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Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.3

Students can Download Maths Chapter 3 Algebra Ex 3.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 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

Identifying the Degree and Leading Coefficient Calculator of Polynomials.

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

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.3

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.

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.3

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.

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.3

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.

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.3

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

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.3

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

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2

Students can Download Maths Chapter 1 Number System Ex 1.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2

Question 1.
Fill in the following place value table.
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2 2
Answer:
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2 1

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2

1/2 as a decimal is 0.5.

Question 2.
Write the decimal numbers from the following place value table.
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2 3
Answer:
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2 4

Question 3.
Write the following decimal numbers in the place value table.
(i) 25.178
(ii) 0.025
(iii) 428.001
(iv) 173.178
(v) 19.54
Solution:
(i) 25.178
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2 5
(ii) 0.025
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2 6
(iii) 428.001
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2 7
(iv) 173.178
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2 8
(v) 19.54
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2 9

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2

Question 4.
Write each of the following as decimal numbers.
(i) 20 + 1 + \(\frac { 2 }{ 10 } \) + \(\frac { 3 }{ 100 } \) + \(\frac { 7 }{ 1000 } \)
(ii) 3 + \(\frac { 8 }{ 10 } \) + \(\frac { 4 }{ 100 } \) + \(\frac { 5 }{ 1000 } \)
(iii) 6 + \(\frac { 0 }{ 10 } \) + \(\frac { 0 }{ 100 } \) + \(\frac { 9 }{ 1000 } \)
(iv) 900 + 50 + 6 + \(\frac { 3 }{ 100 } \)
(v) \(\frac { 6 }{ 10 } \) + \(\frac { 3 }{ 100 } \) + \(\frac { 1 }{ 1000 } \)
Solution:
(i) 20 + 1 + \(\frac { 2 }{ 10 } \) + \(\frac { 3 }{ 100 } \) + \(\frac { 7 }{ 1000 } \) = 21 + 2 × \(\frac { 1 }{ 10 } \) + 3 × \(\frac { 1 }{ 100 } \) + 7 × \(\frac { 1 }{ 1000 } \) = 21.237

(ii) 3 + \(\frac { 8 }{ 10 } \) + \(\frac { 4 }{ 100 } \) + \(\frac { 5 }{ 1000 } \) = 3 + 8 × \(\frac { 1 }{ 10 } \) + 4 × \(\frac { 1 }{ 100 } \) + 5 × \(\frac { 1 }{ 1000 } \) = 3.845

(iii) 6 + \(\frac { 0 }{ 10 } \) + \(\frac { 0 }{ 100 } \) + \(\frac { 9 }{ 1000 } \) = 6 + 0 × \(\frac { 1 }{ 10 } \) + 0 × \(\frac { 1 }{ 100 } \) + 9 × \(\frac { 1 }{ 1000 } \) = 6.009

(iv) 900 + 50 + 6 + \(\frac { 3 }{ 100 } \) = 956 + 0 × \(\frac { 1 }{ 10 } \) + 3 × \(\frac { 1 }{ 100 } \) = 956.03

(v) \(\frac { 6 }{ 10 } \) + \(\frac { 3 }{ 100 } \) + \(\frac { 1 }{ 1000 } \) = 6 × \(\frac { 1 }{ 10 } \) + 3 × \(\frac { 1 }{ 100 } \) = 0.631

Question 5.
Convert the following fractions into decimal numbers.
(i) \(\frac { 3 }{ 10 } \)
(ii) 3 \(\frac { 1 }{ 2 } \)
(iii) 3 \(\frac { 3 }{ 5 } \)
(iv) \(\frac { 3 }{ 2 } \)
(v) \(\frac { 4 }{ 5 } \)
(vi) \(\frac { 99 }{ 100 } \)
(vii) 3 \(\frac { 19 }{ 25 } \)
Solution:
(i) \(\frac { 3 }{ 10 } \) = 0.3
(ii) 3 \(\frac { 1 }{ 2 } \) = \(\frac { 7 }{ 2 } \) = \(\frac{7 \times 5}{2 \times 5}\) = \(\frac { 35 }{ 10 } \) = 3.5
(iii) 3 \(\frac { 3 }{ 5 } \) = \(\frac { 18 }{ 5 } \) = \(\frac{18 \times 2}{5 \times 2}\) = \(\frac { 36 }{ 10 } \) = 3.6
(iv) \(\frac { 3 }{ 2 } \) = \(\frac{3 \times 5}{2 \times 5}\) = \(\frac { 15 }{ 10 } \) = 1.5
(v) \(\frac { 4 }{ 5 } \) = \(\frac{4 \times 2}{5 \times 2}\) = \(\frac { 8 }{ 10 } \) = 0.8
(vi) \(\frac { 99 }{ 100 } \) = 0.99
(vii) 3 \(\frac { 19 }{ 25 } \) = \(\frac { 94 }{ 25 } \) = \(\frac{94 \times 4}{25 \times 4}\) = \(\frac { 376 }{ 100 } \) = 3.76

Question 6.
Write the following decimals as fractions.
(i) 2.5
(ii) 6.4
(iii) 0.75
Solution:
(i) 2.5 = 2 + \(\frac { 5 }{ 10 } \) = \(\frac { 25 }{ 10 } \)
(ii) 6.4 = 6 + \(\frac { 4 }{ 10 } \) = \(\frac { 64 }{ 10 } \)
(iii) 0.75 = 0 + \(\frac { 7 }{ 10 } \) + \(\frac { 5 }{ 100 } \) = \(\frac { 70+5 }{ 100 } \) = \(\frac { 75 }{ 100 } \)

Question 7.
Express the following decimals as fractions in lowest form.
(i) 2.34
(ii) 0.18
(iii) 3.56
Solution:
(i) 2.34 = 2 + \(\frac { 34 }{ 100 } \) = 2 + \(\frac{34 \div 2}{100 \div 2}\) = 2 + \(\frac { 17 }{ 50 } \) = 2\(\frac { 17 }{ 50 } \) = \(\frac { 117 }{ 50 } \)
(ii) 0.18 = 0 + \(\frac { 18 }{ 100 } \) = \(\frac{18 \div 2}{100 \div 2}\) = \(\frac { 9 }{ 50 } \)
(iii) 3.56 = 3 + \(\frac { 56 }{ 100 } \) = 3 + \(\frac{56 \div 4}{100 \div 4}\) = 3 + \(\frac { 14 }{ 25 } \) = 3 \(\frac { 14 }{ 25 } \) = \(\frac { 89 }{ 25 } \)

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2

Objective Questions

Question 8.
3 + \(\frac { 4 }{ 100 } \) + \(\frac { 9 }{ 1000 } \) = ?
(i) 30.49
(ii) 3049 9
(iii) 3.0049
(iv) 3.049
Answer:
(iv) 3.049
Hint: = 3 × 1 + \(\frac { 0 }{ 10 } \) + \(\frac { 4 }{ 100 } \) + \(\frac { 9 }{ 1000 } \) = 3.049

Question 9.
\(\frac { 3 }{ 5 } \) = _______
(i) 0.06
(ii) 0.006
(iii) 6
(iv) 0.6
Answer:
(iv) 0.6
Hint: \(\frac { 3 }{ 5 } \) = \(\frac{3 \times 2}{5 \times 2}\) = \(\frac { 6 }{ 10 } \) = 0.06

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2

Question 10.
The simplest form of 0.35 is
(i) \(\frac { 35 }{ 1000 } \)
(ii) \(\frac { 35 }{ 10 } \)
(iii) \(\frac { 7 }{ 20 } \)
(iv) \(\frac { 7 }{ 100 } \)
Answer:
(iii) \(\frac { 7 }{ 20 } \)
Hint: 0.35 = \(\frac { 35 }{ 100 } \) = \(\frac{35 \div 5}{100 \div 5}\) = \(\frac { 7 }{ 20 } \)

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.5

Students can Download Maths Chapter 1 Number System Ex 1.5 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 1.5

5/8 as a decimal is 0.625.

Question 1.
Write the following decimal numbers in the place value table.
(i) 247.36
(ii) 132.105
Solution:
(i) 247.36
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.5 1

(ii) 132.105
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.5 2

11/16 as a decimal is 0.6875.

Question 2.
Write each of the following as decimal number.
(i) 300 + 5 + \(\frac { 7 }{ 10 } \) + \(\frac { 9 }{ 100 } \) + \(\frac { 2 }{ 100 } \)
(ii) 1000 + 400 + 30 + 2 + \(\frac { 6 }{ 10 } \) + \(\frac { 7 }{ 100 } \)
Solution:
(i) 300 + 5 + \(\frac { 7 }{ 10 } \) + \(\frac { 9 }{ 100 } \) + \(\frac { 2 }{ 100 } \) = 305.792
(ii) 1000 + 400 + 30 + 2 + \(\frac { 6 }{ 10 } \) + \(\frac { 7 }{ 100 } \) = 1432.67

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.5

Question 3.
Which is greater?
(i) 0.888 (or) 0.28
(ii) 23.914 (or) 23.915
Solution:
(i) 0.888 (or) 0.28
The whole number parts is equal for both the numbers.
Comparing the digits in the tenths place we get, 8 > 2.
0.888 > 0.28 ∴ 0.888 is greater.

(ii) 23.914 or 23.915
The whole number part is equal in both the numbers.
Also the tenth place and hundredths place are also equal.
∴ Comparing the thousandths place, we get 5 > 4.
23.915 > 23.914 ∴ 23.915 is greater.

Question 4.
In a 25 m swimming competition, the time taken by 5 swimmers A, B, C, D and E are 15.7 seconds, 15.68 seconds, 15.6 seconds, 15.74 seconds and 15.67 seconds respectively. Identify the winner.
Solution:
The winner is one who took less time for swimming 25 m.
Comparing the time taken by A, B, C, D, E the whole number part is equal for all participants.
Comparing digit in tenths place we get 6 < 7.
∴ Comparing 15.68, 15.6, 15.67, that is comparing the digits in hundredths place we get 15.60 < 15.67 < 15.68
One who took 15.6 seconds is the winner. ∴ C is the winner.

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.5

Question 5.
Convert the following decimal numbers into fractions
(i) 23.4
(ii) 46.301
Solution:
(i) 23.4 = \(\frac { 234 }{ 10 } \) = \(\frac{234 \div 2}{10 \div 2}\) = \(\frac { 117 }{ 5 } \)
(ii) 46.301 = \(\frac { 46301 }{ 1000 } \)

Question 6.
Express the following in kilometres using decimals,
(i) 256 m
(ii) 4567 m
Solution:
1 m = \(\frac { 1 }{ 1000 } \) km = 0.001 Km
(i) 256 m = \(\frac { 256 }{ 1000 } \) km = 0.256 km
(ii) 4567 m = \(\frac { 4567 }{ 1000 } \) km = 4.567 km

Question 7.
There are 26 boys and 24 girls in a class. Express the fractions of boys and girls as decimal numbers.
Solution:
Boys = 26; Girls = 24; Total = 50
Fraction of boys = \(\frac { 26 }{ 50 } \) = \(\frac{26 \times 2}{50 \times 2}\) = \(\frac { 52 }{ 100 } \) = 0.52
Fraction of girls = \(\frac { 24 }{ 50 } \) = \(\frac{24 \times 2}{50 \times 2}\) = \(\frac { 48 }{ 100 } \) = 0.48

Challenge Problems

Question 8.
Write the following amount using decimals.
(i) 809 rupees 99 paise
(ii) 147 rupees 70 paise
Solution:
100 paise = 1 rupee; 1 paise = \(\frac { 1 }{ 100 } \) rupee

(i) 809 rupees 99 paise = 809 rupees + \(\frac { 99 }{ 100 } \) rupees
= 809 + 0.99 rupees = ₹ 809.99

(ii) 147 rupees 70 paise = 147 rupees + \(\frac { 70 }{ 100 } \) rupees
= 147 rupees + 0.70 rupees = ₹ 147.70

Question 9.
Express the following in metres using decimals.
(i) 1328 cm
(ii) 419 cm
Solution:
100 cm = 1 m; 1 cm = \(\frac { 1 }{ 100 } \) m
(i) 1328 cm = \(\frac { 1328 }{ 100 } \) m = 13.28 m
(ii) 419 cm = \(\frac { 419 }{ 100 } \) m = 4.19 m

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.5

Question 10.
Express the following using decimal notation.
(i) 8 m 30 cm in metres
(ii) 24 km 200 m in kilometres
Solution:
(i) 8 m 30 cm in metres
8 m + \(\frac { 30 }{ 100 } \) m = 8 m + 0.30 m = 8.30 m

(ii) 24 km 200 m in kilometres
24 km + \(\frac { 200 }{ 1000 } \) km = 24 km + 0.200 km = 24.200 km

Question 11.
Write the following fractions as decimal numbers.
(i) \(\frac { 23 }{ 10000 } \)
(ii) \(\frac { 421 }{ 100 } \)
(iii) \(\frac { 37 }{ 10 } \)
Solution:
(i) \(\frac { 23 }{ 10000 } \) = 0.0023
(ii) \(\frac { 421 }{ 100 } \) = 4.21
(iii) \(\frac { 37 }{ 10 } \) = 3.7

Question 12.
Convert the following decimals into fractions and reduce them to the lowest form,
(i) 2.125
(ii) 0.0005
Solution:
(i) 2.125 = \(\frac { 2125 }{ 1000 } \) = \(\frac{2125 \div 25}{1000 \div 25}\) = \(\frac { 85 }{ 40 } \) = \(\frac{85 \div 5}{40 \div 5}\) = \(\frac { 17 }{ 8 } \)

(ii) 0.0005 = \(\frac { 5 }{ 1000 } \) = \(\frac{5 \div 5}{10000 \div 5}\) = \(\frac { 1 }{ 2000 } \)

Question 13.
Represent the decimal numbers 0.07 and 0.7 on a number line.
Solution:
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.5 3
0.07 lies between 0.0 and 0.1
The unit space between 0 and 0.1 is divided into 10 equal parts and 7th part is taken. Also 0.7 lies between 0 and 1.
The unit space between 0 and 1 is divided into 10 equal parts, and the 7th part is taken.

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.5

Question 14.
Write the following decimal numbers in words.
(i) 4.9
(ii) 220.0
(iii) 0.7
(iv) 86.3
Solution:
(i) 4.9 = Four and nine tenths
(ii) 220.0 = Two hundred and twenty
(iii) 0.7 = Seven tenths
(iv) 86.3 = Eighty six and three tenths.

Question 15.
Between which two whole numbers the given numbers lie?
(i) 0.2
(ii) 3.4
(iii) 3.9
(iv) 2.7
(v) 1.7
(vi) 1.3
Solution:
(i) 0.2 lies between 0 and 1.
(ii) 3.4 lies between 3 and 4.
(iii) 3.9 lies between 3 and 4.
(iv) 2.7 lies between 2 and 3.
(v) 1.7 lies between 1 and 2.
(vi) 1.3 lies between 1 and 2.

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.5

Question 16.
By how much is \(\frac { 9 }{ 10 } \) km less than 1 km. Express the same in decimal form.
Solution:
Given measures are 1 km and \(\frac { 9 }{ 10 } \) km. i.e., 1 km and 0.9 km.
Difference = 1.0 – 0.9 = 0.1 km.

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.1

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 3 Chapter 1 Number System Ex 1.1

Question 1.
Round each of the following decimals to the nearest whole number.
(i) 8.71
(ii) 26.01
(iii) 69.48
(iv) 103.72
(v) 49.84
(vi) 101.35
(vii) 39.814
(viii) 1.23
Solution.
(i) 8.71
Underlining the digit to be rounded 8.71. Since the digit next to the underlined digit, 7 which is greater than 5, adding 1 to the underlined digit.
Hence the nearest whole number 8.71 rounds to is 9.

(ii) 26.01
Underlining the digit to be rounded 26.01. Since the digit next to the underlined digit, 0 which is less than 5, the underlined digit 6 remains the same.
∴ The nearest whole number 26.01 rounds to is 26.

(iii) 69.48
Underlining the digit to be rounded 69.48. Since the digit next to the underlined digit, 4 which is less than 5, the underlined digit 9 remains the same.
∴ The whole number is 69.48 rounds to is 69.

(iv) 103.72
Underlining the digit to be rounded 103.72 since the digit next to the underlined digit, 7 which is greater than 5, we add 1 to the under lined digit.
Hence the nearest whole number 103.72 rounds to is 104.

(v) 49.84
Underlining the digit to be rounded 49.84. Since the digit next to the underlined digit 8 which is greater than 5, we add 1 to the underlined digit.
Hence the nearest whole number 49.84 rounds to 50.

(vi) 101.35
Underlining the digit to be rounded 101.35. Since the digit next to the underlined digit 3 is less than 5, the underlined digit 1 remains the same.
Hence the nearest whole number 101.35 rounds to is 101.

(vii) 39.814
Underlining the digit to be rounded 39.814. Since the digit next to the underlined digit 8 is greater than 5, we add 1 to the underlined digit.
Hence the nearest whole number 39.814 rounds to is 40.

(viii) 1.23
Underlining the digit to be rounded 1.23. Since the digit next to the underlined digit 2, is less than 5, the underlined digit 1 remains the same.
Hence the nearest whole number 1.23 rounds to is 1.

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.1

1/8 as a decimal is 0.125.

Question 2.
Round each decimal number to the given place value.
(i) 5.992; tenths place
(ii) 21.805; hundredth place
(iii) 35.0014; thousandth place
Solution:
(i) 992; tenths place
Underlining the digit to be rounded 5.992. Since the digit next to the underlined digit is 9 greater than 5, we add 1 to the underlined digit.
Hence the rounded number is 6.0.

(ii) 21.805; hundredth place
Underlining the digit to be rounded 21.805 since the digit next to the underlined digit is 5, we add 1 to the underlined digit.
Hence the rounded number is 21.81.

(iii) 35.0014; thousandth place
Underlining the digit to be rounded 35.0014. Since the digit next to the underlined digit is 4 less than 5 the underlined digit remains the same.
Hence the rounded number is 35.001.

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.1

One Decimal is equal to 435.56 Square Feet.

Question 3.
Round the following decimal numbers upto 1 places of decimal.
(i) 123.37
(ii) 19.99
(iii) 910.546
Solution:
(i) 123.37
Rounding 123.37 upto one places of decimal means round to the nearest tenths place. Underling the digit in the tenths place of 123.37 gives 123.37. Since the digit next to the tenth place value is 7 which is greater than 5, we add 1 to the underlined digit to get 123.4. Hence the rounded value of 123.37 upto one places of decimal is 123.4.

(ii) 19.99
Rounding 19.99 upto one places of decimal means round to the nearest tenth place. Underling the digit in the tenths place of 19.99 gives 19.99. Since the digit next to the tenth place value is 9 which is greater than 5, we add 1 to the underlined digit to get 20.
Hence the rounded value of 19.99 upto one places of decimal is 20.0.

(iii) 910.546
Rounding 910.546 upto one places of decimal means round to the nearest tenths place underlining the digit in the tenths place of 910.546 gives 910.546. Since the digit next to the tenth place value is 4, which is less than 5 the underlined digit remains the same. Hence the rounded value of 910.546 upto one places of decimal is 910.5.

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.1

Rounding to the nearest hundredth is 838.27.

Question 4.
Round the following decimal numbers upto 2 places of decimal.
(i) 87.755
(ii) 301.513
(iii) 79.997
Solution:
(i) 87.755
Rounding 87.755 upto 2 places of decimal means round to the nearest hundredths place. Underlining the digit in the hundredth place of 87.755 gives 87.755. Since the digit next to the hundredth place value is 5, we add 1 to the underlined digit.

Hence the rounded value of 87.755 upto two places of decimal is 87.76.

(ii) 301.513
Rounding 301.51 upto 2 places of decimal means round to the nearest hundredths place. Underlining the digit in the hundredth place of 301.513 gives 301.513. Since the digit next to the underlined digit 3 is less than 5, the underlined digit remains the same.

∴ The rounded value of 301.513 upto 2 places of decimal is 301.51.

(iii) 79.997
Rounding 79.997 upto 2 places of decimal means round to the nearest hundredths place. Underlining the digit in the hundredth place of 79.997 gives 79.997. Since the digit next to the underlined digit 7 is greater than 5, we add 1 to the underlined number.

Hence the rounded value of 79.997 upto 2 places of decimal is 80.00.

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.1

Question 5.
Round the following decimal numbers upto 3 place of decimal
(a) 24.4003
(b) 1251.2345
(c) 61.00203
Solution:
(a) 24.4003
Rounding 24.4003 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandths place of 24.4003 gives 24.4003. In 24.4003 the digit next to the thousandths value is 3 which is less than 5.

∴ The underlined digit remains the same. So the rounded value of24.4003 upto 3 places of decimal is 24.400.

(b) 1251.2345
Rounding 1251.2345 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandths place of 1251.2345 gives 1251.2345, the digit next to the thousandths place value is 5 and so we add 1 to the underlined digit. So the rounded value of 1251.2345 upto 3 places of decimal is 1251.235.

(c) 61.00203
Rounding 61.00203 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandth place of 61.00203 gives 61.00203. In 61.00203, the digit next to the thousandths place value is 0, which is less than 5.

Hence the underlined digit remains the same. So the rounded value of 61.00203 upto 3 places of decimal is 61.002.

When we write a decimal number with three places, we are representing the thousandths place.

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.1

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.
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System 1.1 1
Solution:
(i) Tens 2 ones 2 tenths = 12.2
(ii) Tens 1 ones 3 tenths = 21.3

That’s literally all there is to it! 1/32 as a decimal is 0.03125.

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

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.1

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

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.1

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
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System 1.1 2
(ii) 263.271
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System 1.1 3
(iii) 17.39
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System 1.1 7
(iv) 9.657
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System 1.1 5
(v) 4972.068
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System 1.1 6

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.1

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 × \(\frac { 1 }{ 1000 } \)

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.1

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

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.2

Students can Download Maths Chapter 4 Information Processing Ex 4.2 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 4 Information Processing Ex 4.2

The factors of 70 are 1, 2, 5, 7, 10, 14, 35, and 70.

Question 1.
Choose the correct answer

Question (i)
Common prime factors of 30 and 250 are
(a) 2 x 5
(b) 3 x 5
(c) 2 x 3 x 5
(d) 5 x 5
Answer:
(a) 2 x 5
Hint:
Prime factors of 30 are 2 x 3 x 5
Prime factors of 250 are 5 x 5 x 5 x 2
∴ Common prime factors are 2 x 5

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.2

The Prime Factorization of 84 is 22 × 3 × 7.

Question (ii)
Common prime factors of 36, 60 and 72 are
(a) 2 x 2
(b) 2 x 3
(c) 3 x 3
(d) 3 x 2 x 2
Answer:
(d) 3 x 2 x 2
Hint:
Prime factors of 36 are 2 x 2 x 3 x 3
Prime factors of 60 are 2 x 2 x 3 x 5
Prime factors of 72 are 2 x 2 x 2 x 3 x 3
∴ Common prime factors are 2 x 2 x 3

Question (iii)
Two numbers are said to be co-prime numbers if their HCF is –
(a) 2
(b) 3
(c) 0
(d) 1
Answer:
(d) 1

What is the greatest common factor of 12 and 18? Prealgebra Greatest Common Factor.

Question 2.
Using repeated division method find HCF of the following:

  1. 455 and 26
  2. 392 and 256
  3. 6765 and 610
  4. 184, 230 and 276

1. 455 and 26 divisor
Solution:

  • Step 1: The larger number should be dividend = 455 & smaller number should be divisor = 26
  • Step 2: After 1st division, the remainder becomes new divisor & the previous divisor
  • becomes next dividend.
  • Step 3: This is done till remainder is zero.
  • Step 4: The last divisor is the HCF

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.2 1
Answer:
The HCF is 13.

2. 392 and 256
256 is smaller, so it is the 1st divisor
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.2 2
∴ HCF = 8

3. 6765 and 610
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.2 3
∴ HCF = 5

4. 184, 230 and 276
First let us take 184 & 230
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.2 4
∴ 46 is the Hi of 184, and 230.
Now the HCF of the first two numbers is the dividend for the third number.
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.2 5
Answer:
HCF of 184, 230 & 276 is 46

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.2

Question 3.
Using repeated subtracting method find HCF of the following:

  1. 42 and 70
  2. 36 and 80
  3. 280 and 420
  4. 1014 and 654

1. 42 and 70
Solution:
Let number be m & n
m > n
We do m – n & the result of subtraction becomes new ‘m’. if m becomes less than n, we do n – m and then assign the result as n. We should do this till m = n. When m = n then ‘m’ is the HCF.
42 and 70 now m = 70, n = 42
70 – 42 = 28, now m = 42, n = 28
42 – 28 = 14, now m = 28, n = 14
28 – 14= 14, now m = 14, n = 14
we stop here as m = n
∴ HCF of 42 & 70 is 14

2. 36 and 80
Solution:
36 and 80 m = 80, n = 36
80 – 36 = 44, now n = 44, m = 36
Since n > m, we should do n m
44 – 36 = 8, now n = 8, m = 36
36 – 8 = 28
Similarly, processing, proceeding, we do repeated subtraction
till m = n
28 – 8 = 20
20 – 8 = 12
12 – 8 = 4
8 – 4 = 4 now m = n = 4
∴ HCF is 4

3. 280 and 420
Solution:
Let m = 420, n = 280
m – n = 420 – 280 = 140
now m = 280, n = 140
m – n = 280 – 140 = 140
now m = n = 140
∴ HCF is 140

4. 1014 and 654
Solution:
Let m = 1014, n = 654
m – n = 1014 – 654 = 360
now m = 654, n = 360
m – n = 654 – 360 = 294
now m = 360, n = 294
m – n = 360 – 294 = 66
now m = 294, n = 66
m – n = 294 – 66 = 228
now m = 66, n = 228
n – m = 228 – 66 = 162
now m = 162, n = 66
∴ m – n = 162 – 66 = 96
n – m = 96 – 66 = 30
Similarly 66 – 30 = 36
36 – 30 = 6
30 – 6 = 24
24 – 6 = 18
18 – 6 = 12
12 – 6 = 6
now m = n
∴ HCF of 1014 and 654 is 6

Question 4.
Do the given problems in repeated subtraction method
1. 56 and 12
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.2 6
2. 320,120 and 95
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.2 7
1. 56 and 12
Solution:
56 & 12
Let n = 56 & n = 12
m – n = 56 – 12 = 44
now m = 44, n = 12
m – n = 44 – 12 = 32
m – n = 32 – 12 = 20
m – n = 20 – 12 = 8
n – m = 12 = 8 = 4
m – n = 8 – 44. now m = n
HCF of 56 & 12 is 4

2. 320, 120 and 95
Let us take 320 & 120 first m = 320, n = 120
m – n = 320 – 120 = 200
m = 200, n = 120
∴ m – n = 200 – 120 = 80
120 – 80 = 40
80 – 40 = 40
∴ m = w = 40 → HCF of 320, 120
Now let us find HCF of 40 & 95
m = 95, n = 40
∴ m – n = 95 – 40 = 55
55 – 40 = 15
40 – 15 = 25
25 – 15 = 10
15 – 10 = 5
HCF of 40 & 95 is 5 10 – 5 = 5
HCF of 320, 120 & 95 is 5

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.2

Question 5.
On a school trip, 56 girls and 98 boys went to Kanyakumari. They were divided into as many groups as possible so that there were equal numbers of girls and boys in each group. Find the largest group possible? (To find the HCF using repeated division method)
Solution:
56 girls & 98 boys. Now, we need to find HCF of 56 & 98
Using repeated division method. So, we first divide. 98 by 56
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.2 8
Answer:
14.

Question 6.
Kalai wants to cut identical squares as big as she can, from a piece of paper measuring 168 mm and by 196 mm. What is the length of the side of the biggest square? (To find HCF using repeated subtraction method)
Solution:
Sides are 168 & 196
To find HCF of 168 & 196, we are to use repeated subtraction method.
m = 196, n = 168
m – n = 196 -168 = 28 now n = 28, m = 168
m – n = 168 – 28 = 140 now m = 140, n = = 28
m – n = 140 – 28 = 112 now m = 112, n = 28
m – n = 112 -28 = 84 now m = 84, n = 28
m – n = 84 – 28 = = 56 now m = 56, n = 28
m – n = 56 – 28 = 28
∴ HCF is 28
∴ Length of biggest square is 28

Samacheer Kalvi 12th Commerce Solutions Chapter 11 Employee Selection Process

Students can Download Commerce Chapter 11 Employee Selection Process Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Commerce 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 12th Commerce Solutions Chapter 11 Employee Selection Process

Samacheer Kalvi 12th Commerce Employee Selection Process Textbook Exercise Questions and Answers

I. Choose the Correct Answer

Question 1.
The recruitment and Selection Process aimed at right kind of people.
(a) At right people
(b) At right time
(c) To do right things
(d) All of the above
Answer:
(d) All of the above

Convert CGPA into Percentage out of 4 … You must divide the percentage by 25 to get the percentage

Question 2.
The poor quality of selection will mean extra cost on _________ and supervision.
(a) Training
(b) Recruitment
(c) work quality
(d) None of these
Answer:
(a) Training

Question 3.
_________ refers to the process of identifying and attracting job seekers so as to build a pool of qualified job applicants.
(a) Selection
(b) Training
(c) Recruitment
(d) Induction
Answer:
(c) Recruitment

Samacheer Kalvi 12th Commerce Solutions Chapter 11 Employee Selection Process

Question 4.
Selection is usually considered as a _________ process.
(a) Positive
(b) Negative
(c) Natural
(d) None of these
Answer:
(b) Negative

Question 5.
Which of the following test is used to measure the various characteristics of the candidate?
(a) physical Test
(b) Psychological Test
(c) attitude Test
(d) Proficiency tests
Answer:
(b) Psychological Test

Question 6.
Wfifich of the following orders is followed in a typical selection process? .
(a) application form test and or interview, reference check and physical examination
(b) Application form test and or interview, reference check, and physical examination
(c) Reference check, application form, test and interview and physical examination
(d) physical examination test and on interview application term and reference check.
Answer:
(b) Application form test and or interview, reference check, and physical examination

Question 7.
The purpose of an application blank is to gather information about the
(a) Company
(b) Candidate
(c) Questionnaire or Interview Schedule
(d) Competitors
Answer:
(b) Candidate

Question 8.
Identify the test that acts as an instrument to discover the inherent ability of a candidate.
(a) Aptitude Test
(b) Attitude Test
(c) Proficiency Test
(d) Physical Test
Answer:
(a) Aptitude Test

Question 9.
The process of eliminating unsuitable candidate is called _________
(a) Selection
(b) Recruitment
(c) Interview
(d) Induction
Answer:
(a) Selection

Question 10.
Scrutiny of application process is the _________
(a) Last step in Selection process
(b) First step in Selection process
(c) Third step in Selection Process Selection process
(d) None of the above
Answer:
(b) First step in Selection process

Samacheer Kalvi 12th Commerce Solutions Chapter 11 Employee Selection Process

Question 11.
Scrutiny of application process is the
(a) Locating candidates
(b) Determinining the suitable of the candidates
(c) preparing employees for training
(d) None of the above
Answer:
(b) Determinining the suitable of the candidates

Question 12.
The process of placing the right man on the right job is called _________
(a) Training
(b) Placement
(c) Promotion
(d) Transfer
Answer:
(b) Placement

Question 13.
Probation/Trial period signifies _________
(a) one year to two years
(b) One year to three years
(c) Two years to four years
(d) None of the above
Answer:
(a) one year to two years

Question 14.
Job first man next is one of the principles of _________
(a) Test
(b) Interview
(c) Training
(d) placement
Answer:
(d) placement

II. Very Short Answer Questions

Question 1.
What is selection?
Answer:
Selection is the process of choosing the most suitable person for the vacant position in the organization.

Question 2.
What is an interview?
Answer:
According to Scott and others “an interview is a purpose full exchange of ideas, the answering of questions and communication between two or more persons.”

Question 3.
What is intelligence test?
Answer:
Intelligence tests are one of the psychological tests, that is designed to measure a variety of mental ability, individual capacity of a candidate.

Question 4.
What do you mean by test?
Answer:
Several tests are conducted in the selection process to ensure whether the candidate possesses the necessary qualification to fit into various positions in the organization.

Samacheer Kalvi 12th Commerce Solutions Chapter 11 Employee Selection Process

Question 5.
What do you understand about bio data?
Answer:
Most of the Public sector undertakings Recruitment boards supply applications forms for selection of jobs. In this form, the candidate fill the details about family background, educational qualifications, experience, co-curricular activities.

Question 6.
What do you mean by placement?
Answer:
Placement is a process of assigning a specific job to each and every candidate selected. It includes initial assignment of new employees and promotion, transfer or demotion of present employees.

III. Short Answer Questions

Question 1.
What is stress interview?
This type of interview is conducted to test the temperament and emotional balance of the candidate interviewed. Interviewer deliberately creates stressful situation and tries to assess the suitability of the candidate by observing his reaction and response to the stressful situations.

Question 2.
What is structured interview?
Answer:
Under this method, a series of questions to be asked by the interviewer are pre-prepared by the interviewer and only these questions are asked in the interview.

Question 3.
Name the types of selection test?
Answer:
Selection tests are of two types: Ability Tests and Personality Tests. Ability tests can further be divided into: aptitude test, achievement test, intelligence test, and judgement test. Personality tests can further be divided into: interest test, personality inventory test, projective test or thematic appreciation test, and attitude test.

Question 4.
What do you mean by achievement test?
Answer:
This test measures a candidate’s capacity to achieve in a particular field. In other words this test measures a candidate’s level of skill in certain areas, accomplishment and knowledge in a particular subject. It is also called proficiency test.

Samacheer Kalvi 12th Commerce Solutions Chapter 11 Employee Selection Process

Question 5.
Why do you think the medical examinations of a candidate is necessary?
Answer:
The last technique used in selection process is medical examination. This is the most important step in the selection because a person of poor health cannot work competently and any investment on him may go waste, if he/she is unable to discharge duties efficiently on medical grounds.

Question 6.
What is aptitude test?
Answer:
Aptitude test is a test to measure suitability of the candidates for the post/role. It actually measures whether the candidate possess a set of skills required to perform a given job. It helps in predicting the ability and future performance of the candidate.

Question 7.
How is panel interview conducted?
Answer:
Where a group of people interview the candidate, it is called panel interview. Usually panel comprises chair person, subject expert, psychological experts, representatives of minorities/ underprivileged groups, nominees of higher bodies and so on. All panel members ask different types of questions on general areas of specialization of the candidate.

Question 8.
List out the various selection interviews.
Answer:
Interview represents a face to face interaction between the interviewer and interviewee

  1. Preliminary Interview
  2. Structured Interview
  3. Unstructured Interview
  4. In-depth Interview
  5. Panel Interview
  6. Stress Interview
  7. Telephone Interview
  8. Online Interview
  9. Group interview
  10. Video Conference Interview

Question 9.
List out the significance of placement.
Answer:
The significance of the placement is as follows:

  1. It improves employee morale.
  2. It helps in reducing employee turnover.
  3. It helps in reducing conflict rates or accidents.
  4. It avoids misfit between the candidates and the job.
  5. It helps the candidate to work as per the predetermined objectives of the organization.

IV. Long Answer Questions

Question 1.
Briefly explain the various types of tests.
Answer:
Several tests are conducted in the selection process to ensure whether the candidate possesses the necessary qualifications.

A. Ability Test: Ability test may be divided into:

  • Aptitude Test: Aptitude test is a test to measure suitability of the candidates for the post.
  • Achievement Test: This test measures a candidate’s capacity to achieve in a particular field.
  • Intelligence Test: Intelligence test is designed to measure a variety of mental ability, individual capacity of a candidate.
  • Judgment Test: This test is conducted to test the presence of mind and reasoning capacity of the candidates.

B. Personality Test: It refers to the test conducted to find out the non-intellectual traits of a candidate. It can be further divided into:

  • Interest Test: Interest test measures a candidate’s extent of interest in a particular area.
  • Projective Test: This test measures the candidate’s values, personality of the candidate.
  • Attitude Test: measures candidate’s tendencies towards the people, situation, action and related things.

Question 2.
Explain the important methods of interview.
Interview means a face to face interaction between the interviewer and interviewee. Interview may be of various types:-

  1. Preliminary Interview: It is conducted to know the general suitability of the candidates who have applied for the job.
  2. Structured Interview: In this method, a series of questions is to be asked by the interviewer. The questions may be pre-prepared.
  3. In depth Interview: This interview is conducted to test the level of knowledge of the interviewee in a particular field.
  4. Panel Interview: Where a group of people interview the candidate. The panel usually comprises the chair person, subject expert, psychological experts and so on.
  5. Stress Interview: This type of interview is conducted to test the temperament and emotional balance of the candidate.
  6. Online Interview: Due to tremendous growth in information and communication technology, interviews are conducted by means of internet via Skype, Google duo, Whatsapp.

Samacheer Kalvi 12th Commerce Solutions Chapter 11 Employee Selection Process

Question 3.
Explain the principles of placement.
Answer:
The following are the principles of placement:

  1. Job First, Man Next: Man should be placed on the job according to the requirements of the job.
  2. Job Offer: The job should be offered to the man based on his qualification..
  3. Terms and conditions: The employee should be informed about the terms and conditiqns of the organisation.
  4. Aware about the Penalties: The employee should also be made aware of the penalties if he / she commits a mistake.
  5. Loyalty and Co-operation: When placing a person in a new job, an effort should be made to develop a sense of loyalty and co-operation in him.

Samacheer Kalvi 12th Commerce Employee Selection Process Additional Questions and Answers

I. Choose the Correct Answer

Question 1.
The types of Aptitude Test are
(i) Numerical Reasoning Test
(ii) Attitude Test
(iii) Vocabulary Test
(iv) Interest Test
(a) (i) and (ii)
(b) (i) and (iii)
(c) (i) and (iv)
(d) (ii) and (iii)
Answer:
(b) (i) and (iii)

Question 2.
Which one of the following is not correctly matched?

(a) In-depth interview Skype
(b) Online interview Group of people interview the candidate
(c) Video conferencing interview Level of knowledge
(d) Telephone interview Face to face interview

Answer:
(d) Telephone interview – Face to face interview

II. Very Short Answer Questions

Question 1.
What is meant by personality test?
Answer:
Personality test refers to the test conducted to find out the non-intellectual traits of a candidate namely temperament, emotional response, capability and stability.

Samacheer Kalvi 12th Commerce Solutions Chapter 11 Employee Selection Process

Question 2.
Write a note on attitude test.
Answer:
Attitude test measures candidate’s tendencies towards the people, situation, action and related things. For example: morale study, values study, etc.

III. Short Answer Questions

Question 1.
What do you mean by online interview?
Answer:
Due to tremendous growth in information and communication technology, these days interviews are conducted by means of internet via Skype. We chat, Google duo, Viber, Whatsapp or Video chat applications. This enables the interviewers to conduct interview with the candidates living in faraway places.

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1

Students can Download Maths Chapter 1 Life Mathematics Ex 1.1 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.
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 1 Life Mathematics Ex 1.1

Question 1.
Fill in the blanks:

Question (i)
If 30% of x is 150, then x is ………….
Answer:
500
Hint:
Given 30% of x is 150
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 1

Question (ii)
2 minutes is ………… % to an hour.
Answer:
3\(\frac{1}{3}\)%
Hint:
Let 2 min be x% of an hour
and 1 hr = 60 min
x% = \(\frac{2}{60}\) x 100 = \(\frac{200}{3}\) = \(\frac{10}{3}\) = 3\(\frac{1}{3}\)
x = 3\(\frac{1}{3}\)

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1

Question (iii)
If x % of x = 25, then x = …………
Answer:
50
Hint:
Given that x % of x is 25
∴ \(\frac{x}{100}\) = 25
x2 = 25 x 100 = 2500
∴ x = \(\sqrt{2500}\) = 50

Question (iv)
In a school of 1400 students, there are 420 girls. The percentage of boys in the school is ………….
Answer:
70
Hint:
Given total number of students in school = 1400
Number of girls in school = 420
Number of boys in school = 1400 – 420 = 980
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 2
% of boys = 70%

Question (v)
0.5252 is ………….. %.
Answer:
52.52%
Hint:
Given a number, and to express as a percentage, we need to multiply by 100
∴ to express 0.5252 as percentage, we should multiply by 100
∴ 0.5252 x 100 = 52.52%

The Percentage Difference Calculator (% difference calculator) will find the percent difference between two positive numbers greater than 0.

Question 2.
Rewrite each underlined portion using percentage language.

Question (i)
One half of the cake is distributed to the children.
Answer:
Answer:
50% of the cake is distributed to the children
Hint:
One half is nothing but \(\frac{1}{2}\) as percentage, we need to multiply by 100
\(\frac{1}{2}\) x 100 = 50%

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1

Question (ii)
Aparna scored 7.5 points out of 10 in a competition.
Answer:
Aparna scored 75% in a competition
Hint:
7.5 points out of 10 is \(\frac{7.5}{10}\) = 0.75
For percentage, we need to multiply by 100
∴ We get 0.75 x 100 = 75%

Question (iii)
The statue was made of pure silver.
Answer:
96% students participated in sports
Hint:
Pure silver means there are no other metals
so, 100 out of 100 parts is made of silver = \(\frac{100}{100}\)
∴ to express as percentage, \(\frac{100}{100}\) x 100% = 100%

Question (iv)
48 out of 50 students participated in sports
Answer:
96% students participated in sports
Hint:
48 out of 50 students in fraction form is \(\frac{48}{50}\).
As a percentage, we need to multiply by 100
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 3

Question (v)
Only 2 persons out of 3 will be selected in the interview
Answer:
Only 66\(\frac{2}{3}\)% persons will be selected in the interview.
Hint:
2 out of 3 in fraction form is \(\frac{2}{3}\) to express as percentage, we need to multiply by 100
\(\frac{2}{3}\) x 100 = \(\frac{200}{3}\) = 66\(\frac{2}{3}\)%

Online percentage off calculator tool makes the calculation faster, and it displays the percentage off of the product in a fraction of seconds.

Question 3.
48 is 32% of what number?
Solution:
Let the number required to be found be ‘x’
Given that 32% of x is 48
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 4

Question 4.
A bank pays ₹ 240 as interest for 2 years for a sum of ? 3000 deposited as savings. Find the rate of interest given by the bank.
Solution:
The formula for simple interest is given by
Interest (I) = \(\frac{PxRxT}{100}\) 100
Where P is principal amount; R is rate of interest in percentage; t is time period Substituting given values in formula, we get
P = 3000, t = 2 yrs, I = Rs 240
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 5

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1

Question 5.
A Welfare Association has a sports club where 30% of the members play cricket, 28% play volleyball, 22% play badminton and the rest play indoor games. If 30 member play indoor games.

  1. How many members are there in the sports club?
  2. How many play cricket, volleyball and badminton?

Solution:
Given data
Members who play Cricket = 30% …..(1)
Members who play Vollyball = 28% …..(2)
Members who play Batminton = 22% ……(3)
Members who play outdoor games = (1) + (2) + (3) = 30
Members who play indoor games = 100 – 80 = 20%
Also given 30 members play indoor games
Let total no of members be ‘x’
∴ \(\frac{30}{x}\) = 20
∴ x = \(\frac{30×100}{20}\) = 15

1. Total number of members is 150

2. Members who play Cricket = 30% of total = \(\frac{30}{100}\) x 150
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 6

Question 6.
What is 25% of 30% of 400?
Solution:
Required to find 25% of 30% of 400
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 7

Question 7.
If the difference between 75% of a number and 60% of the same number is 82.5, then find 20% of the number.
Solution:
Given that 75% of number less 60% of number is 82.5. Let the number be ‘x’
∴ \(\frac{75}{100}\) × x = 82.5
∴ 0.75 – 0.6.60x = 82.5
∴ 0.15x = 82.5
∴ x = \(\frac{82.5}{0.15}\) = \(\frac{8250}{15}\) = 550
Required to find 20% of number ie 20% of x.
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 8

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1

Question 8.
If a car is sold for ₹ 2,00,000 from its original price of? 3,00,000 find the percentage for decrease in the value of the car.
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 9
Solution:
Original price of car = ₹ 3,00,000
actual selling price of car = ₹ 2,00,000
Decrease in amount from original = 3,00,000 – 2,00,000 = 1,00,000
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 10

Question 9.
A number when increased by 18% gives 236. Find the number.
Solution:
Let the number be x. Given that when it is increased by 18%, we get 236.
x + \(\frac{18}{100}\) = 236
\(\frac{100x+18x}{100}\) = 236
\(\frac{118}{100}\)x = 236
The number = x = \(\frac{236×100}{118}\) = 200

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1

Question 10.
A number when decreased by 20% gives 80. Find the number.
Solution:
Let the number be x. Given that when it is increased by 20% we get 80.
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 11

Question 11.
A number is increased by 25% and then decreased by 20%. Find the change in that number.
Solution:
Method 1:
Let the number be x. First it is increased by 25%
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 12
Method 2:
[to understand, let us assume that number is 100]
So, first when we increase by 25%, we get
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 13
We get back 100 ⇒ No change

Question 12.
If the numerator of a fraction is increased by 25% and the denominator is increased by 10%, it becomes \(\frac{2}{5}\). Find the original fraction.
Solution:
Let the fraction be \(\frac{N}{D}\) where N is numerator & D is denominator
It is given that numerator is include by 25%
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 14

Question 13.
A fruit vendor bought some mangoes of which 10% were rotten. He sold 33 \(\frac{1}{3}\)% of the rest. Find the total number of mangoes bought by him initially, if he still has 240 mangoes with him.
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 15
Solution:
Let the number of mangoes bought by fruit seller initially be x.
Given that 10% of mangoes were rotten
∴ Number of rotten mangoes = \(\frac{10}{100}\) × x
Number of good mangoes = x – no. of rotten mangoes
= x – \(\frac{10}{100}\)x = \(\frac{100x-10x}{100}\) = \(\frac{90}{100}\)x …(1)
Number of mangoes sold = 33\(\frac{1}{3}\)% of good mangoes = \(\frac{100}{3}\)%
∴ Mangoes sold = \(\frac{100}{3}\) x \(\frac{90}{100}\)x × \(\frac{1}{100}\) = \(\frac{30}{100}\)x
Number of mangoes remaining = No. of good mangoes – No. of mangoes sold
From (1) and (2)
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 16

Question 14.
A student gets 31% marks in an examination but fails by 12 marks. If the pass percentage is 35%, find the maximum marks of the examination.
Solution:
Let the maximum marks in the exam be ‘x’. Pass percentage is given as 35%
∴ Pass mark = \(\frac{35}{100}\) × x = \(\frac{35}{100}\)x
Student gets 31 % marks = \(\frac{31}{100}\) × x = \(\frac{31}{100}\)x
But student fails by 12 marks → meaning his mark is 12 less than pass mark.
∴ \(\frac{31}{100}\)x = \(\frac{35}{100}\)x – 12
∴ \(\frac{35}{100}\)x – \(\frac{31}{100}\)x = 12
∴ \(\frac{35x-31x}{100}\) = 12 ⇒ \(\frac{4x}{100}\) = 12
∴ \(\frac{12×100}{4}\) = 300

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1

Question 15.
The ratio of boys and girls in a class is 5:3. If 16% of boys and 8% of girls failed in an examination, then find the percentage of passed students.
Solution:
Let number of boys be ‘6’ and number of girls be ‘g’
Ratio of boys and girls is given as 5:3
b:g = 5:3 ⇒ \(\frac{b}{g}\) = \(\frac{5}{3}\) …..(A)
Failure in boys = 16% = \(\frac{16}{100}\) x b = \(\frac{16b}{100}\)
Failure in girls = 8% = \(\frac{8}{100}\) x g = \(\frac{8g}{100}\)
Pass in boys = 100 – 16% = 84% \(\frac{84}{100}\)b …..(1)
Pass in girls = 100 – 8% = 92% = \(\frac{92}{100}\)g
From A, we have \(\frac{b}{g}\) = \(\frac{5}{3}\), = adding 1 on both sides, we get
\(\frac{b}{g}\) + 1 = \(\frac{5}{3}\) + 1
\(\frac{b+g}{g}\) = \(\frac{5+3}{3}\) = \(\frac{8}{3}\)
∴ g = \(\frac{3}{8}\)(b + g)
Similarly b = \(\frac{5}{8}\)(b + g)
Total pass = Pass in girls + Pass in boys
= (1) + (2) = \(\frac{84}{100}\)b + \(\frac{92}{100}\)g
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 17

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1

Objective Type Questions

Question 16.
12% of 250 litres is the same as ………. of 150 liters
(a) 10%
(b) 15%
(c) 20%
(d) 30%
Answer:
(c) 20%
Hint:
12% of 250 = \(\frac{12}{100}\) x 250 = 30 lit
Percentage : \(\frac{30}{150}\) x 100 = 20%

Question 17.
If three candidates A, B and C is 3 in a school election got 153, 245 and 102 votes respectively, the percentage of votes for the winner is ……..
(a) 48%
(b) 49%
(c) 50%
(d) 45%
Answer:
(b) 49%
Hint:
Candidate 1: 153
Candidate 2: 245 – winner [as maximum votes]
Candidate 3: 102
Total votes = 1 + 2 + 3 = 153 + 245 + 102 = 500
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 18

Question 18.
15% of 25% of 10000 = ………..
(a) 375
(b) 400
(c) 425
(d) 475
Answer:
(a) 375
Hint:
15% of 25% of 10000 is
First let us do 25% of 10,000, which is
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 19
Next 15% of the above is \(\frac{15}{100}\) x 2500 = 375

Question 19.
When 60 is subtracted from 60% of a number to give 60, the number is –
(a) 60
(b) 100
(c) 150
(d) 200
Answer:
(d) 200
Hint:
Let the number be ‘x’.
60% of the number is \(\frac{60}{100}\) × x = \(\frac{60x}{100}\)
Given that when 60 is subtracted from 60%, we get 60
i.e \(\frac{60}{100}\)x – 60 = 60
∴ \(\frac{60}{100}\)x = 60 + 60 = 120
∴ x = \(\frac{120×100}{60}\) = 200

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1

Question 20.
If 48% of 48 = 64% of x, then x =
(a) 64
(b) 56
(c) 42
(d) 36
Answer:
(d) 36
Hint:
Given that 48% of 48 48 = 64% of x
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1 20

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.5

Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.5 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.5

To convert CGPA to percentage, all you need to do is multiply your CGPA by 9.5.

Miscellaneous Practice Problems

Question 1.
When Mathi was buying her flat she had to put down a deposit of \(\frac { 1 }{ 10 } \) of the value of the flat. What percentage was this?
Solution:
Percentage of \(\frac { 1 }{ 10 } \) = \(\frac { 1 }{ 10 } \) × 100 % = 10 %
Mathi has to put down a deposit of 10 % of the value of the flat.

Question 2.
Yazhini scored 15 out of 25 in a test. Express the marks scored by her in percentage.
Solution:
Yazhini’s score = 15 out of 25 = \(\frac { 15 }{ 25 } \)
Score in percentage = \(\frac { 15 }{ 25 } \) × 100% = 60%

Question 3.
Out of total 120 teachers of a school 70 were male. Express the number of male teachers as percentage.
Solution:
Total teachers of the school = 120
Number of male teachers = 70
∴ Percentage of male teacher = \(\frac { 70 }{ 120 } \) × 100 % = \(\frac { 700 }{ 12 } \) %
Score in percentage = 58.33%
Percentage of male teachers = 58.33%

The percentage difference calculator is here to help you compare two numbers.

Question 4.
A cricket team won 70 matches during a year and lost 28 matches and no results for two matches. Find the percentage of matches they won.
Solution:
Number of Matches won = 70
Number of Matches lost = 28
“No result” Matches = 2
Total Matches = 70 + 28 + 2 = 100
Percentage of Matches won = \(\frac { 70 }{ 100 } \) × 100 % = 70 %
The won 70% of the matches

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.5

The Percentage Difference Calculator (% difference calculator) will find the percent difference between two positive numbers greater than 0.

Question 5.
There are 500 students in a rural school. If 370 of them can swim, what percentage of them can swim and what percentage cannot?
Solution:
Total number of students = 500
Number of students who can swim = 370
Percentage of students who can swim = \(\frac { 370 }{ 500 } \) × 100 % = 74 %
Number of students who cannot swim = 500 – 370 = 130
Percentage of students who cannot swim = \(\frac { 130 }{ 500 } \) × 100 % = 26 %
i.e. 74% can swim and 26% cannot swim

Question 6.
The ratio of Saral’s income to her savings is 4 : 1. What is the percentage of money saved by her?
Solution:
Total parts of money = 4 + 1 = 5
Part of money saved = 1
∴ Percentage of money saved = \(\frac { 1 }{ 5 } \) × 100% = 20%
∴ 20% of money is saved by Saral

Question 7.
A salesman is on a commission rate of 5%. How much commission does he make on sales worth ₹ 1,500?
Solution:
Total amount on sale = ₹ 1,500
Commission rate = 5 %
Commission received = 5 % of ₹ 1,500 = \(\frac { 5 }{ 100 } \) × 1500 = ₹ 75
∴ Commission received = ₹ 75

Question 8.
In the year 2015 ticket to the world cup cricket match was ₹ 1,500. This year the price has been increased by 18%. What is the price of a ticket this year?
Solution.
Price of a ticket in 2015 = ₹ 1500
Increased price this year = 18% of price in 2015
= 18 % of ₹ 1500 = \(\frac { 18 }{ 100 } \) × 1500
= ₹ 270
Price of ticket this year = last year price + increased price
= ₹ 1500 + ₹ 270 = ₹ 1770
Price of ticket this year = ₹ 1770

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.5

Question 9.
2 is what percentage of 50?
Solution:
Let the required percentage be x
x% of 50 = 2
\(\frac { x }{ 100 } \) × 50 = 2
x = \(\frac{2 \times 100}{50}\) = 4 %
∴ 4 % of 50 is 2

Question 10.
What percentage of 8 is 64?
Solution:
Let the required percentage be x
So x % of 8 = 64
\(\frac { x }{ 100 } \) × 8 = 64
x = \(\frac{64 \times 100}{8}\) = 800
∴ 800 % of 8 is 64

Question 11.
Stephen invested ₹ 10,000 in a savings bank account that earned 2% simple interest. Find the interest earned if the amount was kept in the bank for 4 years.
Solution:
Principal (P) = ₹ 10,000
Rate of interest (r) = 2%
Time (n) = 4 years
∴ Simple Interest I = \(\frac { pnr }{ 100 } \)
= \(\frac{10000 \times 4 \times 2}{100}\)
= ₹ 800
Stephen will earn ₹ 800

Question 12.
Riya bought ₹ 15,000 from a bank to buy a car at 10% simple interest. If she paid ₹ 9,000 as interest while clearing the loan, find the time for which the loan was given.
Solution:
Here Principal (P) = ₹ 15,000
Rate of interest (r) = 10 %
Simple Interest (I) = ₹ 9000
I = \(\frac { pnr }{ 100 } \)
9000 = \(\frac{15000 \times n \times 10}{100}\)
n = \(\frac{9000 \times 100}{15000 \times 10}\)
n = 6 years
∴ The loan was given for 6 years

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.5

Question 13.
In how much time will the simple interest on ₹ 3,000 at the rate of 8% per annum be the same as simple interest on ?4,000 at 12% per annum for 4 years?
Solution:
Let the required number of years be x
Simple Interest I = \(\frac { pnr }{ 100 } \)
Principal P1 = ₹ 3000
Rate of interest (r) = 8 %
Time (n1) = n1 years
Simple Interest I1 = \(\frac{3000 \times 8 \times n_{1}}{100}\) = 240 n1
Principal (P2) = ₹ 4000
Rate of interest (r) = 12 %
Time n2 = 4 years
Simple Interest I2 = \(\frac{4000 \times 12 \times 4}{100}\)
I2 = 1920
If I1 = I2
240 n1 = 1920
n1 = \(\frac { 1920 }{ 240 } \) = 8
∴ The required time = 8 years

Challenge Problems

Question 14.
A man travelled 80 km by car and 320 km by train to reach his destination. Find what percent of total journey did he travel by car and what per cent by train?
Solution:
Distance travelled by car = 80 km.
Distance travelled by train = 320 km
Total distance = 80 + 320 km = 400 km
Percentage of distance travelled by car = \(\frac { 80 }{ 400 } \) × 100 % = 20 %
Percentage of distance travelled by train = \(\frac { 320 }{ 800 } \) × 100 % = 40 %

Question 15.
Lalitha took a math test and got 35 correct and 10 incorrect answers. What was the percentage of correct answers?
Solution:
Number of correct answers = 35
Number of incorrect answers = 10
Total number of answers = 35 + 10 = 45
Percentage of correct answers = \(\frac { 35 }{ 45 } \) × 100 %
= 77.777 % = 77.78 %

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.5

Question 17.
The population of a village is 8000. Out of these, 80% are literate and of these literate people, 40% are women. Find the percentage of literate women to the total population?
Solution:
Population of the village = 8000 people
literate people = 80 % of population
= 80 % of 8000 = \(\frac { 80 }{ 100 } \) × 8000
literate people = 6400
Percentage of women = 40 %
Number of women = 40 % of literate people
= \(\frac { 40 }{ 100 } \) × 6400 = 2560
∴ literate women : Total population
= 8000 : 2560
= 25 : 8

Question 18.
A student earned a grade of 80% on a math test that had 20 problems. How many problems on this test did the student answer correctly?
Solution:
Total number of problems in the test = 20
Students score = 80 %
Number of problem answered = \(\frac { 80 }{ 100 } \) × 20 = 16

Question 19.
A metal bar weighs 8.5 kg. 85% of the bar is silver. How many kilograms of silver are in the bar?
Solution:
Total weight of the metal = 8.5 kg
Percentage of silver in the metal = 85%
Weight of silver in the metal = 85% of total weight
= \(\frac { 85 }{ 100 } \) × 8.5 kg
= 7.225 kg
7.225 kg of silver are in the bar.

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.5

Question 20.
Concession card holders pay ₹ 120 for a train ticket. Full fare is ₹ 230. What is the percentage of discount for concession card holders?
Solution:
Train ticket fare = ₹ 230
Ticket fare on concession = ₹ 120
Discount = Ticket fare – concession fare = 230 – 120 = ₹ 110
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.5 1
Percentage of discount = 47.83%

Question 21.
A tank can hold 200 litres of water. At present, it is only 40% full. How many litres of water to fill in the tank, so that it is 75 % full?
Solution:
Capacity of the water tank = 200 litres
Percentage of water in the tank = 40%
Percentage of water to fill = Upto 75%
Difference in percentage = 75 % – 40 % = 35 %
∴ Volume of water to be filled = Percentage of difference × total capacity
= \(\frac { 35 }{ 100 } \) × 200 = 70 l
70 l of water to be filled

The Percentage Difference Calculator (% difference calculator) will find the percent difference between two positive numbers greater than 0.

Question 22.
Which is greater 16 \(\frac { 2 }{ 3 } \) or \(\frac { 2 }{ 5 } \) or 0.17 ?
Solution:
16 \(\frac { 2 }{ 3 } \) = \(\frac { 50 }{ 30 } \)
= \(\frac { 50 }{ 30 } \) × 100 % = 1666.67 %
⇒ \(\frac { 2 }{ 5 } \)
= \(\frac { 2 }{ 5 } \) × 100 = 40 %
0.17 = \(\frac { 17 }{ 100 } \) = 17 %
∴ 1666.67 is greater
∴ 16 \(\frac { 2 }{ 3 } \) is greater

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.5

Question 23.
The value of a machine depreciates at 10% per year. If the present value is ₹ 1,62,000, what is the worth of the machine after two years.
Solution:
Present value of the machine = ₹ 1,67,000
Rate of depreciation = 10 % Per annum
Time (n) = 2 years
For 1 year depreciation amount = \(\frac{1,62,000 \times 1 \times 10}{100}\) = ₹ 16,200
Worth of the machine after one year = Worth of Machine – Depreciation
= 1,67,000 – 16,200 = 1,45,800
Depreciation of the machine for 2nd year = 145800 × 1 × \(\frac { 10 }{ 100 } \) = 14580
Worth of the machine after 2 years = 1,45,800 – 14,580 = 1,31,220
∴ Worth of the machine after 2 years = ₹ 1,31,220

Question 24.
In simple interest, a sum of money amounts to ₹ 6,200 in 2 years and ₹ 6,800 in 3 years. Find the principal and rate of interest.
Solution:
Let the principal P = ₹ 100
If A = 6200
⇒ Principal + Interest for 2 years = 6200
A = ₹ 7400
⇒ Principal + Interest for 3 years = 7400
∴ Difference gives the Interest for 1 year
∴ Interest for 1 year = 7400 – 6200
I = 1200
\(\frac { pnr }{ 100 } \) = 1200 ⇒ \(\frac{P \times 1 \times r}{100}\) = 1200
If the Principal = 10,000 then
\(\frac{10,000 \times 1 \times r}{100}\) = 1200 ⇒ r = 12 %
Rate of interest = 12 % Per month

Question 25.
A sum of ₹ 46,900 was lent out at simple interest and at the end of 2 years, the total amount was ₹ 53,466.Find the rate of interest per year.
Solution:
Here principal P = ₹ 46900
Time n = 2 years
Amount A = ₹ 53466
Let r n be the rate of interest per year p
Intrest I = \(\frac { pnr }{ 100 } \)
A = P + I
53466 = 46900 + \(\frac{46900 \times 2 \times r}{100}\)
53466 – 46900 = \(\frac{46900 \times 2 \times r}{100}\)
6566 = 469 × 2 × r
r = \(\frac{6566}{2 \times 469}\) % = 7 %
Rate of interest = 7 % Per Year

Question 26.
Arun lent ₹ 5,000 to Balaji for 2 years and ₹ 3,000 to Charles for 4 years on simple interest at the same rate of interest and received ₹ 2,200 in all from both of them as interest. Find the rate of interest per year.
Solution:
Principal lent to Balaji P1 = ₹ 5000
Time n1 = 2 years
Let r be the rate of interest per year
Simple interest got from Balaji = \(\frac { pnr }{ 100 } \) ⇒ I1 = \(\frac{5000 \times 25 \times r}{100}\)
Again principal let to Charles P2 = ₹ 3000
Time (n2) = 4 years
Simple interest got from Charles (I2) = \(\frac{3000 \times 4 \times r}{100}\)
Altogether Arun got ₹ 2200 as interest.
∴ I1 + I2 = 2200
\(\frac{5000 \times 2 \times r}{100}+\frac{3000 \times 4 \times r}{100}\) = 2200
100r + 120r = 2200
220r = 2200 = \(\frac { 2200 }{ 220 } \)
r = 10 %
Rate of interest per year = 10 %

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.5

Question 27.
If a principal is getting doubled after 4 years, then calculate the rate of interest. (Hint: Let P = ₹ 100)
Solution:
Let the principal P = ₹ 100
Given it is doubled after 4 years
i.e. Time n = 4 years
After 4 years A = ₹ 200
∴ A = P + I
A – P = I
200 – 100 = I
After 4 years interest I = 100
I = \(\frac { pnr }{ 100 } \) ⇒ 100 = \(\frac{100 \times 4 \times r}{100}\)
4r = 100 ⇒ r = 25 %
Rate of interest r = 25 %

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.2

Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.2

The percentage difference calculator is here to help you compare two numbers.

Question 1.
Write each of the following percentage as decimal.
(i) 21 %
(ii) 93.1 %
(iii) 151 %
(iv) 65 %
(v) 0.64 %
Solution:
(i) 21 %
= \(\frac { 21 }{ 100 } \) = 0.21

(ii) 93.1 %
= \(\frac { 93.1 }{ 100 } \) = 0.931

(iii) 151 %
= \(\frac { 151 }{ 100 } \) = 1.51

(iv) 65 %
= \(\frac { 65 }{ 100 } \) = 0.65

(v) 0.64 %
= \(\frac { 0.64 }{ 100 } \) = 0.0064

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.2

Question 2.
Convert each of the following decimal as percentage
(i) 0.282
(ii) 1.51
(iii) 1.09
(iv) 0.71
(v) 0.858
Solution:
(i) 0.282
= 0.282 × 100% = \(\frac { 282 }{ 1000 } \) × 100 %
= 28.2 %

(ii) 1.51
= \(\frac { 151 }{ 100 } \) × 100 %
= 151 %

(iii) 1.09
= \(\frac { 109 }{ 100 } \) × 100 %
= 109 %

(iv) 0.71
= \(\frac { 71 }{ 100 } \) × 100 %
= 71 %

(v) 0.858
= \(\frac { 858 }{ 1000 } \) × 100 %
= 85.8 %

Question 3.
In an examination a student scored 75% of marks. Represent the given the percentage in decimal form?
Solution:
Student’s Score = 75% = \(\frac { 75 }{ 100 } \) = 0.75

Question 4.
In a village 70.5% people are literate. Express it as a decimal.
Solution:
Percentage of literate people = 70.5%
= \(\frac { 70.5 }{ 100 } \)
= 0.705

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.2

Question 5.
Scoring rate of a batsman is 86%. Write his strike rate as decimal.
Solution:
Scoring rate of the batsman = 86%
= \(\frac { 86 }{ 100 } \)
= 0.86

Question 6.
The height of a flag pole in school is 6.75m. Write it as percentage.
Solution:
Height of flag pole = 6.75m
= \(\frac { 675 }{ 100 } \)
= 6.75%

Question 7.
The weights of two chemical substances are 20.34 g and 18.78 g. Write the difference in percentage?
Solution:
Weight of substance 1 = 20.34g
Percentage of substance 1 = \(\frac { 2034 }{ 100 } \) = 2034 %
Weight of substance 2 = 18.78g
Percentage of substance 2 = \(\frac { 1878 }{ 100 } \) = 1878 %
Their difference = 2034 – 1878 = 156%

Percentage decrease calculator finds the decrease from one amount to another as a percentage of the first amount.

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.2

Question 8.
Find the percentage of shaded region in the following figure.
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.2 1
Solution:
Total region = 4 parts
Shaded region = 1 part
Fraction of shaded region = \(\frac { 1 }{ 4 } \)
Percentage of shaded region = \(\frac { 1 }{ 4 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 1 }{ 4 } \) × 100 %
= 25 %

Objective Type Questions

Question 1.
Decimal value of 142.5% is
(i) 1.425
(ii) 0.1425
(iii) 142.5
(iv) 14.25
Hint:
142.5 % = \(\frac { 1425 }{ 10 } \) %
= \(\frac { 1425 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= 1.425
Answer:
(i) 1.425

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.2

Question 2.
The percentage of 0.005 is
(i) 0.005 %
(ii) 5 %
(iii) 0.5 %
(iv) 0.05 %
Hint:
0.005 = \(\frac { 5 }{ 1000 } \)
= \(\frac { 5 }{ 1000 } \) × \(\frac { 100 }{ 100 } \)
= 0.5 %
Answer:
(iii) 0.5 %

Question 3.
The percentage of 4.7 is
(i) 0.47 %
(ii) 4.7 %
(iii) 47 %
(iv) 470 %
Hint:
4.7 = \(\frac { 47 }{ 10 } \)
= \(\frac { 47 }{ 10 } \) × \(\frac { 100 }{ 100 } \)
= 470 %
Answer:
(iv) 470 %