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NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.2.

• Direct and Indirect Proportions Class 8 Ex 13.1

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.2

Question 1.
Which of the following are in inverse proportion?
(i) The number of workers on a job and the time to complete the job.
(ii) The time is taken for a journey and the distance traveled in a uniform speed.
(iii) Area of cultivated land and the crop harvested.
(iv) The time is taken for a fixed journey and the speed of the vehicle.
(v) The population of a country and the area of land per person.
Solution.
(i) The number of workers on jobs and the time to complete the job are in inverse proportion.
(ii) The time is taken for a journey and the distance traveled in a uniform speed are not in inverse proportion.
(iii) Area of cultivated land and the crop harvested are not in inverse proportion.
(iv) The time taken for a fixed journey and the speed of the vehicle are in inverse proportion.
(v) The population of a country and the area of land per person are in inverse proportion.

Question 2.
In a Television game show, the prize money of  1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners.

Solution.

Question 3.
Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table.


(i) Are the number of spokes and the angles formed between the pairs of consecutive spokes in verse proportion?
(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?
Solution.
Let the angle (in degree) between a pair of consecutive spokes be \({ y }_{ 3 }\), \({ y }_{ 4 }\) and \({ y }_{ 5 }\) respectively. Then,

Question 4.
If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?
Solution.
Suppose that each would get \({ y }_{ 2 }\) sweets.
Thus, we have the following table.

Question 5.
A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?
Solution.
Suppose that the food would last for \({ y }_{ 2 }\) days. We have the following table:

Question 6.
A contractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?
Solution.
Suppose that they take \({ y }_{ 2 }\) days to complete the job. We have the following table

Question 7.
A batch of bottles was packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?

Solution.
Suppose that \({ y }_{ 2 }\) boxes would be filled. We have the following table:

Question 8.
A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?
Solution.
Suppose that \({ x }_{ 2 }\) machines would be required. We have the following table:

Question 9.
A car takes 2 hours to reach a destination by traveling at a speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?
Solution.
Let it take \({ y }_{ 2 }\) hours. We have the following table:
sol.

Question 10.
Two persons could fit new windows in a house in 3 days.
(i) One of the persons fell ill before the work started. How long would the job take now?
(ii) How many persons would be needed to fit the windows in one day?
Solution.
(i) Let the job would take \({ y }_{ 2 }\) days. We have the following table:

Clearly, more the number of persons, lesser would be the number of days to do the job. So, the number of persons and number of days vary in inverse proportion.
So, 2 x 3 = 1 x \({ y }_{ 2 }\)
⇒ \({ y }_{ 2 }\) = 6
Thus, the job would now take 6 days.

(ii) Let \({ y }_{ 2 }\) persons be needed. We have the following table:

Clearly, more the number of persons, lesser would be the number of days to do the job. So, the number of persons and number of days vary in inverse proportion.
So, 3 x 2 = 1 x \({ y }_{ 3 }\)
⇒ \({ T }_{ 2 }\) = 6
Thus, 6 persons would be needed.

Question 11.
A school has 8periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?
Solution.
Let each period be \({ y }_{ 2 }\) minutes long.
We have the following table:

We note that more the number of periods, lesser would be the length of each period. Therefore, this is a case of inverse proportion.
So, 8 x 45 = 9 x \({ y }_{ 2 }\)
⇒ \({ y }_{ 2 }=\frac { 8\times 45 }{ 9 } \)
⇒ \({ y }_{ 2 }\) = 40
Hence, each period would be 40 minutes long.

We hope the NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.2 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Ex 12.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Ex 12.2.

• Exponents and Powers Class 8 Ex 12.1

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Ex 12.2

Question 1.
Express the following numbers in standard form:
(i) 0.0000000000085
(ii) 0.00000000000942
(iii) 6020 000 000 000 000
(iv) 0.00000000837
(v) 31860000000
Solution.

Question 2.
Express the following numbers in usual form:

Solution.

Question 3.
Express the number appearing in the following statements in standard form:
(i) 1 micron is equal to \(\frac { 1 }{ 1000000 } \)m.
(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.
(iii) Size of a bacteria is 0.0000005 m
(iv) Size of a plant cell is 0.00001275 m
(v) Thickness of a thick paper is 0.07 mm.
Solution.

Question 4.
In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack ?
Solution.
Total thickness of books
= 5 x 20 mm = 100 mm
Total thickness of paper sheets
= 5 x 0.016 mm = 0.080 mm
∴ Total thickness of the stack
= Total thickness of books + Total thickness of paper sheets
= 100 mm + 0.080 mm
= (100 + 0.080) mm
= 100.080 mm
= 1.0008 x \({ 10 }^{ 2 }\) mm.

We hope the NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Ex 12.2 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Ex 12.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.6 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.6.

• Basic Geometrical Ideas Class 6 Ex 4.1
• Basic Geometrical Ideas Class 6 Ex 4.2
• Basic Geometrical Ideas Class 6 Ex 4.3
• Basic Geometrical Ideas Class 6 Ex 4.4
• Basic Geometrical Ideas Class 6 Ex 4.5

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.6

Question 1.
From the figure identify
(a) the center of the circle
(b) three radii
(c) a diameter
(d) a chord

(e) two points in the interior
(f) a point in the exterior
(g) a sector
(h) a segment.
Solution :
(a) O is the centre of the circle.
(b) \(\bar { OA }\), \(\bar { OB }\), \(\bar { OC }\) are three radii of the circle.
(c) \(\bar { AC }\) is a diameter of the circle.
(d) \(\bar { ED }\) is a chord of the circle,
(e) O and P are two points in the interior.
(f) Q is a point in the exterior.
(e) OAB (shaped portion) is a sector of the circle.
(f) Shaded portion of the circular region enclosed by line segment ED and the corresponding arc.

Question 2.
(a) Is every diameter of a circle also a chord?
(b) Is every chord of a circle also a diameter?
Solution :
(a) Yes! every diameter of a circle is also a chord.
(b) No ! every chord of a circle is not also a

Question 3.
Draw any circle and mark:
(a) it’s center
(b) a radius
(c) a diameter
(d) a sector
(e) a segment
(f) a point in its interior
(g) a point in its exterior
(h) an arc.
Solution :
(a) O is the centre.
(b) \(\bar { OA }\) is a radius.
(c) \(\bar { AB }\) is a diameter.
(d) OBC is a sector.
(e) AGD is a segment.
(f) P is a point in its interior.
(g) Q is a point in its exterior.

(h) \(\hat { EF }\)  is an arc.

Question 4.
Say true or false:
(a) Two diameters of a circle will necessarily intersect.
(b) The center of a circle is always in its interior.
Solution :
(a) True
(b) True.

We hope the NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.6 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.6, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.5 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.5.

• Basic Geometrical Ideas Class 6 Ex 4.1
• Basic Geometrical Ideas Class 6 Ex 4.2
• Basic Geometrical Ideas Class 6 Ex 4.3
• Basic Geometrical Ideas Class 6 Ex 4.4
• Basic Geometrical Ideas Class 6 Ex 4.6

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.5

Question 1.
Draw a rough sketch of a quadrilateral PQRS. Draw its diagonals. Name them. Is the meeting point of the diagonal in the interior or exterior of the quadrilateral?
Solution :

The meeting point O of the diagonals PR and QS of the quadrilateral PQRS is in the interior of the quadrilateral PQRS.

Question 2.
Draw a rough sketch of a quadrilateral KLMN. State :
(a) two pairs of opposite sides.
(b) two pairs of opposite angles.
(c) two pairs of adjacent sides.
(d) two pairs of adjacent angles.
Solution :
(a) \(\bar { KL }\), \(\bar { NM }\) and \(\bar { KN }\), \(\bar { ML }\)
(b) ∠K, ∠M and ∠N, ∠L
(c) \(\bar { KL }\) , \(\bar { KN }\) and \(\bar { NM }\), \(\bar { ML }\) or \(\bar { KL }\) , \(\bar { LM }\) and \(\bar { NM }\), \(\bar { ML }\)

(d) ∠K, ∠L, and ∠M. ∠N or ∠K. ∠L and ∠L, ∠M etc.

Question 3.
Investigate:
Use strips and fasteners to make a triangle and a quadrilateral. Try to push inward at any one vertex of the triangle. Do the same to the quadrilateral. Is the triangle distorted? Is the quadrilateral distorted? Is the triangle rigid? Why is it that structures like electric towers make use of triangular shapes and not quadrilaterals?
Solution :
On pushing inward at any one vertex of the triangle, the triangle is not distorted, However, the quadrilateral is distorted. Hence, a triangle is a rigid figure. This is why structures like electric towers make use of triangular shapes and not quadrilaterals.

We hope the NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.5 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.4 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.4.

• Basic Geometrical Ideas Class 6 Ex 4.1
• Basic Geometrical Ideas Class 6 Ex 4.2
• Basic Geometrical Ideas Class 6 Ex 4.3
• Basic Geometrical Ideas Class 6 Ex 4.5
• Basic Geometrical Ideas Class 6 Ex 4.6

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.4

Question 1.
Draw a rough sketch of a triangle ABC. Mark a point P in its interior and a point Q in its exterior. Is the point A in its exterior or in its exterior?
Solution :

The point A is neither in the exterior nor in the interior of triangle ABC. It is on the triangle ABC.

Question 2.
(a) Identify three triangles in the figure.
(b) Write the names of seven angles,
(c) Write the names of the six line segments,
(d) Which two triangles have ∠B as common?
Solution :
(a) Three triangles
Triangle ABC,
Triangle ABD,
Triangle ADC

(b) Seven Angles
∠ABC. ∠ACB, ∠BAC, ∠BAD, ∠CAD, ∠ADB and ∠ADC
(c) Six line segments
\(\bar { AB }\), \(\bar { AC }\), \(\bar { BC }\), \(\bar { AD }\), \(\bar { BD }\), \(\bar { DC }\)
(d) ΔABC. ΔABD.

We hope the NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.4 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.3.

• Basic Geometrical Ideas Class 6 Ex 4.1
• Basic Geometrical Ideas Class 6 Ex 4.2
• Basic Geometrical Ideas Class 6 Ex 4.4
• Basic Geometrical Ideas Class 6 Ex 4.5
• Basic Geometrical Ideas Class 6 Ex 4.6

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.3

Question 1.
Name the angles in the given figure.

Solution :
∠A or ∠DAB; ∠B or ∠ABC; ∠C or ∠BCD; ∠Dor ∠CDA.

Question 2.
In the given diagram, name the point (s)
(a) in the interior of ∠DOE
(b) in the exterior of∠EOF
(c) on ∠EOF.

Solution :
(a) A
(b) C, A, D
(c) E, B,0, F.

Question 3.
Draw rough diagrams of two angles such that they have
(a) One point in common
(b) Two points in common
(c) Three points in common
(d) Four points in common
(e) One ray in common.
Solution :
(a)

∠AOB and ∠BOC have one point O in common.
(b) ∠AOB and ∠OBC have two points O and B in common.

(c) Not possible
(d) Not possible
(e)

∠AOB and ∠BOC have one ray \(\overrightarrow { OB }\) in common

We hope the NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.2.

• Basic Geometrical Ideas Class 6 Ex 4.1
• Basic Geometrical Ideas Class 6 Ex 4.3
• Basic Geometrical Ideas Class 6 Ex 4.4
• Basic Geometrical Ideas Class 6 Ex 4.5
• Basic Geometrical Ideas Class 6 Ex 4.6

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.2

Question 1.
Classify the following curves as
(i) open
(ii) closed.

Solution :
(i) a, c
(ii) b,d,e.

Question 2.
Draw rough diagrams to illustrate the following :
(a) Open curve
(b) Closed curve.
Solution :
(a)

(b)

Question 3.
Draw any polygon and shade its interior.
Solution :

Question 4.
Consider the given figure and answer the questions:
(a) Is it a curve?

(b) Is it closed?
Solution :
(a) Yes! It is a curve
(b) Yes! It is closed.

Question 5.
Illustrate, if possible, each one of the following with a rough diagram :
(a) A closed curve that is not a polygon
(b) An open curve made up entirely of line segments.
(c) A polygon with two sides.
Solution :
(a)

(b)

(c) Not possible.

We hope the NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4.

• Mensuration Class 8 Ex 11.1
• Mensuration Class 8 Ex 11.2
• Mensuration Class 8 Ex 11.3

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4

Question 1.
Given a cylindrical tank, in which situation will you find the surface area and in which situation volume?
(a) To find how much it can hold.
(b) Number of cement bags required to plaster it.
(c) To find the number of smaller tanks that can be filled with water from it.
Solution.
(a) Volume
(b) Surface area
(c) Volume.

Question 2.
The diameter of cylinder A is 7 cm, and the height is 14 cm. The diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has a greater surface area?

Solution.

Question 3.
Find the height of a cuboid whose base area is 180 \({ cm }^{ 2 }\) and volume is 900 \({ cm }^{ 3 }\).
Solution.
Height of the cuboid
= \(\frac { Volume\quad of\quad the\quad cuboid }{ Base\quad area\quad of\quad the\quad cuboid } \)
= \(\frac { 900 }{ 180 } \)
= 5 cm

Question 4.
A cuboid is of dimensions 60 cm x 54 cm x 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
Solution.
A volume of the cuboid

Question 5.
Find the height of the cylinder whose volume is 1.54 \({ m }^{ 3 }\) and diameter of the base is 140 cm.
Solution.

Question 6.
A milk tank is in the form of a cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in liters that can be stored in the tank.
Solution.

Question 7.
If each edge of a cube is doubled,
(i) how many times will its surface area increase?
(ii) how many times will its volume increase?
Solution.
Let the original edge of the cube be a cm.
Then, its new edge = 2a cm

Question 8.
Water is pouring into a cuboidal reservoir at the rate of 60 liters per minute. If the volume of the reservoir is 108 m3, find the number of hours it will take to fill the reservoir.
Solution.

We hope the NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3.

• Visualising Solid Shapes Class 8 Ex 10.1
• Visualising Solid Shapes Class 8 Ex 10.2

NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3

Question 1.
Can a polyhedron have for its faces
(i) 3 triangles?
(ii) 4 triangles?
(iii) a square and four triangles?
Solution.
(i) No
(ii) Yes
(iii) Yes

Question 2.
Is it possible to have a polyhedron with any given number of faces? (Hint: Think of a pyramid).
Solution.
Possible, only if the number of faces is greater than or equal to 4.

Question 3.
Which are prisms among the following?


Solution.
We know that a prism is a polyhedron whose base and top faces are congruent and parallel and other (lateral) faces are parallelograms in shape. So, only (ii) and (iv) are prisms.

Question 4.
(i) How are prisms and cylinders alike?
(ii) How are pyramids and cones alike?
Solution.
(i) The prisms and cylinder, both, have their base and top faces as congruent and parallel to each other. Also, a prism becomes a cylinder as the number of sides of its base becomes larger and larger.

(ii) The pyramids and cones are alike in the sense that their lateral faces meet at a point (called vertex). Also, a pyramid becomes a cone as the number of sides of its base becomes larger and larger.

Question 5.
Is a square prism same as a cube ? Explain.
Solution.
No; not always as it can be a cuboid also.

Question 6.
Verify Euler’s formula for these solids

Solution.
(i)
F = 7
V= 10
E = 15
F + V = 7 + 10 = 17
E + 2 = 15 + 2 = 17
So, F + V = E + 2
Hence, Euler’s Formula is verified,

(ii)
F = 9
V = 9
E = 16
F + V = 9 + 9 = 18
E + 2 = 16 + 2 = 18
So, F + V = E + 2
Hence, Euler’s Formula is verified

Question 7.
Using Euler’s formula find the unknown.

Solution.
(i)
F + V = E + 2
⇒ F + 6 = 12 + 2
⇒ F + 6 = 14
⇒ F = 14 – 6 = 8

(ii)
F + V = E + 2
⇒ 5 + V = 9 + 2
⇒ 5 + V = 11
⇒ V= 11-5 = 6

(iii)
F + V = E + 2
⇒ 20 + 12 = E + 2
⇒ 32 = E + 2
⇒ E = 32 – 2
⇒ E = 30

Question 8.
Can a polyhedron have 10 faces, 20 edges, and 15 vertices?
Solution.
Here F = 10
E = 20
V= 15
So, F + V = 10 + 15 = 25
E + 2 = 20 + 2 = 22
∵ F + V ≠ E + 2
∴ Such a polyhedron is not possible.

We hope the NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.2.

• Visualising Solid Shapes Class 8 Ex 10.1
• Visualising Solid Shapes Class 8 Ex 10.3

NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.2

Question 1.
Look at the given map of a city.
Answer the following.
(a) Colour the map as follows: Bluewater, Red-fire station, Orange-Library, Yellow-schools, Green-Parks, Pink-Community Centre, Purple-Hospital, Brown-Cemetery.
(b) Mark a green X’ at the intersection of Road ‘C’ and Nehru Road, Green Y’ at the intersection of Gandhi Road and Road A.
(c) In red, draw a short street route from Library to the bus depot.
(d) Which is further east, the city park or the market?
(e) Which is further south, the primary school or the Sr. Secondary School?

Solution.
(a) Please color yourself.
(b) See the above figure.
(c) See the above figure.
(d) City Park.
(e) Senior Secondary school.

Question 2.
Draw a map of your classroom using a proper scale and symbols for different objects.
Solution.
Please draw yourself.

Question 3.
Draw a map of your school compound using a proper scale and symbols for various features like playground main building, garden etc.
Solution.
Please draw yourself.

Question 4.
Draw a map giving instructions to your friend so that she reaches your house without any difficulty.
Solution.
Please draw yourself.

We hope the NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.2 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.7 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.7.

• Playing With Numbers Class 6 Ex 3.1
• Playing With Numbers Class 6 Ex 3.2
• Playing With Numbers Class 6 Ex 3.3
• Playing With Numbers Class 6 Ex 3.4
• Playing With Numbers Class 6 Ex 3.5
• Playing With Numbers Class 6 Ex 3.6

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.7

Question 1.
Renu purchases two bags of fertilizer of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertilizer e×act number of times.
Solution :
Factors of 75 are 1, 3, 5, 15, 25 and 75.
Factors of 69 are 1, 3, 23 and 69.
∴ Common factors of 75 and 69 are 1 and 3.
Highest of these common factors is 3.
∴ H.C.F. of 75 and 69 is 3.
Hence, the maximum value of weight which can measure the weight of the fertilizer e×act number of times is 3 kg.

Question 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 :

∴ L.C.M. of 63, 70 and 77
= 2 × 3 × 3 × 5 × 7 × 11 = 6930.
Hence, the minimum distance each should cover so that all can cover the distance in complete steps is 6930 cm.

Question 3.
The length, breadth, and height of a room are 825 cm, 675 cm, and 450 cm, respectively. Find the longest tape which can measure the three dimensions of the room exactly.
Solution :
Factors of 825 are 1, 3, 5, 11, 15, 25, 33, 55,75,165, 275 and 825.
Factors of 675 are 1, 3, 5,9,15, 25, 27,45,75, 135, 225 and 675.
Factors of 450 are 1,2,3,5,6,9,10,15,18,25, 30,45, 50, 75, 90, 150, 225 and 450.
∴ Common factors of 825, 675 and 450 are 1,3,5,15, 25 and 75.
Highest of these common factors is 75.
Hence, the length of the longest tape which can measure the three dimensions of the room exactly is 75 cm.

Question 4.
Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.
Solution :

∴ L.C.M. of 6, 8 and 12 = 2 × 2 × 2 × 3 = 24. Multiples of 24 are 24,48,72,96,120,144,
Hence, the smallest 3-digit number which is exactly divisible by 6, 8 and 12 is 120.

Question 5.
Determine the largest 3-digit number exactly divisible by 8, 10 and 12.
Solution :

∴ L.C.M. of 8,10 and 12 = 2×2×2×3×5
= 120.
Multiples of 120 are :
120 × 1 = 120,120 × 2 = 240,120 × 3 = 360,120 × 4 = 480,120 × 5 = 600,120 × 6 = 720,120 × 7 = 840,
120 × 8 = 960,120 × 9 = 1080,
Hence, the largest 3-digit number exactly divisible by 8, 10 and 12 is 960.

Question 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 am at what time will they change simultaneously again?
Solution :

∴ L.C.M. of 48,72 and 108 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432.
432 seconds = 7 min 12 seconds.
Hence, they will change simultaneously again after 7 min 12 seconds from 7 a.m.

Question 7.
Three tankers contain 403 liters, 434 liters and 465 liters of diesel respectively. Find the ma×imum capacity of a container that can measure the diesel of the three containers e×act a number of times.
Solution :
Factors of 403 are 1, 13, 31 and 403. Factors of 434 are 1, 2, 7, 14, 31, 62, 217 and 434.
Factors of 465 are 1, 3, 5, 15, 31, 93, 155 and 465.
Common factors of 403,434 and 465 are 1 and 31.
Highest of these common factors is 31.
∴ H.C.F. of 403. 434 and 465 is 31.
Hence, the maximum capacity of the container that can measure the diesel of the three containers an e×act number of times is 31 litres.

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

∴ L.C.M. of 6, 15 and 18 = 2 × 3 × 3 × 5 = 90.
Hence, the required number is 90 + 5 i.e., 95.

Question 9.
Find the smallest four digit number which is divisible by 18, 24 and 32.
Solution :

∴ L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288.
Multiples of 288 are :
288 × 1 = 288, 288 × 2 = 576, 288 × 3 = 864, 288×4= 1152,
Hence, the smallest four digit number which is divisible by 18, 24 and 32 is 1152.

Question 10.
Find the L.C.M. of the following numbers:
(a) 9 and 4
(b) 12 and 5
(c) 6 and 5
(d) 15 and 4.
Observe a common property in the obtained L.C.M.s. Is L.C.M. the product of two numbers in each case?
Solution :
(a) 9 and 4

∴ L.C.M. of 9 and 4 = 2 × 2 × 3 × 3 = 36 (= 9 × 4).

(b) 12 and 5

∴ L.C.M. of 12 and 5 = 2×2×3×5 = 60 (= 12 × 5).

(c) 6 and 5

∴ L.C.M. of 6 and 5 = 2×3×5 = 30 (= 6 × 5).

(d) 15 and 4

∴ L.C.M. of 15 and 4 = 2 × 2 × 3 × 5 = 60 (=15×4).
We observe a common property in the obtained L.C.M.’s that L.C.M. is the product of two numbers in each case.

Question 11.
Find the L.C.M. of the following numbers in which one number is the factor of the other.
(a) 5, 20
(b) 6, 18
(c) 12, 48
(d) 9, 45.
What do you observe in the results obtained?
Solution :
(a) 5, 20
Prime factorisations of 5 and 20 are as follows: 5 = 5
20 = 2 × 2 × 5 ∴ L.C.M. of 5 and 20
=2×2×5 = 20.

(b) 6, 18
Prime factorisations of 6 and 18 are as follows:
6 = 2×3 18 = 2×3×3
∴ L.C.M. of 6 and 18 = 2×3×3 = 18.

(c) 12, 48
Prime factorisations of 12 and 48 are as follows:
12 = 2 × 2 × 3
48 = 2×2×2×2×3
∴ L.C.M. of 12 and 48 = 2 × 2 × 2 × 2 × 3 = 48.

(d) 9, 45
Prime factorisations of 9 and 45 are as follows:
9 = 3×3
45 = 3 × 3 × 5
∴ L.C.M. of 9 and 45 = 3 × 3 × 5 = 45.
In the results obtained, we observe that L.C.M. of the two numbers in which one number is the factor of the other is the greater number.

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