Problem Set 43 Class 5 Maths Chapter 10 Measuring Time Question Answer Maharashtra Board

Measuring Time Class 5 Problem Set 43 Question Answer Maharashtra Board

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

Std 5 Maths Chapter 10 Measuring Time

Question 1.
Write the time shown in each clock in the box given below it.
(1) Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 1
Answer:
25 minutes past 2

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

(2) Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 2
Answer:
50 minutes past 7

(3) Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 3
Answer:
5 minutes past 8

(4) Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 4
Answer:
40 minutes past 4

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

Question 2.
Draw the hands of the clock to show the time given in the box.
(1) Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 5
Four-thirty
Answer:
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 19

(2) Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 6
Quarter past nine
Answer:
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 20

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

(3) Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 7
Quarter to five
Answer:
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 21

(4) Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 8
20 minutes past 11
Answer:
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 22

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

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

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

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:

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

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.

Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 10 20 minutes and 10 seconds past 7
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 11 15 minutes and 40 seconds past 10

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

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.
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 12

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

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.
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 13

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

The timetables of some trains going from Badnera to Nagpur are given below. Observe the use of the 24 hour clock in the timetable.
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 14

Measuring Time Problem Set 43 Additional Important Questions and Answers

Write the time shown in each clock in the box given below it.
(1) Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 15
Answer:
15 minutes past 6

(2) Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 16
Answer:
30 minutes past 9

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

Draw the hands of the clock to sho the time given in the box.
(1) 5 minutes to four
Answer:
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 17

(2) 35 minutes past to 2
Answer:
Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 18

(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.

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

(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.

Class 5 Maths Solution Maharashtra Board

Problem Set 28 Class 5 Maths Chapter 7 Circles Question Answer Maharashtra Board

Circles Class 5 Problem Set 28 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 7 Circles

Question 1.
Draw circles with the radii given below.

(1) 2 cm
Answer:
Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28 7

Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28

(2) 4 cm
Answer:
Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28 8
(3) 3 cm
Answer:
Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28 9

Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28

Question 2.
Draw a circle of any radius. Show one diameter, one radius and one chord on that circle.
Answer:
In a circle, P is the centre.
AB is a diameter.
PQ is a radius
MN is a chord
Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28 12

Relationship between radius and diameter
Study the circle given alongside. Think over the following questions.
Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28 1

  • Which are the radii in the circle?
  • How many radii make up diameter AB?
  • If the length of one radius is 3 cm, what is the length of the diameter?
  • How long is the diameter as compared to the radius?

Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28

The diameter of a circle is twice the length of its radius.

  • If another diameter CD is drawn on the same circle, will its length be the same as that of AB?

All the diameters of a circle are of the same length.

Test 1 :
Measure the diameters and radii of the circles given below with a ruler and verify the relationship between their lengths.
Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28 2

Test 2 :
1. Draw a circle on a piece of paper and cut it out.
Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28 3

Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28

2. Name the centre of the circle P.
Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28 4
3. Draw the diameter of the circle and name it AB. Note that PA and PB are radii of the circle.
Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28 5
4. Fold the circular paper along AB as shown in the picture.
Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28 6
Fold the paper at P in such a way that point B will fall on point A. Radius PB falls exactly on radius PA. In other words, they coincide.

From this, we can see that every radius of a circle is half the length of its diameter.

Circles Problem Set 28 Additional Important Questions and Answers

Question 1.
Draw circles with the radii given below:

(1) 1.5 cm
Answer:
Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28 10

Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28

(2) 2.3 cm
Answer:
Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28 11

Question 2.
Which one of the following statement is true:

(1) All chords are diameters.
(2) All diameters are chords.
Answer:
Statement (2) is true.

Class 5 Maths Solution Maharashtra Board

Problem Set 41 Class 5 Maths Chapter 9 Decimal Fractions Question Answer Maharashtra Board

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:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 1

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

(2) ‘Five and three quarter rupees’ and ‘seven and a quarter rupees’
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 2

(3) ‘Six and a half metres’ and ‘three and three quarter metres’.
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 3

Question 2.
(1) 23.4 + 87.9
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 6

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

(2) 35.74 + 816.6
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 7

(3) 6.95 + 74.88
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 8

(4) 41.03 + 9.98
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 9

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

Question 3.
(1) 51.4 cm + 68.5 cm
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 12

(2) 94.7 m + 1738.45 m
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 13

(3) 5158.75 + `841.25
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 14

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

Subtraction of decimal fractions

Study the subtraction of decimal fractions given below.
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 41 1

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:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 4

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

(2) ‘Three quarters’ and ‘a half’
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 5

Add the following:

(1) 37.84 + 12.16
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 10

(2) 328.69 + 84.84
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 11

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

Solve the following:

(1) 304.86 m + 70.94 m
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 15

(2) 79.56 cm + 19.65 cm
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 16

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

(3) ₹ 64.79 + ₹  49.5
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 17

Class 5 Maths Solution Maharashtra Board

Problem Set 8 Class 5 Maths Chapter 3 Addition and Subtraction Question Answer Maharashtra Board

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
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8 12
Answer:
65,78,635

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8

(2) 3,17,529 + 8,04,613
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8 11
Answer:
11,22,142

(3) 12,42,746 + 4,83,748
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8 10
Answer:
17,26,494

(4) 24,12,636 + 23,19,058
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8 9
Answer:
47,31,694

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8

(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
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8 1

(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
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8 2

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8

(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
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8 3

(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
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8 4

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8

(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
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8 5

(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
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8 6

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8

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
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8 7

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8

(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
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 8 8

Class 5 Maths Solution Maharashtra Board

Problem Set 47 Class 5 Maths Chapter 11 Problems on Measurement Question Answer Maharashtra Board

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

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

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

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

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

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 1
That is 1 km in 15 minutes Hence, 3
km in 15 x 3 = 45 min

∴ 3 km in 15 x 3 = 45 min

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

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:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 2

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

Question 9.
If one shirt needs 2 m 50cm of cloth, how much cloth do we need for 5 shirts?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 3
∴ 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

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

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:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 4
∴ 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
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 5
∴ 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

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

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
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 6
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:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 7
bought from 1 shop
bought from another shop
500 + 800 = 1300 gm
= 1000 + 300
= 1 kg 300 gm

∴ 94 kg 300 gm bought altogether

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

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:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 8
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 9

∴ Weight of onions is 9 kg 200 gm

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

Question 18.
One day, Gurminder Singh walked 3 km 750m and Parminder Singh walked 2km 825m. Who walked farther and by how much?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 11

∴ 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:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 12

∴ 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

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

(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.
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 16
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 15

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:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 14
∴ 77 l 210 ml total milk

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

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

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

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

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

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)

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

(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

Problem Set 12 Class 5 Maths Chapter 3 Addition and Subtraction Question Answer Maharashtra Board

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,

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 10

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12

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.

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 11

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.

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 12

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12

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

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 13

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

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 14

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12

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

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 1
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 3

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

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 2
Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 4

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?

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12

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

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 8

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

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 6

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

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 7

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12

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

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 5

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

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 15

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12

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

Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 16

Class 5 Maths Solution Maharashtra Board

Problem Set 20 Class 5 Maths Chapter 5 Fractions Question Answer Maharashtra Board

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:
\(\frac{1}{5}+\frac{3}{5}=\frac{1+3}{5}=\frac{4}{5}\)

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 20

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

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 20

\(\text { (8) } \frac{4}{9}+\frac{1}{9}\)
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.

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 20

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.

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 20

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:
\(\frac{1}{2}+\frac{1}{2}=\frac{1+1}{2}=\frac{2}{2}=1\)

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 20

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

Problem Set 27 Class 5 Maths Chapter 6 Angles Question Answer Maharashtra Board

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.

Maharashtra Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 27

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.
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 18
Answer:
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 1

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:
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 2

Maharashtra Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 27

Question 2.
Complete the following:
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 3
Answer:
(1) ZDEF or ZFED, Vertex E, arms are ED
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 16
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 17

Question 3.
Which of the following figures show angle?
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 4
Answer:
Figure 3

Maharashtra Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 27

Question 4.
(A) Measure the angles given below and write the measure in the given boxes:
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 5
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 6
Answer:
(1) 30°
(2) 135°
(3) 90°
(4) 50°
(5) 90°
(6) 150°

Maharashtra Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 27

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:
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 7
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 8
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 9
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 10
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 11
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 12
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 13
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 14
Answer:
Students to draw angles.

Maharashtra Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 27

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:
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 25 15
Answer:
(1) Parallel lines
(2) Perpendicular lines
(3) Perpendicular lines
(4) Parallel lines

Maharashtra Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 27

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

Problem Set 19 Class 5 Maths Chapter 5 Fractions Question Answer Maharashtra Board

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.

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 1
Answer:
=

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 2
Answer:
>

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 3
Answer:
<

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 4
Answer:
=

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 5
Answer:
>

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 6
Answer:
>

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 7
Answer:
>

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 8
Answer:
>

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 9
Answer:
>

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 10
Answer:
>

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 11
Answer:
=

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 12
Answer:
=

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 13
Answer:
>

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 14
Answer:
>

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 15
Answer:
<

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 16
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, \(\frac{3}{7}+\frac{2}{7}=\frac{3+2}{7}=\frac{5}{7}\)
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 17

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19

Example (2) Add : \(\frac{3}{8}+\frac{2}{8}+\frac{1}{8}\)
The total coloured part is \(\frac{3}{8}+\frac{2}{8}+\frac{1}{8}=\frac{3+2+1}{8}=\frac{6}{8}\)
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 18

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.
\(\frac{5}{5}=\frac{5 \div 5}{5 \div 5}=\frac{1}{1}=1\)

Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19

Fractions Problem Set 19 Additional Important Questions and Answers

Question 1.
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 19
Answer:
>

Question 2.
Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 20
Answer:
=

Class 5 Maths Solution Maharashtra Board

Problem Set 51 Class 5 Maths Chapter 13 Three Dimensional Objects and Nets Question Answer Maharashtra Board

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.
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 1
Answer:
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 10
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 18

Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51

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:
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 12
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 13

Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51

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.
    Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 2
  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.
    Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 3

Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51

Activity :
Draw the nets shown below on card sheet. Cut out the shapes and find out the shapes of the boxes they form.
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 4

A five-square net (Pentomino)

In the figure alongside, five squares of the same size are placed together with their sides joined.
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 5
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.

Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 6

Activity :
Some five-square nets are given below. Draw these nets on a card sheet. Make open boxes from these nets.
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 7
Try to find out other five-square nets that can be used to make open boxes.

Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51

A riddle
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 8
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?
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 9

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:
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 16

Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51

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:
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 17

Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51

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:
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 14
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 15

Class 5 Maths Solution Maharashtra Board