Practice Set 1 Class 7 Answers Chapter 1 Geometrical Constructions Maharashtra Board

Geometrical Constructions Class 7 Maths Chapter 1 Practice Set 1 Solutions Maharashtra Board

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

Std 7 Maths Practice Set 1 Solutions Answers

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.
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 1
Line AB is the perpendicular bisector of seg PQ.

ii.
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 2
Line UV is the perpendicular bisector of seg ST.

iii.
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 3
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°
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 4

ii. 55°
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 5

iii. 90°
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 6

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:
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 7
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:
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 8
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:
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 9

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

7th Std Maths Solution Maharashtra Board

Practice Set 31 Class 7 Answers Chapter 7 Joint Bar Graph. Maharashtra Board

Joint Bar Graph Class 7 Maths Chapter 7 Practice Set 31 Solutions Maharashtra Board

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

Std 7 Maths Practice Set 31 Solutions Answers

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 SaplingAlmondKaranjNeemAshokGulmohar
Nutan Vidyalaya4060721542
Bharat Vidyalaya4238602540

Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 7 Joint Bar Graph Practice Set 31 1

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\FruitsSweet LimeOrangeApplePineapple
Saturday43305640
Sunday59657867

Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 7 Joint Bar Graph Practice Set 31 2

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.12345
Men200270560820850
Women700240340640470

Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 7 Joint Bar Graph Practice Set 31 3

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

City\TemperatureDelhiMumbaiKolkataNagpurKapurthala
Maximum temperature3532374137
Minimum temperature2625262926

Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 7 Joint Bar Graph Practice Set 31 4

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\VaccineD.P.T. (Booster)Polio (Booster)MeaslesHepatitis
Solapur65606563
Pune89878886

Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 7 Joint Bar Graph Practice Set 31 5

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\Year19711981199120012011
Maharashtra4657657783
Gujarat4045616979

Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 7 Joint Bar Graph Practice Set 31 6

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?

Maharashtra Board Class 7 Maths Solutions Chapter 7 Joint Bar Graph Practice Set 31 7

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)
Maharashtra Board Class 7 Maths Solutions Chapter 7 Joint Bar Graph Practice Set 31 8

  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
Maharashtra Board Class 7 Maths Solutions Chapter 7 Joint Bar Graph Practice Set 31 9
ii. Line graph
Maharashtra Board Class 7 Maths Solutions Chapter 7 Joint Bar Graph Practice Set 31 10
iii. Pie chart
Maharashtra Board Class 7 Maths Solutions Chapter 7 Joint Bar Graph Practice Set 31 11
Solution:
(Students should attempt the above activities on their own.)

Class 7 Maths Solution Maharashtra Board

Practice Set 30 Class 7 Answers Chapter 6 Indices Maharashtra Board

Indices Class 7 Maths Chapter 6 Practice Set 30 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 30 Answers Solutions Chapter 6 Indices.

Std 7 Maths Practice Set 30 Solutions Answers

Question 1.
Find the square root:
i. 625
ii. 1225
iii. 289
iv. 4096
v. 1089
Solution:
i. 625
Maharashtra Board Class 7 Maths Solutions Chapter 6 Indices Practice Set 30 1
∴ 625 = 5 x 5 x 5 x 5
∴ √625 = 5 x 5 = 25

ii. 1225
Maharashtra Board Class 7 Maths Solutions Chapter 6 Indices Practice Set 30 2
∴ 1225 = 5 x 5 x 7 x 7
∴ √1225 = 5 x 7 = 35

iii. 289
Maharashtra Board Class 7 Maths Solutions Chapter 6 Indices Practice Set 30 3
∴ 289 = 17 x 17
∴ √289 = 17

iv. 4096
Maharashtra Board Class 7 Maths Solutions Chapter 6 Indices Practice Set 30 4
∴ 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
Maharashtra Board Class 7 Maths Solutions Chapter 6 Indices Practice Set 30 5
∴ 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)
Maharashtra Board Class 7 Maths Solutions Chapter 6 Indices Practice Set 30 6
Solution:
1 x 10100

Class 7 Maths Solution Maharashtra Board

Practice Set 29 Class 7 Answers Chapter 6 Indices Maharashtra Board

Indices Class 7 Maths Chapter 6 Practice Set 29 Solutions Maharashtra Board

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

Std 7 Maths Practice Set 29 Solutions Answers

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
= 34×(-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
= 65×4
= 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. \(\frac{1}{y^{4}}\)

Class 7 Maths Solution Maharashtra Board

Practice Set 28 Class 7 Answers Chapter 6 Indices Maharashtra Board

Indices Class 7 Maths Chapter 6 Practice Set 28 Solutions Maharashtra Board

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

Std 7 Maths Practice Set 28 Solutions Answers

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

ii. m5 ÷ m8
= m5-8
= m-3

iii. p3 ÷ p13
= p3-13
= p-10

iv. x10 ÷ x10
= x10-10
= 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
= 75-3
= 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
= 47-5
= 42
= 16

Class 7 Maths Solution Maharashtra Board

Practice Set 27 Class 7 Answers Chapter 6 Indices Maharashtra Board

Indices Class 7 Maths Chapter 6 Practice Set 27 Solutions Maharashtra Board

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

Std 7 Maths Practice Set 27 Solutions Answers

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

Class 7 Maths Solution Maharashtra Board

Practice Set 26 Class 7 Answers Chapter 6 Indices Maharashtra Board

Indices Class 7 Maths Chapter 6 Practice Set 26 Solutions Maharashtra Board

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

Std 7 Maths Practice Set 26 Solutions Answers

Question 1.
Complete the table below:

Sr. No.Indices (Numbers in index form)BaseIndexMultiplication formValue
i.34343 x 3 x 3 x 381
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)BaseIndexMultiplication formValue
i.34343 x 3 x 3 x 381
ii.163 16316 x 16 x 164096
iii.(-8)²(-8)2-8 x -864

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 -134(-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. \(\left(-\frac{1}{2}\right)^{4}\)
\(=\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right)=\frac{1}{16}\)

Class 7 Maths Solution Maharashtra Board

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

Operations on Rational Numbers Class 7 Maths Chapter 5 Practice Set 25 Solutions Maharashtra Board

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

Std 7 Maths Practice Set 25 Solutions Answers

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.

Class 7 Maths Solution Maharashtra Board

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

Operations on Rational Numbers Class 7 Maths Chapter 5 Practice Set 24 Solutions Maharashtra Board

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

Std 7 Maths Practice Set 24 Solutions Answers

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. \(\frac { 13 }{ 4 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 1

ii. \(\frac { -7 }{ 8 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 2

iii. \(7\frac { 3 }{ 5 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 3

iv. \(\frac { 5 }{ 12 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 4

v. \(\frac { 22 }{ 7 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 5

vi. \(\frac { 4 }{ 3 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 6

vii. \(\frac { 7 }{ 9 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 7

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.

Class 7 Maths Solution Maharashtra Board

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

Operations on Rational Numbers Class 7 Maths Chapter 5 Practice Set 23 Solutions Maharashtra Board

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

Std 7 Maths Practice Set 23 Solutions Answers

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.

Class 7 Maths Solution Maharashtra Board