Category: Class 5

Perimeter and Area Class 5 Problem Set 50 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 12 Perimeter and Area

Question 1.
The length of the side of each square is given below. Find its area.

(1) 12 metres
Solution:
Area of a square
= side x side
= 12 x 12

(2) 6 cm
Solution:
Area of a square
= side x side
= 6 x 6
= 36 sq.cm.

(3) 25 metres
Solution:
Area of a square
= side x side
= 25 x 25
= 625 sq.m.

(4) 18 cm
Solution:
Area of a square
= side x side
= 18 x 18
= 324 sq.cm.

Question 2.
If the cost of 1 sq m of a plot of land is 900 rupees, find the total cost of a plot of land that is 25 m long and 20 m broad.
Solution:
Area of the rectangular plot
= length x breadth
= 25 x 20
= 500 sq.m.

Cost of the plot of land
= Area of the plot x rate
= 500 x 900
= 4,50,000 rupees

Question 3.
The side of a square is 4 cm. The length of a rectangle is 8 cm and its width is 2 cm. Find the perimeter and area of both figures.
Solution:
Perimeter of a square = 4 x side
= 4 x 4
= 16 cm

Area of a square = side x side
= 4 x 4
= 16 sq.cm.

Perimeter of a rectangle
= 2 x length + 2 x breadth
= 2 x 8 + 2 x 2
= 16 + 4
= 20 cm

Area of a rectangle
= length x breadth
= 8 x 2
= 16 sq.cm.

Question 4.
What will be the labour cost of laying the floor of an assembly hall that is 16 m long and 12 m wide if the cost of laying 1 sq m is 80 rupees?
Solution:
Area of rectangular floor
= length x breadth
= 16 x 12
= 192 sq.cm.

The cost of laying 1 sq.m, is 80 rupees.
Hence, the cost of laying 192 sq.m.
= 192 x 80
= 15,360 rupees.
∴ ₹ 15,360

Question 5.
The picture alongside shows some squares. Find out how many squares with the same measures will fit in the empty space in the figure.

Solution:
length of the empty space = 4 – 1 = 3 cm
breadth of the empty space = 3 – 1 = 2 cm
square in empty space
= length x breadth
= 3 x 2 = 6 sq.cm.

∴ 6 squares will fit in the empty space in the figure

Question 6.
Divide the figure given alongside into four parts in such a way that the area and shape of each part is the same. Colour the parts with different colours.

Fair and square

As shown in the figure alongside, a square plot of land owned by the government contains four houses and a well right in the centre. The government has to divide the houses and the land between four poor persons according to the following conditions.
(1) Each person must get only one house.
(2) The shape and area of the land must be the same.
(3) Each person must be able to use the well without trespassing on any one else’s land.

Show the appropriate divisions in four different colours.

Activity
Using a graph paper, find out the area of different rectangles and squares.

Perimeter and Area Problem Set 50 Additional Important Questions and Answers

Question 1.
15 cm
Solution:
Area of a square
= side x side
= 15 x 15
= 225 sq.cm.

Question 2.
21 cm
Solution:
Area of a square = side x side
= 21 x 21
= 441 sq.cm.

Solve the following:

Question 1.
The side of a square hall is of length 8 m. If it is tiled with a tile of length 4 m and breadth 2 m., how many tiles will be required?
Solution:
Area of the square hall
= side x side
= 8 x 8
= 64 sq.m.

Area of 1 rectangular tile
= length x breadth
= 4 x 2

∴ 8 sq.m.

For 8 sq.m., 1 tile is required,
but for 64 sq.m = \(\frac{64}{8}\) = 8 tiles required

∴ 8 tiles required

Question 2.
Perimeter of a square is 16 cm. What is the length of each side? What is the area of the square?
Solution:
Perimeter of square = 4 x side
16 = 4 x side

∴ side of a square = \(\frac{16}{4}\) = 4 cm
Area of the square
= side x side
= 4 x 4
= 16 sq.cm

∴ Side of square is 4 cm and area of the sqaure is 16 sq.cm

Question 3.
Write the perimeter of each figure in the box given below it.

Answer:
24 cm

Answer:
18 cm

Answer:
21 cm

Answer:
20 cm

Question 4.
Two squares of side 2 cm is cut out of two corners of a larger square with side 5 cm (see the figure). What will be the perimeter of the remaining shape?

Answer:
20 cm

Question 5.
Match the columns ‘A’ and ‘B’

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

Question 6.
Solve the following word problems:
(1) What is the perimeter of a rectangle having length 9 cm and its breadth 6 cm?
(2) The sides of a rectangular field are having length 150 m and breadth 100 m. Find the perimeter of field.
(3) If each side of a square is 8 cm then what is the perimeter of the square?
(4) A rectangular garden of length 650 m and breadth 350 m. Mohan makes four rounds daily. How many kilometres does he walk everyday?
(5) One square field is having one side of it is of 225 m, Soham makes 6 rounds of the square field daily. How much distance is covered by him? Write it in km and m.
(6) A rectangular field whose length is 58 m and breadth is 32 m. Fencing the field by 4 rounds with a wire, what length of wire is required? If the cost of 1 m wire is ? 75, then what is the expenditure of fencing the field?
(7) A length of a rectangular classroom is 8 m and its breadth is 5 m. A wooden strip is to be fitted along the four walls to hang charts and pictures. What is the length of the wooden strip required?
(8) The side of a square table is 1.5 m. To fit a strip of tin sheet around the table, how many metres of strip is required?
(9) What is the perimeter of the triangle whose sides are 13.8 cm, 17.6 cm and 10.6 cm?
(10) The sides of some squares are given below. Find their areas.
(i) 11 cm (ii) 23 cm (iii) 9 cm
(11) Find the perimeter and area of the following:
(i) square of side 6 cm
(ii) Rectangle: length 12 cm and breadth 6 cm
(12) A rectangular land is having its length 25 m and breadth 16 m. If the cost of 1 sq.m, land is ? 1,500, then what will be cost of land?
(13) The side of a square plot is 10 metre. It is tiled at the rate of 50 rupees per sq.m. What will be cost of tiling the floor?
(14) A wall of length 25 m and breadth 12 m. It is painted at the rate of 60 rupees per sq.m. What will be cost of painting the wall?
Answer:
(1) 30 cm
(2) 500 m
(3) 32 cm
(4) 8 km
(5) 5 km 400 m
(6) ₹ 54,000
(7) 26 m
(8) 6 m
(9) 42 cm

(10) (i) 121 sq. cm
(ii) 529 sq.cm
(iii) 81sq.cm.

(11) (i) Perimeter = 24 cm, Area = 36 sq.cm,
(ii) Perimeter = 36 cm, Area = 72 sq. cm

(12) 6,00,000 rupees
(13) 5,000 rupees
(14) t 18,000

Question 7.
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.

Perimeter of Rectangle
(1) XYZW = [ ] cm
(2) CDEF = [ ] cm
(3) JKLM = [ ] cm
(4) NOPQ = [ ] cm
Answer:
(1) 12 cm
(2) 10 cm
(3) 10 cm
(4) 8 cm

Question 8.
Find the area of the. following figures. (All small squares are having side 1 cm)

Answer:
5 sq.cm

Answer:
4 sq.cm

Answer:
9 sq.cm

Answer:
16 sq.cm.

Question 9.
The pictures below shows some squares. Find out how many squares with the same measures will fit in the empty space in the figures.

Answer:
(1) 9
(2) 9

Question 10.
Fill in the blanks:

Answer:
(1) Area = 24 sq.cm., Perimeter = 22 cm
(2) Breadth = 4 cm, Perimeter = 20 cm
(3) Length = 2 cm, Perimeter = 8 cm

Maharashtra Board Class 5 Maths Solutions

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

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)
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:

∴ 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:

∴ 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

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

∴ 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:

∴ 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:

∴ Speech was for 1 hr. 25 min.

Maharashtra Board Class 5 Maths Solutions

• Measuring Time Problem Set 43 Class 5 Maths Solutions
• Measuring Time Problem Set 44 Class 5 Maths Solutions
• Measuring Time Problem Set 45 Class 5 Maths Solutions

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:

(2) 4 cm
Answer:

(3) 3 cm
Answer:

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

Relationship between radius and diameter
Study the circle given alongside. Think over the following questions.

• 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?

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.

Test 2 :
1. Draw a circle on a piece of paper and cut it out.

2. Name the centre of the circle P.

3. Draw the diameter of the circle and name it AB. Note that PA and PB are radii of the circle.

4. Fold the circular paper along AB as shown in the picture.

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:

(2) 2.3 cm
Answer:

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.

Maharashtra Board Class 5 Maths Solutions

• Circles Problem Set 28 Class 5 Maths Solutions
• Circles Problem Set 29 Class 5 Maths Solutions
• Circles Problem Set 30 Class 5 Maths Solutions
• Circles Problem Set 31 Class 5 Maths Solutions

Decimal Fractions Class 5 Problem Set 41 Question Answer Maharashtra Board

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

Std 5 Maths Chapter 9 Decimal Fractions

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

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

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

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

Question 2.
(1) 23.4 + 87.9

(2) 35.74 + 816.6
Solution:

(3) 6.95 + 74.88
Solution:

(4) 41.03 + 9.98
Solution:

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

(2) 94.7 m + 1738.45 m
Solution:

(3) 5158.75 + `841.25
Solution:

Subtraction of decimal fractions

Study the subtraction of decimal fractions given below.

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

Decimal Fractions Problem Set 41 Additional Important Questions and Answers

Convert the following into decimal fractions and add them.

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

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

Add the following:

(1) 37.84 + 12.16
Solution:

(2) 328.69 + 84.84
Solution:

Solve the following:

(1) 304.86 m + 70.94 m
Solution:

(2) 79.56 cm + 19.65 cm
Solution:

(3) ₹ 64.79 + ₹  49.5
Solution:

Maharashtra Board Class 5 Maths Solutions

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

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

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

Std 5 Maths Chapter 3 Addition and Subtraction

Question 1.
Add the following:

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

Answer:
65,78,635

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

Answer:
11,22,142

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

Answer:
17,26,494

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

Answer:
47,31,694

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

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

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

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

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

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

Study the following word problem.

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

Addition and Subtraction Problem Set 8 Additional Important Questions and Answers

Question 1.
Add the following:

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

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

Maharashtra Board Class 5 Maths Solutions

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

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

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

Std 5 Maths Chapter 11 Problems on Measurement

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

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:

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

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:

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:

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

(B) Match the following:

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

Maharashtra Board Class 5 Maths Solutions

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

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

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

Std 5 Maths Chapter 3 Addition and Subtraction

Question 1.
Solve the following word problems:

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

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

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

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

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

Mixed examples
Study the following solved examples.

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

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

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

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

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

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

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

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

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

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

Addition and Subtraction Problem Set 12 Additional Important Questions and Answers

Solve the following word problems:

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

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

Maharashtra Board Class 5 Maths Solutions

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

Fractions Class 5 Problem Set 20 Question Answer Maharashtra Board

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

Std 5 Maths Chapter 5 Fractions

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

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

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

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

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

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

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

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

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

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

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

Subtraction of like fractions

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

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

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

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

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

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

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

Addition and Subtraction Problem Set 13 Additional Important Questions and Answers

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

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

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

Solve the following word problems:

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

Maharashtra Board Class 5 Maths Solutions

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

Angles Class 5 Problem Set 27 Question Answer Maharashtra Board

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

Std 5 Maths Chapter 6 Angles

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

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

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

Answer:

Angles Problem Set 27 Additional Important Questions and Answers

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

Question 2.
Complete the following:

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

Question 3.
Which of the following figures show angle?

Answer:
Figure 3

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


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

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

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








Answer:
Students to draw angles.

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

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

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

Maharashtra Board Class 5 Maths Solutions

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

Roman Numerals Class 5 Problem Set 1 Question Answer Maharashtra Board

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

Std 5 Maths Chapter 1 Roman Numerals

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

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

Question 3.
Fill in the empty boxes.

Answer:

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

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

Answer:

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

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

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

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

Roman Numerals Problem Set 1 Additional Important Questions and Answers

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

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

Question 3.
Fill in the empty boxes:
Answer:

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

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

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

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

Question 8.
Fill in the empty boxes:

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

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

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

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

Question 12.
Match the columns:

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

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

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

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

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

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

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

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

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

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

Class 5 Maths Solution Maharashtra Board

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