Maharashtra Board Practice Set 28 Class 6 Maths Solutions Chapter 11 Ratio-Proportion

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 11 Ratio-Proportion Class 6 Practice Set 28 Answers Solutions.

6th Standard Maths Practice Set 28 Answers Chapter 11 Ratio-Proportion

6th Standard Maths Practice Set 28 Question 1.
In each example below, find the ratio of the first number to the second:
i. 24, 56
ii. 63,49
iii. 52, 65
iv. 84, 60
v. 35, 65
vi. 121, 99
Solution:
i. 24, 56
\(\frac{24}{56}=\frac{24 \div 8}{56 \div 8}=\frac{3}{7}\)
= 3:7

ii. 63,49
\(\frac{63}{49}=\frac{63 \div 7}{49 \div 7}=\frac{9}{7}\)
= 9:7

iii. 52, 65
\(\frac{52}{65}=\frac{52 \div 13}{65 \div 13}=\frac{4}{5}\)
= 4:5

iv. 84, 60
\(\frac{84}{60}=\frac{84 \div 12}{60 \div 12}=\frac{7}{5}\)
= 7:5

v. 35, 65
\(\frac{35}{65}=\frac{35 \div 5}{65 \div 5}=\frac{7}{13}\)
= 7:13

vi. 121, 99
\(\frac{121}{99}=\frac{121 \div 11}{99 \div 11}=\frac{11}{9}\)
= 11:9

6th Maths Practice Set 28 Question 2.
Find the ratio of the first quantity to the second.
i. 25 beads, 40 beads
ii. Rs 40, Rs 120
iii. 15 minutes, 1 hour
iv. 30 litres, 24 litres
v. 99 kg, 44000 grams
vi. 1 litre, 250 ml
vii. 60 paise, 1 rupee
viii. 750 grams, \(\frac { 1 }{ 2 }\) kg
ix. 125 cm, 1 metre
Solution:
i. Required Ratio = \(\frac{25}{40}=\frac{25 \div 5}{40 \div 5}=\frac{5}{8}\)

ii. Required Ratio = \(\frac{40}{120}=\frac{40 \div 40}{120 \div 40}=\frac{1}{3}\)

iii. 1 hour = 60 minutes
Required Ratio = \(\frac{15}{60}=\frac{15 \div 15}{60 \div 15}=\frac{1}{4}\)

iv. Required Ratio = \(\frac{30}{24}=\frac{30 \div 6}{24 \div 6}=\frac{5}{4}\)

v. 99 kg = 99 x 1000 grams = 99000 grams
Required Ratio = \(\frac{99000}{44000}=\frac{99000 \div 1000}{44000 \div 1000}=\frac{99}{44}\)
= \(\frac{99}{44}=\frac{99 \div 11}{44 \div 11}=\frac{9}{4}\)

vi. 1 litre, 250 ml
1 litre = 1000 ml
Required Ratio = \(\frac{1000}{250}=\frac{1000 \div 10}{250 \div 10}=\frac{100}{25}\)
= \(\frac{100}{25}=\frac{100 \div 25}{25 \div 25}=\frac{4}{1}\)

viii. 750 grams, \(\frac { 1 }{ 2 }\) kg
\(\frac { 1 }{ 2 }\) kg = \(\frac { 1000 }{ 2 }\) grams = 500 grams
Required Ratio = \(\frac{750}{500}=\frac{750 \div 10}{500 \div 10}=\frac{75}{50}\)
= \(\frac{75}{50}=\frac{75 \div 25}{50 \div 25}=\frac{3}{2}\)

ix. 125 cm, 1 metre
1 metre = 100 cm
Required Ratio = \(\frac{125}{100}=\frac{125 \div 25}{100 \div 25}=\frac{5}{4}\)

6th Std Maths Practice Set 28 Question 3.
Reema has 24 notebooks and 18 books. Find the ratio of notebooks to books.
Solution:
Ratio of notebooks to books
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 28 1
∴ The ratio of notebooks to books with Reema is \(\frac { 4 }{ 3 }\)

Practice Set 28 Question 4.
30 cricket players and 20 kho-kho players are training on a field. What is the ratio of cricket players to the total number of players?
Solution:
Total number of players = Cricket players + Kho-kho players
= 30 + 20 = 50
Ratio of cricket players to the total number of players
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 28 2
∴ The ratio of cricket players to the total number of players is \(\frac { 3 }{ 5 }\).

Question 5.
Snehal has a red ribbon that is 80 cm long and a blue ribbon 2.20 m long. What is the ratio of the length of the red ribbon to that of the blue ribbon?
Solution:
1 metre =100 cm
Length of the red ribbon = 80 cm
Length of the blue ribbon = 2.20 m = 2.20 x 100 cm
\(=\frac{220}{100} \times \frac{100}{1}=\frac{220 \times 100}{100 \times 1}\)
= 220 cm
∴ Ratio of length of the red ribbon to that of the blue ribbon
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 28 3
∴ The ratio of the length of the red ribbon to that of the blue ribbon is \(\frac { 4 }{ 11 }\).

11 Ratio Question 6.
Shubham’s age today is 12 years and his father’s is 42 years. Shubham’s mother is younger than his father by 6 years. Find the following ratios.
i. Ratio of Shubham’s age today to his mother’s age today.
ii. Ratio of Shubham’s mother’s age today to his father’s age today.
iii. The ratio of Shubham’s age to his mother’s age when Shubham was 10 years old.
Solution:
Shubham’s age today = 12 years
Shubham’s father’s age = 42 years
Shubham’s mother age = Shubham’s father’s age – 6 years
= 42 years – 6 years = 36 years

i. Ratio of Shubham’s age today to his mother’s age today
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 28 4
∴ The ratio of Shubham’s age today to his mother’s age today is \(\frac { 1 }{ 3 }\).

ii. Ratio of Shubham’s mother age today to his father’s age today
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 28 5
∴ The ratio of Shubham’s mother’s age today to his father’s age today is \(\frac { 6 }{ 7 }\).

iii. Shubham’s age today is 12 years and his mothers age is 36 years.
Hence when Shubham’s age was 10 years, his mother’s age was 34 years (i.e. 36 – 2 years).
Ratio of Shubham’s age to his mother’s age when Shubham was 10 years old
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 28 6
∴ The ratio of Shubham’s age to his mother’s age when Shubham was 10 years old is \(\frac { 5 }{ 17 }\)

Maharashtra Board Class 6 Maths Chapter 11 Ratio-Proportion Practice Set 28 Intext Questions and Activities

Question 1.
In the figure, colour some squares with any colour you like and leave some blank. (Textbook pg. no. 57)
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 28 7
i. Count all the boxes and write the number.
ii. Count the colored ones and write the number.
iii. Count the blank ones and write the number.
iv. Find the ratio of the colored boxes to the blank ones.
v. Find the ratio of the colored boxes to the total boxes.
vi. Find the ratio of the blank boxes to the total boxes.
Solution:
i. The number of all boxes is 16.
ii. The number of colored boxes is 7.
iii. The number of blank boxes is 9.
iv. Ratio of the colored boxes to the blank ones
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 28 8
v. Ratio of the colored boxes to the total boxes
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 28 9
vi. Ratio of the blank boxes to the total boxes
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 28 10

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

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

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

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.

Maharashtra Board Practice Set 5 Class 6 Maths Solutions Chapter 3 Integers

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 3 Integers Class 6 Practice Set 5 Answers Solutions.

6th Standard Maths Practice Set 5 Answers Chapter 3 Integers

Question 1.
Add:

  1. 8 + 6
  2. 9 + (-3)
  3. 5 + (-6)
  4. – 7 + 2
  5. – 8 + 0
  6. – 5 + (-2)

Solution:

1. 8 + 6 = (+8) + (+6) = +14 2. 9 + (-3) = (+9) + (- 3) = +6 3. 5 + (-6) = (+5) + (-6) = -1
4. -7 + 2 = (-7) + (+2) = -5 5. -8 + 0 = (-8) + 0 = -8 6. -5 + (-2) = (-5) + (-2) = -7

Question 2.
Complete the table given below:

+ 8 4 -3 -5
-2 -2 + 8 = +6
6
0
-4

Solution:

+ 8 4 -3 -5
-2 (-2) + (+8) = +6 (-2) +(+4) = 2 (-2) +(-3) =-5 (-2) +(-5) =-7
6 (+6) + (+8) = 14 (+6) + (+4) = 10 (+6) + (-3) = 3 (+6) + (-5) = 1
0 0 + (+8) = 8 0 + (+4) = 4 0 + (-3) = -3 0 + (-5) = -5
-4 (-4) + (+8) = 4 (-4) +(+4) = 0 (-4) + (-3) = -7 (-4) + (-5) = -9

Maharashtra Board Class 6 Maths Chapter 3 Integers Practice Set 5 Intext Questions and Activities

Question 1.
On the playground, mark a timeline showing the years from 2000 to 2024. With one child standing at the position of the 2017, ask the following questions: (Textbook pg. no. 15)

  1. While playing this game, what is his/her age?
  2. Five years ago, which year was it? And what was his / her age then?
  3. In which year will he / she go to Std X? How old will he / she be then?

The child should find answers to such questions by walking the right number of units and in the right direction on the timeline.
[Assume child born year is 2009]
Solutions:

  1. Age of child is 8 years.
  2. Five years ago, year was 2012. His/her age is 3 years.
  3. In 2024, he/she will go the Std X. His/her age is 15 years.

Question 2.
On a playground mark a timeline of 100 years. This will make it possible to count the years from 0 to 2100 on it. Important historical events can then be shown in proper centuries. (Textbook pg. no. 16)
Solution:
(Students should attempt this activity on their own)

Question 3.
Observe the figures and write appropriate number in the boxes given below. (Textbook pg. no. 16 and 17)
i.

Maharashtra Board Class 6 Maths Solutions Chapter 3 Integers Practice Set 5 1
a. At first the rabbit was at the number ____
b. It hopped ___ units to the right.
c. It is now at the number ___
Solution:
i.
a. +1
b. 5
c. +6

ii.
Maharashtra Board Class 6 Maths Solutions Chapter 3 Integers Practice Set 5 2
a. At first the rabbit was at the number ___
b. It hopped ____ units to the right.
c. It is now at the number ____
Solution:
ii.
a. -2
b. 5
c. +3

iii.
Maharashtra Board Class 6 Maths Solutions Chapter 3 Integers Practice Set 5 3
a. At first the rabbit was at the number ___
b. It hopped ____ units to the left.
c. It is now at the number ___
Solution:
iii.
a. -3
b. 4
c. -7

iv.
Maharashtra Board Class 6 Maths Solutions Chapter 3 Integers Practice Set 5 4
a. At first the rabbit was at the number ___
b. It hopped___units to the left.
c. It is now at the number ____
Solution:
iv.
a. +3
b. 4
c. -1

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

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. \(\frac { 13 }{ 4 }\)
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.

Maharashtra Board Practice Set 39 Class 6 Maths Solutions Chapter 17 Geometrical Constructions

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 17 Geometrical Constructions Class 6 Practice Set 39 Answers Solutions.

6th Standard Maths Practice Set 39 Answers Chapter 17 Geometrical Constructions

Question 1.
Draw line l. Take any point P on the line. Using a set square, draw a line perpendicular to line l at the point P.
Solution:
Step 1:
Maharashtra Board Class 6 Maths Solutions Chapter 17 Geometrical Constructions Practice Set 39 1

Step 2:
Maharashtra Board Class 6 Maths Solutions Chapter 17 Geometrical Constructions Practice Set 39 2
line PQ ⊥ line l

Question 2.
Draw a line AB. Using a compass, draw a line perpendicular to AB at point B.
Solution:
Step 1:
Maharashtra Board Class 6 Maths Solutions Chapter 17 Geometrical Constructions Practice Set 39 3

Step 2:
Maharashtra Board Class 6 Maths Solutions Chapter 17 Geometrical Constructions Practice Set 39 4

Step 3:
Maharashtra Board Class 6 Maths Solutions Chapter 17 Geometrical Constructions Practice Set 39 5
line BC ⊥ line AB.

Question 3.
Draw line CD. Take any point M on the line. Using a protractor, draw a line perpendicular to line CD at the point M.
Solution:
Step 1:
Maharashtra Board Class 6 Maths Solutions Chapter 17 Geometrical Constructions Practice Set 39 6

Step 2:
Maharashtra Board Class 6 Maths Solutions Chapter 17 Geometrical Constructions Practice Set 39 7
line MN ⊥ line CD

Maharashtra Board Class 6 Maths Chapter 17 Geometrical Constructions Practice Set 39 Questions and Activities

Question 1.
When constructing a building, what is the method used to make sure that a wall is exactly upright? What does the mason in the picture have in his hand? What do you think is his purpose for using it? (Textbook pg. no. 87)
Maharashtra Board Class 6 Maths Solutions Chapter 17 Geometrical Constructions Practice Set 39 8
Solution:
When constructing a building, a weight (usually with a pointed tip at the bottom) suspended from a string called as plummet or plump bob is aligned from the top of the wall to make sure that the wall is built exactly upright.
The mason in the picture is holding a plumb bob.
The string of the plumb bob is suspended from the top of the wall, such that plumb bob hangs freely. By observing whether the vertical wall is parallel to the string we can check if the constructed wall is vertical.

Question 2.
Have you looked at lamp posts on the roadside? How do they stand? (Textbook pg. no. 87)
Solution:
The lamp posts on the road side are standing erect or vertical.

Question 3.
For the above explained construction, why must we take a distance greater than half of the length of AB? What will happen if we take a smaller distance? (Textbook pg. no. 88)
Solution:
For the above construction, in step-3 we take distance greater than half of the length of AB, so that the arcs drawn by keeping the compass on points A and B intersect each other at point Q.
If the distance in compass is less than half of the length of AB, then the arcs drawn by keeping the compass at A and B will not intersect each other.

Maharashtra Board Practice Set 18 Class 6 Maths Solutions Chapter 6 Bar Graphs

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 6 Bar Graphs Class 6 Practice Set 18 Answers Solutions.

6th Standard Maths Practice Set 18 Answers Chapter 6 Bar Graphs

Question 1.
This bar graph shows the maximum temperatures in degrees Celsius in different cities on a certain day in February. Observe the graph and answer the questions:
Maharashtra Board Class 6 Maths Solutions Chapter 6 Bar Graphs Practice Set 18 1

  1. What data is shown on the vertical and the horizontal lines?
  2. Which city had the highest temperature?
  3. Which cities had equal maximum temperatures?
  4. Which cities had a maximum temperature of 30 °C?
  5. What is the difference between the maximum temperatures of Panchgani and Chandrapur?

Solution:

  1. Temperature is shown on the vertical line and cities are shown on the horizontal line.
  2. The city Chandrapur had the highest temperature.
  3. Pune and Nashik had the equal maximum temperature of 30°C and Panchgani and Matheran had the equal maximum temperature of 25°C.
  4. Pune and Nashik had a maximum temperature of 30 °C.
  5. The difference between the maximum temperatures of Panchgani and Chandrapur can be calculated as Difference in temperature = Temperature of Chandrapur – Temperature of Panchgani
    = 35°C – 25°C
    = 10°C

Maharashtra Board Class 6 Maths Chapter 6 Bar Graphs Practice Set 18 Intext Questions and Activities

Question 1.
Observe the picture alongside: (Textbook pg. no. 35)

  1. To which sport is this data related?
  2. How many things does the picture tell us about?
  3. What shape has been used in the picture to represent runs?

Maharashtra Board Class 6 Maths Solutions Chapter 6 Bar Graphs Practice Set 18 2
Ans:

  1. The given data is related to cricket.
  2. The picture tells about runs scored in different overs by India and Srilanka. The represents the wickets fallen in that over.
  3. To represent runs, rectangular or bar shape is used.

Question 2.
A pictogram of the types and numbers of vehicles in a city is given below.
Taking 1 picture = 5 vehicles, write the numbers in the pictogram. (Textbook pg. no.35)
Maharashtra Board Class 6 Maths Solutions Chapter 6 Bar Graphs Practice Set 18 3
Solution:
Maharashtra Board Class 6 Maths Solutions Chapter 6 Bar Graphs Practice Set 18 4
Drawing pictograms is time consuming.
Sometimes, it is practically not possible to draw pictures for the given values (for example population of villages etc). In such cases, representing the data by making use of graphs can serve the purpose. Such data can be represented by using graphs.

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

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.

Maharashtra Board Practice Set 4 Class 6 Maths Solutions Chapter 3 Integers

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 3 Integers Class 6 Practice Set 4 Answers Solutions.

6th Standard Maths Practice Set 4 Answers Chapter 3 Integers

Question 1.
Classify the following numbers as positive numbers and negative numbers.
-5, +4, -2, 7, +26, -49, -37, 19, -25, +8, 5, -4, -12, 27
Solution:

Positive Numbers +4, 7, +26, 19, +8, 5, 27
Negative Numbers -5, -2, -49, -37, -25, -4, -12

Question 2.
Given below are the temperatures in some cities. Write them using the proper signs.

Place Shimla Leh Delhi Nagpur
Temperature 7 °C below 0° 12 °C below 0° 22 °C above 0° 31 °C above 0°

Solution:

Place Shimla Leh Delhi Nagpur
Temperature with proper sign -7 °C -12 °C +22 °C +31 °C

Question 3.
Write the numbers in the following examples using the proper signs.

  1. A submarine is at a depth of 512 meters below sea level.
  2. The height of Mt Everest, the highest peak in the Himalayas, is 8848 meters.
  3. A kite is flying at a distance of 120 meters from the ground.
  4. The tunnel is at a depth of 2 meters under the ground.

Solution:

  1. A submarine is at a depth of -512 meters from sea level.
  2. The height of Mt Everest, the highest peak in the Himalayas is +8848 meters.
  3. A kite is flying at a distance of +120 meters from the ground.
  4. The tunnel is at a depth of -2 meters from the ground.

Maharashtra Board Class 6 Maths Chapter 3 Integers Practice Set 4 Intext Questions and Activities

Question 1.
Take warm water in one beaker, some crushed ice in another and a mixture of salt and crushed ice in a third beaker. Ask your teacher for help in measuring the temperature of the substance in each of the beakers using a thermometer. Note the temperatures. (Textbook pg. no. 13)
Maharashtra Board Class 6 Maths Solutions Chapter 3 Integers Practice Set 4 1
Solution:
( Students should attempt this activity on their own)

Question 2.
Look at the picture of the kulfi man. Why do you think he keeps the kulfi moulds in a mixture of salt and ice? (Textbook pg. no. 14)
Maharashtra Board Class 6 Maths Solutions Chapter 3 Integers Practice Set 4 2
Solution:
Kulfi man keeps the kulfi moulds in a mixture of salt and ice because such a mixture helps in keeping the kulfi cool for a longer period of time. The kulfi kept in the said mixture relatively takes more time to melt. This mixture is Considered ideal as it has the temperature of -4°C as against the temperature of ice i.e. 0°C.

Question 3.
My class, i.e. Std. VI, is a part of my school. My school is in my town. My town is a part of a taluka. In the same way, the taluka is a part of a district, and the district is a part of Maharashtra State. In the same way, what can you say about these groups of numbers? Textbook pg. no. 15)
Maharashtra Board Class 6 Maths Solutions Chapter 3 Integers Practice Set 4 3
Solution:
By observing the above given groups of numbers, we can infer that natural numbers are a part of whole numbers. In turn, whole numbers are a part of integers.

Maharashtra Board Practice Set 21 Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 21 Answers Solutions Chapter 4 Angles and Pairs of Angles.

Angles and Pairs of Angles Class 7 Practice Set 21 Answers Solutions Chapter 4

Question 1.
∠ACD is an exterior angle of ∆ABC. The measures of ∠A and ∠B are equal. If m∠ACD = 140°, find the measures of the angles ∠A and ∠B.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 21 1
Solution:
Let the measures of ∠A be x°.
m∠A = m∠B = x°
∠ACD is the exterior angle of ∆ABC
∴ m∠ACD = m∠A + m∠B
∴ 140 = x + x
∴ 140 = 2x
∴ 2x = 140
∴ x = \(\frac { 140 }{ 2 }\)
= 70
∴ The measures of the angles ∠A and ∠B is 70° each.

Question 2.
Using the measures of the angles given in the figure alongside, find the measures of the remaining three angles.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 21 2
Solution:
m∠EOD = m∠AOB = 8y ….(vertically opposite angles)
∠FOL, ∠EOD and ∠COD form a straight angle.
∴ m∠FOE + m∠EOD + m∠COD = 180°
∴ 4y + 8y + 6y = 180
∴ 18y = 180
∴ y = \(\frac { 180 }{ 18 }\)
∴ y = 10
m∠EOD = 8y = 8 x 10 = 80°
m∠AOF = m∠COD ….(Vertically opposite angles)
= 6y = 6 x 10 = 60°
m∠BOC = m∠FOE ….(Vertically opposite angles)
= 4y = 4 x 10 = 40°
∴ The measures of ∠EOD, ∠AOF and ∠BOC are 80°, 60° and 40° respectively.

Question 3.
In the isosceles triangle ABC, ∠A and ∠B are equal. ∠ACD is an exterior angle of ∆ABC. The measures of ∠ACB and ∠ACD are (3x – 17)° and (8x + 10)° respectively. Find the measures of ∠ACB and ∠ACD. Also find the measures of ∠A and ∠B.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 21 3
Solution:
Let the measure of ∠A be y°. A
∴ m∠A = m∠B = y°
∠ACB and ∠ACD form a pair of linear angles.
∴ m∠ACB + m∠ACD = 180°
∴ (3x – 17) + (8x + 10) = 180
∴ 3x + 8x – 17 + 10 = 180
∴ 11x – 7 = 180
∴ 11x – 7 + 7 = 180 + 7 …(Adding 7 on both sides.)
∴ 11x = 187
∴ x = \(\frac { 187 }{ 11 }\) = 17
m∠ACB = 3x – 17 = (3 x 17) – 17 = 51 – 17 = 34°
m∠ACD = 8x + 10 = 8 x 17 + 10 = 136 + 10 = 146°
Here ∠ACD is the exterior angle of ∆ABC and ∠A and ∠B are its remote interior angles.
∴ m∠ACD = m∠A + m∠B
∴ 146 = y + y
∴ 146 = 2y
∴ 2y = 146
∴ y = \(\frac { 146 }{ 2 }\) = 73
∴ The measures of ∠ACB, ∠ACD, ∠A and ∠B are 34°, 146°, 73° and 73° respectively.

Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 21 Intext Questions and Activities

Question 1.
Use straws or sticks to make all the kinds of angles that you have learnt about. (Textbook pg. no. 29)
Solution:
(Student should attempt the activity on their own)

Question 2.
Observe the table given below and draw your conclusions (Textbook pg. no. 31)
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 21 4
Solution:
i. 180°
ii. 360°
iii. 540°
iv. 720°
v. 180° x 5 = 900°
vi. Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 21 5 , 180° x 6 = 1080°