Tag: NCERT Solutions for Class 6 Maths

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.5 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.5.

• Understanding Elementary Shapes Class 6 Ex 5.1
• Understanding Elementary Shapes Class 6 Ex 5.2
• Understanding Elementary Shapes Class 6 Ex 5.3
• Understanding Elementary Shapes Class 6 Ex 5.4
• Understanding Elementary Shapes Class 6 Ex 5.6
• Understanding Elementary Shapes Class 6 Ex 5.7
• Understanding Elementary Shapes Class 6 Ex 5.8
• Understanding Elementary Shapes Class 6 Ex 5.9

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 5
Chapter Name	Understanding Elementary Shapes
Exercise	Ex 5.5
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.5

Question 1.
Which of the following are models for perpendicular lines :
(a) The adjacent edges of a table top.
(b) The lines of a railway track.
(c) The line segments forming the letter “L”.
(d) The letter V.
Solution :
(a) and (c) are models for perpendicular lines.

Question 2.
Let \(\overline { PQ }\) be the perpendicular to the line segment \(\overline { XY }\) . Let \(\vec { PQ }\) and \(\overline { XY }\) intersect in the point A. What is the measure of ∠PAY ?
Solution :
The measure of ∠PAY is 90°.

Question 3.
There are two “set-squares ” in your box. What are the measures of the angles that are formed at their comers? Do they have any angle measure that is common?
Solution :
One is a 30° – 60° – 90° set square; the other is a 45° – 45° – 90° set square. The angle of measure 90° (i.e., a right angle) is common between them.

Question 4.
Study the diagram. The line l is perpendicular to line m.

(a) Is CE = EG?
(b) Does PE bisect CG?
(c) Identify any two line segments for which PE is the perpendicular bisector.
(d) Are these true?
(i) AC > EG
(ii) CD = GH
(iii) BC < EH.
Solution :
(a) CE= CD + DE
= 1 + 1=2 units
EG = EF + FG
= 1 + 1=2 units
∴ CE= EG

(b) ∴ CE = EG
∴ E is the mid-point of CG.
∴ Line PE bisects line segment CG

(c) ∴ DE = EF = 1 unit
∴ PE is the perpendicular bisector of line segment DF
∴ BE = EH = 3 units
∴ PE is the perpendicular bisector of BH

(d)
(i) true
(ii) true
(iii) true

 

We hope the NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.5 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.4 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.4.

• Understanding Elementary Shapes Class 6 Ex 5.1
• Understanding Elementary Shapes Class 6 Ex 5.2
• Understanding Elementary Shapes Class 6 Ex 5.3
• Understanding Elementary Shapes Class 6 Ex 5.5
• Understanding Elementary Shapes Class 6 Ex 5.6
• Understanding Elementary Shapes Class 6 Ex 5.7
• Understanding Elementary Shapes Class 6 Ex 5.8
• Understanding Elementary Shapes Class 6 Ex 5.9

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 5
Chapter Name	Understanding Elementary Shapes
Exercise	Ex 5.4
Number of Questions Solved	11
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.4

Question 1.
What is the measure of
(i) a right angle
(ii) a straight angle?
Solution :
(i) The measure of a right angle is 90°.
(ii) The measure of a straight angle is 180°.

Question 2.
Say True or False:
(a) The measure of an acute angle < 90°.
(b) The measure of an obtuse angle of < 90°.
(c) The measure of reflex angle > 180°.
(d) The measure of one complete revolution = 360°.
(e) If m∠A = 53° and m∠B = 35°, then m∠A > m∠B.
Solution :
(a) True
(b) False
(c) True
(d) True
(e) True.

Question 3.
Write down the measures of
(a) some acute angles.
(b) some obtuse angles. (give at least two examples of each).
Solution :
(a) 23°, 89°
(b) 91°, 179°.

Question 4.
Measure the angles given below, using the Protractor and write down the measure.

Solution :
(a) 45°
(b) 125°
(e) 90°
(d) ∠l = 40°, ∠2 = 125° and ∠3 = 95°.

Question 5.
Which angle has a large measure? First, estimate and then measure.

Measure of angle A =
Measure of angle B =
Solution :
Measure of angle A = 40°
Measure of angle B = 65°
The angle B has a larger measure.

Question 6.
From these two angles which have larger measure? Estimate and then confirmed by the measuring them.

Solution :
Measure of first angle = 45°
Measure of second angle = 60°
The second angle has larger measure.

Question 7.
Fill in the blanks with acute, obtuse, right or straight:
(a) An angle whose measure is less than that of a right angle is. ____________
(b) An angle whose measure is greater than that of a right angle is. ____________
(c) An angle whose measure is the sum of the measures of two right angles is. ____________
(d) When the sum of the measures of two angles is that of a right angle, then each one of them is. ____________
(e) When the sum of the measures of two angles is that of a straight angle and if one of them is acute than the other should be ____________.
Solution :
(a) An angle whose measure is less than that of a right angle is an acute angle.
(b) An angle whose measure is greater than that of a right angle is the obtuse angle (if the angle is less than 180°)
(c) An angle whose measure is the sum of the measures of two right angles is the straight angle.
(d) When the sum of the measures of two angles is that of a right angle, then each one of them is acute.
(e) When the sum of the measures of two angles is that of a straight angle and if one of them is acute than the other should be an obtuse angle.

Question 8.
Find the measure of the angle shown in each figure. (First, estimate with your eye, and then find the actual measurements with a protractor).
(a)

(b)

(c)

(d)

Solution :
Estimation with eye
(a) 45°
(b) 125°
(c) 60°
(d) 135° actual measure with a protractor
Solution :
(a) 40°
(b) 130°
(c) 65°
(d) 135°

Question 9.
Find the angle measure between the hands of the clock in each figure :

Solution :

• 90°
• 30°
• 180°.

Question 10.
Investigate :

In the given figure, the angle measures 30°. Look at the same figure through a magnifying glass. Does the angle become larger? Does the size of the angle change?
Solution :
No!; No!

Question 11.
Measure and classify each angle :

Angle	Measure	Type
∠AOC
∠AOB
∠BOC
∠DOC
∠DOA
∠DOB

Solution :
Angle ∠AOB ∠AOC ∠BOC ∠DOC ∠DOA ∠DOB Measure 40° 120° 80° 105° 140° 180° Type acute obtuse acute obtuse straight.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.4 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.3.

• Understanding Elementary Shapes Class 6 Ex 5.1
• Understanding Elementary Shapes Class 6 Ex 5.2
• Understanding Elementary Shapes Class 6 Ex 5.4
• Understanding Elementary Shapes Class 6 Ex 5.5
• Understanding Elementary Shapes Class 6 Ex 5.6
• Understanding Elementary Shapes Class 6 Ex 5.7
• Understanding Elementary Shapes Class 6 Ex 5.8
• Understanding Elementary Shapes Class 6 Ex 5.9

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 5
Chapter Name	Understanding Elementary Shapes
Exercise	Ex 5.3
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.3

Question 1.
Match the following :
(i) Straight angle (a) Less than one-fourth a revolution.
(ii) Right angle    (b) More than half a revolution
(iii) Acute angle   (c) Half of a revolution
(iv) Obtuse angle (d) One-fourth a revolution
(v) Reflex angle    (e) Between \(\frac { 1 }{ 4 }\) and \(\frac { 1 }{ 2 }\) of a revolution (f) One complete revolution.
Solution :
(i) Straight angle ↔ (c) Half of a revolution
(ii) Right angle ↔ (d) One-fourth a revolution
(iii) Acute angle ↔ (a) Less than one-fourth a revolution
(iv) Obtuse angle ↔ (e) Between 1/4 and 1/2 of a revolution
(v) Reflex angle ↔ (b) More than half a revolution.

Question 2.
Classify each one of the following angles as right, straight, acute, obtuse or reflex :

Solution :
(a) Acute
(b) Obtuse
(c) Right
(d) Reflex
(e) Straight
(f) Acute.

We hope the NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.2.

• Understanding Elementary Shapes Class 6 Ex 5.1
• Understanding Elementary Shapes Class 6 Ex 5.3
• Understanding Elementary Shapes Class 6 Ex 5.4
• Understanding Elementary Shapes Class 6 Ex 5.5
• Understanding Elementary Shapes Class 6 Ex 5.6
• Understanding Elementary Shapes Class 6 Ex 5.7
• Understanding Elementary Shapes Class 6 Ex 5.8
• Understanding Elementary Shapes Class 6 Ex 5.9

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 5
Chapter Name	Understanding Elementary Shapes
Exercise	Ex 5.2
Number of Questions Solved	7
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.2

Question 1.
What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from
(a) 3 to 9
(b) 4 to 7
(c) 7 to 10
(d) 12 to 9
(e) 1 to 10
(f) 6 to 3?
Solution :
(a) \(\frac { 1 }{ 2 }\) of a revolution
(b) \(\frac { 1 }{ 4 }\) of a revolution
(c) \(\frac { 1 }{ 4 }\) of a revolution
(d) \(\frac { 3 }{ 4 }\) of a revolution
(e) \(\frac { 3 }{ 4 }\) of a revolution
(f) \(\frac { 3 }{ 4 }\) of a revolution.

Question 2.
Where will the hand of a clock stop, if it
(a) starts at 12 and makes 1/2 of a revolution, clockwise?
(b) starts at 2 and makes 1/2 of a revolution, clockwise?
(c) starts at 5 and makes 1/4 of a revolution, clockwise?
(d) starts at 5 and makes 3/4 of a revolution, clockwise?
Solution :
(a) at 6
(b) at 8
(c) at 8
(d) at 2.

Question 3.
Which direction will you face if you start facing
(a) east and make 1/2 of a revolution clockwise?
(b) east and make 1/2 of a revolution clockwise?
(c) west and make 3/4 of a. revolution anti-clockwise?
(d) south and make one full revolution? (Should we specify clockwise or anti-clockwise for this last question? Why not?)
Solution :
(a) West
(b) West
(c) North
(d) South.
To answer (d), it is immaterial whether we turn clockwise or anti-clockwise because one full revolution will bring us back to the original position.

Question 4.
What part of a revolution have you turned through if you stand to face
(a) east and turn clockwise to face north?
(b) south and turn clockwise to face east?
(c) west and turn clockwise to face east?
Solution :
(a) \(\frac { 3 }{ 4 }\) of a revolution.
(b) \(\frac { 3 }{ 4 }\) of a revolution.
(c) \(\frac { 1 }{ 2 }\) of a revolution.

Question 5.
Find the number of right angles turned through by the hour hand of a clock when it goes from
(a) 3 to 6
(b) 2 to 8
(c) 5 to 11
(d) 10 to 1
(e) 12 to 9
(f) 12 to 9
Solution :
(a) 1 right angle
(b) 2 right angles
(c) 2 right angles
(d) 1 right angle
(e) 3 right angles
(f) 2 right angle

Question 6.
Now many right angles do you make if you start facing
(a) south and turn clockwise to the west?
(b) north and turn anti-clockwise to the east?
(c) west and turn to west?
(d) south and turn to the north?
Solution :
(a) 1 right angle
(b) 3 right angles
(c) 4 right angles
(d) 2 right angles
(clockwise or anticlockwise).

Question 7.
Where will the hour’s hand of a clock stop if it starts
(a) from 6 and turns through 1 right angle?
(b) from 8 and turns through 2 right angles?
(c) from 10 and turns through 3 right angles?
(d) from 7and turns through 2 straight angles?
Solution :
(a) 9
(b) 2
(c) 7
(d) 7.
(We should consider only clockwise direction here).

 

We hope the NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.2, 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

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 4
Chapter Name	Basic Geometrical Ideas
Exercise	Ex 4.6
Number of Questions Solved	4
Category	NCERT Solutions

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

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 4
Chapter Name	Basic Geometrical Ideas
Exercise	Ex 4.5
Number of Questions Solved	3
Category	NCERT Solutions

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

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 4
Chapter Name	Basic Geometrical Ideas
Exercise	Ex 4.4
Number of Questions Solved	2
Category	NCERT Solutions

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

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 4
Chapter Name	Basic Geometrical Ideas
Exercise	Ex 4.3
Number of Questions Solved	3
Category	NCERT Solutions

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

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 4
Chapter Name	Basic Geometrical Ideas
Exercise	Ex 4.2
Number of Questions Solved	5
Category	NCERT Solutions

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 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

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 3
Chapter Name	Playing With Numbers
Exercise	Ex 3.7
Number of Questions Solved	11
Category	NCERT Solutions

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.

We hope the NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.7 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.7, 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 3.6 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 3.6.

• 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 3.7

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 3
Chapter Name	Playing With Numbers
Exercise	Ex 3.6
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6

Question 1.
Find the H.C.F. of the following numbers,
(a) 18, 48
(b) 30, 42
(c) 18, 60
(d) 27, 63
(e) 36, 84
(f) 34, 102
(g) 70, 105, 175
(h) 91, 112, 49
(i) 18, 54, 81
(j) 12 45 75
Solution :
(a) 18,48
Factors of 18 are 1, 2, 3, 6, 9 and 18.
Factors of 48 are 1, 2, 3,4, 6, 8, 12, 16, 24 and
∴ Common factors of 18 and 48 are 1, 2, 3
Highest of these common factors is 6. ∴ H.C.F. of 18 and 48 is 6.

(b) 30,42
Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.
Factors of 42 are 1, 2, 3. 6. 7, 14, 21 and 42. ,
∴ Common factors of 30 and 42 are 1,2, 3 and 6.
Highest of these common factors is 6.
∴ H.C.F. of 30 and 42 is 6.

(c) 18,60
Factors of 18 are 1, 2, 3, 6, 9 and 18. Factors of 60 are 1,2, 3,4, 5,6,10 12,15, 20, 30 and 60.
∴ Common factors of 18 and 60 are 1, 2, 3 and 6.
Highest of these common factors is 6.
∴ H.C.F. of 18 and 60 is 6.

(d) 27,63
Factors of 27 are 1, 3, 9 and 27.
Factors of 63 are 1, 3, 7, 9, 21 and 63.
Common factors of 27 and 63 are 1, 3 and 9.
Highest of these common factors is 9.
∴ H.C.F. of 27 and 63 is 9.

(e) 36,84
Factors of 36 are 1, 2, 3,4, 6, 9,12, 18 and 36.
Factors of 84 are 1,2, 3,4,6,7, 12, 14,21, 28, 42 and 84.
Common factors of 36 and 84 are 1,2, 3,4, 6 and 12.
Highest of these common factors is 12.
∴ H.C.F. of 36 and 84 is 12. if) 34,102 • ‘
Factors of 34 are 1,2, 17 and 34.
Factors of 102 are 1, 2, 3,6,17, 34. 51 and 102.
∴ Common factors of 34 and 102 are 1, 2, 17 and 34.
Highest of these common factors is 34.
∴ H.C.F. of 34 and 102 is 34.

(g) 70,105,175
Factors of 70 are 1. 2, 5. 7, 10. 14, 35 and 70.
Factors of 105 are 1, 3, 5. 7. 15. 21. 35 and 105.
Factors of 175 are 1. 5, 7. 25. 35 and 175. .’. Common factors of 70, 105 and 175 are 1, 5 and 35.
Highest of these common factors is 35.
∴ H.C.F. of 70. 105 and 175 are 35.

(h) 91,112,49
Factors of 91 are 1,7, 13 and 91.
Factors of 112 are 1,2. 4. 7, 8, 14. 16. 28, 56 and 112.
Factors of 49 are 1.7 and 49.
Common factors of 91,112 and 49 are 1 and 7.
Highest of these common factors is 7.
∴ H.C.F. of 91, 112 and 49 is 7.

(i) 18,54,81
Factors of 18 are 1. 2, 3. 6, 9 and 18. Factors of 54 are 1, 2. 3, 6. 9, 18. 27 and 54.
Factors of 81 are 1. 3, 9, 27 and 81.
∴ Common factors of 18,54 and 81 are 1, 3 and 9.
Highest of these common factors is 9.
∴ H.C.F. of 18, 54 and 81 is 9.
(j) 12, 45, 75
Factors of 12 are 1, 2, 3, 4, 6 and 12. Factors of 45 are 1, 3, 5, 9, 15 and 45.
: Factors of 75 are 1, 3, 5, 15, 25 and 75.
∴ Common factors of 12,45 and 75 are 1 and 3.
Highest of these common factors is 3.
H.C.F. of 12. 45 and 75 is 3.

Question 2.
What is the H.C.F. of two consecutive :
(a) numbers?
(b) even numbers?
(c) odd numbers?
Solution :
(a) The H.C.F. of two consecutive numbers is 1.
(b) The H.C.F. of two consecutive even numbers is 2.
(c) The H.C.F. of two consecutive odd numbers is 1.

Question 3.
H. C.F. of co-prime numbers 4 and 15 was found as follows factorization: 4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so H.C.F. of 4 and 15 is 0. Is the answer correct? If not, what is the correct H.C.F?
Solution :
No, the answer is not correct. The correct answer is as follows :
H.C.F. of 4 and 15 is 1.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6, drop a comment below and we will get back to you at the earliest.