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Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 12 Percentage Class 6 Practice Set 30 Answers Solutions.

6th Standard Maths Practice Set 30 Answers Chapter 12 Percentage

Question 1.
Shabana scored 736 marks out of 800 in her exams. What was the percentage she scored?
Solution:
Total marks of the examination = 800
Marks scored by Shabana = 736
Suppose Shabana scored A% marks.

∴ A = 92%
∴ Shabana scored 92% marks.

Question 2.
There are 500 students in the school in Dahihanda village. If 350 of them can swim, what percent of them can swim and what percent cannot?
Solution:
Total number of students in the school = 500
Number of students who can swim = 350
Suppose A% students can swim.

∴ A = 70%
Percentage of students who cannot swim = 100% – Percentage of students who can swim .
= 100% – 70% = 30%
∴ 70% of the students can swim and 30% cannot swim.

Question 3.
If Prakash sowed jowar on 75% of the 19500 sq. m. of his land, on how many sq. m. did he actually plant jowar?
Solution:
Total area of the land = 19500 sq. m.
Percentage of area in which Prakash sowed jowar = 75%
Suppose Prakash planted jowar in A sq. m.

∴ A = 14,625 sq. m.
∴ Prakash planted jowar in 14,625 sq.m.

Question 4.
Soham received 40 messages on his birthday. Of these, 90% were birthday greetings. How many other messages did he get besides the greetings?
Solution:
Total messages received by Soham on his birthday = 40
Percentage of messages received for birthday greetings = 90%
Suppose Soham got A number of birthday greetings.

∴ A = 36
∴ Number of messages received other than birthday greetings
= total messages received – total number of birthday greetings
= 40 – 36 = 4
∴ The number of messages received other than birthday greetings is 4.

Question 5.
Of the 5675 people in a village 5448 are literate. What is the percentage of literacy in the village?
Solution:
Number of people in the village = 5675
Number of people who are literate = 5448
Suppose the percentage of literacy in the village is A%.

∴ A = 96%
∴ The percentage of literacy in the village is 96%.

Question 6.
In the elections, 1080 of the 1200 women in Jambhulgaon cast their vote, while 1360 of the 1700 in Wadgaon cast theirs. In which village did a greater proportion of women cast their votes?
Solution:
Total number of women in Jambhulgaon = 1200
Number of women in Jambhulgaon who voted = 1080
Suppose A% women cast their vote in Jambhulgaon village.


∴ A = 90%
In Jambhulgaon, the percentage of women who voted in the elections was 90%.
Total number of women in Wadgaon = 1700 Number of women in Wadgaon who voted = 1360
Suppose B% women cast their vote in Wadgaon.

∴ B = 80%
∴ In Wadgaon, the percentage of women who voted in the elections was 80%.
∴ A greater proportion of women cast their votes in Jambhulgaon.

Maharashtra Board Class 6 Maths Chapter 12 Percentage Practice Set 30 Intext Questions and Activities

Question 1.
There are 9 squares in the figure alongside. The letters ABCDEFGHI are written in squares. Give each of the letters a unique number from 1 to 9 so that every letter has a different number.
Besides, A + B + C = C + D + E = E + F + G = G + H + I should also be true. (Textbook pg. no. 64)

Solution:

(This is one of the possible solutions of the above riddle. There are more solutions possible.)

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 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.

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.

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.

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.

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

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 6 Maths Solutions covers the Std 6 Maths Chapter 17 Geometrical Constructions Class 6 Practice Set 40 Answers Solutions.

6th Standard Maths Practice Set 40 Answers Chapter 17 Geometrical Constructions

Question 1.
Draw line l. Take point P anywhere outside the line. Using a set square draw a line PQ perpendicular to line l.
Solution:
Step 1:

Step 2:

line PQ ⊥ line l.

Question 2.
Draw line AB. Take point M anywhere outside the line. Using a compass and ruler, draw a line MN perpendicular to line AB.
Solution:
Step 1:

Step 2:

Step 3:

line MN ⊥ line AB.

Question 3.
Draw a line segment AB of length 5.5 cm. Bisect it using a compass and ruler.
Solution:
Step 1:

Step 2:

line MN is the perpendicular bisector of seg AB.

Question 4.
Take point R on line XY. Draw a perpendicular to XY at R, using a set square.
Solution:
Step 1:

Step 2:

line TR ⊥ line XY.

Maharashtra Board Class 6 Maths Chapter 17 Geometrical Constructions Practice Set 40 Questions and Activities

Question 1.
In the above construction, why must the distance in the compass be kept constant? (Textbook pg. no. 90)
Solution:
The point N is at equal distance from points P and Q.
If we change the distance of the compass while drawing arcs from points P and Q, we will not get a point which is at equal distance from P and Q. Hence, the distance in the compass must be kept constant.

Question 2.
The Perpendicular Bisector. (Textbook pg. no. 90)

1. A wooden ‘yoke’ is used for pulling a bullock cart. How is the position of the yoke determined?
2. To do that, a rope is used to measure equal distances from the spine/midline of the bullock cart. Which geometrical property is used here?
3. Find out from the craftsmen or from other experienced persons, why this is done.

Solution:

1. For the bullock cart to be pulled in the correct direction by the yoke, its Centre O should be equidistant from the either sides of the cart.
2. The property of perpendicular bisector is used to make the point equidistant from both the ends
3. A rope is used just like a compass to get equal distances from the spine/midline of bullock cart.

Question 3.
Take a rectangular sheet of paper. Fold the paper so that the lower edge of the paper falls on its top edge, and fold it over again from right to left. Observe the two folds that have formed on the . paper. Verify that each fold is a perpendicular bisector of the other. Then measure the following distances. (Textbook pg. no. 91)
i. l(XP)
ii. l(XA)
iii. l(XB)
iv. l(YP)
v. l(YA)

You will observe that l(XP) = l(YP), l(XA) = l(YA) and l(XB) = l(YB)
Therefore we can conclude that all points on the vertical fold (perpendicular bisector) are equidistant from the endpoints of the horizontal fold.
Solution:
[Note: Students should attempt this activity on their own.]

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 6 Bar Graphs Class 6 Practice Set 19 Answers Solutions.

6th Standard Maths Practice Set 19 Answers Chapter 6 Bar Graphs

Question 1.
The names of the heads of some families in a village and the quantity of drinking water their family consumes in one day are given below. Draw a bar graph for this data.
(Scale: On Y axis. 1 cm = 10 liters of water)

Name	Ramesh	Shobha	Ayub	Julie	Rahul
Liters of water Used	30 L	60 L	40 L	50L	55 L

Solution:

Question 2.
The names and numbers of animals in a certain zoo are given below. Use the data to make a bar graph. (Scale: On Y axis, 1 cm = 4 animals).

Animals	Deer	Tiger	Monkey	Rabbit	Peacock
Number	20	4	12	16	8

Solution:

Question 3.
The table below gives the number of children who took part in the various items of the talent show as part of the annual school gathering. Make a bar graph to show this data.
(Scale: On Y-axis, 1 cm = 4 children)

Programme	Theater	Dance	Vocal music	Instrumental music	One-act plays
Number of Children	24	40	16	8	4

Solution:

Question 4.
The number of customers who came to a juice centre during one week is given in the table below. Make two different bar graphs to show this data.
(On Y-axis, 1 cm = 10 customers, 1 cm = 5 customers)

Type of juice	Orange	Pineapple	Apple	Mango	Pomegranate
Number of customers	50	30	25	65	10

Solution:

Question 5.
Students planted trees in 5 villages of Sangli district. Make a bar graph of this data. (Scale: On Y-axis, 1 cm = 100 trees).

Name of Place	Dudhgaon	Bagni	Samdoli	Ashta	Kavathepiran
Number of Trees Planted	500	350	600	420	540

Solution:

Question 6.
Yashwant gives different amounts of time as shown below, to different exercises he does during the week. Draw a bar graph to show the details of his schedule using an appropriate scale.

Type of exercise	Running	Yogasanas	Cycling	Mountaineering	Badminton
Time	35 minutes	50 minutes	1 hr 10 min	\(1\frac { 1 }{ 2 }\) hours	45 minutes

Solution:
1 hour = 60 minutes
∴ 1 hour 10 minutes = 1 hour + 10 minutes = 60 minutes +10 minutes = 70 minutes
and \(1\frac { 1 }{ 2 }\) hours = 1 hour + \(\frac { 1 }{ 2 }\) hour = 60 minutes + 30 minutes = 90 minutes
The given table can be written as follows:

Type of Exercise	Running	Yogasanas	Cycling	Moutaineering	Badminton
Time	35 minutes	50 minutes	70 minutes	90 minutes	45 minutes

Question 7.
Write the names of four of your classmates. Beside each name, write his/her weight in kilograms. Enter this data in a table like the above and make a bar graph.
Solution:

Name of classmates	Weight (kg)
Rohan	32
Laxmi	28
Rakesh	40
Riya	36

Scale: On Y-axis, 1 cm = 4 kg [Note: Students can take their own examples]

Maharashtra Board Class 6 Maths Chapter 6 Bar Graphs Practice Set 19 Intext Questions and Activities

Question 1.
Collect bar graphs from newspapers or periodicals showing a variety of data. (Textbook pg. no. 38)
Solution:
(Student should attempt the activities on their own.)

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 6 Maths Solutions covers the Std 6 Maths Chapter 3 Integers Class 6 Practice Set 7 Answers Solutions.

6th Standard Maths Practice Set 7 Answers Chapter 3 Integers

Question 1.
Write the proper signs >, < or = in the boxes below:

1. -4 __ 5
2. 8 __ -10
3. +9 __ +9
4. -6 __ 0
5. 7 __ 4
6. 3 __ 0
7. -7 __ 7
8. -12 __ 5
9. -2 __ -8
10. -1 __ -2
11. 6 __ -3
12. -14 __ -14

Solution:

1. -4 < 5
2. 8 > -10
3. +9 = +9
4. -6 < 0
5. 7 > 4
6. 3 > 0
7. -7 < 7
8. -12 < 5
9. -2 > -8
10. -1 > -2
11. 6 > -3
12. -14 = -14

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

6th Standard Maths Practice Set 6 Answers Chapter 3 Integers

Question 1.
Write the opposite number of each of the numbers given below.

Number	47	+52	-33	-84	-21	+16	-26	80
Opposite number

Solution:

Number	47	+52	-33	-84	-21	+16	-26	80
Opposite number	-47	-52	+33	+84	+21	-16	+26	-80