Class 6

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 13 Profit-Loss Class 6 Practice Set 32 Answers Solutions.

6th Standard Maths Practice Set 32 Answers Chapter 13 Profit-Loss

Question 1.
From a wholesaler, Santosh bought 400 eggs for Rs 1500 and spent Rs 300 on transport. 50 eggs fell down and broke. He sold the rest at Rs 5 each. Did he make a profit or a loss? How much?
Solution:
Cost price of 400 eggs = Rs 1500
Transportation cost = Rs 300
∴ Total cost price of 400 eggs = Cost price of 400 eggs + Transportation cost
= 1500 + 300 = Rs 1800
50 eggs fell and broke
∴ Remaining eggs = 400 – 50 = 350
Selling price of 1 egg = Rs 5
∴ Selling price of 350 eggs = 5 x 350 = Rs 1750
Total cost price is greater than the selling price.
∴ Santosh suffered a loss.
Loss = Total cost price – Selling price
= 1800 – 1750
= Rs 50
∴ Santosh incurred a loss of Rs 50.

Question 2.
Abraham bought goods worth Rs 50000 and spent Rs 7000 on transport and octroi. If he sold the goods for Rs 65000, did he make a profit or a loss? How much?
Solution:
Cost price of goods = Rs 50000
Transportation cost and octroi = Rs 7000
∴ Total cost price for buying goods = Cost price of goods + Transportation cost and octroi
= 50000 + 7000 = Rs 57000
Selling price of goods = Rs 65000
Selling price is greater than the total cost price
∴ Abraham made a profit.
Profit = Selling price – Total cost price
= 65000 – 57000
= Rs 8000
∴ Abraham made a profit of Rs 8000.

Question 3.
Ajit Kaur bought a 50 kg sack of sugar for Rs 1750, but as sugar prices fell, she had to sell it at Rs 32 per kg. How much loss did she incur?
Solution:
Cost price of 50 kg sugar = Rs 1750
Selling price of 1 kg sugar = Rs 32
∴ Selling price of 50 kg sugar = 50 x 32 = Rs 1600
Loss = Total cost price – Selling price
= 1750 – 1600 = Rs 150
∴ Ajit Kaur incurred a loss of Rs 150.

Question 4.
Kusumtai bought 80 cookers at Rs 700 each. Transport cost her Rs 1280. If she wants a profit of Rs 18000, what should be the selling price per cooker?
Solution:
Cost price of one cooker = Rs 700
∴ Cost price of 80 cookers = 700 x 80 = Rs 56000
Transportation cost = Rs 1280
∴ Total cost price = Cost price of 80 cookers + Transportation cost
= 56000 + 1280
= Rs 57280
Profit = Rs 18000
Profit = Selling Price – Total Cost Price
∴ Required selling price = Total cost price + profit
= 57280 + 18000
= Rs 75280
∴ Selling price of 80 cookers = Rs 75280
∴ Selling price of 1 cooker = \(\frac { 75280 }{ 80 }\) = Rs 941
∴ The selling price per cooker should be Rs 941.

Question 5.
Indrajit bought 10 refrigerators at Rs 12000 each and spent Rs 5000 on transport. For how much should he sell each refrigerator in order to make a profit of Rs 20000?
Solution:
Cost price of 1 refrigerator = Rs 12000
Cost price of 10 refrigerator = 10 x 12000 = Rs 120000
Transportation cost = Rs 5000
∴ Total cost price of 10 refrigerators = Cost price of 10 refrigerators + Transportation cost
= 120000 + 5000 = Rs 125000
Profit = Rs 20000
Profit = Selling Price – Total Cost Price
∴ Required selling price = Total cost price + Profit
= 125000 + 20000 = Rs 145000
∴ Selling price of 10 refrigerators = Rs 145000
∴ Selling price of 1 refrigerator = \(\frac { 145000 }{ 10 }\) = Rs 14500
∴ Indrajit must sell each refrigerator at Rs 14500 to make a profit of Rs 20000.

Question 6.
Lalitabai sowed seeds worth Rs 13700 in her field. She had to spend Rs 5300 on fertilizers and spraying pesticides and Rs 7160 on labor. If, on selling her produce, she earned Rs 35400 what was her profit or her loss?
Solution:
Cost price of seeds = Rs 13700
Cost of fertilizers and pesticides = Rs 5300
Labor cost = Rs 7160
∴ Total cost price = Cost price of seeds + Cost of fertilizers and pesticides + Labor cost
= 13700 + 5300 + 7160
= Rs 26160
Selling price = Rs 35400
Selling price is greater than the total cost price.
∴ Lalitabai made a profit.
Profit = Selling price – Cost price
= 35400 – 26160
= Rs 9240
∴ Lalitabai made a profit of Rs 9240.

Maharashtra Board Class 6 Maths Chapter 13 Profit-Loss Practice Set 32 Intext Questions and Activities

Question 1.
At Diwali, in a certain school, the students undertook a Design a Diya project. They bought 1000 diyas for Rs 1000 and some paint for Rs 200. To bring the diyas to the school, they spent Rs 100 on transport. They sold the painted lamps at Rs 2 each. Did they make a profit or incur a loss? (Textbook pg. no. 67 and 68)

i. Is Anju right?
ii. What about the money spent on paints and transport?
iii. How much money was actually spent before the diyas could be sold?
iv. How much actual profit was made in this project of colouring the diyas and selling them?

Ans:
i. No, Anju is wrong.
Cost price of diyas also includes the painting and transportation cost.
∴ Total cost price of diyas = Cost of diyas + Cost of paint + Transportation cost
= 1000 + 200+ 100
= Rs 1300
ii. The cost of paint was Rs 200 and that for transportation was Rs 100. These costs are also to be added to the cost price of diyas.
iii. Rs 1300 was actually spent before the diyas could be sold.
iv. Total Cost Price of 1000 Diyas = Rs 1300
Selling Price of 1 Diya = Rs 2
∴ Selling Price of 1000 Diyas = 2 x 1000 = Rs 2000
∴ Profit = Selling Price – Total Cost Price
= 2000 – 1300
= Rs 700
∴ The profit made by coloring the diyas and selling them was Rs 700.

Question 2.
A farmer sells what he grows in his fields. How is the total cost price calculated? What does a farmer spend on his produce before he can sell it? What are the other expenses besides seeds, fertilizers and transport? (Textbook pg. no. 68)
Solution:
The farmer, in order to calculate the total cost price of his produce, needs to consider all the expenses associated with the growing and selling of his produce.

Following are the things on which farmer spends money before he can sell it.

1. Time and energy
2. Ploughing and tilling
3. Irrigation and electricity cost
4. Harvesting and cleaning
5. Packing

As given above, there are a multiple of costs to be included besides seeds, fertilizers and transport for the farmer to price its produce appropriately.

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 4 Operations on Fractions Class 6 Practice Set 10 Answers Solutions.

6th Standard Maths Practice Set 10 Answers Chapter 4 Operations on Fractions

Question 1.
Add:
i. \(6 \frac{1}{3}+2 \frac{1}{3}\)
ii. \(1 \frac{1}{4}+3 \frac{1}{2}\)
iii. \(5 \frac{1}{5}+2 \frac{1}{7}\)
iv. \(3 \frac{1}{5}+2 \frac{1}{3}\)
Solution:
i. \(6 \frac{1}{3}+2 \frac{1}{3}\)

ii. \(1 \frac{1}{4}+3 \frac{1}{2}\)

iii. \(5 \frac{1}{5}+2 \frac{1}{7}\)

iv. \(3 \frac{1}{5}+2 \frac{1}{3}\)

Question 2.
Subtract:
i. \(3 \frac{1}{3}-1 \frac{1}{4}\)
ii. \(5 \frac{1}{2}-3 \frac{1}{3}\)
iii. \(7 \frac{1}{8}-6 \frac{1}{10}\)
iv. \(7 \frac{1}{2}-3 \frac{1}{5}\)
Solution:
i. \(3 \frac{1}{3}-1 \frac{1}{4}\)

ii. \(5 \frac{1}{2}-3 \frac{1}{3}\)

iii. \(7 \frac{1}{8}-6 \frac{1}{10}\)

iv. \(7 \frac{1}{2}-3 \frac{1}{5}\)

Question 3.
Solve:
i. Suyash bought \(2\frac { 1 }{ 2 }\) kg of sugar and Ashish bought \(3\frac { 1 }{ 2 }\) kg. How much sugar did they buy altogether? If sugar costs Rs 32 per kg, how much did they spend on the sugar they bought?

ii. Aradhana grows potatoes in \(\frac { 2 }{ 5 }\) part of her garden, greens in \(\frac { 1 }{ 3 }\) part and brinjals in the remaining part. On how much of her plot did she plant brinjals?

iii. Sandeep filled water in \(\frac { 4 }{ 7 }\) of an empty tank. After that, Ramakant filled \(\frac { 1 }{ 4 }\) part more of the same tank. Then Umesh used \(\frac { 3 }{ 14 }\) part of the tank to water the garden. If the tank has a maximum capacity of 560 litres, how many litres of water will be left in the tank?
Solution:
i. Sugar bought by Suyash = \(2\frac { 1 }{ 2 }\) kg
Sugar bought by Ashish = \(3\frac { 1 }{ 2 }\) kg
∴ Total sugar bought by both

Cost of 1 kg of sugar = Rs 32
∴ Cost of 6 kg of sugar = 32 x 6
= Rs 192
∴ They bought 6 kg sugar altogether and the total money spent on sugar is Rs 192.

ii. Part of garden occupied by potatoes = \(\frac { 2 }{ 5 }\)
Part of garden occupied by greens = \(\frac { 1 }{ 3 }\)
Since brinjals are planted in the remaining part,
∴ (Part occupied by potatoes) + (part occupied by greens) + (part occupied by brinjals) = 1 entire garden.
∴ Part of garden occupied by brinjals = 1 – (part of garden occupied by potatoes + part of garden occupied by greens)

∴ Aradhana planted brinjals on \(\frac { 4 }{ 15 }\) part of her plot.

iii. Part of tank filled by Sandeep = \(\frac { 4 }{ 7 }\)
Part of tank filled by Ramakant = \(\frac { 1 }{ 4 }\)

Since maximum capacity of tank is 560 litres
∴ Quantity of water left in tank = \(\frac { 17 }{ 28 }\times560\) = 340 litres
∴ The quantity of water left in the tank is 340 litres.

Maharashtra Board Class 6 Maths Chapter 4 Operations on Fractions Practice Set 10 Intext Questions and Activities

Question 1.
How to do this subtraction: \(4 \frac{1}{4}-2 \frac{1}{2}\) ? Is it same as \(\left[4-2+\frac{1}{4}-\frac{1}{2}\right]\) ? (Textbook pg. no. 23)
Solution:
\(4 \frac{1}{4}-2 \frac{1}{2}\)

\(\left[4-2+\frac{1}{4}-\frac{1}{2}\right]\)

The subtraction \(4 \frac{1}{4}-2 \frac{1}{2}\) is the same as \(\left[4-2+\frac{1}{4}-\frac{1}{2}\right]\).

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 13 Profit-Loss Class 6 Practice Set 31 Answers Solutions.

6th Standard Maths Practice Set 31 Answers Chapter 13 Profit-Loss

Question 1.
The cost price and selling price are given in the following table. Find out whether there was a profit or a loss and how much it was.

Ex.	Cost price (in Rs)	Selling price (in Rs)	Profit or Loss	How much?
i.	4500	5000
ii.	4100	4090
iii.	700	799
iv.	1000	920

Solution:

i. Cost price = Rs 4500
Selling price = Rs 5000
Selling price is greater than cost price.
∴ There is a profit.
∴ Profit = Selling price – Cost price
= 5000 – 4500
Profit = Rs 500

ii. Cost price = Rs 4100
Selling price = Rs 4090
Cost price is greater than selling price.
∴ There is a loss.
∴ Loss = Cost price – Selling price
= 4100 – 4090
∴ Loss = Rs 10

iii. Cost price = Rs 700
Selling price = Rs 799
Selling price is greater than cost price.
∴ There is a profit.
∴ Profit = Selling price – Cost price
= 799 – 700
∴ Profit = Rs 99

iv. Cost price = Rs 1000
Selling price = Rs 920
Cost price is greater than selling price.
∴ There is a loss.
∴ Loss = Cost price – Selling price
= 1000 – 920
∴ Loss = Rs 80

Ex.	Cost price (in Rs)	Selling price (in Rs)	Profit or Loss	How much?
i.	4500	5000	Profit	Rs 500
ii.	4100	4090	Loss	Rs 10
iii.	700	799	Profit	Rs 99
iv.	1000	920	Loss	Rs 80

Question 2.
A shopkeeper bought a bicycle for Rs 3000 and sold the same for Rs 3400. How much was his profit?
Solution:
Cost price = Rs 3000, Selling price = Rs 3400
∴ Profit = Selling price – Cost price
= 3400 – 3000
= Rs 400
The shopkeeper’s profit was Rs 400.

Question 3.
Sunandabai bought milk for Rs 475. She converted it into yogurt and sold it for Rs 700. How much profit did she make?
Solution:
∴ Cost price = Rs 475, Selling price = Rs 700
∴ Profit = Selling price – Cost price
= 700 – 475
= Rs 225
∴ Sunandabai made a profit of Rs 225.

Question 4.
The Jijamata Women’s Saving Group bought raw materials worth Rs 15000 for making chakalis.
They sold the chakalis for Rs 22050. How much profit did the WSG make?
Solution:
Cost price of raw materials = Rs 15000
Selling price of chakalis = Rs 22050
∴ Profit = Selling price – Cost price
= 22050 – 15000
= Rs 7050
∴ The Women’s Saving Group made a profit of Rs 7050.

Question 5.
Pramod bought 100 bunches of methi greens for Rs 400. In a sudden downpour, 30 of the bunches on his handcart got spoil. He sold the rest at the rate of Rs 5 each. Did he make a profit or a loss? How much?
Solution:
Cost price of 100 bunches of methi green = Rs 400
Since, 30 bunches got spoil,
∴ Remaining bunches of methi green = 100 – 30 = 70
Selling price of 1 bunch of methi green = Rs 5
∴ Selling price of 70 bunches of methi green = 5 x 70 = Rs 350
Cost price is greater than selling price
∴ Pramod suffered a loss.
Loss = Cost price – Selling price
= 400 – 350
= Rs 50
∴ Pramod suffered a loss of Rs 50.

Question 6.
Sharad bought one quintal of onions for Rs 2000. Later he sold them all at the rate of Rs 18 per kg. Did he make a profit or incur a loss? How much was it?
Solution:
Cost price of one quintal onions = Rs 2000
Selling price of 1 kg onions = Rs 18
Since, 1 quintal = 100 kg
∴ Selling price of 1 quintal (100 kg) onions = 18 x 100 = Rs 1800
Cost price is greater than selling price
∴ Sharad suffered a loss.
∴ Loss = Cost price – Selling price
= Rs 2000 – Rs 1800
= Rs 200
∴ Sharad incurred a loss of Rs 200.

Question 7.
Kantabai bought 25 saris from a wholesale merchant for Rs 10000 and sold them all at Rs 460 each. How much profit did Kantabai get in this transaction?
Solution:
Cost price of 25 saris = Rs 10000
Selling price of 1 sari = Rs 460
∴ Selling price of 25 saris = 460 x 25 = Rs 11500
Selling price is greater than cost price.
∴ Profit = Selling price – Cost price
= 11500 – 10000
= Rs 1500
∴ Kantabai made a profit of Rs 1500.

Maharashtra Board Class 6 Maths Chapter 13 Profit-Loss Practice Set 31 Intext Questions and Activities

Question 1.
Pranav and sarita had set up stalls in a fun fair. Study the data given below and answer the questions. (Textbook pg. no. 65)


Solution:
Total amount invested by Pranav = 70 + 25 + 45 + 14 + 20 = Rs 174
Amount gained through sale = Rs 160
∴ Selling price is less than invested price.
∴ Pranav incurred a loss in his Pav Bhaji business. Hence, he is disappointed.

Total amount invested by Sarita = 20 + 10 + 30 + 50 + 20 + 60 = Rs 190
Amount gained by selling = Rs 230
∴ Selling price is more than invested price.
∴ Sarita made profit in her business. Hence, she is happy.

Question 2.
For the above example,

1. If Sarita had bought twice as much, would she have gained twice as much?
2. What should Pranav do the next time he sets up a stall to sell more pav bhaji and make more gains? (Textbook pg. no. 66)

Solution:

1. If Sarita would have bought twice as much, she would have prepared double quantity of food items. Hence, she would have gained twice as much.
2. Next time Pranav sets a stall, he must sell pav bhaji at a higher cost than he had sold earlier in order to make more gains.

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 8 Divisibility Class 6 Practice Set 22 Answers Solutions.

6th Standard Maths Practice Set 22 Answers Chapter 8 Divisibility

Question 1.
There are some flowering trees in a garden. Each tree bears many flowers with the same number printed on it. Three children took a basket each to pick flowers. Each basket has one of the numbers, 3, 4 or 9 on it. Each child picks those flowers which have numbers divisible by the number on his or her basket. If He / She takes only 1 flower from each tree. Can you tell which numbers the flowers in each basket will have?

Solution:
Each child will have flowers bearing the following numbers:
Girl with basket number 3: 111, 369, 435, 249, 666, 450, 960, 432, 999, 72, 336, 90, 123, 108
Boy with basket number 4: 356, 220, 432, 960, 72, 336, 108
Girl with basket number 9: 369, 666, 450, 432, 999, 72, 90, 108

Maharashtra Board Class 6 Maths Chapter 8 Divisibility Practice Set 22 Intext Questions and Activities

Question 1.
Read the numbers given below. Which of these numbers are divisible by 2, by 5, or by 10? Write them in the empty boxes. 125,364,475,750,800,628,206,508,7009,5345,8710. (Textbook pg. no. 43)

Divisible by 2	Divisible by 5	Divisible by 10

Solution:

Divisible by 2	Divisible by 5	Divisible by 10
364,750, 800, 628, 206, 508, 8710	125,475, 750, 800, 5345, 8710	750, 800, 8710

Question 2.
Complete the following table: (Textbook pg. no. 43)

Number	Sum of digits in the number	Is the sum divisible by 3?	Is the given number divisible by 3?
63	6 + 3 = 9	✓	✓
872	17	X	X
91
552
9336
4527

Solution:

Number	Sum of digits in the number	Is the sum divisible by 3?	Is the given number divisible by 3?
63	6 + 3 = 9	✓	✓
872	8 + 7 + 2 = 17	X	X
91	9 + 1 = 10	X	X
552	5 + 5 + 2 = 12	✓	✓
9336	9 + 3 + 3 + 6 = 21	✓	✓
4527	4 + 5 + 2 + 7 = 18	✓	✓

Question 3.
Complete the following table: (Textbook pg. no. 44)

Number	Divide the number by 4. Is it completely divisible?	The number formed by the digits in the tens and units places.	Is this number divisible by 4?
992	✓	92	✓
7314
6448
8116
7773
3024

Solution:

Number	Divide the number by 4. Is it completely divisible?	The number formed by the digits in the tens and units places.	Is this number divisible by 4?
992	✓	92	✓
7314	X	14	X
6448	✓	48	✓
8116	✓	16	✓
7773	X	73	X
3024	✓	24	✓

Question 4.
Complete the following table: (Textbook pg. no. 44)

Number	Divide the number by 9. Is it completely divisible?	Sum of the digits in the number.	Is the sum divisible by 9?
1980	✓	1 + 9 + 8 + 0 = 18	✓
2999	X	2 + 9 + 9 + 9 = 29	X
5004
13389
7578
69993

Solution:

Number	Divide the number by 9. Is it completely divisible?	Sum of the digits in the number.	Is the sum divisible by 9?
1980	✓	1 + 9 + 8 + 0 = 18	✓
2999	X	2 + 9 + 9 + 9 = 29	X
5004	✓	5 + 0 + 0 + 4 = 9	✓
13389	X	1 + 3 + 3 + 8 + 9 = 24	X
7578	✓	7 + 5 + 7 + 8 = 27	✓
69993	✓	6 + 9 + 9 + 9 + 3 = 36	✓

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 4 Operations on Fractions Class 6 Practice Set 9 Answers Solutions.

6th Standard Maths Practice Set 9 Answers Chapter 4 Operations on Fractions

Question 1.
Convert into improper fractions:
i. \(7 \frac{2}{5}\)
ii. \(5 \frac{1}{6}\)
iii. \(4 \frac{3}{4}\)
iv. \(2 \frac{5}{9}\)
v. \(1 \frac{5}{7}\)
Solution:
i. \(7 \frac{2}{5}\)

ii. \(5 \frac{1}{6}\)

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

iv. \(2 \frac{5}{9}\)

v. \(1 \frac{5}{7}\)

Question 2.
Convert into mixed numbers:
i. \(\frac { 30 }{ 7 }\)
ii. \(\frac { 7 }{ 4 }\)
iii. \(\frac { 15 }{ 12 }\)
iv. \(\frac { 11 }{ 8 }\)
v. \(\frac { 21 }{ 4 }\)
v. \(\frac { 20 }{ 7 }\)
Solution:
i. \(\frac { 30 }{ 7 }\)

ii. \(\frac { 7 }{ 4 }\)

iii. \(\frac { 15 }{ 12 }\)

iv. \(\frac { 11 }{ 8 }\)

v. \(\frac { 21 }{ 4 }\)

v. \(\frac { 20 }{ 7 }\)

Question 3.
Write the following examples using fraction:
i. If 9 kg rice is shared among 5 people, how many kilograms of rice does each person get?
ii. To make 5 shirts of the same size, 11 meters of cloth is needed. How much cloth is needed for one shirt?
Solution:
i. Total quantity of rice = 9 kg
Number of people = 5
∴ Kilograms of rice received by each person = \(\frac { 9 }{ 5 }\)
∴ Each person will get \(\frac { 9 }{ 5 }\) kg of rice.

ii. Total meters of cloth = 11 meters
Number of shirts to be made = 5
Meters of cloth needed to make 1 shirt = \(\frac { 11 }{ 5 }\)
∴ Cloth needed to make 1 shirt is \(\frac { 11 }{ 5 }\) meters.

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 7 Symmetry Class 6 Practice Set 21 Answers Solutions.

6th Standard Maths Practice Set 21 Answers Chapter 7 Symmetry

Question 1.
Along each figure shown below, a line l has been drawn. Complete the symmetrical figures by drawing a figure on the other side such that the line l becomes the line of symmetry.

Solution:

Maharashtra Board Class 6 Maths Chapter 7 Symmetry Practice Set 21 Intext Questions and Activities

Question 1.
In the figures below, the line l divides the figure in two parts. Do these parts fall on each other? Verify? (Textbook pg. no. 42)

Solution:
Yes, the two parts of both the figures fall on each other on folding along the line l.

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 6 Maths Solutions covers the Std 6 Maths Chapter 3 Integers Class 6 Practice Set 8 Answers Solutions.

6th Standard Maths Practice Set 8 Answers Chapter 3 Integers

Question 1.
Subtract the numbers in the top row from the numbers in the first column and write the proper number in each empty box:

–	6	9	-4	-5	0	+7	-8	-3
3	3 – 6 = -3
8				8 – (-5) = 13
-3
-2

Solution:

–	6	9	-4	-5
3	(+3) + (-6) = -3	(+3) + (-9) = -6	(+3) + (+4) = 7	(+3) + (+5) = 8
8	(+8) + (-6) = +2	(+8) + (-9) = -1	(+8) + (+4) = 12	(+8) + (+5) = 13
-3	(-3) + (-6) = -9	(-3) + (-9) = -12	(-3) + (+4) = 1	(-3) + (+5) = 2
-2	(-2) + (-6) = -8	(-2) + (-9) = -11	(-2) + (+4) = 2	(-2) + (+5) = 3

–	0	+7	-8	-3
3	(+3) – 0 = 3	(+3) + (-7) = -4	(+3) + (+8) = 11	(+3) + (+3) = 6
8	(+8) – 0 = 8	(+8) + (-7) = 1	(+8) + (+8) = 16	(+8) + (+3) = 11
-3	(-3) – 0 = -3	(-3) + (-7) = -10	(-3) + (+8) = 5	(-3) + (+3) = 0
-2	(-2) – 0 = -2	(-2) + (-7) = -9	(-2) + (+8) = 6	(-2) + (+3) = 1

Maharashtra Board Class 6 Maths Chapter 3 Integers Practice Set 8 Intext Questions and Activities

Question 1.
A Game of Integers. (Textbook pg. no. 20)
The board for playing this game is given in the back cover of the textbook. Place your counters before the number 1. Throw the dice. Look at the number you get. It is a positive number. Count that many boxes and move your counter forward. If a problem is given in that box, solve it. If the answer is a positive number, move your counter that many boxes further. It it is negative, move back by that same number of boxes.

Suppose we have reached the 18th box. Then the answer to the problem in it is -4 + 2 = -2. Now move your counter back by 2 boxes to 16. The one who reaches 100 first, is the winner.

Solution:
(Students should attempt this activity on their own)

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 7 Symmetry Class 6 Practice Set 20 Answers Solutions.

6th Standard Maths Practice Set 20 Answers Chapter 7 Symmetry

Question 1.
Draw the axes of symmetry of each of the figures below. Which of them has more than one axis of symmetry?

Solution:

Figures (i), (ii) and (iv) have more than one axis of symmetry.

Question 2.
Write the capital letters of the English alphabet in your notebook. Try to draw their axes of symmetry. Which ones have an axis of symmetry? Which ones have more than one axis of symmetry?
Solution:
Alphabets having axis of symmetry:

Alphabets having more than one axis of symmetry:

Question 3.
Use color, a thread and a folded paper to draw symmetrical shapes.
Solution:
Take any color, a thread and a folded square paper.
Step 1:
Take a folded square paper which is folded along one of its axis of symmetry.

Step 2:
Open the paper. Draw a square in one comer. Place the thread in the square drawn and apply colour on it as shown in the figure.

Step 3:
Remove the thread. You will see a white patch where the thread was.

Step 4:
Fold the paper and press it along the axis of symmetry. When you unfold the paper, you will see an imprint on the other side of the fold which is identical to the color patch you had made earlier.

Question 4.
Observe various commonly seen objects such as tree leaves, birds in flight, pictures of historical buildings, etc. Find symmetrical shapes among them and make a collection of them.
Solution:
Some of the symmetrical objects seen in daily life are shown below:

Maharashtra Board Class 6 Maths Chapter 7 Symmetry Practice Set 20 Intext Questions and Activities

Question 1.
Do you recognize this picture?
Why do you think the letters written on the front of the vehicle are written the way they are? Copy them on a paper. Hold the paper in front of a mirror and read it.
Do you see letters written like this anywhere else?
(Textbook pg. no. 40)

Solution:

1. The name written in reverse alphabets on the vehicle reads
   as ‘AMBULANCE’ when viewed in the mirror.
   In the case of an emergency, it helps a driver to quickly notice an ambulance by looking into his rear view mirror and read the reverse alphabets which appear perfectly normal in a mirror
2. Other than ambulance, we see letters written in reverse on school bus.

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.]