Sainavle

Mensuration Exercise – 10.1

Question 1. Find the perimeter of each of the following figures:

Solution:

1. Perimeter = Sum of all the sides = 4 cm + 2 cm +1 cm + 5 cm = 12 cm
2. Perimeter = Sum of all the sides = 23 cm + 35 cm + 40 cm + 35 cm = 133 cm
3. Perimeter = Sum of all the sides = 15 cm + 15 cm + 15 cm + 15 cm = 60 cm
4. Perimeter = Sum of all the sides =4 cm+4 cm+4 cm+4 cm+ 4 cm=20 cm
5. Perimeter = Sum of all the sides =1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm = 15 cm
6. Perimeter = Sum of all the sides = 4cm +lcm + 3cm + 2cm + 3cm + 4cm +lcm + 3cm + 2cm + 3cm + 4cm + lcm + 3cm + 2cm + 3cm + 4cm +lcm + 3cm + 2cm + 3cm = 52 cm

Question 2. The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Solution:

Total length of tape required

= Perimeter of rectangle

= 2 (length + breadth)

= 2(40 +10) cm = 2 x 50 cm = 100 cm =1 m

Thus, the total length of tape required is 100 cm or 1 m.

Question 3. A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the tabletop?
Solution:

Length of table-top = 2m 25 cm = 2.25 m

Breadth of table-top =1 m 50 cm = 1.50 m

Perimeter of table-top = 2 x (length + breadth)

= 2 x (2.25 + 1.50) m

= 2 x 3.75 m = 7.50 m

Thus, the perimeter of the table-top is 7.5 m

Question 4. What Is the length of the wooden strip required to frame a photograph of length and breadth of 32 cm and 21 cm respectively?
Solution:

Length of the wooden strip

= Perimeter of photograph

= 2 x (length + breadth)

= 2 (32 + 21) cm = 2 x 53 cm = 106 cm

Thus, the length of the wooden strip required is 106 cm

Question 5. A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Solution:

Since 4 rows of wires are needed.

Therefore, the total length of the wire is equal to 4 times the perimeter of land.

Perimeter of land = 2 x (length + breadth)

= 2 x (0.7 + 0.5) km = 2 x 1.2 km = 2.4 km

= 2.4 x 1000 m = 2400 m

Thus, the length of wire = 4 x 2400 m = 9600 m = 9.6 km

Question 6. Find the perimeter of each of the following shapes :

1. A triangle of sides 3 cm, 4 cm and 5 cm.
2. An equilateral triangle of side 9 cm.
3. An isosceles triangle with equal sides 8 cm each and third side 6 cm.

Solution:

1. Perimeter of ΔABC

=AB+BC+CA

= 3cm+5cm+4cm

= 12 cm

2. Perimeter of equilateral

ΔABC

= 3 x side

= 3 x 9 cm

= 27 cm

3. Perimeter of ΔABC

= AB + BC + CA

= 8cm + 6cm + 8cm

= 22 cm

Question 7. Find the perimeter of a triangle with sides measuring 1 0 cm, 1 4 cm and 1 5 cm.
Solution:

Perimeter of triangle = Sum of all three sides

= 10 cm + 14 cm + 15 cm = 39 cm

Thus, the perimeter of the triangle is 39 cm.

Question 8. Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution:

The perimeter of a regular hexagon

= 6 x length of one side

= 6 x 8m = 48m

Thus, the perimeter of a regular hexagon is 48 m.

Question 9. Find the side of the square whose perimeter is 20 m,
Solution:

The perimeter of the square = 4 x side

20 m = 4 x side ⇒ side = \(=\frac{20}{4} \mathrm{~m}=5 \mathrm{~m}\)

Thus, the side of the square is 5 m.

Question 10. The perimeter of a regular pentagon is 1 00 cm. How long is it on each side?
Solution:

The perimeter of a regular pentagon = 5 x side

100 cm = 5 x side ⇒ side = \(\frac{100}{5} \mathrm{~cm}=20 \mathrm{~cm}\)

Thus, the side of the regular pentagon is 20 cm.

Question 11. A piece of string is 30 cm long. What will be the length of each side if the string is used to form:

1. A square?
2. An equilateral triangle?
3. A regular hexagon?

Solution:

Length of string= Perimeter of each shape

The perimeter of the square = 4 x side

30 cm = 4 x side ⇒ side = \(\frac{30}{4} \mathrm{~cm}=7.5 \mathrm{~cm}\)

Thus, the length of each side of the square will be 7.5 cm

The perimeter of equilateral triangle = 3 x side

30 cm = 3 x side ⇒ side =\(\frac{30}{3} \mathrm{~cm}=10 \mathrm{~cm}\)

Thus, each side of the equilateral triangle will be 10 cm long.

The perimeter of a regular hexagon = 6 x side

30 cm = 6 x side ⇒ side = \(\frac{30}{6} \mathrm{~cm}=5 \mathrm{~cm}\)

Thus, each side of a regular hexagon will be 5 cm long.

Question 12. Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Solution:

Let the length of the third title be x cm. The length of the other two sides are 12 cm and 14 cm

Now, the perimeter of the triangle = 36 cm

12 + 14 + x = 36

26 + x = 36

X = 36- 26 ⇒ x = 10

Thus, the length of the third side is 10 cm

Question 13. Find the cost of fencing a square park of side 250 m at the rat of? 20 per metre.
Solution:

Side of square park = 250 m

The perimeter of the square park = 4 x side

= 4 x 250 m = 1000 m

Sin ce, cost of fencing for metre = ₹ 20

Therefore, the cost of fencing for 1000 metres

= ₹ 20×1000 = ₹ 20,000

Question 14. Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of ₹ 12 per metre
Solution:

Length of rectangular park = 175 m

The breadth of the rectangular park = 125 m

Perimeter of park = 2 x (length + breadth)

= 2 x (175 + 125) m

= 2 x 300 m = 600 m

Since, the cost of fencing a park for metre = ? 12

Therefore, the cost of fencing the park for 600 m

= ₹ 12 x 600 = ₹ 7,200

Question 15. Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with a leaf of 60 m a and breadth of h 45 m. Who covers less distance?
Solution:

Distance covered by Sweety

= Perimeter of square park = 4 x side

= 4 x 75 m = 300 m

Thus, the distance covered by Sweety is 300 m.

Nthe ow, distance covered by Bulbul

= Perimeter of a rectangular park

= 2 x (length + breadth)

= 2 x (60 + 45) m

= 2 x 105 m = 210 m

Thus, Bulbul covers a distance of 210 m.

So, Bulbul covers less distance.

Question 16. What is the perimeter of each of the following figures? What do you infer from the answers?

Solution:

Perimeter of square = 4 x side = 4 x 25 cm = 100 cm

Perimeter of rectangle = 2 x (length + breadth)

= 2 x (40 + 10) cm = 2 x 50 cm = 100 cm

Perimeter of rectangle = 2 x (length + breadth)

= 2 x (30 + 20) cm = 2 x 50 cm = 100 cm

Perimeter of triangle = Sum of all sides

= 30 cm + 30 cm + 40 cm

= 100 cm

Thus, all the figures have the same perimeter

Question 17. Avneet buys 9 square paving slabs, each with a side of \(\frac{1}{2}\) m. He lays them in the form of a square.

1. What is the perimeter of his arrangement?
2. Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement? ent.
3. Which a has greater perimeter?
4. Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be broken.)

Solution:

1. Side of one small square = \(\frac{1}{2}\) m

Side of given square ⇒ \(\frac{3}{2}\) perimetereter of square = 4 x side

⇒\(4 \times \frac{3}{2} \mathrm{~m}=6 \mathrm{~m}\)

2. The perimeter of the given figure

⇒ Sum of all sides = \(20 \times \frac{1}{2} \mathrm{~m}=10 \mathrm{~m}\)

3. The arrangement crosses a greater perimeter.

4. It is not possible to determine the arrangement with a perimeter greater than 10 m.

Mensuration Exercise – 10.2

Question 1. Find the areas of the following figures by countsquaresuare:

Solution:

1. Number of filled squares = 9

∴ Area covered by filled squares

= (9 x 1) sq units = 9 sq units

2. Number of filled squares = 5

∴ Area covered by filled squares

= (5 x 1) sq units = 5 sq units

3. Number of fully-filled squares = 2

Number of half-filled squares = 4

∴ Area covered by fully-filled squares

= (2 x 1) sq units = 2 sq units

Area covered by half-filled squares

⇒ \(\left(4 \times \frac{1}{2}\right) \text { sq units }=2 \text { sq units }\)

Total area = (2 + 2) sq units = 4 sq units

4. Number of filled squares = 8

∴ Area covered filled squares

= (8 x 1) sq units = 8 sq units

5. Number of filled squares = 10

∴ Area covered by filled squares

= (10 x 1) sq units = 10 sq units

6. Number of fully-filled squares = 2

Number of half-filled squares = 4

∴ Area covered by fully-filled squares

= (2 x 1) sq units = 2 sq units

Area covered by half-filled squares

⇒ \(\left(4 \times \frac{1}{2}\right) \text { sq units }=2 \text { sq units }\)

Total area = (2 + 2) sq units = 4 sq units

7. Number of fully-filled squares = 4

Number of half-filled squares = 4

∴ Area covered by fully-filled squares

= (4 x 1) sq units = 4 sq units

Area covered by half-filled squares

⇒ \(\left(4 \times \frac{1}{2}\right) \text { sq units }=2 \text { sq units }\)

∴ Total area = (4 + 2) sq units = 6 sq units

8. Number of filled squares = 5

∴ Area covered by filled squares

= (5 x 1) sq units = 5 sq units

9. Number of filled squares = 9

∴ Area covered by filled squares

= (9 x 1) sq units = 9 sq units

10. Number of fully-filled squares = 2

Number of half-filled squares = 4

∴ Area covered by fully-filled squares

= (2 x 1) sq units = 2 sq units

Area covered by half-filled squares

⇒ \(\left(4 \times \frac{1}{2}\right) \text { sq units }=2 \text { sq units }\)

∴ Total area = (2 + 2) sq units = 4 sq units

11. Number of fully-filled squares = 4

Number of half-filled squares = 2

∴ Area covered by fully-filled squares

= (4 x 1) sq units = 4 sq units

Area covered by half-filled squares

⇒ \(\left(2 \times \frac{1}{2}\right) \text { sq units }=1 \text { sq units }\)

Total area = (4+1) sq units = 5 sq units

12. Number of fully-filled squares = 3,

Number of half-filled squares* 2,

Number of more than half-filled

squares = 4

and number of less than half-filled squares = 4.

Now, estimated area covered by fully-filled squares = 3 sq units, half-filled squares = \(\left(2 \times \frac{1}{2}\right) \text { sq units }\)

=1 sq unit

more than half-filled squares = 4 sq units and less than half-filled squares = 0 sq unit

∴ Total area- (3 +1 + 4 + 0) sq units = 8 sq units

13. Number of fully-filled squares = 7,

Number of more than half-filled

squares = 7

The number of less than half-filled

squares = 5

Estimated area covered by

fully-filled squares = 7 sq units,

more than half-filled squares = 7 sq units

and less than half-filled squares = 0 sq unit

∴ Total area = (7 + 7 + 0) sq units = 14 sq units

14. Number of fully-filled squares = 10,

Number of more than half-filled squares = 8

and number of less than half-filled squares = 5

Estimated area covered by

fully-filled squares = 10 sq units,

more than half-filled squares = 8 sq units

less than half-filled squares = 0 squint

∴ Total area= (10 + 8 + 0) sq units = 18 sq units

Mensuration Exercise – 10.3

Question 1. Find the areas of the rectangles whose sides are:

1. 3 cm and 4 cm
2. 12 m and 21 m
3. 2 km and 3 km
4. 2 m and 70 cm

Solution:

1. Area of rectangle = length x breadth = 3 cm x 4 cm = 12 cm²
2. Area of rectangle- length x breadth = 12 m x 21 m = 252 m²
3. Area of rectangle = length x breadth = 2 km x 3km = 6 km²
4. Area of rectangle = length x breadth = 2 m x 70 cm = 2 m x 0.7 m = 1.4 m²

Question 2. Find the areas of the squares whose sides are:

1. 10 cm
2. 14 cm
3. 5 m

Solution:

1. Area of square = side x side = 10 cm x 10 cm = 100 cm²
2. Area of square = side x side = 14 cm x 14 cm = 196 cm²
3. Area of square = side x side =5mx5m = 25m²

Question 3. The length and breadth  of the three rectangles are as given below:

1. 9 m and 6 m
2. 17 m and 3 m
3. 4 m and 14 m

Which one has the largest area and which one has the smallest?
Solution:

1. Area of rectangle = length x breadth =9mx6m= 54m²
2. Area of rectangle = length x breadth = 17m x 3 m = 51 m²
3. Area of rectangle = length x breadth = 4 m x 14 m = 56 m²

Thus, rectangle (c) has the largest area, therefore 56 m2 and rectangle (b) has the smallest area, therefore, 51 m²

Question 4. The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
Solution:

Length of rectangle = 50 m

Area of rectangle = 300 m²

Since, area other rectangle = length x breadth

Therefore, breadth  \(=\frac{\text { area of rectangle }}{\text { length }}\)

⇒ \(=\frac{300}{50} \mathrm{~m}=6 \mathrm{~m}\)

Thus, the breadth of the garden is 6 m

Question 5. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq m?
Solution:

Length of land = 500 m

Breadth of land = 200 m

Area of land = length x breadth = 500 m x 200 m = 1,00,000 sqm

Cost of tiling 100 sq m of land = ₹ 8

Cost of tiling 1,00,000 sq m of land

⇒ \(₹ \frac{8 \times 100000}{100}=₹ 8000\)

Question 6. A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Solution:

Length of table-top = 2 m

Breadth of table-top =1 m 50 cm = 1.50m

Area of table-top = length x breadth = 2 m x 1.50 m = 3 m²

Question 7. A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?
Solution:

Length of room = 4 m

And breadth of room = 3 m 50 cm = 3.50 m

Area of carpet = length x breadth = 4 m x 3.50 m = 14 m²

Question 8. A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Solution:

Length of floor = 5 m

The d breadth of the floor = 4 m

Area of floor = length x breadth = 5mx4m = 20m²

Now, side the f square carpet = 3 m

Area of square carpet = side x side = 3mx3m = 9m²

Area of floor that is not carpeted = 20 m²- 9 m² = 11 m²

Question 9. Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?
Solution:

Side of square flower bed =1 m

Area of square flower bed = side x side =lm x lm = lm²

Area of 5 square flower beds = (1×5) m²

= 5 m²

Now, length of land = 5 m

The d breadth of land = 4 m

Area of land = length x breadth – 5 m x 4 m

=20 m²

Area of remaining part

= Area of land- Area of 5 flower beds

= 20 m²- 5 m² = 15 m²

Question 10. By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Solution:

1. we have,

Area of square HKLM = 3×3 cm²= 9 cm²

Area of rectangle IJGH-1×2 cm² = 2 cm²

Area of square FEDG = 3×3 cm² = 9 cm²

Area of rectangle ABCD = 2×4 cm² = 8 cm²

Total area of the figure = (9 + 2 + 9 + 8) cm² = 28 cm²

2. we have,

Area of rectangle ABCD =3×1 cm² = 3 cm²

Area of rectangle BJEF = 3×1 cm² = 3 cm²

Area of rectangle FGHI = 3×1 cm² = 3 cm²

Total area of the figure = (3 + 3 + 3) cm² = 9 cm²

Question 11. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres).

Solution:

1.  We have,

Area of rectangle ABCD = 2 x 10 cm² = 20 cm²

Area of rectangle DEFG = 10 x 2 cm² = 20 cm²

Total area of the figure = (20 + 20) cm²

= 40 cm²

2.  We have,

There are 5 squares each of side 7 cm.

Area of one square = 7×7 cm² = 49 cm²

Area of 5 squares = 5 x 49 cm² = 245 cm²

3.  We have,

Area of rectangle ABCD = 5 x 1 = 5 cm²

Area of rectangle EFGH = 4 x 1 = 4 cm²

Total area of the figure = (5+4) cm²

= 9 cm²

Question 12. How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively:

1. 100 cm and 144 cm
2. 70 cm and 36 cm.

Solution:

1.  The area of the rectangular region

=length x breadth= 100 cm x 144 cm= 14400 cm²

Area of one tile = 12 cm x 5 cm = 60 cm²

Number of tiles = \(=\frac{\begin{array}{r}
\text { Area of rectangular region }
\end{array}}{\text { Area of one tile }}\)

\(=\frac{14400}{60}=240\)

Thus, 240 tiles are required.

2.  The area of the rectangular region

= length x breadth = 70 cm x 36 cm = 2520 cm²

Area of one tile = 12 cm x 5 cm- 60 cm²

Number of tiles \(=\frac{\text { Area of rectangular region }}{\text { Area of one tile }}=\frac{2520}{60}=42\)

Thus, 42 tiles are required.

Data Handling

Data Handling Exercise – 9.1

Question 1. In a Mathematics test, the following marks were obtained by 40 students. Arrange these marks in a table using tally marks.

1. Find how many students obtained marks equal to or more than 7.
2. How many students obtained marks below 4?

Solution:

5 + 4 + 3 = 12 students obtained marks equal to or more than 7.

2 + 3 + 3 = 8 students obtained marks below 4.

Question 2. Following is the choice of sweets for students of Class VI.

Ladoo, Barfi, Ladoo, Jalebi, Ladoo, Rasgulla, Jalebi, Ladoo, Barfi, Rasgulla, Ladoo, Jalebi, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Ladoo, Ladoo, Barfi, Rasgulla, Rasgulla, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo.

1. Arrange the names of sweets on a table using tally marks.
2. Which sweet is preferred by most of the students?

Solution:

Ladoo is preferred by most of the students.

Question 3. Catherine threw a die 40 times and noted the number appearing each time as shown below:

Make a table and enter the data using tally marks. Find the number that appeared

1. The minimum number of times.
2. Tire maximum number of times.
3. Find those numbers that appear an equal number of times.

Solution:

1. 4 appeared a minimum number of times.
2. 5 appeared the maximum number of times
3. 1 and 6 appeared an equal number of times.

Question 4. The following pictograph shows the number of tractors in five villages.

Observe the pictograph and answer the following questions.

1. Which village has the minimum number of tractors?
2. Which village has the maximum number of tractors?
3. How many more tractors does village C have as compared to village B?
4. What is the total number of tractors in all the five villages?

Solution:

1. VillageD has the minimum number of tractors
2. Village C has the maximum number of tractors.
3. Village C has 8-5=3 more tractors than village B.
4. Total number of tractors =6+5+8+3+6 = 28

Question 5. The number of female students in each class of a co-educational middle school is depicted by the pictograph :

1. Observe this pictograph and answer the following questions:
2. Which class has the minimum number of girl students?
3. Is the number of girls in Class 6 less than the number of girls in Class 5?
4. How many girls are there in Class 7?

Solution:

1. Class 6 has the minimum number of female students.
2. No, the number of girls in Class 6 is greater than the number of girls in Class 5.
3. There are 12 girls in Class 7

Question 6. The sale of electric bulbs on different days of the week is shown below:

Observe the pictograph and answer the following questions:

1. How many bulbs were sold on Friday?
2. On which day were the maximum number of bulbs sold?
3. On which of the day’s same number of bulbs were sold?
4. On which of the day is a minimum number of bulbs sold?
5. If one big carton can hold 9 bulbs. How many cartons were needed in the given week?

Solution:

1. Number of bulbs sold on Friday is 14.
2. A maximum number of bulbs were sold on Sunday.
3. Same number of bulbs were sold on Wednesday and Saturday.
4. A minimum number of bulbs were sold on Wednesday and Saturday.
5. The long number of bulbs sold in the given week – 86

Number of cartons required for 9 bulbs- 1

∴ Number of cartons required for 86 bulbs = 86 + 9 = 9.55 = 10

Therefore, 10 cartons were needed in the given week

Question 7. In a village, six fruit merchants sold the following number of fruit baskets in a particular season:

Observe this pictograph and answer the following questions:

1. Which merchant sold the maximum number of baskets?
2. How many fruit baskets were sold by Anwar?
3. The merchants who have sold 600 or more baskets are planning to buy a godown for the next season. Can you name them?

Solution:

1. Martin sold the maximum number of baskets.
2. 700 fruit baskets were sold by Anwar.
3. Anwar, Martin, and Ranjit Singh have sold more than 600 baskets.

Class 6 Maths Chapter 3 Playing With Numbers Exercises

1. Write all the factors of the following numbers:

(1) 24
(2) 15
(3) 21
(4) 27
(5) 12
(6) 20
(7) 18
(8)23
(9) 36

Solution: (1) 24 =1×24 = 2×12 = 3 x 8 = 4 x 6 = 6 x4

Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24

(2) 15 =1×15 = 3×5 = 5×3

Factors of 15 are 1, 3, 5 and 15

(3) 21 = 1×21 = 3×7 = 7×3

Factors of 21 are 1, 3, 7 and 21

(4) 27 =1×27 = 3×9 = 9×3

Factors of 27 are 1, 3, 9 and 27

(5) 12 =1×12 = 2×6 =3×4 = 4×3 = 6×2

Factors of 12 are 1, 2, 3, 4, 6 and 12

(6) 20 =1×20 = 2×10 = 4×5 = 5×4

Factors of 20 are 1, 2, 4, 5, 10 and 20

(7) 18 = 1×18 = 2×9 = 3×6 = 6×3 = 9×2

Factors of 18 are 1, 2, 3, 6, 9 and 18

(8)23 = 1×23

Factors of 23 are1 and 23

(9) 36 =1×36 = 2×18 = 3×12 = 4×9 =6×6

Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36

2. Write first five multiples of:

(1) 5
(2) 8
(3) 9

Solution: (1) 5 x 1 = 5, 5 x 2 = 10, 5 x 3 = 15, 5 x 4 = 20, 5 x 5 = 25

First five multiples of 5 are 5, 10, 15, 20, 25.

(2) 8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 4 = 32, 8 x 5 = 40

First five multiples of 8 are 8, 16, 24, 32, 40.

(3) 9 x 1 = 9, 9 x 2 = 18, 9 x 3 = 27, 9 x 4 = 36, 9×5 = 45

First five multiples of 9 are 9, 18, 27, 36, 45

3. Match the items in column 1 with the items in column 2.

Column 1                                  Column 2

(i) 35                                    (1) Multiple of 8
(ii) 15                                   (2) Multiple of 7
(iii) 16                                  (3) Multiple of 70
(iv) 20                                  (4) Factor of 30
(v) 25                                   (5) Factor of 50
(6) Factor of 20

Solution:  (i) -> (2); (ii) -> (4); (iii) (1); (iv) -> (6); (v) -> (5)

(1) Multiples of 8 are 8, 16, 24, 32, 40,
(2) Multiples of 7 are 7, 14, 21, 28, 35,
(3) Multiples of 70 are 70, 140, 210,
(4) Factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30.
(5) Factors of 50 are 1, 2, 5, 10, 25.
(6) Factors of 20 are 1, 2, 4, 5, 10, 20.

4. Find all the multiples of 9 upto 100.

Solution: Multiples of 9 upto 100 are 9, 18, 27, 36,45, 54, 63, 72, 81, 90, 99

Exercise – 3.2

1. What is the sum of any two
(1) Odd numbers?
(2) Even numbers?

Solution: (1) The sum of any two odd numbers is an even number.

As like, 1+3 = 4, 3 + 5 = 8

(2) The sum of any two even numbers is an even number.

As like, 2 + 4 = 6, 6 + 8 = 14

2. State whether the following statements are True or False:

(1) The sum of three odd numbers Is even.
(2) The sum of two odd numbers and one even number is even.
(3) The product of three odd numbers is odd.
(4) If an even number is divided by 2, the quotient is always odd.
(5) All prime numbers are odd.
(6) Prime numbers do not have any factors.
(7) Sum of two prime numbers is always even.
(8)2 is the only even prime number.
(9) All even numbers are composite numbers.
(10) The product of two even numbers is always even.

Solution: (1) False

Since, sum of two odd numbers is even and sum of one odd number and one

(2) True

Since, sum of two odd numbers is even and sum of two even numbers is always even.

(3) True

(4) False

If an even number is divided by 2, then the quotient is either odd or even.

(5) False
Since, prime number 2 is even.

(6) False

Factors of prime numbers are1 and the number itself.

(7) False

Sum of two prime numbers is either even or odd.

(8)True

(9) False

Since, even number 2 is prime i.e., not composite.

(10) True

3. The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers upto 100.

Solution: Pairs of prime numbers having same digits upto 100 are 17 and 71; 37 and 73; 79 and 97

4. Write down separately the prime and composite numbers less than 20.

Solution: Prime numbers less than 20 are 2, 3, 5,7, 11, 13, 17, 19

Composite numbers less than 20 are 4, 6, 8, 9,10, 12, 14, 15, 16, 18

5. What is the greatest prime number between 1 and 10?

Solution: The greatest prime number between 1 and 10 is 7.

6. Express the following as the sum of two odd primes.

(1) 44
(2) 36
(3) 24
(4) 18

Solution: (1) 44 = 3 + 41

(2) 36 = 5 + 31

(3) 24 = 7 + 17

(4) 18 = 7 +11

7. Give three pairs of prime numbers whose difference is 2. [Remark: Two prime numbers whose difference is 2 are called twin primes].

Solution: Three pairs of prime numbers whose difference is 2 are 3 and 5; 5 and 7; 11 and 13.

8. Which of the following numbers are prime?
(1) 23
(2) 51
(3) 37
(4) 26

Solution: 23 and 37 are prime numbers and 51 and 26 are composite numbers. Thus, numbers in option (1) and (3) are prime.

9. Write seven consecutive composite numbers less than 1 00 so that there is no prime number between them.

Solution: Seven consecutive composite numbers less than 100 are 90, 91, 92, 93, 94, 95, 96

10. Express each of the following numbers as the sum of three odd primes:

(1) 21
(2) 31
(3) 53
(4) 61

Solution: (1) 21 = 3 + 7 + 11

(2) 31 = 3 + 11 + 17

(3) 53 = 13 + 17 + 23

(4) 61 = 13 + 19 + 29

11. Write five pairs of prime numbers less than 20 whose sum is divisible by 5. (Hint: 3 + 7 = 10)

Solution: Since, 2 + 3 = 5; 7 + 13 = 20; 3 + 17 = 20; 2 + 13 = 15; 5 + 5 = 10 and 5, 10, 15, 20 all are divisible by 5.

So, five pairs of prime numbers less than 20 whose sum is divisible by 5 are 2, 3; 2, 13; 3,17; 7, 13; 5, 5.

12. Fill in the blanks:

(1) A number which has only two factors is called a____________.

Solution: Prime number

(2) A number which has more than two factors is called a__________.

Solution: Composite number

(3) 1 is neither__________ nor________.

Solution: Prime number, composite number

(4) The smallest prime number is________.

Solution: 2

(5) The smallest composite number is__________.

Solution: 4

(6) The smallest even number is________.

Solution: 2

Exercise – 3.3

1. Using divisibility tests, determine which of the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 1 0; by 11 (say, yes or no):

Solution:

 

2. Using divisibility tests, determine which of the following numbers are divisible by 4; by 8:

(1) 572
(2) 726352
(3) 5500
(4) 6000
(5) 12159
(6) 14560
(7) 21084
(8)31795072
(9) 1700
(10) 2150

Solution: (1) 572 is divisible by 4 as its last two digits are divisible by 4, butit is not divisible by 8 as its last three digits are not divisible by 8.

(2) 726352 is divisible by 4 as its last two digits are divisible by 4 and it is also divisible by 8 as its last three digits are divisible by 8.

(3) 5500 is divisibleby 4 asits last two digits are divisible by 4, but it is not divisible by 8 as its last three digits are not divisible by 8.

(4) 6000 is divisibleby 4 asits last two digits are divisible by 4 and it is also divisible by 8 as its last three digits are divisible by 8.

(5) 12159 is not divisible by 4 and 8 as it is an odd number.

(6) 14560 is divisible by 4 as its last two digits are divisible by 4 and it is also divisible by 8 as its last three digits are divisible by 8.

(7) 21084 is divisible by 4 as its last two digits are divisible by 4, but it is not divisible by 8 as its last three digits are not divisible by 8.

(8)31795072 is divisible by 4 as its last two digits are divisible by 4 and it is also divisible by 8 as its last three digits are divisible by 8.

(9) 1700 is divisibleby 4 as its last two digits are divisible by 4, but it is not divisible by 8 as its last three digits are not divisible by 8.

(10) 2150 is not divisible by 4 as its last two digits are not divisible by 4 and it is not divisible by 8 as its last three digits are not divisible by 8

3. Using divisibility tests, determine which of following numbers are divisible by 6:

(1) 297144
(2) 1258
(3) 4335
(4) 61233
(5) 901352
(6) 438750
(7) 1790184
(8)12583
(9) 639210
(10) 17852

Solution: (1) 297M4 is divisible by 2 as its ones place is aneven number and it is also divisible by 3 as sum of its digits (= 27) is divisible by 3.

Since, the number is divisible by both 2 and 3. Therefore, it is also divisible by 6.

(2) 1258 is divisible by 2 as its ones place is an even number, but it is not divisible by 3 as sum of its digits (= 16) is not divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore,itisnot divisible by 6.

(3) 4335 isnot divisibleby 2 as its ones place is not an even number, butit is divisible by 3 as sum ofits digits (=15) is divisible
by 3.

Since, the number is not divisible by both 2 and 3. Therefore,itisnot divisible by 6.

(4) 61233 is not divisible by 2 as its place is not an even number, but it is divisibleby 3 as sum ofits digits (= 15) is divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore,itisnot divisible by 6.

(5) 901352 is divisible by 2 as its ones place is an even number, butit is not divisible by 3 as sum of its digits (= 20) is not divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore,itisnot divisible by 6.

(6) 438750 is divisible by 2 as its ones place is an evennumber anditis also divisible by 3 as sum ofits digits (= 27) is divisible by 3.

Since, the number is divisible by both 2 and 3. Therefore, it is also divisible by 6.

(7) 1790184 is divisibleby 2 as its ones place is an even number anditis also divisible by 3 as sum ofits digits (= 30) is divisible
by 3.

Since, the number is divisible by both 2 and 3. Therefore, it is also divisible by 6.

(8)12583 is not divisible by 2 as its ones place is not an even number and it is alsonot divisibleby 3 as sum ofits digits
(= 19) is not divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore, it is not divisible by 6.

9. 639210 is divisible by 2 as its ones place is an even number and it is also divisible by 3 as sum ofits digits (= 21) is divisible
by 3.

Since, the number is divisible by both 2 and 3. Therefore, it is divisible by 6.

10. 17852 is divisible by 2 as its ones place is an evennumber, butit is not divisible by 3 as sum of its digits (= 23) is not
divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore,itisnot divisible by 6.

4. Using divisibility tests, determine which of the following numbers are divisible by 11:

(1) 5445
(2) 10824
(3) 7138965
(4) 70169308
(5) 10000001
(6) 901153

Solution: (1) In 5445, sum of the digits at odd places = 5 + 4 = 9

Sum of the digits at even places = 4 + 5=9

Difference ofboth sums =9-9 = 0

Since the difference is 0. Therefore, the number is divisible by 11.

(2) In 10824, sum of the digits at odd places = 4 + 8 +1 = 13

Sum of the digits at even places = 2 + 0=2

Difference of both sums = 13- 2 = 11

Since the difference is divisible by 11.

Therefore, the number is divisibleby 11.

(3) In 7138965, sum of the digits at odd places = 5 + 9 + 3 + 7 = 24

Sum of the digits at even places = 6 + 8 +1 = 15

Difference ofboth sums = 24- 15 = 9

Since the difference is neither 0 nor divisible by 11. Therefore, the number is not divisible by 11.

(4) In 70169308, sum of the digits at odd places = 8 + 3 + 6 + 0 = 17

Sum of the digits at even places =0+9+1+7=17

Difference ofboth sums = 17- 17 = 0

Since the difference is 0. Therefore, the number is divisible by 11.

(5) In 10000001, sum of the digits at odd places =l + 0 + 0 + 0 =l

Sum of the dibits at even places = 0 + 0 + 0+1=1

Difference of both sums =1-1=0

Since the difference is 0. Therefore, the number is divisible by 1 1.

(6) In 901 153, sum of the digits at odd places =3+1+ 0=4

Sum of the digits at even places = 5 + 1+9 = 15

Difference of both sums = 15- 4 = 11

Since the difference is 11. Therefore, the number is divisible by 11.

5. Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3:

(1) _6724
(2) 4765 _2

Solution: We know that a number is divisible by 3 if the sum of all digits is divisible by 3.

(1) The smallest digit will be 2.

Tire number formed is 26724 and 2 + 6 + 7 + 2 + 4 = 21, which is divisible by 3.

And the greatest digit will be 8.

The number formed is 86724 and 8 + 6 + 7 + 2 + 4 = 27, which is divisible by 3.

(2) The smallest digit will be 0.

The number formed is 476502 and 4 + 7 + 6 + 5 + 0 + 2 = 24, which is divisible by 3.

And the greatest digit will be 9.

The number formed is 476592 and 4 + 7+ 6 + 5 + 9 + 2 = 33, which is divisible by 3.

6. Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11 :

(1) 92 _ 389
(2) 8_ 9484

Solution: (1) We know that a number is divisible by 11 if the difference of the sum of the digits at odd places and that of even places is either 0 or divisible by 11.

The number formed is 928389

Sum of digits at odd places = 9 + 3 + 2 = 14

Sum of digits at even places = 8 + 8 + 9 = 25

Their difference = 25 – 14 = 11, which is divisible by 11.

(2) We know that a number is divisible by 11 if the difference of the sum of the digits at odd places and that of even places Is either 0
or divisible by II.

The number formed is 869484

Sum of digits at odd places – 4 + 4 + 6- 14

Sum of digits at even places – 8 + 9 + 8 = 25

Their difference – 25 – 14 – 11, which is divisible by 11.

Exercise – 3.4

1. Find the common factors of:

(1) 20 and 28
(2) 15 and 25
(4) 56 and 120
(3) 35 and 50

Solution: (1) Factors of 20 are 1, 2, 4, 5, 10 and 20

Factors of 28 are 1, 2, 4, 7, 14 and 28

Common factors of 20 and 28 are 1, 2 and 4

(2) Factors of 15 are 1, 3, 5 and 15

Factors of 25 are 1, 5 and 25

Common factors of 15 and 25 are1 and 5

(3) Factors of 35 are 1, 5, 7 and 35

Factors of 50 are 1, 2, 5, 10, 25 and 50

Common factors of 35 and 50 are 1 and 5

(4) Factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56

Factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12,15, 20, 24, 30, 40, 60 and 120

Common factors of 56 and 120 are 1, 2, 4 and 8

2. Find the common factors of: (1) 4, 8 and 12 (2) 5, 15 and 25

Solution: (1) Factors of 4 are 1, 2 and 4

Factors of 8 are 1, 2, 4 and 8

Factors of 12 are 1, 2, 3, 4, 6 and 12

Common factors of 4, 8 and 12 are 1, 2 and 4

(2) Factors of 5 are1 and 5

Factors of 15 are 1, 3, 5 and 15

Factors of 25 are 1, 5 and 25

Common factors of 5, 15 and 25 are 1 and 5

3. Find first three common multiples of: (1) 6 and 8

Solution: (1) Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72,………..

Multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, (2) 12 and 18 72,……….

First three common multiples of 6 and 8 are 24, 48 and 72

(2) Multiples of 12 are 12, 24, 36, 48, 60, 72,84, 96, 108, 120,……….

Multiples of 18 are 18, 36, 54, 72, 90, 108, 126,……….

First three common multiples of 12 and 18 are 36, 72 and 108

4. Write all the numbers less than 1 00 which are common multiples of 3 and 4.

Solution: Multiples of 3 are 3, 6, 9, 12, 15, 18, 21,24. 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60,63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99,…………

Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36,40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88,92, 96, 100,………..

Common multiples of 3 and 4 which are less than 100 are 12, 24, 36, 48, 60, 72, 84 and 96

5. Which of the following numbers are co-prime?

(1) 18 and 35
(2) 15 and 37
(4) 17 and 68
(6) 81 and 16
(3) 30 and 415
(5) 216 and 215

Solution: (1) Factors of 18 are 1, 2, 3, 6, 9 and 18

Factors of 35 are 1, 5, 7 and 35

Common factor of 18 and 35 is1

Since, both have only one common factor,i.e., 1. Therefore, 18 and 35 are co-prime numbers.

(2) Factors of 15 are 1, 3, 5 and 15

Factors of 37 are 1 and 37

Common factor of 15 and 37 is 1

Since, both have only one common factor, i.e., 1. Therefore, 15 and 37 are co-prime numbers.

(3) Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30 ……, 83 and 415

Factors of 415 are 1, 5,

Common factors of 30 and 415 are 1 and 5 ?

Since, bothhave more than one common factor. Therefore, 30 and 415 are not co-prime numbers.

(4) Factors of 17 are 1 and 17

Factors of 68 are 1, 2, 4, 17, 34 and 68

Common factors of 17 and 68 are1 and 17

Since, both have more than one common factor. Therefore, 17 and 68 are not co-prime numbers.

(5) Factors of 216 are 1, 2, 3, 4, 6, 8, 9, 12, 18,24, 27, 36, 54, 72, 108 and 216

Factors of 215 are 1, 5, 43 and 215

Common factor of 216 and 215 is1

Since, both have only one common factor, i.c., I. Therefore, 216 and 215 are co-prime numbers.

(6) Factors of 81 are 1, 3, 9, 27 and 81

Factors of 16 are 1, 2, 4, 8 and 16

Common factor of 81 and 16 is 1

Since, both have only one common factor, i.e., 1. Therefore, 81 and 16 are co-prime numbers.

6. A number is divisible by both 5 and 12. By which other number will that number be always divisible?

Solution: Since 5 x 12 = 60. The number divisible by both 5 and 12, must also be divisible by 60.

7. A number is divisible by 12. By what other numbers will that number be divisible?

Solution: Factors of 12 are 1, 2, 3, 4, 6 and 12.

Therefore, the number divisible by 12, will also be divisible by 1, 2, 3, 4 and 6.

Exercise – 3.5

1. Here are two different factor trees for 60. Write the missing numbers.

Solution:

Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

(1) Since 6 = 2*3 and 10 = 5 * 2

The missing numbers are 3 and 2.

(2) Since, 60- 30 x 2 30- 10 x 3 and 10- 5 x 2

2. Which factors are not included in the prime factorisation of a composite number?

Solution:1 and the number it self are not included in the prime factorisation of a composite number.

3. Write the greatest 4-digit number and express it in terms of its prime factors.

Solution: The greatest four-digit number is 9999.

9999 = 3x3x 11 x101.

4. Write the smallest 5-digit number and express it in the form ofits prime factors.

Solution: The smallest five digitnumber is 10000.

10000 = 2x2x2x2x5x5x5x5.

5. Find all the prime factors of 1 729 and arrange them in ascending order. Now state the relation, if any; between two consecutive
prime factors.

Solution:

1729 = 7 x 13 x 19.

The difference of two consecutive prime factors is 6. (v 13- 7 = 6 and 19- 13 = 6)

6. The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.

Solution: Among the three consecutive numbers, there must be atleast one even number and one multiple of 3. Thus, the product must be divisible by 6.

For example: (1) 2 x 3 x 4 = 24
(2) 4x5x6 = 120, where both 24 and 120 are divisible by 6.

7. The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.

Solution: The sum of two consecutive odd numbers is divisible by 4.

For example : 3 + 5 = 8 and 8 is divisible by 4.

5 + 7 = 12 and 12 is divisible by 4.

7 + 9 = 16 and 16 is divisible by 4.

9 + 11 = 20 and 20 is divisible by 4.

8. In which of the following expressions, prime factorisation has been done?

(1) 24 = 2x3x4
(2) 56 = 7x2x2x2
(3) 70 = 2x5x7
(4) 54 = 2x3x9

Solution: In expressions (2) and (3), prime factorisation has been done.

9. 18 is divisible byboth 2 and 3.It is also divisible by 2 x 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 x 6 = 24? Ifnot, give an example to justify your answer.

Solution: No. The number 12 is divisible by both 6 and 4, but 12 is not divisible by 24.

A number divisible by both 4 and 6 may or may not be divisible by 4 x 6 = 24.

10. I am the smallest number, having four different prime factors. Can you find me?

Solution: Since, 2 x 3 x 5 x 7= 210

210 is the smallest number, having 4 different prime factors i.e, 2, 3, 5 and 7.

Exercise – 3.6

1. Find the HCF of the following numbers:

(1) 18,48
(2) 30,42
(3) 18,60
(4) 27,63
(5) 36,84
(6) 34,102
(7) 70,105,175
(8) 91,112,49
(9) 18,54,81
(10) 12,45,75

Solution: (1) The prime factorisation of 18 and

48 are; 18 = 2 x 3 x 3

48=2x2x2x2x3

HCF (18, 48) = 2×3 = 6

(2) The prime factorisation of 30 and 42 are; 30 = 2 x 3 x 5

42 = 2x3x7

HCF (30, 42) =2×3 = 6

(3) The prime factorisation of 18 and 60 are; 18 = 2 x 3 x 3

60 = 2x2x3x5

HCF (18, 60) = 2×3 = 6

(4) The prime factorisation of 27 and 63 are; 27 = 3 x 3 x 3

63 =3x3x7

HCF (27, 63) = 3×3 = 9

(5) The prime factorisation of 36 and 84 are; 36 = 2x2x3x3

84 = 2x2x3 x7

HCF (36, 84) = 2 x 2 x 3 = 12

(6) The prime factorisation of 34 and 102 are; 34 = 2 x 17

102 = 2x3x17

HCF (34, 102) = 2×17 = 34

(7) The prime factorisation of 70, 105 and 175 are; 70 = 2 x 5 x 7

105 =3x5x7

175 =5x5x7

HCF (70, 105, 175) = 5 x 7 = 35

(8)The prime factorisation of 91, 112 and

49 are; 91 = 7 x 13

112 = 2x2x2x2x7

49 = 7 x 7

HCF (91, 112, 49) = 7

(9) The prime factorisation of 18, 54 and 81 are; 18 = 2 x 3 x 3

54 = 2x3x3x3

81=3x3x3x3

HCF (18, 54, 81) = 3×3 = 9

(10) The prime factorisation of 12, 45 and 75 are; 12 = 2 *2 *3

45 = 3 x 3 x 5

75 = 3 x 5 x 5

HCF (12, 45, 75) = 3

2. What is the HCF of two consecutive

(1) numbers?
(2) even numbers?
(3) odd numbers?

Solution:(1) HCF of two consecutive numbers is 1.

(2) HCF of two consecutive even numbers is 2.

(3) HCF of two consecutive odd numbers is 1.

3. HCF of co-prime numbers 4 and 1 5 was found as follows by factorisation: 4 = 2×2 and 15 = 3×5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?

Solution: No. The correct HCF of 4 and 15 is 1.

Exercise – 3.7

1. Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.

Solution: For finding maximum weight, we have to find HCF of 75 and 69.

The prime factorisation of 75 and 69 are;

75 = 3x5x5

69 = 3×23

So, HCF of 75 and 69 = 3

Therefore, the required weight is 3 kg.

2. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?

Solution: For finding minimum distance, we have to find LCM of 63, 70, 77.

LCM of 63, 70 and 77 = 2x3x3x 5 x7x 11 = 6930

Therefore, the minimum distance three boys should cover is 6930 cm.

3. The length, breadth and height of a room are S25 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.

Solution: The measurement of the longest tape = HCF of 825 cm, 675 cm and 450 cm.

The prime factorisation of 825, 675 and 450 are;

825 =3x5x5x 11

675 =3x3x3x5x5

450 = 2x3x3x5x5

Now, HCF of 825, 675 and 450 = 3 x 5 x 5 = 75

Therefore, the measurement of longest tape is 75 cm.

4. Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 1 2.

Solution: The smallest 3-digit number = 100

LCM of 6, 8 and 12 =2x2x2x3 = 24

To find the number, we have to divide 100 by 24.

Therefore, the required number

= 100 + (24 -4) = 120.

5. Determine the greatest 3-digit number exactly divisible by 8, 1 0 and 1 2.

Solution: The greatest three digit number = 999.

LCM of 8, 10 and 12=2x2x2x3x5=120

Now, to find the number, we have to divide 999 by 120.

Therefore, the required number = 999- 39 = 960

6. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?

Solution:

LCM of 48, 72 and 108 = 2 x 2 x 2 x 2 x 3 x 3 x 3 = 432

After 432 seconds, the traffic lights change simultaneously.

432 seconds = 7 minutes 12 seconds

Therefore, the required time = 7 a.m. + 7 minutes 12 seconds i.e., 7 minutes 12 seconds past 7 a.m.

7. Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.

Solution: The maximum capacity of a container = HCF (403, 434, 465)

The prime factorisation of 403, 434 and 465 are;

403 = 13 x 31

434 = 2x7x31

465 = 3x5x31

HCF of 403, 434,465 = 31

Therefore, a container of capacity 31 litres is required to measure the diesel of the three containers.

8. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case

Solution:

LCM of 6, 15 and 18 = 2 x 3 x 3 x 5 = 90

Therefore, the required number = 90 + 5 = 95

9. Find the smallest 4-digit number which is divisible by 18, 24 and 32.

Solution: The smallest four digit number = 1000

LCM of 18, 24 and 32 = 2x2x2x2x2x3x3 = 288

Now,

Therefore, the required number is 1000 + (288- 136) = 1152.

10. Find the LCM of the following numbers:

(1) 9 and 4
(2) 12 and 5
(3) 6 and 5
(4) 15 and 4

Observe a common property in the obtained LCMs. Is LCM the product of two numbers in each case?

Solution: (1) We have,

LCM of 9 and 4 = 2x2x3x3 = 36

(2) We have

LCM of 12 and 5 = 2 x 2 x 3 x 5 = 60

(3) We have

LCM of 6 and 5 = 2x3x5 = 30

(4) We have

LCM of 15 and 4 = 2x2x3x5 = 60

Yes, the LCM is equal to the product of two numbers in each case and all LCMs are also the multiple of 3.

11. Find the LCM of the following numbers in which one number is the factor of the other.

(1) 5,20
(2) 6,18
(3) 12,48
(4) 9,45

What do you observe in the results obtained?

Solution:(1) We have

LCM of 5 and 20 = 2 x 2 x 5 = 20

(2) We have,

LCM of 6 and 18 = 2 x 3 x 3 = 18

(3) We have,

LCM of 12 and 48 = 2 x 2 x 2 x 2 x 3 = 48

(4) We have,

LCM of 9and 45 = 3x3x5 = 45

From above, we observe that the LCM of the given numbers in each case is the larger of the two numbers.

Decimals Exercise – 8.1

Question 1. Which is greater?

1. 0.3 or 0.4
2. 0.07 or 0.02
3. 3 or 0.8
4. 0.5 or 0.05
5. 1.23 or 1.2
6. 0.099 or 0.1 9
7. 1.5 or 1.50
8. 1.431 or 1.490
9. 3.3 or 3.300
10. 5.64 or 5.603

Solution: Before comparing, we write both terms in like decimals:

1. 0.3 <0.4
2. 0.07 >0.02
3. 3.0 or 0.8⇒ 3.0 > 0.8
4. 0.50 or 0.05 ⇒ 0.50 > 0.05
5. 1.23 or 1.20 ⇒ 1.23 >1.20
6. 0.099 or 0.190 ⇒ 0.099 < 0.190
7. 1.50 or 1.50 ⇒ 1.50 = 1.50
8. 1.431 < 1.490
9. 3.300 or 3.300 ⇒ 3.300 = 3.300
10. 5.640 or 5.603 ⇒ 5.640 >5.603

Question 2. Make five more examples and find the greater number from them.
Solution: Five examples are :

1. 1.8 or 1.82
2. 1.0009 or 1.09
3. 10.01 or 100.1
4. 5.1 or 5.01
5. 4.213 or 421.3

Before comparing, we write both terms in decimals

1. 1.80 or 1.82 ⇒ 1.82 is greater than 1.8
2. 1.0009 or 1.0900 ⇒ 1.09 is greater than 1.0009
3. 10.01 or 100.10 ⇒ 100.1 is greater than 10.01
4. 5.10 or 5.01 ⇒ 5.1 is greater than 5.01
5. 4.213 or 421.300 ⇒ 421.3 is greater than 4.213

Decimals Exercise – 8.2

1. Express as rupees using decimals.

1. 5 paise
2. 75 paise
3. 20 paise
4. 50 rupees 90 paise
5. 725 paise

Solution:

1 paisa \(=₹ \frac{1}{100}\)

5 paise \(=₹\left(\frac{1}{100} \times 5\right)=₹ 0.05\)

1 paisa \(=₹ \frac{1}{100}\)

75 paise \(=₹\left(\frac{1}{100} \times 75\right)=₹ 0.75\)

1 paisa \(=₹ \frac{1}{100}\)

20 paise \(=₹\left(\frac{1}{100} \times 20\right)=₹ 0.20\)

1 paisa \(=₹ \frac{1}{100}\)

₹ 50 + 90 paise \(=₹\left(50+\frac{1}{100} \times 90\right)\)

= ₹ 50.90

1 paisa \(=₹ \frac{1}{100}\)

725 paise \(=₹\left(\frac{1}{100} \times 725\right)=\frac{725}{100}=₹ 7.25\)

Question 2. Express as metres using decimals.

1. 15 cm
2. 6 cm
3. 2 m 45 cm
4. 9 m 7 cm
5. 419 cm

Solution:

1 cm \(=\frac{1}{100} \mathrm{~m}\)

15 cm \(=\frac{1}{100} \times 15 \mathrm{~m}=0.15 \mathrm{~m}\)

1 cm \(=\frac{1}{100} \mathrm{~m}\)

6 cm \(=\frac{1}{100} \times 6 \mathrm{~m}=0.06 \mathrm{~m}\)

1 cm \(=\frac{1}{100} \mathrm{~m}\)

2 m 45 cm \(2 \mathrm{~m} 45 \mathrm{~cm}=\left(2+\frac{1}{100} \times 45\right) \mathrm{m}=2.45 \mathrm{~m}\)

1 cm \(=\frac{1}{100} \mathrm{~m}\)

9 m 7 cm \(=\left(9+\frac{1}{100} \times 7\right) \mathrm{m}=9.07 \mathrm{~m}\)

1 cm \(=\frac{1}{100} \mathrm{~m}\)

419 cm \(=\frac{1}{100} \times 419 \mathrm{~m}=\frac{419}{100}=4.19 \mathrm{~m}\)

Question 3. Express as cm using decimals.

1. 5 mm
2. 60 mm
3. 164 mm
4. 9 cm 8 mm
5. 93 mm
6. Solution:

1 mm \(=\frac{1}{10} \mathrm{~cm}\)

5 mm \(=\frac{1}{10} \times 5 \mathrm{~cm}=0.5 \mathrm{~cm}\)

1 mm \(=\frac{1}{10} \mathrm{~cm}\)

60 mm \(=\frac{1}{10} \times 60 \mathrm{~cm}=6 \mathrm{~cm}\)

1 mm \(=\frac{1}{10} \mathrm{~cm}\)

164 mm \(=\frac{1}{10} \times 164 \mathrm{~cm}=16.4 \mathrm{~cm}\)

1 mm \(=\frac{1}{10} \mathrm{~cm}\)

9 cm 8 mm \(=\left(9+\frac{1}{10} \times 8\right) \mathrm{cm}=9.8 \mathrm{~cm}\)

1 mm \(=\frac{1}{10} \mathrm{~cm}\)

93 mm \(=\frac{1}{10} \times 93 \mathrm{~cm}=9.3 \mathrm{~cm}\)

Question 4. Express as km using decimals.

1. 8 m
2. 88 m
3. 8888 m
4. 70 km 5 m

Solution:

1 m \(=\frac{1}{1000} \mathrm{~km}\)

8 m \(=\frac{1}{1000} \times 8 \mathrm{~km}=0.008 \mathrm{~km}\)

1 m \(=\frac{1}{1000} \mathrm{~km}\)

88 m\(=\frac{1}{1000} \times 88 \mathrm{~km}=0.088 \mathrm{~km}\)

1 m \(=\frac{1}{1000} \mathrm{~km}\)

8888 m \(=\frac{1}{1000} \times 8888 \mathrm{~km}=8.888 \mathrm{~km}\)

1 m \(=\frac{1}{1000} \mathrm{~km}\)

70 km 5 m \(=\left(70+\frac{1}{1000} \times 5\right) \mathrm{km}=70.005 \mathrm{~km}\)

Question 5. Express as kg using decimals.

1. 2 g
2. 100 g
3. 3750 g
4. 5 kg 8 g
5. 26 kg 50 g

Solution:

1 g \(=\frac{1}{1000} \mathrm{~kg}\)

2 g \(=\frac{1}{1000} \times 2 \mathrm{~kg}=0.002 \mathrm{~kg}\)

1 g \(=\frac{1}{1000} \mathrm{~kg}\)

100 g \(=\frac{1}{1000} \times 100 \mathrm{~kg}=0.1 \mathrm{~kg}\)

1 g \(=\frac{1}{1000} \mathrm{~kg}\)

3750 g \(=\frac{1}{1000} \times 3750 \mathrm{~kg}=3.750 \mathrm{~kg}\)

1 g \(=\frac{1}{1000} \mathrm{~kg}\)

5 kg 8 g \(=\left(5+\frac{1}{1000} \times 8\right) \mathrm{kg}=5.008 \mathrm{~kg}\)

1 g \(=\frac{1}{1000} \mathrm{~kg}\)

26 kg 50 g \(=\left(26+\frac{1}{1000} \times 50\right) \mathrm{kg}=26.050 \mathrm{~kg}\)

Decimals Exercise – 8.3

Question 1. Find the sum in each of the following:

1. 0.007 + 8.5 + 30.08
2. 15 + 0.632 + 13.8
3. 27.076 + 0.55 + 0.004
4. 25.65 + 9.005 + 3.7
5. 0.75 + 10.425 + 2
6. 280.69 + 25.2 + 38

Solution:

1. \(\begin{array}{r}
0.007 \\
8.500 \\
+30.080 \\
\hline 38.587 \\
\hline
\end{array}\)

2. \(+\begin{array}{r}
15.000 \\
0.632 \\
13.800 \\
\hline 29.432 \\
\hline
\end{array}\)

3. \(\begin{array}{r}
27.076 \\
0.550 \\
+0.004 \\
\hline \underline{27.630} \\
\hline
\end{array}\)

4. \(\begin{array}{r}
25.650 \\
9.005 \\
+3.700 \\
\hline 38.355 \\
\hline
\end{array}\)

5. \(\begin{array}{r}
0.750 \\
10.425 \\
+2.000 \\
\hline 13.175 \\
\hline
\end{array}\)

6. \(\begin{array}{r}
280.69 \\
25.20 \\
+38.00 \\
\hline 343.89 \\
\hline
\end{array}\)

Question 2. Rashid spent ₹ 35.75 for the Maths book and ₹ 32.60 for the Science book. Find the total amount spent by Rashid.
Solution:

Money spent formats books =₹ 35.75

Money spent on Science book = ₹ 32.60

Total money spent = ₹ 35.75 + ? 32.60 = ₹ 68.35

Therefore, the total money spent by Rashid is ₹ 68.35

Question 3. Radhika’s mother gave her ₹ 10.50 and her father gave her ₹ 1 5.80, find the total amount given to Radhika by the parents.
Solution:

Money given by mother = ₹ 10.50

Money given by father ₹ 15.80

Total money received by Radhika = ₹ 10.50 + ₹ 15.80 = ₹ 26.30

Therefore, the total money received by Radhika is ₹ 26.30

Question 4. Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for her trouser. Find the total length of clothes bought by her.
Solution:

Cloth bought for shirt = 3 m 20 cm = 3.20 m

Cloth bought for trouser = 2 m 5 cm = 2.05 m

Total length of cloth bought by Nasreen = 3.20 m + 2.05 m = 5.25 m

Therefore, the total length of cloth bought by Nasreen is 5.25 m

Question 5. Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all?
Solution:

Distance travelled in the morning = 2 km 35 m = 2.035 km

Distance travelled in the evening =1 km 7m = 1.007 km

Total distance travelled = (2.035 + 1.007) km- 3.042 km

Therefore, the total distance travelled by Naresh is 3.042 km.

Question 6. Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot to reach her school. How far is her school from her residence?
Solution:

Distance travelled by bus = 15 km 268 m = 15.268 km

Distance travelled by car = 7 km 7m = 7.007 km

Distance travelled on foot = 500 m = 0.500 km

Total distance travelled = (15.268 + 7.007 + 0.500) km = 22.775 km

Therefore, the total distance travelled by Sunita is 22.775 km

Question 7. Ravi purchased 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850g flour. Find the total weight of his purchases.
Solution:

Weight of Rice = 5 kg 400 g = 5.400 kg

Weight of Sugar = 2 kg 20 g = 2.020 kg

Weight flour = 10 kg 850 g = 10.850 kg

Total weight = (5.400 + 2.020 + 10.850) kg = 18.270 kg

Therefore, the total weight of Ravi’s purchases is 18.270 kg

Decimals Exercise – 8.4

Question 1. Subtract:

1. ₹ 18.25 from ₹ 20.75
2. 202.54 m from 250 m
3. ₹ 5.36 from ₹ 8.40
4. 2.051 km from 5.206 km
5. 0.314 kg from 2.107 kg

Solution:

1. \(\begin{array}{r}
20.75 \\
-18.25 \\
\hline 02.50 \\
\hline
\end{array}\)

₹ 20.75- ₹ 18.25 2.50

2. \(\begin{array}{r}
250.00 \\
-202.54 \\
\hline 47.46 \\
\hline
\end{array}\)

250 m- 202.54 m = 47.46 m

3. \(\begin{array}{r}
8.40 \\
-5.36 \\
\hline 3.04 \\
\hline
\end{array}\)

₹ 8.40 – ₹ 5.36 = ? 3.04

4. \(\begin{array}{r}
5.206 \\
-2.051 \\
\hline 3.155 \\
\hline
\end{array}\)

5.206 km- 2.051 km = 3.155 km

5. \(\begin{array}{r}
2.107 \\
-0.314 \\
\hline 1.793
\end{array}\)

2.107 kg – 0.314 kg = 1.793 kg

Question 2. Find the value of:

1. 9.756-6.28
2. 21.05-15.27
3. 18.5-6.79
4. 11.6-9.847

Solution:

1. \(\begin{array}{r}
9.756 \\
-6.28 \\
\hline 3.476 \\
\hline
\end{array}\)

2. \(\begin{array}{r}
21.05 \\
-15.27 \\
\hline 05.78 \\
\hline
\end{array}\)

3. \(\begin{array}{r}
18.50 \\
-6.79 \\
\hline 11.71 \\
\hline
\end{array}\)

4. \(\begin{array}{r}
11.600 \\
-9.847 \\
\hline 1.753 \\
\hline
\end{array}\)

Question 3. Raju bought a book for ₹ 35.65. He gave ₹ 50 to the shopkeeper. How much money did he get back from the shopkeeper?
Solution:

The total amount is given to shopkeeper=₹ 50

Cost of book = ₹ 35.65

Amount left = ₹ 50.00- ₹ 35.65 = ₹ 14.35

Therefore, Raju got back ₹ 14.35 from the shopkeeper.

Question 4. Rani had ₹ 18.50. She bought one ice cream for ₹ 1 1.75. How much money does she have now?
Solution:

Total money = ₹ 18.50

Cost of ice-cream = ₹ 11.75

Amount left = ₹ 18.50- ₹ 11.75 = ₹ 6.75

Therefore, Rani has ₹ 6.75 now

Question 5. Tina had 20 m 5 cm long cloth. She cuts 4m 50 cm length of cloth from this for making a curtain. How many clothes are left with her?
Solution:

Total length of cloth = 20 m 5 cm = 20.05 m

Length of cloth used = 4 m 50 cm = 4.50 m

Remaining cloth = 20.05 m- 4.50 m = 15.55 m

Therefore, 15.55 m of cloth is left with Tina

Question 6. Namita travels 20 km 50 m every day. Out of this, she travels 10 km 200 m by bus and the rest by auto. How much distance does she travel by auto?
Solution:

Total Distance travelled = 20 km 50 m

= 20.050 km

Distance travelled by auto = (20.050- 10.200) km = 9.850 km

Therefore, 9.850 km distance is travelled by auto.

Question 7. Aakash bought vegetables weighing 10 kg. Out of this, 3 kg 500 g is onions, 2 kg 75 g is tomatoes and the rest is potatoes. What is the weight of the potatoes?
Solution:

Weight of onions=3kg 500 g = 3.500kg

Weight of tomatoes = 2 kg 75 g = 2.075 kg

Total weight of onions and tomatoes = (3.500 + 2.075) kg = 5.575 kg

Therefore, weight of potatoes = (10.000- 5.575) kg = 4.425 kg

Thus, the weight of potatoes is 4.425 kg.

Class 6 Maths Chapter 2 Whole Numbers

Exercise – 2.1

1. Write the next three natural numbers after 10999.

Solution: 10,999 +1 = 11,000

11,000 +1 = 11,001

11,001 +1 = 11,002

The next three natural numbers after 10,999 are 11,000; 11,001; 11,002

2. Write the three whole numbers occurring just before 10001.

Solution: 10,001-1 = 10,000

10,000-1 = 9,999

9,999-1 = 9,998

The three whole numbers occurring just before 10,001 are 10,000; 9,999; 9,998

3. Which is the smallest whole number?

Solution: 0 is the smallest whole number.

4. How many whole numbers are there between 32 and 53?

Solution: Since, 53- 32-1 = 20

There are 20 whole numbers between 32 and 53.

5. Write the successor of:

(1) 2440701
(2) 100199
(3) 1099999
(4) 2345670

Solution: (1) Successor of 2440701 is 2440701 +1 = 2440702

(2) Successor of 100199 is 100199 +1 = 100200

(3) Successor of 1099999 is 1099999 +1 = 1100000

(4) Successor of 2345670 is 2345670 +1 = 2345671

6. Write the predecessor of:

(1) 94
(2) 10000
(3) 208090
(4) 7654321

Solution:

(1) The predecessor of 94 is 94-1 = 93

(2) The predecessor of 10000 is 10000-1=9999

(3) The predecessor of 208090 is 208090-1 = 208089

(4) The predecessor of 7654321 is 7654321 -1 = 7654320

7. In each of the following pairs of numbers, state which whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them.

(1) 530,503
(2) 370,307
(3) 98765,56789
(4) 9830415,10023001

Solution: (1) 503 appears on left side of 530 on the number line. So, 530 > 503

(2) 307 appears on left side of 370 on the number line. So, 370 > 307

(3) 56789 appears on left side of 98765 on the number line. So, 98765 > 56789

(4) 9830415 appears on left side of 10023001 on the number line.

So, 9830415 < 10023001

8. Which of the following statements are true (T) and which are false (F)?

(1) Zero is the smallest natural number.
(2) 400 is the predecessor of 399.
(3) Zero is the smallest whole number.
(4) 600 is the successor of 599.
(5) All natural numbers .ire whole numbers.
(6) All whole numbers are natural numbers.
(7) The predecessor of a two digit number is never a single digit number.
(8) 1 is the smallest whole number.
(9) The natural number 1 has no predecessor.
(10) The whole number 1 has no predecessor.
(11) The whole number 1 3 lies between 11 and 12.
(12) The whole number 0 has no predecessor.
(13) The successor of a two digit number is always a two digit number.

Solution:

(1) False Since, zero is not a natural number.

(2) False Since, 400 is the successor of 399.

(3) True

(4) True

(5) True

(6) False

Since, all whole numbers are not natural numbers as 0 is a whole number but not a natural number.

(7) False

Since, predecessor of a two digit number i.e., 10 is a single digit number i.e., 9.

(8) False

Since, 0 is the smallest whole number.

(9) True

(10) False

Since, 0 is the predecessor of1 in whole numbers.

(11) False

Since, 13 lies between 12 and 14.

(12) True

(13) False

Since, successor of two digitnumber i.e.,

99 is a three digit number i.e., 100

Class 6 Maths Chapter 1 Knowing Our Numbers

Exercise -1.1

1. Fill in the blanks:

(1) 1 lakh =________ ten thousand.

Answer: 10: 1 lakh = 1,00,000

= 10 x 10,000 = 10 ten thousand

(2) 1 million=_______ hundred thousand.

Answer: 10:1 million = 1,000,000

= 10 x 100,000 = 10 hundred thousand

(3) 1 crore =__________ ten lakh.

Answer: 10: 1 crore = 1,00,00,000

= 10 x 10,00,000 = 10 ten lakh

(4) 1 crore =__________ million.

Answer: 10: 1 crore = 10 million

(5) 1 million =_________lakh.

Answer: 10: 1 million = 10 lakh

2. Place commas correctly and write the numerals:

(1) Seventy three lakh seventy five thousand three hundred seven.

Answer: 73,75,307

(2) Nine crore five lakh forty one.

Answer: 9,05,00,041

(3) Seven crore fifty two lakh twenty one thousand three hundred two.

Answer: 7,52,21,302

(4) Fifty eight million four hundred twenty three thousand two hundred two.

Answer: 58,423,202

(5) Twenty three lakh thirty thousand ten.

Answer: 23,30,010

3. Insert commas suitably and write the according to Indian System of Numeration:

(1) 87595762

Answer: 8,75,95,762 -> Eight crore seventy five lakh ninety-five thousand seven hundred sixty-two

(2) 8546283

Answer: 85,46,283 -> Eighty-five lakh forty-six thousand two hundred eighty-three.

(3) 99900046

Answer: 9,99,00,046 -> Nine crore ninety-nine lakh forty-six

(4) 98432701

Answer: 9,84,32,701 -» Nine crore eighty-four lakh thirty-two thousand seven hundred one.

4. Insert commas suitably and write the names according to International System of Numeration:

(1) 78921092

Answer: 78,921,092 -> Seventy-eight million nine hundred twenty-one thousand ninetytwo

(2) 7452283

Answer: 7,452,283 —> Seven million four hundred fifty-two thousand two hundred eighty three

(3) 99985102

Answer: 99,985,102 —> Ninety-nine million nine hundred eighty-five thousand one hundred two

(4) 48049831

Answer: 48,049,831 —> Forty-eight million fortynine thousand eight hundred thirty-one

Exercise – 1.2

1. A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.

Solution: Number of tickets sold on first day = 1,094

Number of tickets sold on second day = 1,812

Number of tickets sold on third day = 2,050

Number of tickets sold on fourth day = 2,751

Total tickets sold = 1,094 + 1,812 + 2,050 + 2,751 = 7,707

Therefore, 7,707 tickets were sold on all the four days.

2. Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?

Solution:  Number of runs to achieve = 10,000

Number of runs scored = 6,980

Number of runs required = 10,000- 6,980 = 3,020

Therefore, Shekhar needs 3,020 more runs.

2. Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?

Solution: Number of runs to achieve = 10,000

Number of runs scored = 6,980

Number of runs required = 10,000- 6,980 = 3,020

Therefore, Shekhar needs 3,020 more runs.

3. In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?

Solution: Number of votes secured by successful candidate = 5,77,500

Number of votes secured by his nearest rival = 3,48,700

Margin between them = 5,77,500- 3,48,700 = 2,28,800

Therefore, the successful candidate won by a margin of 2,28,800 votes.

4. Kirti bookstore sold books worth 2,85,891 in the first week of June and books worth 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?

Solution: Worth of books sold in first week = 2,85,891

Worth of books sold in second week = 4,00,768

Total worth of books sold = (2,85,891 +4,00,768) = 6,86,659

Since, 4,00,768 > 2,85,891

Therefore, sale of second week is greater than that of first week by (4,00,768- 2,85,891) = 1,14,877

5. Find the difference between the greatest and the least 5 – digit number that can be written using the digits 6, 2, 7, 4, 3 each only once

Solution: Greatest five-digit number using digits 6,2,7,4,3 = 76432

Smallest five-digit number using digits 6,2,7,4,3 – 23467

Difference = 76432- 23467 52965

Therefore, the difference is 52,965.

6. A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?

Solution: Number of screws manufactured in one day = 2,825

Number of screws manufactured in the month of January (31 days) = 2,825 x 31 = 87,575

Therefore, the machine produced 87,575 screws in the month of January.

7. A merchant had 78,592 with her. She placed an order for purchasing 40 radio sets at 1200 each. How much money will remain with her after the purchase?

Solution: Cost of one radio set = 1200

Cost of 40 radio sets = (1200 x 40) = 48,000

Now, total money with merchant = 78,592

Money left with her =(78,592- 48,000) = 30,592

Therefore,30,592 will remain with her after the purchase

8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer? (Hint: Do you need to do both the multiplications?)

Solution:

Difference in answers = 470340- 405216 = 65,124

9. To stitch a shirt, 2 m 1 5 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain? (Hint: convert data in cm.)

Solution: Cloth required to stitch one shirt = 2 m 15 cm = 2 x 100 cm + 15 cm = 215 cm

Length of cloth = 40 m = 40 * 100 cm = 4000 cm

Number of shirts can be stitched = 4000 / 215

Therefore, 18 shirts can be stitched and 130 cm (1 m 30 cm) cloth will remain.

10. Medicine is packed in boxes, each weighing 4 kg 500 g. How many such boxes can be loaded in a van which cannot carry beyond
800 kg?

Solution: The weight of one box = 4 kg 500 g

= 4 x 1000 g + 500 g = 4500 g

Maximum load can be loaded in a van = 800 kg = 800 x 1000 g = 800000 g

Number of boxes = 800000 ÷ 4500

Therefore, 177 boxes can be loaded in the van.

11. The distance between the school and a student’s house is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days.

Solution: Distance between the school and house = 1 km 875 m = 1 km + 875/1000km=1.875km

Total distance covered in one day = (1.875 x 2)km = 3.750 km

Distance covered in six days = (3.750 x 6) km = 22.500 km

Therefore, a student covered 22 km 500 m distance in six days.

12. A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?

Solution:  Quantity of curd in a vessel = 4 litres 500 ml = 4×1000 ml + 500 ml = 4500 ml

Capacity of one glass = 25 ml

Number of glasses can be filled = 4500 ÷ 25

Therefore, 180 glasses can be filled by curd

• Chapter 1 Knowing Our Numbers
• Chapter 2 Whole Numbers
• Chapter 3 Playing With Numbers
• Chapter 4 Basic Geometrical Ideas
• Chapter 5 Understanding Elementary Shapes

• Chapter 6 Integers
• Chapter 7 Fractions
• Chapter 8 Decimals
• Chapter 9 Data Handling
• Chapter 10 Mensuration
• Chapter 11 Algebra
• Chapter 12 Ratio and Proportion