
= (122 × 94) – (122 × 94) m2 = 668 m2
∴ Area of the path = 668 m2

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.4

Question 4.
The cost of decorating the circumference of a circular lawn of a house at the rate of ₹55 per metre is ₹16940. What is the radius of the lawn?
Solution:
Cost of decorating the circumference = ₹ 16,940
Cost of decorating per meter = ₹ 55
∴ Length of the circumference = \(\frac { 16940 }{ 55 } \) m = 308 m
Circumference of the circular lawn = 308 m
2 × πr = 308 m
2 × \(\frac { 22 }{ 7 } \) × r = 308 m
r = \(\frac{308 \times 7}{2 \times 22}\)
r = 49 m
Radius of the lawn = 49 m

Question 5.
Four circles are drawn side by side in a line and enclosed by a rectangle as shown below.
If the radius of each of the circles is 3 cm, then calculate:
(i) The area of the rectangle.
(ii) The area of each circle.
(iii) The shaded area inside the rectangle.
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System 2.4 1
Solution:
Given radius of a circle r = 3 cm
Diameter of the circle = 2r = 2 × 3 = 6 cm
Breadth of the rectangle = Diamter of the circle
B = 6cm
Length of the rectangle L = 4 × diameter of a circle
L = 4 × 6
L = 24cm

(i) Area of the rectangle = L × B sq. units
= 24 × 6 cm2
Area of the rectangle = 144 cm2

(ii) Area of the circle = πr2 sq. units
= \(\frac { 22 }{ 7 } \) × 3 × 3 cm2
= \(\frac { 198 }{ 7 } \) cm2
= 28.28 cm2

(iii) Area of the shaded area = Area of the rectangle – Area of the 4 circles
= 144 – (4 × \(\frac { 198 }{ 7 } \)) cm2 = 144 – \(\frac { 792 }{ 7 } \) cm2
= 144 – 113.14 cm2 = 30.85 cm2

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.4

Challenge Problems

Question 6.
A circular path has to be constructed around a circular lawn. If the outer and inner circumferences of the path are 88 cm and 44 cm respectively, find the width and area of the path.
Solution:
Outer circumference of the circular lawn = 88 cm
2πR = 88 cm
Inner circumference of the lawn 2πr = 44 cm
2πR – 2πr = 88 – 44
2 × \(\frac { 22 }{ 7 } \) (R – r) = 44
(R – r) = \(\frac{44 \times 7}{2 \times 22}\)
Outer radius – Inner radius = 7 cm
∴ Width of the lawn = 7 cm
Also 2πR + 2πr = 88 + 44
2π (R + r) = 132
π (R + r) = \(\frac { 132 }{ 2 } \) = 66 cm
Area of the path = πR2 – πr2 sq. units
= π (R + r) (R – r) = 66 × 7
Area of the path = 462cm2

Question 7.
A cow is tethered with a rope of length 35 m at the centre of the rectangular field of length 76 m and breadth 60 m. Find the area of the land that the cow cannot graze?
Solution:
Length of the field l = 76 m
Breadth of the field b = 60m
Area of the field A = l × b sq. units = 76 × 60 m2
Area of the field A = 4560 m2
Length of the rope = 35m
Radius of the land that the cow can graze = 35m
Area of the land tha the cow can graze = circle of radius 35 m = πr2 sq.units
π × 35 × 35 m2 = \(\frac { 22 }{ 7 } \) × 35 × 35 m2
= 3850 m2
Area of the land the cow cannot graze = Area of the field – Area that the cow can graze
= 4560 – 3860 m2 = 710 m2
Area of the land that the cow cannot graze = 710 m2

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.4

Question 8.
A path 5 m wide runs along the inside of the rectangular field. The length of the rectangular field is three times the breadth.of the field. If the area of the path is 500 m2 then find the length and breadth of the field.
Solution:
Let the length of the rectangular field = ‘L’ m
Breadth of the rectangular field = = ‘B’ m
Area of the rectangular field = (L × B) m2
Also given length = 3 × Breadth
L = 3B
Width of the path (W) = 5m
Lenth of the inner rectangle = L – 2W = l – 2(5)
= 3B – 10m
Breadth of the inner rectangle = B – 2W
= B – 2(5)
= B – 10 m
Area of the inner rectangle = (3B – 10) (B – 10)
= 3B2 – 10B – 30B + 100
Area of the path = Area of outer rectangle
– Area of inner rectangle
= (L × B) – (3B2 – 10B – 30B + 100)
3B × B – (3B2 – 40B + 100)
= 3B2 – 3B2 + 40B – 100
Area of the path = 40B – 100
Given area of the path = 500 m2
40B – 100 = 500
40B = 500 + 100 = 600
B = \(\frac { 600 }{ 40 } \)
B = 15m
Length of the field = 45 m; Breadth of the field = 15 m

Question 9.
A circular path has to be constructed around a circular ground. 1f the areas of the outer and inner circles are 1386 m2 and 616 m2 respectively, find the width and area of the path.
Solution:
Area of the outer circle = 1386 m2
πR2 = 1386m2
Area of the inner circle = 616 m2
πr2 = 616m2
Area of the path = Area of outer circle – Area of the inner circle
1386 m2 – 616 m2
Area of the path = 770m2
Also πR2 = 1386
R2 = \(\frac{1386 \times 7}{22}\)
R2 = 63 × 7
R2 = 9 × 7 × 7
R2 = 32 × 72
R = 3 × 7
Outer Radius R = 21 m
Again πr2 = 616
\(\frac { 22 }{ 7 } \) × r2 = 616
r2 = 28 × 7
r2 = 4 × 7 × 7
r2 = 22 × 72
r = 2 × 7
Inner radius r = 14m
Width of the path = Outer radius – Inner radius = 21 – 14
Width of the path = 7m

Question 10.
A goat is tethered with a rope of length 45 m at the centre of the circular grass land whose radius is 52 m. Find the area of the grass land that the goat cannot graze.
Solution:
Length of the rope = 45 m = Radius of the inner circle
∴ Area of the circular area that the goat graze = πr2 sq. units
= \(\frac { 22 }{ 7 } \) × 45 × 45 m2 = 6364.28 m2
Radius of the gross land = 52 m
Area of the grass land = \(\frac { 22 }{ 7 } \) × 52 × 52 = 8,498.28 m2
Area that the goat cannot graze
= Area of the outer circle – Area of the inner circle
= 8498.28 – 6364.28 = 2134 m2
Area of the goat cannot grass = 2134 m2

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.4

Question 11.
A strip of 4 cm wide is cut and removed from all the sides of the rectangular cardboard with dimensions 30 cm × 20 cm. Find the area of the removed portion and area of the remaining cardboard.
Solution:
Area of the outer rectangular cardboard
= L × B sq.units = 30 × 20 cm2 = 600 cm2
Width of the stip = 4 cm
Length of the inner rectangle = L – 2W
l = 30 – 2(4) = 30 – 8
l = 22cm
Breadth of the inner rectangle B = 2W = 20 – 2(4) = 20 – 8
b = 12cm
Area of the inner rectangle = l × b sq.units = 22 × 12 cm2 = 264 cm2
Area of the remaining cardboard = 264 cm2
Area of the removed portion = Area of outer rectangle
– Area of the inner rectangle
= 600 – 264 cm2
Area of the removed portion = 336 cm2

Question 12.
A rectangular field is of dimension 20 m × 15 m. Two paths run parallel to the sides of the rectangle through the centre of the field. The width of the longer path is 2m and that of the shorter path is 1 m. Find (i) the area of the paths (ii) the area of the remaining portion of the field (iii) the cost of constructing the roads at the rate of ₹ 10 per sq.m.
Solution:
Length of the rectangular field L = 20 m
Breadth B = 15m
Area = L × B
20 × 15 m2
Area of outer rectangle = 300 m2
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System 2.4 2
Area of inner small rectangle = \(\frac { 19 }{ 2 } \) × \(\frac { 13 }{ 2 } \) = 61.75 cm2

(i) Area of the path = Area of the outer rectangle
– Area of 4 inner small rectangles
= 300 – 4(61.75) = 300 – 247 = 53 m2
Area of the paths = 53 m2

(ii) Area of the remaining portion of the field
= Area of the outer rectangle – Area of the paths
= 300 – 53 m2 = 247 m2
Area of the remaining portion = 247 m2

(iii) Cost of constructing 1 m2 road = ₹10
∴ Cost of constructing 53 m2 road = ₹10 × 53 = ₹530
∴ Cost of constructing road = ₹530

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Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.4

Students can Download Maths Chapter 1 Number System Ex 1.4 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 2 Chapter 1 Number System Ex 1.4

Question 1.
Write the decimal numbers represented by the points P, Q, R and S on the given number line.
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.4 1
Solution:
The unit length between 1 and 2 is divided into 10 equal parts and the third part is taken as Q.
∴ Q represents 1 + 0.3 = 1.3
The unit lenth between 3 and 4 is divided into 10 equal parts and the 6th part is taken as P.
∴ P represents 3 + 0.6 = 3.6
The unit length between 4 and 5 is divied into 10 equal parts and the second part is taken as S.
∴ S represents 4 + 0.2 = 4.2
The unit length between 6 and 7 is divided into 10 equal parts and the 8th part is taken.
∴ R represents 6 + 0.8 = 6.8
P(3.6), Q(1.3), R(6.8), S(4.2).

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.4

Question 2.
Represent the following decimal numbers on the number line.
(i) 1.7
(ii) 0.3
(iii) 2.1
Solution:
(i) 1.7
We know that 1.7 is more than 1, but less than 2.
There are one ones and 7 tenths in it. Divide the unit length between 1 and 2 on the number line into 10 equal parts and take 7 parts which represents 1.7 = 1 + 0.7
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.4 2

(ii) 0.3
We know that 0.3 is more than 0, but less than 1.
There are 3 tenths in it. Divide the unit lenght between 0 and 1 on the number line into 10 equal parts and take 3 parts, which represent 0.3.
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.4 3

(iii) 2.1
We knowthat 2.1 is more than 2 and less than 3.
There are 2 ones and 1 tenths in it.
Divide the unit length between 2 and 3 into 10 equal parts and take 1 part, which represemt 2.1 = 2 + 0.1
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.4 4

Question 3.
Between which two whole numbers, the following decimal numbers lie?
(i) 3.3
(ii) 2.5
(iii) 0.9
Solution:
(i) 3.3 – 3.3 lies between 3 and 4.
(ii) 2.5 – 2.5 lies between 2 and 3.
(iii) 0.9 – 0.9 lies between 0 and 1.

Question 4.
Find the greater decimal number in the following.
(i) 2.3,3.2
(ii) 5.6,6.5
(iii) 1.2,2.1
Solution:
(i) 2.3, 3.2
Comparing the whole number parts of 2.3 and 3.2 we get 3 > 2.
3.2 > 2.3 – Greater number is 3.2

(ii) 5.6, 6.5
Comparing the whole number parts of 5.6 and 6.5, we get 6 > 5.
6.5 > 5.6 – Greater number is 6.5

(iii) 1.2, 2.1
Comparing the whole number parts of 1.2 and 2.1, we get 2 > 1.
2.1 > 1.2 – Greater number is 2.1

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.4

Question 5.
Find the smaller decimal number in the following.
(i) 25.3,25.03
(ii) 7.01,7.3
(iii) 5.6,6.05
Solution:
(i) 25.3, 25.03
The whole number parts of both the numbers are equal.
∴ Comparing the digits at tenths place we get 0 < 3.
∴ 25.03 < 25.3 – Smaller number 25.03

(ii) 7.01,7.3
The whole number parts of both the numbers are equal.
Comparing the digits at tenths place we get 0 < 3.
∴ 7.01 < 7.3 – Smaller number is 7.01.

(iii) 5.6, 6.05
Comparing the whole number parts, we get 5 < 6.
∴ 5.6 < 6.05 – Smaller number is 5.6

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.4

Objective Question

Question 6.
Between which two whole numbers 1.7 lie?
(i) 2 and 3
(ii) 3 and 4
(iii) 1 and 2
Answer:
(iii) 1 and 2

Question 7.
The decimal number which lies between 4 and 5 is _______
(i) 4.5
(ii) 2.9
(iii) 1.9
Answer:
(i) 4.5

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Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2

Students can Download Maths Chapter 2 Measurements Ex 2.2 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 2 Chapter 2 Measurements Ex 2.2

Question 1.
Find the area of the dining table whose diameter is 105 cm.
Solution:
Diameter of the dinig table (d) = 105 cm
∴ Radius r = \(\frac { d }{ 2 } \) = \(\frac { 105 }{ 2 } \) cm
Area of the circle = π r2 = \(\frac { 22 }{ 7 } \) × \(\frac { 105 }{ 2 } \) × \(\frac { 105 }{ 2 } \) = 8662.5 sq.cm
Area of the dinning table = 8662.5 cm2

Question 2.
Calculate the area of the shotput ring whose diameter is 2.135 m.
Solution:
Radius of the shotput ring r = \(\frac { d }{ 2 } \) = \(\frac { 2.135 }{ 2 } \) m
Area of the circle = π r2
= \(\frac { 22 }{ 7 } \) × \(\frac { 2.135 }{ 2 } \) × \(\frac { 2.135 }{ 2 } \)
= \(\frac { 25.07 }{ 7 } \) = 3.581 m2
∴ Area of the shotput ring = 3.581 m2

Question 3.
A sprinkler placed at the centre of a flower garden sprays water covering a circular area. If the area watered is 1386 cm2, find its radius and diameter.
Solution:
Area of the Circle = π r2 sq.units
Area of the circular portion watered = 1386 cm2
π r2 = 1386
\(\frac { 22 }{ 7 } \) × r2 = 1386
r2 = 1386 × \(\frac { 7 }{ 22 } \) = 63 × 7 = 9 × 7 × 7
r2 = 32 × 72
r = 3 × 7
Radius (r) = 21 cm
Diameter (d) = 2 r = 2 × 21 cm
Diameter (d) = 42 cm

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2

Question 4.
The circumference of a circular park is 352 m. Find the area of the park.
Solution:
Circumference of a Circle = 2 π r units
Given circumference of a circular park = 352 m
2 π r = 352
2 × \(\frac { 22 }{ 7 } \) × r = 352
r = 352 × \(\frac { 7 }{ 22 } \) × \(\frac { 1 }{ 2 } \) = 56 m
Area of the park = π r2 = \(\frac { 22 }{ 7 } \) × 56 × 56 sq.units
= 22 × 8 × 56 = 9856 m2
∴ Area of the Circular park = 9856 m2

Question 5.
In a grass land, a sheep is tethered by a rope of length 4.9 m. Find the maximum area that the sheep can graze.
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System 2.2 1
Solution:
Length of the rope = 4.9 m
Area that the sheep can graze = Area of circle with radius 4.9m
Area of the circle = π r2 sq.units
= \(\frac { 22 }{ 7 } \) × 4.9 × 4.9 = 22 × 0.7 × 4.9 = 75.46
∴ Area that the sheep can graze = 75.46 m2

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2

Question 6.
Find the length of the rope by which a bull must be tethered in order that it may be able to graze an area of 2464 m2.
Solution:
If the bull is tethered by a rope then the area it can graze is a circular area of radius
= length of the rope
Area of the circle = 2464 m2
π r2 = 2464 m2
\(\frac { 22 }{ 7 } \) × r2 = 2464
r2 = 2464 × \(\frac { 7 }{ 22 } \) = 122 × 7 = 16 × 7 × 7
r2 = 42 × 72
r = 4 × 7 = 28 m
length of the rope r = 28 m

Question 7.
Lalitha wants to buy a round carpet of radius is 63 cm for her hall. Find the area that will be covered by the carpet.
Solution:
Radius of the round carpet = 63 cm
Area covered by the round carpet = πr2 sq units
A = \(\frac { 22 }{ 7 } \) × 63 × 63 = 22 × 9 × 63 = 12474 cm2
Area covered by the round carpet = 12,474 cm2

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2

Question 8.
Thenmozhi wants to level her circular flower garden whose diameter is 49 m at the rate of ₹150 per m2 Find the cost of levelling.
Solution:
Diamter of the circular garden d = 49 m
Radius r = \(\frac { d }{ 2 } \) = \(\frac { 49 }{ 2 } \) m
Area of the circular garden = πr2 sq units
= \(\frac { 22 }{ 7 } \) × \(\frac { 49 }{ 2 } \) × \(\frac { 49 }{ 2 } \) m2 = 1,886.5 m2
Cost of levelling a m2 area = ₹ 150
∴ Cost of levelling 1886.5 m2 = ₹ 150 × 1886.5 = ₹ 2,82,975
Cost of levelling the flower garden = ₹ 2,82,975

Question 9.
The floor of the circular swimming pool whose radius is 7 m has to be cemented at the rate of ₹ 18 per m2. Find the total cost of cementing the floor.
Solution:
Radius of the circular swimming pool r = 7 m
Area of the circular swimming pool A = πr2 sq. units
= \(\frac { 22 }{ 7 } \) × 7 × 7 m2 = 154 m2
Cost of cementing a m2 floor = ₹ 18.
Cost of cementing 154 m2 floor = ₹ 18 × 154 = ₹ 2,772

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2

Objective Type Questions

Question 10.
The formula used to find the area of the circle is
(i) 47πr2
(ii) πr2
(iii) 2πr2
(iv) πr2 + 2r
Answer:
(ii) πr2

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2

Question 11.
The ratio of the area of a circle to the area of its semicircle is
(i) 2 : 1
(ii) 1 : 2
(iii) 4 : 1
(iv) 1 : 4
Answer:
(i) 2 : 1

Question 12.
Area of circle of radius ‘n’ units is
(i) 2πrp sq.units
(ii) πm2 sq.units
(iii) πr2 sq.units
(iv) πn2 sq.units
Answer:
(iv) πn2 sq.units

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Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.1

Students can Download Maths Chapter 2 Measurements Ex 2.1 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 2 Chapter 2 Measurements Ex 2.1

Question 1.
Find the missing values in the following table for the circles with radius (r), diameter (d) and Circumference (C).
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System 2.1 1
Solution:
(i) Given radius r = 15cm
∴ diameter d = 2 × 15 = 30 cm
Circumference C = π d units
= \(\frac { 22 }{ 7 } \) × 30 = \(\frac { 660 }{ 7 } \) = 94.28 cm

(ii) Given circumference C = 1760 cm
2πr = 1760
2 × \(\frac { 22 }{ 7 } \) × r = 1760
r = \(\frac{1760 \times 7}{2 \times 22}\) = \(\frac{160 \times 7}{2 \times 2}\) = 40 × 7 = 280 cm
diameter = 2 × r
= 2 × 280 = 560 cm

(iii) diameter d = 24m
radius r = \(\frac { d }{ 2 } \) = \(\frac { 24 }{ 2 } \) = 12 m
Circumference C = 2 π r units
= 2 × \(\frac { 22 }{ 7 } \) × 12 = \(\frac { 528 }{ 7 } \) = 75.4 m

Tabulating the results
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System 2.1 2

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.1

Question 2.
Diameters of different circles are given below. Find their circumference (Take π = \(\frac { 22 }{ 7 } \) )
(i) d = 70cm
(ii) d = 56m
(iii) d = 28mm
Solution:
(i) Diameter d = 70 cm
Circumference C = π d units = \(\frac { 22 }{ 7 } \) × 70 = 22 × 10 = 220 cm

(ii) Diameter d = 56 m
Circumference = π d units
= \(\frac { 22 }{ 7 } \) × 56 = 22 × 8 = 176 m

(iii) Diameter d = 28 mm
Circumference C = π d units = \(\frac { 22 }{ 7 } \) × 28 = 22 × 4 = 88 mm

Question 3.
Find the circumference of the circles whose radii are given below.
(i) 49 cm
(ii) 91 mm
Solution:
Radius r = 49 cm
Circumference C = 2 π r units = 2 × \(\frac { 22 }{ 7 } \) × 49 = 2 × 22 × 7
= 44 × 7 = 308 cm

(ii) Radius r = 91 mm
Circumference C = 2 π r units
= 2 × \(\frac { 22 }{ 7 } \) × 91 = 2 × 22 × 13 = 44 × 13 = 572 mm

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.1

Question 4.
The diameter of a circular well is 4.2 m. What is its circumference?
Solution:
Given the diameter d = 4.2 m
Circumference C = π d units = \(\frac { 22 }{ 7 } \) × 4.2 m = 22 × 0.6 = 13.2 m

Question 5.
The diameter of the bullock cart wheel is 1.4 m. Find the distance covered by it in 150 rotations?
Solution:
Diameter of the bullock cart wheel d= 1.4 m
Distance covered in 1 rotation = Its circumference
= π d units = \(\frac { 22 }{ 7 } \) × 1 .4 m = 22 × 0.2 = 4.4 m
Distance covered in one rotation = 4.4 m
Distance covered in 150 rotations = 4.4 × 150 = 660.0
Distance covered in 150 rotations = 660 m

Question 6.
A ground is in the form of a circle whose diameter is 350 m. An athlete makes 4 revolutions. Find the distance covered by the athlete.
Solution:
Diameter of the ground d = 350 m
Distance covered in 1 revolution = Circumference of the circle
= π d units = \(\frac { 22 }{ 7 } \) × 350 m = 22 × 50 = 1100 m
Distance covered in 1 rotation = 1100 m
Distance covered in 4 revolutions = 1100 × 4 = 4400 m

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.1

Question 7.
A wire of length 1320 cm is made into circular frames of radius 7 cm each. How many frames can be made?
Solution:
Length of the wire = 1320 cm
Radius of each circular frame = 7cm
Circumference of the frame 2 π r units = 2 × \(\frac { 22 }{ 7 } \) × 7 cm = 2 × 22 = 44 cm
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System 2.1 3
30 frames can be made.

Question 8.
A Rose garden is in the form of circle of radius 63 m. The gardener wants to fence it at the rate of ₹ 150 per metre. Find the cost of fencing?
Solution:
Radius of the garden r = 63 m
Circumference of the garden = 2 π r units = 2 × \(\frac { 22 }{ 7 } \) × 63 m = 2 × 22 × 9 = 396 m
Cost of fencing 1 meter = ₹ 150
Cost of fencing 396 meter = ₹ 150 × 396 = ₹ 59,400
∴ Cost of fencing the garden = ₹ 59,400

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.1

Objective Type Questions

Question 9.
Formula used to find the circumference of a circle is
(i) 2πr units
(ii) πr2 + 2r units
(iii) πr2 sq. units
(iv) πr3 cu. units
Answer:
(i) 2πr units

Question 10.
In the formula, C = 2πr, ‘r’ refers to
(i) circumference
(ii) area
(iii) rotation
(iv) radius
Answer:
(iv) radius

Question 11.
If the circumference of a circle is 82π, then the value of ‘r’ is
(i) 41cm
(ii) 82 cm
(iii) 21cm
(iv) 20 cm
Answer:
(i) 41cm

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.1

Question 12.
Circumference of a circle is always
(i) three times of its diameter
(ii) more than three times of its diameter
(iii) less than three times of its diameter
(iv) three times of its radius
Answer:
(ii) more than three times of its diameter

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Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.3

Students can Download Maths Chapter 1 Number System Ex 1.3 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 2 Chapter 1 Number System Ex 1.3

Question 1.
Compare the following decimal numbers and find out the smaller number.
(i) 2.08,2.086
(ii) 0.99,1.9
(iii) 3.53,3.35
(iv) 5.05,5.50
(v) 123.5,12.35
Solution:
(i) 2.08, 2.086
Let us take 2.080, 2.086.
Comparing 2.08 and 2.086 the whole number part, tenths place digit and the digit in the hundredths place are equal.
Comparing the digits at thousandths place we get 0 < 6.
Therefore 2.08 < 2.086.
Smallest number is 2.08

(ii) 0.99, 1.9
Comparing 0.99 and 1.9.
First when we compare the digit in the whole number parts we get 0 < 1.
∴ 0.99 < 1.9 Smallest number is 0.99

(iii) 3.53,3.35
Comparing 3.53 and 3.35
Here the whole number parts of the given two numbers are equal.
Comparing the digits at tenths place, we get 3 < 5
∴ 3.35 < 3.53
Smallest number is 3.35

(iv) 5.05, 5.50
Comparing 5.05 and 5.50
Here the whole number parts of the given two numbers are equal.
Comparing the digits at tenths place, we get 0 < 5.
∴ 5.5 < 5.50
Smallest number is 5.05

(v) 123.5, 12.35
Comparing 123.5 and 12.35.
Comparing the whole number parts, we get 12 < 123
∴ 12.35 < 123.5
Smallest number is 12.35

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.3

Question 2.
Arrange the following in ascending order.
(i) 2.35, 2.53, 5.32, 3.52, 3.25
(ii) 123.45, 123.54,125.43, 125.34,125.3
Solution:
(i) 2.35, 2.53, 5.32, 3.52, 3.25
Comparing the whole number parts of all the numbers 5 is the greatest and 5 > 3 > 2.
∴ Greatest number is 5.32
Next 3.52 and 3.25 are equal in their whole number.
So comparing their digits in tenths place, we get 5 > 2
So 3.52 > 3.25
Now comparing 2.35 and 2.53 their whole number parts also equal.
∴ Comparing the digit in tenths place we get
2.53 > 2.35 ……(2)
Ascending order :
2.35 < 2.53 < 3.25 < 3.52 < 5.32

(ii) 123.45, 123.54, 125.43, 125.34, 125.3
Comparing the whole number parts we have 123 is the smallest number and two numbers 123.45 and 123.54 have same whole number part.
So in 123.45 and 123.54 comparing their digits in the tenths place we get 4 < 5
∴ 123.45 < 123.54 …(1)
Now comparing the remaining numbers
125.43, 125.34, 125.3 they all have the same whole number part.
Comparing the numbers in the tenths place we have 3 < 4
∴ 125.43 is the greatest …(2)
Also tenths place value 3 = 3 in 125.34 and 125.3
Again comparing the hundredths place value in 125.34 and 125.3, we get
125.3 < 125.34 …(3)
From (1), (2) and (3) we have,
123.45 < 123.54 < 125.3 < 125.34 < 125.43

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.3

Question 3.
Compare the following decimal numbers and find the greater number.
(i) 24.5,20.32
(ii) 6.95,6.59
(iii) 17.3,17.8
(iv) 235.42,235.48
(v) 0.007,0.07
(vi) 4.571,4.578
Solution:
(i) 24.5, 20.32
Comparing the whole number part we get 24 > 20
∴ 24.5 > 20.32
greater number is 24.5

(ii) 6.95,6.59
Here the whole number part of given two numbers are equal.
Comparing the digits at tenths place we get 9 > 5.
∴ 6.95 > 6.59
Greater number is 6.95

(iii) 17.3,17.8
Here the whole number part of given two numbers are equal.
Comparing the digits at tenths place we get 8 > 3.
∴ 17.8 > 17.3
Greater number is 17.8

(iv) 235.42,235.48
Here the whole number part of given two numbers are equal.
Also the digits at tenths place also equal.
Comparing the digits at the hundredths place we get 8 > 4.
∴ 235.48 > 235.42
Greater number is 235.48

(v) 0.007,0.07
Here the whole number part of given two numbers are equal.
Also the digits at the tenths place also equal.
∴ Comparing the the digits at the hundredths place we get 7 > 0.
0. 07 > 0.007
greater number is 0.07.

(vi) 4.571,4.578
Here the whole number part of given two numbers are equal.
Also the digits at the tenths place and the hundredths place also equal.
Again comparing the digits in the thousandths place we get 8 > 1.
∴ 4.578 > 4.571
∴ Greater number is 4.578

Question 4.
Arrange the given decimal numbers in descending order.
(i) 17.35, 71.53, 51.73, 73.51,37.51
(ii) 456.73, 546.37, 563.47, 745.63 457.71
Solution:
(i) 17.35,71.53,51.73,73.51,37.51
Comparing the whole number parts of the given numbers we get
73 > 71 > 51 > 37 > 17.
Descending order:
73.51,71.53,51.73,37.51, 17.35

(ii) 456.73, 546.37, 563.47, 745.63 457.71
Comparing the whole number parts of the given numbers from left to right we get
745 > 563 > 546 > 457 > 456
Descending Order: 745.63, 563.47, 546.37, 457.71, 456.73

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.3

Objective Question

Question 5.
0.009 is equal to
(i) 0.90
(ii) 0.090
(iii) 0.00900
(iv) 0.900
Answer:
(iii) 0.00900

Question 6.
37.70 [ ] 37.7
(i) =
(ii) <
(iii) >
(iv) ≠
Answer:
(i) =

Question 7.
78.56 [ ] 78.57
(i) <
(ii) >
(iii) =
(iv) ≠
Answer:
(i) <

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Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions

Students can Download Maths Chapter 1 Number System 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 2 Chapter 1 Number System Intext Questions

Exercise 1.1
(Try These Text book Page No. 2)

Question 1.
Observe the following and write the fraction of the shaded portion and mention in decimal form also.
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 1
Solution:
(i) Total parts = 8
Shaded parts = 4
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 2

(ii)
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 3

(iii)
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 5

Question 2.
Represent the following fractions in decimal form by converting denominator into ten or powers of 10.
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 4
Solution:
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 6

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions

Question 3.
Give any two life situations where we use decimal numbers.
Solution:
(i) Measuring weight of gold.
(ii) Weighing our height

(Try These Text book Page No. 3)

Question 1.
Represent the following decimal numbers pictorially.
(i) 5 ones and 3 tenths
(ii) 6 tenths
(iii) 7 ones and 9 tenths
(iv) 6 ones and 4 tenths
(v) Seven tenths
Solution:
(i) 5 ones and 3 tenths
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 21
(ii) 6 tenths
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 8
(iii) 7 ones and 9 tenths
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 9
(iv) 6 ones and 4 tenths
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 10
(v) Seven tenths
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 11

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions

(Try These Text book Page No. 5 & 6)

Question 1.
Express the following decimal numbers in an expanded form and place value grid form.
(i) 56.78
(ii) 123.32
(iii) 354.56
Solution:
(i) 56.78
(a) Expanded form
56.78 = 5 × 101 + 6 × 100 + 7 × 10-1 + 8 × 10-2

(b) Place value grid
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 12

(ii) 123.32
(a) Expanded form
123.32 = 1 × 102 + 2 × 101 + 3 × 100 + 3 × 10-1 +2 × 10-2

(b) Place value grid
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 13

(iii) 354.56
(a) Expande form
354.56 = 3 × 102 + 5 × 101 + 4 × 100+ 5 × 10-1 + 6 × 10-2

(b) Place value grid
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 14

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions

Question 2.
Express the following measurements in terms of metre and in decimal form. One is done for you.
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 22
Solution:
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 16

Question 3.
Write the following numbers in the place value grid and find the place value of the underlined digits.
(i) 36.37
(ii) 267.06
(iii) 0.23
(iv) 27.69
(v) 53.27
Solution:
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 17
(i) Place value of 3 in 36.37 is Tenths.
(ii) Place value of 6 in 267.06 is Hundredths.
(iii) Place value of 2 in 0.23 is Tenths.
(iv) Place value of 9 in 27.69 is Hundredths.
(v) Place value of 2 in 53.27 is Tenths.

Exercise 1.2

(Try These Text book Page No. 10)

Question 1.
Convert the following fractions into the decimal numbers.
(i) \(\frac { 16 }{ 1000 } \)
(ii) \(\frac { 638 }{ 10 } \)
(iii) \(\frac { 1 }{ 20 } \)
(iv) \(\frac { 3 }{ 50 } \)
Solution:
(i) \(\frac { 16 }{ 1000 } \) = 0.016
(ii) \(\frac { 638 }{ 10 } \) = 63.8
(iii) \(\frac { 1 }{ 20 } \) = \(\frac{1 \times 5}{20 \times 5}\) = \(\frac { 5 }{ 100 } \) = 0.05
(iv) \(\frac { 3 }{ 50 } \) = \(\frac{3 \times 2}{50 \times 2}\) = \(\frac { 6 }{ 100 } \) = 0.06

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions

Question 2.
Write the fraction for each of the following:
(i) 6 hundreds + 3 tens + 3 ones + 6 hundredths + 3 thousandths.
(ii) 3 thousands + 3 hundreds + 4 tens + 9 ones + 6 tenths.
Solution:
(i) 6 hundreds + 3 tens + 3 ones + 6 hundreds + 3 thousandths.
= 6 × 100 + 3 × 10 + 3 × 1 + 0 × \(\frac { 1 }{ 10 } \) + 6 × \(\frac { 1 }{ 100 } \) + 3 × \(\frac { 1 }{ 1000 } \)
= 600 + 30 + 3 + 0 + \(\frac { 6 }{ 100 } \) + \(\frac { 3 }{ 1000 } \)
= 633 + 0.06 + 0.003
= 633.063

(ii) 3 thousands + 3 hundreds + 4 tens + 9 ones + 6 tenths.
= 3 × 1000 + 3 × 100 + 4 × 10 + 9 × 1 + 6 × \(\frac { 1 }{ 10 } \)
= 3000 + 300 + 40 + 9 + \(\frac { 6 }{ 10 } \)
= 3349 + 0.6
= 3349.6

Question 3.
Convert the following decimals into fractions.
(i) 0.0005
(ii) 6.24
Solution:
(i) 0.0005 = \(\frac { 5 }{ 10000 } \) = \(\frac{5 \div 5}{10000 \div 5}\) = \(\frac { 1 }{ 2000 } \)
(ii) 6.24 = \(\frac { 624 }{ 100 } \) = \(\frac{624 \div 4}{100 \div 4}\) = \(\frac { 156 }{ 25 } \)

Exercise 1.4

(Try These Text book Page No. 22)

Question 1.
Mark the following decimal numbers on the number line.
(i) 0.3
(ii) 1.7
(iii) 2.3
Solution:
(i) 0.3
We know that 0.3 is more than 0, but less than 1.
There are 3 tenths in it. Divide the unit lenght between O and 1 on the number line
into 10 equal parts and take 3 parts, which represent 0.3.
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 18

(ii) 1.7
We know that 1.7 is more than 1, but less than 2.
There are one ones and 7 tenths in it. Divide the unit length between 1 and 2 on the number line into 10 equal parts and take 7 parts which represents 1.7 = 1 + 0.7
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 19

(iii) We know that 2.3 is more than 2 and less than 3.
There are 2 ones and 3 tenths in it. Divide the unit length between 2 and 3 into 10 equal parts and take 3 parts, which represents 2.3 = 2 + 0.3
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions 20

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions

Question 2.
Identify any two decimal numbers between 2 and 3.
Solution:
2.5 and 2.9

Question 3.
Write any decimal number which is greater than 1 and less than 2.
Solution:
1.7, 1.9, 1.6, ………………..

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Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions

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 1 Chapter 3 Algebra Additional Questions

Additional Questions and Answers

Exercise 3.1

Question 1.
Write any three expressions each having 4 terms:
Solution:
(i) 2x3 – 3x2 + 3xy + 8
(ii) 7x3 + 9y2 – 2xy2 – 6
(iii) 9x2 – 2x + 3xy – 1

Question 2.
Identify the co-efficients of the terms of the following expressions
(i) 2x – 2y
(ii) x + y +3
Solution:
(i) 2x – 2y
The co-efficient of x in 2x is 2
The co-efficient of y in – 2y is – 2

(ii) x + y + 3
The co-efficient of x is 1
The co-efficient ofy is 1
The constant term is 3

SamacheerKalvi.Guru

Question 3.
Group the like terms together from the following: 6x, 6, -5x, – 5, 1, x, 6y, y, 7y, 16x, 3
Solution:
We have 6x, -5x, x, 16x are like terms
6y, y, 7y, are like terms
6, – 5, 1, 3 are like terms

Question 4.
Give the algebraic expressions for the following cases:
(i) One half of the sum of a and b.
(ii) Numbers p and q both squared and added
Solution:
(i) \(\frac{1}{2}\) (a + b)
(ii) p2 + q2

Exercise 3.2

Question 1.
If A = 2a2 – 4b – 1 ; B = 5a2 + 3b – 8 and C = 2a2 – 9b + 3 then find the value of A – B + C.
Solution:
Given A = 2a2 – 4b – 1 ; B = 5a2 + 3b – 8 ; C = 2a2 – 9b + 3
A – B + C = (2a2 – 4b – 1) – (5a2 + 3b – 8) + (2a2 – 9b + 3)
= 2a2 – 4b – 1 + (-5a2 – 3b + 8) + 2a2 – 9b + 3
= 2a2 – 4b – 1 – 5a2 – 3b + 8 + 2a2 – 9b + 3
= 2a2 – 5a2 + 2a2 – 4b – 3b – 9b – 1 + 8 + 3
= (2 – 5 + 2) a2 + (-4 – 3 – 9) 6 + (-1 + 8 + 3)
= -a2 – 16b + 10

Question 2.
How much 2x3 – 2x2 + 3x + 5 is greater than 2x3 + 7x2 – 2x + 7?
Solution:
The required expression can be obtained as follows.
= 2x3 – 2x2 + 3x + 5 – (2x3 + 7x2 – 2x + 7)
= 2x3 – 2x2 + 3x + 5 + (-2x3 – 7x2 + 2x – 7)
= 2x3– 2x2 + 3x + 5 – 2x3 – 7x2 + 2x – 7
= (2 – 2) x3 + (-2 – 7) x2 + (3 + 2) x + (5 – 7)
= 0x3 + (-9x2) + 5x – 2 = -9x2 + 5x – 2
∴ 2x3 – 2x2 + 3x + 5 is greater than 2x3 + 7x2 – 2x + 7 by -9x2 + 5x – 2

SamacheerKalvi.Guru

Question 3.
What should be added to 2b2 – a2 to get b2 – 2a2
Solution:
The required expression is obtained by subtracting 2b2 – a2 from b2 – 2a2
b2 – 2a2 – (2b2 – a2) = b2 – 2a2 + (-2b2 + a2)
= b2 – 2a2 – 2b2 + a2
= (1 – 2) b2 + (-2 + 1) a2 = -b2 – a2
So -b2 – a2 must be added

Exercise 3.3

Question 1.
Length of one side of an equilateral triangle is 3x – 4 units. Find the perimeter.
Solution:
Equilateral triangle has three sides equal.
Perimeter = Sum of three sides
= (3x – 4) + (3x – 4) + (3x – 4) = 3x – 4 + 3x – 4 + 3x – 4
= (3 + 3 + 3)x + [(-4) + (-4) + (-4)] = 9x + (-12) = 9x – 12
∴ Perimeter = 9x – 12 units.

Question 2.
Find the perimeter of a square whose side is y – 2 units.
Solution:
Perimeter = (y – 2) + (y – 2) + (y – 2) + (y – 2)
= y – 2 + y – 2 + y – 2 + y – 2 = 4y – 8
Perimeter of the square = 4y – 8 units.

SamacheerKalvi.Guru

Question 3.
Simplify 3x – 5 – x + 9 if x = 3
Solution:
3x – 5 – x + 9 = 3(3) – 5 – 3 + 9
= 9 – 5 – 3 + 9 = 18 – 8 = 10

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Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.4

Students can Download Maths Chapter 3 Algebra Ex 3.4 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 1 Chapter 3 Algebra Ex 3.4

Miscellaneous Practice Problems

Question 1.
Subtract – 3ab – 8 from 3ab – 8. Also subtract 3ab + 8 from -3ab – 8.
Solution:
Subtracting -3ab – 8 from 3ab + 8
= 3ab + 8 – (-3ab – 8) = 3ab + 8 + (3ab + 8)
= 3ab + 8 + 3ab + 8 = (3 + 3) ab + (8 + 8)
= 6ab + 16
Also subtracting 3 ab + 8 from – 3ab – 8
= – 3ab – 8 – (3ab + 8) = – 3ab – 8 + (-3ab – 8) = – 3ab – 8 – 3 ab – 8
= [(-3) + (- 3)] ab + [(-8) + (-8)] = – 6ab + (- 16)
= -6ab – 16

Question 2.
Find the perimeter of a triangle whose sides are x + 3y, 2x + y, x – y.
Solution:
Perimeter of a triangle = Sum of three sides
= (x + 3y) + (2x + y) + (x – y)
= x + 3y + 2x + y + x – y
= (1 + 2 + 1)x + (3 + 1 + (-1))y = 4x + 3y
∴ Perimeter of the triangle = 4x + 3y

Question 3.
Thrice a number when increased by 5 gives 44. Find the number.
Solution:
Let the required number be x.
Thrice the number = 3x.
Thrice the number increased by 4 = 3x + 5
Given 3x + 5 = 44
3x + 5 – 5 = 44 – 5
3x = 39
\(\frac{3 x}{3}=\frac{39}{3}\)
x = 13
∴ The required number = 13

Question 4.
How much smaller is 2ab + 4b – c than 5ab – 3b + 2c.
Solution:
To find the answer we have to find the difference.
Here greater number 5ab – 3ab + 2c.
∴ Difference = 5ab – 3b + 2c – (2ab + 4b – c) = 5ab – 3b + 2c + (- 2ab -4b + c)
= 5ab – 3b + 2c – 2ab – 4b + c
= (5 – 2) ab + (-3 – 4) b + (2 + 1) c = 3ab + (-7)b + 3c
= 3ab – 7b + 3c
It is 3ab – 7b + 3c smaller.

SamacheerKalvi.Guru

Question 5.
Six times a number subtracted from 40 gives – 8. Find the number.
Solution:
Let the required number be x. Six times the number = 6x.
Given 40 – 6x = – 8
-6x + 40 – 40 = -8 – 40
– 6x = – 48
\(\frac{-6 x}{-6}=\frac{-48}{-6}\)
x = 8
∴ The required number is 8.

Challenge Problems

Question 6.
From the sum of 5x + 7y -12 and 3x – 5y + 2, subtract the sum of 2x – 7y – 1 and – 6x + 3y + 9.
Solution:
Sum of 5x + 7y – 12 and 3x – 5y + 2 .
= 5x + 7y- 12 + 3x – 5y + 2 = (5 + 3) x + (7 – 5) y + ((- 12) + 2)
= 8x + 2y – 10.
Again Sum of 2x – 7y – 1 and – 6x + 3y + 9
= 2x – 7y – 1 + (- 6x + 3y + 9) = 2x – 7y – 1 – 6x + 3y + 9
= (2 – 6) x + (- 7 + 3) y + (- 1 + 9)
= – 4x – 4y + 8
Now 8x + 2y – 10 – (-4x – 4y + 8)
= 8x + 2y – 10 + (4x + 4y – 8)
= 8x + 2y – 10 + 4x + 4y – 8
= (8 + 4) x + (2 + 4) y + ((- 10) + (- 8))
= 12x + 6y – 18

Question 7.
Find the expression to be added with 5a – 3b – 2c to get a – 4b – 2c?
Solution:
To get the required expression we must subtract 5a – 3b + 2c from a – 4b – 2c.
∴ a – 4b – 2c – (5a – 3b + 2c) = a – 4b – 2c + (- 5a + 3b – 2c)
= a – 4b – 2c – 5a + 3b -2c
= (1 – 5) a + (- 4 + 3) b + (- 2 – 2) c
= – 4a – b – 4c.
∴ -4a – b – 4c must be added.

Question 8.
What should be subtracted from 2m + 8n + 10 to get – 3m + 7n + 16?
Solution:
To get the expression we have to subtract – 3m + 7n + 16 from 2m + 8n + 10.
(2m + 8n + 10) – (-3m + 7n + 16) = 2m + 8n + 10 + 3m – 7n – 16
= (2 + 3) m + (8 – 7) n + (10 – 16)
= 5m + n – 6

SamacheerKalvi.Guru

Question 9.
Give an algebraic equation for the following statement:
“The difference between the area and perimeter of a rectangle is 20”.
Solution:
Let the length of a rectangle = l and breadth = b then Area = lb; Perimeter = 2(1 + b)
Area – Perimeter = 20
∴ lb – 2(l + b)

Question 10.
Add : 2a + b + 3c and a + \(\frac{1}{3}\)b + \(\frac{2}{5}\)c
Solution:
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.4 1

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Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Students can Download Maths Chapter 3 Algebra 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 1 Chapter 3 Algebra Intext Questions

Exercise 3.1

Try These (Text Book Page No. 51)

Question 1.
Identify the variable and constants among the following terms.
a, 11 – 3x, xy, -89, -m, -n, 5, 5ab, -5 3y, 8pqr, 18, -9t, -1, -8
Solution:
Variable : a, -3x, xy, -m, -n, 5ab, 3y, -9t, 8pqr
Constants : 11, -89, 5, -5, 18, -1, -8

Question 2.
Complete the following table.
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions 80
Solution:
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions 2

Try this (Text book Page No. 53)

Question 1.
Can we use the operations multiplication and division to combine terms?
Solution:
No, We can use addition and subtraction to combine terms.
If we use multiplication or division to combine then it become a single term.
Eg : xy, \(\frac{x}{y}\) are monomials.

Try This (Text book Page No. 54)

Question 1.
Complete the following table by forming expressions using the terms given. One is done for you.
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions 85
Solution:
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions 3

Try this (Text book Page No. 56)

Question 1.
Identify the like terms among the following and group them.
7xy, 19x, 1, 5y, x, 3yx, 15, -13y, 6x, 12xy, -5, 16y, -9x, 15xy, 23, 45y, -8y, 23x, -y, 11
Solution:
7xy, 3yx, 12xy, 15xy, are like terms
19x, x, 6x,-9x, 23x, are like terms
5y, -13y, 16y, 45y, -8y, -y, are like terms
1, 15, -5, 23, 11, are like terms.

Try This (Text book Page No. 57)

Question 1.
Try to find the value of the following expressions if p = 5 and q = 6.
(i) p + q
(ii) q – p
(iii) 2p + 2 > q
(iv) pq – p – q
(v) 5pq – 1
Solution:
(i) Given p = 5; q = 6
p + q = 5 + 6 = 11
(ii) q – p = 6 – 5 = 1
(iii) 2p + 2 > q = 2(5) + 3(6) = 10 + 18 = 28
(iv) pq – p – q = (5) (6) – 5 – 6 = 30 – 5 – 6 = 25 – 6 = 19

Exercise 3.2

Try These (Text book Page No. 59)

Question 1.
Add the terms
(i) 3p, 14p
(ii) m, 12m, 21m
(iii) 11abc, 5abc
(iv) 12y, -y
(v) 4x, 2x, -7x.
Solution:
(i) 3p + 14p = 17p
(ii) m + 12m + 21m = (1 + 12 + 21 )m
= 34 m
(iii) 11abc + 5abc = (11 + 5) abc
= 16 abc
(iv) 12y + (-y) = (12 + (-1))y
= (12 – 1 )y
= 11y
(v) 4x + 2x + (-7x) = (4 + 2+(-7))x
= (6 + (-7))x
= -1x

Ty this (Text Book Page No. 60)

Question 1.
3x; + (y – x) = 3x + y – x, but 3x – (y – x) ≠ 3x – y – x. why ?
Solution:
In the first case
LHS = 3x + (y – x) = 3x + y – x = 3x – x + y = (3 – 1)x + y
= 2x + y
RHS = 3x + y – x = 2x + y
LHS = RHS ⇒ 3x + (y – x) = 3x + y – x
But in the second case
LHS = 3x – (y – x) = 3x – y + x
= (3 + 1)x – y = 4x – y
RHS = 3x – y – x = 3x – x – y
LHS ≠ RHS
∴ 3x – (y – x) ≠ 3x – y – x

Try this (Page No. 1)

Question 1.
What will you get if twice a number is subtracted from thrice the same number?
Solution:
Let the unknown number be x.
Twice the number = 2x.
Thrice the number = 3x.
Twice the number is subtracted from thrice the number = 3x – 2x = (3 – 2)x = x

Exercise 3.3

Try These (Text book Page No. 65)

Question 1.
Try to construct algebraic equations for the following verbal statements.

Question 1.
One third of a number plus 6 to 10.
Solution:
\(\frac{1}{3}\) + 6 = 10

Question 2.
The sum of five times of x and 3 is 28
Solution:
5 (x + 3) = 28

Question 3.
Taking away 8 from y gives 11
Solution:
y – 8 = 11

Question 4.
Perimeter of a square with side a is 16 cm.
Solution:
4 × a = 16

Question 5.
Venkat’s mother’s age is 7 years more than 3 times venkat’s age. His mother’s age is 43 years.
Solution:
3x + 7 = 43, where x is venkat’s age.

Try this (Text book Page No. 65)

Question 1.
Why should we subtract 5 and not some other number ? why don’t we add 5 on both sides? Discuss.
Solution:
Given x + 5 = 12
(i) Our aim is to find the value of x. Which means we have to eliminate the other values from LHS. Since 5 is given with x it should be subtracted.
(ii) If we add 5 on both sides we cannot eliminate the numbers from LHS and we get x + 10.

Try this (Text book Page No. 66)

Question 1.
If the dogs, cats and parrots represents unknown find them. Substitute each of the values so obtained in the equations and verify the answers.
Solution:
(i) 1 dog + 1 dog + 1 dog = 24
3 dog = 24
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions 95
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions 62

(ii) 1 dog + 1 cat + 1 cat = 14
1 dog + 2cat = 14
8 + 2cat = 14
2cat = 14 – 8
2 cat = 6
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions 63ditional Questions 63″ width=”107″ height=”87″ />

(iii) 1 dog + 1 cat – 1 parrot = 9
8 + 3 – 1 parrot = 9
8 + 3 – 9 = 1 parrot
11 – 9 = 1 parrot
2 = 1 parrot
1 parrot = 2

(iv) 1 dog + 1 cat + 1 parrot = ?
8 + 3 + 2 = 13
Verification:
(i) 8 + 8 + 8 = 24
(ii) 8 + 3 + 3 = 14
(iii) 8+ 3 – 2 = 9
(iv) 8 + 3 + 2 = 13

Try These (Text book Page No. 68)

Question 1.
Kandhan and kaviya are friends. Both of them are having some pen. Kandhan: If you give me one pen then, we will have equal number of pens. Will you? Kaviya: But, if you give me one of your pens, then mine will become twice as yours. Will you?
Construct algebraic equations for this situation, can you guess and find the actual number of pens, they have?
Solution:
Let the number of pens initially Kandhan and Kaviya had be x and y respectively.

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Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.3

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 1 Chapter 3 Algebra Ex 3.3

Question 1.
Fill in the blanks.
(i) An expressions equated to another expression is called _______.
(ii) If a = 5, the value of 2a + 5 is _______.
(iii) The sum of twice and four times of the variable x is ______.
Solution:
(i) an equation
(ii) 15
(iii) 6x

Question 2:
Say True or False
(i) Every algebraic expression is an equation.
(ii) The expression 7x + 1 cannot be reduced without knowing the value of x.
(iii) To add two like terms, its coefficients can be added.
Solution:
(i) False
(ii) True
(iii) True

Question 3.
Solve (i) x + 5 = 8
(ii) p – 3 = 1
(iii) 2x = 30
(iv) \(\frac{m}{6}\) = 5
(v) 7x + 10 = 80
Solution:
(i) Given x + 5 = 8 ; Subtracting 5 on both the sides
x + 5 – 5 = 8 – 5
x = 3

(ii) Given p – 3 = 7 ; Adding 3 on both the sides,
p – 3 + 3 = 7 + 3
p = 10

(iii) Given 2x = 30 ; Dividing both the sides by 2,
\(\frac{2 x}{2}=\frac{30}{2}\)
x = 15

(iv) Given \(\frac{m}{6}\) = 5 ; Multiplying both the sides by 6,
\(\frac{m}{6}\) × 6 = 5 × 6
m = 30

(v) Given 7x + 10 = 80 ; Subtracting 10 from both the sides,
7x + 10 – 10 = 80 – 10
7x = 70
Dividing both sides by 7,
\(\frac{7 x}{7}=\frac{70}{7}\)
x = 10

Question 4.
What should be added to 3x + 6y to get 5x + 8y?
Solution:
To get the expression we should subtract 3x + 6y from 5x + 8y
5x + 8y – (3x + 6y) = 5x + 8y + (-3x – 6y)
= 5x + 8y – 3x – 6y = (5 – 3) x + (8 – 6) y
= 2x + 2y
So 2x + 2y should be added.

SamacheerKalvi.Guru

Question 5.
Nine added to thrice a whole number gives 45. Find the number
Solution:
Let the whole number required be x.
Thrice the whole number = 3x
Nine added to it = 3x + 9
Given 3x + 9 = 45
3x + 9 – 9 = 45 – 9 [Subtracting 9 on both sides]
3x = 36
\(\frac{3 x}{3}=\frac{36}{3}\)
x = 12
∴ The required whole number is 12

Question 6.
Find the two consecutive odd numbers whose sum is 200
Solution:
Let the two consecutive odd numbers be x and x + 2
∴ Their sum = 200
x + (x + 2) = 200
x + x + 2 = 200
2x + 2 = 200
2x + 2 – 2 = 200 – 2 [∵ Subtracting 2 from both sides]
2x = 198
\(\frac{2 x}{2}=\frac{198}{2}\) [Dividing both sides by 2]
x = 99
The numbers will be 99 and 99 + 2.
∴ The numbers will be 99 and 101.

Question 7.
The taxi charges in a city comprise of a fixed charge of ₹ 100 for 5 kms and ₹ 16 per km for ever additional km. If the amount paid at the end of the trip was ₹ 740, find the distance traveled.
Solution:
Let the distance travelled by taxi be ‘x’ km
For the first 5 km the charge = ₹ 100
For additional kms the charge = ₹ 16(x – 5)
∴ For x kms the charge = 100 + 16(x – 5)
Amount paid = ₹ 740
∴ 100 + 16 (x – 5) = 740
100 + 16 (x – 5) – 100 = 740- 100
16 (x – 5) = 640
\(\frac{16(x-5)}{16}=\frac{640}{16}\)
x – 5 = 40
x – 5 + 5 = 45 + 5
x = 45
x = 45 km
∴ Total distance travelled = 45 km

Objective Type Questions

Question 8.
The generalization of the number pattern 3, 6, 9, 12, …………. is
(i) n
(ii) 2n
(iii) 3n
(iv) 4n
Solution:
(iii) 3n

Question 9.
The solution of 3x + 5 = x + 9 is t
(i) 2
(ii) 3
(iii) 5
(iv)4
Solution:
(i) 2
Hint: 3x + 5 = x + 9 ⇒ 3x – x = 9 – 5 ⇒ 2x = 4 ⇒ x = 2

SamacheerKalvi.Guru

Question 10.
The equation y + 1 = 0 is true only when y is
(i) 0
(ii) -1
(iii) 1
(iv) – 2
Solution:
(ii) -1

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