## Maharashtra Board Practice Set 12 Class 6 Maths Solutions Chapter 4 Operations on Fractions

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 4 Operations on Fractions Class 6 Practice Set 12 Answers Solutions.

## 6th Standard Maths Practice Set 12 Answers Chapter 4 Operations on Fractions

Question 1.
Multiply:
i. $$\frac{7}{5} \times \frac{1}{4}$$
ii. $$\frac{6}{7} \times \frac{2}{5}$$
iii. $$\frac{5}{9} \times \frac{4}{9}$$
iv. $$\frac{4}{11} \times \frac{2}{7}$$
v. $$\frac{1}{5} \times \frac{7}{2}$$
vi. $$\frac{9}{7} \times \frac{7}{8}$$
vii. $$\frac{5}{6} \times \frac{6}{5}$$
viii. $$\frac{6}{17} \times \frac{3}{2}$$
Solution:
i. $$\frac{7}{5} \times \frac{1}{4}$$

ii. $$\frac{6}{7} \times \frac{2}{5}$$

iii. $$\frac{5}{9} \times \frac{4}{9}$$

iv. $$\frac{4}{11} \times \frac{2}{7}$$

v. $$\frac{1}{5} \times \frac{7}{2}$$

vi. $$\frac{9}{7} \times \frac{7}{8}$$

vii. $$\frac{5}{6} \times \frac{6}{5}$$

viii. $$\frac{6}{17} \times \frac{3}{2}$$

Question 2.
Ashokrao planted bananas on $$\frac { 2 }{ 7 }$$ of his field of 21 acres. What is the area of the banana plantation?
Solution:
Area of banana plantation is $$\frac { 2 }{ 7 }$$ of 21
∴ Area of banana plantation

∴ Area of banana plantation is 6 acres

Question 3.
Of the total number of soldiers in our army, $$\frac { 4 }{ 9 }$$ are posted on the northern border and one-third of them on the north-eastern border. If the number of soldiers in the north is 5,40,000, how many are posted in the north-east?
Solution:
Number of soldiers posted on northern border = 5,40,000
Since, number of soldiers in north-east = one third of the soldiers on northern border
∴ Number of soldiers in the north-east

∴ The number of soldiers in the north-east is 1,80,000.

## Maharashtra Board Practice Set 25 Class 6 Maths Solutions Chapter 9 HCF-LCM

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 9 HCF-LCM Class 6 Practice Set 25 Answers Solutions.

## 6th Standard Maths Practice Set 25 Answers Chapter 9 HCF-LCM

Question 1.
Find out the LCM of the following numbers.
i. 9,15
ii. 2,3,5
iii. 12,28
iv. 15,20
v. 8,11
Solution:
i. Multiples of 9 = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90
Multiples of 15 = 15, 30, 45
∴ LCM of 9 and 15 = 45

ii. Multiples of 2 = 2, 4,6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30
Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Multiples of 5 = 5, 10, 15, 20, 25, 30
∴ LCM of 2,3 and 5 = 30

iii. Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120
Multiples of 28 = 28, 56, 84
∴ LCM of 12 and 28 = 84

iv. Multiples of 15 = 15, 30, 45, 60, 75, 90, 105, 120
Multiples of 20 = 20, 40, 60
∴ LCM of 15 and 20 = 60

v. Multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96
Multiples of 11 = 11, 22, 33, 44, 55, 66, 77, 88
∴ LCM of 8 and 11 = 88

Question 2.
Solve the following problems:
i. On the playground, if the children are made to stand for drill either 20 to a row or 25 to a row, all rows are complete and no child is left out. What is the lowest possible number of children in that school?

ii. Veena has some beads. She wants to make necklaces with an equal number of beads in each. If She makes necklaces of 16 or 24 or 40 beads, there is no bead left over. What is the least number of beads with her?

iii. An equal number of laddoos have been placed in 3 different boxes. The laddoos in the first box were distributed among 20 children equally, the laddoos in the second box among 24 children and those in the third box among 12 children. Not a single laddoo was left over. What was the minimum number of laddoos in the three boxes altogether?

iv. We observed the traffic lights at three different squares on the same big road. They turn green every 60 seconds, 120 seconds and 24 seconds. When the signals are switched on at 8 o’clock in the morning, all the lights were green. How long after that will all three signals turn green simultaneously again?

v. Given the fractions $$\frac { 13 }{ 45 }$$ and $$\frac { 22 }{ 75 }$$. Write their equivalent fractions with same denominators and add the fractions.
Solution:
i. The lowest possible number of children is equal to the lowest common multiple of 20 and 25.
Multiples of 20 = 20, 40, 60, 80, 100, 120, 140, 160, 180, 200
Multiples of 25 = 25, 50, 75, 100
∴ LCM of 20 and 25 = 100
∴ The least number of students in the school is 100.

ii. The least number of beads with Veena is equal to the lowest common multiple of 16,24 and 40.
Multiples of 16 = 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, 240, 256, 272, 288
Multiples of 24 = 24, 48, 72, 96, 120, 144, 168, 192, 216, 240
Multiples of 40 = 40, 80, 120, 160, 200, 240
∴ LCM of 16, 24 and 40 = 240
∴ The least number of beads with Veena are 240.

iii. The lowest common multiple of 20,24 and 12 gives the minimum number of laddoos in one box.
Multiples of 20 = 20, 40, 60, 80, 100,120, 140, 160, 180, 200
Multiples of 24 = 24, 48, 72, 96, 120
Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120
∴ LCM of 20, 24 and 12 = 120
∴ Minimum number of ladoos in 1 boxes =120
∴ Minimum number of ladoos in 3 boxes = 3 x 120 = 360
∴ The minimum number of ladoos in 3 boxes are 360.

iv. All three signals will turn green for lowest common multiple of 60 seconds, 120 seconds and 24 seconds.
Multiples of 60 = 60, 120, 180, 240, 300, 360, 420, 480
Multiples of 120 = 120, 240, 360
Multiples of 24 = 24, 48, 72, 96, 120
LCM of 60, 120 and 24 = 120
Since, 60 seconds = 1 minute
∴ 120 seconds = 2 minutes
∴ The signals will turn green simultaneously again after 120 seconds i.e. 2 minutes.

v. The lowest common multiple of 45 and 75 gives the same denominator.
Multiples of 45 = 45, 90, 135, 180, 225, 270, 315, 360, 405, 450
Multiples of 75 = 75, 150, 336
∴ LCM of 45 and 75 = 225

#### Maharashtra Board Class 6 Maths Chapter 9 HCF-LCM Practice Set 25 Intext Questions and Activities

Question 1.
Pravin, Bageshri and Yash are cousins who live in the same house. Pravin is an Army Officer. Bageshri is studying in a Medical College in another city. Yash lives in a nearby town in a hostel. Pravin can come home every 120 days.
Bageshri comes home every 45 days and Yash, every 30 days. All three of them left home at the same time on the 15th of June 2016. Their parents said, “We shall celebrate like a festival the day you all come home together.” Mother asked Yash, “What day will that be?”
Yash said, “The number of days after which we come back together must be divisible by 30, 120. That means we shall be back together on the 10th of June next year. That will certainly be a for us!”
How did Yash find the answer? (Textbook pg. no. 49)

Solution:
The day when Pravin, Bageshri and Yash come back together is lowest common multiple of 30, 45 and 120.
Multiples of 30: 30, 60, 90, 120, 150, 180,210, 240, 270, 300, 330, 360
Multiples of 45: 45, 90, 135, 180, 225, 270, 315, 360
Multiples of 120: 120, 240, 360
∴ They will come together after 360 days
Day when they left home = 15th June
∴ Day when they come back together = 15th June + 360 days
= 10th June next year
∴ Pravin, Bageshri and Yash will come back together on 10th June next year.

Question 2.
A Maths Riddle! (Textbook pg. no. 50)
We have four papers. On each of them there is a number on one side and some information on the other. The numbers on the papers are 7, 2, 15, 5. The information on the papers is given below in random order.
i. A number divisible by 7
ii. A prime number
iii. An odd number
iv. A number greater than 100
If the number on every paper is mismatched with the information on its other side, what is the number on the paper which says ‘A number greater than 100?
Solution:

 Analysis Reason Outcome The paper having information (iii) ‘an odd number’ can be mismatched with the number ‘2’ from the other available options. Only the number ‘2’ is an even number, while the rest are odd numbers. The number ‘2’ and (iii) ‘an odd number’ will appear on the opposite sides of the same paper. Now, we are left with the numbers 7, 15 and 5. The paper having information (i) ‘a number divisible by 7′ can be mismatched with the number ‘5’. The number ‘5’ is not divisible by 7. The number ‘5’ and (i) ‘a number divisible by 7’ will appear on the opposite sides of the same paper. Now, we are left with the numbers 7 and 15. The paper having information (ii) ‘a prime number’ can be mismatched with the number ‘15’. The number ‘15’ is not a prime number. Hence, the number ‘15’ and (ii) ‘a prime number’ will appear on the opposite sides of the same paper.

## Maharashtra Board Practice Set 11 Class 6 Maths Solutions Chapter 4 Operations on Fractions

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 4 Operations on Fractions Class 6 Practice Set 11 Answers Solutions.

## 6th Standard Maths Practice Set 11 Answers Chapter 4 Operations on Fractions

Question 11.
What fractions do the points A and B show on the number lines below?

Solution:
(1) Each unit is divided in 6 parts
A is 5th division from 0
∴ $$A=\frac { 5 }{ 6 }$$

B is 10th division from 0
∴ $$B=\frac { 10 }{ 6 }$$

(2) Each unit is divided in 5 parts
A is 3rd division from 0
∴ $$A=\frac { 3 }{ 5 }$$

B is 7th division from 0
∴ $$B=\frac { 7 }{ 5 }$$

(3) Each unit is divided in 7 parts
A is 10th division from 0
∴ $$A=\frac { 10 }{ 7 }$$

B is 3rd division from 0
∴ $$B=\frac { 3 }{ 7 }$$

Question 2.
Show the following fractions on the number line:
i. $$\frac{3}{5}, \frac{6}{5}, 2 \frac{3}{5}$$
ii. $$\frac{3}{4}, \frac{5}{4}, 2 \frac{1}{4}$$
Solution:
i. $$\frac{3}{5}, \frac{6}{5}, 2 \frac{3}{5}$$

ii. $$\frac{3}{4}, \frac{5}{4}, 2 \frac{1}{4}$$

Maharashtra Board Class 6 Maths Chapter 4 Operations on Fractions Practice Set 11 Intext Questions and Activities

Question 1.
If we want to show the fractions $$\frac{3}{10}, \frac{9}{20}, \frac{19}{40}$$ on the number line, how big should the unit be? (Textbook pg. no. 24)
Solution:
The denominators of the given fractions are not equal.
The numbers in the denominators 10, 20 and 40 have common multiple 40.
∴ Making the denominators equal, we get

∴ To represent these fractions on the numbers line, each main unit should be divided into 40 equal sub-units.

Therefore,
$$\frac{3}{10}=\frac{12}{40}$$ is represented on 12th mark from 0.
$$\frac{9}{20}=\frac{18}{40}$$ is represented on 18th mark from 0 and
$$\frac { 19 }{ 40 }$$ is represented on 19 mark from 0.

## Maharashtra Board Practice Set 20 Class 6 Maths Solutions Chapter 7 Symmetry

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.

## Maharashtra Board Practice Set 23 Class 6 Maths Solutions Chapter 9 HCF-LCM

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 9 HCF-LCM Class 6 Practice Set 23 Answers Solutions.

## 6th Standard Maths Practice Set 23 Answers Chapter 9 HCF-LCM

Question 1.
Write all the factors of the given numbers and list their common factors:
i. 12, 16
ii. 21, 24
iii. 25, 30
iv. 24, 25
v. 56, 72
Solution:
i. Factors of 12 = 1, 2, 3, 4, 6, 12
Factors of 16 = 1, 2, 4, 8, 16
∴ Common factors of 12 and 16 = 1, 2, 4

ii. Factors of 21 = 1, 3, 7, 21
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
∴ Common factors of 21 and 24 = 1, 3

iii. Factors of 25 = 1, 5, 25
Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30
∴ Common factors of 25 and 30 = 1, 5

iv. Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 25 = 1,5, 25
∴ Common factor of 24 and 25 = 1

v. Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56
Factors of 72 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
∴ Common factors of 56 and 72 = 1, 2, 4, 8

#### Maharashtra Board Class 6 Maths Chapter 9 HCF-LCM Practice Set 23 Intext Questions and Activities

Question 1.
In the empty boxes, write the proper words: dividend, divisor, quotient, remainder. (Textbook pg. no. 46)

Solution:

When we divide 36 by 4, the remainder is zero. Therefore, 4 is a factor of 36 and 36 is a multiple of 4. But, when we divide 65 by 9, the remainder is not zero. Therefore, 9 is not a factor of 65. Also, 65 is not a multiple of 9.

Question 2.
Write all the factors of the numbers 36 and 48. Also, list their common factors. (Textbook pg. no. 46)
Solution:
36 = 1 × 36
= 2 × 18
= 3 × 12
= 4 × 9
= 6 × 6

48 = 1 × 48
= 2 × 24
= 3 × 16
= 4 × 12
= 6 × 8

∴ Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Common factors of 36 and 48: [1] ,[2], [3], [4], [6], [12]

## Maharashtra Board Practice Set 32 Class 6 Maths Solutions Chapter 13 Profit-Loss

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
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.
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.

## Maharashtra Board Practice Set 10 Class 6 Maths Solutions Chapter 4 Operations on Fractions

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.
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 $$\left[4-2+\frac{1}{4}-\frac{1}{2}\right]$$.

## Maharashtra Board Practice Set 31 Class 6 Maths Solutions Chapter 13 Profit-Loss

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
∴ 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.

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.

## Maharashtra Board Practice Set 22 Class 6 Maths Solutions Chapter 8 Divisibility

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 ✓

## Maharashtra Board Practice Set 21 Class 6 Maths Solutions Chapter 7 Symmetry

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.