Class 7

Samacheer Kalvi 7th Books Solutions Guide

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

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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 English Book Answers Solutions Guide

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Samacheer Kalvi 7th Maths Book Answers Solutions Guide

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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.