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Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 46 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 46

Question 1.
Add :

(1) ₹ 9, 50 paise + ₹ 14, 60 paise
Solution:

₹	Paise
1
9 + 14	5 0 6 0
2 4	1 0

50 paise + 60 paise
= 110 paise
= 1 ₹ 10 paise
∴ ₹ 24, 10 paise

(2) 6 cm 5 mm + 7 cm 9 mm
Solution:

cm	mm
1
6 + 7	5 9
1 4	4

5 mm + 9 mm
= 14 mm 14 mm
= 1 cm 4 mm
∴ 14 cm 4 mm

(3) 22 m 50 cm + 25 m 75 cm
Solution:

m	cm
1
2 2 + 2 5	5 0 7 5
4 8	2 5

50 cm + 75 cm
= 125 cm
= 1 m 25 cm
∴ 48 m 25 cm

(4) 15 km 740 m + 13 km 950 m
Solution:

km	m
1
1 5 + 13	7 4 0 9 5 0
2 9	6 9 0

740 m + 950 m
= 1690 m 1690 m
= 1km 690 m
∴ 29 km 690 m

(5) 25 kg 650 g + 29 kg 770 g
Solution:

kg	gm
1
2 5 + 29	6 5 0 7 7 0
5 5	4 2 0

650 gm + 770 gm
= 1420 gm
= 1 kg 420 gm
∴ 55 kg 420 gm

(6) 19l 840ml + 25l 250ml
Solution:

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

840 ml + 250 ml
= 1090 ml
= 11 + 90 ml
∴ 45 l 90 ml

Question 2.
Subtract :

(1) ₹ 19, 50 paise – ₹ 12, 60 paise
Solution:

₹	Paise
1 8	1 5 0
1 9 – 1 2	5 0 6 0
6	9 0

We cannot subtract 60 paise from 50 paise. So convert 1 ₹ into 100 paise.
₹ 6, 90 paise

∴ ₹ 6, 90 paise

(2) 24 cm 2 mm – 3 cm 8 mm
Solution:

cm	mm
2 3	1 2
2 4 – 3	2 8
2 0	4

We cannot subtract 8 mm from 2 mm. So, convert 1 cm = 10 mm

∴ 20 cm 4 mm

(3) 20 m 30 cm – 17 m 60 cm
Solution:

m	cm
1 9	1 3 0
2 0 – 1 7	3 0 6 0
2	. 7 0

We cannot subtract 60 cm from 30 cm. So, convert 1 m = 100 cm

∴ 2 m 70 cm

(4) 40 km 255 m – 17 km 960 m
Solution:

km	m
3 9	12 2 5
4 0 -1 7	2 2 5 9 6 0
2 2	2 6 5

We cannot subtract 960 m from 225 m. So, convert 1 km = 1000 m

∴ 22 km 265 m

(5) 35 kg 150 g – 26 kg 470 g
Solution:

kg	gm
3 4	1 1 5 0
3 5 – 2 6	1 5 0 4 7 0
8	6 8 0

We cannot subtract 470 gm from 150 gm. So, convert I kg= 1000gm

∴ 8 kg 680 gm

(6) 46 l 200 ml – 38 l 750 ml
Solution:

l	ml
4 5	1 2 0 0
4 6 – 3 8	2 0 0 7 5 0
7	4 5 0

We cannot subtract 750 ml from 200 ml. So, convert 1 l = 1000 ml

∴ 7 l 450 ml

Word problems

Study the following examples.

Example (1) If a shopkeeper has 150 kg 500 g of rice and sells 75 kg 750 g, how much rice will be left?

74 kg 750 g of rice is left.

Example (2) A can of milk has 20 l 450 ml of milk. Another can has 18 l 800 ml. How much milk is there in the two cans altogether?

The total quantity of milk is 39l 250ml.

Example (3) At a speed of 90 km per hour, what distance will a train cover in two and a half hours?

The speed of the train is 90 kmph. That is, it travels 90 km in one hour. It travels 90 more km in the second hour.
In the next half an hour, 90 ÷ 2 = 45 km
The total distance travelled is 90 + 90 + 45 = 225 km.

Example (4) If one dress requires 3 m 25 cm of cloth, how much do 4 dresses need?

Manju’s method :
3 m 25 cm for the 1st dress
+ 3 m 25 cm for the 2nd dress
+ 3 m 25 cm for the 3rd dress
3 m 25 cm for the 4th dress
_________
12 m 100 cm
1 m is 100 cm, therefore 12 + 1 = 13 m

Example (5)
If a wire that is 9 m 50 cm long is cut into pieces of 5 cm each, how many pieces will be made?
9 m 50 cm = (900 + 50) cm
To find out how many pieces of 5 cm can be made from a wire 950 cm long, let us use division.
190 pieces will be made.

Example (6) A play started at 30 minutes past 6 in the evening and finished two and three quarter hours later. What time did the play get over?

The play got over at 15 minutes past 9 at night.

Note : The units for length, mass and capacity are written in decimal form. This makes it easy to carry out addition and subtraction of length, mass and capacity.

Units of measuring time are not in decimal form. It is a little more difficult to carry out additions and subtractions of those quantities.

Problems on Measurement Problem Set 46 Additional Important Questions and Answers

Add the following:

(1) 12 km 880 m + 7 km 620 m
Solution:

km	m
1
1 2 + 7	8 8 O 6 2 0
2 0	5 0 0

880m + 620 m = 1500 m
= 1km 500 m
∴ 20 km 500 m

(2) ₹ 62, 45 paise + ₹ 37, 55 paise
Solution:

₹	Paise
1
6 2 + 3 7	4 5 5 5
1 0 0	0 0

45 paise + 55 paise
100 paise = 1 ₹
∴ 100 rupees

Subtract the following:

(1) 15 m 15 cm – 4 m 65 cm
Solution:

kg	gm
1 4	1 1 5
1 5 – 4	1 5 6 5
1 0	5 0

We cannot subtract 65 cm from 15 cm. So, convert l m = 100 cm
∴ 10 m 50 cm

(2) 29 kg 880 gm – 8 kg 900 gm
Solution:

kg	gm
2 8	1 8 8 0
2 9 – 8	8 8 0 9 0 0
2 0	9 8 0

We cannot subtract 900 gin from 880 gm. So, convert 1 kg = 1000 gm
∴ 20 kg 980 gm

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 49 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 49

Question 1.
How much wire will be needed to make a rectangle 7 cm long and 4 cm wide?
Solution:
Perimeter of a rectangle
= 2 x length + 2 x breadth
= 2 x 7 + 2 x 4
= 14 + 8
= 22 cm

∴ 22 cm wire will be needed to make a rectangle

Question 2.
If the length of a rectangle is 20 m and its width is 12m, what is its perimeter?
Solution:
Perimeter of a rectangle
= 2 x length + 2 x breadth
= 2×20 + 2×12
= 40 + 24
= 64 m
∴ Perimeter is 64 m

Question 3.
Each side of a square is 9 m long. Find its perimeter.
Solution:
Perimeter of a square
= 4 x length of one side
= 4 x 9
= 36 m
∴ Perimeter is 36 m

Question 4.
If we take 4 rounds around a field that is 160 m long and 90 m wide, what is the distance we walk in kilometres?
Solution:
Perimeter of a rectangular field
= 2 x length + 2 x breadth
= 2 x 160 + 2 x 90
= 320 + 180
= 500 m

In one round distance walked is 500 m, hence, distance walked in 4 rounds
= 500 x 4
= 2000 m
= 2km
∴ The distance walked in 4 rounds is 2 km

Question 5.
Sanju completes 12 rounds around a square park every day. If one side of the park is 120 m, find out in kilometres and metres the distance that Sanju covers daily.
Solution:
Perimeter of a square
= 4 x length of one side
= 4 x 120
= 480 m

So, in one round the distance can be covered is 480 m, hence in 12 rounds the distance can be covered is
= 480 x 12
= 5760 m
= 5000 m + 760 m

∴ Sanju covers 5 km 760 m daily

Question 6.
The length of a rectangular plot of land is 50 m and its width is 30 m. A triple fence has to be put along its edges. If the wire costs 60 rupees permetre, what will be the total cost of the wire needed for the fence?
Solution:
Perimeter of a rectangular plot
= 2 x length + 2 x breadth
= 2 x 50 + 2 x 30
= 100 + 60 – 160 m
For a triple fence, wire needed
= 3 x 160 = 480 m

Cost of the wire needed
= wire needed x rate
= 480 x 60
= 28800 rupees
∴ The total cost of the wire needed for the fence is ₹ 28,800

Question 7.
A game requires its players to run around a square playground. Each side of the playground is 20 m long. One player took 5 rounds around the playground. How many metres did he run altogether?
Solution:
Perimeter of a square
= 4 x length of one side
= 4 x 20
= 80 m

In one round 80 m.
So in 5 round
= 80 x 5
= 400
= 400 m

∴ He runs altogether = 400 m

Question 8.
Four rounds of wire fence have to be put around a field. If the field is 60 m long and 40 m wide, how much wire will be needed?
Solution:
Perimeter of rectangular field
= 2 x length + 2 x breadth
= 2 x 60 + 2 x 40
= 120 + 80
= 200 m
Hence, wire required for 4 rounds
= 200 x 4
= 800 m

∴ Wire required for 4 rounds
= 800 m

Question 9.
The sides of a triangle are 24.7cm, 20.4 cm and 10.5 cm respectively. What is the perimeter of the triangle?
Solution:
Perimeter of triangle
= 24.7 + 20.4 + 10.5
= 55.6

∴ The perimeter of a triangle
= 55.6 cm

Question 10.
Look at the figures on the sheet of graph paper. Measure their sides with the help of the lines on the graph paper. Write the perimeter of each in the right box.

(1) Perimeter of
rectangle ABCD
= [ ] cm
(2) Perimeter of
rectangle EFGH
= [ ] cm
(3) Perimeter of
square PQRS
= [ ] cm
(4) Perimeter of
rectangle STUV
= [ ] cm
Solution:

(1) Perimeter of a rectangle ABCD
= 2 x length + 2 x breadth
= 2 x 3.5 + 2 x 2.5
= 7 + 5
= 12 cm

∴ 12 cm

(2) Perimeter of a rectangle EFGH
= 2 x length + 2 x breadth
= 2 x 3.8 + 2 x 1.3
= 7.6 + 2.6
= 10.2 cm

∴ 10.2 cm

(3) Perimeter of a rectangle PQRS
= 2 x length + 2 x breadth
= 2 x 2.4 + 2 x 2.4
= 4.8+ 4.8
= 9.6 cm

∴ 9.6 cm

(4) Perimeter of a rectangle STUV
= 2 x length + 2 x breadth
= 2 x 3 + 2 x 2
= 6 + 4
= 10 cm

∴ 10 cm

(5) Perimeter of a triangle LMN
= 1.5 + 2.5 + 2
= 6 cm

∴ 6 cm

Area : Revision

Of the figures given above, figure ABCD has six squares of 1 cm each inside it. It means that its area is 6 sq cm.

In the same way, count the squares in each figure and write its area.
(1) Area of MNRS = [ ] sq cm
(2) Area of EFGH = [ ] sq cm
(3) Area of PQRS = [ ] sq cm
(4) Area of IJKL = [ ] sq cm

Atul : Sir, why is the unit for area written as sq cm? We measure the sides in centimetres.

Teacher : Centimetre is a standard unit of length. In order to measure area, we need a standard unit of area. For this, a square with a side 1 cm is taken as the standard unit. The area of this square is 1 square centimetre. That is why this unit is written as sq cm, in short.

To measure large areas like fields, parks and playgrounds, a square with side 1 m, that is, an area of 1 sq m, is taken as the standard unit.

To measure the areas oftalukas or districts, a square with side 1km, or 1sq km is the standard unit used.

Formula for the area of a rectangle

(1) In the rectangle ABCD given alongside, 1 cm divisions were marked off on each side. The points on opposite sides were joined as shown in the figure. The length of the sides of each square thus created is 1cm. Therefore, the area of each square is 1 sq cm.

In ABCD, 3 rows with 5 squares each have been created.
The number of squares in rectangle ABCD is 3 × 5 = 15.
Therefore, the area of rectangle ABCD is 15 sq cm.
Here, the length of the figure is 5 cm and its breadth is 3 cm.
Note that the product of 3 and 5 is 15.

(2) In the rectangle with sides 4 cm and 2 cm, make squares of 1 sq cm each as shown above. Count the number of squares.

Note that here too, the number of squares formed are the same as the product of the length and width of the rectangle.

Therefore, The area of a rectangle = length × breadth

Formula for the area of a square

(1) Look at the square given alongside. The side of the square is 3 cm long. 9 squares of 1 cm each are formed within this square.

Therefore, the area of this square is 9 sq cm.

Here, there are 3 rows with 3 squares each, i.e., there are 3 × 3 = 9 squares.
The length of each side of the square is 3 cm.
The product of two sides of the square is 3 × 3 = 9.

(2) Measure the area of a square with side 5 cm, in the same way.
The answer will be 25 sq cm.
Note that 5 × 5 = 25

Therefore, The area of a square = length of a side × length of a side

It is not necessary to divide a square or rectangle into small squares every time you calculate their area. The advantage of a formula is that you can calculate the area simply by substituting the appropriate values.

Word problems
Example (1) What is the area of a rectangle of length 20 cm and width 15 cm?
Area of a rectangle = length × breadth
= 20 × 15 = 300.
Therefore, the area of the rectangle is 300 sq cm.

Example (2) A wall that is 4 m long and 3 m wide has to be painted. If the labour charges are ₹ 25 per sq m, what is the cost of labour for painting this wall?

First let us calculate the area of the wall to be painted.
Area of the wall = length of the wall × breadth of the wall = 4 × 3 = 12
Thus, the area of the wall is 12 sq m.
Labour cost of 1 sq m is 25 rupees.
So the labour cost for 12 sq m will be = 12 × 25 = 300
The cost of labour for painting the wall will be 300 rupees.

Example (3) What will be the area of a square with sides 15 cm?
Area of a square = length of side × length of side
= 15 × 15 = 225
The area of the square is 225 sq cm.

Example (4) One side of a square room is 4 m. If the cost of labour for laying 1 sq m of the floor is 35 rupees, what will be the total cost of labour?
First we must find the area of the square room.
Area of the square room = length of side × length of side = 4 × 4 = 16
Therefore, the area of the square room is 16 sq m.
The labour cost of laying 1 sq m of flooring is 35 rupees.
Therefore, the cost of laying 16 sq m of flooring is 16 × 35 = 560 rupees.

Perimeter and Area Problem Set 49 Additional Important Questions and Answers

Question 1.
Devendra walks five rounds of a square garden everyday. If the side of the garden is 150 m, how many kilometres does Devendra walk every morning?
Solution:
Perimeter of a square garden
= 4 x one side of the garden
= 4 x 150
= 600 m

In 5 rounds walking
= 5 x 600
= 3000 m
= 3 km
3 km

Question 2.
The length of a rectangular play ground is 75 m and its breadth is 50 m. Rupali walks four rounds. How many kilometres did she walk?
Solution:
Perimeter of rectangle
= 2 x length + 2 x breadth
= 2 x 75 + 2 x 50
= 150 + 100
= 250 m

In 4 rounds walking
= 4 x 250
= 1000 m
= 1 km

∴ 1 km

Question 3.
Length of the rectangle is 10 cm and its breadth is 8 cm and one square is side 9 cm. Whose perimetre is more? By how much?
Solution:
Perimeter of a rectangle
= 2 x length + 2 x breadth
= 2 x 10 + 2 x 8
= 20 + 16
= 36 cm ……………….. (i)

Perimeter of a square
= 4 x length of side
= 4 x 9
36 cm ……………….. (ii)
From (i) and (ii) perimeter of both is equal.

∴ perimeter of both is equal

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

Maharashtra State Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

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)

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 44 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 44

Question 1.
The time below is given by the 12 hour clock. Write the same by the 24 hour clock.
(1) 30 minutes past 10 in the morning –
(2) 10 minutes past 8 in the morning –
(3) 20 minutes past 1 in the afternoon –
(4) 40 minutes past 5 in the evening –
Answer:
(1) [10:30]
(2) [8:10]
(3) [13:20]
(4) [17:40]

Question 2.
Match the following.
12 hour clock – 24 hour clock
(1) 9:10 am – 23:10
(2) 2:10 pm – 7:25
(3) 5:25 pm – 14:10
(4) 11:10 pm – 9:10
(5) 7:25 am – 17:25
Answer:
(1) – d
(2) – c
(3) – e
(4) – a
(5) – b

Examples of time measurement

Example (1) If Abdul started working on the computer at 11 in the morning and finished his work at 3:30 in the afternoon, how long did he work?

Method 1 :
From 11 in the morning to 12 noon, it is 1 hour. From 12 noon to 3:30 in the afternoon, it is 3 hours and 30 minutes. Therefore, the total time is 4 hours and 30 minutes.

Method 2 :
According to the 24 hour clock, 11’o’clock in the morning is 11:00 and 3:30 in the afternoon is 15:30.

Hr	Min
15 – 11	30 00
4	30

Abdul worked for a total of 4 hours and thirty minutes, or four and a half hours.

Example (2) Add : 4 hours 30 min + 2 hours 45 min

Example (3) Subtract : 5 hr 30 min – 2 hr 45 min

45 minutes cannot be subtracted from 30 minutes. Therefore, we borrow 1 hour and convert it into 60 minutes for the subtraction.

Example (4) Amruta travelled by bus for 3 hours 40 minutes and by motorcycle for 1 hour 45 minutes. How long did she spend travelling?

(60 + 25) minutes are 85 minutes, that is, 1 hour and 25 minutes.
Let us add this 1 hour to 4 hours.

Therefore, Amruta travelled for a total of 5 hours and 25 minutes.

Measuring Time Problem Set 44 Additional Important Questions and Answers

The time is given by the 12-hour clock. Write the same by the 24-hour clock.

(1) 15 minutes past 9 in the evening –
(2) 12 midnight –
Answer:
[21:15]
[0o:00]

The time below is given by the 24-hour clock. Write the same by the 12-hour clock.

(1) 20:20 =
(2) 9:30 =
(3) 23:00 =
(4) 4:00 =
(5) 12:00 =
(6) 00:00 =
Answer:
[8:20 pm]
[9:30 am]
[11 pm]
[4 am]
[12 noon] or [12:00]
[12 midnight]

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43

Question 1.
Write the time shown in each clock in the box given below it.
(1)
Answer:
25 minutes past 2

(2)
Answer:
50 minutes past 7

(3)
Answer:
5 minutes past 8

(4)
Answer:
40 minutes past 4

Question 2.
Draw the hands of the clock to show the time given in the box.
(1)
Four-thirty
Answer:

(2)
Quarter past nine
Answer:

(3)
Quarter to five
Answer:

(4)
20 minutes past 11
Answer:

Question 3.
If a bus that leaves Nashik at 5 o’clock in the morning reaches Pune that same day at ten-thirty in the morning, how long does the journey take?
Solution:

Hrs.	Min.
10 – 5	30 00
5	30

∴ Bus took 5 hrs 30 min

Question 4.
A play that was to start at nine fifteen at night was delayed by half an hour because of a power outage. What time did the play start?
Solution:

Hrs.	Min.
9 +	15 30
9	45

∴ Play started 9:45 at night

Question 5.
If a train leaves Mumbai at ten-fifteen at night and reaches Nagpur at one forty the next afternoon, how long does the journey take?
Solution:
Ten fifteen at night to 12 mid night is

Hrs.	Min.
1 2 -10	0 0 1 5
1	4 5

12 mid night to next 1:40 afternoon = 13 hours 40 minutes
Total times

Hrs.	Min.
1 + 13	45 40
15	25

∴ Total time of the journey is 15 hrs. 25 min:

Learning about seconds

This clock is showing 5 minutes past 3. We know this because of the position of the hour and minute hands. There is another hand in the picture called the second hand. This hand moves swiftly. The second is a very small unit used to measure time less than a minute.

The face of a clock is a circle divided into 60 equal parts. When the second-hand moves one part, it takes one second. When it completes one round of the clock face, it moves across all 60 parts. This takes 60 seconds. In the same time, the minute hand moves one place, which means that one minute is over.

It means that, 1 minute is equal to 60 seconds.

1 minute = 60 seconds

The clock in the picture above shows 5 minutes and 50 seconds past 3.

20 minutes and 10 seconds past 7
15 minutes and 40 seconds past 10

Seconds are used on various occasions such as measuring temperature with a thermometer, measuring heartbeats or timing a race.

Ante meridiem and post meridiem

Shripati was sitting at home at night, tired. There were guests at home. They asked, “You must have worked very hard in the fields today. How long were you working?”

Shripati said, “I was in the field from six o’clock to eight o’clock.” Someone asked, “You’re this tired even though you were in the field for only two hours?”

Shripati said, “No, no, I was in the field from 6 in the morning till 8 at night! Now tell me how many hours I spent in the field.”

The guests had not understood what Shripati said at first. To avoid such mistakes, it has been internationally agreed that as the clock strikes 12 midnight, one day ends and the next day begins. From that moment on, the clock shows the time for the next day. When one hour passes after 12 midnight, it is 1’o’clock. After that, it is 2, 3, 4, …, 12 o’clock in serial order. After 12 noon, again it is 1, 2, 3, …, 12 o’clock in serial order. The time before 12 noon is stated as ante meridiem or am. The time after 12 noon is stated as post meridiem or pm.

This method of measuring time is called the 12 hour clock.

Shripati was in the field from 6 am to 8 pm or for 14 hours.

The 24 hour clock
The 24 hour clock is used to avoid this division of the day into ante meridiem and post meridiem. This method is used in timetables for trains, planes, buses and long distance boat journeys. In this method, instead of starting again from 1, 2, 3 after 12 noon, we continue with 13, 14, 15,…,24. In a 24 hour digital watch, time is shown only in the form of numbers. It does not have hands. In such a clock, 20 minutes past 6 in the morning is shown as ‘6:20’ and 20 minutes past 6 in the evening is shown as ‘18:20’.

23:59 means 59 minutes after 23 and one minute later, 24 hours will be complete. The digital clock will show this time as 00:00 at midnight and the day will change. At that time, a 12 hour clock shows 12 midnight.

Study the following table to see how different times of the day are shown in the 12 hour and 24 hour clocks.

The timetables of some trains going from Badnera to Nagpur are given below. Observe the use of the 24 hour clock in the timetable.

Measuring Time Problem Set 43 Additional Important Questions and Answers

Write the time shown in each clock in the box given below it.
(1)
Answer:
15 minutes past 6

(2)
Answer:
30 minutes past 9

Draw the hands of the clock to sho the time given in the box.
(1) 5 minutes to four
Answer:

(2) 35 minutes past to 2
Answer:

(3) Sujata left home 6:30 and returned at 11. How much time did she spend away from home?
Solution:

Hrs.	Min*
11 – 6	00 30
4	30

∴ 4 hrs. 30min. spent away from home.

(4) A speech that started at 4:20 in the afternoon ended at 5:45. How long was the speech?
Solution:

Hrs.	Min.
5 + 4	4.5 20
1	25

∴ Speech was for 1 hr. 25 min.

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40

Write the following fractions as decimal fractions.

(1) Two and a half
Answer:
2 \(\frac{1}{2}\) = 2.5 = 2.50

(2) Two and a quarter
Answer:
2 \(\frac{1}{4}\) = 2.25

(3) Two and three quarters
Answer:
2 \(\frac{3}{4}\) = 2.75

(4) Ten and a half
Answer:
10 \(\frac{1}{2}\) = 10.5 = 10.50

(5) Fourteen and three quarters
Answer:
14 \(\frac{3}{4}\) = 14.75

(6) Sixteen and a quarter
Answer:
16 \(\frac{1}{4}\) = 16.25

(7) Twenty-eight and a half
Answer:
28 \(\frac{1}{2}\) = 28.50 = 28.5

Adding decimal fractions

Sir : If the cost of one pencil is two and a half rupees and the cost of a pen is four and half rupees, what is the total cost?

Sumit : Two and a half rupees means two rupees and one half rupee. Similarly, four and a half rupees means four rupees and one-half rupee. 4 rupees and 2 rupees make 6 rupees and two half rupees make one rupee, so both objects together cost 6 + 1 = 7 rupees.

Sir : Correct ! Now, see how this is done using decimals.
The sum of the 0’s in the hundredths place is 0.
0.5 + 0.5 is the same as
\(\frac{5}{10}+\frac{5}{10}=\frac{5+5}{10}=\frac{10}{10}=\frac{1}{1}=1\)

This 1 is carried over to the units place. There is nothing in the tenths place, so we put a zero there. In the units place, 2 + 4 = 6 plus the carried over 1 makes 7.

So 2.50 rupees and 4.50 rupees add up to 7 rupees.

We use the decimal system to write whole numbers. We extend the same method to write fractions; therefore, we can add in the same way as we add whole numbers.

I will now show some more additions. Watch carefully.

Sumit : There is no carried over number in the first sum, but there are carried over numbers in the second and third sums.

Rekha : While adding whole numbers, we add units first. Similarly, here, tenths are added first. In the second example, the sum of the tenths place is 13. 13 tenths are 10 tenths + 3 tenths = 1 unit + 3 tenths.

Sumit : That is why, in the sum, 3 stayed in the tenths place and 1 was carried over to the units place. 6 + 5 plus 1 carried over makes 12.

Sir : Your observations are absolutely correct. We write digits one below the other according to their place values while adding whole numbers. We do the same thing here. Remember that while writing down an addition problem and the total, the decimal points should always be written one below the other.

Study the following additions. (Note that: 10 tenths = 1 unit. 10 hundredths = 1 tenth)

Example (1) Add : 7.09 + 54.93
First, add the digits in the 100ths place. 9 + 3 = 12.

The 1 from the sum 12 in the hundredths place is carried over to the tenths place and 2 is written in the hundredths place. Adding 1 + 9 gives 10 tenths or 1 unit. This 1 is carried over to the units place. 0 is left in the tenths place. Then, the addition is completed in the usual way.

Example (2) Add : 45.83 + 167.4
4 5 . 8 3 We arrange the numbers so that the places and
+
1 6 7. 4 decimal points come one below the other.

\(\frac{4}{10}=\frac{4 \times 10}{10 \times 10}=\frac{40}{100}\) Therefore, to make the denominators of the fractions equal, 167.4 is written as 167.40 and then the fractions are added.

As usual, the digits with the smallest place values are added first and then those with bigger place values are added serially.

Example (3) 10.46 Rupees

Example (4) 48.80 m

Example (5) 7.5 cm

Decimal Fractions Problem Set 40 Additional Important Questions and Answers

(1) Thirty and a quarter
Answer:
30 \(\frac{1}{4}\) = 30.25

(2) Thirty and a half
Answer:
30 \(\frac{1}{2}\) = 30.50 = 30.5

(3) Thirty and three quarters
Answer:
30 \(\frac{3}{4}\) = 30.75

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 39 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 39

Question 1.
Write how many rupees and how many paise.

(1) ₹ 58.43
Answer:
58 rupees 43 paise.

(2) ₹ 9.30
Answer:
9 rupees 30 paise.

(3) ₹ 2.30
Answer:
2 rupees 30 paise.

(4) ₹ 2.3
Answer:
2 rupees 30 paise.

Question 2.
Write how many rupees in decimal form.

(1) 6 rupees 25 paise
Answer:
₹ 6.25

(2) 15 rupees 70 paise
Answer:
₹ 15.70

(3) 8 rupees 5 paise
Answer:
₹ 8:05

(4) 22 rupees 4 paise
Answer:
₹ 22.04

(5) 720 paise
Answer:
₹ 7.20

Question 3.
Write how many metres and how many centimetres.

(1) 58.75 m
Answer:
58 m 75 cm

(2) 9.30 m
Answer:
9 m 30 cm

(3) 0.30 m
Answer:
30 cm

(4) 0.3 m
Answer:
30 cm

(5) 1.62 m
Answer:
1 m 62 cm

(6) 91.4 m
Answer:
91 cm 40 cm

(7) 7.02 m
Answer:
7 m 2 cm

(8) 0.09 m
Answer:
9 cm

Question 4.
Write how many metres in decimal form.

(1) 1 m 50 cm
Answer:
1.5 m

(2) 50 m 40 cm
Answer:
50.40 m

(3) 50 m 4 cm
Answer:
50.04 m

(4) 734 cm
Answer:
7.34 m

(5) 10 cm
Answer:
0.1 m

(6) 2 cm
Answer:
0.02 m

Question 5.
Write how many centimetres and how many millimetres.

(1) 6.9 cm
Answer:
6 cm 9 mm

(2) 20.4 cm
Answer:
20 cm 4 mm

(3) 0.8 cm
Answer:
8 mm

(4) 0.5 cm
Answer:
5 mm

Question 6.
Write how many centimetres in decimal form.
(1) 7 cm 1 mm
Answer:
7.1 cm

(2) 16 mm
Answer:
1.6 cm

(3) 144 mm
Answer:
14.4 cm

(4) 8 mm
Answer:
0.8 cm

Writing half, quarter, three-quarters and one and a quarter in decimal form

‘Half’ is usually written as \(\frac{1}{2}\). To convert this fraction into decimal form, the denominator of \(\frac{1}{2}\) must be converted into an equivalent fraction with denominator 10.

\(\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\) so the decimal form of \(\frac{1}{2}\) will be \(\frac{5}{10}\) or 0.5 Just as \(\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\) = 0.5, note that \(\frac{1}{2}=\frac{1 \times 50}{2 \times 50}=\frac{50}{100}\) = 0.50

Therefore, ‘half’ is written as ‘0.5’ or 0.50’. ‘Quarter’ and ‘three quarters’ are written in fractions as \(\frac{1}{4}\) and \(\frac{3}{4}\) respectively. Let us convert them into decimal fractions. 10 is not divisible by 4. Therefore, the denominators of \(\frac{1}{4}\) and \(\frac{3}{4}\) cannot be made into fractions with multiples of 10. However, 4 × 25 = 100, so the denominator can be 100.

A quarter \(=\frac{1}{4}=\frac{1 \times 25}{4 \times 25}=\frac{25}{100}=0.25\)
and Three quarters \(=\frac{3}{4}=\frac{3 \times 25}{4 \times 25}=\frac{75}{100}=0.75\)
One and a quarter = 1 \(\frac{1}{4}\) = 1.25
One and a half = 1 \(\frac{1}{2}\) = 1.50 = 1.5
One and three quarters = 1 \(\frac{3}{4}\) = 1.75
Seventeen and a half = 17 \(\frac{1}{2}\) = 17.50 = 17.5

Decimal Fractions Problem Set 37 Additional Important Questions and Answers

Question 1.
Write how many rupees and how many paise.

(1) ₹ 147.5
Answer:
1 hundred and 47 rupees 50 paise.

(2) ₹ 40.4
Answer:
40 rupees and 40 paise.

Question 2.
Write how many rupees in decimal form.

(1) 105 paise
Answer:
₹ 1.05

(2) 6 rupees 6 paise
Answer:
₹ 6.06

(3) 20 rupees 20 paise
Answer:
₹ 20.2

Question 3.
Write how many metres and how many centimetres.

(1) 1.1m =
Answer:
1 m 10 cm

(2) 120 cm =
Answer:
1 m 20 cm

(3) 24.8 m =
Answer:
24 m 80 cm

(4) 0.5 m =
Answer:
50 cm

Question 4.
Write how many metres in decimal form.

(1) 110 cm =
Answer:
1.1m

(2) 60 cm =
Answer:
0.6 m

Question 5.
Write how many centimetres and how many millimetres.

(1) 0.1 cm =
Answer:
1 mm

(2) 10.5 cm =
Answer:
10 cm 5 mm

Question 6.
Write how form. many centimetres in decimal
(1) 1 mm =
Answer:
0.1 cm

(2) 100 mm =
Answer:
10.0 cm

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 42 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 42

Question 1.
Subtract the following :

(1) 25.74 – 13.42
Solution:

(2) 206.35 – 168.22
Solution:

(3) 63.4 – 31.8
Solution:

(4) 63.43 – 31.8
Solution:

(5) 63.4 – 31.83
Solution:

(6) 8.23 – 5.45
Solution:

(7) 18.23 – 9.45
Solution:

(8) 78.03 – 41.65
Solution:

Question 2.
Vrinda was 1.48 m tall. After a year, her height became 1.53 m. How many centimeters did her height increase in a year?
Answer:

∴ 5 cm height has increased in a year.

Something more

Decimals used for measurement

We need to measure distance, mass (weight) and volume every day. We use suitable units for these measurements. Kilometre, metre and centimeter for distance; litre, millilitre for volume and kilogram and gram for mass are the units that are used most of the time.

All these units are decimal units. In this method, gram, metre and litre are taken as the basic units for mass, distance and volume respectively. Units larger than these increase 10 times at every step and smaller units become \(\frac{1}{10}\) of the previous unit at each step.

Look at the table of these units given below.

The origin of the terms kilo, hecto… milli is in the Greek or Latin language. Their English equivalents are given in brackets along with the terms.

Decimal Fractions Problem Set 42 Additional Important Questions and Answers

Subtract the following:

(1) 304.17 – 95.28
Solution:

(2) 72.84 – 36.96
Solution:

(3) 9.17 – 5.88
Solution:

(4) 100 – 49.99
Solution:

(5) Atul has 56.25 and Anup has 65. Whose amount is more? How much?
Solution:

∴ Anup’s amount is more by ₹ 8.75

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

Maharashtra State Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 41

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: