Class 7

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 1 Answers Solutions Chapter 1 Geometrical Constructions.

Geometrical Constructions Class 7 Practice Set 1 Answers Solutions Chapter 1

Question 1.
Draw line segments of the lengths given below and draw their perpendicular bisectors:
i. 5.3 cm
ii. 6.7 cm
iii. 3.8 cm
Solution:
i.

Line AB is the perpendicular bisector of seg PQ.

ii.

Line UV is the perpendicular bisector of seg ST.

iii.

Line ST is the perpendicular bisector of seg LM.

Question 2.
Draw angles of the measures given below and draw their bisectors:
i. 105°
ii. 55°
iii. 90°
Solution:
i. 105°

ii. 55°

iii. 90°

Question 3.
Draw, an obtuse-angled triangle and a right-angled triangle. Find the points of concurrence of the angle bisectors of each triangle. Where do the points of concurrence lie?
Solution:

The points of concurrence of the angle bisectors of both the triangles lie in the interior of the triangles.

Question 4.
Draw a right-angled triangle. Draw the perpendicular bisectors of its sides. Where does the point of concurrence lie?
Solution:

The point of concurrence of the perpendicular bisectors of the sides of the right angled triangle lies on the hypotenuse.

Question 5.
Maithili, Shaila and Ajay live in three different places in the city. A toy shop is equidistant from the three houses. Which geometrical construction should be used to represent this? Explain your answer.
Solution:
Since, Maithili, Shaila and Ajay live in three different places, lines joining their houses will form a triangle.
The position of the toy shop which is equidistant from three houses can be found out by drawing the perpendicular bisector of the sides of the triangle joining the three houses.
The shop will be at the point of concurrence of the perpendicular bisectors.

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 1 Intext Questions and Activities

Question 1.
Draw a line segment PS of length 4cm and draw its perpendicular bisector. (Textbook pg. no. 1)

1. How will your verify that CD is the perpendicular bisector? m∠CMS = __°
2. Is l(PM) = l(SM)?

Solution:

1. Here, m∠CMS = 90°
2. Also, l(PM) = l(SM) = 2cm
   ∴ line CD is the perpendicular bisector of seg PS.

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 30 Answers Solutions Chapter 6 Indices.

Indices Class 7 Practice Set 30 Answers Solutions Chapter 6

Question 1.
Find the square root:
i. 625
ii. 1225
iii. 289
iv. 4096
v. 1089
Solution:
i. 625

∴ 625 = 5 x 5 x 5 x 5
∴ √625 = 5 x 5 = 25

ii. 1225

∴ 1225 = 5 x 5 x 7 x 7
∴ √1225 = 5 x 7 = 35

iii. 289

∴ 289 = 17 x 17
∴ √289 = 17

iv. 4096

∴ 4096 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
∴ √4096 = 2 x 2 x 2 x 2 x 2 x 2
= 64

v. 1089

∴ 1089 = 3 x 3 x 11 x 11
∴ √1089 = 3 x 11
= 33

Maharashtra Board Class 7 Maths Chapter 6 Indices Practice Set 30 Intext Questions and Activities

Question 1.
Try to write the following numbers in the standard form. (Textbook pg. no. 48)
i. The diameter of Sun is 1400000000 m.
ii. The velocity of light is 300000000 m/sec.
Solution:
i. 1400000000 m = 1.4 x 10⁹ m
ii. 300000000 m/s = 3.0 x 10⁸ m/sec.

Question 2.
The box alongside shows the number called Googol. Try to write it as a power of 10. (Textbook pg. no. 48)

Solution:
1 x 10¹⁰⁰

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 29 Answers Solutions Chapter 6 Indices.

Indices Class 7 Practice Set 29 Answers Solutions Chapter 6

Question 1.
Simplify:
i. \(\left[\left(\frac{15}{12}\right)^{3}\right]^{4}\)
ii. (3⁴)^-2
iii. \(\left[\left(\frac{1}{7}\right)^{-3}\right]^{4}\)
iv. \(\left[\left(\frac{2}{5}\right)^{-2}\right]^{-3}\)
v. (6⁵)⁴
vi. \(\left[\left(\frac{6}{7}\right)^{5}\right]^{2}\)
vii. \(\left[\left(\frac{2}{3}\right)^{-4}\right]^{5}\)
viii. \(\left[\left(\frac{5}{8}\right)^{3}\right]^{-2}\)
ix. \(\left[\left(\frac{3}{4}\right)^{6}\right]^{7}\)
x. \(\left[\left(\frac{2}{5}\right)^{-3}\right]^{2}\)
Solution:
i. \(\left[\left(\frac{15}{12}\right)^{3}\right]^{4}\)
\(=\left(\frac{15}{12}\right)^{3 \times 4}=\left(\frac{15}{12}\right)^{12}\)

ii. (3⁴)^-2
= 3^4×(-2)
= 3^-8

iii. \(\left[\left(\frac{1}{7}\right)^{-3}\right]^{4}\)
\(=\left(\frac{1}{7}\right)^{(-3) \times 4}=\left(\frac{1}{7}\right)^{-12}\)

iv. \(\left[\left(\frac{2}{5}\right)^{-2}\right]^{-3}\)
\(=\left(\frac{2}{5}\right)^{(-2) \times(-3)}=\left(\frac{2}{5}\right)^{6}\)

v. (6⁵)⁴
= 6^5×4
= 6²⁰

vi. \(\left[\left(\frac{6}{7}\right)^{5}\right]^{2}\)
\(=\left(\frac{6}{7}\right)^{5 \times 2}=\left(\frac{6}{7}\right)^{10}\)

vii. \(\left[\left(\frac{2}{3}\right)^{-4}\right]^{5}\)
\(=\left(\frac{2}{3}\right)^{(-4) \times 5}=\left(\frac{2}{3}\right)^{-20}\)

viii. \(\left[\left(\frac{5}{8}\right)^{3}\right]^{-2}\)
\(=\left(\frac{5}{8}\right)^{3 \times(-2)}=\left(\frac{5}{8}\right)^{-6}\)

ix. \(\left[\left(\frac{3}{4}\right)^{6}\right]^{7}\)
\(=\left(\frac{3}{4}\right)^{6 \times 1}=\left(\frac{3}{4}\right)^{6}\)

x. \(\left[\left(\frac{2}{5}\right)^{-3}\right]^{2}\)
\(=\left(\frac{2}{5}\right)^{(-3) \times 2}=\left(\frac{2}{5}\right)^{-6}\)

Question 2.
Write the following numbers using positive indices:
i. \(\left(\frac{2}{7}\right)^{-2}\)
ii. \(\left(\frac{11}{3}\right)^{-5}\)
iii. \(\left(\frac{1}{6}\right)^{-3}\)
iv. \((y)^{-4}\)
Solution:
i. \(\left(\frac{7}{2}\right)^{2}\)
ii. \(\left(\frac{3}{11}\right)^{5}\)
iii. \(6^{3}\)
iv. \(\frac{1}{y^{4}}\)

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 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 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.