Class 5: Maths Chapter 11 solutions. Complete Class 5 Maths Chapter 11 Notes.
Contents
Maharashtra Board Solutions Class 5-Maths (Problem Set 46) – Part 2: Chapter 11- Problems on Measurement
Maharashtra Board 5th Maths Chapter 11, Class 5 Maths Chapter 11 solutions
Important Questions and Answers.
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:
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 2 | 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 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 |
– 1 7 | 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 |
-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 |
– 2 6 | 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 |
– 3 8 | 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.
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 |
– 4 | 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 |
– 8 | 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
Download PDF
Maharashtra Board Solutions Class 5-Maths (Problem Set 46) – Part 2: Chapter 11- Problems on Measurement
Chapterwise Maharashtra Board Solutions Class 5 Maths :
Part 1
- Chapter 1- Roman Numerals (Problem Set 1)
- Chapter 2- Number Work (Problem Set 2)
- Chapter 2- Number Work (Problem Set 3)
- Chapter 2- Number Work (Problem Set 4)
- Chapter 2- Number Work (Problem Set 5)
- Chapter 2- Number Work (Problem Set 6)
- Chapter 3- Addition and Subtraction (Problem Set 7)
- Chapter 3- Addition and Subtraction (Problem Set 8)
- Chapter 3- Addition and Subtraction (Problem Set 9)
- Chapter 3- Addition and Subtraction (Problem Set 10)
- Chapter 3- Addition and Subtraction (Problem Set 11)
- Chapter 3- Addition and Subtraction (Problem Set 12)
- Chapter 3- Addition and Subtraction (Problem Set 13)
- Chapter 4- Multiplication and Division (Problem Set 14)
- Chapter 4- Multiplication and Division (Problem Set 15)
- Chapter 4- Multiplication and Division (Problem Set 16)
- Chapter 5- Fractions (Problem Set 17)
- Chapter 5- Fractions (Problem Set 18)
- Chapter 5- Fractions (Problem Set 19)
- Chapter 5- Fractions (Problem Set 20)
- Chapter 5- Fractions (Problem Set 21)
- Chapter 5- Fractions (Problem Set 22)
- Chapter 5- Fractions (Problem Set 23)
- Chapter 6- Angles (Problem Set 24)
- Chapter 6- Angles (Problem Set 25)
- Chapter 6- Angles (Problem Set 26)
- Chapter 6- Angles (Problem Set 27)
- Chapter 7- Circles (Problem Set 28)
- Chapter 7- Circles (Problem Set 29)
- Chapter 7- Circles (Problem Set 30)
- Chapter 7- Circles (Problem Set 31)
Part 2.
- Chapter 8- Multiples and Factors (Problem Set 32)
- Chapter 8- Multiples and Factors (Problem Set 33)
- Chapter 8- Multiples and Factors (Problem Set 34)
- Chapter 8- Multiples and Factors (Problem Set 35)
- Chapter 9- Decimal Fractions (Problem Set 36)
- Chapter 9- Decimal Fractions (Problem Set 37)
- Chapter 9- Decimal Fractions (Problem Set 38)
- Chapter 9- Decimal Fractions (Problem Set 39)
- Chapter 9- Decimal Fractions (Problem Set 40)
- Chapter 9- Decimal Fractions (Problem Set 41)
- Chapter 9- Decimal Fractions (Problem Set 42)
- Chapter 10- Measuring Time (Problem Set 43)
- Chapter 10- Measuring Time (Problem Set 44)
- Chapter 10- Measuring Time (Problem Set 45)
- Chapter 11- Problems on Measurement (Problem Set 46)
- Chapter 11- Problems on Measurement (Problem Set 47)
- Chapter 12- Perimeter and Area (Problem Set 48)
- Chapter 12- Perimeter and Area (Problem Set 49)
- Chapter 12- Perimeter and Area (Problem Set 50)
- Chapter 13- Three Dimensional Objects and Nets (Problem Set 51)
- Chapter 14- Pictographs (Problem Set 52)
- Chapter 15- Patterns (Problem Set 53)
- Chapter 16- Preparation for Algebra (Problem Set 54)
- Chapter 16- Preparation for Algebra (Problem Set 55)
- Chapter 16- Preparation for Algebra (Problem Set 56)
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