# Class 7

## 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 _______
(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.
(i) True
(ii) False
(iii) False
(iv) True

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Objective Type Question

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

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

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

1/2 as a decimal is 0.5.

Question 2.
Write the decimal numbers from the following place value table.

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

(ii) 0.025

(iii) 428.001

(iv) 173.178

(v) 19.54

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 $$\frac { 14 }{ 25 }$$ = $$\frac { 89 }{ 25 }$$

Objective Questions

Question 8.
3 + $$\frac { 4 }{ 100 }$$ + $$\frac { 9 }{ 1000 }$$ = ?
(i) 30.49
(ii) 3049 9
(iii) 3.0049
(iv) 3.049
(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
(iv) 0.6
Hint: $$\frac { 3 }{ 5 }$$ = $$\frac{3 \times 2}{5 \times 2}$$ = $$\frac { 6 }{ 10 }$$ = 0.06

Question 10.
The simplest form of 0.35 is
(i) $$\frac { 35 }{ 1000 }$$
(ii) $$\frac { 35 }{ 10 }$$
(iii) $$\frac { 7 }{ 20 }$$
(iv) $$\frac { 7 }{ 100 }$$
(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

(ii) 132.105

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

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.

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

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:

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.

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.

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.

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.

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.

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.

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.

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

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

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

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

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

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

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

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

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

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

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

(ii) 263.271

(iii) 17.39

(iv) 9.657

(v) 4972.068

Objective Type Questions

Question 6.
The place value of 3 in 85.073 is _____
(i) tenths
(ii) hundredths
(iii) thousands
(iv) thousandths
(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
(ii) 1000
Hint: 85.073 = 8 × 10 + 5 × 1 + 0 × $$\frac { 1 }{ 10 }$$ + 7 × $$\frac { 1 }{ 100 }$$ + 3 × $$\frac { 1 }{ 1000 }$$

Question 8.
The decimal representation of 30 kg and 43 g is ____ kg.
(i) 30.43
(ii) 30.430
(iii) 30.043
(iv) 30.0043
(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
(ii) 2.64
Hint: 264 cm = $$\frac { 264 }{ 100 }$$ m = 2.64 m

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

Students can Download Maths Chapter 1 Number System Ex 1.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

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

Question 1.
Fill in the blanks.
(i) -80 × ____ = -80
(ii) (-10) × ____ = 20
(iii) 100 × ___ = -500
(iv) ____ × (-9) = -45
(v) ___ × 75 = 0
Solution:
(i) 1
(ii) -2
(iii) -5
(iv) 5
(v) 0

Question 2.
Say True or False:
(i) (-15) × 5 = 75
(ii) (-100) × 0 × 20 = 0
(iii) 8 × (-4) = 32
Solution:
(i) False
(ii) True
(iii) False

Question 3.
What will be the sign of the product of the following:
(i) 16 times of negative integers.
(ii) 29 times of negative integers.
Solution:
(i) 16 is an even interger.
If negative integers are multiplied even number of times, the product is a positive integer.
∴ 16 times a negative integer is a positive integer.

(ii) 29 times negative integer.
If negative integers are multiplied odd number of times, the product is a negative integer. 29 is odd.
∴ 29 times negative integers is a negative integer.

Question 4.
Find the product of
(i) (-35) × 22
(ii) (-10) × 12 × (-9)
(iii) (-9) × (-8) × (-7) × (-6)
(iv) (-25) × 0 × 45 × 90
(v) (-2) × (+50) × (-25) × 4
Solution:
(i) 35 × 22 = -770
(ii) (-10) × 12 × (-9) = (-120) × (-9) = +1080
(iii) (-9) × (-8) × (-7) × (-6) = (+72) × (-7) × (-6) = (-504) × (-6) = +3024
(iv) (-25) × 0 × 45 × 90 = 0 × 45 × 90 = 0 × 90 = 0
(v) (-2) × (+50) × (-25) × 4 = (-100) × -25 × 4 = 2500 × 4 = 10,000

Question 5.
Check the following for equality and if they are equal, mention the property.
(i) (8 – 13) × 7 and 8 – (13 × 7)
Solution:
Consider (8 – 13) × 7 = (-5) × 7 = -35
Now 8 – (13 × 7) = 8 – 91 = -83
∴ (8 – 13) × 7 ≠ 8 – (13 × 7)

(ii) [(-6) – (+8)] × (-4) and (-6) – [8 × (-4)]
Solution:
[(-6) – (+8)] × (-4) = [(-6) + (-8)] × (-4) = (-14) × (-4) = +56
Now (-6) – [8 × (-4)] = (-6) – (-32)
= (-6) + (+32) = +26
∴ [(-6) – (+8)] × (-4) ≠ (-6) – [8 × (-4)]

(iii) 3 × [(-4) + (-10)] and [3 × (-4) + 3 × (-10)]
Solution:
Consider 3 × [(-4) + (-10)] = 3 × -14 = -42
Now [3 × (-4) + 3 × (-10)] = (-12) + (-30) = -42
Here 3 × [(-4) + (-10)] = [3 × (-4) + 3 × (-10)]
It is the distributive property of multiplication over addition.

Question 6.
During summer, the level of the water in a pond decreases by 2 inches every week due to evaporation. What is the change in the level of the water over a period of 6 weeks?
Solution:
Level of water decreases a week = 2 inches.
Level of water decreases in 6 weeks = 6 × 2 = 12 inches

Question 7.
Find all possible pairs of integers that give a product of -50.
Solution:
Factor of 50 are 1, 2, 5, 10, 25, 50.
Possible pairs of integers that gives product -50:
(-1 × 50), (1 × (-50)), (-2 × 25), (2 × (-25)), (-5 × 10), (5 × (-10))

What is the factor of 50?​. Answer: The factors of 50 are 1, 2, 5,10, 25, and 50.

Objective Type Questions

Question 8.
Which of the following expressions is equal to -30.
(i) -20 – (-5 × 2)
(ii) (6 × 10) – (6× 5)
(iii) (2 × 5)+ (4 × 5)
(iv) (-6) × (+5)
Solution:
(iv) (-6) × (+5)
Hint:
(i) -20 + (10) = -10
(ii) 60 – 30 = 30
(iii) 10 + 20 = 30
(iv) (-6) × (+5) = – 30

Question 9.
Which property is illustrated by the equation: (5 × 2) + (5 × 5) = 5 × (2 + 5)
(i) commutative
(ii) closure
(iii) distributive
(iv) associative
Solution:
(iii) distributive

Question 10.
11 × (-1) = _____
(i) -1
(ii) 0
(iii) +1
(iv) -11
Solution:
(iv) -11

Question 11.
(-12) × (-9) =
(i) 108
(ii) -108
(iii) +1
(iv) -1
Solution:
(i) 108

## 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

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

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

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 %

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.

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

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

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 %

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

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

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.

Question 8.
Find the percentage of shaded region in the following figure.

Solution:
Total region = 4 parts
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
(i) 1.425

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 %
(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 %
(iv) 470 %

## Samacheer Kalvi 7th Science Solutions Term 1 Chapter 1 Measurement

Students can Download Science Chapter 1 Measurement Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Science 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 Science Solutions Term 1 Chapter 1 Measurement

### Samacheer Kalvi 7th Science Measurement Textual Evaluation

Question 1.
Which of the following is a derived unit?
(a) mass
(b) time
(c) area
(d) length
(c) area

Question 2.
Which of the following is correct?
(a) 1L=lcc
(b) 1L= l0cc
(c) 1L= l00cc
(d) 1L= l000cc
(d) 1L = 1000cc

Question 3.
SI unit of density is
(a) kg/m2
(b) kg/m3
(c) kg/m
(d) g/m3
(b) kg/m3

Question 4.
Two spheres have equal mass and volume in the ratio 2:1. The ratio of their density is
(a) 1:2
(b) 2:1
(c) 4:1
(d) 1:4
(b) 2:1

Question 5.
Light year is the unit of
(a) Distance
(b) time
(c) density
(d) both length and time
(a) Distance

II. Fill in the blanks:

1. Volume of irregularly shaped objects are measured using the law of ___________
2. One cubic metre is equal to ___________ cubic centimetre.
3. Density of mercury is ___________
4. One astronomical unit is equal to ___________
5. The area of a leaf can be measured using a ___________

1. Archimedes
2. 10,00,000 or 1066
3. 13,600 kg/m3
4. 1.496×1011m
5. graph sheet

III. State whether the following statements are true or false.

Question 1.
The region covered by the boundary of the plane figure is called its volume.
(False) Correct statement: The region covered by the boundary of plane figure is called its area.

Question 2.
Volume of liquids can be found using measuring containers.
True

Question 3.
Water is denser than kerosene.
True

Question 4.
A ball of iron floats in mercury.
True

Question 5.
A substance which contains less number of molecules per unit volume is said to be denser.
False. Correct statement: A substance which contains more number of molecules per unit volume is said to be denser.

IV. Match the items in column – I to the items in column – II :

Question 1.

 Column -1 Column – II i. Area (a) light year ii. Distance (b) m3 iii. Density (c) m2 iv. Volume (d) kg V. Mass (e) kg/ m3

1. i
2. ii
3. iii
4. iv
5. v

Question 2.

 Column -1 Column – II i. Area (a) g / cm3 ii. Length (b) measuring jar iii. Density (c) amount of a substance iv. Volume (d) rope V. Mass (e) plane figures

1. i
2. ii
3. iii
4. iv
5. v

V. Arrange the following in correct sequence :

Question 1.
1L, 100 cc, 10 L, 10 cc
10 cc, 100 cc, 1L, 10L

Question 2.
Copper, Aluminium, Gold, Iron
Aluminium, Iron, Copper, Gold

VI. Use the analogy to fill in the blank:

Question 1.
Area: M2 :: Volume : _________
M3

Question 2.
Liquid : Litre :: Solid : _________
cm3

Question 3.
Water: Kerosene :: ______ : Aluminium
Iron

VII. Assertion and reason type questions:
Mark the correct choice as

(a) Both assertion and reason are true and reason is the correct explanation of assertion
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion
(c) If assertion is true but reason is false
(d) Assertion is false but reason is true.

Question 1.
Assertion (A) : Volume of a stone is found using a measuring cylinder.
Reason (R) : Stone is an irregularly shaped object.
(a) If both assertion and reason are true and reason is the correct explanation of assertion

Question 2.
Assertion (A) : Wood floats in water.
Reason (R) : Water is a transparent liquid.
(b) If both assertion and reason are true, but reason is not the correct explanation of assertion
Correct explanation: Density of water is more than the density of wood.

Question 3.
Assertion (A) : Iron ball sinks in water.
Reason (R) : Water is denser than iron.
(b) If both assertion and reason are true, but reason is not the correct explanation of assertion
Correct explanation : Density of iron is more than that of water.

Question 1.
Name some of the derived quantities.
Area, volume, density.

Question 2.
Give the value of one light year.
One light year = 9.46 x 1015m

Question 3.
Write down the formula used to find the volume of a cylinder.
Volume of a cylinder = πr2 h

Question 4.
Give the formula to find the density of objects.

Question 5.
Name the liquid in which an iron ball sinks.
Iron ball sinks in water. The density of an iron ball is more than that of water so it sinks in water.

Question 6.
Name the unit used to measure the distance between celestial objects.
Astronomical unit and light year are the units used to measure the distance between celestial objects.

Question 7.
What is the density of gold?
Density of gold is 19,300 kg/m3

Question 1.
What are derived quantities?
The physical quantities which can be obtained by multiplying, dividing or by mathematically combining the fundamental quantities are known as derived quantities.
(or)
The physical quantities which are expressed is terms of fundamental quantities are called

Question 2.
Distinguish between the volume of liquid and capacity of a container.

 S.No Volume of liquid Capacity of a container 1. Volume is the amount of space taken up by a liquid Capacity is the measure of an objects ability to hold a substance like solid, liquid or gas 2. It is measured in cubic units. It is measured in litres, gallons, pounds, etc. 3. It is calculated by multiplying the length, width and height of an object. It’s measurement is cc or ml.

Question 3.
Define the density of objects.
Density of a substance is defined as the mass of the substance contained in unit volume

Question 4.
What is one light year?
One light year is the distance travelled by light in vacuum during the period of one year.
1 Light year = 9.46 x 1015m.

Question 5.
Define one astronomical unit?
One astronomical unit is defined as the average distance between the earth and the sun.
1 AU = 1.496 5 106km = 1.496 × 1011m.

Question 1.
Describe the graphical method to find the area of an irregularly shaped plane figure.
To find the area of an irregularly shaped plane figure, we have to use graph paper.

1. Place a piece of paper with an irregular shape on a graph paper and draw its outline.
2. To find the area enclosed by the outline, count the number of squares inside it (M).
3. You will find that some squares lie partially inside the outline.
4. Count a square only if half (p) or more of it (N) lies inside the outline.
5. Finally count the number of squares, that are less than half. Let it be

For the shape in figure we have the following:
M = 50
N = 7
P = 4
Q = 4
Now, the approximate area of the can be calculated using the following formula.

Area of the leaf = M+($$\frac { 3 }{ 4 }$$) N + ($$\frac { 1 }{ 2 }$$) P+$$\frac { 1 }{ 4 }$$ Qsq.cm
= 52 + 5.25 = 58.25 sq.mm = 0.5825 sq.cm

Question 2.
How will you determine the density of a stone using a measuring jar?
Determination of density of a stone using a measuring cylinder.

1. In order to determine the density of a solid, we must know the mass and volume of the stone.
2. The mass of the stone is determined by a physical balance very accurately. Let it be ‘m’ grams.
3. In order to find the volume, take a measuring cylinder and pour in it some water.
4. Record the volume of water from the graduations marked on measuring cylinder. Let it be 40 cm3.
5. Now tie the given stone to a fine thread and lower it gently in the measuring cylinder, such that it is completely immersed in water.
6. Record the new level of water. Let it be 60 cm3

∴Volume of the solid = (60-40) cm3
= 20 cm3 = V cm3 (assume)
Knowing the mass and the volume of the stone, the density can be calculate by the formula:

XI. Questions based on Higher er Thinking skills:

Question 1.
There are three spheres A, B, C as shown below :

Sphere A and B are made of the same material. Sphere C is made of a different material. Spheres A and C have equal radii. The radius of sphere B is half that of A. Density of A is double that of C.

Question 2.

1. Find the ratio of masses of spheres A and B.
2. Find the ratio of volumes of spheres A and B.
3. Find the ratio of masses of spheres A and C.

i. Ratio of masses of spheres A and B
MA : MB
D × VA : D × VB
Let the mass of sphere A = MA
Let the mass of sphere B = MB
Mass = Density × Volume
MA = DA × VA
MB = DB × VB (Density is same)
Volume of Sphere A =$$\frac { 4 }{ 3 }$$πr3

ii. Ratio of volumes of spheres A and B
VA : VB
8 : 1 (As mass is directly proportional to volume)

iii. Ratio of masses of spheres A and C.

[∴ Density of A is double that of C ]

XII. Numerical problems:

Question 1.
A circular disc has a radius 10 cm. Find the area of the disc in m2. (Use n = 3.14)
Given radius = 10 cm = 0.1m
π= 3.14
Area of a circular disc A = ?
Formula : Area of a circle A = πr2
= 3.14 × 0.1 × 0.1
Solution : A = 0.0314 m2

Question 2.
The dimension of a school playground is 800 m x 500 m. Find the area of the ground.
Given :The dimension of a school
Playground = l x b = 800 m x 500 m

Formula :Area of the ground A = l x b
= 800 x 500 = 4,00,000
A = 4,00,000 m2

Question 3.
Two spheres of same size are made from copper and iron respectively. Find the ratio between their masses. Density of copper 8,900 kg/m and iron 7,800 kg/m3
Given : Density Copper Dc = 8900 kg/m3
Density of Iron D1 = 7800 kg/m3
Volume of Copper sphere = Volume of Iron sphere
To find : Ratio of Masses of Copper (MC) and Iron (MI)
Solution: Mass = Density x Volume
MC = DC x V, M1 = D1 x V
MC = 8900 V, M1 = 7,800 V
MC = M1
8900 V : 7800 V
= 1.14: 1

Question 4.
A liquid having a mass of 250 g fills a space of lOOOcc. Find the density of the liquid.
Given : Mass of a liquid M = 250 g
Volume V = l000cc
Density of the liquid D = ?

Solution: Density of the liquid = 0.25 g/cc

Question 5.
A sphere of radius 1cm is made from silver. If the mass of the sphere is 33 g, find the density of silver (Take π = 3.14)
Given : radius of a sphere r = 1cm
Volume of the sphere V = ?
Mass of the sphere M = 33 g
Density of silver D = ?

Solution: Density of silver sphere = 7.889 g/cc.

XIII. Cross word puzzle:

Clues – Across
1. SI unit of temperature
2. A derived quantity
3. Mass per unit volume
4. Maximum volume of liquid a container can hold

Clues – Down
a. A derived quantity
b. SI unit of volume
c. A liquid denser than iron
d. A unit of length used to measure very long distances

Clues – Across
1. KELVIN
2. VOLUME
3. DENSITY
4. CAPACITY

Clues – Down
a. VELOCITY
b. CUBIC METRE
c. MERCURY
d. LIGHT YEAR

### Samacheer Kalvi 7th Science Measurement lntext Activities

Students can practice CBSE Class 7 Science MCQs Multiple Choice Questions with Answers to score good marks in the examination.

Activity -1

Take a leaf from any one of trees in your neighborhood.
Place the leaf on a graph sheet and draw the outline of the leaf with a pencil. Remove the leaf. You can see the outline of the leaf on the graph sheet.

1. Now, count the number of whole squares enclosed within the outline of the leaf. Take it to be M.
2. Then, count the number of squares that are more than half. Take it as N.
3. Next, count the number of squares which are half of a whole square. Note it to be P.
4. Finally, count the number of squares that are less than half. Let it be Q.
5. M = _______;N = _______; P = _______; Q = _______

Now, the approximate area of the leaf can be calculated using the following formula:
Approximate area of the leaf = M +($$\frac { 3 }{ 4 }$$) N+($$\frac { 1 }{ 2 }$$) P+($$\frac { 1 }{ 4 }$$) Q square cm
Area of the leaf =________.
This formula can be used to calculate the area of any irregularly shaped plane figures.
M = 50
N = 7
P = 4
Q = 4

Activity – 2

Draw the following regularly shaped figures on a graph sheet and find their area by the graphical method. Also, find their area using appropriate formula. Compare the results obtained in two methods by tabulating them.
(a) A rectangle whose length is 12 cm and breadth is 4 cm.
(b) A square whose side is 6 cm.
(c) A circle whose radius is 7 cm.
(d) A triangle whose base is 6 cm and height is 8 cm.

Activity – 3

Take a measuring cylinder and pour some water into it (Do not fill the cylinder completely). Note down the volume of water from the readings of the measuring cylinder. Take it as V . Now take a small stone and tie it with a thread. Immerse the stone inside the water by holding the thread. This has to be done such that the stone does not touch the walls of the measuring cylinder. Now, the level of water has raised. Note down the volume of water and take it to be V . The volume of the stone is equal to the raise in the volume of water.

V1= _______
V2=_______
Volume of stone = v2 – v1 =_______.
V1 = 30 cc, V2 = 40 cc; Volume of stone =v2 – v1 = 40cc – 30cc = 10cc

Activity – 4
(a) Take an iron block and a wooden block of same mass (say 1 kg each). Measure their volume. Which one of them has more volume and occupies more volume?
(b) Take an iron block and a wooden block of same size. Weigh them and measure their mass. Which one of them has more mass?
(a) Wooden block has more volume and occupies more volume. (As the molecules of wood are loosely packed)
(b) Iron block has more mass. (In iron block, molecules are closely packed).

### Samacheer Kalvi 7th Science Measurement Additional Questions

Question 1.
The unit of volume is _____
(a) m3
(b) m3
(c) cm3
(d) km
(a) m3

Question 2.
Physical quantities are classified into _____ type
(a) three
(b) two
(c) four
(d) none of the above
(b) two

Question 3.
The SI unit of speed is _____
(a) m/s2
(b) m/s
(c) km/h
(d) m2/s
(a) m/s2

Question 4.
1 litre = ______ cc
(a) 100
(b) 1000
(c) 10
(d) 0.1
1000

Question 5.
The formula to calculate area of a rectangle is _____
(b) side × side
(d) none of the above

Question 6.
_____ is a derived quantity.
(a) length
(b) mass
(c) time
(d) area
(d) area

Question 7.
The amount of space occupied by a three dimensional object is known as its _____
(a) density
(b) volume
(c) Area
(d) mass
(b) volume

Question 8.
The maximum volume of liquid that a continer can hold is _____
(a) area
(b) volume
(c) capacity
(d) density
(c) capacity

Question 9.
The shortest distance between the earth and the sun is called as _____ position.
(a) Light year
(b) normal
(c) perihelion
(d) aphelion
(c) Perihelion

Question 10.
The largest distance between the earth and the sun is called as _____ position.
(a) normal
(b) perihelion
(c) aphelion
(d) none of the above
(c) aphelion

II. Fill in the Blanks.

1. The materials with higher density are called ______
2. The materials with lower density are called ______
3. The area of irregularly shaped figures can be calculated with the help of a ______
4. The SI unit of volume is ______
5. The SI unit of density is ______
6. The CGS unit of density is ______
7. If the density of a solid is lower than that of a liquid it ______ is that liquid
8. If the density of a solid is higher than that of a liquid, it ______ is that liquid.
9. The total number of seconds in one year = ______
10. The average distance between the earth and the sun is about ______ million kilometre.

1. denser
2. rarer
3. graph sheet
4. cubic metre or m3
5. kg/m3
6. g/cm3
7. floats
8. sinks
9. 3.15 3 x 107 second
10. 149.6

III. True or False – if false give the correct statement.

Question 1.
One square metre is the area enclosed inside a square of side 2 metre.
(False) Correct Statement: One square metre is the area enclosed inside a square of side 1 metre.

Question 2.
Area is a derived quantity as we obtain by multiplying twice of the fundamental physical quantity length.
True.

Question 3.
Density of water is 100 kg/m3
(False) Correct statement: Density of water is 1000 kg/m3

Question 4.
Density is defined as the mass of the substance contained in unit volume.
True.

Question 5.
The lightness or heaviness of a body is due to volume
(False) Correct statement: The lightness or heaviness of a body is due to density.

Question 6.
Neptune is 30 AU away from sun.
True.

Question 7.
The nearest star to our solar system is proxima centauri.
True.

Question 8.
The volume of a figure is the region covered by the boundary of the figure.
(False) Correct statement: The area of a figure is the region covered by the boundary of the figure.

Question 9.
1 Light year = 9.46 x 105 m.
True.

Question 10.
One light year is defined as the distance travelled by light inW vacuum during the period of one year.
True.

IV. Match the following :
Question 1.

 1 Length (a) ampere (A) 2 time (b) kelvin (K) 3 Mass (c) metre (M) 4 Temperature (d) second (S) 5 Electric current (e) kilogram (K)

1. c
2. d
3. e
4. b
5. a

Question 2.

1. c
2. d
3. a
4. b

V. Assertion and Reason.

Mark the correct choice as
(a) Both A and R are true but R is not the correct reason.
(b) Both A and R are true and R is the correct reason.
(c) A is true but R is false.
(d) A is false but R is true

Question 1.
Assertion (A): The distance between two celestial bodies is measured by the unit of light year.
Reason (R) : The distance travelled by the light in one year in vacuum is called one light year.
(a) Both A and R are true but R is not the correct reason.

Question 2.
Assertion (A): It is easier to swim in sea water than in river water.
Reason (R) : Density of sea water is more than that of river water
(a) Both A and R are true but R is not the correct reason.
(b) Both A and R are true and R is the correct reason.
(c) A is true but R is false.
(d) A is false but R is true.
(b) Both A and R are true and R is the correct reason.

Question 1.
Write the SI unit of speed.
m/s

Question 2.
What is the fundamental unit of amount of substance?
mole (mol)

Question 3.
What are the types of physical quantity?

1. Fundamental quantity
2. Derived quantity.

Question 4.
What is the SI unit of electric charge?
Coulomb (C)

Question 5.
Mention the formula to calculate area of a circle?
n × r2 = πr2.

Question 6.
How do you find the area of irregularly shaped figures?
Graphical method.

Question 7.
How will you determine the volume of a liquid?
By using measuring cylinder.

Question 8.
What are the other units used to measure the volume of liquids?
Gallon, ounce and quart.

Question 9.
Which one of the following has more volume. Iron block or a wooden block of same mass.
Wooden block.

Question 10.
Which one of the following has more density. Water or cooking oil.
Water

Question 1.
What is fundamental quantity? Give examples.
A set of physical quantities which cannot be expressed in terms of any other quantities are known as fundamental quantities. Ex: Length, mass, time.

Question 2.
Define mass Mention its unit.
Mass is the amount of matter contained in a body. It’s unit is kilogram (kg).

Question 3.
What are the multiples and sub multiples of mass?
The multiples of mass are quintal and metric tonne.
The sub-multiples of mass are gram and milligrams.

Question 4.
What is physical quantity? give example.
A quantity that can be measured is called a physical quantity. For example, the length of a piece of cloth, the time at which school begins.

Question 5.
What do you mean by ‘unit’?
The known measure of a physical quantity is called the unit of measurement.

Question 6.
What is measurement?
Comparison of an unknown quantity with a standard quantity is called measurement.

Question 7.
What is meant by area?
Area is the measure of the region inside a closed line.

Question 8.
What is capacity of a container?
The volume of liquid which a container can hold is called its capacity.

Question 9.
What is the relation between density, volume and mass?

Question 10.
Define astronomical unit.
One astronomical unit is defined as the average distance between the earth and the sun. 1AU = 1.496 x 1011 m or 149.6 x 106 m

Question 1.
How will you find the volume of an irregularly shaped object (stone) by using measuring cylinder?

1. Take a measuring cylinder and pour some water into it.
2. Note down the volume of water from the readings of the measuring cylinder.
3. Take it as V1
4. Now take Q small stone and tie it with a thread.
5. Immerse the stone inside the water by holding the thread.
6. This has to be done such that the stone does not touch the walls of the measuring cylinder.
7. Now the level of water has raised.
8. Note down the volume of water and take it to be V2

The volume of the stone is equal to the raise in the volume of water.
V1 = 30cc, V2 = 40cc
Volume of stone = V2 – Vj = 40 – 30 = 10cc

IX. Problems for practice:

Question 1.
A piece of iron weighs 230 g and has a volume of 20 cm3. Find the density of iron.
Solution:
Mass of iron (m) = 230 g
Volume of iron (v) = 20 cm3

Question 2.
Find the mass of silver of volume 50 cm3 and density 10.5 g / cm3
Solution:
Mass of silver (M) = ?
Volume of silver (V) =50 cm3
Density of silver D = 10.5 g/cm3

mass (M) = Density × Volume
= 10.5 x 50 = 525 g

X. Creative questions: HOTS

Question 1.
Why does an iron needle sink in water, but not an iron ship?
Iron needle is compact and its density is 7.6 g/cm3 Thus, as the density of iron needle is more than 1 g/cm3 therefore, it sinks in water. However, the iron ship is constructed in such a way that it is mostly hollow from within, thus, the volume of iron ship becomes very large as compared to its mass and hence its density is less than lg/cm3 . As the density of iron ship is less than 1g/cm3, therefore it floats in water.

## Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.1

Students can Download Maths Chapter 4 Geometry Ex 4.1 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 4 Geometry Ex 4.1

Question 1.
Can 30°, 60° and 90° be the angles of a triangle?
Solution:
Given angles 30°, 60° and 90°
Sum of the angles = 30° + 60° + 90° = 180°
∴ The given angles form a triangle.

Question 2.
Can you draw a triangle with 25°, 65° and 80° as angles?
Solution:
Given angle 25°, 65° and 80°.
Sum of the angles = 25° + 65° + 80° = 170° ≠ 180
∴ We cannot draw a triangle with these measures.

Find the value of x in each of the given triangles.

Question 3.
In each of the following triangles, find the value of x.

Solution:
(i) Let ∠G = x
By angle sum property we know that,
∠E + ∠F + ∠G = 180°
80° + 55° + x = 180°
135° + x = 180°
x = 45°

(ii) Let ∠M = x
By angle sum property of triangles we have
∠M + ∠M + ∠O = 180°
x + 96° + 22° = 180°
x + 118° = 180°
X = 180° – 118° = 620

(iii) Let ∠Z = (2x + 1)° and ∠Y = 90°
By the sum property of triangles we have
∠x + ∠y + ∠z = 180°
29° + 90° + (2x + 1)° = 180°
119° + (2x + 1)° = 180°
(2x + 1)° = 180° – 119°
2x + 1° = 61°
2x = 61° – 1°
2x = 60°
x = $$\frac{60^{\circ}}{2}$$
x = 30°

(iv) Let ∠J = x and ∠L – 3x.
By angle sum property of triangles we have
∠J + ∠K + ∠L = 180°
x + 112° + 3x = 180°
4x = 180° – 112°
x = 68°
x = $$\frac{68^{\circ}}{4}$$
x = 17°

(v) Let ∠S = 3x°
Given $$\overline{\mathrm{RS}}$$ = Given $$\overline{\mathrm{RT}}$$ = 4.5 cm
Given ∠S = ∠T = 3x° [∵ Angles opposite to equal sides are equal]
By angle sum property of a triangle we have,
∠R + ∠S + ∠T = 180°
72° + 3x + 3x = 180°
72° + 6x = 180°
x = $$\frac{108^{\circ}}{6}$$
x = 18°

(vi) Given ∠X = 3x; ∠Y = 2x; ∠Z = ∠4x
By angle sum property of a triangle we have
∠X + ∠Y + ∠Z = 180°
3x + 2x + 4x = 180°
∴ 9x = 180°
x = $$\frac{180^{\circ}}{9}$$ = 20°

(vii) Given ∠T = (x – 4)°
∠U = 90°
∠V = (3x – 2)°
By angle sum property of a triang we have
∠T + ∠U + ∠V = 180°
(x – 4)° + 90° + (3x – 2)° = 180°
x – 4° + 90° + 3x – 2° = 180°
x + 3x + 90° – 4° – 2° = 180°
4x + 84° = 180°
4x = 180° – 84°
4x = 96°
x = $$\frac{96^{\circ}}{4}$$ = 24°
x = 24°

(viii) Given ∠N = (x + 31)°
∠O = (3x – 10)°
∠P = (2x – 3)°
By angle sum property of a triangle we have
∠N + ∠O + ∠P = O
(x + 31)° + (3x – 10)° + (2x – 3)° = 180°
x + 31°+ 3x – 10° + 2x – 3° = 180°
x + 3x + 2x + 31° – 10° – 3° = 180°
6x + 18° = 180°
6x = 180° + 18°
6x = 162°
x = $$\frac{162^{\circ}}{6}$$ = 27°
x = 27°

Question 4.
Two line segments $$\overline{A D}$$ and $$\overline{B C}$$ intersect at O. Joining $$\overline{A B}$$ and $$\overline{D C}$$ we get two triangles, ∆AOB and ∆DOC as shown in the figure. Find the ∠A and ∠B.

Solution:
In ∆AOB and ∆DOC,
∠AOB = ∠DOC [∵ Vertically opposite angles are equal]
Let ∠AOB = ∠DOC = y
By angle sum property of a triangle we have
∠A + ∠B + ∠AOB = ∠D + ∠C + ∠DOC = 180°
3x + 2x + y = 70° + 30° + y = 180°
5x + y = 100° + y = 180°
Here 5x + y = 100° + y
5x = 100° + y – y
5x = 100°
x = $$\frac{100^{\circ}}{5}$$ = 20°
∠A = 3x = 3 × 20 = 60°
∠B = 2x = 2 × 20 = 40°
∠A = 60°
∠B = 40°

Question 5.
Observe the figure and find the value of
∠A + ∠N + ∠G + ∠L + ∠E + ∠S.

Solution:
In the figure we have two triangles namely ∆AGE and ∆NLS.
By angle sum property of triangles,
Sum of angles of ∆AGE = ∠A + ∠G + ∠E = 180° …(1)
Also sum of angles of ∆NLS = ∠N + ∠L + ∠S = 180° … (2)
(1) + (2) ∠A + ∠G + ∠E + ∠N + ∠L + ∠S = 180° + 180°
i.e., ∠A + ∠N + ∠G + ∠L + ∠E + ∠S = 360°

Question 6.
If the three angles of a triangle are in the ratio 3 : 5 : 4, then find them.
Solution:
Given three angles of the triangles are in the ratio 3 : 5 : 4.
Let the three angle be 3x, 5x and 4x.
By angle sum property of a triangle, we have
3x + 5x + 4x = 180°
12x = 180°
x = $$\frac{180^{\circ}}{12}$$
x = 15°
∴ The angle are 3x = 3 × 15° = 45°
5x = 5 × 15° = 75°
4x = 4 × 15° = 60°
Three angles of the triangle are 45°, 75°, 60°

Question 7.
In ∆RST, ∠S is 10° greater than ∠R and ∠T is 5° less than ∠S , find the three angles of the triangle.
Solution:
In ∆RST. Let ∠R = x.
Then given S is ∠10° greater than ∠R
∴ ∠S = x + 10°
Also given ∠T is 5° less then ∠S.
So ∠T = ∠S – 5° = (x + 10)° – 5° = x + 10° – 5°
By angle sum property of triangles, sum of three angles = 180°.

∠R + ∠S + ∠T = 180°
x + x + 10° + x + 5° = 180°
3x + 15° = 180°
3x = 180° – 15°
x = $$\frac{165^{\circ}}{3}$$ = 55°
∠R = x = 55°
∠S = x + 10° = 55° + 10° = 65°
∠T = x + 5° = 55° + 5° = 60°
∴ ∠R = 55°
∠S = 65°
∠T = 60°

Question 8.
In ∆ABC , if ∠B is 3 times ∠A and ∠C is 2 times ∠A, then find the angles.
Solution:
In ABC, Let ∠A = x,
then ∠B = 3 times ∠A = 3x
∠C = 2 times ∠A = 2x
By angle sum property of a triangles,
Sum of three angles of ∆ABC =180°.
∠A + ∠B + ∠C = 180
x + 3x + 2x = 180°
x (1 + 3 + 2) = 180°
6x = 180°

x = $$\frac{180^{\circ}}{6}$$ = 30°
∠A = x = 30°
∠B = 3x = 3 × 30° = 90°
∠C = 2x = 2 × 30° = 60°
∴ ∠A = 30°
∠B = 90°
∠C = 60°

Question 9.
In ∆XYZ, if ∠X : ∠Z is 5 : 4 and ∠Y = 72°. Find ∠X and ∠Z.
Solution:
Given in ∆XYZ, ∠X : ∠Z = 5 : 4
Let ∠X = 5x; and ∠Z = 4x given ∠Y = 72°
By the angle sum property of triangles sum of three angles of a triangles is 180°.
∠X + ∠Y + ∠Z = 180°
5x + 72 + 4x = 180°
5x + 4x = 180° – 72°
9x = 108°
x = $$\frac{108^{\circ}}{9}$$ = 12°
∠X = 5x = 5 × 12° = 60°
∠Z = 4x = 4 × 12° = 48°
∴ ∠X = 60°
∠Z = 48°

Question 10.
In a right angled triangle ABC, ∠B is right angle, ∠A is x + 1 and ∠C is 2x + 5. Find ∠A and ∠C.
Solution:
Given in ∆ABC ∠B = 90°
∠A = x + 1
∠B = 2x + 5

By angle sum property of triangles
Sum of three angles of ∆ABC = 180°
∠A + ∠B + ∠C = 180°
(x + 1) + 90° + (2x + 5) = 180°
x + 2x + 1° + 90° + 5° = 180°
3x + 96° = 180°
3x = 180° – 96° = 84°
x = $$\frac{84^{\circ}}{3}$$ = 28°
∠A = x + 1 = 28 + 1 = 29
∠C = 2x + 5 = 2 (28) + 5 = 56 + 5 = 61
∴ ∠A = 29°
∠C = 61°

Question 11.
In a right angled triangle MNO, ∠N = 90°, MO is extended to P. If ∠NOP = 128°, find the other two angles of ∆MNO.
Solution:
Given ∠N = 90°
MO is extended to P, the exterior angle ∠NOP = 128°
Exterior angle is equal to the sum of interior opposite angles.

∴ ∠M + ∠N = 128°
∠M + 90° = 128°
∠M = 128° – 90°
∠M = 38°
By angle sum property of triangles,
∴ ∠M + ∠N + ∠O = 180°
38° + 90° + ∠O = 180°
∠O = 180° – 128°
∠O = 52°
∴ ∠M = 38° and ∠O = 52°

Question 12.
Find the value of x in each of the given triangles.

Solution:
(i) In ∆ABC, given B = 65°,
AC is extended to L, the exterior angle at C, ∠BCL = 135°
Exterior angle is equal to the sum of opposite interior angles.
∠A + ∠B = ∠BCL
∠A + 65° = 135°
∠A = 135° – 65°
∴ ∠A = 70°
x + ∠A = 180° [∵ linear pair]
x + 70° = 180° [∵ ∠A = 70°]
x = 180° – 70°
∴ x = 110°

(ii) In ∆ABC, given B = 3x – 8°
∠XAZ = ∠BAC [∵ vertically opposite angles]
8x + 7 + ∠BAC
i.e., In ∆ABC, ∠A = 8x + 7
Exterior angle ∠XCY = 120°
Exterior angle is equal to the sum of the interior opposite angles.
∠A + ∠B = 120°
8x + 7 + 3x – 8 = 120°
8x + 3x = 120° + 8 – 7
11x = 121°
x = $$\frac{121^{\circ}}{11}$$ = 11°

Question 13.
In ∆LMN, MN is extended to O. If ∠MLN = 100 – x, ∠LMN = 2x and ∠LNO = 6x – 5, find the value of x.
Solution:
Exterior angle is equal to the sum of the opposite interior angles.
∠LNO = ∠MLN + ∠LMN

6x – 5 = 100° – x + 2x
6x – 5 + x – 2x = 100°
6x + x – 2x = 100° + 5°
5x = 105°
x = $$\frac{105^{\circ}}{5}$$ = 21°
x = 21°

Question 14.
Using the given figure find the value of x.

Solution:
In ∆EDC, side DE is extended to B, to form the exterior angle ∠CEB = x.
We know that the exterior angle is equal to the sum of the opposite interior angles
∠CEB = ∠CDE + ∠ECD
x = 50° + 60°
x = 110°

Question 15.
Using the diagram find the value of x.

Solution:
Given triangle is an equilateral triangle as the three sides are equal. For an equilateral triangle all three angles are equal and is equal to 60° Also exterior angle is equal to sum of opposite interior angles.
x = 60° + 60°.
x = 120°

Objective Type Questions

Question 16.
The angles of a triangle are in the ratio 2:3:4. Then the angles are
(i) 20,30,40
(ii) 40, 60, 80
(iii) 80, 20, 80
(iv) 10, 15, 20
(ii) 40, 60, 80

Question 17.
One of the angles of a triangle is 65°. If the difference of the other two angles is 45°, then the two angles are
(i) 85°, 40°
(ii) 70°, 25°
(iii) 80°, 35°
(iv) 80° , 135°
(iii) 80°,35°

Question 18.
In the given figure, AB is parallel to CD. Then the value of b is

(i) 112°
(ii) 68°
(iii) 102°
(iv) 62° A
(ii) 68°

Question 19.
In the given figure, which of the following statement is true?

(i) x + y + z = 180°
(ii) x + y + z = a + b + c
(iii) x + y + z = 2(a + b + c)
(iv) x + y + z = 3(a + b + c)
Ans :
(iii) x + y + z = 2(a + b + c)]

Question 20.
An exterior angle of a triangle is 70° and two interior opposite angles are equal. Then measure of each of these angle will be
(i) 110°
(ii) 120°
(iii) 35°
(iv) 60°
(iii) 35°

Question 21.
In a ∆ABC, AB = AC. The value of x is _____.

(i) 80°
(ii) 100°
(iii) 130°
(iv) 120°