Laxmi

Students can Download Maths Chapter 2 Algebra Ex 2.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th 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 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.2

Question 1.
Fill in the blanks:

Question (i)
The solution of the equation ax + b = 0 is ………
Answer:
– \(\frac{b}{a}\)
Solution:
ax + b = 0
ax = – b
∴ x = – \(\frac{b}{a}\)

Question (ii)
If a and b are positive integers then the solution of the equation ax = b has to be always ………
Answer:
positive
Hint:
Since a & b are positive integers,
The solution to the equation ax = b is x = – \(\frac{b}{a}\) is also positive.

Question (iii)
One-sixth of a number when subtracted from the number itself gives 25. The number is ……..
Answer:
30
Hint:
Let the number be x.
As per question, when one sixth of number is subtracted from itself it gives 25
x – \(\frac{x}{6}\) = 25
∴ \(\frac{6x-x}{6}\) = 25
∴ \(\frac{5x}{6}\) = 25
∴ x = \(\frac{25×6}{3}\) = 5 x 6 = 30

Question (iv)
If the angles of a triangle are in the ratio 2 : 3 : 4 then the difference between the greatest and the smallest angle is
Answer:
40°
Hint:
Given angles are in the ratio 2 : 3 : 4
Let the angles be 2x, 3x & 4x
Since sum of the angles of a triangle is 180°,
We get
2x + 3x + 4x = 180
∴ 9x = 180
∴ X = \(\frac{180}{9}\) = 20°
∴ The angles are 2x = 2 x 20 = 40°
3x = 3 x 20 = 60°
4x = 4 x 20 = 80°
∴ Difference between greatest & smallest angle is
80° – 40° = 40°

Question (v)
In an equation a + b = 23. The value of a is 14 then the value of b is ……..
Answer:
b = 9
Hint:
Given equation is a + b = 23
a = 14
14 + b = 23
b = 23 – 14 = 9
b = 9

Question 2.
Say True or False

Question (i)
“Sum of a number and two times that number is 48” can be written as y + 2y = 48
Answer:
True
Hint:
Let the number be ‘y’
Sum of number & two times that number is 48
Can be written as y + 2y = 48 – True

Question (ii)
5(3x + 2) = 3(5x – 7) is a linear equation in one variable.
Answer:
True
Hint:
5 (3x + 2) = 3 (5x – 7) is a linear equation in one variable – ‘x’ – True

Question (iii)
x = 25 is the solution of one third of a number is less than 10 the original number.
Answer:
False
Hint:
One third of number is 10 less than original number.
Let number be ‘x’ Therefore let us frame the equation
\(\frac{x}{3}\) = x – 10
∴ x = 3x – 30
3x – x = 30
2x = 30
x = 15 is the solution

Question 3.
One number is seven times another. If their difference is 18, find the numbers.
Solution:
Let the numbers be x & y
Given that one number is 7 times the other & that the difference is 18.
Let x = 7y
also, x – y = 18 (given)
Substituting for x in the above
We get 7y – y = 18
∴ 6y = 18
y = \(\frac{18}{6}\) = 3
x = 7y = 7 x 3 = 21
The number are 3 & 21

Question 4.
The sum of three consecutive odd numbers is 75. Which is the largest among them?
Solution:
Given sum of three consecutive odd numbers is 75
Odd numbers are 1, 3, 5,1,9, 11, 13,……..
∴ The difference between 2 consecutive odd numbers is always 2. or in other words, if one odd number is x, the next odd number would be x + 2 and the next number would be x + 2 + 2 = x + 4
i.e x + 4
Since sum of 3 consecutive odd nos is 75
∴ x + x + 2 + x + 4 = 75
3x + 6 = 75 ⇒ 3x = 75 – 6
3x = 69
x = \(\frac{69}{3}\) = 23
The odd numbers are 23, 23 + 2, 23 + 4
i.e 23, 25, 27
∴ Largest number is 27.

Question 5.
The length of a rectangle is \(\frac{1}{3}\)rd of its breadth. If its perimeter is 64 m, then find the length and breadth of the rectangle.
Solution:

Let length & breadth of rectangle be l and ‘6’ respectively
Given that length is \(\frac{1}{3}\) of breadth,
∴ l = \(\frac{1}{3}\) x b ⇒ l = \(\frac{b}{3}\) ⇒ b = 3l …..(1)
Also given that perimeter is 64 m
Perimeter = 2 x (l + b)
2 x l + 2 x b = 64
Substituting for vahie of b from (1), we get
2l + 2(3l) = 64
∴ 2l + 6l = 64
8l = 64
∴ l = \(\frac{64}{8}\) = 8 m
b = 3l = 3 x 8 = 24 m
length l = 8 m & breadth h = 24 m

Question 6.
A total of 90 currency notes, consisting only of X5 and ?10 denominations, amount to ₹ 500. Find the number of notes in each denomination.
Solution:
Let the number of ₹ 5 notes be ‘x’
And number of ₹ 10 notes be ly ’
Total numbers of notes is x + y = 90 (given)
The total value of the notes is 500 rupees.
Value of one ₹ 5 rupee note is 5
Value of x ₹ 5 rupee notes is 5 × x = 5x
Value ofy ₹ 10 rupee notes is 10 × y = 10y
∴ The total value is 5x + 10y which is 500
we have 2 equations:
x + y = 90
5x + 1oy = 500
Multiplying both sides of (1) by 5, we get
5 × x + 5 x y = 90 x 5
5x + 5y = 450
Subtracting (3) from (2), we get

∴ y = \(\frac{50}{5}\) = 10
Substitute y = 10 in equation (1)
x + y = 90 ⇒ x + 10 = 90
⇒ x = 90 – 10 ⇒ x = 80
There are ₹ 5 denominations are 80 numbers and ₹ 10 denominations are 10 numbers

Question 7.
At present, Thenmozhi’s age is 5 years more than that of Murali’s age. Five years ago, the ratio of Thenmozhi’s age to Murali’s age was 3:2. Find their present ages.
Solution:
Let present ages ofThenmozhi & Murali be l’ & ‘m’
Given that at present
Thenmozhi’s age is 5 years more than Murali
∴ t = m + 5
5 years ago, Thenmozhi’s age would be t – 5
& Murali’s age would be m – 5
Ratio of their ages is given as 3 : 2
∴ \(\frac{t-5}{m-5}\) = \(\frac{3}{2}\) [∴ By cross multiplication]

2(t – 5) = 3(m – 5)
2 x t – 2 x 5 = 3 x m – 3 x 5 ⇒ 2t – 10 = 3m – 15
Substituting for t from (1)
2(m + 5) – 10 = 3m – 15
2m + 10 – 10 = 3m – 15
2m = 3m – 15
3m – 2m = 15
m = 15
t = m + 5 = 15 + 5 = 20
∴ Present ages of Thenmozhi & Murali are 20 & 15

Question 8.
A number consists of two digits whose sum is 9. If 27 is subtracted from the original number, its digits are interchanged. Find the original number.
Solution:
Let the units/digit of a number be ‘u’ & tens digit of the number be ‘t’
Given that sum of it’s digits is 9
∴ t + u = 9 ….(1)
If 27 is subtracted from original number, the digits are interchanged
The number is written as 10t + u
[Understand: Suppose a 2 digit number is 21,
it can be written as 2 x 10 + 1
∴ 32 = 3 x 10 + 2
45 = 4 x 10 + 5
tu = t x 10 + u = 10t + u]
Given that when 27 is subtracted, digits interchange
10t + u – 27 = 10u + t (number with interchanged digits)
∴ By transposition & bringing like variables together
10t – t + 10u =27
∴ 9t – 9u = 27
Dividing by ‘9’ throughout, we get
\(\frac{9t}{9}\) – \(\frac{9u}{9}\) = \(\frac{27}{9}\) ⇒ t – u = 3 ….(2)
Solving (1) & (2)

t = \(\frac{12}{2}\) = 6
∴ u = 3
t = 6 substitute in (1)
t + u = 9
⇒ 6 + u = 9
⇒ u = 9 – 6 = 3
Hence the number is 63.

Question 9.
The denominator of a fraction exceeds its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, we get \(\frac{3}{2}\). Find the original fraction.
Solution:
Let the numerator & denominator be ‘n’ & ‘d’
Given that denominator exceeds numerator by 8
∴ d = n + 8
If numerator increased by 17 & denominator decreased by 1,
it becomes (n + 17) & (d – 1), fraction is \(\frac{3}{2}\).
i.e = \(\frac{n+17}{d-1}\) = \(\frac{3}{2}\) by cross multiplying, we get
\(\frac{n+17}{d-1}\) = \(\frac{3}{2}\)
2(n + 17) = 3(d – 1)
2n + 2 x 17 = 3d – 3

∴ 34 + 3 = 3d – 2n
∴ 3d – 2n = 37
Substituting eqn. (1) in (2), we get,
3 x (n + 8) – 2n = 37
3n + 3 x 8 – 2n = 37

∴ n = 37 – 24 = 13
d = n + 8 = 13 + 8 = 21
The fraction is \(\frac{n}{d}\) = \(\frac{13}{21}\)

Question 10.
If a train runs at 60 km/hr it reaches its destination late by 15 minutes. But, if it runs at 85 kmph it is late by only 4 minutes. Find the distance to be covered by the train.
Solution:
Let the distance to be covered by train be ‘d’
Using the formula, time take (t) = \(\frac{Distance}{Speed}\)
Case 1:
If speed = 60 km/h
The time taken is 15 minutes more than usual (t + \(\frac{15}{60}\))
Let usual time taken be ‘t’ hrs.
Caution:
Since speed is given in km/hr, we should take care to maintain all units such as time should be in hour and distance should be in km.
Given that in case 1, it takes 15 min. more
15 m = \(\frac{15}{60}\) hr = \(\frac{1}{4}\) hr.
∴ Substituting in formula
\(\frac{Distance}{Speed}\) = time
∴ \(\frac{d}{60}\) = t + \(\frac{1}{4}\)
Since usually it takes ‘t’ hr, but when running at 60 km/h, it takes 15 min (\(\frac{1}{4}\)) extra.
Multiplying by 60 on both sides
d = 60 x t + 60 x \(\frac{1}{4}\) = 60t + 15 …..(1)
Case 2:
Speed is given as 85 km/h
Time taken is only 4 min (\(\frac{4}{60}\) hr) more than usual time
time taken = (t + \(\frac{1}{15}\)) hr. (\(\frac{4}{60}\) = \(\frac{1}{15}\))
Using the formula,
\(\frac{Distance}{Speed}\) = time
∴ \(\frac{d}{85}\) = t + \(\frac{1}{15}\)
Multiplying by 85 on both sides
\(\frac{d}{85}\) x 85 = 85 x t + 85 x \(\frac{1}{15}\)
∴ d = 85 t + \(\frac{17}{3}\)
From (1) & (2), we will solve for ‘t’
Equating & eliminating ‘d we get

By transposing, we get
15 – \(\frac{17}{3}\) = 85t – 60t
\(\frac{45-17}{3}\) = 25t
∴ 25t = \(\frac{28}{3}\)
∴ t = \(\frac{28}{3×25}\) = \(\frac{28}{75}\) hr (\(\frac{28}{75}\) x 60 = 22.4 min)
Substituting this value of ‘t’ in eqn. (1), we get
d = 60t+ 15 = 60 x \(\frac{28}{75}\) + 15 = \(\frac{1680}{75}\) + 15 = 22.4 + 15
= 37.4 km

Objective Type Questions

Question 11.
Sum of a number and its half is 30 then the number is ……..
(a) 15
(b) 20
(c) 25
(d) 40
Answer:
(b) 20
Hint:
Let number be V
half of number is \(\frac{x}{2}\)
Sum of number and it’s half is given by
x + \(\frac{x}{2}\) = 30 [Multiplying by 2 on both sides]
2x + x = 30 x 2
3x = 60
x = \(\frac{60}{3}\) = 20

Question 12.
The exterior angle of a triangle is 120° and one of its interior opposite angle 58°, then the other opposite interior angle is ………
(a) 62°
(b) 72°
(c) 78°
(d) 68°
Answer:
(a) 62°

Hint:
As per property of ∆, exterior angle is equals to sum of interior opposite angles Let the other interior angle to be found be ‘x’
∴ x + 58 = 120°
∴ x = 120 – 58 = 62°

Question 13.
What sum of money will earn ₹ 500 as simple interest in 1 year at 5% per annum?
(a) 50000
(b) 30000
(c) 10000
(d) 5000
Answer:
(c) 10000
Hint:
Let sum of money be ‘P’
Time period (n) is given as 1 yr.
Rate of simple interest (r) is given as 5% p.a
∴ As per formula for simple interest.
S.I = \(\frac{Pxrxn}{100}\) = \(\frac{px5x1}{100}\)
P x 5 x 1 = 500 x 100
∴ P = \(\frac{500×100}{5}\) = 100 x 100 = 10,000

Question 14.
The product of LCM and HCF of two numbers is 24. If one of the number is 6, then the other number is ………
(a) 6
(b) 2
(c) 4
(d) 8
Answer:
(c) 4
Hint:
Product of LCM & HCF of 2 numbers is always product of the numbers. [this is property]
Product of LCM & HCF is given as 24.
∴ Product of the 2 nos. is 24
Given one number is 6.
Let other number be ‘x’
∴ 6 × x = 24
∴ x = \(\frac{24}{6}\) = 4

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.4

Miscellaneous Practice Problems

Question 1.
6² × 6^m = 6⁵, find the value of ‘m’
Solution:
6² × 6^m = 6⁵
6^2+m = 6⁵ [Since a^m × aⁿ= a^m+n]
Equating the powers, we get
2 + m = 5
m = 5 – 2 = 3

Question 2.
Find the unit digit of 124¹²⁸ × 126¹²⁴
Solution:
In 124¹²⁸, the unit digit of base 124 is 4 and the power is 128 (even power).
Therefore, unit digit of 124¹²⁸ is 4.
Also in 126¹²⁴, the unit digit of base 126 is 6 and the. power is 124 (even power).
Therefore, unit digit of 126¹²⁴ is 6.
Product of the unit digits = 6 × 6 = 36
∴ Unit digit of the 124¹²⁸ × 126¹²⁴ is 6.

Question 3.
Find the unit digit of the numeric expression: 16²³ + 71⁴⁸ + 59⁶¹
Solution:
In 16²³, the unit digit of base 16 is 6 and the power is 23 (odd power).
Therefore, unit digit of 16²³ is 6.
In 71⁴⁸, the unit digit of base 71 is 1 and the power is 48 (even power).
Therefore, unit digit of 71⁴⁸ is 1.
Also in 59⁶¹, the unit digit of base 59 is 9 and the power is 61 (odd power).
Therefore, unit digit of 59⁶¹ is 9.
Sum of the unit digits = 6 + 1 + 9 = 16
∴ Unit digit of the given expression is 6.

Question 4.
Find the value of

Solution:

Question 5.
Identify the degree of the expression, 2a³be + 3a³b + 3a³c – 2a²b²c²
Solution:
The terms of the given expression are 2a³bc, 3a³b + 3a³c – 2a²b²c²
Degree of each of the terms: 5,4,4,6.
Terms with the highest degree: – 2a²b²c²
Therefore degree of the expression is 6.

Question 6.
If p = -2, q = 1 and r = 3, find the value of 3p²q²r.
Solution:
Given p = -2; q = 1; r = 3
∴ 3p²q²r = 3 × (-2)² × (1)² × (3)
= 3 × (-2 × 1)² × (3) [Since a^m × b^m = (a × b)^m]
= 3 × (-2)² × (3)
= 3 × (-1)² × 2² × 3
= 3¹⁺¹ × 1 × 4 [Since a^m × aⁿ = a^m+n]
= 3² × 4 = 9 × 4
∴ 3p²q²r = 36

Challenge Problems

Question 7.
LEADERS is a WhatsApp group with 256 members. Every one of its member is an admin for their own WhatsApp group with 256 distinct members. When a message is posted in LEADERS and everybody forwards the same to their own group, then how many members in total will receive that message?
Solution:
Members of the groups LEADERS = 256
Members is individual groups of the members of LEADERS = 256
Total members who receive the message

= 256 × 256 = 2⁸ × 2⁸
2⁸⁺⁸ = 2¹⁶
= 65536
Totally 65536 members receive the message.

Question 8.
Find x such that 3^x+2 = 3^x + 216.
Solution:
Given 3^x+2 = 3^x + 216 ; 3^x+2 = 3^x + 216
Dividing throught by 3^x, we get

Equating the powers of same base

Question 9.
If X = 5x² + 7x + 8 and Y = 4x² – 7x + 3, then find the degree of X + Y.
Solution:
Given x = 5x² + 7x + 8
X + Y = 5x² + 7x + 8 + (4x² – 7x + 3)
= (5x² + 4x²) + (7x – 7x) + (8 + 3)
= x² (5 + 4) + x(7 – 7) + (8 + 3) = 9x² + 11
Degree of the expression is 2.

Question 10.
Find the degree of (2a² + 3ab – b²) – (3a² -ab- 3b²)
Solution:
(2a² + 3ab – b²) – (3a² – ab – 3b²)
= (2a² + 3ab – b²) + (- 3a² + ab + 3b²)
= 2a² + 3ab – b² – 3a² + ab + 3b²
= 2a² – 3a² + 3ab + ab + 3b² – b²
= 2a² – 3a² + ab (3 + 1) + b²(3 – 1)
= – a² + 4 ab + 2b²
Hence degree of the expression is 2.

Question 11.
Find the value of w, given that x = 4, y = 4, z = – 2 and w = x² – y² + z² – xyz.
Solution:
Given x = 3; y = 4 and z = -2.
w = x² – y² + z² – xyz
w = 3² – 4² + (-2)² – (3)(3)(-2)
w = 9 – 16 + 4 + 24
w = 37 – 16
w = 21

Question 12.
Simplify and find the degree of 6x² + 1 – [8x – {3x² – 7 – (4x² – 2x + 5x + 9)}]
Solution:
6x² + 1 – [8x – (3x² – 7 – (4x² – 2x + 5x + 9)}]
= 6x² + 1 – [8x – {3x² – 7 – 4x² – 2x + 5x + 9}]
= 6x² + 1 – [8x – 3x² + 7 + 4x² – 2x + 5x + 9}]
= 6x² – 1 – [8x + 3x² – 7 – 4x² + 2x – 5x – 9]
= 6x² + 3x² – 4x² – 8x + 2x – 5x – 1 – 7 – 9]
= x²(6 + 3 – 4) + x(8 + 2 – 5) – 15
= 5x² – 11x – 15
Degree of the expression is 2.

Question 13.
The two adjacent sides of a rectangle are 2x² – 5xy + 3z² and 4xy – x² – z². Find the perimeter and the degree of the expression.
Solution:
Let the two adjacent sides of the rectangle as
l = 2x² – 5xy + 3z² and b = 4xy – x²y + 3z²

Perimeter of the rectangle
= 2(l + b) = 2(2x² – 5xy + 3z² + 4xy – x² – z²)
= 4x² – 10xy + 6z² + 8xy – 2x² – 2z²
= 4x² – 2x² – 10xy + 8xy + 6z² – 2z²
= x²(4 – 2) + xy (-10 + 8) + z² (6 – 2z²)
Perimeter = 2x² – 2xy + 4z²
Degree of the expression is 2.

Students can Download Maths Chapter 2 Algebra Ex 2.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th 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 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.1

Question 1.
Fill in the blanks

Question (i)
The value of x in the equation x +5 12 ¡s ……….
Answer:
7
Hint:
Given,
x + 5 = 12
x = 12 – 5 = 7 (by transposition method)
Value of x is 7

Question (ii)
The value ofy in the equation y – 9 = (-5) + 7 is ……….
Answer:
11
Hint:
Given,
y – 9 = (- 5) + 7
y – 9 = 7 – 5 (re-arranging)
y – 9 = 2
∴ y = 2 + 9 = 11 (by transposition method)

Question (iii)
The value of m in the equation 8m = 56 is ………
Answer:
7
Hint:
Given,
8m = 56
Divided by 8 on both sides
\(\frac{8xm}{8}\) = \(\frac{56}{8}\)
∴ m = 7

Question (iv)
The value ofp in the equation \(\frac{2p}{3}\) = 10 is ……….
Answer:
1
Hint:
Given,
\(\frac{2p}{3}\) = 10
Multiplying by 3 on both sides,

∴ p = 15

Question (v)
The linear equation in one variable has ……… Solution.
Answer:
one.

Question 2.
Say True or False.

Question (i)
The shifting of a number from one side of an equation to other is called transposition.
Answer:
True

Question (ii)
Linear equation in one variable has only one variable with power 2.
Answer:
False
[Linear equation in one variable has only one variable with power one – correct statement]

Question 3.
Match the following :

(A) (i), (ii), (iv), (iii), (v)
(B) (iii), (iv), (i), (ii), (v)
(C) (iii), (i), (iv), (v), (ii)
(D) (iii), (i), (v), (iv), (ii)
Answer:
(C) (iii),(i), (iv), (v), (ii)
Hint:
a. \(\frac{x}{2}\) = 10,
multiplying by 2 on both sides, we get
\(\frac{x}{2}\) x 2 = 10 x 2 ⇒ x = 20

b. 20 = 6x – 4
by transposition ⇒ 20 + 4 = 6x
6x = 24
dividing by 6 on both sides,
\(\frac{6x}{6}\) = \(\frac{24}{6}\) ⇒ x = 4

c. 2x – 5 = 3 – x
By transposing the variable ‘x’, we get
2x – 5 + x = 3
by transposing – 5 to other side,
2x + x = 3 + 5
∴ 3x = 8

∴ x = \(\frac{8}{3}\)

d. 7x – 4 – 8x = 20
by transposing – 4 to other side,
7x – 8x = 20 + 4
– x = 24
∴ x = – 24
\(\frac{4}{11}\) – x = \(\frac{-7}{11}\)
Transposing \(\frac{4}{11}\) to other side,
– x = \(\frac{-7}{11}\)\(\frac{-4}{11}\) = \(\frac{-7-4}{11}\) = \(\frac{-11}{11}\) = – 1
∴ – x = – 1 ⇒ x = 1

Question 4.
Find x:

Question (i)
\(\frac{2x}{3}\) – 4 = \(\frac{10}{3}\)
Solution:
Transposing -4 to other side, it becomes +4
∴ \(\frac{2x}{3}\) = \(\frac{10}{3}\) + 4
Taking LCM & adding,
\(\frac{2x}{3}\) = \(\frac{10}{3}\) + \(\frac{4}{1}\) = \(\frac{10+12}{3}\) = \(\frac{22}{3}\)
\(\frac{2x}{3}\) = \(\frac{22}{3}\)
Multiplying by 3 on both sides

⇒ 2x = 22
dividing by 2 on both sides,
We get \(\frac{2x}{2}\) = \(\frac{22}{2}\)
∴ x = 11

Question (ii)
y + \(\frac{1}{6}\) – 3y = \(\frac{2}{3}\)
Solution:
Transposing \(\frac{1}{6}\) to the other side,
y – 3y = \(\frac{2}{3}\) – \(\frac{1}{6}\)
Taking LCM,
– 2y = \(\frac{2}{3}\) – \(\frac{1}{6}\) = \(\frac{2×2-1}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ – 2y = \(\frac{1}{2}\) ⇒ 2y = – \(\frac{1}{2}\)
dividing by 2 or both sides.

Question (iii)
\(\frac{1}{3}\) – \(\frac{x}{3}\) = \(\frac{7x}{12}\) + \(\frac{5}{4}\)
Transposing \(\frac{-x}{3}\) to the other side, it becomes + \(\frac{x}{3}\)
∴ \(\frac{1}{3}\) = \(\frac{7x}{12}\) + \(\frac{5}{4}\) + \(\frac{x}{3}\)
Transposing \(\frac{5}{4}\) to the other side, it becomes \(\frac{-5}{4}\)
\(\frac{1}{3}\) + \(\frac{5}{4}\) = \(\frac{7x}{12}\) + \(\frac{x}{3}\)
Multiply by 12 throughout
[we look at the denominators 3,4, 12, 3 and take the LCM, which is 12]

4 – 15 = 7x + x × 4
-11 = 7x + 4x
11x = – 11
x = -1

Question 5.
Find x:

Question (i)
-3(4x + 9) = 21
Solution:

Expanding the bracket,
-3 × 4x + (-3) × 9 = 21
-12x + (-27) = 21
-12x – 27 = 21
Transposing – 27 to other side, it becomes +27
-12x = 21 + 27 = 48
12x = 48 ⇒ 12x = -48
Dividing by 12 on both sides

⇒ x = – 4

Question (ii)
20 – 2 ( 5 – p) = 8
Solution:

Expanding the bracket,
20 – 2 x 5 – 2 x (-p) = 8
20 – 10 + 2 + p = 8 (-2 x -P = 2p)
10 + 2p = 8 transporting 10 to other side
2P = 8 – 10 = -2
∴ 2p = -2
∴ p = -1

Question (iii)
(7x – 5) – 4(2 + 5x) = 10(2 – x)
Solution:

Expanding the brackets,
7x – 5 – 4 × 2 – 4 × 5x = 10 × 2 + 10 × (-x)
7x – 5 – 8 – 20x = 20 – 10x
7x – 13 – 20x = 20 – 10x
Transposing 10x & -13, we get
7x – 13 – 20x + 10x = 20
7x – 20x + 10x = 20 + 13,
Simplifying,
-3x = 33
∴ 3x = -33
x = \(\frac{-33}{3}\) = -11
x = -11

Question 6.
Find x and m:

Question (i)
\(\frac{3x-2}{4}\) – \(\frac{(x-3)}{5}\) = -1
Solution:
\(\frac{3x-2}{4}\) – \(\frac{(x-3)}{5}\)
Taking LCM on LHS, [LCM of 4 & 5 is 20]

∴ 11x + 2 = -20
∴ 11x = – 20 – 2 = – 22
x = \(\frac{-22}{11}\) = -2
x = -2

Question (ii)
\(\frac{m+9}{3m+15}\) = \(\frac{5}{3}\)
Solution:
\(\frac{m+9}{3m+15}\) = \(\frac{5}{3}\)
Cross multiplying, we get

∴ (m + 9) x 3 = 5 x (3m + 15)
m x 3 + 9 x 3 = 5 x 3m + 5 x 15

Transporting 3m & 75, we get
27 – 75 = 15m – 3m
-48 = 12m

⇒ m = -4

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 2 Chapter 3 Algebra Ex 3.3

Question 1.
Fill in the blanks.

(i) The degree of the term a³b²c⁴d² is _______
(ii) Degree of the constant term is _______
(iii) The coefficient of leading term of the expression 3z²y + 2x – 3 is _______
Answers:
(i) 11
(ii) 0
(iii) 3

Question 2.
Say True or False.

(i) The degree of m² n and mn² are equal.
(ii) 7a²b and -7ab² are like terms.
(iii) The degree of the expression -4x² yz is -4
(iv) Any integer can be the degree of the expression.
Answers:
(i) True
(ii) False
(iii) False
(iv) True

Question 3.
Find the degree of the following terms.
(i) 5x²
(ii) -7 ab
(iii) 12pq² r²
(iv) -125
(v) 3z
Solution:
(i) 5x²
In 5x², the exponent is 2. Thus the degree of the expression is 2.

(ii) -7ab
In -7ab, the sum of powers of a and b is 2. (That is 1 + 1 = 2).
Thus the degree of the expression is 2.

(iii) 12pq² r²
In 12pq² r², the sum of powers of p, q and r is 5. (That is 1 +2 + 2 = 5).
Thus the degree of the expression is 5.

(iv) -125
Here – 125 is the constant term. Degree of constant term is 0.
∴ Degree of -125 is 0.

(v) 3z
The exponent is 3z is 1.
Thus the degree of the expression is 1.

Question 4.
Find the degree of the following expressions.
(i) x³ – 1
(ii) 3x² + 2x + 1
(iii) 3t⁴ – 5st² + 7s²t²
(iv) 5 – 9y + 15y2 – 6y3
(v) u⁵ + u⁴v + u³v² + u₂v³ + uv⁴
Solution:
(i) x³ – 1
The terms of the given expression are x³, -1
Degree of each of the terms: 3,0
Terms with highest degree: x³.
Therefore, degree of the expression is 3.

(ii) 3x² + 2x + 1
The terms of the given expression are 3x², 2x, 1
Degree of each of the terms: 2, 1, 0
Terms with highest degree: 3x²
Therefore, degree of the expression is 2.

(iii) 3t⁴ – 5st² + 7s²t²
The terms of the given expression are 3t⁴, – 5st², 7s³t²
Degree of each of the terms: 4, 3, 5
Terms with highest degree: 7s²t²
Therefore, degree of the expression is 5.

(iv) 5 – 9y + 15y² – 6y³
The terms of the given expression are 5, – 9y , 15y², – 6y³
Degree of each of the terms: 0, 1, 2, 3
Terms with highest degree: – 6y³
Therefore, degree of the expression is 3.

(v) u⁵ + u⁴v + u³v² + u₂v³ + uv⁴
The terms of the given expression are u⁵, u⁴v , u³v², u²v³, uv⁴
Degree of each of the terms: 5, 5, 5, 5, 5
Terms with highest degree: u⁵, u⁴v , u³v², u²v³, uv⁴
Therefore, degree of the expression is 5.

Question 5.
Identify the like terms : 12x³y²z, – y³x²z, 4z³y²x, 6x³z²y, -5y³x²z
Solution:
-y³ x²z and -5y³x²z are like terms.

Question 6.
Add and find the degree of the following expressions.
(i) (9x + 3y) and (10x – 9y)
(ii) (k² – 25k + 46) and (23 – 2k² + 21 k)
(iii) (3m²n + 4pq²) and (5nm² – 2q²p)
Solution:
(i) (9x + 3y) and (10x – 9y)
This can be written as (9x + 3y) + (10x – 9y)
Grouping the like terms, we get
(9x + 10x) + (3y – 9y) = x(9 + 10) + y(3 – 0) = 19x + y(-6) = 19x – 6y
Thus degree of the expression is 1.

(ii) (k² – 25k + 46) and (23 – 2k² + 21k)
This can be written as (k² – 25k + 46) + (23 – 2k² + 21k)
Grouping the like terms, we get
(k² – 2k²) + (-25 k + 21 k) + (46 + 23)
= k² (1 – 2) + k(-25 + 21) + 69 = – 1k² – 4k + 69
Thus degree of the expression is 2.

(iii) (3m²n + 4pq²) and (5nm² – 2q²p)
This can be written as (3m²n + 4pq²) + (5nm² – 2q²p)
Grouping the like terms, we get
(3m²n + 5m²n) + (4pq² – 2pq²)
= m²n(3 + 5) + pq²(4 – 2) = 8m²n + 2pq²
Thus degree of the expression is 3.

Question 7.
Simplify and find the degree of the following expressions.
(i) 10x² – 3xy + 9y² – (3x² – 6xy – 3y²)
(ii) 9a⁴ – 6a³ – 6a⁴ – 3a² + 7a³ + 5a²
(iii) 4x² – 3x – [8x – (5x² – 8)]
Solution:
(i) 10x² – 3xy + 9y² – (3x² – 6xy – 3y²)
= 10x² – 3xy + 9y² + (-3x² + 6xy + 3y²)
= 10x² – 3xy + 9y² – 3x² + 6xy + 3y²
= (10x² – 3x²) + (- 3xy + 6xy) + (9y² + 3y²)
= x²(10 – 3) + xy(-3 + 6) + y²(9 + 3)
= x²(7) + xy(3) + y²(12)
Hence, the degree of the expression is 2.

(ii) 9a⁴ – 6a³ – 6a⁴ – 3a² + 7a³ + 5a²
= (9a⁴ – 6a⁴) + (- 6a³ + 7a³) + (- 3a² + 5a²)
= a⁴(9-6) + a³ (- 6 + 7) + a²(-3 + 5)
= 3a⁴ + a³ + 2a²
Hence, the degree of the expression is 4.

(iii) 4x² – 3x – [8x – (5x² – 8)]
= 4x² – 3x – [8x + -5x² + 8)]
= 4x² – 3x – [8x – 5x² – 8]
= 4x² – 3x – 8x + 5x² – 8
(4x² + 5x²) + (- 3x – 8x) – 8
= x²(4+ 5) + x(-3-8) – 8
= x²(9) + x(- 11) – 8
= 9x² – 11x – 8
Hence, the degree of the expression is 2.

Objective Type Question

Question 8.
3p² – 5pq + 2q² + 6pq – q² +pq is a
(i) Monomial
(ii) Binomial
(iii) Trinomial
(iv) Quadrinomial
Answer:
(iii) Trinomial

Question 9.
The degree of 6x⁷ – 7x³ + 4 is
(i) 7
(ii) 3
(iii) 6
(iv) 4
Answer:
(i) 7

Question 10.
If p(x) and q(x) are two expressions of degree 3, then the degree of p(x) + q(x) is
(i) 6
(ii) 0
(iii) 3
(iv) Undefined
Answer:
(iii) 3

Students can Download Maths Chapter 1 Life Mathematics Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Additional Questions

Additional Questions and Answers

Question 1.
Fill in the blanks

Question (a)
Percent means ……….
Answer:
Per hundred or out of hundred.

Question (b)
Percent is useful in ………
Answer:
Comparing quantities easily

Question (c)
The formula to find the increased quantity ………
Answer:
I = (1 + \(\frac{x}{100}\))

Question (d)
The formula to find the decreased quantity ………
Answer:
D = (1 + \(\frac{x}{100}\))

Question (e)
Gain or profit % ……..
Answer:
(\(\frac{Profit}{C.P}\) x 100)%

Question (f)
Loss % = ………..
Answer:
(\(\frac{Loss}{C.P}\) x 100)%

Question (g)
S.P = ………. (if gain % is given)
Answer:

Question (h)
C.P = ……… (if gain % is given)
Answer:

Question (f)
S.P = ………. (if loss % is given)
Answer:

Question (h)
C.P = ………. (if loss % is given)
Answer:

Question (k)
Selling price = Marked price – …………
Answer:
Discount

Question (l)
Cost price = Cost price + ……….
Answer:
Over head expenses

Question 2.
If y% of ₹ 1000 is 600, find the value ofy.
Solution:
y% of 1000 = 600
\(\frac{y}{100}\) x 1000 = 600
y = \(\frac{600}{10}\)
y = 60

Question 3.
A number when decreased by 10% becomes 900. Then find the number.
Solution:
Let the number be ‘x’
Given x – \(\frac{10}{100}\)x = 900
\(\frac{100x-10x}{100}\) = 900
\(\frac{90x}{100}\) = 900
x = \(\frac{900×100}{90}\) = 1000

Question 4.
If the population in a city has increased from 5,00,000 to 7,00,000 in a year, find the percentage increase in population.
Solution:
Increase in population = 7,00,000 – 5,00,000 = 2,00,000
Percentage increase in population = \(\frac{2,00,000}{5,00,000}\) x 100 = 40%

Question 5.
If the selling price of a refrigerator is equal to \(\frac{10}{8}\) of its cost price, then find the gain/ profit percent.
Solution:
Let the C.P of the refrigerator be x
S.P = \(\frac{10}{8}\)x
Profit = S.P – C.P = \(\frac{10}{8}\)x – x = \(\frac{10x-8x}{8}\) = \(\frac{2x}{8}\)

Question 6.
Karnan bought a dishwasher for ₹ 32,300 and paid ₹ 2700 for its transportation. Then he sold it for X 38,500. Find his gain or loss percent.
Solution:
Total C.P of the dishwasher = C.P + Overhead expenses.
= ₹ 32300 + ₹ 2700 = ₹ 35000
S.P = ₹ 38500
Therefore, we find S.P > C.P

Question 7.
The value of a car 2 years ago was ₹ 1,40,000. It depreciates at the rate of 4% p.a. Find its present value.
Solution:
Depreciated value = P(1 + \(\frac{r}{100}\))ⁿ = 1,40,000(1 – \(\frac{4}{100}\))²
= 1,40,000(\(\frac{96}{100}\)) x (\(\frac{96}{100}\)) = ₹ 1,29,024

Question 8.
Find the difference in C.I and S.I for P = ₹ 10,000, r = 4% p.a, n = 2 years.
Solution:
C.I – S.I = P(\(\frac{r}{100}\))² = 10,000(\(\frac{4}{100}\))²
= 10,000 x \(\frac{4}{100}\) x \(\frac{4}{100}\) = ₹ 16

Question 9.
Find the C.I for the given Principal = ₹ 8,000, r = 5% p.a, n = 2 years
Amount A = P(1 + \(\frac{r}{100}\))ⁿ = 8000(1 + \(\frac{5}{100}\))²
= 8000 x \(\frac{105}{100}\) x \(\frac{105}{100}\)
= 8000 x \(\frac{21}{20}\) x \(\frac{21}{20}\)
A = ₹ 8820
Cl = A – P = 8820 – 8000 = 820

Question 10.
Find the S.I for the principal P = ₹ 16,000, r = 5% p.a, n = 3 years
Solution:
P = ₹ 16,000, n= 3 years, r = 5%
S.I = \(\frac{Pnr}{100}\) = \(\frac{16000x3x5}{100}\) = ₹ 2400

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 2 Chapter 3 Algebra Ex 3.1

Question 1.
Fill in the blanks.

1. The exponential form 14⁹ should be read as ______
2. The expanded form of p³ q² is ______
3. When base is 12 and exponent is 17, its e×ponential form is _____
4. The value of (14 × 21)⁰ is _____

Answers:

1. 14 Power 9
2. p × p × p × q × q
3. 12¹⁷
4. 1

Question 2.
Say True or False.

1. 2³ × 3² = 65
2. 2⁹ × 3² = (2 × 3)^9×2
3. 3⁴ × 3⁷= 3¹¹
4. 2⁰ × 1000⁰
5. 2³ < 3²

Answers:

1. False
2. False
3. True
4. True
5. True

Question 3.
Find the value of the following.

1. 2⁶
2. 11²
3. 5⁴
4. 9³

Solution:

1. 2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64
2. 11² = 11 × 11 = 121
3. 5⁴ = 5 × 5 × 5 × 5 = 625
4. 9³ = 9 × 9 × 9 = 729

Question 4.
Express the following in e×ponential form.

1. 6 × 6 × 6 × 6
2. t × t
3. 5 × 5 × 7 × 7 × 7
4. 2 × 2 × a × a

Solution:

1. 6 × 6 × 6 × 6 = 6^1+1+1+1 = 64 [Since a^m × aⁿ = a^m+n]
2. t × t = t¹⁺¹ = t²
3. 5 × 5 × 7 × 7 × 7 = 5¹⁺¹ × 7¹⁺¹⁺¹ = 5² × 7³
4. 2 × 2 × a × a = 2¹⁺¹ × a¹⁺¹ = 2² × a² = (2a)²

Question 5.
E×press each of the following numbers using e×ponential form,
(i) 512
(ii) 343
(iii) 729
(iv) 3125
Solution:
(i) 512

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2⁹ [Using product rule]

(ii) 343

343 = 7 × 7 × 7 = 7¹⁺¹⁺¹
= 7³ [Using product rule]

(iii) 729

729 = 3 × 3 × 3 × 3 × 3 × 3
= 3⁶ [Using product rule]

(iv) 3125

3125 = 5 × 5 × 5 × 5 × 5
= 5⁵ [Using product rule]

Question 6.
Identify the greater number in each of the following.
(i) 6³ or 3⁶
(ii) 5³ or 3⁵
(iii) 2⁸ or 8²
Solution:
(i) 6³ or 3⁶
6³ = 6 × 6 × 6 = 36 × 6 = 216
3⁶ = 3 × 3 × 3 × 3 × 3 × 3 = 729
729 > 216 gives 3⁶ > 6³
∴ 36 is greater.

(ii) 5³ or 3⁵
5³ = 5 × 5 × 5 = 125
3⁵ = 3 × 3 × 3 × 3 × 3 = 243
243 > 125 gives 3⁵ > 5³
∴ 3⁵ is greater.

(iii) 2⁸ or 8²
2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
8² = 8 × 8 = 64
256 > 64 gives 2⁸ > 8²
∴ 2⁸ is greater.

Question 7.
Simplify the following
(i) 7² × 3⁴
(ii) 3² × 2⁴
(iii) 5² × 10⁴
Solution:
(i) 7² × 3⁴ = (7 × 7) × (3 × 3 × 3 × 3)
= 49 × 81 = 3969

(ii) 3² × 2⁴ = (3 × 3) × (2 × 2 × 2 × 2)
= 9 × 16 = 144

(iii) 5² × 10⁴ = (5 × 5) × (10 × 10 × 10 × 10)
= 25 × 10000 = 2,50,000

Question 8.
Find the value of the following.
(i) (-4)²
(ii) (-3) × (-2)³
(iii) (-2)³ × (-10)³
Solution:
(i) (-4)² = (-1)² × (4)² [since a^m × b^m = (a × b)^m]
= 1 × 16 = 16 [since (-1)ⁿ = 1 if n is even]

(ii) (-3) × (-2)³ = (-1) × (-3) × (-1)³ × (-2)³
= (-1)⁴ × 24 [Grouping the terms of same base]
= 24

(iii) (-2)³ × (-10)³ = (-1)³ × (-2)³ × (-1)³ × (-10)³
= (-1)³⁺³ × 2³ × 10³ [Grouping the terms of same base]
= (-1)⁶ × (2 × 10)³
[∵ a^m × b^m = (a × b)^m]
= 1 × 20³ [since (-1)ⁿ = 1 if n is even]
= 8000

Question 9.
Simplify using laws of exponents.
(i) 3⁵ × 3⁸
(ii) a⁴ × a¹⁰
(iii) 7x × 7²
(iv) 2⁵ ÷ 2³
(v) 18⁸ ÷ 18⁴
(vi) (6⁴)³
(vii) (x^m)⁰
(viii) 9⁵ × 3⁵
(ix) 3^y × 12^y
(x) 25⁶ × 5⁶
Solution:

Question 10.
If a = 3 and b = 2, then find the value of the following.
(i) a^b + b^a
(ii) a^a – b^b
(iii) (a + b)^b
(iv) (a – b)^a
Solution:
(i) a^b + b^a
a = 3 and b = 2
we get 3² + 2³ = (3 × 3) + (2 × 2 × 2) = 9 + 8 = 17

(ii) (a^a – b^b)
Substituting a = 3 and b = 2
we get 3² – 2² = (3 × 3 × 3) – (2 × 2) = 27 – 4 = 23

(iii) (a + b)^b
Substituting a = 3 and b = 2
we get (3 + 2)² = 5² = 5 × 5 = 25

(iv) (a – b)^a
Substituting a = 3 and b = 2
we get (3 – 2)³ = 1³ = 1 × 1 × 1 = 1

Question 11.
Simplify and express each of the following in exponential form:
(i) 4⁵ × 4² × 4⁴
(ii) (3² × 3³)⁷
(iii) (5² × 5⁸) ÷ 5^s
(iv) 2⁰ × 3⁰ × 4⁰
(v) \(\frac{5^{5} \times a^{8} \times b^{3}}{4^{3} \times a^{5} \times b^{2}}\)
Solution:

Objective Type Questions

Question 12.
a × a × a × a × a equal to
(i) a⁵
(ii) 5 a
(iii) 5a
(iv) a + 5
Answer:
(i) a⁵

Question 13.
The exponential form of 72 is
(i) 72
(ii) 27
(iii) 2² × 3³
(iv) 2³ × 3²
Answer:
(iv) 2³ × 3²

Question 14.
The value of x in the equation a¹³ = x³ × a¹⁰ is
(i) a
(ii) 13
(iii) 3
(iv) 10
Answer:
(i) a

Question 15.
How many zeros are there in 100¹⁰ ?
(i) 2
(ii) 3
(iii) 100
(iv) 20
Answer:
(iv) 20

Question 16.
2⁴⁰ + 2⁴⁰ is equal to
(i) 4⁴⁰
(ii) 2⁸⁰
(iii) 2⁴¹
(iv) 4⁸⁰
Answer:
(iii) 2⁴¹

Students can Download Maths Chapter 1 Life Mathematics Ex 1.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th 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 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.4

Miscellaneous Practise Problems

Question 1.
Nanda’s marks in 3 Math tests Tl, T2 and T3 were 38 out of 40,27 out of 30 and 48 out of 50. In which test did he do well? Find his overall percentage in all the 3 tests.
Solution:
Nanda’s marks are as follows
In Test 1 → T₁ \(\frac{38}{40}\)
Test 2 → T₂ \(\frac{27}{30}\)
Test 3 → T₃ \(\frac{48}{50}\)
For percentage, multiply by 100
T₁ = \(\frac{38}{40}\) x 100 = 95%
T₂ = \(\frac{27}{30}\) x 100 = 90%
T₃ = \(\frac{48}{50}\) x 100 = 96%
Hence, he has scored highest percentage in test 3
∴ He is done well in T3
Overall percentage is the average of the 3 percentages.
i.e \(\frac{90+95+96}{3}\) = \(\frac{281}{50}\) = 93\(\frac{2}{3}\)%

Question 2.
Sultana bought the following things from a general store. Calculate the total bill amount to be paid by her.
(i) Medicines costing ₹ 800 with GST @ 5% ………..

(ii) Cosmetics costing ₹ 650 with GST @ 12% ………….

(iii) Cereals costing ₹ 900 with GST @ 0% …………

(iv) Sunglass costing ₹ 1750 with GST @ 18 % ………..

(v) Air Conditioner costing ₹ 28500 with GST @ 28% ………..

Solution:

(i) Medicine : bill amount is 800(1 + \(\frac{5}{100}\)) = 800 x \(\frac{105}{100}\) = 840
(ii) Cosmetics: Bill amount is 650(1 + \(\frac{12}{100}\)) = 650 x \(\frac{112}{100}\) = 728
(iii) Cereals : Bill amount is 900(1 + \(\frac{0}{100}\)) = 900
(iv) Sunglass: bill amount is 1750(1 + \(\frac{18}{100}\)) = 1750 x \(\frac{118}{100}\) = 2065
(v) AC : Bill amount is 28500(1 + \(\frac{28}{100}\)) = 28500 x \(\frac{128}{100}\) = 36480
∴ Total bill amount = 840 + 728 + 900 + 2065 + 36480
= ₹ 41,013 (total bill amount)

Question 3.
P’s income is 25% more than that of Q. By what percentage is Q’s income less than P’s?
Solution:
Let Q’s income be 100.
P’s income is 25% more than that of Q
∴ P’s income = 100 + \(\frac{25}{100}\) x 100 = 125
Q’s income is 25 less than that of P
In percentage terms, Q’s income is less than P’s with respect to P’s income is
\(\frac{P-Q}{P}\) x 100 = \(\frac{125-100}{125}\) x 100 = \(\frac{25}{100}\) x 100 = 20%

Question 4.
Gopi sold a laptop at 12% gain. If it had been sold for ₹ 1200 more, the gain would have been 20%. Find the cost price of the laptop.

Solution:
Let the cost price of the laptop be ‘x’
Gain = 12%

If the selling price was 1200 more
i.e \(\frac{112}{100}\)x +1200, the gain is 20%
i.e new selling price = x(1 + \(\frac{2}{100}\))
= \(\frac{112}{100}\)x + 1200 = x(100 + \(\frac{20}{100}\)) = \(\frac{112}{100}\)x
∴ 1200 = \(\frac{120}{100}\)x = \(\frac{112}{100}\)x = \(\frac{8}{100}\)x
∴ x = \(\frac{120×100}{8}\) = 15000
Cost price of the laptop is ₹ 15000

Question 5.
Vaidegi sold two sarees for ₹ 2200 each. On one she gains 10% and on the other she loses 12%. Calculate her gain or loss percentage in the sales.

Solution:
Saree 1:
The selling price is ₹ 2200, let cost price be CP₁, gain is 10%
Cost price ₹ Using the formula

Saree 2:
The selling price is 2200, let cost price be CP₂, loss is given as 12%. We need to find CP₂ using the formula as before,

Cost price of both together is CP₁ + CP₂
= 2000 + 2500 = 4500 ……(1)
Selling price of both together is 2 x 2200 = 4400 …….(2)
Since net selling price is less than net cost price, there is a loss.
loss% = \(\frac{loss}{cost price}\) x 100
Loss = Net cost price – Net selling price
(1) – (2) = 4500 – 4400 = 100
∴ loss% = \(\frac{100}{4500}\) x 100 = \(\frac{100}{45}\) = \(\frac{20}{9}\) = 2\(\frac{2}{9}\)%
= 2\(\frac{2}{9}\)% loss

Question 6.
A sum of money becomes ₹ 18000 in 2 years and ₹ 40500 in 4 years on compound interest. Find the sum.
Solution:
Let the sum of money be ‘P’
Given that sum ‘P’ becomes 18000 in 2yrs.
i.e Amount (A) = P(1 + \(\frac{r}{100}\))ⁿ
18000 = P(1 + \(\frac{r}{100}\))²
(1 + \(\frac{r}{100}\))² = \(\frac{18000}{P}\) ⇒ (1 + \(\frac{r}{100}\))⁴ = (\(\frac{18000}{P}\))² ……(1)
Also given that sum ‘P’ becomes46500 in 4 yrs.
i.e Amount (A) = P(1 + \(\frac{r}{100}\))ⁿ
40500 = P(1 + \(\frac{r}{100}\))⁴
Substituting (1) in (2), we get

Question 7.
Find the difference in the compound interest on ₹ 62500 for 1\(\frac{1}{2}\) years at 8% p.a compounded annually and when compounded half-yearly.
Solution:
Case 1:
P = ₹ 62500
n = 1\(\frac{1}{2}\)yrs. (a\(\frac{b}{c}\)) formula
r = 8% Compound annully
CI = A – P

CI = A – P = 70200 – 62500 = 7700 ……(1)
CAse 2:
P = ₹ 62500
n = 1\(\frac{1}{2}\)yrs.
r = 8% p.a when compound half yearly
r = 4% compound half yearly

= 70304 – 62500 = ₹ 7804
Difference between case 1 & case 2 is (2) – (1)
∴ (2) – (1) = 7804 – 7700 = ₹ 104

Challenging Problems

Question 8.
If the first number is 20% less than the second number and the second number is 25% more than 100, then find the first number.
Solution:
Second number is 25% more than 100
∴ 2nd number is 100 + \(\frac{25}{100}\) x 100= 125
First number is 20% less than second no.

1st number is 100

Question 9.
A shopkeeper gives two successive discounts on an article whose marked price is ? 180 and selling price is ? 108. Find the first discount percent if the second discount is 25%.
Solution:
Marked price is given as ? 180
Let 1^st discount be d₁% = ? (to find)
2nd discount be d₂% = 25%
Selling price is 108 (given)
Price after 1^st discount = 180(1 – \(\frac { d_{ 1 } }{ 100 } \)) = P₁
Price after 2^nd discount = P₁(1 – \(\frac { d_{ 2 } }{ 100 } \)) = 108
Substituting for P₁ from (1), we get

∴ 1 – \(\frac { d_{ 1 } }{ 100 } \) = \(\frac{4}{5}\)
∴ \(\frac { d_{ 1 } }{ 100 } \) = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
∴ d₁ = \(\frac{1}{5}\) x 100 = 20%

Question 10.
A man bought an article on 30% discount and sold it at 40% more than the marked price. Find the profit made by him.
Solution:
Let marked price be ‘P’
Discounted price = P(1 – \(\frac{d}{100}\)) where d is the discount %
∴ Discounted price = P(1 – \(\frac{30}{100}\)) = \(\frac{70}{100}\)P → this is the cost price.
Selling price = 40% more than marked price

Question 11.
Find the rate of compound interest at which a principal becomes 1.69 times itself in 2 years.
Solution:
Let principal be ‘P’
Amount is given to be 1.69 times principal
i.e 1.69 P
Time period is 2yrs. = (n)
Rate of interest = r = ? (required)
Applying the formula,
Amount = Principal (1 + \(\frac{r}{100}\))ⁿ
Substituting, 1.69 P = P (1 + \(\frac{r}{100}\))²
∴ (1 + \(\frac{r}{100}\))² = \(\frac{1.69P}{P}\)
Taking square root on both sides, we get
\(\sqrt{1.69}\) = 1 + \(\frac{r}{100}\)
∴ 1 + \(\frac{r}{100}\) = 1.3
∴ \(\frac{r}{100}\) = 1.3
r = 30%
∴ rate of compound interst is 30%

Question 12.
The simple interest on a certain principal for 3 years at 10% p.a is ₹ 300. Find the compound interest accrued in 3 years.
Solution:
Let principal be ‘P’
Rate of interest is 10% p.a (r)
Time period (n) = 3 yrs.
Formula for simple interest
∴ P = \(\frac{300×100}{10×3}\) = 1000
Compound Interest = CI = A – P
Amount (A) = P (1 + \(\frac{r}{100}\))ⁿ
= 1000(1 + \(\frac{10}{100}\))³ = 1000 x (\(\frac{10o+10}{100}\))³
= 1000 x \(\frac{110}{100}\) x \(\frac{110}{100}\) x \(\frac{110}{100}\) = 1331
Compound Interest (Cl) = Amount – Principal
= 1331 – 1000 = 331
Compound Interest = ₹ 331

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

You can Download Samacheer Kalvi 9th Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Science Solutions Chapter 8 Sound

Samacheer Kalvi 9th Science Sound Textbook Exercises

I. Choose the correct answer:

Question 1.
Which of the following vibrates when a musical note is produced by the cymbals in a orchestra?
(a) stretched strings
(b) stretched membranes
(c) air columns
(d) metal plates
Answer:
(a) stretched strings

Question 2.
Sound travels in air:
(a) if there is no moisture in the atmosphere.
(b) if particles of medium travel from one place to another.
(c) if both particles as well as disturbance move from one place to another.
(d) if disturbance moves.
Answer:
(b) if particles of medium travel from one place to another.

Question 3.
A musical instrument is producing continuous note. This note cannot be heard by a person having a normal hearing range. This note must then be passing through ………….
(a) wax
(b) vacuum
(c) water
(d) empty vessel
Answer:
(d) empty vessel

Question 4.
The maximum speed of vibrations which produces audible sound will be in ……………..
(a) seawater
(b) ground glass
(c) dry air
(d) Human blood
Answer:
(a) seawater

Question 5.
The sound waves travel faster
(a) in liquids
(b) in gases
(c) in solids
(d) in vacuum
Answer:
(c) in solids

II. Fill in the blanks.

1. Sound is a ………… wave and needs a material medium to travel.
2. Number of vibrations produced in one second is …………..
3. The velocity of sound in solid is …………. than the velocity of sound in air.
4. Vibration of object produces …………..
5. Loudness is proportional to the square of the …………..
6. …………… is a medical instrument used for listening to sounds produced in the body.
7. The repeated reflection that results in persistence of sound is called ……………..

Answer:

1. longitudinal
2. frequency of wave
3. faster
4. Sound
5. amplitude
6. ECG
7. reverberation

III. Match the following.

Answer:
(a) (v)
(b) (i)
(c) (iv)
(d) (ii)
(e) (iii)

IV. Answer in brief.

Question 1.
Through which medium sound travels faster, iron or water? Give reason.
Answer:
Sound travels faster through iron as solids are packed together tighter than liquids and gases.

Question 2.
Name the physical quantity whose SI unit is ‘hertz’. Define.
Answer:
The SI unit of frequency is hertz (Hz). The number of vibrations (complete waves or cycles) produced in one second is called frequency of the wave.

Question 3.
What is meant by supersonic speed?
Answer:
When the speed of any object exceeds the speed of sound in air (330 m s-1) it is said to be travelling at supersonic speed.

Question 4.
How does the sound produced by a vibrating object in a medium reach your ears?
Answer:
When an object vibrates, it forces the neighbouring particles of the medium to vibrate. These vibrating particles then force the particles adjacent to them to vibrate. In this way vibrations produced by an object are transferred till it reaches the ear.

Question 5.
You and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer:
No, as there is no medium on moon for the sound to travel.

V. Answer in detail.

Question 1.
Describe with diagram, how compressions and rarefactions are produced.
Answer:
Sound is also a longitudinal wave. Sound can travel only when there are particles which can be compressed and rarefied. Compressions are the regions where particles are crowded together. Rarefactions are the regions of low pressure where particles are spread apart. A sound wave is an example of a longitudinal mechanical wave. Below figure represents the longitudinal nature of sound wave in the medium.

Question 2.
Verify experimentally the laws of reflection of sound.
Answer:
The laws of reflection are:

• The angle in which the sound is incident is equal to the angle in which sound is reflected.
• Direction of incident sound, direction of the reflected sound and the normal are in the same plane.

Take two metal tubes A and B. Keep one end of each tube on a metal plate as shown in figure. Place a wrist watch c at the open end of the tube A and interpose a cardboard between A and B. Now at a particular inclination of the tube B with the cardboard, ticking of the watch is clearly heard. The angle of reflection made by the tube B with the cardboard is equal to the angle of incidence made by the tube A with the cardboard.
For example, if the angle of incidence is 20° on the left side of a protractor (see figure arrangement), the angle of reflection at which we are able to hear the sound clearly will also be at 20° on the right side of the protractor.
∴ ∠ i = ∠ r    ➝ verifying Law I
We will also further observe that pipe 1, pipe 2 i.e, the incident ray, reflected ray and the normal lie on the same plane, verifying law II.

Question 3.
List the applications of sound.
Applications of ultrasound
Answer:

• Ultra sound can be used in cleaning technology. Minute foreign particles can be removed from objects placed in a liquid bath through which ultrasound is passed.
• Ultrasounds can also be use d to detect cracks and flaws in metal blocks.
• Ultrasonic waves are made to reflect from various parts of the heart and form the image of the heart. This technique is called ‘echo cardiography’.
• Ultrasound may be employed to break small ‘stones’ formed in the kidney into fine grains. These grains later get flushed out with urine.

Question 4.
Explain how does SONAR work?
Answer:
SONAR stands for Sound Navigation And Ranging. Sonar is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects. Sonar consists of Science – 9 (Physics)
a transmitter and a detector and is installed at the bottom of boats and ships. The transmitter produces and transmits ultrasonic waves. These waves travel through water and after striking the object on the seabed, get reflected back and are sensed by the detector.

The detector converts the ultrasonic waves into electrical signals which are appropriately interpreted. The distance of the object that reflected the sound wave can be calculated by knowing the speed of sound in water and the time interval between transmission and reception of the ultrasound. Sonar technique is used to determine the depth of the sea and to locate underwater hills, valleys, submarine, icebergs etc.

VI. Numerical problem.

Question 1.
The frequency of a source of sound is 600 Hz. How many times does it vibrate in a minute?
Answer:
The number of vibrations (complete waves or cycles) produced in one second is called frequency of the wave.
Hence, in one minute 600 x 60 vibrations are produced = 36000 (Need clarification)

Question 2.
A stone is dropped from the top of a tower 750 m high into a pond of water at the base of the tower. When is the splash heard at the top?
(Given g = 10 m s^{– 2} and speed of sound = 340 ms^{– 1})
Height = 750m
h = ut + 0.5 gt²
The initial velocity is 0
750 = 0.5 × 10 × t²
t = 10sec
Speed of sound is 340m/sec
So, time taken to travel 750m upwards is,
\(\frac { 750 }{ 340 }\) = 2.20s
time taken = 10 + 2.20 = 12.2 sec.

Samacheer Kalvi 9th Science Sound In Text Problems

Question 1.
A sound wave has a frequency of 2 kHz and wavelength of 15 cm. How much time will it take to travel 1.5 km?
Solution:
v = nλ
n = 2 kHz = 2000Hz
λ = 15 cm = 0.15 m
0.15 × 2000 = 300ms^{– 1}
Time(t) = \(\frac{\text { Distance (d) }}{\text { Velocity (v) }}\)
The sound will take 5 s to travel a distance of 1.5 km.

Question 2.
What is the wavelength of a sound wave in air at 20° C with a frequency of 22 MHz?
Solution:
λ = v/n
Here, v = 344 m s^{– 1}
n = 22 MHz = 22 × 10⁶ Hz
λ = \(\frac { 344 }{ 22 }\) × 10⁶ = 15.64 × 10⁶ m = 15.64 µm.

Question 3.
A man fires a gun and hears its echo after 5 s. The man then moves 310 m towards the hill and fires his gun again. If he hears the echo after 3 s, calculate the speed of sound.
Solution:
Distance (d) = velocity (v) × time (t)
Distance travelled by sound when gun fires first time, 2d = v × 5 …………. (1)
Distance travelled by sound when gun fires second time, 2d – 620 = v × 3 …………… (2)
Rewriting equation (2) as,
2d = (v × 3) + 620 ……………. (3)
Equating (1) and (3), 5v = 3v + 620
2v = 620 .
Velocity of sound, v = 310 m s^{– 1}

Question 4.
A ship sends out ultrasound that returns from the seabed and is detected after 3.42 s.
If the speed of ultrasound through sea water is 1531m s^{– 1}, what is the distance of the seabed from the ship?
Solution:
We know, distance = speed × time
2d = speed of ultrasound × *time
2d = 1531 × 3.42
∴ d = \(\frac{5236}{2}\) = 2618 m
Thus, the distance of the seabed from the ship is 2618 m or 2.618 km.

Samacheer Kalvi 9th Science Sound Additional Questions

I. Answer the following questions.

Question 1.
“Sound needs a medium for propagation”. Justify with an experiment.
Answer:

Sound needs a material like air, water, steel, etc. for its propagation. It cannot travel through vacuum. This can be demonstrated by the Bell-Jar experiment.

An electric bell and an airtight glass jar are taken. The electric bell is suspended inside the airtight jar. The jar is connected to a vacuum pump. If the bell is made to ring, we will be able to hear the sound of the bell. Now when the jar is evacuated with the vacuum pump, the air in the jar is pumped out gradually and the sound becomes feebler and feebler. We will not hear any sound, if the air is fully removed (if the jar has vacuum).

Question 2.
Name the characteristics that describe a sound.
Answer:
A sound wave can be described completely by five characteristics viz – amplitude, frequency, time period, wavelength and velocity or speed.

Question 3.
What does the intensity of sound heard at a place depend on?
Answer:
The intensity of sound heard at a place depends on the following factors;

1. Amplitude of the source
2. Distance of the observer
3. Surface area of the source
4. Density of the medium
5. Frequency of the source.

Question 4.
Why is sound wave called longitudinal wave?
Answer:
As the vibration of the particles of the medium are along the direction of wave propagation, sound waves are longitudinal waves.

Question 5.
What is sound and how is it produced?
Answer:
Sound is a form of energy which produces the sensation of hearing. It is produced due to vibrations of different objects, eg. stretched vibrating wires and vibrating drums, etc.

Question 6.
How can two whales in the sea, hundreds of kilometers apart, communicate?
Answer:
Sound travels about 5 times faster in water than in air. Therefore, two whales in the sea, hundreds of kilometers apart can communicate easily.

Question 7.
What is a stethescope? What is the principle on which it works?
Answer:
Stethescope is a medical instrument used for listening to sounds produced in the body. In stethescopes, the sounds reach doctor’s ears by multiple reflections that happen in the connecting tube.

Question 8.
What is an ECG? How does it help in the field of medicine?
Answer:
The electrocardiogram (ECG) is one of the simplest and oldest cardiac investigations available. In ECG, the sound variations produced by heart is converted into electric signals. Thus an ECG is simply a representation of the electrical activity of the heart muscle as it changes with time. Usually it is printed on paper for easy analysis. The sum of this electrical activity, when amplified and recorded for just a few seconds is known as an ECG.

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