RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2

RD Sharma Class 8 Solutions Chapter 25 Data Handling III (Pictorial Representation of Data as Pie Charts or Circle Graphs) Ex 25.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2

Other Exercises

Question 1.
The pie-chart given in figure represents the expenditure on different items in constructing a flat in Delhi. If the expenditure incurred on cement is Rs 1,12,500, find the following:
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 1
(i) Total cost of the flat,
(ii) Expenditure incurred on labour.
Solution:
Expenditure on cement = Rs 1,12,500
and its central angle = 75°
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 2
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 3

Question 2.
The pie-chart given in the figure shows the annual agricultural production of an Indian state. If the total production of all the commodities is 81000 tonnes, find the production (in tonnes) of:
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 4
(i) Wheat
(ii) Sugar
(iii) Rice
(iv) Maize
(v) Gram
Solution:
Total production = 81000 tonnes
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 5

Question 3.
The following pie-chart shows the number of students admitted in different faculties of a college. If 1000 students are admitted in Science answer the following:
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 6

(i) What is the total number of students ?
(ii) What is the ratio of students in science and arts ?
Solution:
Students admitted in science = 1000
Central angle = 100°
(i) Total number of students
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 7
∴ Ratio in science and arts = 1000 : 1200 = 5:6

Question 4.
In the figure, the pie-chart shows the marks obtained by a student in an examination. If the student secures 440 marks in all, calculate his marks in each of the given subjects.
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 8
Solution:
Total marks secured by a student = 440
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 9

Question 5.
In the figure, the pie-charts shows the marks obtained by a student in various subjects. If the student scored 135 marks in mathematics, find the total marks in all the subjects. Also, find his score in individual subjects.
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 10
Solution:
Marks obtained in mathematics =135
Central angle = 90°
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 11

Question 6.
The following pie-chart shows the monthly expenditure of Shikha on various items. If she spends Rs 16,000 per month, answer the following questions:
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 12
(i) How much does she spend on rent ?
(ii) How much does she spend on education ?
(iii) What is the ratio of expenses on food and rent ?
Solution:
Total expenditure per month = Rs 16,000
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 13
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 14

Question 7.
The pie-chart (as shown in the figure) represents the amount spent on different sports by a sports club in a year. If the total money spent by the club on sports is Rs 1,08,000, find the amount spent on each sport.
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 15
Solution:
total amount spent on sports = Rs 1,08,000
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 16

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RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4A

RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4A.

Other Exercises

Question 1.
Solution:
(i) A decrease of 8
(ii) A gain of Rs. 7
(iii) Loosing a weight of 5 kg
(iv) 10 km below sea level
(v) 5°C above the freezing point
(vi) A withdrawal of Rs. 100
(vii) Spending Rs. 500
(viii) Going 6 m to the west
(ix) – 24
(x) 34

Question 2.
Solution:
(i) + Rs. 600
(ii) – Rs. 800
(iii) – 7°C
(iv) – 9
(v) + 2 km
(vi) – 3 km
(vii) + Rs. 200
(viii) – Rs. 300

Question 3.
Solution:
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4A 3.1

Question 4.
Solution:
(i) 0
(ii) – 3
(iii) 2
(iv) 8
(v) – 365
(vi) 8

Question 5.
Solution:
(i) – 7
(ii) – 1
(iii) – 27
(iv) – 26
(v) – 603
(vi) – 777

Question 6.
Solution:
(i) The integers between 0 and 6 are
1, 2, 3, 4, 5.
(ii) The integers between – 5 and 0 are
– 4, – 3, – 2, – 1.
(iii) The integers between – 3 and 3 are
– 2, – 1, 0, 1, 2.
(iv) The integer between – 7 and – 5 is – 6.

Question 7.
Solution:
(i) 0 < 7
(ii) 0 > – 3
(iii) – 5 < – 2
(iv) – 15 < 13
(v) – 231 < – 132
(vi) – 6 < 6

Question 8.
Solution:
(i) – 7, – 2, 0, 5, 8
(ii) – 100, – 23, – 6, – 1, 0, 12
(iii) – 501, – 363, – 17, 15, 165
(iv) – 106, – 81, – 16, – 2, 0, 16, 21.

Question 9.
Solution:
(i) 36, 7, 0, – 3, – 9, – 132
(ii) 51, 0, – 2, – 8, – 53
(iii) 36, 0, – 5, – 71, – 81
(iv) 413, 102, – 7, – 365, – 515.

Question 10.
Solution:
(i) We want to write an integer 4 more than 6. So, we start from 6 and proceed 4 steps to the right to obtain 10, as shown below:
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4A 10.1
∴ 4 more than 6 is 10.
(ii) We want to write an integer 5 more than – 6. So, we start from – 6 and proceed 5 steps to the right to obtain – 1, as shown below :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4A 10.2
∴ 5 more than – 6 is – 1.
(iii) We want to write an integer 6 less than 2. So we start from 2 and come back to the left by 6 steps to obtain – 4, as shown below:
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4A 10.3
∴ 6 less than 2 is – 4.
(iv) We want to write an integer 2 less than – 3. So we start from – 3 and come back to the left by 2 steps to obtain – 5, as shown below :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4A 10.4
∴ 2 less than – 3 is – 5.

Question 11.
Solution:
(i) False, as zero is greater than every negative integer.
(ii) False, as zero is an integer.
(iii) True, as zero is neither positive nor negative.
(iv) False, as – 10 is to the left of – 6 on a number line.
(v) False, as absolute value of an integer is always equal to the integer.
(vi) True, as 0 is to right of every negative integer, on a number line.
(vii) False, as every natural number is positive. False, the successor is – 186
(viii) False, the predecessor is – 216

Question 12.
Solution:
(i) | – 9 | = 9
(ii) | 36 | = 36
(iii) | 0 | = 0
(iv) | 15 | = 15
(v) – | – 3 | = – 3
(vi) 7 + | – 3 | = 7 + 3 = 10
(vii) |7 – 4| = | 3 | = 3
(viii) 8 – | – 7| = 8 – 7 = 1

Question 13.
Solution:
The required integers are – 6, – 5, – 4, – 3, – 2.
The required integers are – 21, – 22, – 23, – 24, – 25.
The required integers are – 21, – 22, – 23, – 24, – 25.

 

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RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3F

RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3F

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3F.

Other Exercises

Objective questions
Mark against the correct answer in each of the following :

Question 1.
Solution:
The smallest whole number is 0 (b)

Question 2.
Solution:
The least 4-digit number = 1000
On dividing 1000 by 9, we get
Remainder = 1
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3F 1.1
Least 4-digit number which is
Divisible by 9 = 1000 – 1 + 9
= 1000 + 8
= 1008 (d)

Question 3.
Solution:
The largest 6-digit number = 999999
On dividing by 16, we get
Remainder =15
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3F 3.1
The greatest 6-digit number divisible by 16
= 999999 – 15
= 999984 (c)

Question 4.
Solution:
On dividing 10004 by 12, we get remainder = 8
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3F 4.1
8 is to be subtracted from 10004 (c)

Question 5.
Solution:
On dividing 10056 by 23 We get remainder =12
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3F 5.1
The least number to be added = 23 – 5
= 18 (b)

Question 6.
Solution:
On dividing 457 by 11
We get remainder = 6
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3F 6.1
Which is greater than half of 11
The number nearest to 457 which is divisible 11 will be = 457 – 6 + 11
= 457 + 5
= 462 (d)

Question 7.
Solution:
Whole number between 1018 and 1203 are 1019 to 1202 are 1202 – 1018
= 184 (c)

Question 8.
Solution:
Divisor = 46
Quotient =11
Remainder =15
Number = Divisor x Quotient + Remainder
= 46 x 11 + 15
= 506 + 15
= 521 (b)

Question 9.
Solution:
Dividend = 199
Quotient =16
Remainder = 7
Divisor = \(\\ \frac { 199-7 }{ 16 } \) = \(\\ \frac { 192 }{ 16 } \)
= 12 (c)

Question 10.
Solution:
7589 – ? = 3434
Required number = 7589 – 3434
= 4155 (c)

Question 11.
Solution:
587 x 99 = 587 x (100 – 1)
= 587 x 100 – 587 x 1
= 58700 – 587
= 58113 (c)

Question 12.
Solution:
4 x 538 x 25 = 538 x 4 x 25
= 538 x 100
= 53800 (c)

Question 13.
Solution:
24679 x 92 + 24679 x 8
= 24679 x (92 + 8)
= 24679 x 100
= 2467900 (c)

Question 14.
Solution:
1625 x 1625 – 1625 x 625
= 1625 (1625 – 625)
= 1625 x 1000
= 1625000 (a)

Question 15.
Solution:
1568 x 185 – 1568 x 85
= 1568 (185 – 85)
= 1568 x 100
= 156800 (c)

Question 16.
Solution:
888 + 111 + 555 = 111 x ?
= 11 (8 + 7 + 5)
= 111 x 20 (c)

Question 17.
Solution:
The sum of two odd number is an even number (b)

Question 18.
Solution:
The product of two odd numbers is an odd number (a)

Question 19.
Solution:
If a is a whole number such that
a + a = a, then a = 0
as 0 + 0 = 0
None of these (d)

Question 20.
Solution:
The predecessor of 10000 is 10000 – 1
= 9999 (b)

Question 21.
Solution:
The successor of 1001 is 1001 + 1
= 1002 (b)

Question 22.
Solution:
The smallest even whole number is 2 (b)

Hope given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3F are helpful to complete your math homework.

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Value Based Questions in Science for Class 10 Chapter 6 Life Processes

Value Based Questions in Science for Class 10 Chapter 6 Life Processes

These Solutions are part of Value Based Questions in Science for Class 10. Here we have given Value Based Questions in Science for Class 10 Chapter 6 Life Processes

Question 1.
What do we get from environment and what do we give out to environment ? Should we degrade it ?
Answer:
We obtain nutrients, water and oxygen from the environment. In return we give out undigested materials, CO2 and waste products. Environment has been metabolising our wastes for regenerating nutrients and other materials required by us. However, we are producing so many articles for our use which are non-biodegradable. We also generate a lot more garbage, sewage, effluents, polluting gases and particulate matter than the capacity of the environment to metabolise them. They are becoming source of contaminations, infections and disorders. Therefore, we should take immediate measures to stop degrading our environment.

More Resources

Question 2.
Two processes are basic to life on earth. What are they ? How do these processes sustain life ?
Answer:
Photosynthesis and respiration are two processes that are basic to life on earth. Without them, life will disappear from earth. Photosynthesis traps solar energy, converts it into chemical energy and manufactures organic matter by reducing CO2. The organic matter produced in photosynthesis not only supports plants but all types of life, helping them in building their body and supplying energy.
Respiration is the process of release of energy from organic matter. This food or chemical energy is essential for maintenance, growth and working of the bodies of the living beings including the plants which trap solar energy.

Question 3.
Soil air is essential for growth and functioning of roots. However, roots do not have pores for exchange of gases. Explain.
Answer:
Roots require a lot of energy for active absorption of minerals and good growth so as to form new root hairs for continued absorption of water. This is possible only with the help of aerobic respiration for which good soil air is required. It should remain connected with atmospheric air for renewal. In soil, exchange of gases occurs at the surface of young roots through diffusion because the external walls of epiblema cells and root hairs are uncutinised. Therefore, despite the absence of pores, roots are always exchanging metabolic gases with the soil air.

Question 4.
In breathing, which ones are active and which ones are passive processes — expansion of thorax, passage of air from outside to lungs and exhalation.
Answer:

  1. Expansion of Thorax: It is an active process which involves contraction of phrenic and external intercostal muscles.
  2. Passage of Air from Outside into Lungs: It is a passive process which occurs due to development of negative pressure in the lungs when they expand with the expansion of thorax.
  3. Exhalation: It is a passive process which occurs due to relaxation of phrenic and external intercostal muscles resulting in contraction of thorax and hence lungs.

Hope given Value Based Questions in Science for Class 10 Chapter 6 Life Processes are helpful to complete your science homework.

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RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2

RD Sharma Class 8 Solutions Chapter 22 Mensuration III (Surface Area and Volume of a Right Circular Cylinder) Ex 22.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2

Other Exercises

Question 1.
Find the volume of a cuboid whose
(i) r = 3.5 cm, h = 40 cm
(ii) r = 2.8 m, h = 15 m
Solution:
(i) Radius (r) = 3.5 cm
Height (h) = 40 cm
Volume of cylinder = πr2h
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 1

Question 2.
Find the volume of a cylinder, if the diameter (d) of its base and its altitude (h) are:
(i) d= 21 cm, h = 10 cm
(ii) d = 7 m, h = 24 m.
Solution:
(i) Diameter (d) = 21 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 2

Question 3.
The area of the base of a right circular cylinder is 616 cm3 and its height is 25 cm. Find the volume of the cylinder.
Solution:
Base area of cylinder = 616 cm2
Height (h) = 25 cm.
∴ Volume = Area of base x height
= 616 x 25 cm3 = 15400 cm3

Question 4.
The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find the volume of the cylinder.
Solution:
Circumference of the base of cylinder = 88 cm
Let r be the radius
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 3

Question 5.
A hollow cylindrical pipe is 21 dm long. Its outer and inner diameters are 10 cm and 6 cm respectively. Find the volume of the copper used in making the pipe.
Solution:
Length (Height) of hollow cylindrical pipe = 21 dm = 210 cm
Inner diameter = 6 cm
Outer diameter = 10 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 4
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 5

Question 6.
Find the (i) curved surface area (ii) total surface area and (iii) volume of a right circular cylinder whose height is 15 cm and the radius of the base is 7 cm
Solution:
Radius of the cylider (r) = 7 cm
and height (h) = 15 cm
(i) Curved surface area = 2πrh
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 6

Question 7.
The diameter of the base of a right circular cylinder is 42 cm and its height is 10 cm. Find the volume of the cylinder.
Solution:
Diameter of the base of cylinder = 42 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 7

Question 8.
Find the volume of a cylinder, the diameter of whose base is 7 cm and height being 60 cm. Also, find the capacity of the cylinder in litres.
Solution:
Diameter of cylinder = 7 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 8

Question 9.
A rectangular strip 25 cm x 7 cm is rotated about the longer side. Find the volume of the solid, thus generated.
Solution:
Size of rectangular strip = 25 cm x 7 cm
By rotating, about the longer side, we find a right circular cylinder,
then the radius of cylinder (r) = 7 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 9

Question 10.
A rectangular sheet of paper 44 cm x 20 cm, is rolled along its length to form a cylinder. Find the volume of the cylinder so formed.
Solution:
Size of sheet of paper = 44 cm x 20 cm
By rolling along length, a cylinder is formed in which circumference of its base = 20 cm
and height (h) = 44 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 10

Question 11.
The volume and the curved surface area of a cylinder are 1650 cm3 and 660 cm2 respectively. Find the radius and height of the cylinder.
Solution:
Volume of cylinder = 1650 cm3
and curved surface area = 660 cm2
Let r be the radius and h be the height of the cylinder, then
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 11
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 12

Question 12.
The radii of two cylinders are in the ratio 2 :3 and their heights are in the ratio 5 : 3. Calculate the ratio of their volumes.
Solution:
Ratio in radii = 2:3
and in heights = 5:3
Let r1,h1 and r2, h2 are the radii and heights of two cylinders respectively.
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 13

Question 13.
The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the volume of the cylinder, if its total surface area is 616 cm2.
Solution:
Ratio in curved surface are and total surface area =1 : 2
Let r be the radius and h be the height.
Then 2πrh : 2πr (h + r)= 1 : 2
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 14
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 15

Question 14.
The curved surface area of a cylinder is 1320 cm2 and its base has diameter 21 cm. Find the volume of the cylinder.
Solution:
Curved surface area of a cylinder = 1320 cm2
Diameter of base = 21 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 16

Question 15.
The ratio between the radius of the base and the height of a cylinder is 2 : 3. Find the total surface area of the cylinder, if its volume is 1617 cm3.
Solution:
Ratio between radius and height of a cylinder = 2:3
Volume = 1617 cm3
Let r be the radius and A be the height of the cylinder, then
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 17

Question 16.
The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the diameter and the height of the pillar.
Solution:
Let r be the radius and A be the height of the cylinder, then
2πrh = 264 …(i)
and πr2h= 924 …(ii)
Dividing (ii) by (i),
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 18
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 19

Question 17.
Two circular cylinders of equal volumes have their heights in the ratio 1 : 2. Find the ratio of their radii.
Solution:
Volumes of two cylinders are equal.
Let r1, r2 are the radii and h1, h2 are their heights, then
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 20
Now, volume of first cylinder = πr12h1
and Volume of second cylinder = πr22h2
Their volumes are equal
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 21

Question 18.
The height of a right circular cylinder is 10.5 m. Three times the sum of the areas of its two circular faces is twice the area of the curved surface. Find the volume of the cylinder.
Solution:
Height of cylinder (A) = 10.5 m.
Let r be the radius, then
Sum of areas of two circular faces = 2π2
and curved surface area = 2πrh
According to the condition,
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 22

Question 19.
How many cubic metres of earth must be dug-out to sink a well 21 m deep and 6 m diameter ?
Solution:
Diameter of well = 6 m
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 23

Question 20.
The trunk of a tree is cylindrical and its circumference is 176 cm. If the length of the trunk is 3 m, find the volume of the timber that can be obtained from the trunk.
Solution:
Circumference of cylindrical trunk = 176 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 24

Question 21.
A well is dug 20 m deep and it has a diameter of 7 m. The earth which is so dug out is spread out on a rectangular plot 22 m long and 14 m broad, what is the height of the platform so formed ?
Solution:
Diameter of the well = 7 m
∴ Radius (r) = \(\frac { 7 }{ 2 }\) m
Depth (h) = 20 m
= Volume of earth dug out = πr2h
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 25

Question 22.
A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of the well = 14 m
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 26
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 27

Question 23.
A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm x 22 cm x 14 cm. Find the rise in the level of the water when the solid is completely submerged.
Solution:
Diameter of the base of a cylindrical container = 56 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 28
Dimensions of the rectangular solid = 32 cm x 22 cm x 14 cm
∴ Volume of solid = 32 x 22 x 14 cm3 = 9856 cm3
∴Volume of water rose up = 9856 cm3
Let h be the height of water, then πr2h = 9856
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 29

Question 24.
A rectangular sheet of paper 30 cm x 18 cm can be transformed into the curved surface of a right circular cylinder in two ways Le., either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed.
Solution:
Size of paper = 30 cm and 18 cm.
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 30
(i) By rolling length wise,
The circumference of base = 30 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 31
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 32
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 33

Question 25.
The rain which falls on a roof 18 m long and 16.5 m wide is allowed to be stored in a cylindrical tank 8 m in diameter. If it rains 10 cm on a day, what is the rise of water level in the tank due to it.
Solution:
Length of roof (l) = 18 m
Breadth (b) = 16.5 m
Height of water on the roof = 10 cm
∴ Volume of water collected
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 34

Question 26.
A Piece of ductile metal is in the form of a cylinder of diameter 1 cm and length 5 cm. It is drawnout into a wire of diameter 1 mm. What will be the length of the wire so formed ?
Solution:
Diameter of ductile metal = 1 cm
and length (h) = 5 cm
and radius (r) = \(\frac { 1 }{ 2 }\) cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 35

Question 27.
Find the length of 13.2 kg of copper wire of diameter 4 mm when 1 cubic cm of copper weighs 8.4 gm.
Solution:
Weight of wire = 13.2 kg
Diameter of wire = 4 mm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 36
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 37

Question 28.
2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire.
Solution:
Volume of brass = 2.2 cu.dm
= 2.2 x 10 x 10 x 10 = 2200 cm2
Diameter of wire = 0.25 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 38

Question 29.
The difference between inside and outside surfaces of a cylindrical tube 14 cm long is 88 sq. cm. If the volume of the tube is 176 cubic cm, find the inner and outer radii of the tube.
Solution:
Let r and R be the radii of inner and outer surfaces of a cylindrical tube,
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 39
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 40

Question 30.
Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 metres per second into a cylindrical tank, the radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes.
Solution:
Diameter of pipe = 2 cm
∴ Radius (r) = \(\frac { 2 }{ 2 }\) = 1 cm = 0.01 m
Length of flow of water in 1 second = 6 m Length of flow in 30 minutes = 6 x 30 x 60 m = 10800 m
∴ Volume of water = πr2h
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 41

Question 31.
A cylindrical tube, open at both ends, is to made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal tube is 8 mm everywhere. Calculate the volume of the metal.
Solution:
Length of metal tube (h) = 25 cm
Internal diameter = 10.4 cm
∴ Internal radius (r) = \(\frac { 10.4 }{ 2 }\) = 5.2 cm
Thickness of metal = 8 mm = 0.8 cm
∴ Outer radius = 5.2 + 0.8 = 6 cm
Now volume of the metal
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 42

Question 32.
From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second. Find the volume in litres of water delivered by the pipe in one hour.
Solution:
Inner radius of pipe (r) = 0.75 cm
Rate of water flow = 7 m per second
∴ Length of water flow in 1 hour (h)
= 7 x 3600 m = 25200 m
∴ Volume of water in 1 hour
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 43

Question 33.
A cylindrical water tank of diameter 1. 4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 metre per second. In how much time the tank will be filled ?
Solution:
Diameter of cylindrical tank = 1.4 m
∴ Radius (r) = \(\frac { 1.4 }{ 2 }\) = 0.7 m
Height (h) = 2.1m.
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 44

Question 34.
A rectangular sheet of paper 30 cm x 18 cm can be transformed into the curved surface of a right circular cylinder in two ways Le. either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinder thus formed.
Solution:
In first case,
By rolling the paper along its length,
the circumference of the base = 30 cm
and height (h) = 18 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 45
Note: See Q.No. 24 this exercise

Question 35.
How many litres of water flow out of a pipe having an area of cross-section of 5 cm2 in one minute, if the speed of water in the pipe is 30 cm/sec ?
Solution:
Speed of water = 30 cm/sec
∴ Water flow in 1 minute = 30 cm x 60 = 1800 cm
Area of cross-section = 5 cm2
∴ Volume of water = 1800 x 5 = 9000 cm3
Capacity of water in litres = 9000 x l ml
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 46

Question 36.
A solid cylinder has a total surface area of 231 cm2. Its curved surface area is \(\frac { 2 }{ 3 }\) of the total surface area. Find the volume of the cylinder.
Solution:
Total surface area of a solid cylinder = 231 cm2
Curved surface area = \(\frac { 2 }{ 3 }\) of 231 cm2 = 2 x 77 = 154 cm2
and area of two circular faces = 231-154 = 77 cm2
Let r be the radius, then
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 47

Question 37.
Find the cost of sinking a tubewell 280 m deep, having diameter 3 m at the rate of Rs 3.60 per cubic metre. Find also the cost of cementing its inner curved surface at Rs 2.50 per square metre.
Solution:
Diameter of well = 3 m
∴ Radius (r) = \(\frac { 3 }{ 2 }\) m
and depth (h) = 280 m
(i) Volume of earth dug out = πr2h
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 48
Rate of sinking the well = Rs 3.60 per m3
∴ Total cost of sinking = Rs 1980 x 3.60 = Rs 7,128
(ii) Inner curved surface area = 2πrh
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 49
Rate of cementing = Rs 2.50 per m2
∴ Total cost of cementing = Rs 2.50 x 2640
= Rs 6,600

Question 38.
Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm.
Solution:
Note : See Q.No. 27 of this exercise

Question 39.
2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire.
Solution:
Note: See Q.No. 28 of this exercise.

Question 40.
A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of the well = 10 m
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 50
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 51

Question 41.
A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4 cm thick iron. Find the volume of the iron.
Solution:
Width of roller (h) = 63 cm.
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 52
Outer circumference of roller = 440 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 53
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 54

Question 42.
What length of a solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of length 16 cm, external diameter 20 cm and thickness 2.5 mm ?
Solution:
Length of hollow cylinder (h) = 16 cm
External diameter = 20 cm
∴ External radius (R) = \(\frac { 20 }{ 2 }\) = 10 cm
Thickness of iron = 2.5 mm
∴ Internal radius (r) = 10 – 0.25 = 9.75 cm
∴ Volume of iron = πh (R2 – r2)
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 55

Question 43.
In the middle of rectangular field measuring 30 m x 20 m, a well of 7 m diameter and 10 m depth is dug. The earth so removed is evenly spread over the remaining part of the field. Find the height through which the level of the field is raised.
Solution:
Diameter of well = 7 m
∴ Radius (r) = \(\frac { 7 }{ 2 }\) m
Depth (h) = 10 m.
∴ Volume of earth dug out = πr2h
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 56
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 57

Hope given RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D

RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2D.

Other Exercises

Find the H.C.F. of the numbers in each of the following, using the prime factorization method :

Question 1.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 1.1
84 = 2 x 2 x 3 x 7
= 22 x 3 x 7
98 = 2 x 7 x 7 = 2 x 72
∴H.C.F. =2 x 7 = 14.

Question 2.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 2.1
So, 170 = 2 x 5 x 17
238 = 2 x 7 x 17
∴ H.C.F. of 170 and 238 = 2 x 17 = 34

Question 3.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 3.1
So, 504 = 2 x 2 x 2 x 3 x 3 x 7 = 23 x 32 x 7
980 = 2 x 2 x 5 x 7 x 7 = 22 x 5 x 72
∴ H.C.F. of 504 and 980 = 22 x 7
= 4 x 7 = 28.

Question 4.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 4.1
So, 72 = 2 x 2 x 2 x 3 x 3 = 23 x 32
108 = 2 x 2 x 3 x 3 x 3 = 22 x 33
180 = 2 x 2 x 3 x 3 x 5 = 22 x 32 x 5
∴ H.C.F. of 72, 108,
180 = 22 x 32
= 4 x 9 = 36

Question 5.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 5.1
84 = 2 x 2 x 3 x 7 = 22 x 3 x 7
120 = 2 x 2 x 2 x 3 x 5 = 23 x 3 x 5
138 = 2 x 3 x 23
∴ H.C.F. of 84, 120 and 138 = 2 x 3 = 6

Question 6.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 6.1
106 = 2 x 53
159 = 3 x 53
371 = 7 x 53
∴ H.C.F. of 106, 159, 371 = 53

Question 7.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 7.1
272 = 2 x 2 x 2 x 2 x 17 = 24 x 17
425 = 5 x 5 x 17
= 52 x 17
∴ H.C.F. of 272 and 425 = 17.

Question 8.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 8.1
So, 144 = 2 x 2 x 2 x 2 x 3 x 3 = 24 x 32
252 = 2 x 2 x 3 x 3 x 7 = 22 x 32 x 7
630 = 2 x 3 x 3 x 5 x 7 = 2 x 32 x 5 x 7
∴ H.C.F. of 144, 252 and 630 = 2 x 32
= 2 x 9 = 18.

Question 9.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 9.1
So, 1197 = 3 x 3 x 7 x 19 = 32 x 7 x 19
5320 = 2 x 2 x 2 x 5 x 7 x 19 = 23 x 5 x 7 x 19
4389 = 7 x 3 x 11 x 19
∴ H.C.F. of 1197, 5320,
4389 = 7 x 19 = 133.

Find the H.C.F. of the numbers in each of the following using division method:

Question 10.
Solution:
By division method, we have :
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 10.1
∴ H.C.F. of 58 and 70 = 2.

Question 11.
Solution:
By division method, we have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 11.1
∴ H.C.F. of 399 and 437 = 19

Question 12.
Solution:
By division method, we have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 12.1
∴ H.C.F. of 1045 and 1520 = 95.

Question 13.
Solution:
By division method, we have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 13.1
∴ H.C.F. of 1965 and 2096 = 131

Question 14.
Solution:
By division method, we have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 14.1
∴ H.C.F. of 2241 and 2324 = 83.

Question 15.
Solution:
First, we find the H.C.F. of 658 and 940
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 15.1
∴ H.C.F. of 658 and 940 is 94.
Now, we find the H.C.F. of 94 and 1128.
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 15.2
∴ H.C.F. of 94 and 1128 = 94
Hence, H.C.F. of 658, 940 and 1128 = 94.

Question 16.
Solution:
First, we find the H.C.F. of 754 and 1508
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 16.1
∴ H.C.F. of 754 and 1508 is 754
Now, we find the H.C.F. of 754 and 1972
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 16.2
∴ H.C.F. of 754 and 1972 = 58
Hence, the H.C.F. of 754,1508 and 1972 = 58.

Question 17.
Solution:
First, we find the H.C.F. of 391 and 425
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 17.1
∴ H.C.F. of 391 and 425 is 17.
Now, we find the H.C.F. of 17 and 527
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 17.2
∴ H.C.F. of 17 and 527 is 17 Hence, H.C.F. of 391, 425 and 527 = 17.

Question 18.
Solution:
First, we find the H.C.F. of 1794 and 2346
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 18.1
H.C.F. of 1794 and 2346 is 138.
Now, we find the H.C.F. of 138 and 4761
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 18.2
∴ H.C.F. of 138 and 4761 is 69.
Hence, the H.C.F. of 1794, 2346 and 4761 = 69.

Show that the following pairs are co-primes :

Question 19.
Solution:
First, we find the H.C.F. of 59, 97.
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 19.1
∴H.C.F. of 59 and 97 is 1.
Hence 59 and 97 are co-prime.

Question 20.
Solution:
First, we find the H.C.F. of 161 and 192.
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 20.1
∴ H.C.F. of 161 and 192 is 1.
Hence 161 and 192 are co-prime.

Question 21.
Solution:
First, we find the H.C.F. of 343 and 432
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 21.1
∴ H.C.F. of 343 and 432 is 1.
Hence 343 and 432 are co-prime.

Question 22.
Solution:
First, we find the H.C.F. of 512 and 945.
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 22.1
∴ H.C.F. of 512 and 945 is 1.
Hence 512 and 945 are co-prime.

Question 23.
Solution:
First, we find the H.C.F. of385 and 621
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 23.1
∴ H.C.F. of 385 and 621 is 1.
Hence the numbers 385 and 621 are co-prime

Question 24.
Solution:
First, we find the H.C.F. of 847 and 1014.
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 24.1
∴ H.C.F. of 847 and 1014 is 1.
Hence 847 and 1014 are co-prime.

Question 25.
Solution:
Clearly, we have to find the greatest number which divides (615 – 6) and (963 – 6) exactly.
So, the required number = H.C.F. of 615 – 6 = 609 and 963 – 6 = 957
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 25.1
The required greatest number = 87.

Question 26.
Solution:
Clearly, we have to find the greatest number which divides 2011 – 9 = 2002 and 2623 – 5 = 2618.
So, the required greatest number = H.C.F. of 2002 and 2618
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 26.1
∴ Required greatest number = 154.

Question 27.
Solution:
Clearly, we have to find the greatest number which divides (445 4), (572 – 5) and (699 – 6). So, the required number = H.C.F. of 441, 567 and 693. First we find the H.C.F. of 441 and 567
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 27.1
∴ H.C.F. of 441 and 567 is 63
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 27.2
So H.C.F. of 63 and 693 is 63
Hence the required number = 63.

Question 28.
Solution:
(i) The given fraction = \(\\ \frac { 161 }{ 207 } \)
First we find the H.C.F. of 161 and 207
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 28.1
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 28.2
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 28.3

Question 29.
Solution:
Lengths of three pieces of timber = 42 metres, 49 metres, 63 metres Greatest possible length of each plank = H.C.F. of 42 metres, 49 metres and 63 metres
First we find the H.C.F. of 42 and 49
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 29.1
∴ H.C.F. of 42 and 49 = 7
Now we find the H.C.F. of 7 and 63
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 29.2
So, the H.C.F. of 7 and 63 is 7
∴ H.C.F. of 42 metres, 49 metres of 63 metres = 7 metres
Hence required possible length of each plank = 7 metres.

Question 30.
Solution:
Quantity of milk in three different containers = 403 L, 434 L and 465 L Clearly, the maximum capacity of the required container = H.C.F. of 403 L, 434 L, 465 L, we have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 30.1
∴ 403 = 13 x 31
434 = 2 x 7 x 31
465 = 5 x 3 x 31
So the H.C.F. of 403 L, 434 L and 465 L = 31 L
Maximum capacity of the required container = 31 L.

Question 31.
Solution:
The given fruits = 527 apples, 646 pears and 748 oranges
Clearly, the greatest number of fruits in each heap = H.C.F. of 527, 646 and 748 we have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 31.1
∴ 527 = 17 x 31
646 = 2 x 17 x 19
748 = 2 x 2 x 11 x 17
So, the H.C.F. of 527, 646 and 748 = 17
∴ Required number of fruits in each heap = 17
Total number of fruits = 527 + 646 + 748 = 1921
Number of heaps = \(\frac { Total\quad number\quad of\quad fruits }{ Number\quad of\quad fruits\quad in\quad one\quad heap } \)
= \( \frac { 1921 }{ 17 } \)
=113

Question 32.
Solution:
The given lengths are :
7 metres = 7 x 100 cm
= 700 cm 3 metres 85 cm
= (3 x 100 + 85) cm
= 385 cm
12 metres 95 cm = (12 x 100 + 95) cm
= 1295 cm
Clearly, the length of the required longest tape = H.C.F. of 700 cm, 385 cm and 1295 cm
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 32.1
So, 700 = 2 x 2 x 5 x 5 x 7
= 22 x 52 x 7
385 = 5 x 7 x 11
1295 = 5 x 7 x 37
∴ H.C.F. of 700, 385 and 1295 = 5 x 7 = 35
∴ The required length of the longest tape
= 35 cm.

Question 33.
Solution:
Length of the courtyard = 18 m 72 cm = (18 x 100 + 72) cm = 1872 cm
Breadth of the courtyard = 13 m 20 cm = (13 x 100 + 20) cm = 1320 cm
Greatest side of each of the square tiles = H.C.F. of 1872 cm and 1320 cm
Now
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 33.1
1872 = 2 x 2 x 2 x 2 x 9 x 13
= 24 x 32 x 13
1320 = 2 x 2 x 2 x 3 x 5 x 11
= 23 x 3 x 5 x 11
So, the H.C.F. of 1872 and 1320
= 23 x 3 = 8 x 3 = 24
Greatest side of the square tile = 24 cm
Now Area of the courtyard = Length x Breadth = 1872 x 1320 cm2
Area of one square tile = Side x Side
= 24 x 24 cm2
∴ Least possible number of such tiles
= \(\frac { Area\quad of\quad the\quad courtyard }{ Area\quad of\quad the\quad tile } \)
= \( \frac { 1872\times 1320 }{ 24\times 24 } \)
= 78 x 55
= 4290

Question 34.
Solution:
Let the two prime numbers be 13 and 17, we find the H.C.F. of 13 and 17 as under
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 34.1
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 34.2

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HOTS Questions for Class 10 Science Chapter 6 Life Processes

HOTS Questions for Class 10 Science Chapter 6 Life Processes

These Solutions are part of HOTS Questions for Class 10 Science. Here we have given HOTS Questions for Class 10 Science Chapter 6 Life Processes

Question 1.
What does the diagram depict ?
HOTS Questions for Class 10 Science Chapter 6 Life Processes image - 1
Answer:
Gaseous exchange in Amoeba.

More Resources

Question 2.
What does diagram depict ? What are A and B ?
HOTS Questions for Class 10 Science Chapter 6 Life Processes image - 2
Answer:
Double Circulation :
(A) Pulmonary Circulation
(B) Systemic Circulation.

Question 3.
How are viruses living when they do not show movements ?
Answer:
Viruses do not show movements outside the host cells. They show movements at the molecular level inside the living cells.

Question 4.
What is
(a) Primary reaction of photosynthesis
(b) Calvin cycle
(c) Krebs cycle
(d) EMP
(e) Oxidative phosphorylation ?
Answer:
(a) Primary Reaction of Photosynthesis: It is the conversion of light energy into chemical energy by chlorophyll a molecules.
(b) Calvin Cycle: It is a cycle of reactions that occur during reduction of CO2 to carbohydrate with the help of ATP and NADPH2 produced during light reaction.
(c) Krebs Cycle: It is a cycle of reactions that occur inside the mitochondria wherein an activated acetyl group is completely oxidised to form CO2, NADH2 and FADH2.
(d) EMP: Embden-Meyerhoff-Parnas pathway, also called glycolysis, is the first step of respiratory breakdown of glucose that occurs in the cytoplasm forming two molecules each of pyruvate, ATP and NADH2.
(e) Oxidative Phosphorylation: It is the process of ATP formation from ADP and inorganic phosphate with the help of energy liberated during oxidation of reduced coenzymes (NADH2, FADH2).

Question 5.
A girdled tree dies if the girdle is wide and is not filled up. Comment.
Answer:
Girdling removes bark containing phloem from the trunk region. Food manufactured by foliage does not reach the roots which requires the same as they are always growing. In the absence of food supply, roots starve and stop absorbing water. The foliage wilts and the plant dies.

Question 6.
Study the diagram. Name the parts “A” and “B”. State one function of each. (CBSE A.I. 2008)
HOTS Questions for Class 10 Science Chapter 6 Life Processes image - 3
Answer:
A—Stomatal pore or stoma Function: Pathway for exchange of gases or photosymthesis and respiration and loss of water vapours in transpiration.
B—Guard cell Function: Two guard cells present in each stoma, create pore when they are turgid and close the same when they are flaccid.

Question 7.
In the experiment “Light is essential for photosynthesis” why does the uncovered part of the leaf turn blue-black after putting iodine solution. (CBSE Foreign 2010)
Answer:
The uncovered part of the leaf exposed to sunlight performs photosynthesis and accumulates starch. Iodine reacts.with starch to give blue-black colouration.

Question 8.
Give one reason why multicellular organisms require special organs for exchange of gases between their body and their environment. (CBSE A.I. 2010)
Answer:
Multicellular organisms require special organs for exchange of gases as most of their cells are internal and are not in direct contact with environment.

Question 9.
Name the green dot like structures in some cells observed by a student when a leaf peel was viewed under a microscope. What is the green colour due to ? (CBSE Delhi 2010)
Answer:
Chloroplasts with green colour due to chlorophyll.

Question 10.
Explain the process of breakdown of glucose in a cell

  1. In the presence of oxygen
  2. In the absence of oxygen. (CBSE foreign 2010)

Answer:
Glucose is first broken down to pyruvic acid during glycolysis. Glycolysis occurs in cell cytoplasm. It produces energy.

  1. Presence of Oxygen: It is aerobic respiration where pyruvic acid is completely oxidised to carbon dioxide and water inside mitochondria releasing a lot of energy.
  2. Absence of Oxygen: Pyruvic acid is metabolised anaerobically in the cell cytoplasm forming either
    1. Ethyl alcohol (ethanol) and carbon dioxide or
    2. lactic acid: Very little energy is released.

Question 11.
Explain the process of digestion of food in mouth, stomach and small intestine in human body. (CBSE Delhi 2010)
Answer:
Mouth: Food is moistened, crushed and acted upon by salivary amylase which converts some starch into maltose and dextrins.
Stomach: Food is acidified, churned and mixed with enzyme pepsin. Pepsin acts in acidic medium over proteins to form soluble components, peptones and proteoses. Small quantities of semi-liquified food called chyme is passed on the duodenum.
Small Intestine: Acidity is neutralised and the food is made alkaline. Food is acted upon by pancreatic juice and succus entericus. Pancreatic juice has trypsin, amylase and lipase enzymes to digest proteins, carbohydrates and fats respectively. Succus entericus has enzymes for breakdown of peptides and disaccharides into amino acids and monosaccharides respectively.

Hope given HOTS Questions for Class 10 Science Chapter 6 Life Processes are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E

RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3E.

Other Exercises

Question 1.
Solution:
(i) By actual division, we have
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 1.1
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 1.2
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 1.3
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 1.4
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 1.5

Question 2.
Solution:
(i) By actual division, we have
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 2.1
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 2.2
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 2.3
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 2.4
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 2.5
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 2.6

Question 3.
Solution:
(i) We know that any number (non-zero) divided by 1 gives the number itself
65007 ÷ 1 = 65007
(ii) We know that 0 divided by any natural number gives 0
0 ÷ 879 = 0
(iii) 981 + 5720 ÷ 10 = 981 + (5720 ÷ 10)
= 981 + 572
= 1553
(iv) 1507 – 625 ÷ 25 = 1507 – (625 ÷ 25)
= 1507 – 25
= 1482
(v) 32277 ÷ (648 – 39)
= 32277 ÷ 609
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 3.1
32277 ÷ (648 – 39) = 53
(vi) 1573 ÷ 1573 – 1573 ÷ 1573
= (1573 ÷ 1573) – (1573 ÷ 1573)
= 1 – 1
= 0

Question 4.
Solution:
We have n ÷ n = n
let n = 1, 1 ÷ 1 = 1
1 = 1
which is true
∴ Hence 1 is the required whole number.

Question 5.
Solution:
Product of two numbers = 504347
One number = 317
Other number = 504347 ÷ 317
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 5.1
∴ Other number = 1591

Question 6.
Solution:
Here Dividend = 59761, Quotient = 189
∴ Remainder = 37
We know that Dividend = Divisor x Quotient + Remainder
59761 = Divisor x 189 + 37
59761 – 37 = Divisor x 189
59724 = Divisor x 189
Divisor x 189 = 59724
∴ Divisor = 59724 ÷ 189
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 6.1
∴ Divisor = 59724 ÷ 189 = 316

Question 7.
Solution:
Here dividend = 55390,
Divisor = 299 and Remainder = 75
By division algorithm, we have
Dividend = Quotient x Divisor + Remainder
55390 = Quotient x 299 + 75
55390 – 75 = Quotient x 299
55315 = Quotient x 299
Quotient x 299 = 55315
Quotient = 55315 ÷ 299
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 7.1
∴ Required quotient = 185

Question 8.
Solution:
On dividing 13601 by 87, we have
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 8.1
It is clear that if we subtract 29 from 13601, the resulting number will be exactly divisible by 87.
∴ The required least number = 29.

Question 9.
Solution:
Here dividend = 1056, Divisor = 23
By actual division, we have
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 9.1
It is clear that if we add 2 to 21, it will become 23 which is divisible by 23.
∴ Required least number = 2.

Question 10.
Solution:
Greatest 4-digit number = 9999
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 10.1
On, dividing by 16, we get remainder as 15
∴ The required largest 4-digit number = 9999 – 15
= 9984

Question 11.
Solution:
Largest number of 5-digits = 99999
On dividing 99999 by 653, we have
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 11.1
∴ Quotient = 153, Remainder = 90
Check : By division algorithm Dividend = Divisor x Quotient + Remainder
= 653 x 153 + 90
= 99909 + 90
= 99999

Question 12.
Solution:
The least 6-digit number = 100000
On dividing 100000 by 83, we have
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 12.1
It is clear that if we add 15 to 68, it will become 83 which is divisible by 83.
∴ Required least 6-digit number = 100000 + 15
= 100015

Question 13.
Solution:
Cost of 1 dozen bananas = Rs. 29
Bananas can be purchase in Rs. 1392
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 13.1
1392 ÷ 29
= 48 dozens

Question 14.
Solution:
Total number of trees = 19625
Total number of rows = 157
Number of trees in each row = 19625 ÷ 157
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 14.1
∴ Number of trees in each row = 125

Question 15.
Solution:
Total population of the town = 517530
Since there is one educated person out of 15
Total number of educated persons in the town = 517530 ÷ 15
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 15.1
∴ Total number of educated persons in the town = 34502.

Question 16.
Solution:
Cost of 23 colour TV sets = Rs. 570055
Cost of 1 colour TV set
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 16.1
∴ Cost of 1 color TV set
= Rs. 570055 ÷ 23
= Rs. 24785

Hope given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3E are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3D

RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3D

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3D.

Other Exercises

Question 1.
Solution:
(i) 246 x 1 = 246
(By multiplicative property of 1)
(ii) 1369 x 0 = 0
(By multiplicative property of 0)
(iii) 593 x 188 = 188 x 593
(By commutative law of multiplication)
(iv) 286 x 753 = 753 x 286
(By commutative law of multiplication)
(v) 38 x (91 x 37) = 91 x (38 x 37)
(By associative law of multiplication)
(vi) 13 x 100 x 1000 = 1300000
(vii) 59 x 66 + 59 x 34 = 59 x (66 + 34)
(By distributive law of multiplication)
(vii) 68 x 95 = 68 x 100 – 68 x 5

Question 2.
Solution:
(i) Commutative law of multiplication
(ii) Closure property
(iii) Associative law of multiplication
(iv) Multiplicative property of 1
(v) Multiplicative property of 0
(vi) Distributive law of multiplication over addition in whole numbers
(vii) Distributive law of multiplication over subtraction in whole numbers

Question 3.
Solution:
Using the law of distribution over addition and subtraction
(i) 647 x 13 + 647 x 7
= 647 x (13 + 7)
= 647 x 20
= 12940
(ii) 8759 x 94 + 8759 x 6
= 8759 x (94 + 6)
= 8759 x 100
= 875900
(iii) 7459 x 999 + 7459
= 7459 x 999 + 7459 x 1
= 7459 x (999 + 1)
= 7459 x 1000
= 7459000
(iv) 9870 x 561 – 9870 x 461
= 9870 x (561 – 461)
= 9870 x 100
= 987000
(v) 569 x 17 + 569 x 13 + 569 x 70
= 569 x (17 + 13 + 70)
= 569 x 100
= 56900
(vi) 16825 x 16825 – 16825 x 6825
= 16825 x (16825 – 6825)
= 16825 x 10000
= 168250000

Question 4.
Solution:
(i) 2 x 1658 x 50 = 1658 x (2 x 50) (Associative law of multiplication)
= 1658 x 100
= 165800
(ii) 4 x 927 x 25 = 927 x (4 x 25) (Associative law of multiplication)
= 927 x 100
= 92700
(iii) 625 x 20 x 8 x 50
(By associative law of multiplication)
(625 x 8) x (20 x 50)
= 5000 x 1000
= 5000000
(iv) 574 x (625 x 16)
= 574 x 10000
= 5740000
(v) 250 x 60 x 50 x 8
= (250 x 8) x (60 x 50)
(By associative law)
= 2000 x 3000
= 6000000
(vi) 8 x 125 x 40 x 25
= (8 x 125) x (40 x 25)
= 1000 x 1000
= 1000000

Question 5.
Solution:
Using distributive law of multiplication over addition or subtraction,
(i) 740 x 105
= 740 x (100 + 5)
= 740 x 100 + 740 x 5
= 74000 + 3700
= 77700
(ii) 245 x 1008
= 245 x (1000 + 8)
= 245 x 1000 + 245 x 8
= 245000 + 1960
= 246960
(iii) 947 x 96
= 947 x (100 – 4)
= 947 x 100 – 947 x 4
= 94700 – 3788
= 90912
(iv) 996 x 367
= 367 x (1000 – 4)
= 367 x 1000 – 367 x 4
= 367000 – 1468
= 365532
(v) 472 x 1097
= 472 x (1100 – 3)
= 472 x 1100 – 472 x 3
= 519200 – 1416
= 517784
(vi) 580 x 64
= 580 x (60 + 4)
= 580 x 60 + 580 x 4
= 34800 + 2320
= 37120
(vii) 439 x 997
= 437 x (1000 – 3)
= 439 x 1000 – 439 x 3
= 439000 – 1317
= 437683
(viii) 1553 x 198
= 1553 x (200 – 2)
= 1553 x 200 – 1553 x 2
= 310600 – 3106
= 307494

Question 6.
Solution:
(i) 3576 x 9 = 3576 x (10 – 1)
= 3576 x 10 – 3576 x 1
= 35760 – 3576
= 32184
(ii) 847 x 99 = 84 x (100 – 1)
= 847 x 100 – 847 x 1
= 84700 – 847
= 83853
(iii) 2437 x 999 = 2437 x (1000 – 1)
= 2437 x 1000 – 2437 x 1
= 2437000 – 2437
= 2434563

Question 7.
Solution:
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3D 7.1

Question 8.
Solution:
Largest 3-digit number = 999
Largest 5-digit number = 99999
Required product = 99999 x 999
= 99999 x (1000 – 1)
= 99999 x 1000-99999 x 1
= 99999000 – 99999
= 9,98,99,001

Question 9.
Solution:
Speed of car = 75 km per hour
In 1 hour, distance covered by a car = 75 km
.’. In 98 hours, distance will be covered = 75 x 98
= 75 x (100 – 2)
= 75 x 100 – 75 x 2
= 7500 – 150
= 7350 km

Question 10.
Solution:
Cost of 1 set of VCR = Rs. 24350
Cost of 139 sets of VCR = Rs. 24350 x 139
= Rs. 3384650
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3D 10.1

Question 11.
Solution:
Cost of 1 house = Rs. 450000
Cost of 197 houses = Rs. 450000 x 197
= Rs. 450000 x (200 – 3)
= Rs. (450000 x 200 – 450000 x 3)
= Rs. (90000000 – 1350000)
= Rs. 88650000

Question 12.
Solution:
Cost of each chair = Rs. 1065
Cost of 50 chairs = Rs. 1065 x 50
= Rs. 53250
Cost of each blackboard = Rs. 1645
Cost of 30 blackboards = Rs. 1645 x 30
= Rs. 49350
Total cost of 50 chairs and 30 blackboards = Rs. 53250 + 49350
= Rs. 102600

Question 13.
Solution:
Number of students in 1 section = 45
Number of students in 6 sections = 45 x 6 = 270
Monthly charges of 1 student = Rs. 1650
Total monthly incomes from the class VI = Rs. 270 x 1650
= Rs. (300 – 30) x 1650
= Rs. (1650 x 300 – 1650 x 30)
= Rs. (495000 – 49500)
= Rs. 445500

Question 14.
Solution:
Since the product of two whole numbers is zero
.’. From multiplicative property of zero, we conclude that one of the whole numbers is zero.

Question 15.
Solution:
(i) Sum of two odd numbers is an even number.
(ii) Product of two odd numbers is an odd number.
(iii) a x a = a => a = 1 as 1 x 1 = 1

Hope given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3D are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 26 Data Handling IV (probability) Ex 26.1

RD Sharma Class 8 Solutions Chapter 26 Data Handling IV (probability) Ex 26.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 26 Data Handling IV (probability) Ex 26.1

Question 1.
The probability that it will rain to morrow is 0.85. What is the probability that it will not rain tomorrow ?
Solution:
Total number of possible events = 1
∴ P (\(\bar { A }\) ) = 0.85
∴ P (\(\bar { A }\) ) = 1-0.85 = 0.15

Question 2.
A die is thrown. Find the probability of getting (i) a prime number (ii) 2 or 4 (iii) a multiple of 2 or 3.
Solution:
Total number of possible events = 6 (1 to 6)
(i) Let A be the favourable occurrence which are prime number i.e., 2,3,5
∴ P(A) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)
(ii) Let B be the favourable occurrence which are 2 or 4
∴ P(B) = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)
(iii) Let C be the favourable occurrence which are multiple of 2 or 3 i.e., 2, 3, 4, 6.
∴ P(C) = \(\frac { 4 }{ 6 }\) = \(\frac { 2 }{ 3 }\)

Question 3.
In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet, of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) an even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) at least once
(xiii) a number other than 5 on any dice.
Solution:
By throwing of a pair of dice, total number of possible events = 6 × 6 = 36
(i) Let A be the occurrence of favourable events whose sum is 8 i.e. (2,6), (3,5), (4,4), (5,3) , (6,2) which are 5
∴ P(A) = \(\frac { 5 }{ 36 }\)
(ii) Let B be the occurrence of favourable events which are doublets i.e. (1, 1), (2, 2), (3, 3), (4,4), (5, 5) and (6, 6).
∴ P(B) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)
(iii) Let C be the occurrence of favourable events which are doublet of prime numbers which are (2, 2), (3,3), (5, 5)
∴ P(C) = \(\frac { 3 }{ 36 }\) = \(\frac { 1 }{ 12 }\)
(iv) Let D be the occurrence of favourable events which are doublets of odd numbers which are (1, 1), (3, 3) and (5, 5)
∴ P(D) = \(\frac { 3 }{ 36 }\) = \(\frac { 1 }{ 12 }\)
(v) Let E be the occurrence of favourable events whose sum is greater than 8 i.e, (3,6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6) which are 6 in numbers
∴ P(E) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)
(vi) Let F be the occurrence of favourable events in which is an even number is on first i.e (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1),(4,2), (4,3) (4,4), (4,5), (4,6), (6,1), (6,2), (6, 3), (6,4 ), (6, 5), (6,6) which are 18 in numbers.
∴ P(F) = \(\frac { 18 }{ 36 }\) = \(\frac { 1 }{ 2 }\)
(vii) Let G be the occurrence of favourable events in which an even number on the one and a multiple of 3 on the other which are (2,3), (2,6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (3, 4), (3, 6), (6,2), (6,4) = which are 11th number
∴ P(G) = \(\frac { 11 }{ 36 }\)
(viii) Let H be the occurrence of favourable events in which neither 9 or 11 as the sum of the numbers on the faces which are (1,1), (1,2), (1,3) , (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2,4) , (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3,5) , (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (5, 1), (5,2), (5,3), (5,5), (6,1), (6,2), (6,4), (6,6) which are 30
∴ P(H) = \(\frac { 30 }{ 36 }\) = \(\frac { 5 }{ 6 }\)
(ix) Let I be the occurrence of favourable events, such that a sum less than 6, which are (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4,1) which are 10
∴ P(I) = \(\frac { 10 }{ 36 }\) = \(\frac { 5 }{ 18 }\)
(x) Let J be the occurrence of favourable events such that a sum is less than 7, which are
(1.1) , (1,2), (1,3), (1,4), (1,5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5.1) which are 15
∴ P(J) = \(\frac { 15 }{ 36 }\) = \(\frac { 5 }{ 6 }\)
(xi) Let K be the occurrence of favourable events such that the sum is more than 7, which are (2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) which 15
∴ P(K) = \(\frac { 15 }{ 36 }\) = \(\frac { 5 }{ 12 }\)
(xii) Let L be the occurrence of favourable events such that at least P (L) one is black card
∴ P(L) = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)
(xiii) Let M is the occurrence of favourable events such that a number other than 5 on any dice which can be (1,1), (1,2), (1,3), (1,4), (1,6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3,1), (3,2), (3, 3), (3,4), (3,6), (4,1), (4, 2), (4, 3), (4,4), (4,6), (6, 1), (6,2), (6, 3), (6,4), (6, 6) which are 25
∴ P(M) = \(\frac { 25 }{ 36 }\)

Question 4.
Three coins are tossed together. Find the probability of getting:
(i) exactly two heads
(ii) at least two heads
(iii) at least one head and one tail
(iv) no tails
Solution:
Total number of events tossed by 3 coins each having one head and one tail = 2x2x2 = 8
(i) Let A be the occurrence of favourable events which is exactly two heads, which can be 3 in number which are HTH, HHT, THH.
∴ P(A) = \(\frac { 3 }{ 8 }\)
(ii) Let B be the occurrence of favourable events which is at least two heads, which will be 4 which are HHT, HTH, THH, and HHH.
∴ P(B) = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)
(iii) Let C be the occurrence 6f favourable events which is at least one head and one tail which are 6 which can be HHT, HTH, THH, TTH, THT, HTT
∴ P(C) = \(\frac { 6 }{ 8 }\) = \(\frac { 3 }{ 4 }\)
(iv) Let D be the occurrence of favourable events in which there is no tail which is only 1 (HHH)
∴ P(D) = \(\frac { 1 }{ 8 }\)

Question 5.
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
(i) a black king
(ii) either a black card or a king
(iii) black and a king
(iv) a jack, queen or a king
(v) neither a heart nor a king
(vi) spade or an ace
(vii) neither an ace nor a king
(viii) neither a red card nor a queen
(ix) other than an ace
(x) a ten
(xi) a spade
(xii) a black card
(xiii) the seven of clubs
(xiv) jack
(xv) the ace of spades
(xvi) a queen
(xvii) a heart
(xviii) a red card
Solution:
A pack of cards have 52 cards, 26 black and 26 red and four kinds each of 13 cards from 2 to 10, one ace, one jack, one queen and one king.
∴ Total number of possible events = 52
(i) Let A be the occurrence of favourable events which is a black king which are 2.
∴ P(A) = \(\frac { 2 }{ 52 }\) = \(\frac { 1 }{ 26 }\)
(ii) Let B be the occurrence of favourable events such that it is either a black card or a king.
Total = number of black cards = 26 + 2 red kings = 28
∴ P(B) = \(\frac { 28 }{ 52 }\) = \(\frac { 7 }{ 13 }\)
(iii) Let C be the occurrence of favourable events such that it is black and a king which can be 2.
∴ P(C) = \(\frac { 2 }{ 52 }\) = \(\frac { 1 }{ 26 }\)
(iv) Let D be the occurrence of favourable events such that it is a jack, queen or a king which will be4 + 4 + 4 = 12
∴ P(D) = \(\frac { 12 }{ 52 }\) = \(\frac { 3 }{ 13 }\)
(v) Let E be the occurrence of favourable events such that it is neither a heart nor a king.
∴ Number of favourable event will be 13 x 3 -3 = 39 – 3 = 36
∴ P(E) = \(\frac { 36 }{ 52 }\) = \(\frac { 9 }{ 13 }\)
(vi) Let F be the occurrence of favourable events such that it is a spade or an ace.
∴ Number of events = 13 + 3 = 16
∴ P(F) = \(\frac { 16 }{ 52 }\) = \(\frac { 4 }{ 13 }\)
(vii) Let G be the occurrence of favourable events such that it neither an ace nor a king.
∴Number of events = 52 – 4 – 4 = 44
∴ P(G) = \(\frac { 44 }{ 52 }\) = \(\frac { 11 }{ 13 }\)
(viii) Let H be the occurrence of favourable events such that it is neither a red card nor a queen.
∴Number of events = 26 – 2 = 24,
∴ P(H) = \(\frac { 24 }{ 52 }\) = \(\frac { 6 }{ 13 }\)
(ix) Let 1 be the occurrence of favourable events such that it is other than an ace.
∴ Number of events = 52 – 4 = 48
∴ P(I) = \(\frac { 48 }{ 52 }\) = \(\frac { 12 }{ 13 }\)
(x) Let J be the occurrence of favourable event such that it is ten
∴ Number of events = 4
∴ P(J) = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)
(xi) Let K be the occurrence of favourable event such that it is a spade.
∴ Number of events =13
∴ P(K) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
(xii) Let L be the occurrence of favourable event such that it is a black card.
∴ Number of events = 26
∴ P(L) = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)
(xiii) Let M be the occurrence of favourable event such that it is the seven of clubs.
∴ Number of events = 1
∴ P(M) = \(\frac { 1 }{ 52 }\)
(xiv) Let N be the occurrence of favourable event such that it is a jack.
∴ Number of events = 1
∴ P(N) = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)
(xv) Let O be the occurrence of favourable event such that it is an ace of spades.
∴ Number of events = 1
∴ P(O) = \(\frac { 1 }{ 52 }\)
(xvi) Let Q be the occurrence of favourable event such that it is a queen.
∴ VNumber of events = 4
∴ P(P) = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)
(xvii) Let R be the occurrence of favourable event such that it is a heart card.
∴ Number of events =13
∴ P(P) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
(xviii) Let S be the occurrence of favourable event such that it is a red card
∴ Number of events = 26
∴ P(P) = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)

Question 6.
An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.
Solution:
Number of possible events = 10 + 8 = 18
Let A be the occurrence of favourable event such that it is a white ball.
∴ Number of events = 8
∴ P(A) = \(\frac { 8 }{ 18 }\) = \(\frac { 4 }{ 9 }\)

Question 7.
A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) white ? (ii) red ? (iii) black ? (iv) not red ?
Solution:
Number of possible events = 3 + 5 + 4 = 12 balls
(i) Let A be the favourable event such that it is a white ball.
∴ P(A) = \(\frac { 4 }{ 12 }\) = \(\frac { 1 }{ 3 }\)
(ii) Let B be the favourable event such that it is a red ball.
∴ P(B) = \(\frac { 3 }{ 12 }\) = \(\frac { 1 }{ 4 }\)
(iii) Let C be the favourable event such that it is a black ball.
∴ P(C) = \(\frac { 5 }{ 12 }\)
(iv) Let D be the favourable event such that it is not red.
∴ Number of favourable events = 5 + 4 = 9
∴ P(D) = \(\frac { 9 }{ 12 }\) = \(\frac { 3 }{ 4 }\)

Question 8.
What is the probability that a number selected from the numbers 1,2,3,…………, 15 is a multiple of 4 ?
Solution:
Number of possible events =15
Let A be the favourable event such that it is a multiple of 4 which are 4, 8, 12
∴ P(A) = \(\frac { 3 }{ 15 }\) = \(\frac { 1 }{ 5 }\)

Question 9.
A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that ball drawn is not black ?
Solution:
Number of possible events = 6 + 8 + 4 = 18 balls
Let A be the favourable event such that it is not a black
∴ Number of favourable events = 6 + 4=10
∴ P(A) = \(\frac { 10 }{ 18 }\) = \(\frac { 5 }{ 9 }\)

Question 10.
A bag contains 5 white and 7 red balls. One ball is drawn at random, what is the probability that ball drawn is white ?
Solution:
Number of possible events = 5 + 7 = 12
Let A be the favourable event such that it is a while which are 5.
∴ P(A) = \(\frac { 5 }{ 12 }\)

Question 11.
A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is (i) white (if) red (iii) not black (iv) red or white.
Solution:
Number of possible events = 4 + 5 + 6 = 15
(i) Let A be the favourable events such that it is a white.
∴ P(A) = \(\frac { 6 }{ 15 }\) = \(\frac { 2 }{ 5 }\)
(ii) Let B be the favourable event such that it is a red
∴ P(B) = \(\frac { 4 }{ 15 }\)
(iii) Let C be the favourable event such that it is not black.
∴ Number of favourable events = 4 + 6=10
∴ P(C) = \(\frac { 10 }{ 15 }\) = \(\frac { 2 }{ 3 }\)

Question 12.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is: (i) red (ii) black.
Solution:
Number of possible events = 3 + 5 = 8
(i) Let A be the favourable events such that it is red.
∴ P(A) = \(\frac { 3 }{ 8 }\)
(ii) Let B be the favourable event such that it is black.
∴ P(B) = \(\frac { 5 }{ 8 }\)

Question 13.
A bag contains 5 red marbles, 8 white marbles, 4 green marbles. What is the probability that if one marble is taken out of the bag at random, it will be
(i) red
(ii) white
(iii) not green.
Solution:
Total number of possible events = 5 + 8+4=17
(i) Let A be the favourable event such that it is red.
∴ P(A) = \(\frac { 5 }{ 7 }\)
(ii) Let B be the favourable event such that it is white
Then P (B) = \(\frac { 8 }{ 7 }\)
(iii) Let C be the favourable event such that it is not green.
∴ Number of favourable events = 5 + 8 = 13
∴ P(C) = \(\frac { 13 }{ 17 }\)

Question 14.
If you put 21 consonants and 5 vowels in a bag. What would carry greater probability ? Getting a consonant or a vowel ? Find each probability.
Solution:
Total number of possible events = 21 + 5 = 26
(i) Probability of getting a consonant is greater as to number is greater than the other.
(ii) Let A be the favourable event such that it is a consonant.
∴ P(A) = \(\frac { 21 }{ 26 }\)
(iii) Let B be the favourable event such that it is a vowel.
∴ P(B) = \(\frac { 5 }{ 26 }\)

Question 15.
If we have 15 boys and 5 girls in a class which carries a higher probability ? Getting a copy belonging to a boy or a girl ? Can you give it a value ?
Solution:
Number of possible outcome (events) = 15 + 5 = 20
∵ The number of boys is greater than the girls
∴ The possibility of getting a copy belonging to a boy is greater.
Let A be the favourable outcome (event) then
P(A) = \(\frac { 15 }{ 20 }\) = \(\frac { 3 }{ 4 }\)

Question 16.
If you have a collection of 6 pairs of white socks and 3 pairs of black socks. What is the probability that a pair you pick without looking is (i) white ? (if) black ?
Solution:
Total number of possible outcomes =6+3=9
(i) Let A be the favourable outcome which is white pair.
∴ P(A) = \(\frac { 6 }{ 9 }\) = \(\frac { 2 }{ 3 }\)
(ii) Let B be the favourable outcome which is a black pair.
∴ P(B) = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)

Question 17.
If you have a spinning wheel with 3 green sectors, 1-blue sector and 1-red sector, what is the probability of getting a green sector ? Is it the maximum ?
Solution:
Total number of possible outcomes = 3 + 1 + 1 =5
Let A be the favourable outcome which is green sector
∴ P(A) = \(\frac { 3 }{ 5 }\)
∴ Number of green sectors is greater.
∴ It’s probability is greater.

Question 18.
When two dice are rolled :
(i) List the outcomes for the event that the total is odd.
(ii) Find probability of getting an odd total.
(iii) List the outcomes for the event that total is less than 5.
(iv) Find the probability for getting a total less than 5.
Solution:
∵ Every dice has 6 number from 1 to 6.
∴ Total outcomes = 6 x 6 = 36.
(i) List of outcomes for event then the total is odd will be (1,2), (1,4), (1,6), (2,1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6,1), (6, 3), (6, 5)
(ii) Probability of getting an odd total
Let A be the favourable outcomes which are
P(A) = \(\frac { 18 }{ 36 }\) = \(\frac { 1 }{ 2 }\)
(iii) List of outcomes for the event that total is less than 5 are (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2,1), which are 6.
(iv) Probability of getting a total less than 5 Let B be the favourable outcome,
Then P (B) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)

 

Hope given RD Sharma Class 8 Solutions Chapter 26 Data Handling IV (probability) Ex 26.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2B

RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2B.

Other Exercises

Question 1.
Solution:
(i) The given number = 2650
Digit at unit’s place = 0
It is divisible by 2.
(ii) The given number = 69435
Digit at unit’s place = 5
It is not divisible by 2.
(iii) The given number = 59628
Digit at unit’s place = 8
It is divisible by 2.
(iv) The given number = 789403
Digit at unit’s place = 3
It is not divisible by 2.
(v) The given number = 357986
Digit at unit’s place = 6
It is divisible by 2.
(vi) The given number = 367314
Digit at unit’s place = 4
It is divisible by 2.

Question 2.
Solution:
(i) The given number = 733
Sum of its digits = 7 + 3 + 3 = 13,
which is not divisible by 3.
∴ 733 is not divisible by 3.
(ii) The given number = 10038
Sum of its digits = 1 + 0 + 0 + 3 + 8 = 12,
which is divisible by 3
∴ 10038 is divisible by 3.
(iii) The given number = 20701
Sum of its digits = 2 + 0 + 7 + 0 + 1 = 10,
which is not divisible by 3
∴ 20701 is not divisible by 3.
(iv) The given number = 524781
Sum of its digits = 5 + 2 + 4 + 7 + 8 + 1 = 27,
which is divisible by 3
∴ 524781 is divisible by 3.
(v) The given number = 79124
Sum of its digits = 7 + 9 + 1 + 2 + 4 = 23,
which is not divisible by 3
∴ 79124 is not divisible by 3.
(vi) The given number = 872645
Sum of its digits = 8 + 7 + 2 + 6 + 4 + 5 = 32,
which is not divisible by 3
∴ 872645 is not divisible by 3.

Question 3.
Solution:
(i) The given number = 618
The number formed by ten’s and unit’s digits is 18, which is not divisible by 4.
∴ 618 is not divisible by 4.
(ii) The given number = 2314
The number formed by ten’s and unit’s digits is 14, which is not divisible by 4.
∴ 2314 is not divisible by 4.
(iii) The given number = 63712
The number formed by ten’s and unit’s digits is 12, which is divisible by 4
∴ 63712 is divisible by 4.
(iv) The given number = 35056
The number formed by ten’s and unit’s digits is 56, which is divisible by 4.
∴ 35056 is divisible by 4.
(v) The given number = 946126
The number formed by ten’s and unit’s digits is 26, which is not divisible by 4.
∴ 946126 is not divisible by 4.
(vi) The given number = 810524
The number formed by ten’s and unit’s digits is 24, which is divisible by 4.
∴ 810524 is divisible by 4.

Question 4.
Solution:
We know that a number is divisible by 5 if its ones digit is 0 or 5
(i) 4965, (ii) 23590 (iv) 723405 and (vi) 438750 are divisible by 5

Question 5.
Solution:
(i) The given number = 2070
Its unit’s digit = 0
So, it is divisible by 2
Sum of its digits = 2 + 0 + 7 + 0 = 9,
which is divisible by 3
∴ The given number is divisible by 3.
So, 2070 is divisible by both 2 and 3.
Hence it is divisible by 6.
(ii) The given number = 46523
Its unit’s digit = 3
So, it is not divisible by 2
Hence 46523 is not divisible by 6.
(iii) The given number = 71232
Its unit’s digit = 2
So, it is divisible by 2
Sum of its digits = 7 + 1 + 2 + 3 + 2
= 15, which is divisible by 3
∴ 71232 is divisible by both 2 and 3
Hence it is divisible by 6.
(iv) The given number = 934706
Its unit’s digit = 6 So,
it is divisible by 2
Sum of its digits = 9 + 3 + 4 + 7 + 0 + 6 = 29,
which is not divisible by 3
Hence 934706 is not divisible by 6.
(v) The given number = 251780
Its unit’s digit = 0
So, it is divisible by 2
Sum of its digits = 2 + 5 + 1 + 7 + 8 + 0 = 23,
which is not divisible by 3
251780 is not divisible 6
(vi) 872536 is not divisible by 6 as sum of its digits is 8 + 7 + 2 + 5 + 3 + 6 = 31 which is not divisible by 3

Question 6.
Solution:
We know that a number is divisible by 7 if the difference between twice the ones digit and the number formed by the other digits is either 0 or a multiple of 7
(i) 826, 6 x 2 = 12 and 82
Difference between 82 and 12 = 70
Which is divisible by 7
∴ 826 is divisible by 7
(ii) In 117, 7 x 2 = 14, 11
Difference between 14 and 11 = 14 – 11 = 3
Which is not divisible by 7
∴ 117 is not divisible by 7
(iii) In 2345, 5 x 2 = 10 and 234
Difference between 234 – 10 = 224
which is divisible by 7
∴ 2345 is divisible by 7
(iv) In 6021, 1 x 2 = 2, and 602
Difference between 602 and 2 = 600
which is not divisible by 7
∴ 6021 is not divisible by 7
(v) In 14126, 6 x 2 = 12 and 1412
Difference between 1412 – 12 = 1400
which 8 is divisible by 7
∴ 14126 is divisible by 7
(vi) In 25368, 8 x 2 = 16 and 2536
Difference between 2536 and 16 = 2520
which is divisible by 7
∴ 25368 is divisible by 7

Question 7.
Solution:
(i) The given number = 9364
The number formed by hundred’s, ten’s and unit’s digits is 364, which is not divisible by 8.
∴ 9364 is not divisible by 8.
(ii) The given number = 2138
The number formed by hundred’s, ten’s and unit’s digits is 138, which is not divisible by 8.
∴ 2138 is not divisible by 8.
(iii) The given number = 36792
The number formed by hundred’s, ten’s and unit’s digits is 792, which is divisible by 8.
∴ 36792 is divisible by 8.
(iv) The given number = 901674
The number formed by hundred’s, ten’s and unit’s digits is 674, which is not divisible by 8.
∴ 901674 is not divisible by 8.
(v) The given number = 136976
The number formed by hundred’s, ten’s and unit’s digits is 976, which is divisible by 8.
∴ 136976 is divisible by 8.
(vi) The given number = 1790184
The number formed by hundred’s, ten’s and unit’s digits is 184, which is divisible by 8.
∴ 1790184 is divisible by 8.

Question 8.
Solution:
We know that a number is divisible by 9, if the sum of its digits is divisible by 9
(i) In 2358
Sum of digits : 2 + 3 + 5 + 8 = 18
which is divisible by 9
∴ 2358 is divisible by 9
(ii) In 3333
Sum of digits 3 + 3 + 3 + 3 = 12
which is not divisible by 9
∴ 3333 is not divisible by 9
(iii) In 98712
Sum of digits = 9 + 8 + 7 + 1 + 2 = 27
Which is divisible by 9
∴ 98712 is divisible by 9
(iv) In 257106
Sum of digits = 2 + 5 + 7 + 1 + 0 + 6 = 21
which is not divisible by 9
∴ 257106 is not divisible by 9
(v) In 647514
Sum of digits = 6 + 4 + 7 + 5 + 1 + 4 = 27
which is divisible by 9
∴ 647514 is divisible by 9
(v) In 326999
Sum of digits = 3 + 2 + 6 + 9 + 9 + 9 = 38
which is not divisible by 9
∴ 326999 is divisible by 9

Question 9.
Solution:
We know that a number is divisible by 10 if its ones digit is 0
∴(i) 5790 is divisible by 10

Question 10.
Solution:
(i) The given number = 4334
Sum of its digits in odd places = 4 + 3 =7
Sum of its digits in even places = 3 + 4 = 7
Difference of the two sums = 7 – 7 = 0
∴4334 is divisible by 11.
(ii) The given number = 83721
Sum of its digits in odd places = 1 + 7 + 8 = 16
Sum of its digits in even places = 2 + 3 = 5
Difference of the two sums = 16 – 5 = 11,
which is multiple of 11.
∴ 83721 is divisible by 11.
(iii) The given number = 66311
Sum of its digits in odd places = 1 + 3 + 6 = 10
Sum of its digits in even places = 1 + 6 = 7
Difference of the two sums = 10 – 7 = 3,
which is not a multiple of 11.
∴ 66311 is not divisible by 11.
(iv) The given number = 137269
Sum of its digits in odd places = 9 + 2 + 3 = 14
Sum of its digits in even places = 6 + 7 + 1 = 14
Difference of the two sums = 14 – 14 = 0
∴ 137269 is divisible by 11.
(v) The given number = 901351
Sum of its digits in odd places = 1 + 3 + 0 = 4
Sum of its digits in even places = 5 + 1 + 9 = 15
Difference of the two sums = 15 – 4 = 11,
which is a multiple of 11.
∴ 901351 is divisible by 11.
(vi) The given number = 8790322
Sum of its digits in odd places = 2 + 3 + 9 + 8 = 22
Sum of its digits in even places = 2 + 0 + 7 = 9
Difference of the two sums = 22 – 9 = 13,
which is not a multiple of 11.
∴ 8790322 is not divisible by 11.

Question 11.
Solution:
(i) The given number = 27*4
Sum of its digits = 2 + 7 + 4 = 13
The number next to 13 which is divisible by 3 is 15.
∴ Required smallest number = 15 – 13
= 2.
(ii) The given number = 53*46
Sum of the given digits = 5 + 3 + 4 + 6 = 18,
which is divisible by 3.
∴ Required smallest number = 0.
(iii) The given number = 8*711
Sum of the given digits = 8 + 7 + 1 + 1 = 17
The number next to 17,
which is divisible by 3 is 18.
∴ Required smallest number =18 – 17 = 1
(iv) The given number = 62*35
Sum of the given digits = 6 + 2 + 3 + 5 = 16
The number next to 16,
which is divisible by 3 is 18.
∴ Required smallest number =18 – 16 = 2
(v) The given number = 234*17
Sum of the given digits = 2 + 3+ 4 + 1 + 7 = 17
The number next to 17, which is divisible by 3 is 18.
Required smallest number = 18 – 17 = 1.
(vi) The given number = 6* 1054
Sum of the given digits = 6 + 1+ 0 + 5 + 4 = 16
The number next to 16,
which is divisible by 3 is 18.
Required smallest number = 18 – 16 = 2.

Question 12.
Solution:
(i) The given number = 65*5
Sum of its given digits = 6 + 5 + 5 = 16
The number next to 16, which is divisible by 9 is 18.
∴ Required smallest number =18 – 16 = 2
(ii) The given number = 2*135
Sum of its given digits = 2 + 1 + 3 + 5
The number next to 11, which is divisible by 9 is 18.
∴ Required smallest number = 18 – 11 =7.
(iii) The given number = 6702*
Sum of its given digits = 6 + 7 + 0 + 2 = 15
The number next to 15, which is divisible by 9 is 18.
∴ Required smallest number = 18 – 15 = 3
(iv) The given number = 91*67
Sum of its given digits = 9 + 1 + 6 + 7 =23
The number next to 23, which is divisible by 9 is 27.
∴ Required smallest number = 27 – 23 = 4.
(v) The given number = 6678*1
Sum of its given digits= 6 + 6 + 7 + 8 + 1 = 28
The number next to 28, which is divisible by 9 is 36.
∴ Required smallest number = 36 – 28 = 8.
(vi) The given number = 835*86
Sum of its given digits = 8 + 3 + 5 + 8 + 6 = 30
The number next to 30, which is divisible by 9 is 36.
∴ Required smallest number = 36 – 30 = 6.

Question 13.
Solution:
(i) The given number = 26*5
Sum of its digits is odd places = 5 + 6 = 11
Sum of its digits in even places = * + 2
Difference of the two sums = 11 – (* + 2)
The given number will be divisible by 11 if the difference of the two sums = 0.
∴ 11 – (* + 2) = 0
11 = * + 2
11 – 2 = *
9 = *
Required smallest number = 9.
(ii) The given number = 39*43
Sum of its digits in odd places
=3 + * + 3 = * + 6
Sum of its digits in even places = 4 + 9 = 13
Difference of the two sums = * + 6 – 13 = * – 7
The given number will be divisible by 11, if the difference of the two sums = 0.
∴ * – 7 = 0
* = 7
∴ Required smallest number = 7.
(iii) The given number = 86*72
Sum of its digits in odd places
= 2 + * + 8 = * + 10
Sum of its digits in even places = 7 + 6 = 13
Difference of the two sums = * + 10 – 13 = * – 3
The given number will be divisible by 11, if the difference of the two sums = 0.
∴ * – 3 = 0
* = 3
∴ Required smallest number = 3.
(iv) The given number = 467*91
Sum of its digits in odd places = 1 + * + 6 = * + 7
Sum of its digits in even places = 9 + 7 + 4 = 20
Difference of the two sums
= 20 – (* + 7)
= 20 – * – 7 = 13 – *
Clearly the difference of the two sums will be multiple of 11 if 13 – * = 11
∴ 13 – 11 = *
2 = *
* = 2 .
∴ Required smallest number = 2.
(v) The given number = 1723*4
Sum of its digits in odd places = 4 + 3 + 7 = 14
Sum of its digits in even places = * + 2 + 1 = * + 3
Difference cf the two sums = * + 3 – 14 = * – 11
The given number will be divisible by 11,if *- 11 is a multiple of 11,which is possible if * = 0.
Required smallest number = 0.
(vi) The given number = 9*8071
Sum of its digits in odd places = 1 + 0 + * = 1 + *
Sum of its digits in even places = 7 + 8 + 9 = 24
Difference of the two sums = 24 – 1 – * = 23 – *
∴ The given number will be divisible by 11, if 23 – * is a multiple of 11, which is possible if * = 1.
∴ Required smallest number = 1.

Question 14.
Solution:
(i) The given number = 10000001
Sum of its digits in odd places
= 1 + 0 + 0 + 0 = 1
Sum of its digits in even places = 0 + 0 + 0 + 1 = 1
Difference of the two sums = 1 – 1 = 0
∴ The number 10000001 is divisible by 11.
(ii) The given number = 19083625
Sum of its digits in odd places
= 5 + 6 + 8 + 9 = 28
Sum of its digits in even places = 2 + 3 +0 + 1 = 6
Difference of the two sums = 28 – 6 = 22,
which is divisible by 11.
∴ The number 19083625 is divisible by 11.
(iii) The given number = 2134563
Sum of its digits
= 2 + 1 + 3 + 4 + 5 + 6 + 3 = 24,
which is not divisible by 9.
∴ The number 2134563 is not divisible by 9.
(iv) The given number = 10001001
Sum of its digits
= 1 + 0 + 0 + 0 + 1 + 0 + 0 + 1 = 3,
which is divisible by 3.
∴ The number 10001001 is divisible by 3.
(v) The given number = 10203574
The number formed by its ten Is and unit’s digits is 74, which is not divisible by 4.
The number 10203574 is not divisible by 4.
(vi) The given number = 12030624
The number formed by its hundred’s, ten’s and unit’s digits = 624,
which is divisible by 8.
∴ The number 12030624 is divisible by

Question 15.
Solution:
103, 137, 179, 277, 331, 397 are prime numbers.

Question 16.
Solution:
(i) 154
(ii) 612
(iii) 5112, 3816 etc.
(iv) 3426, 5142 etc.

Question 17.
Solution:
(i) False
(ii) True
(iii) False
(iv) True
(v) False
(vi) True
(vii) True
(viii) True

Hope given RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2B are helpful to complete your math homework.

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