Author name: Prasanna

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.1

Solve the following equations.

Question 1.
x – 2 = 7
Solution:
x – 2 = 7
Transposing (-2) to R.H.S., we get
x = 7 + 2
∴ x = 9

Question 2.
y + 3 = 10
Solution:
y + 3 = 10
Transposing 3 to R.H.S., we get
y = 10 – 3
∴ y = 7

Question 3.
6 = z + 2
Solution:
6 = z + 2
Transposing 2 to L.H.S., we get
6 – 2 = z
4 = z
∴ z = 4

Question 4.
\(\frac{3}{7}+x=\frac{17}{7}\)
Solution:
\(\frac{3}{7}+x=\frac{17}{7}\)
Transposing \(\frac{3}{7}\) to R.H.S., we get
x = \(\frac{17}{7}-\frac{3}{7}=\frac{17-3}{7}=\frac{14}{7}=2\)
∴ x = 2

Question 5.
6x = 12
Solution:
6x = 12
Divided by 6 on both sides, we get
\(\frac{6 x}{6}=\frac{12}{6}\)
∴ x = 2

Question 6.
\(\frac{t}{5}=10\)
Solution:
\(\frac{\mathrm{t}}{5}\) = 10
Multiplying both sides by 5, we get
\(\frac{\mathrm{t}}{5}\) × 5 = 10 × 5
∴ t = 50

Question 7.
\(\frac{2 x}{3}=18\)
Solution:
\(\frac{2 x}{3}=18\)
Multiplying both sides, by 3, we get
\(\frac{2 x}{3}\) × 3 = 18 × 3
2x = 18 × 3
x = \(\frac{18 \times 3}{2}\) = 9 × 3 = 27
∴ x = 27

Question 8.
1.6 = \(\frac{\mathrm{y}}{1.5}\)
Solution:
1.6 = \(\frac{\mathrm{y}}{1.5}\)
Multiplying both sides by 1.5, we get
1.6 × 1.5 = \(\frac{\mathrm{y}}{1.5}\) × 1.5
2.4 = y
∴ y = 2.4

Question 9.
7x – 9 = 16
Solution:
7x – 9 = 16
Transposing (-9) to R.H.S., we get
7x = 16 + 9
Dividing both sides by 7, we get
∴ x = \(\frac{25}{7}\)

Question 10.
14y – 8 = 13
Solution:
14y – 8 = 13
Transposing (-8) to R.H.S., we get
14y = 13 + 8
14y = 21
Dividing both sides by 14 we get,
\(\frac{14 y}{14}\) = \(\frac{21}{14}\)
∴ y = \(\frac{3}{2}\)

Question 11.
17 + 6p = 9
Solution:
17 + 6P = 9
Transposing 17 to RHS, we get
6P = 9 – 17
6P = -8
Dividing both sides by 6, we have
\(\frac{6 P}{6}=\frac{-8}{6}\)
∴ P = \(\frac{-4}{3}\)

Question 12.
\(\frac{x}{3}+1=\frac{7}{15}\)
Solution:
\(\frac{x}{3}+1=\frac{7}{15}\)
Transposing 1 to R.H.S, we get
\(\frac{x}{3}=\frac{7}{15}-1\)
\(\frac{x}{3}=\frac{7-15}{15}\)
\(\frac{x}{3}=\frac{-8}{15}\)
Multiplying both sides by 3, we have
\(\frac{x}{3} \times 3=\frac{-8}{15} \times 3\)
∴ x = \(\frac{-8}{5}\)

These NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals InText Questions

NCERT In-text Question Page No. 34
Question 1.
Find:
(a) \(\frac { 1 }{ 2 }\) × 3
(b) \(\frac { 9 }{ 7 }\) × 6
(c) 3 × \(\frac { 1 }{ 8 }\)
(d) \(\frac { 13 }{ 11 }\) × 6
if the product is an improper fraction express it as a mixed fraction
Answer:

Question 2.
Represent pictorially: 2 × \(\frac{2}{5}=\frac{4}{5}\)
Answer:

NCERT In-text Question Page No. 34
Question 1.
(i) 5 × 2 \(\frac{3}{7}\)
(ii) 1 \(\frac{4}{9}\) × 6
Answer:

NCERT In-text Question Page No. 35
Question 1.
Can you tell, what is (i) \(\frac{1}{2}\) of 10?
(i) \(\frac{1}{4}\) of 16? (iii) (i) \(\frac{2}{5}\) of 25?
Answer:

NCERT In-text Question Page No. 39
Question 1.
Fill in these boxes:

NCERT In-text Question Page No. 40
Question 1.
Find:
\(\frac{1}{3} \times \frac{4}{5} ; \frac{2}{3} \times \frac{1}{5}\)
Answer:

Question 2.
Find:
\(\frac{8}{3} \times \frac{4}{7} \times \frac{3}{4} \times \frac{2}{3}\)
Answer:

NCERT In-text Question Page No. 44
Question 1.
(i) Will the reciprocal of a proper fraction be again a proper fraction?
(ii) Will the reciprocal of an improper fraction be again an improper fraction?
Answer:
(i) No, the reciprocal of an improper fraction is a improper fraction.
(ii) No, the reciprocal of an improper fraction is a proper fraction.
Now, we can say that
(a) 1 ÷ \(\frac { 1 }{ 2 }\) = 1 × \(\frac { 2 }{ 1 }\) = 1 × reciprocal of \(\frac { 1 }{ 2 }\)

(b) 3 ÷ \(\frac { 1 }{ 4 }\) = 3 × \(\frac { 4 }{ 1 }\) = 3 × reciprocal of \(\frac { 1 }{ 4 }\)

(c) 3 ÷ \(\frac { 1 }{ 2 }\) = ………….. = …………….. 3 ÷ \(\frac { 1 }{ 2 }\) = 3 × \(\frac { 2 }{ 1 }\) = 3 × reciprocal of \(\frac { 1 }{ 2 }\)
And, 2 ÷ \(\frac { 3 }{ 4 }\) = 2 × reciprocal of \(\frac { 3 }{ 4 }\) = 2 × \(\frac { 4 }{ 3 }\)

(d) 5 ÷ \(\frac { 2 }{ 9 }\) = 5 × …………… = 5 × ……………..

∴ 5 ÷ \(\frac { 2 }{ 9 }\) = 5 × \(\frac { 9 }{ 2 }\) = 5 × reciprocal of \(\frac { 2 }{ 9 }\)

NCERT In-text Question Page No. 45
Question 1.
Find:
(i) 7 ÷ \(\frac { 2 }{ 5 }\)
(ii) 7 ÷ \(\frac { 4 }{ 7 }\)
(iii) 7 ÷ \(\frac { 8 }{ 9 }\)
Answer:

NCERT In-text Question Page No. 45
Question 1.
Find:
(i) 6 ÷ 5\(\frac { 1 }{ 3 }\)
(ii) 7 ÷ 2\(\frac { 4 }{ 7 }\)
Answer:

NCERT In-text Question Page No. 45
Question 1.
Find:
(i) \(\frac{3}{5} \div \frac{1}{2}\)
(ii) \(\frac{1}{2} \div \frac{3}{5}\)
(iii) 2\(\frac{1}{3} \div \frac{3}{5}\)
(iv) 5\frac{1}{6} \div \frac{9}{2}
Answer:

NCERT In-text Question Page No. 50
Question 1.
Find:
(i) 2.7 × 4
(ii) 1.8 × 1.2
(iii) 2.3 × 4.35
Answer:
(i) 27 × 4 = 108 and there is one digit to the right of the decimal point in 27 2.7 × 4 = 10.8
(ii) 18 × 12 = 216 and number of digits to the right of decimal point is (1 + 1) 2 = 2
1.8 × 1.2 = 2.16

(iii) 23 × 435 = 10005 and there are 1 + 2 = 3 digits to the right of decimal point
2.3 × 4.35 = 10.005

Question 2.
Arrange the products obtained in Question in descending order.
Answer:
The products are 10.8, 2.16, 10.005. Comparing 10.8 and 10.005, we have :
10 = 10, 8 > 0, i.e. 10.005 < 10.8
Here, the smallest number = 2.16
and, the largest number = 10.8
Thus the required descending order is : 10.8, 10.005,2.16.

NCERT In-text Question Page No. 51
Question 1.
Find:
(i) 0.3 × 10
(ii) 1.2 × 100
(iii) 56.3 × 1000
Answer:
(i) 0.3 × 10
There is one zero in 10
∵ The decimal point is shifted to the right by one place
Thus, 0.3 × 10 = 3

(ii) There are 2 zeroes in 100
∵ The decimal point is shifted to the right by 2 places
= 1.20 × 100= 120
Thus, 1.2 × 100 = 120

(iii) There are 3 zeros in 1000
∵ The decimal point is shifted to the right by 3 places
Thus, 56.3 × 1000 = 56.300 × 1000 = 56300

NCERT In-text Question Page No. 53
Question 1.
Find:
(i) 235.4 ÷ 10
(ii) 235.4 ÷ 100
(iii) 235.4 ÷ 1000
Answer:
(i) 235.4 ÷ 10
Since, there is one zero in 10.
∴ The decimal point in the quotient is shifted to the left by one place.
∴ 235.4 ÷ 10 = 23.54

(ii) 235.4 ÷ 100
Since, there are two zeros in 100
∴ The decimal point in the quotient is
shifted to the left by three places.
∴ 235.4 H- 1000 = 0.2354

NCERT In-text Question Page No. 53
Question 1.
Find:
(i) 35.7 ÷ 3 = ?
(ii) 25.5 ÷ 3 = ?
Answer:
(i) 35.7 ÷ 3
Since, \(\frac{357}{3}\) = 119 and there is one digit in the decimal part of the given decimal number.
∴ The decimal point is placed in the quotient after one digit from the right most digit.
∴ 35.7 ÷ 3 = 11.9

(ii) 25.5 ÷ 3
Since, 255 ÷ 3 = 85 and there is one digit in the decimal part of the given decimal number.
∴ The decimal is placed in the quotient after one digit from the right most digit.
∴ 25.5 ÷ 3 = 8.5

NCERT In-text Question Page No. 53

Question 1.
(i) 43.15 ÷ 5 = ?
(ii) 82.44 ÷ 6 = ?
Answer:

NCERT In-text Question Page No. 53
Question 1.
Find:
(i) 15.5 ÷ 5
(ii) 126.35 ÷ 7
Answer:
(i) 15.5 ÷ 5
Since 155 ÷ 5 = 31 and there is one digit in the decimal part of the given decimal number.
∴ Place the decimal point in 31 such that there is one digit to its right.
∴ 15.5 + 5 = 3.1

(ii) 126.37 + 7
Since 12635 + 7 = 1805 and there are two digits in the decimal part of the given decimal number.
∴ Place the decimal point in 1805 such that there are two digits to its right.
∴ 126.35 ÷ 7 = 18.05

NCERT In-text Question Page No. 54
Question 1.
Find:
(i) \(\frac{7.75}{0.25}\)
(ii) \(\frac{42.8}{0.02}\)
(iii) \(\frac{5.6}{1.4}\)
Answer:

These NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers InText Questions

Try These (Page No. 4)

Question 1.
Fill in the blanks in the following table.

Solution:
Using the closure property over addition, subtraction, multiplication, and division for rational numbers, integers, whole numbers, and natural numbers, we have:

Try These (Page No. 6)

Question 2.
Complete the following table:

Solution:

Try These (Page No. 9)

Question 3.
Complete the following table:

Solution:

Try These (Page No. 13)

Question 4.
Find using distributivity.

Solution:

Try These (Page No. 17)

Question 5.
Write the rational number for each point labelled with a letter.

Solution:
(i) Here, the rational number for the point A is \(\frac{1}{5}\).
The rational number for the point B is \(\frac{4}{5}\).
The rational number for the point C is \(\frac{5}{5}\).
The rational number for the point D is \(\frac{8}{5}\).
The rational number for the point E is \(\frac{9}{5}\).

(ii) The rational number for:
The point F is \(\frac{-2}{6}\) or \(\frac{-1}{3}\).
The point G is \(\frac{-5}{6}\).
The point H is \(\frac{-7}{6}\).
The point I is \(\frac{-8}{6}\) or \(\frac{-4}{3}\).
The point J is \(\frac{-11}{6}\).

These NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Exercise 1.2

Question 1.
Represent these numbers on the number line:
(i) \(\frac{7}{4}\)
(ii) \(\frac{-5}{6}\)
Solution:
(i) \(\frac{7}{4}\)
We make 7 markings at distance of \(\frac{1}{4}\) each on the right of 0 and starting from 0, the seventh marking represents \(\frac{7}{4}\).

(ii) \(\frac{-5}{6}\)
We make 5 markings at distance of \(\frac{1}{6}\) each on the left of ‘0’ and starting from ‘0’, the fifth marking represents \(\frac{-5}{6}\)

Question 2.
Represent \(\frac{-2}{11}, \frac{-5}{11}, \frac{-9}{11}\) on a number line.
Solution:
We make 9 markings at distance of \(\frac{1}{11}\) each on the left of ‘0’ and starting from ‘0’, the second marking represents \(\frac{-2}{11}\); the fifth marking represents \(\frac{-5}{11}\) and the ninth marking represents \(\frac{-9}{11}\).
The point A represents \(\frac{-2}{11}\), the point B represents \(\frac{-5}{11}\) and the point C represents \(\frac{-9}{11}\).

Question 3.
Write five rational numbers which are smaller than 2.
Solution:
There can be infinite rational numbers smaller than 2.
Five rational numbers are 0, -1, \(\frac{1}{2}\), \(\frac{-1}{2}\), 1

Question 4.
Find ten rational numbers between \(\frac{-2}{5}\) and \(\frac{1}{2}\)
Solution:
Convert \(\frac{-2}{5}\) and \(\frac{1}{2}\) with the same denominators.
\(\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\)
\(\frac{-2}{5}=\frac{2 \times 2}{5 \times 2}=\frac{-4}{10}\)
To get ten rational numbers, multiply both numerator and denominator by 2
\(\frac{5}{10}=\frac{5 \times 2}{10 \times 2}=\frac{10}{20}\)
\(\frac{-4}{10}=\frac{-4 \times 2}{10 \times 2}=\frac{-8}{20}\)
The rational numbers between \(\frac{10}{20}\) and \(\frac{-8}{20}\) are
\(\frac{9}{20}, \frac{8}{20}, \frac{7}{20}, \frac{6}{20}, \frac{5}{20}, \frac{4}{20}, \frac{3}{20}, \ldots \frac{-6}{20}, \frac{-7}{20}\)
We can take any 10 of them.

Question 5.
Find five rational numbers between
(i) \(\frac{2}{3}\) and \(\frac{4}{5}\)
(ii) \(\frac{-3}{2}\) and \(\frac{5}{3}\)
(iii) \(\frac{1}{4}\) and \(\frac{1}{2}\)
Solution:
Convert \(\frac{2}{3}\) and \(\frac{4}{5}\) with the same denominators
\(\frac{2}{3}=\frac{2 \times 5}{3 \times 5}=\frac{10}{15}\)
\(\frac{4}{5}=\frac{4 \times 3}{5 \times 3}=\frac{12}{15}\)
The difference between the numerators should be more than 5
\(\frac{10}{15}=\frac{10 \times 4}{15 \times 4}=\frac{40}{60}\)
\(\frac{12}{15}=\frac{12 \times 4}{15 \times 4}=\frac{48}{60}\)
Five rational numbers between \(\frac{40}{60}\) and \(\frac{48}{60}\) are
\(\frac{41}{60}, \frac{42}{60}, \frac{43}{60}, \frac{44}{60}, \frac{45}{60}\)

(ii) Convert \(\frac{-3}{2}\) and \(\frac{5}{3}\) with the same denominators.
\(\frac{-3}{2}=\frac{-3 \times 3}{2 \times 3}=\frac{-9}{6}\)
\(\frac{5}{3}=\frac{5 \times 2}{3 \times 2}=\frac{10}{6}\)
Five rational numbers between \(\frac{-9}{6}\) and \(\frac{10}{6}\) are \(\frac{9}{6}, \frac{8}{6}, \frac{7}{6}, 0, \frac{-7}{6}\)

(iii) Convert \(\frac{1}{4}\) and \(\frac{1}{2}\) with the same denominators
\(\frac{1}{4}=\frac{1}{4}\)
\(\frac{1}{2}=\frac{1 \times 2}{2 \times 2}=\frac{2}{4}\)
The difference between the numerators should be more than 5
\(\frac{1}{4}=\frac{1 \times 8}{4 \times 8}=\frac{8}{32}\)
\(\frac{2}{4}=\frac{2 \times 8}{4 \times 8}=\frac{16}{32}\)
Five rational numbers between \(\frac{8}{32}\) and \(\frac{16}{32}\) are \(\frac{9}{32}, \frac{10}{32}, \frac{11}{32}, \frac{12}{32}, \frac{13}{32}\)

Question 6.
Write five rational numbers greater than -2.
Solution:
There are infinite rational numbers greater than -2.
The numbers are \(\frac{-3}{2},-1, \frac{-1}{2}, 0, \frac{1}{2}\)

Question 7.
Find ten rational numbers between \(\frac{3}{5}\) and \(\frac{3}{4}\).
Solution:
Convert \(\frac{3}{5}\) and \(\frac{3}{4}\) with the same denominators
\(\frac{3}{5}=\frac{3 \times 4}{5 \times 4}=\frac{12}{20}\)
\(\frac{3}{4}=\frac{3 \times 5}{4 \times 5}=\frac{15}{20}\)
The difference between the numerator should be more than 10.
\(\frac{12}{20}=\frac{12 \times 8}{20 \times 8}=\frac{96}{160}\)
\(\frac{15}{20}=\frac{15 \times 8}{20 \times 8}=\frac{120}{160}\)
Thus, ten rational numbers between