Category: Class 7

Students can Download Maths Chapter 2 Measurements 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 2 Chapter 2 Measurements Additional Questions

Additional Questions and Answers

Exercise 2.1

Question 1.
A circular disc of radius 28 cm is divided into two equal parts. What is the perimeter of each semicircular shape disc? Also find the perimeter of the circular disc.
Answer:
Radius of the circular disc = 28 cm
∴ circumference of the circle = 2 π r units

= 2 × \(\frac { 22 }{ 7 } \) × 28 cm
= 2 × 22 × 4 = 44 × 4 = 176 cm
Perimeter of the circular disc = 176 cm
Now circumference of the semicircular disc = \(\frac { 1 }{ 2 } \) × (2 π r) = π r units
Perimeter of the semicircular disk = π r + r + r = \(\frac { 22 }{ 7 } \) × 28 + 28 + 28
= 22 × 4 + 28 + 28 = 88 + 28 + 28 = 144
Perimeter of each semicircular disc = 144 cm

Question 2.
A gardener wants to fence a circular garden of diameter 14m. Find the length of the rope he needs to purchase if he makes 3 rounds of fence. Also find the cost of the rope, if it costs ₹ 4 per metre.
Solution:
Diameter of the circular garden (d) = 14 m
Circumference = π d = \(\frac { 22 }{ 7 } \) × 14 = 44 m
Since the rope makes 3 rounds of fence,
length of the rope needed = 3 × circumference of the garden
= 3 × 44 m = 132 m
Cost of rope per meter = ₹ 4 × 132 = ₹ 528

Question 3.
Latha wants to put a lace on the edge of a circular table cover of diameter 3 m. Find the length of the lace required and also find the cost if one metre of the lace costs ₹ 30 (Take π = 3.15 )
Solution:
Diameter of the circular table cover = 3m
Circumference C = π d units = 3.15 × 3 m = 9.45 m
Length of the lace required = 9.45 m
Cost of lace per meter = ₹ 30
∴ Cost of 9.45m lace = ₹ 30 × 9.45 = ₹ 283.50

Exercise 2.2

Question 1.
The circumference of two circles are on the ratio 5 : 6. Find the ratio of their areas.
Solution:
Let the radii of the given circles be r₁ and r₂
Let their circumference be C₁ and C₂ respectively
C₁ = 2πr₁ and C₂ = 2πr₂

Now let the area of the given circles the A₁ and A₂

∴ The areas of two circles are in the ratio 25 : 36.

Question 2.
Find the area of the circle whose circumference is 88 cm.
Solution:
The circumference of the circle = 88 cm
2πr = 88 cm
2 × \(\frac { 22 }{ 7 } \) × r = 88 cm
r = \(\frac{88 \times 7}{2 \times 22}\) = 14 cm
r = 14 cm
Area of the circle A = πr²
= \(\frac { 22 }{ 7 } \) × 14 × 14 cm²
= 22 × 2 × 14 cm²
= 616 cm²

Question 3.
Find the cost of polishing a circular table top of diameter 16 dm if the rate of polishing is ₹ 15 per dm².
Solution:
Diameter of the circular table top = 16 dm
Radius (r) = \(\frac { 16 }{ 2 } \) = 8cm
Area of the circular table top
= πr² sq.units
= \(\frac { 22 }{ 7 } \) × 8 × 8 dm² = \(\frac { 1408 }{ 7 } \) dm²
= 201.14 dm²
Rate of the polishing per dm² = ₹ 15
∴ Rate of the polishing 201.14 dm²
= ₹ 15 × 201.14
= ₹ 3,017

Exercise 2.3

Question 1.
From a circular sheet of radius 5 cm a circle of radius 3 cm is removed. Find the area of the remaining sheet.
Solution:
Radius of the outer circle R = 5 cm
Radius of the inner circle r = 3 cm
Area of the remaining sheet = Area of the outer circle
– Area of the inner circle
= πR² – πr² sq. units = π(R² – r²) sq. units
= \(\frac { 22 }{ 7 } \) (5² – 3²) cm² = \(\frac { 22 }{ 7 } \) (5² – 3²) cm²
= \(\frac { 22 }{ 7 } \) × (5 + 3) (5 – 3) = \(\frac { 22 }{ 7 } \) × (8) (2)
= \(\frac { 352 }{ 7 } \) = 50.28 cm²

Question 2.
A picture is painted on a cardborad 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
Solution:
Length of the outer rectangle L = 8 cm
Breadth of the outer rectangle B = 5 cm
Area of the outer rectangle = L × B sq. units = 8 × 5 cm² = 40 cm²
Length of the inner rectangle l = L – 2W = 8 – 2(1.5)cm = 8 – 3cm
l = 5cm
Breadth of thqnner rectangle b = B – 2W = 5 – 2(1.5) cm
= 5 – 3cm = 2cm
∴ Area of the margin = Area of the outer rectangle
– Area of the inner rectangle
= (40 – 10) cm² = 30cm²

Question 3.
A circular piece of radius 2 cm is cut from a rectangle sheet of length 5 cm and breadth 3 cm. Find the area left in the sheet.
Solution:
Radius of the portion removed r = 2 cm
Area of the circular sheet = πr² sq.units
= \(\frac { 22 }{ 7 } \) × 2 × 2 cm² = \(\frac { 88 }{ 7 } \) cm² = 12.57 cm²
Length of the rectangular sheet L = 5 cm
Breadth of the rectangular sheet B = 3 cm
Area of the rectangle = L × B sq. units = 5 × 3 cm² = 15 cm²
Area of the sheet left over = Area of the rectangle – Area of the circle
= 15 cm² – 12.57 cm² = 2.43 cm²

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

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.

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

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

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

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

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

Students can Download Maths Chapter 2 Measurements 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 2 Measurements Intext Questions

Exercise 2.1

Try These (Text book Page No. 23)

Question 1.
A few real life examples of circular shapes are given below.

Can you give three more examples.
Solution:

1. One and Two rupee coins
2. Bangles
3. Mouth of Bottle

Question 2.
Find the diameter of your bicycle wheel?
Solution:
Diameter of my bicycle wheel is 700 mm

Question 3.
If the diameter of the circle is 14cm, what will be it’s radius?
Solution:
diameter d = 14 cm
radius = \(\frac { d }{ 2 } \) = \(\frac { 14 }{ 2 } \) = 7cm

Question 4.
If the radius of a bangle is 2 inches then find the diameter.
Solution:
Given radius of the bangle = 2 inches
Diameter = 2 × radius = 2 × 2 = 4 inches

Exercise 2.2

Try These (Text book Page No. 33)

Question 1.
Draw circles of different radii on a graph paper. Find the area by counting the number of squares covered by the circle. Also find the area by using the formula.
(i) Find the area of the circle, if the radius is 4.2 cm.
(ii) Find the area of the circle if the diameter is 28 cm.
Solution:
(i) Radius of the circle r = 4.2 cm
Area of the circle A = π r² sq.units
= \(\frac { 22 }{ 7 } \) × 4.2 × 4.2 cm² = 5.44 cm²

(ii) Diameter of the circle d = 28 cm
radius r = \(\frac { d }{ 2 } \) = \(\frac { 28 }{ 2 } \) = 14 cm
Area of the circle A = π r² sq.units
= \(\frac { 22 }{ 7 } \) × 14 × 14 cm² = 616 cm²

Exercise 2.3

Try These (Text book Page No. 35)

Question 1.
If the outer radius and inner radius of the circles are respectively 9 cm and 6 cm, find the width of the circular pathway.
Solution:
Radius of the outer circle R = 9 cm
Radius of the inner circle r = 6 cm
Width of the circular pathway = Radius of the outer circle
– Radius of the inner circle
= (9 – 6) cm = 3 cm
Width of the circular pathway = 3 cm

Question 2.
If the area of the circular pathway is 352 sq.cm and the outer radius is 16 cm, find the inner radius.
Solution:
Given outer radius R = 16 cm
Area of the circular pathway = πR² = πr²
Area of the circular pathway = 352 sq. cm
πR² – πr² = 352 cm²
π(R² – r²) = 352
16² – r² = \(\frac{352 \times 7}{22}\)
16² – r² = 16 × 7
16² – r² = 112
16² – 112 = r²
r² = 256 – 112
r² = 144
r = 12 cm
Inner radius r = 12 cm

Question 3.
If the area of the inner rectangular region is 15 sq.cm and the area covered by the outer rectangular region is 48 sq.cm, find the area of the rectangular pathway. Area of the outer rectangle Area of the inner rectangle Area of the rectangular pathway
Solution:
Area of the outer rectangle = 48 sq.cm
Area of the inner rectangle = 15 sq.cm
Area of the rectangular pathway = Area of the outer rectangle
– Area of the inner rectangle
= 48 – 15 = 33 cm²

Students can Download Maths Chapter 2 Measurements Ex 2.3 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 2 Measurements Ex 2.3

Question 1.
Find the area of a circular pathway whose outer radius is 32 cm and inner radius is 18 cm.
Solution:
Radius of the outer circle R = 32 cm
Radius of the inner circle r = 18 cm
Area of the circular pathway = π (R² – r²) sq. units = \(\frac { 22 }{ 7 } \) (32² – 18²) cm²
= \(\frac { 22 }{ 7 } \) × (32 + 18) × (32 – 18) cm²
= \(\frac { 22 }{ 7 } \) × 50 × 14 cm² = 2,200 cm²
Area of the circular pathway = 2,200 cm²

Question 2.
There is a circular lawn of radius 28 m. A path of 7 m width is laid around the lawn. What will be the area of the path?
Solution:
Radius of the circular lawn r = 28 m
Radius of the lawn with path = 28 + 7 m = 35 m
Area of the circular path = π (R² – r²) sq. units
Area of the path = \(\frac { 22 }{ 7 } \) (35² – 28²) m² = \(\frac { 22 }{ 7 } \) × (35 + 28) (35 – 28) m²
= \(\frac { 22 }{ 7 } \) × 63 × 7 m² = 1386 m²
Area of the path = 1386 m²

Question 3.
A circular carpet whose radius is 106 cm is laid on a circular hall of radius 120 cm. Find the area of the hall uncovered by the carpet.
Solution:
Radius of the circular hall R = 120 cm
Radius of the circular carpet r = 106 cm
Area of the hall uncovered = Area of the hall – Area of the carpet
= π (R² – r²) cm²
= \(\frac { 22 }{ 7 } \) × (120² – 106²) cm²
= \(\frac { 22 }{ 7 } \) × (120 + 106) × (120 – 106) cm²
= \(\frac { 22 }{ 7 } \) × 226 × 14 cm² = 9,944 cm²
Area of the hall uncovered = 9, 944 cm²

Question 4.
A school ground is in the shape of a circle with radius 103 m. Four tracks each of 3 m wide has to be constructed inside the ground for the purpose of track events. Find the cost of constructing the track at the rate of ₹ 50 per sq.m.

Solution:
Radius of the ground R = 103 m
Width of a track W = 3 m
Width of 4 tracks = 4 × 3 = 12 m
Radius of the ground without track
r = (103 – 12)m
r = 91 m
Area of 4 tracks = Area of the ground
– Area of the ground without crack
= πR² – πr² sq.units
= π(R² – r²) sq.units
= \(\frac { 22 }{ 7 } \) [103² – 91²]
= \(\frac { 22 }{ 7 } \) [103 + 91] [103 – 91]m²
= \(\frac { 22 }{ 7 } \) × 194 × 12 = \(\frac { 51216 }{ 7 } \) = 7316.57 m²
∴ Area of 4 tracks = 7316.57 m²
Cost of constructing 7316.57 m² = ₹ 50
∴ Cost of constructing 7316.57 m² = ₹ 50 × 7316.57 = ₹ 3,65,828,57
Cost of constructing the track ₹ 3,65,828,57

Question 5.
The figure shown is the aerial view of the pathway. Find the area of the pathway.

Solution:
Area of the rectangle = (Lenght × Breadth) sq. units
Area of the outer rectangle = (L × B) sq. units
Length of the outer rectangle L = 80 m
Breadth of the outer rectangle B = 50 m
Length of the inner rectangle l = 70 m
Breadth of the inner rectangle b = 40 m
Area of the outer rectangle = 80 × 50 m² = 4000 m²
Area of the inner rectangle = l × b sq. unit = 70 × 40 m² = 2800 m²
Area of the pathway = Area of the outer rectangle
– Area of the inner rectangle
= 4000 – 2800 m² = 1200 m²
Area of the pathway = 1200 m²

Question 6.
A rectangular garden has dimensions 11 m × 8 m. A path of 2 m wide has to be constructed along its sides. Find the area of the path.
Solution:
Area of the rectangular garden L × B = 11 m × 8 m = 88 m²
Length of the inner rectangle L = L – 2 W = 11 – 2(2) = 11 – 4 = 7 m
Breadth of the inner rectangle b = B – 2W = 8 – 2(2) = 8 – 4 = 4 m
Area of the inner rectangle = l × b sq. units = 7 × 4 m² = 28 m²
Area of the path = Area of the outer rectangular garden
– Area of the inner rectangle
= 88 m² – 28m² = 60 m²
Area of the path = 60 m²

Question 7.
A picture is painted on a ceiling of a marriage hall whose length and breadth are 18 m and 7 m respectively. There is a border of 10 cm along each of its sides. Find the area of the border.
Solution:
Length of the ceiling L = 18 m
Breadth of the ceiling B = 7 m
Area of the ceiling = L × B sq. units = 18 × 7 m² = 126 m²
Width of the boarder W = 10 cm = \(\frac { 10 }{ 100 } \) m = 0.1 m
Length of the ceiling without border = L – 2W = 18 – 2(0.1) m
= 18 – 0.2 m = 17.8 m
Breadth of the ceiling without border = B – 2W = 7 – 2 (0.1) m
= 7 – 0.2 m = 6.8 m
Area of the ceiling without border = l × b sq.units
= 17.8 × 6.8 m² = 121.04 m²
∴ Area of the border = Area of the ceiling
– Area of the ceiling without border
= 126 – 121.04 m² = 4.96 m²
Area of the border = 4.96 m²

Question 8.
A canal of width 1 m is constructed all along inside the field which is 24 m long and 15 m wide. Find (i) the area of the canal (ii) the cost of constructing the canal at the rate of ₹ 12 per sq.m.
Solution:
Length of the field L = 24 m
Width (Breadth) of the field B = 15 m
(i) Area of the field = L × B sq. units = 24 × 15 m² = 360 m²
(ii) Width of the canal (W) = 1 m
Length of the field without canal (l) = L – 2(W) = 24 – 2(1) m
= 24 – 2 m = 22 m
Width of the field without canal (b) = B – 2W = 15 – 2(1) m
= 15 – 2 m = 13 m
Area of the field without canal = l × b sq. units = 22 × 13 m² = 286 m²
Area of the canal = 360 – 286 = 74 m²
Cost of constructing 1 m² canal = ₹ 12
Cost of the constructing 74 m² canal = ₹ 12 × 74 = ₹ 888

Objective Type Question

Question 9.
The formula to find the area of the circular path is
(i) π(R² – r²) sq. units
(ii) πr² sq. units
(iii) 2πr² sq. units
(iv) πr² + 2r sq. units
Answer:
(i) π(R² – r2) sq. units

Question 10.
The formula used to find the area of the rectangular path is
(i) p(R² – r²) sq. units
(ii) (L × B) – (l × b) sq. units
(iii) LB sq. units
(iv) lb sq. units
Answer:
(ii) (L × B) – (l × b) sq. units

Question 11.
The formula to find the width of the circular path is
(i) (L – l) units
(ii) (B – b) units
(iii) (R – r) units
(iv) (r – R) units
Answer:
(iii) (R – r) units

Students can Download Maths Chapter 1 Number System Ex 1.1 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 1 Number System Ex 1.1

Question 1.
Write the decimal numbers for the following pictorial representation of numbers.

Solution:
(i) Tens 2 ones 2 tenths = 12.2
(ii) Tens 1 ones 3 tenths = 21.3

Question 2.
Express the following in cm using decimals.
(i) 5 mm
(ii) 9 mm
(iii) 42 mm
(iv) 8 cm 9 mm
(v) 375 mm
Solution:
(i) 5 mm
1 mm = \(\frac { 1 }{ 10 } \) cm = 0.1 cm
5 mm = \(\frac { 5 }{ 10 } \) = 0.5 cm

(ii) 9 mm
1 mm = \(\frac { 1 }{ 10 } \) cm = 0.1 cm
9 mm = \(\frac { 9 }{ 10 } \) cm = 0.9 cm

(iii) 42 mm
1 mm = \(\frac { 1 }{ 10 } \) cm = 0.1 cm
42 mm = \(\frac { 42 }{ 10 } \) cm = 4.2 cm

(iv) 8 cm 9 mm
1 mm = \(\frac { 1 }{ 10 } \) cm = 0.1 cm
8 cm 9 mm = 8 cm + \(\frac { 9 }{ 10 } \) cm = 8.9 cm

(v) 375 mm
1 mm = \(\frac { 1 }{ 10 } \) cm = 0.1 cm
375 mm = \(\frac { 375 }{ 10 } \) cm = 37.5 cm

Question 3.
Express the following in metres using decimals.
(i) 16 cm
(ii) 7 cm
(iii) 43 cm
(iv) 6 m 6 cm
(v) 2 m 54 cm
Solution:
(i) 16 cm
1 cm = \(\frac { 1 }{ 100 } \) cm = 0.01 m
16 cm = \(\frac { 16 }{ 100 } \) m = 0.16 m

(ii) 7 cm
1 cm = \(\frac { 1 }{ 100 } \) cm = 0.01 m
1 cm = \(\frac { 7 }{ 100 } \) m = 0.07 m

(iii) 43 cm
1 cm = \(\frac { 1 }{ 100 } \) cm = 0.01 m
43 cm = \(\frac { 43 }{ 100 } \) m = 0.43 m

(iv) 6 m 6 cm
1 cm = \(\frac { 1 }{ 10 } \) m = 0.01 m
6 m 6 cm = 6 m + \(\frac { 6 }{ 100 } \) m = 6 m + 0.06 m = 6.06 m

(v) 2 mm 54 cm
1 cm = \(\frac { 1 }{ 100 } \) cm = 0.01 m
2 m 54 cm = 2 m + \(\frac { 54 }{ 100 } \) m = 2 m + 0.54 m = 2.54 m

Question 4.
Expand the following decimal numbers.
(i) 37.3
(ii) 658.37
(iii) 237.6
(iv) 5678.358
Solution:
(i) 37.3 = 30 + 7 + \(\frac { 3 }{ 10 } \) = 3 × 10¹ + 7 × 10⁰ + 3 × 10^-1

(ii) 658.37 = 600 + 50 + 8 + \(\frac { 3 }{ 10 } \) + \(\frac { 7 }{ 100 } \)
= 6 × 10² + 5 × 10¹ + 8 × 100 + 3 × 10^-1 + 7 × 10^-2

(iii) 237.6 = 200 + 30 + 7 + \(\frac { 6 }{ 10 } \)
= 2 × 10² + 3 × 10¹ + 7 × 10⁰ + 6 × 10^-1

(iv) 5678.358 = 5000 + 600 + 70 + 8 + \(\frac { 3 }{ 10 } \) + \(\frac { 5 }{ 100 } \) + \(\frac { 8 }{ 1000 } \)
= 5 × 10³ + 6 × 10² + 7 × 10¹ + 8 × 10⁰ + 3 × 10^-1 + 5 × 10^-2 + 8 × 10^-3

Question 5.
Express the following decimal numbers in place value grid and write the place value of the underlined digit.
(i) 53.61
(ii) 263.271
(iii) 17.39
(iv) 9.657
(v) 4972.068
Solution:
(i) 53.61

(ii) 263.271

(iii) 17.39

(iv) 9.657

(v) 4972.068

Objective Type Questions

Question 6.
The place value of 3 in 85.073 is _____
(i) tenths
(ii) hundredths
(iii) thousands
(iv) thousandths
Answer:
(iv) thousandths
Hint: 1000 g = 1 kg; 1 g = \(\frac { 1 }{ 1000 } \) kg

Question 7.
To convert grams into kilograms, we have to divide it by
(i) 10000
(ii) 1000
(iii) 100
(iv) 10
Answer:
(ii) 1000
Hint: 85.073 = 8 × 10 + 5 × 1 + 0 × \(\frac { 1 }{ 10 } \) + 7 × \(\frac { 1 }{ 100 } \) + 3 × \(\frac { 1 }{ 1000 } \)

Question 8.
The decimal representation of 30 kg and 43 g is ____ kg.
(i) 30.43
(ii) 30.430
(iii) 30.043
(iv) 30.0043
Answer:
(iii) 30.043
Hint: 30 kg and 43 g = 30 kg + \(\frac { 43 }{ 1000 } \) kg = 30 + 0.043 = 30.043

Question 9.
A cricket pitch is about 264 cm wide. It is equal to _____ m.
(i) 26.4
(ii) 2.64
(iii) 0.264
(iv) 0.0264
Answer:
(ii) 2.64
Hint: 264 cm = \(\frac { 264 }{ 100 } \) m = 2.64 m

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

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

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.

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

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.

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

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:

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.

Solution:

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.

Solution:

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

(ii) 1 dog + 1 cat + 1 cat = 14
1 dog + 2cat = 14
8 + 2cat = 14
2cat = 14 – 8
2 cat = 6
ditional 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.

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.

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

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

Students can Download Maths Chapter 3 Algebra Ex 3.1 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 1 Chapter 3 Algebra Ex 3.1

Question 1.
Fill in the blanks
(i) The variable in the expression 16x – 7 is _____
(ii) The constant term of the expression 2y – 6 is _____
(iii) In the expression 25m + 14M, the type of the terms are ______ terms
(iv) The number of terms in the expression 3ab + 4c – 9 is _____
Hint: Terms are 3ab, 4c – 9.
(v) The numerical co-efficient of the term -xy is ______
Hint: -x,y = (- 1 )xy.
Solution:
(i) x
(ii) -6
(iii) unlike
(iv) three
(v) -1

Question 2.
Say true or False
(i) x + (-x) = 0.
(ii) The co-efficient of ab in the term 15 abc is 15.
Hint: Coefficient of ab is 15c
(iii) 2pq and – 7qp are like terms.
(iv) When y = -1, the value of the expression 2y – 1 is 3.
Hint: 2(-1) – 1 = -2 – 1 = – 3
Solution:
(i) True
(ii) False
(iii) True
(iv) False

Question 3.
Fing the numerical co-efficient of each of the following terms: -3yx, 12k, y, 121bc, -x, 9pq, 2ab.
Solution:
(i) Numerical co-efficient of-3yx is – 3
(ii) Numerical co-efficient of 12k is 12
(iii) Numerical coefficient of y is 1
(iv) Numerical co-efficient of 1216c is 121
(v) Numerical co-efficient of – x is – 1
(vi) Numerical co-efficient of 9pq is 9
(vii) Numerical co-efficient of 2ab is 2

Question 4.
Write the variables, constants and terms of the following expressions,
(i) 18 + x – y
(ii) 7p – 4q + 5
(iii) 29x + 13y
(iv) b + 2
Solution:

Question 5.
Identify the like terms among the following 7x, 5y, -8x, 12y, 6z, z, -12x, -9y, 11 z
Solution:

Question 6.
If x = 2 andy = 3, then find the value of the following expressions,
(i) 2x – 3y
(ii) x + y
(iii) 4y – x
(iv) x + 1 – y
Solution:
Given x = 2; y = 3.
(i) 2x – 3y = 2 (2) – 3 (3) = 4 – 9
= 4 + (Additive inverse of 9)
= 4 +(-9) = -5
(ii) x + y = 2 + 3 = 5
(iii) 4y – x = 4 (3) – 2 = 12 – 2 = 10
(iv) x + 1 – y = 2 + 1 – 3 = 3 – 3 = 0

Objective Type Questions

Question 1.
An algebraic statement which is equivalent to the verbal statement “Three times the sum of ‘x’ and ‘y’ is
(i) 3 (x + y)
(ii) 3 + x + y
(iii) 3x + y
(iv) 3 + xy
Solution:
(i) 3 [(x + y)]

Question 2.
The numerical co-efficient of -7mn is
(i) 7
(ii) -7
(iii) p
(iv) -p
Solution:
(ii) -7

Question 3.
Choose the pair of like terms
(i) 7p, 7x
(ii) 7r, 7x
(iii) – 4x, 4
(iv) – 4x, 7x
Solution:
(iv) -4x, 7x

Question 4.
The value of 7a – 4b when a = 3, b = 2 is
(i) 21
(ii) 13
(iii) 8
(iv) 32
Solution:
(ii) 13
Hint: 7(3) – 4(2) = 21 – 8 = 13