Category: Class 7

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 12 Answers Solutions Chapter 3 HCF and LCM.

HCF and LCM Class 7 Practice Set 12 Answers Solutions Chapter 3

Question 1.
i. 25, 40
ii. 56, 32
iii. 40, 60, 75
iv. 16, 27
v. 18, 32,48
vi. 105, 154
vii. 42, 45, 48
viii. 57, 75, 102
ix. 56, 57
x. 777, 315, 588
Solution:
i. 25, 40

∴ 25 = 5 × 5
40 = 2 × 2 × 2 × 5
∴ HCF of 25 and 40 = 5

ii. 56, 32

∴ 56 = 2 × 2 × 2 × 7
32 = 2 × 2 × 2 × 2 × 2
∴ HCF of 56 and 32 = 2 × 2 × 2
∴ HCF of 56 and 32 = 8

iii. 40, 60, 75

∴ 40 = 2 × 2 × 2 × 5
60 = 2 × 2 × 3 × 5
75 = 3 × 5 × 5
∴ HCF of 40, 60 and 75 = 5

iv. 16, 27

∴ 16 = 2 × 2 × 2 × 2 × 1
27 = 3 × 3 × 3 × 1
∴ HCF of 16 and 27 = 1

v. 18, 32,48

∴ 18 = 2 × 3 × 3
32 = 2 × 2 × 2 × 2 × 2
48 = 2 × 2 × 2 × 2 × 3
∴ HCF of 18, 32 and 48 = 2

vi. 105, 154

∴ 105 = 3 × 5 × 7
154 = 2 × 2 × 11
∴ HCF of 105 and 154 = 7

vii. 42, 45, 48

∴ 42 = 2 × 3 × 7
45 = 3 × 3 × 5
48 = 2 × 2 × 2 × 2 × 3
∴ HCF of 42,45 and 48 =3

viii. 57, 75, 102

∴ 57 = 3 × 19
75 = 3 × 5 × 5
102 = 2 × 3 × 17
∴ HCF of 57, 75 and 102 = 3

ix. 56, 57

∴ 56 = 2 × 2 × 2 × 7 × 1
57 = 3 × 19 × 1
∴ HCF of 56 and 57 = 1

x. 777, 315, 588

∴ 777 = 3 × 7 × 37
315 = 3 × 3 × 5 × 7
588 = 2 × 2 × 3 × 7 × 7
∴ HCF of 777, 315 and 588 = 3 × 7
HCF of 777, 315 and 588 = 21

Question 2.
Find the HCF by the division method and reduce to the simplest form:
i. \(\frac { 275 }{ 525 }\)
ii. \(\frac { 76 }{ 133 }\)
iii. \(\frac { 161 }{ 69 }\)
Solution:
i. \(\frac { 275 }{ 525 }\)

ii. \(\frac { 76 }{ 133 }\)

iii. \(\frac { 161 }{ 69 }\)

Maharashtra Board Class 7 Maths Chapter 3 HCF and LCM Practice Set 12 Intext Questions and Activities

Question 1.
In each of the following examples, write all the factors of the numbers and find the greatest common divisor. (Textbook pg. no. 17)
i. 28, 42
ii. 51, 27
iii. 25, 15, 35
Solution:
i. Factors of 28 = 1,2,4, 7, 14, 28
Factors of 42 = 1,2, 3, 6, 7, 14, 21, 42
∴ HCF of 28 and 42 = 14

ii. Factors of 51 = 1, 3, 17, 51
Factors of 27 = 1, 3, 9, 27
∴ HCF of 51 and 27 = 3

iii. Factors of 25 = 1, 5, 25
Factors of 15 = 1, 3, 5, 15
Factors of 35 = 1, 5, 7, 35
∴ HCF of 25, 15 and 35 = 5

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 31 Answers Solutions Chapter 7 Joint Bar Graph.

Joint Bar Graph Class 7 Practice Set 31 Answers Solutions Chapter 7

Question 1.
The number of saplings planted by schools on World Tree Day is given in the table below. Draw a joint bar graph to show these figures.

School Name\Name of Sapling	Almond	Karanj	Neem	Ashok	Gulmohar
Nutan Vidyalaya	40	60	72	15	42
Bharat Vidyalaya	42	38	60	25	40

Solution:

Question 2.
The table below shows the number of people who had the different juices at a juice bar on a Saturday and a Sunday. Draw a joint bar graph for this data.

Days\Fruits	Sweet Lime	Orange	Apple	Pineapple
Saturday	43	30	56	40
Sunday	59	65	78	67

Solution:

Question 3.
The following numbers of votes were cast at 5 polling booths during the Gram Panchayat elections. Draw a joint bar graph for this data.

Persons\Booth No.	1	2	3	4	5
Men	200	270	560	820	850
Women	700	240	340	640	470

Solution:

Question 4.
The maximum and minimum temperatures of five Indian cities are given in °C. Draw a joint bar graph for this data.

City\Temperature	Delhi	Mumbai	Kolkata	Nagpur	Kapurthala
Maximum temperature	35	32	37	41	37
Minimum temperature	26	25	26	29	26

Solution:

Question 5.
The numbers of children vaccinated in one day at the government hospitals in Solapur and Pune are given in the table. Draw a joint bar graph for this data:

City\Vaccine	D.P.T. (Booster)	Polio (Booster)	Measles	Hepatitis
Solapur	65	60	65	63
Pune	89	87	88	86

Solution:

Question 6.
The percentage of literate people in the states of Maharashtra and Gujarat are given below. Draw a joint bar graph for this data.

State\Year	1971	1981	1991	2001	2011
Maharashtra	46	57	65	77	83
Gujarat	40	45	61	69	79

Solution:

Maharashtra Board Class 7 Maths Chapter 7 Joint Bar Graph Practice Set 31 Intext Questions and Activities

Question 1.
Observe the graph shown below and answer the following questions. (Textbook pg. no. 51)

1. In which year did Ajay and Vijay both produce equal quantities of wheat?
2. In year 2014, who produced more wheat?
3. In year 2013, how much wheat did Ajay and Vijay each produce?

Solution:

1. Both produced equal quantities of wheat in the year 2011.
2. Ajay produced more wheat in the year 2014.
3. Ajay’s wheat production in 2013 = 40 quintal.
   Vijay’s wheat production in 2013 = 30 quintal.

Question 2.
The minimum and maximum temperature in Pune for five days is given. Read the joint bar graph and answer the questions below: (Textbook pg. no. 52)

1. What data is shown on X- axis?
2. What data is shown on Y- axis?
3. Which day had the highest temperature?
4. On which day is the minimum temperature the highest?
5. On Thursday, what is the difference between the minimum and maximum temperature?
6. On which day is the difference between the minimum and maximum temperature the greatest?

Solution:

1. Five days of a week are shown on X – axis.
2. Temperature in the city of Pune is shown on Y – axis.
3. Monday had the highest temperature.
4. The minimum temperature was highest on Wednesday.
5. Maximum temperature = 29.5° C
   Minimum temperature = 15° C
   ∴ Difference in temperature = 29.5° C – 15° C = 14.5 ° C
6. The difference in minimum and maximum temperature is greatest on Thursday.

Question 3.
Collect various kinds of graphs from newspapers and discuss them. (Textbook pg. no. 53)
i. Histogram

ii. Line graph

iii. Pie chart

Solution:
(Students should attempt the above activities on their own.)

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 25 Answers Solutions Chapter 5 Operations on Rational Numbers.

Operations on Rational Numbers Class 7 Practice Set 25 Answers Solutions Chapter 5

Question 1.
Simplify the following expressions.
i. 50 x 5 ÷ 2 + 24
ii. (13 x 4) ÷ 2 – 26
iii. 140 ÷ [(-11) x (-3) – (-42) ÷ 14 – 1)]
iv. {(220 – 140) + [10 x 9 + (-2 x 5) ]} – 100
v. \(\frac{3}{5}+\frac{3}{8} \div \frac{6}{4}\)
Solution:
i. 50 x 5 ÷ 2 + 24 = 250 ÷ 2 + 24
= 125 + 24
= 149

ii. (13 x 4) = 2 – 26
= 52 ÷ 2 – 26
= 26 – 26
= 0

iii. 140 ÷ [(-11) x (-3) – (-42) ÷ 14 – 1)]
= 140 ÷ [33 + 42 ÷ 14 – 1]
= 140 ÷ [33 + 3 – 1]
= 140 ÷ 35
= 4

iv. {(220 – 140) + [10 x 9 + (-2 x 5) ]} – 100
= {80 + [90 – 10]} – 100
= {80 + 80} – 100
= 160 – 100
= 60

v. \(\frac{3}{5}+\frac{3}{8} \div \frac{6}{4}\)
\(=\frac{3}{5}+\frac{3}{8} \times \frac{4}{6}\)
\(=\frac{3}{5}+\frac{1}{4}\)
\(=\frac{12}{20}+\frac{5}{20}=\frac{12+5}{20}=\frac{17}{20}\)

Maharashtra Board Class 7 Maths Chapter 5 Operations on Rational Numbers Practice Set 25 Intext Questions and Activities

Question 1.
Use the signs and numbers in the boxes and form an expression such that its value will be 112. (Textbook pg. no. 42)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
[+ x ÷ -]
Solution:
{3 + (6 x 7) + (9 ÷ 3)} + {- 8 + 8 x 9}
Note: The above problem has many solutions. Students may write solution other than the one given.

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 18 Answers Solutions Chapter 4 Angles and Pairs of Angles.

Angles and Pairs of Angles Class 7 Practice Set 18 Answers Solutions Chapter 4

Question 1.
Name the pairs of opposite rays in the figure alongside.

Solution:

1. Ray PL and ray PM
2. Ray PN and ray PT

Question 2.
Are the ray PM and PT opposite rays? Give reasons for your answer.

Solution:
No.
Ray PM and Ray PT do not form a straight line and hence are not opposite rays.

Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 18 Intext Questions and Activities

Question 1.
Observe the adjacent figure and answer the following questions. (Textbook pg. no. 28)

1. Name the rays in the figure alongside.
2. Name the origin of the rays
3. Name the angle in the given figure

Solution:

1. Ray BA and ray BC
2. Point B
3. ∠ABC or ∠CBA

Question 2.
Observe the adjacent figure and answer the following questions. (Textbook pg. no. 28)

1. Name the angle in the figure alongside.
2. Name the rays whose origin is point B

Solution:

1. ∠ABC or ∠CBA
2. Ray BA and ray BC

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 28 Answers Solutions Chapter 6 Indices.

Indices Class 7 Practice Set 28 Answers Solutions Chapter 6

Question 1.
Simplify:
i. a⁶ ÷ a⁴
ii. m⁵ ÷ m⁸
iii. p³ ÷ p¹³
iv. x¹⁰ ÷ x¹⁰
Solution:
i. a⁶ ÷ a⁴
= a^6-4
= a²

ii. m⁵ ÷ m⁸
= m^5-8
= m^-3

iii. p³ ÷ p¹³
= p^3-13
= p^-10

iv. x¹⁰ ÷ x¹⁰
= x^10-10
= x⁰
= 1

Question 2.
Find the value of:
i. (-7)¹² ÷ (-7)¹²
ii. 7⁵ ÷ 7³
iii. \(\left(\frac{4}{5}\right)^{3} \div\left(\frac{4}{5}\right)^{2}\)
iv. 4⁷ ÷ 4⁵
Solution:
i. (-7)¹² ÷ (-7)¹²
= (-7)^12-12
= (-7)⁰
= 1

ii. 7⁵ ÷ 7³
= 7^5-3
= 7²
= 49

iii. \(\left(\frac{4}{5}\right)^{3} \div\left(\frac{4}{5}\right)^{2}\)
\(=\left(\frac{4}{5}\right)^{3-2}=\frac{4}{5}\)

iv. 4 ÷ 4
= 4^7-5
= 4²
= 16

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 27 Answers Solutions Chapter 6 Indices.

Indices Class 7 Practice Set 27 Answers Solutions Chapter 6

Question 1.
Simplify:
i. 7⁴ × 7²
ii. (-11)⁵ × (-11)²
iii. \(\left(\frac{6}{7}\right)^{3} \times\left(\frac{6}{7}\right)^{5}\)
iv. \(\left(-\frac{3}{2}\right)^{5} \times\left(-\frac{3}{2}\right)^{3}\)
v. a¹⁶ × a⁷
vi. \(\left(\frac{\mathrm{P}}{5}\right)^{3} \times\left(\frac{\mathrm{P}}{5}\right)^{7}\)
Solution:
i. 7⁴ × 7²
= 7⁴⁺²
= 7⁶

ii. (-11)⁵ × (-11)²
= (-11)⁵⁺²
= (-11)⁷

iii. \(\left(\frac{6}{7}\right)^{3} \times\left(\frac{6}{7}\right)^{5}\)
\(=\left(\frac{6}{7}\right)^{3+5}=\left(\frac{6}{7}\right)^{8}\)

iv. \(\left(-\frac{3}{2}\right)^{5} \times\left(-\frac{3}{2}\right)^{3}\)
\(=\left(-\frac{3}{2}\right)^{5+3}=\left(-\frac{3}{2}\right)^{8}\)

v. a¹⁶ × a⁷
= a¹⁶⁺⁷
= a²³

vi. \(\left(\frac{\mathrm{P}}{5}\right)^{3} \times\left(\frac{\mathrm{P}}{5}\right)^{7}\)
\(=\left(\frac{\mathrm{P}}{5}\right)^{3+7}=\left(\frac{\mathrm{P}}{5}\right)^{10}\)

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 26 Answers Solutions Chapter 6 Indices.

Indices Class 7 Practice Set 26 Answers Solutions Chapter 6

Question 1.
Complete the table below:

Sr. No.	Indices (Numbers in index form)	Base	Index	Multiplication form	Value
i.	34	3	4	3 x 3 x 3 x 3	81
ii.	163
iii.		(-8)	2
iv.				\(\frac{3}{7} \times \frac{3}{7} \times \frac{3}{7} \times \frac{3}{7}\)	\(\frac { 81 }{ 2401 }\)
v.	(-13)4

Solution:

Sr. No.	Indices (Numbers in index form)	Base	Index	Multiplication form	Value
i.	34	3	4	3 x 3 x 3 x 3	81
ii.	163	16	3	16 x 16 x 16	4096
iii.	(-8)²	(-8)	2	-8 x -8	64
iv.	\(\left(\frac{3}{7}\right)^{4}\)	\(\frac { 7 }{ 7 }\)	4	\(\frac{3}{7} \times \frac{3}{7} \times \frac{3}{7} \times \frac{3}{7}\)	\(\frac { 81 }{ 2401 }\)
v.	(-13)4	-13	4	(-13) x (-13) x (-13) x (-13)	28561

Question 2.
Find the value of.
i. 2¹⁰
ii. 5³
iii. (-7)⁴
iv. (-6)³
v. 9³
vi. 8¹
vii. \(\left(\frac{4}{5}\right)^{3}\)
viii. \(\left(-\frac{1}{2}\right)^{4}\)
Solution:
i. 2¹⁰
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 1024

ii. 5³
= 5 × 5 × 5
= 125

iii. (-7)⁴
= (-7) × (-7) × (-7) × (-7)
= 2401

iv. (-6)³
= (-6) × (-6) × (-6)
= -216

v. 9³
= 9 × 9 × 9
= 729

vi. 8¹
= 8

vii. \(\left(\frac{4}{5}\right)^{3}\)
\(=\frac{4}{5} \times \frac{4}{5} \times \frac{4}{5}=\frac{64}{125}\)

viii. \(\left(-\frac{1}{2}\right)^{4}\)
\(=\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right)=\frac{1}{16}\)

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 13 Answers Solutions Chapter 3 HCF and LCM.

HCF and LCM Class 7 Practice Set 13 Answers Solutions Chapter 3

Question 1.
Find the LCM:
i. 12, 15
ii. 6, 8, 10
iii. 18, 32
iv. 10, 15, 20
v. 45, 86
vi. 15, 30, 90
vii. 105, 195
viii. 12,15,45
ix. 63,81
x. 18, 36, 27
Solution:
i. 12, 15

∴ LCM of 12 and 15 = 3 x 2 x 2 x 5
= 60

ii. 6, 8, 10

∴ LCM of 6, 8 and 10 = 2 x 2 x 3 x 2 x 5
= 120

iii. 18, 32

∴ LCM of 18 and 32 = 2 x 2 x 2 x 2 x 3 x 3 x 2
= 288

iv. 10, 15, 20

∴ LCM of 10, 15 and 20 = 5 x 2 x 3 x 2
= 60

v. 45, 86

∴ LCM of 45 and 86 = 2 x 3 x 3 x 5 x 43
= 3870

vi. 15, 30, 90

∴ LCM of 15,30 and 90 = 3 x 5 x 2 x 3
= 90

vii. 105, 195

∴ LCM of 105 and 195 = 5 x 3 x 7 x 13
= 1365

viii. 12, 15, 45

∴ LCM of 12, 15 and 45 = 3 x 3 x 2 x 5 x 2
= 180

ix. 63, 81

∴ LCM of 63 and 81 = 3 x 3 x 3 x 7 x 3
= 567

x. 18, 36, 27

∴ LCM of 18, 36 and 27 = 3 x 3 x 2 x 2 x 3
= 108

Question 2.
Find the HCF and LCM of the numbers given below. Verify that their product is equal to the product of the given numbers:
i. 32, 37
ii. 46, 51
iii. 15, 60
iv. 18, 63
v. 78, 104
Solution:
i. 32 = 2 x 16
= 2 x 2 x 8
= 2 x 2 x 2 x 4
= 2 x 2 x 2 x 2 x 2 x 1
37 = 37 x 1
∴ HCF of 32 and 37 =1
LCM of 32 and 37 = 2 x 2 x 2 x 2 x 2 x 37
= 1184
HCF x LCM = 1 x 1184
= 1184
Product of the given numbers = 32 x 37
= 1184
∴ HCF x LCM = Product of the given numbers.

ii. 46 = 2 x 23 x 1
51 = 3 x 17 x 1
∴ HCF of 46 and 51 = 1
LCM of 46 and 51 = 2 x 23 x 3 x 17
= 2346
HCF x LCM = 1 x 2346
= 2346
Product of the given numbers = 46 x 51
= 2346
∴ HCF x LCM = Product of the given numbers.

iii. 15 = 3 x 5
60 = 2 x 30
= 2 x 2 x 15
= 2 x 2 x 3 x 5
∴ HCF of 15 and 60 = 3 x 5
= 15
LCM of 15 and 60 = 3 x 5 x 2 x 2
= 60
HCF x LCM = 15 x 60
= 900
Product of the given numbers = 15 x 60
= 900
∴ HCF x LCM = Product of the given numbers.

iv. 18 = 2 x 9
= 2 x 3 x 3
63 = 3 x 21
= 3 x 3 x 7
∴ HCF of 18 and 63 = 3 x 3
= 9
LCM of 18 and 63 = 3 x 3 x 2 x 7
= 126
HCF x LCM = 9 x 126
= 1134
Product of the given numbers = 18 x 63
= 1134
∴ HCF x LCM = Product of the given numbers.

v. 78 = 2 x 39
= 2 x 3 x 13
104 = 2 x 52
= 2 x 2 x 26
= 2 x 2 x 2 x 13
∴ HCF of 78 and 104 = 2 x 13
= 26
LCM of 78 and 104 = 2 x 13 x 3 x 2 x 2
= 312
HCF x LCM = 26 x 312
= 8112
Product of the given numbers = 78 x 104
= 8112
∴ HCF x LCM = Product of the given numbers.

Maharashtra Board Class 7 Maths Chapter 3 HCF and LCM Practice Set 13 Intext Questions and Activities

Question 1.
Write the tables of the given numbers and find their LCM. (Textbook pg. no. 19)
i. 6, 7
ii. 8, 12
iii. 5, 6, 15
Solution:
i. Multiples of 6 : 6, 12, 18, 24, 30, 36, 42
Multiples of 7 : 7, 14, 21, 28, 35, 42, 49
∴ LCM of 6 and 7 = 42

ii. Multiples of 8 : 8, 16, 24, 32, 40
Multiples of 12 : 12, 24, 36, 48
∴ LCM of 8 and 12 = 24

iii. Multiples of 5 : 5, 10, 15, 20, 25, 30, 35
Multiples of 6 : 6, 12, 18, 24, 30, 36
Multiples of 15 : 15, 30, 45, 60
∴ LCM of 5,6 and 15 = 30

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 24 Answers Solutions Chapter 5 Operations on Rational Numbers.

Operations on Rational Numbers Class 7 Practice Set 24 Answers Solutions Chapter 5

Question 1.
Write the following rational numbers in decimal form.
i. \(\frac { 13 }{ 4 }\)
ii. \(\frac { -7 }{ 8 }\)
iii. \(7\frac { 3 }{ 5 }\)
iv. \(\frac { 5 }{ 12 }\)
v. \(\frac { 22 }{ 7 }\)
vi. \(\frac { 4 }{ 3 }\)
vii. \(\frac { 7 }{ 9 }\)
Solution:
i. \(\frac { 13 }{ 4 }\)

ii. \(\frac { -7 }{ 8 }\)

iii. \(7\frac { 3 }{ 5 }\)

iv. \(\frac { 5 }{ 12 }\)

v. \(\frac { 22 }{ 7 }\)

vi. \(\frac { 4 }{ 3 }\)

vii. \(\frac { 7 }{ 9 }\)

Maharashtra Board Class 7 Maths Chapter 5 Operations on Rational Numbers Practice Set 24 Intext Questions and Activities

Question 1.
Without using division, can we tell from the denominator of a fraction, whether the decimal form of the fraction will be a terminating decimal? Find out. (Textbook pg. no. 40)
Solution:
If the prime factorization of the denominator of a fraction has only factors as 2 or 5 or a combination of 2 and 5 then the decimal form of that fractional will be a terminating decimal form.
Consider the fractions \(\frac { 17 }{ 20 }\) and \(\frac { 19 }{ 6 }\)
Now, 20 = 2 x 2 x 5, and 6 = 2 x 3
∴ \(\frac { 17 }{ 20 }\) is terminating decimal form while \(\frac { 19 }{ 6 }\) is recurring decimal form.

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 23 Answers Solutions Chapter 5 Operations on Rational Numbers.

Operations on Rational Numbers Class 7 Practice Set 23 Answers Solutions Chapter 5

Question 1.
Write three rational numbers that lie between the two given numbers.
i. \(\frac{2}{7}, \frac{6}{7}\)
ii. \(\frac{4}{5}, \frac{2}{3}\)
iii. \(-\frac{2}{3}, \frac{4}{5}\)
iv. \(\frac{7}{9},-\frac{5}{9}\)
v. \(\frac{-3}{4}, \frac{+5}{4}\)
vi. \(\frac{7}{8}, \frac{-5}{3}\)
vii. \(\frac{5}{7}, \frac{11}{7}\)
viii. \(0, \frac{-3}{4}\)
Solution:
i. \(\frac{2}{7}, \frac{6}{7}\)
The three numbers lying between \(\frac { 2 }{ 7 }\) and \(\frac { 6 }{ 7 }\) are \(\frac{3}{7}, \frac{4}{7}, \frac{5}{7}\)

ii. \(\frac{4}{5}, \frac{2}{3}\)
\(\frac{4}{5}=\frac{24}{30}, \frac{2}{3}=\frac{20}{30}\)
The three numbers between \(\frac { 4 }{ 5 }\) and \(\frac { 2 }{ 3 }\) are \(\frac{21}{30}, \frac{22}{30}, \frac{23}{30}\)

iii. \(-\frac{2}{3}, \frac{4}{5}\)
\(\frac{-2}{3}=\frac{-10}{15}, \frac{4}{5}=\frac{12}{15}\)
The three numbers between \(\frac { -2 }{ 3 }\) and \(\frac { 4 }{ 5 }\) are \(\frac{-9}{15}, \frac{-7}{15}, \frac{4}{15}\)

iv. \(\frac{7}{9},-\frac{5}{9}\)
The three numbers between \(\frac { 7 }{ 9 }\) and \(\frac { -5 }{ 9 }\) are \(\frac{6}{9}, 0, \frac{-4}{9}\)

v. \(\frac{-3}{4}, \frac{+5}{4}\)
The three numbers between \(\frac { -3 }{ 4 }\) and \(\frac { +5 }{ 4 }\) are \(\frac{-2}{4}, \frac{-1}{4}, \frac{3}{4}\)

vi. \(\frac{7}{8}, \frac{-5}{3}\)
\(\frac{7}{8}=\frac{21}{24}, \frac{-5}{3}=\frac{-40}{24}\)
The three numbers between \(\frac { 7 }{ 8 }\) and \(\frac { -5 }{ 3 }\) are \(\frac{17}{24}, \frac{11}{24}, \frac{-13}{24}\)

vii. \(\frac{5}{7}, \frac{11}{7}\)
The three numbers between \(\frac { 5 }{ 7 }\) and \(\frac { 11 }{ 7 }\) are \(\frac{6}{7}, \frac{8}{7}, \frac{9}{7}\)

viii. \(0, \frac{-3}{4}\)
The three numbers between 0 and \(\frac { -3 }{ 4 }\) are \(\frac{-1}{8}, \frac{-2}{8}, \frac{-5}{8}\)

Maharashtra Board Class 7 Maths Chapter 5 Operations on Rational Numbers Practice Set 23 Intext Questions and Activities

Question 1.
Answer the following questions: (Textbook pg. no. 36)

1. Write all the natural numbers between 2 and 9.
2. Write all the integers between -4, and 5.
3. Which rational numbers are there between \(\frac { 1 }{ 2 }\) and \(\frac { 3 }{ 4 }\) ?

Solution:

1. 3, 4, 5, 6, 7, 8
2. -3, -2, -1, 0, 1, 2, 3, 4
3. \(\frac{1}{2}=\frac{1 \times 2}{2 \times 2}=\frac{2}{4}=\frac{2 \times 10}{4 \times 10}=\frac{20}{40}\)
   \(\frac{3}{4}=\frac{3 \times 10}{4 \times 10}=\frac{30}{40}\)
   ∴ The rational numbers between \(\frac { 1 }{ 2 }\) and \(\frac { 3 }{ 4 }\) are \(\frac{21}{40}, \frac{22}{40}, \frac{25}{40}, \frac{27}{40}\) etc.