Class 5

Decimal Fractions Class 5 Problem Set 41 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 41 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 9 Decimal Fractions

Question 1.
Convert the following into decimal fractions and add them.

(1) ‘One and a half metre’ and ‘two and a half metres’
Solution:

(2) ‘Five and three quarter rupees’ and ‘seven and a quarter rupees’
Solution:

(3) ‘Six and a half metres’ and ‘three and three quarter metres’.
Solution:

Question 2.
(1) 23.4 + 87.9

(2) 35.74 + 816.6
Solution:

(3) 6.95 + 74.88
Solution:

(4) 41.03 + 9.98
Solution:

Question 3.
(1) 51.4 cm + 68.5 cm
Solution:

(2) 94.7 m + 1738.45 m
Solution:

(3) 5158.75 + `841.25
Solution:

Subtraction of decimal fractions

Study the subtraction of decimal fractions given below.

8 hundredths cannot be subtracted from 1 hundredth, so 1 tenth (or 10 hundredths) from 4 tenths are borrowed. The borrowed 10 hundredths and the original one hundredth make 11 hundredths. 11 hundredths minus 8 hundredths are 3 hundredths. They are written in the hundredths place under the line. The rest of the subtraction is carried out using the same method.

Decimal Fractions Problem Set 41 Additional Important Questions and Answers

Convert the following into decimal fractions and add them.

(1) ‘Fourteen and a half rupees’ and ‘Fifteen and a half rupees’.
Solution:

(2) ‘Three quarters’ and ‘a half’
Solution:

Add the following:

(1) 37.84 + 12.16
Solution:

(2) 328.69 + 84.84
Solution:

Solve the following:

(1) 304.86 m + 70.94 m
Solution:

(2) 79.56 cm + 19.65 cm
Solution:

(3) ₹ 64.79 + ₹  49.5
Solution:

Class 5 Maths Solution Maharashtra Board

• Decimal Fractions Problem Set 36 Class 5 Maths Solutions
• Decimal Fractions Problem Set 37 Class 5 Maths Solutions
• Decimal Fractions Problem Set 38 Class 5 Maths Solutions
• Decimal Fractions Problem Set 39 Class 5 Maths Solutions
• Decimal Fractions Problem Set 40 Class 5 Maths Solutions
• Decimal Fractions Problem Set 41 Class 5 Maths Solutions
• Decimal Fractions Problem Set 42 Class 5 Maths Solutions

Addition and Subtraction Class 5 Problem Set 8 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 3 Addition and Subtraction

Question 1.
Add the following:

(1) 42,311 + 65,36,624

Answer:
65,78,635

(2) 3,17,529 + 8,04,613

Answer:
11,22,142

(3) 12,42,746 + 4,83,748

Answer:
17,26,494

(4) 24,12,636 + 23,19,058

Answer:
47,31,694

(5) 2,654 + 71,209 + 5,03,789
1 1 2
2 6 5 4
+
7 1 2 0 9
+
5 0 3 7 8 9
Answer:
5 7 7 6 5 2

(6) 29 + 726 + 51,36,274
1 1 1
_______________
2 9
+
7 2 6
+
5 1 3 6 2 7 4
Answer:
5 1 3 7 0 2 9

(7) 14,02,649 + 524 + 28,13,749
1 1 1 2
_____________
1 4 0 2 6 4 9
+
5 2 4
+
2 8 1 3 7 4 9
Answer:
4 2 1 6 9 2 2

(8) 23,45,678 + 9,87,654
1 1 1 1 1 1
_____________
2 3 4 5 6 7 8
+
9 8 7 6 5 4
Answer:
3 3 3 3 3 3 2

(9) 22 + 6,047 + 3,84,527
1 1
2 2
+
6 4 0 7
+
3 8 4 5 2 7
Answer:
3 9 0 9 5 6

(10) 2,345 + 65,432 + 76,54,369
1 1 1 1 1
_________
2 3 4 5
+
6 5 4 3 2
+
7654369
Answer:
7 7 2 2 1 4 6

Study the following word problem.

During the polio eradication campaign, 3,17,658 children were given the polio vaccine in one District and 2,04,969 children in another. Altogether, how many children got the vaccine?
3 1 7 6 5 8
+
2 0 4 9 6 9
___________
5 2 2 6 2 7
___________
Altogether, 5,22,627 children got the vaccine.

Addition and Subtraction Problem Set 8 Additional Important Questions and Answers

Question 1.
Add the following:

(1) 4,506 + 3,82,459 + 6,12,999
Solution:
1 1 2
_________
4 5 0 6
+
3 8 2 4 5 9
+
6 1 2 9 9 9
Answer:
9 9 9 9 6 4

(2) 983 + 4,50,703 + 5,48,313
Solution:
1
________
9 0 0 3
+
4 5 0 7 0 3
+
5 4 00 3 1 3
Answer:
9 9 9 9 9 9

Class 5 Maths Solution Maharashtra Board

• Addition and Subtraction Problem Set 7 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 8 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 9 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 10 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 11 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 12 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 13 Class 5 Maths Solutions

Problems on Measurement Class 5 Problem Set 47 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 11 Problems on Measurement

Question 1.
For his birthday, Ajay gave 20 l 450 ml of milk to the children in an Ashramshala and 28 l 800 ml to the children in an orphanage. How much milk did Ajay donate?
Solution:

l	ml
1
2 0 + 2 8	4 5 0 8 0 0
4 9	2 5 0

450 ml + 800 ml
= 1250 ml
= 11 + 250 ml

∴ Ajay donated 49 l 250 ml milk

Question 2.
Under the Rural Cleanliness Mission, college students cleaned 1 km 750m of a village road that is 2 km 575m long. How much remained to be cleaned?
Solution:

km	m
1	1 5 7 5
2 – 1	5 7 5 7 5 0
0	8 2 5

750 m cannot be subtracted from 575 m. So, convert 1 km = 1000 m.
∴ 825 m remained to be cleaned

Question 3.
Babhulgaon used 21,250 liters of treated waste water in the fields. Samvatsar used 31,350 litres of similar water. How much treated waste water was used in all?
Solution:
2 1 2 5 0 litres Babhulgaon used
+ 3 1 3 5 0 litres Samvatsar used
___________
5 2 6 0 0
___________

∴ 52,600 litres of waste water used in all

Question 4.
If half a litre of milk costs 22 rupees, how much will 7 litres cost?
Solution:
\(\begin{array}{l}\frac{1}{2}+\frac{1}{2}=\frac{1+1}{2}=\frac{2}{2}=1 \text { litre } \\ 22+22=₹ 44\end{array}\)
That is, 1 litre cost ₹ 44
∴ 7 litres costs 44 x 7 = ₹ 308
∴ 7 litres costs ₹ 308

Question 5.
If the speed of a motorcycle is 40 km per hour, how far will it travel in an hour and a quarter?
Solution:
Hour and quarter = 1 + \(\frac{1}{4}\) hours
= 40 km + \(\frac{1}{4}\) x 40 km
= 40 km + 10 km
= 50 km
∴ Motorcycle will travel in a hour and a quarter 50 km

Question 6.
If a man walks at a speed of 4 kmph, how long will it take him to walk 3 km?
Solution:
1 km = 1000 m
4 km in 1 hour, 4 km in 60 minutes
That is
2 km in 30 minutes
+ 1 km in 15 minutes
_______________________
3 km in 45 minutes

That is 1 km in 15 minutes Hence, 3
km in 15 x 3 = 45 min

∴ 3 km in 15 x 3 = 45 min

Question 7.
If a rickshaw travels at a speed of 30 kmph, how far will it travel in three quarters of an hour?
Solution:
30 kmph means
In 60 minutes 30 km and 30 minutes 15 km
and 15 minutes \(\frac{15}{2}=\frac{15 \times 5}{2 \times 5}=\frac{75}{10}\) = 7.5 km
∴ In 45 minutes 15 km + 7.5 km = 22.5 km

Question 8.
During Cleanliness Week, children cleaned the public park in their town. They collected three quarter kilograms of plastic bags and five and a half kilograms of other garbage. How much garbage did they collect in all?
Solution:

Question 9.
If one shirt needs 2 m 50cm of cloth, how much cloth do we need for 5 shirts?
Solution:

∴ 12 m 50 cm cloth needs

Question 10.
If a car travels 60 km in an hour, how far will it travel in
(1) 2 hours?
(2) 15 minutes?
(3) half an hour?
(4) three and a half hours?
Solution:
60 kmph
In 60 minutes 60 km
Hence, 1 minute 1 km
(1) 2 hours = 2 x 60 = 120 km
(2) In 15 minutes = 15 km
(3) In half an hour 60 ÷ 2 = 30 km
(4) In three and half hours
= 3 x 60 + 30
= 180 + 30
= 210 km

∴ (1) 120 km
(2) 15 km
(3) 30 km
(4) 210 km

Question 11.
If one gold bangle is made from 12 grams 250 milligrams of gold, how much gold will be needed to make 8 such bangles? (1000mg = 1 g)
Solution:

∴ 98 grams gold needed

Question 12.
How many pouches of 20g cloves each can be made from 1 kg 240g of cloves?
Solution:
1 kg 240 gm
= 1000 gm + 240 gm
= 1240 gm

∴ pouches can be made

Question 13.
Seema’s mother bought 2m 70cm of cloth for a kurta and 2 m 40cm for a shirt. How much cloth did she buy in all?
Solution:
70 cm + 40 cm
= 110 cm
= 1 m 10 cm

m	cm
1
2 + 2	7 0 4 0
5	1 0

cloth for Kurta
cloth for Shirt

∴ 5 m 10 cm cloth in all

Question 14.
A water tank holds 125 l of water. If 97 l 500 ml of the water is used, how much water remains in the tank?
Solution:
1 litre = 1000 ml

water tank holds
water used
water remain

∴ 27 l 500 ml water remain in tank

Question 15.
Harminder bought 57 kg 500g of wheat from one shop and 36 kg 800 g of wheat from another shop. How much wheat did he buy altogether?
Solution:

bought from 1 shop
bought from another shop
500 + 800 = 1300 gm
= 1000 + 300
= 1 kg 300 gm

∴ 94 kg 300 gm bought altogether

Question 16.
Renu took part in a 100m race. She tripped and fell after running 80 m 50 cm. How much distance did she have left to run?
Solution:

m	cm
9 9	1 0 0
1 0 0 – 8 0	0 0 5 0
1 9	5 0

Borrow l m = 100 cm
So, 100 m = 99 m + 100 cm
Total distance to run
Distance covered
Distance left to run

∴ 19 m 50 cm distance left to run

Question 17.
A sack had 40kg 300 grams of vegetables. There were 17kg 700 g potatoes, 13 kg 400g cabbage and the rest were onions. What was the weight of the onions?
Solution:

∴ Weight of onions is 9 kg 200 gm

Question 18.
One day, Gurminder Singh walked 3 km 750m and Parminder Singh walked 2km 825m. Who walked farther and by how much?
Solution:

∴ Gurminder walked more by 925 metres

Question 19.
Suresh bought 3kg 250g of tomatoes, 2 kg 500g of peas and 1kg 750g of cauliflower. How much was the total weight of the vegetables he bought?
Solution:

∴ Total weight 7 kg 500 gm

Question 20.
Jalgaon, Bhusawal, Akola, Amravati and Nagpur lie serially on a certain route. The distances between Akola and these other places are given below.

Use them to make word problems and solve the problems.
Amravati – 95 km, Bhusawal – 154 km,
Nagpur – 249 km, Jalgaon – 181 km
Solution:
(1) What is the distance between Bhusaval and Nagpur?
249 km – 154 km = 95 km

∴ The distance between Bhusaval and Nagpur is 95 km

(2) What is the distance between Amravati and Jalgaon?
181 km – 95 km = 86 km

∴ The distance between Amravati and Jalgaon is 86 km.

Question 21.
Complete the following table and prepare the total bill.

Solution:

Activity

• You have 1 kg of potatoes. Find out which other ingredients you will need to make potato vadas and approximately how much of each ingredient you will need. Also find out approximately how much each ingredient will cost and how many vadas you will be able to make.
• Fix a 1 m long stick in an open field. Measure the shadow of the stick at 9:00 in the morning, at 12:00 noon, at 3:00 in the afternoon and at 5:00 in the evening. Observe at which time of the day the shadow is shortest and at what time, it is longest.
• Measure the length of a pen refill.

Problems on Measurement Problem Set 47 Additional Important Questions and Answers

Question 1.
One can contains 30 l 560 ml of milk, while second contains 251890 ml of milk and third one contains 20 l 760 ml of milk. How much milk is there in the three cans together?
Solution:

∴ 77 l 210 ml total milk

Question 2.
Add the following:
(1) ₹ 13, 85 paise + ₹ 16, 40 paise
(2) 15 kg 280 gm + 18 kg 920 gm
(3) 24 l 690 ml + 25 l 780 ml
(4) 22 km 750 m + 27 km 500 m
(5) 17 m 40 cm + 19 m 85 cm
(6) 38 cm 8 mm + 17 cm 2 mm
(7) 10 km 950 m + 15 km 125 m
(8) 83 kg 468 gm + 109 kg 532 gm
Answer:
(1) ₹ 30, 25 paise
(2) 34 kg 200 gm
(3) 50 1 470 ml
(4) 50 km 250 m
(5) 37 m 25 cm
(6) 56 cm
(7) 26 km 75 m
(8) 193 kg

Question 3.
Subtract the following:
(1) ₹ 21, 30 paise – ₹ 13, 80 paise
(2) 16 kg 130 gm – 9 kg 250 gm
(3) 9 l 350 ml – 5 l 470 ml’
(4) 41 m 10 cm – 14 m 40 cm
(5) 38 km 175 m – 20 km 365 m
(6) 27 cm 5 mm – 11 cm 8 mm
(7) 28 km 725 m – 13 km 590 m
(8) 380 kg – 232 kg 730 gm
Answer:
(1) ₹ 7, 50 paise
(2) 6 kg 880 gm
(3) 51 880 ml
(4) 26 m 30 cm
(5) 17 km 810 m
(6) 15 cm 7 mm
(7) 15 km 135 m
(8) 147 kg 270 gm

Question 4.
Fill in the blanks:
(1) 1250 m = …………………… km …………………… m
(2) 2.5 m = …………………… m …………………… cm
(3) 3 l 50 ml = …………………… ml
(4) ₹ 2.5 = …………………… paise
Answer:
(1) 1 km 250 m
(2) 2 m 50 cm
(3) 3050 ml
(4) 250 paise

Question 5.
(A) Match the following:

‘A’	‘B’
(1) Potato 3.5 kg, rate per kg ₹ 12	(a) ₹ 40
(2) Onion 2 kg, rate per kg ₹ 20.50	(b) ₹ 42
(3) Vegetables 2.5 kg, rate per kg ₹ 16	(c) ₹ 39
(4) Others 6.5 kg, rate per kg ₹ 6	(d) ₹ 41

Answer:
(1 – b),
(2 – d),
(3 – a),
(4 – c)

(B) Match the following:

‘A’	‘B’
(1) Half metre	(a) 5 cm
(2) Half kilometre	(b) 50 cm
(3) 50 millimetre	(c) 500 cm
(4) 5 kilometre	(d) 500 m
(5) 5 metre	(e) 5000 m

Answer:
(1 – b),
(2 – d),
(3 – a),
(4 – e),
(5 – c)

Class 5 Maths Solution Maharashtra Board

• Problems on Measurement Problem Set 46 Class 5 Maths Solutions
• Problems on Measurement Problem Set 47 Class 5 Maths Solutions
• Perimeter and Area Problem Set 48 Class 5 Maths Solutions
• Perimeter and Area Problem Set 49 Class 5 Maths Solutions
• Perimeter and Area Problem Set 50 Class 5 Maths Solutions

Addition and Subtraction Class 5 Problem Set 12 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 3 Addition and Subtraction

Question 1.
Solve the following word problems:

1. Prathamesh wants to buy a laptop worth 27,450 rupees. He has 22,975 rupees. What is the amount he still needs to be able to buy the laptop?
Solution:
₹ 2 7 4 5 0 Laptop worth
–
₹ 2 2 9 7 5 Prathmesh has
0 4 4 7 5 Require more
Answer:
₹ 04,475 amount require to buy the laptop,

2. A company produced 44,730 scooters in a certain year and 43,150 in the next. How many more scooters did they produce in the previous year?
Solution:
4 4 7 3 0 In previous year
–
4 3 1 5 0 In next year
0 1 5 8 0
Answer:
1,580 scooters produced more in the previous year.

3. In a certain city, the number of men is 16,37,856 and the number of women is 16,52,978. By how many does the number of women exceed the number of men?
Solution:
1 6 5 2 9 7 8 Women
–
1 6 3 7 8 5 6 Men
0 0 1 5 1 2 2
Answer:
15,122 women more than number of men.

4. An organization decided to collect 25,00,000 rupees for a certain project. They collected 26,57,340 through donations and other kinds of aid. By how much did they exceed their target?
Solution:
2 6 5 7 3 4 0 collected
–
2 5 0 0 0 0 0 decided to collect
0 1 5 7 3 4 0 collected more
Answer:
1,57,340

5. Use the numbers 23,849 and 27,056 to make a subtraction problem. Solve the problem.
Solution:
In a certain shop the price of a computer was ₹ 23,849 and that of TV set is ₹ 27,056. Price of TV set is how much more than that of a computer.
₹ 2 7 0 5 6 Price of T.V. sets
₹ 2 3 8 4 9 Price of computer
0 3 2 0 7 Price is more
Answer:
Price of TV set is more than computer by ₹ 3,207

Mixed examples
Study the following solved examples.

Example (1)
4,13,758 + 2,09,542 – 5,16,304
4,13,758 + 2,09,542 – 5,16,304 = 1,06,996

Example (2)
345678 – 162054 + 600127
345678 – 162054 + 600127 = 7,83,751

In these examples, both operations, addition and subtraction, have to be done. They are done in the order in which they are given. In actual cases, we need to consider the specific problem to decide which operation must be done first.

Example (3)
The total amount spent on building a certain house was ₹ 87,14,530. Of this amount, ₹ 24,72,615 were spent on buying the plot of land, ₹ 50,43,720 on the construction material and the rest on labour charges. What was the amount spent on labour?

Method : 1
8 7 1 4 5 3 0 → Total amount spent
–
2 4 7 2 6 1 5 → Cost of plot
6 2 4 1 9 1 5 → Cost of material and labour

6 2 4 1 9 1 5 → Cost of material and labour
–
5 0 4 3 7 2 0 → Cost of material
1 1 9 8 1 9 5 → Amount spent on labour

Method : 2
2 4 7 2 6 1 5 Cost of plot
+
5 0 4 3 7 2 0 Cost of material
7 5 1 6 3 3 5 Cost of plot and material

8 7 1 4 5 3 0 Total amount spent
–
7 5 1 6 3 3 5 Cost of plot and material
1 1 9 8 1 9 5 Amount spent on labour

Let us verify our answer.
2 4 7 2 6 1 5 Cost of plot
+
5 0 4 3 7 2 0 Cost of material
+
1 1 9 8 1 9 5 Amount spent on labour
8 7 1 4 5 3 0 Total cost

The sum total of all the amounts spent tallies with the given total cost. It means that our answer is correct.

Addition and Subtraction Problem Set 12 Additional Important Questions and Answers

Solve the following word problems:

Question 1.
Jethalal purchased goods for ₹ 53,25,675 sold it for ₹ 62,14,563. How much he obtained more in this transaction?
Solution:
6 2 1 4 5 6 3 Sold price
5 3 2 5 6 7 5 Purchased price
8 8 8 8 8 8 Obtained more
Answer:
₹ 8,88,888

Question 2.
In an election candidate A got 13,90,211 votes and candidate B got 8,57,143 votes. By how many votes the winner A defeated the looser B?
Solution:
1 3 9 0 2 1 1 Votes obtained by A
–
8 5 7 1 4 3 Votes obtained by B
0 5 3 3 0 6 8 Votes more obtained by A
Answer:
5,33,068

Class 5 Maths Solution Maharashtra Board

• Addition and Subtraction Problem Set 7 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 8 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 9 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 10 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 11 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 12 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 13 Class 5 Maths Solutions

Fractions Class 5 Problem Set 20 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 20 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 5 Fractions

Question 1.
Add the following
\(\text { (1) } \frac{1}{5}+\frac{3}{5}\)
Answer:
\(\frac{1}{5}+\frac{3}{5}=\frac{1+3}{5}=\frac{4}{5}\)

\(\text { (2) } \frac{2}{7}+\frac{4}{7}\)
Answer:
\(\frac{2}{7}+\frac{4}{7}=\frac{2+4}{7}=\frac{6}{7}\)

\(\text { (3) } \frac{7}{12}+\frac{2}{12}\)
Answer:
\(\frac{7}{12}+\frac{2}{12}=\frac{7+2}{12}=\frac{9}{12}\)

\(\text { (4) } \frac{2}{9}+\frac{7}{9}\)
Answer:
\(\frac{2}{9}+\frac{7}{9}=\frac{2+7}{9}=\frac{9}{9}=1\)

\(\text { (5) } \frac{3}{15}+\frac{4}{15}\)
Answer:
\(\frac{3}{15}+\frac{4}{15}=\frac{3+4}{15}=\frac{7}{15}\)

\(\text { (6) } \frac{2}{7}+\frac{1}{7}+\frac{3}{7}\)
Answer:
\(\frac{2}{7}+\frac{1}{7}+\frac{3}{7}=\frac{2+1+3}{7}=\frac{6}{7}\)

\(\text { (7) } \frac{2}{10}+\frac{4}{10}+\frac{3}{10}\)
Answer:
\(\frac{2}{10}+\frac{4}{10}+\frac{3}{10}=\frac{2+4+3}{10}=\frac{9}{10}\)

\(\text { (8) } \frac{4}{9}+\frac{1}{9}\)
Answer:
\(\frac{4}{9}+\frac{1}{9}=\frac{4+1}{9}=\frac{5}{9}\)

\(\text { (9) } \frac{5}{8}+\frac{3}{8}\)
Answer:
\(\frac{5}{8}+\frac{3}{8}=\frac{5+3}{8}=\frac{8}{8}=1\)

Question 2.
Mother gave \(\frac{3}{8}\) of one guava to Meena and \(\frac{2}{8}\) of the guava to Geeta. What part of the guava did she give them altogether?
Solution:
\(\frac{3}{8}+\frac{2}{8}=\frac{3+2}{8}=\frac{5}{8}\) given altogether
Answer:
\(\frac{5}{8}\) part of guava given altogether

Question 3.
The girls of Std V cleaned \(\frac{3}{4}\) of a field while the boys cleaned \(\frac{1}{4}\). What part of the field was cleaned altogether?
Solution:
Girls cleaned + Boys cleaned
\(\frac{3}{4}+\frac{1}{4}=\frac{3+1}{4}=\frac{4}{4}=1\)
Answer:
Full whole field cleaned altogether.

Subtraction of like fractions

A figure is divided into 5 equal parts and 4 of them are colored. That is, \(\frac{4}{5}\) part of the figure is coloured.

Now, we remove the colour from one of the coloured parts. That is, we subtract \(\frac{1}{5}\) from \(\frac{4}{5}\). The remaining coloured part is \(\frac{3}{5}\). Therefore, \(\frac{4}{5}\) – \(\frac{1}{5}\) = \(\frac{4-1}{5}\) = \(\frac{3}{5}\).

When subtracting a fraction from another like fraction, we write the difference between the numerators in the numerator and the common denominator in the denominator.

Example (1) Subtract : \(\frac{7}{13}\) – \(\frac{5}{13}\)

These two fractions have a common denominator. So, we shall subtract the second numerator from the first and write the denominator as it is.
\(\frac{7}{13}-\frac{5}{13}=\frac{7-5}{13}=\frac{2}{13}\)

Example (2) If Raju got \(\frac{5}{12}\) part of a sugarcane and Sanju got \(\frac{3}{12}\) part, how much was the extra part that Raju got?

To find out the difference, we must subtract.
\(\frac{5}{12}-\frac{3}{12}=\frac{5-3}{12}=\frac{2}{12}\). Thus, Raju got \(\frac{2}{12}\) extra.

Addition and Subtraction Problem Set 13 Additional Important Questions and Answers

\(\text { (1) } \frac{3}{6}+\frac{2}{6}+\frac{1}{6}\)
Answer:
\(\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{3+2+1}{6}=\frac{6}{6}=1\)

\(\text { (2) } \frac{4}{10}+\frac{1}{10}+\frac{3}{10}+\frac{2}{10}\)
Answer:
\(\frac{4}{10}+\frac{1}{10}+\frac{3}{10}+\frac{2}{10}=\frac{4+1+3+2}{10}=\frac{10}{10}=1\)

\(\text { (3) } \frac{1}{2}+\frac{1}{2}\)
Answer:
\(\frac{1}{2}+\frac{1}{2}=\frac{1+1}{2}=\frac{2}{2}=1\)

Solve the following word problems:

Question 1.
of journey travelled by A and of journey travelled by B. What part of the journey travelled by both field was cleaned altogether?
Solution:
Travelled by A + Travelled by B
\(\frac{3}{5}+\frac{2}{5}=\frac{3+2}{5}=\frac{5}{5}=1\)
Answer:
Full (whole) journey travelled by both.

Class 5 Maths Solution Maharashtra Board

• Fractions Problem Set 17 Class 5 Maths Solutions
• Fractions Problem Set 18 Class 5 Maths Solutions
• Fractions Problem Set 19 Class 5 Maths Solutions
• Fractions Problem Set 20 Class 5 Maths Solutions
• Fractions Problem Set 21 Class 5 Maths Solutions
• Fractions Problem Set 22 Class 5 Maths Solutions
• Fractions Problem Set 23 Class 5 Maths Solutions

Angles Class 5 Problem Set 27 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 27 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 6 Angles

Question 1.
Give two examples of parallel lines you can see in your environment.
Answer:
(i) Bars on the window.
(ii) Horizontal lines in the notebook are the examples of parallel lines.

Question 2.
Give two examples of perpendicular lines you can see in your environment.
Answer:
(i) The angles formed by a pole and its shadow on the ground.
(ii) The adjacent sides of a notebook.

Question 3.
Look at the pictures given below. Decide whether the lines given in each picture are parallel or perpendicular to each other and write the answer in the box.

Answer:

Angles Problem Set 27 Additional Important Questions and Answers

Question 1.
Give some examples of perpendicular lines from capital letters of the English alphabet.
Answer:

Question 2.
Complete the following:

Answer:
(1) ZDEF or ZFED, Vertex E, arms are ED

Question 3.
Which of the following figures show angle?

Answer:
Figure 3

Question 4.
(A) Measure the angles given below and write the measure in the given boxes:


Answer:
(1) 30°
(2) 135°
(3) 90°
(4) 50°
(5) 90°
(6) 150°

B) Classify the above figures according to types of the angles.
Answer:
Acute angles: (1) and (4),
Right angles: (3) and (5),
Obtuse angles: (2) and (6)

Question 5.
Draw and name the following angles with the help of a protractor:








Answer:
Students to draw angles.

Question 6.
Look at the pictures given below. Decide whether the lines given in each picture are parallel or perpendicular to each other and write the answer in the box:

Answer:
(1) Parallel lines
(2) Perpendicular lines
(3) Perpendicular lines
(4) Parallel lines

Question 7.
Say, true or false of the following:
(1) Parallel lines do not intersect each other.
(2) Pole and its shadow on the ground makes a cute angle.
(3) Angle between two parallel lines is 90°.
(4) Angle between two intersecting lines may or may not be 90°.
Answer:
(1) True
(2) False
(3) False
(4) True

Class 5 Maths Solution Maharashtra Board

• Angles Problem Set 24 Class 5 Maths Solutions
• Angles Problem Set 25 Class 5 Maths Solutions
• Angles Problem Set 26 Class 5 Maths Solutions
• Angles Problem Set 27 Class 5 Maths Solutions

Roman Numerals Class 5 Problem Set 1 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 1 Roman Numerals Problem Set 1 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 1 Roman Numerals

Question 1.
Write all the numbers from 1 to 20 using Roman numerals.
Answer:

Question 2.
Write the following numbers using international numerals.
(1) V
(2) VII
(3) X
(4) XIII
(5) XIV
(6) XVI
(7) XVIII
(8) IX
Answer:
(1) 5
(2) 7
(3) 10
(4) 13
(5) 14
(6) 16
(7) 18
(8) 9

Question 3.
Fill in the empty boxes.

Answer:

Question 4.
Write the numbers using Roman numerals.
(1) 9
(2) 2
(3) 17
(4) 4
(5) 11
(6) 18
Answer:
(1) IX
(2) II
(3) XVII
(4) IV
(5) XI
(6) XVIII

Question 5.
In the table below, each given number is written in international numerals and then again in Roman numerals. If it is written correctly in Roman numerals, put ‘/’ in the box under it. If not, put ‘X’ and correct it.

Answer:

Something more : L, C, D and M are also used as Roman numerals.

Activity : Apart from clocks and watches, where else do we see Roman numerals?

The decimal system of writing numbers
It is not easy to read or write numbers using Roman numerals. It also makes calculations difficult. You have learnt to write numbers using the ten digits, 0 to 9. In that system, the value of a digit depends upon its place in the number. This system of writing numbers is called the ‘decimal system’.

Mathematicians of ancient India invented the decimal system of writing numbers and began to use it. Later, this system was accepted in all parts of the world because it was simple and convenient.

Roman Numerals Problem Set 1 Additional Important Questions and Answers

Question 1.
Write all the numbers from 21 to 30 using roman numerals.
Answer:

Question 2.
Write the following numbers using international numerals.
(1) XXV
(2) XXIX.
Answer:
(1) 25
(2) 29

Question 3.
Fill in the empty boxes:
Answer:

Question 4.
Write the numbers using Roman Numerals:
(1) 1
(2) 21
(3) 10
Answer:
(1) I
(2) XXI
(3) X

Question 5.
In the table below, each given number is written in international numerals and then again in Roman numerals. If it is written correctly in Roman numerals put ‘S’ in the box under it. if not, put ‘x’ and correct it.
Answer:

Question 6.
Write all the numbers from 31 to 40 using Roman numerals.
Answer:

Question 7.
Write the following numbers using international numerals.
(1) VI
(2) XI
(3) XIV
(4) XVII
Answer:
(1) 6
(2) 11
(3) 14
(4) 17

Question 8.
Fill in the empty boxes:

Answer:
Four = IV, 16 = XVI, Twenty = XX, XV =15

Question 9.
Write the numbers using Roman numerals in the given boxes.
(1) 3 [ ]
(2) 8 [ ]
(3) 13 [ ]
(4) 16 [ ]
Answer:
(1) III
(2) VIII
(3) XIII
(4) XVI

Question 10.
State true or false of the following. If the statement is false then correct it in Roman numerals.
(1) 3 is written as IIV
(2) 14 is written as XIV
(3) 16 is written as XVI
(4) 19 is written as IXX
Answer:
(1) False (III)
(2) True
(3) True
(4) False (XIX)

Question 11.
Write the answers Roman numerals.
(1) III + IV = [ ]
(2) IV + V = [ ]
(3) IX – X = [ ]
(4) VI + X = [ ]
(5) III + II = [ ]
(6) V + V = [ ]
(7) XXX + XX = [ ]
(8) L + L = [ ]
Answer:
(1) VII
(2) IX
(3) XIX
(4) XVI
(5) V
(6) X
(7) L
(8) C

Question 12.
Match the columns:

Answer:
(1 – b),
(2 – a),
(3 – d),
(4 – c)

Question 13.
Circle the correct international numeral of the following:
(1) XXIV
(a) 14
(b) 20
(c) 24
Answer:
(c) 24

(2) VIII
(a) 8
(b) 207
(c) 19
Answer:
(a) 8

(3) XXIX
(a) 31
(b) 29
(c) 20
Answer:
(b) 29

(4) XXVII
(a) 27
(b) 28
(c) 26
Answer:
(a) 27

Question 14.
Circle the correct Roman numeral of the following:
(1) 36
(a) XXVI
(b) XXXVI
(c) XXXXVI
Answers:
(b) XXXVI

(2) 27
(a) XXXVI
(b) XXVII
(c) XXVI
Answers:
(c) XXVII

(3) 18
(a) XVIII
(b) XVI
(c) XVII
Answers:
(a) XVIII

(4) 605
(a)DCV
(b)CDV
(c) XXXXXV
Answers:
(a) DCV

Question 15.
Put >, < or = in the box.
(1) CVI [          ] CLX
(2) CLXIX [          ] CLXXI
(3) XLIX [          ] XXXIX
(4) MMXVI [          ] MMXXVI
(5) LXX VII [          ] LVIII
(6) MCMXL [          ] MCMXLV
Answers:
(1) <
(2) <
(3) >
(4) <
(5) >
(6) <

Class 5 Maths Solution Maharashtra Board

• Roman Numerals Problem Set 1 Class 5 Maths Solutions
• Number Work Problem Set 2 Class 5 Maths Solutions
• Number Work Problem Set 3 Class 5 Maths Solutions
• Number Work Problem Set 4 Class 5 Maths Solutions
• Number Work Problem Set 5 Class 5 Maths Solutions 
• Number Work Problem Set 6 Class 5 Maths Solutions

Fractions Class 5 Problem Set 19 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 5 Fractions

Write the proper symbol from < , > , or = in the box.

Answer:
=

Answer:
>

Answer:
<

Answer:
=

Answer:
>

Answer:
>

Answer:
>

Answer:
>

Answer:
>

Answer:
>

Answer:
=

Answer:
=

Answer:
>

Answer:
>

Answer:
<

Answer:
>

Addition of like fractions
Example (1) 3/7 + 2/7 = ?
Let us divide a strip into 7 equal parts. We shall colour 3 parts with one colour and 2 parts with another.
The part with one colour is 3/7, and that with the other colour is 2/7.
The total coloured part is shown by the fraction 5/7.
It means that, \(\frac{3}{7}+\frac{2}{7}=\frac{3+2}{7}=\frac{5}{7}\)

Example (2) Add : \(\frac{3}{8}+\frac{2}{8}+\frac{1}{8}\)
The total coloured part is \(\frac{3}{8}+\frac{2}{8}+\frac{1}{8}=\frac{3+2+1}{8}=\frac{6}{8}\)

When adding like fractions, we add the numerators of the two fractions and write the denominator as it is.
Example (3) Add : 2/6 + 4/6 \(\frac{2}{6}+\frac{4}{6}=\frac{2+4}{6}=\frac{6}{6}\)
However, we know that 6/6 means that all 6 of the 6 equal parts are taken. That is, 1 whole figure is taken. Therefore, 6/6 = 1.

Note that:
If the numerator and denominator of a fraction are equal, the fraction is equal to one.
That is why, \(\frac{7}{7}=1 ; \frac{10}{10}=1 ; \frac{2}{5}+\frac{3}{5}=\frac{2+3}{5}=\frac{5}{5}=1\)
Remember that, if we do not divide a figure into parts, but keep it whole, it can also be written as 1.
This tells us that \(1=\frac{1}{1}=\frac{2}{2}=\frac{3}{3}\) and so on.
You also know that if the numerator and denominator of a fraction have a common divisor, then the fraction obtained by dividing them by that divisor is equivalent to the given fraction.
\(\frac{5}{5}=\frac{5 \div 5}{5 \div 5}=\frac{1}{1}=1\)

Fractions Problem Set 19 Additional Important Questions and Answers

Question 1.

Answer:
>

Question 2.

Answer:
=

Class 5 Maths Solution Maharashtra Board

• Fractions Problem Set 17 Class 5 Maths Solutions
• Fractions Problem Set 18 Class 5 Maths Solutions
• Fractions Problem Set 19 Class 5 Maths Solutions
• Fractions Problem Set 20 Class 5 Maths Solutions
• Fractions Problem Set 21 Class 5 Maths Solutions
• Fractions Problem Set 22 Class 5 Maths Solutions
• Fractions Problem Set 23 Class 5 Maths Solutions

Three Dimensional Objects and Nets Class 5 Problem Set 51 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 13 Three Dimensional Objects and Nets

Question 1.
The first column shows a structure made of blocks. The other columns show different views of the structure in two dimensions. Say whether each view is from the front, from a side or from above.

Answer:

Question 2.
Draw three pictures of each of these three-dimensional objects – a table, a chair and a water bottle as viewed from the front, from a side and from above.
Answer:

Nets
Last year we saw that cutting some edges of a box and laying it out flat gives us the net from which it was made.
The two dimensional shape from which a three dimensional object can be made by folding is called the ‘net’ of that object.

1. By folding the cardboard shown below, along the lines shown in it, we get a three dimensional object (box). In this shape, all surfaces are square.
   An object of this shape is called a cube.
   
2. The net of another cardboard box is shown in the figure below. By folding along the lines in this net and joining the edges to each other, we can see that a three dimensional box is formed. The surfaces of this box are rectangular in shape.
   An object of this shape is called a cuboid.
   

Activity :
Draw the nets shown below on card sheet. Cut out the shapes and find out the shapes of the boxes they form.

A five-square net (Pentomino)

In the figure alongside, five squares of the same size are placed together with their sides joined.

Such an arrangement of five squares is called a ‘five-square net’ or a ‘pentomino’.

By folding along the edges of such a five-square net, an open box is formed.

Activity :
Some five-square nets are given below. Draw these nets on a card sheet. Make open boxes from these nets.

Try to find out other five-square nets that can be used to make open boxes.

A riddle

The net of a cube-shaped dice is given alongside. If a dice is made of this net, which of the following shapes will it definitely not resemble?

Chapter 12 Perimeter and Area Problem Set 49 Additional Important Questions and Answers

Question 1.
Draw the pictures of each of these three dimensional objects – Mobile, Oil tin as viewed from the front, from a side and from above.
Answer:

Question 2.
The three dimensional figure of block formation is shown in the figure along side. Draw as view from the front, from a side and from above (fig. drawn in answer part)
Answer:

Question 3.
Draw the nets shown below on card sheet. Cut out the shapes and find out the shapes of the boxes they form.
Answer:

Class 5 Maths Solution Maharashtra Board

• Three Dimensional Objects and Nets Problem Set 51 Class 5 maths Solutions
• Pictographs Problem Set 52 Class 5 maths Solutions
• Patterns Problem Set 53 Class 5 maths Solutions
• Preparation for Algebra Problem Set 54 Class 5 maths Solutions
• Preparation for Algebra Problem Set 55 Class 5 maths Solutions
• Preparation for Algebra Problem Set 56 Class 5 maths Solutions

Decimal Fractions Class 5 Problem Set 38 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 38 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 9 Decimal Fractions

Read the decimal fraction and write down the place value of each digit.

(1) 6.13
Answer:
Six point one three.
Here 6 is in the units place. Hence, its place value is 6 x 1 = 6
1 is in tenths place. Hence, its place value is
\(1 \times \frac{1}{10}=0.1\)
3 is in the hundredths place. Its place value is 3 x \(\frac{1}{100}\) = 0.03 100

(2) 48.84
Answer:
Fourty eight point eight four
Place value of 4 is 4 x 10 = 40 and of 8, it is 8 x 1 = 8
Place value of 8 is 8 x \(\frac{1}{10}=\frac{8}{10}\) = 0.8 and
Place value of 4 is 4 x \(\frac{1}{100}=\frac{4}{100}\) = 0.04

(3) 72.05
Answer:
Seventy two point zero five
Place value of 7 is 7 x 10 = 70 and of 2 is, it is 2 x 1 = 2
Place value of 5 is 5 x \(\frac{1}{100}=\frac{5}{100}\) = 0.05

(4) 3.4
Answer:
Three point four.
Place value of 3 is 3 x 1 = 3
Place value of 4 is 4 x \(\frac{1}{10}\) = 0.4

(5) 0.59
Answer:
Zero point five nine.
Place value of 5 is 5 x \(\frac{1}{10}=\frac{5}{10}\) = 0.5 and
Place value of 9 is 9 x \(\frac{1}{100}=\frac{9}{100}\) = 0.09

Use of decimal fractions

Sir : Now we will see how 24.50 equals 24 rupees and 50 paise. How many rupees is one paisa?
Sumit : 100 paise make one rupee, therefore, 1 paisa is one-hundredth of a rupee or 0.01 rupee.
Sir : And 50 paise are?
Sumit : 50 hundredths of a rupee, or 0.50 rupees, so 24.50 rupees is 24 rupees and 50 paise.
Sir : When a large unit of a certain quantity is divided into 10 or 100 parts to make smaller units, it is more convenient to write them in decimal form. As we just saw, 100 paise = 1 rupee. Similarly, 100 cm = 1 metre, so 75 cm = 0.75 m. 10 mm = 1 cm, so 1 mm = 0.1cm. 3 mm are 0.3 cm. 6.3 cm are 6 cm and 3 mm.

Now study the following table.

Decimal Fractions Problem Set 38 Additional Important Questions and Answers

(1) 12.34
Answer:
Twelve point three four.
Place value of 1 is 1 x 10 = 10
Place value of 2 is 2 x 1 = 2
Place value of 3 is 3 x \(\frac{1}{10}\) = 0.3
Place value of 4 is 4 x \(\frac{1}{100}\) = 0.04

(2) 369,58
Answer:
Three hundred sixty nine point five eight. Place value of 3, which is in the hundreds place is 3 x 100 = 300
Place value of 6, which is in the tens place is 6 x 10 = 60
Place value of 9, which is in the units place is 9 x 1 = 9
Place value of 5, which is in the tenths place is 5 x \(\frac{1}{10}\) = 0.5
Place value of 8, which is in the hundredths place is 8 x \(\frac{1}{100}\) = 0.08

(3) 5.5
Answer:
Five point five.
Place value of 5, which is in the units place is 5 x 1 = 5
Place value of 5, which is in the tenths place is 5 x \(\frac{1}{10}\) = 0.5

Class 5 Maths Solution Maharashtra Board

• Decimal Fractions Problem Set 36 Class 5 Maths Solutions
• Decimal Fractions Problem Set 37 Class 5 Maths Solutions
• Decimal Fractions Problem Set 38 Class 5 Maths Solutions
• Decimal Fractions Problem Set 39 Class 5 Maths Solutions
• Decimal Fractions Problem Set 40 Class 5 Maths Solutions
• Decimal Fractions Problem Set 41 Class 5 Maths Solutions
• Decimal Fractions Problem Set 42 Class 5 Maths Solutions