CBSE Sample Papers for Class 12 Physics Paper 1

CBSE Sample Papers for Class 12 Physics Paper 1 are part of CBSE Sample Papers for Class 12 Physics. Here we have given CBSE Sample Papers for Class 12 Physics Paper 1.

CBSE Sample Papers for Class 12 Physics Paper 1

Board CBSE
Class XII
Subject Physics
Sample Paper Set Paper 1
Category CBSE Sample Papers
Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 1 of Solved CBSE Sample Paper for Class 12 Physics is given below with free PDF download solutions.
  • Time Allowed : 3 Hours
    Max. Marks : 70
    General Instructions 
    • All questions are compulsory. There are 26 questions in all.
    • This question paper has five sections: Section A, Section B, Section C, Section D and Section E.
    • Section A contains five questions of 1 mark each. Section B contains five questions of 2 marks each. Section C contains twelve questions of 3 marks each. Section D contains one value based question of 4 marks and Section E contains three questions of 5 marks each.
    • There is no overall choice. However, an internal choice has been provided in 1 question of 2 marks, 1 question of 3 marks and all the 3 questions of 5 marks weightage. You have to attempt only 1 of the choices in such questions.
    • You may use the following values of physical constants wherever necessary :

    CBSE Sample Papers for Class 12 Physics Paper 1 image 1

CBSE Sample Papers for Class 12 Physics Paper 1 image 2

Questions :
SECTION : A

Question 1.
Define the term self-inductance of a coil. Write its S.I. unit

Question 2.
Which basic mode of communication is used for telephonic communication?

Question 3.
Why do the electrostatic field lines not form closed loops?

Question 4.
1-V graph for a metallic wire at two different temperatures, T1 and T2 is as shown in the figure. Which of the two temperature is lower and why?
CBSE Sample Papers for Class 12 Physics Paper 1 image 3

Question 5.
Why does bluish colour predominate in a clear sky?

SECTION : B

Question 6.
Use Kirchhoff’s rules to determine the potential difference between the points A and D when no current flows in the arm BE.
CBSE Sample Papers for Class 12 Physics Paper 1 image 4

Question 7.
When an electron in hydrogen atom jumps from the excited state to the ground state, how would the de-Broglie wavelength associated with the electron change? Justify your answer.

Question 8.
You are given two converging lenses of focal lengths 1.25 cm and 5 cm to design a compound microscope. If it is desired to have a magnification of 30, find out the separation between the objective and the eyepiece.

OR

A small telescope has an objective lens of focal length 150 cm and eyepiece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant objects in normal adjustment? If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?

Question 9.
Calculate the shortest wavelength in the Balmer series of hydrogen atom. In which region (infra-red, visible, ultraviolet) of hydrogen spectrum does this wavelength lie?

Question 10.
Write two factors which justify the need of modulation.

SECTION : C

Question 11.
Draw a circuit diagram of a C.E. transistor amplifier. Briefly explain its working and write the expression for
(i) current gain
(ii) voltage gain of the amplifier.

Question 12.
State clearly how an unpolarised light gets linearly polarized when passed through a Polaroid.
(i) Unpolarised light of intensity I0 is incident on a Polaroid P1 which is kept near another Polaroid P2 whose pass axis is parallel to that of P1. How will the intensities of light, I1 and I2, transmitted by the Polaroids P1 and P2 respectively, change on rotating P1 without disturbing P2?
(ii) Write the relation between the intensities  I1 and I2.

Question 13.
Define modulation index. Why is its value kept, in practice, less than one? A carrier wave of frequency 1.5 MHz and amplitude 50 V is modulated by a sinusoidal wave of frequency 10 kHz producing 50% amplitude modulation. Calculate the amplitude of the AM wave and frequencies of the side bands produced.

Question 14.
A uniform magnetic field B is set up along the positive X – axis. A particle of charge ‘q’ mass ‘m’ moving with a velocity v enters the field at the origin in X – Y plane such that it has velocity components both along and perpendicular to the magnetic field B. Trace, giving reason, the trajectory followed by the particle. Find out the expression for the distance moved by the particle along the magnetic field in one rotation.

Question 15.
Find the relation between drift velocity and relaxation time of charge carriers in a conductor. A conductor of length L is connected to a d.c. source of emf ‘E’. If the length of the conductor is tripled by stretching it, keeping ‘E’ constant, explain how its drift velocity would be affected.

Question 16.
(a) Determine the value of phase difference between the current and the voltage in the given series LCR circuit.
CBSE Sample Papers for Class 12 Physics Paper 1 image 5
(b) Calculate the value of the additional capacitor which may be joined suitably to the capacitor C that would make the power factor of the circuit unity.

Question 17.
Write the expression for the generalized form of Amperes Circuital law. Discuss its significance and describe briefly how the concept of displacement current is explained through charging/ discharging of a capacitor in an electric circuit. 

Question 18.
(a) Describe briefly three experimentally observed features in the phenomenon of photoelectric effect.
(b) Discuss briefly how wave theory of light cannot explain these features.

OR

(a) Write the important properties of photons which are used to establish Einstein’s photoelectric equation.
(b) Use this equation to explain the concept
(i) threshold frequency and
(ii) stopping potential.

Question 19.
Use Huygens principle to show how a plane wavelength propagates from a denser to rarer medium. Hence verify Snell’s law of refraction.

Question 20.
Identify the gates P and Q shown in the figure . Write the truth table for the combination of the gates shown.
CBSE Sample Papers for Class 12 Physics Paper 1 image 6
Name the equivalent gate representing this circuit and write its logic symbol. 

Question 21.
(a) Write three characteristic properties of nuclear force.
(b) Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from the graph.

Question 22.
Calculate the potential difference and the energy stored in the capacitor C2 in the circuit shown in the figure. Given potential at A is 90 V, C1 = 20 µF, C2 = 30 µF and C3 = 15 µF.
CBSE Sample Papers for Class 12 Physics Paper 1 image 7

SECTION : D

Question 23.
One morning an old lady tried bare-footed to replace the fuse wire fitted with power supply mains for her house. Suddenly she cried and collapsed on the floor. Her family members cried loudly for help. Her neighbour heard the cries and rushed to the spot wearing rubber soul shoes on. He took a wooden stick and used it to switch off the main supply.
Answer the folio wing questions :

  1. What are the voltage and frequency of mains supply in India?
  2. These days most of the electrical devices we use require a.c. voltage. Why?
  3.  Can a transformer be used to step up d.c. voltage ?
  4. Write two qualities displayed by neighbour by his action.

SECTION : E

Question 24.
(a) Define electric flux. Write its S.I. unit. “Gauss’s law in electrostatics is true for any closed surface, no matter what its shape or size is”. Justify this statement with the help of a suitable example.
(b) Use Gauss’s law to prove that electric field inside a uniformly charged spherical shell is zero.

OR

(a) Derive the expression for the energy stored in a parallel plate capacitor. Hence obtain the expression for the energy density of the electric field.
(b) A fully charged parallel plate capacitor is connected across an uncharged identical capacitor. Show that the energy stored in the combination is less than that stored initially in the single capacitor.

Question 25.
Explain, using a labelled diagram, the principle and working of a moving coil galvanometer. What is the function of
(i) uniform radial magnetic field,
(ii) soft iron core?
Define the terms :
(i) current sensitivity and
(ii) voltage sensitivity of a galvanometer. Why does increasing the current sensitivity not necessarily increase voltage sensitivity?

OR

CBSE Sample Papers for Class 12 Physics Paper 1 image 8

Question 26.
(a) Consider two coherent sources S1 and S2 producing monochromatic waves to produce interference pattern. Let the displacement of the wave produced by S1 be given by y1 = a cos mt and the displacement by S2 be y2 = a cos (ωt + Φ). Find out the expression for the amplitude of the resultant displacement at a point and show that the intensity at that point will be I = 4a2 cos2 Φ/2. Hence establish the conditions for constructive and destructive interference.
(b) What is the effect on the interference fringes in Young’s double slit experiment when
(i) the width of the source slit is increased,
(ii) the monochromatic source is replaced by a source of white light.

OR

CBSE Sample Papers for Class 12 Physics Paper 1 image 9
(a) A ray ‘PQ’ of light is incident on the face AB of a glass prism ABC (as shown in the figure) and emerges out of the face AC. Trace the path of the ray. Show that ∠i + ∠e = ∠A + ∠δ. Where δ and e denote the angle of deviation and angle of emergence respectively.
Plot a graph showing the variation of the angle of deviation as a function of angle of incidence. State the condition under which ∠δ is minimum.
(b) A ray of light passing through an equilateral triangular prism from air undergoes minimum deviation when angle of incidence is 3/4th of the angle of prism. Calculate the speed of light in the prism.

 Answers :
SECTION : A

Answer 1.
L = Φ / I . If I = 1 then L = Φ i.e self inductance L of a coil is numerically equal to the amount of magnetic flux (Φ) linked with the coil when unit current flows through the coil. The S.I. unit of self inductance is henry (H) or weber per ampere.
CBSE Sample Papers for Class 12 Physics Paper 1 image 10

Answer 2.
The basic mode of communication used in telephony is the point-to-point communications mode, which takes place through a link between a single transmitter and receiver.

Answer 3.
Electrostatic field lines originate from positive charge and terminate at negative charge.

Answer 4.
For the given figure the slope of a V – I graph is I/V = 1/R . Here the smaller the slope larger is the resistance . As the resistance of a metal increases with the increase in temperature, so resistance at T2 is more and T1 is lower.

Answer :
The light is scattered by air molecules. According to Lord Rayleigh the intensity of scattered light I ∝ 1/λ4. As λblue < λred Therefore blue colour scattered more in sky.

SECTION : B

Answer 6.
CBSE Sample Papers for Class 12 Physics Paper 1 image 11
According to Kirchhoff’s junction law at B
i3 = i1 + i2
i3 = i1
As  i3 = 0  (given)
Applying second law to loop AFEB, electronic network shown in the figure.

∴   i3 x 2 + i3 x 3 + i2 R1 = 1+3 + 6
∴   i3 = i1 = 2A

From A to D along AFD
∴  VAD = 2 i3 – 1 + 3 x i3
= (4 – 1 + 6)V
= 9V

Answer 7.
In an hydrogen atom, the negatively charged electron revolves in a circular orbit around the heavy positively charged nucleus. The centripetal force required by the electron is produced by the attractive force exerted by the nucleus on it. The energy is radiated, when an electron jumps from higher to lower energy orbit and the energy is absorbed, when jumps from lower to higher energy orbit.
CBSE Sample Papers for Class 12 Physics Paper 1 image 12
It is called Bohr’s frequency condition.

Answer 8.
Given, f0 = 5 cm , fe = 1.25 cm and m = 30
Let S be tube length (distance between the objective and the eyepiece).
CBSE Sample Papers for Class 12 Physics Paper 1 image 13
Thus, the height of the image would be 31.58 m.

Answer 9.
The formula for wavelength (λ) by Balmer series is :
CBSE Sample Papers for Class 12 Physics Paper 1 image 14
CBSE Sample Papers for Class 12 Physics Paper 1 image 15

Answer 10.
Two factors that justify the need for modulation of low frequency signals to high frequency signals are :

  • increase in range of communication base band signals suffer from attenuation and so cannot be transmitted over long distances, and
  • the antenna height and aperture is inversely proportional to the radiated signal frequency, implying that higher frequencies result in smaller antennas.

SECTION : C

Answer 11.
CBSE Sample Papers for Class 12 Physics Paper 1 image 16
The purpose of this circuit is to amplify a small a.c input signal, such as an audio or radio signal. A small a.c. voltage is applied to the input through a coupling capacitor. The ratio of the a.c. component of the output to the a.c. component of the input, is known as gain.
CBSE Sample Papers for Class 12 Physics Paper 1 image 17

Answer 12.
Polaroid filters are made of a special material that is capable of blocking of the two planes of vibration of an electromagnetic wave. In this sense, a Polaroid serves as a device that filter out one-half of the vibrations upon transmission of the light through the filter. When unpolarised light is transmitted through a Polaroid filter, it emerge with one-half the intensity and with vibrations in a single plane; it emerges as polarized light.
CBSE Sample Papers for Class 12 Physics Paper 1 image 18

Answer 13.
The modulation index p, can be defined as the ratio between the amplitude of the modulating signal and the amplitude of the carrier signal. A modulation index above 1 causes over-modulation, causing the carrier to experience 180° phase reversals. These reversals give rise to additional side and audio signal will be distorted during reception.
Given,
carrier wave frequency = 1.5 MHz, amplitude (V m) = 50 V
sinusoidal wave frequency = 10 kHz
amplitude modulation = 50%
CBSE Sample Papers for Class 12 Physics Paper 1 image 19
Thus, amplitude of the AM wave = 250 volt
Now,
USB = 1.5 x 1000+ 10 kHz
⇒ 1500 + 10 = 1510 kHz
LSB = 1.5 x 1000 -10 = 1490 kHz

Answer 14.
Field at an axial point of a circular coil.
CBSE Sample Papers for Class 12 Physics Paper 1 image 20
Where N is the number of turns of the circular coil. If M =iA is the magnetic moment of the coil, then
CBSE Sample Papers for Class 12 Physics Paper 1 image 21
∵ a = 0, For small loop,
CBSE Sample Papers for Class 12 Physics Paper 1 image 22
Where p is electric dipole  moment.

Answer 15.
The drift velocity of electrons can be formulated as :
CBSE Sample Papers for Class 12 Physics Paper 1 image 23

Answer 16.
CBSE Sample Papers for Class 12 Physics Paper 1 image 24
CBSE Sample Papers for Class 12 Physics Paper 1 image 25

Answer 17.
Generalized form of Ampere Circuital Law :
CBSE Sample Papers for Class 12 Physics Paper 1 image 26
It signifies that the source of magnetic field is not just due to the conduction of electric current due to flow of charge but also due to the time rate of change of electric field called displacement current.

Displacement current :
During charging and discharging of a capacitor the electric field between the plates will change so there will be a change of electric flux (displacement current) between the plates.

Answer 18.
(a)
(i) The photoelectric effect will not occur when the frequency of the incident light is less than the threshold frequency. Different materials have different threshold frequencies and most elements have threshold frequencies in the ultraviolet region of the electromagnetic spectrum.
(ii) The maximum KE of a stream of photo electrons increases linearly with the frequency of the incident light above the threshold frequency.
(iii) The rate at which photo electrons are emitted from a photosensitive surface is directly proportional to the intensity of incident light when the frequency is constant.

(b) Classical wave theory cannot explain :
(i)
The existence of threshold frequency because it predicts that electrons would absorb enough energy to escape and there would not be any threshold frequency.
(ii) The almost immediate emission of photo electrons as, according to this theory, electrons require a period of time before sufficient energy is absorbed by it to escape from the metal; however such a thing does not happen practically; and
(iii) The independence of KE of photo electrons on intensity and dependence on frequency because it cannot explain why maximum KE is dependent on frequency and independent of intensity.

OR

(i) The important property of photons that is useful in establishing Einstein’s photoelectric equation is their ability to hold on to the electrons of an atom by their forces of attraction.
(ii) Einstein’s photoelectric equation states that :
CBSE Sample Papers for Class 12 Physics Paper 1 image 27

Answer 19.
Medium 1
CBSE Sample Papers for Class 12 Physics Paper 1 image 28
CBSE Sample Papers for Class 12 Physics Paper 1 image 29

Answer 20.
Logic gate P is AND gate and logic Q is NOT gate.
CBSE Sample Papers for Class 12 Physics Paper 1 image 30
The combination is NAND
CBSE Sample Papers for Class 12 Physics Paper 1 image 31

Answer 21.
(a)
(i) Nuclear forces are short range forces.
(ii) Nuclear forces are primarily attractive and extremely strong.
(iii) Nuclear forces are charge independent
CBSE Sample Papers for Class 12 Physics Paper 1 image 32
(b) Nuclear forces is of attractive nature when separation between the nuclei is greater than 1 fm and of repulsive nature when separation is less than 1 fm.

Answer 22.
CBSE Sample Papers for Class 12 Physics Paper 1 image 33
CBSE Sample Papers for Class 12 Physics Paper 1 image 34

SECTION : D

Answer 23.

  1. The voltage and frequency of mains supply in India are 220 V and 50 Hz respectively.
  2. Most electrical devices require a.c. voltage because a.c. is available by default through the mains supply and also because d.c., is actually a one-way current, is available only through batteries.
  3. No, a transformer cannot be used to set-up d.c. voltage because a transformer works on the magnetic effect of current and d.c. voltage does not produces any magnetic field.
  4. The two qualities displayed by neighbour are :
    (a) preventing himself from being the earthing conductor by putting on his shoes and
    (b) switching off the mains connection to prevent the old lady from acting as an earthing conductor

SECTION : E

Answer 24.
(a) The electric flux through an area is defined as the electric field multiplied by the area of the surface projected on a plane, perpendicular to the field. Its S.I. unit is volt-metres (Vm) or Newton metres square per coulomb (Nm2 C-1 ). The given statement is justified because while measuring the flux, the surface area is more important than its volume or its size.
(b) To prove that the electric field inside a uniformly charged spherical shell is zero, we place a single positive point charge ‘q’ at the centre of an imaginary spherical surface with radius R. The field lines of this point radiate outside equally in all directions. The magnitude E of the electric field at every point on the surface is given by
CBSE Sample Papers for Class 12 Physics Paper 1 image 35

OR

(a) When charging parallel plate capacitor, whose plate area is A and distance between the plate is d. When charge on capacitor is q, potential difference increase 0 to V. Hence average
CBSE Sample Papers for Class 12 Physics Paper 1 image 36
(b) Let the capacitance of the charged capacitor C1 by and the voltage be ∆V1
CBSE Sample Papers for Class 12 Physics Paper 1 image 37

Answer 25.
The basic principle of a moving coil galvanometer is that when a current carrying coil is placed in a magnetic field, it experiences a torque.
CBSE Sample Papers for Class 12 Physics Paper 1 image 38
When the current I is passed through the coil, the torque experienced is given by
τ = NIAB sinθ
Where
N = no. of turns of the coil,
A =  area of the coil
B = magnetic field and
θ = angle between normal of coil and magnetic field
(i) The uniform radial magnetic field allows the coil to rotate and
(ii) The soft iron core makes the magnetic field linked with the coil to be radial.
The current sensitivity is defien as the deflection produced in the galvanometer, while passing a current of 1 ampere (1amp).
Thus,
CBSE Sample Papers for Class 12 Physics Paper 1 image 39
The voltage sensitivity is defined as the deflection produced in the galvanometer when a potential difference of IV is applied to the coil.
Thus,
 CBSE Sample Papers for Class 12 Physics Paper 1 image 40
Where, R is the resistance.
Increasing the current sensitivity does not necessarily increase the voltage sensitivity as there is an increase in the resistance as well.

OR

CBSE Sample Papers for Class 12 Physics Paper 1 image 41
The direction of dB is perpendicular to the plane formed by dl and r. It has an x-component dBx. When the components perpendicular to the x-axis are summed over, they cancel out and we obtain null result. Thus only the x-component survives.
CBSE Sample Papers for Class 12 Physics Paper 1 image 42
If the coil has N turns, then each turn will contribute equally to B. Then
CBSE Sample Papers for Class 12 Physics Paper 1 image 43
CBSE Sample Papers for Class 12 Physics Paper 1 image 44
Where μ0 in the permeability of free space. Ampere’s circuital law in electromagnetism is analogous to Gauss’ law in electrostatics.

Answer 26.
(a) Let S be a narrow slit illuminated by a monochromatic source of light, and Sj and S2 two similar parallel slits very close together and equidistant from S.
CBSE Sample Papers for Class 12 Physics Paper 1 image 45
Displacement of the wave produced by Sis given by
y1 = a cos ωt and
the displacement of the Wave produced by S2 is given by
y2 = a cos (ωt + Φ)
The resultant displacement is given by
y = y1 + y2 = a cos(ωt + Φ) + a cos (ωt + Φ)
⇒ y = 2a cos Φ/2 cos (ωt + Φ/2)
The amplitude of the resultant displacement is 2a cos Φ/2. The intensity of light is directly proportional to the square of amplitude of the wave.
The resultant intensity in given by
I = 4a cos2 Φ/2
CBSE Sample Papers for Class 12 Physics Paper 1 image 46
(b)  (i) As the width of the slits is increased, the fringe width decreases. It is because,
CBSE Sample Papers for Class 12 Physics Paper 1 image 47
(ii) The different colours of white light will produce different interference patterns but the central bright fringes due to all colours are at the same positions. Therefore, the central bright fringe is white in colour. Since the wavelength of the blue light is smallest, the fringe close to the either side of the central white fringe is blue and farthest is red.
OR
CBSE Sample Papers for Class 12 Physics Paper 1 image 48
(a) Let the incident ray meet refracting face AB of the prism at point P. Ray PQ is the refracted ray inside the prism and δ1 and r1, are the angle of the deviation and refraction at interface AB. At interface AC the ray goes out of the prism. Let e be the angle of emergence. The angle of deviation at point Q is δ2 as shown in figure. Using geometry, we see that at point P,
CBSE Sample Papers for Class 12 Physics Paper 1 image 49
CBSE Sample Papers for Class 12 Physics Paper 1 image 50
CBSE Sample Papers for Class 12 Physics Paper 1 image 51
So angle of deviation produced by a prism depends upon the angle of incidence, refracting angle of prism, and the material of the prism.
CBSE Sample Papers for Class 12 Physics Paper 1 image 53
CBSE Sample Papers for Class 12 Physics Paper 1 image 52

We hope the CBSE Sample Papers for Class 12 Physics Paper 1 help you. If you have any query regarding CBSE Sample Papers for Class 12 Physics Paper 1, drop a comment below and we will get back to you at the earliest.   

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 14
Chapter Name Practical Geometry
Exercise  Ex 14.1
Number of Questions Solved 5
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1

Question 1.
Draw a circle of radius 3.2 cm.
Solution :
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 1
Steps of Construction

  • Open the compasses for the required radius 3.2 cm, by putting the pointer on 0 and opening the pencil upto 3.2 cm.
  • Draw a point with a sharp pencil and marks it as O in the centre.
  • Place the pointer of the compasses where the centre has been marked.
  • Turn the compasses slowly to draw the circle.

Question 2.
With the same centre O, draw two circles of radii 4 cm and 2.5 cm.
Solution :
Steps of Construction
1. For circle of radius 4 cm

  • Open the compasses for the required radius 4 cm, by putting the pointer on 0 and opening the pencil upto 4 cm.
  • Place the pointer of the compasses at O.
  • Turn the compasses slowly to draw the circle.

2. For circle of radius 2.5 cm

  • Open the compasses for the required radius 2.5 cm. by putting the pointer on 0 and opening the pencil upto 2.5 cm.
  • Place the pointer of the compasses at O.
  • Turn the compasses slowly to draw the circle.

Question 3.
Draw a circle and any two of its diameters. If you join the ends of these diameters, what is the figure obtained? What figure is obtained if the diameters are perpendicular to each other? How do you check your answer?
Solution :
(i) On joining the ends of any two diameters of the circle, the figure obtained is a rectangle.
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 2
(ii) On joining the ends of any two diameters of the circle, perpendicular to each other, the figure obtained is a square.
To check the answer, we measured the sides and angles of the figure obtained.
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 3
Question 4.
Draw any circle and mark points A, B and C such that:
(a) A is on the circle.
(b) B is in the interior of the circle.
(c) C is in the exterior of the circle.
Solution :
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 4
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 5

Question 5.
Let A, B be the centres of two circles of equal radii; draw them so that each one of them passes We need a ruler here. through the centre of the other. Let them intersect at C and D. Examine whether \(\overline { AB }\) and \(\overline { CD }\) are at right angles.
Solution :
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 6
Yes! \(\overline { AB }\) and \(\overline { CD }\) are at right angles.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.1

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.1.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 7
Chapter Name Fractions
Exercise  Ex 7.1
Number of Questions Solved 11
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.1

Question 1.
Write the fraction representing the shaded portion.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 1
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 2
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 3
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 4

Question 2.
Color the part according to the given fraction.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 5
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 6

Question 3.
Identify the error, if any
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 7
Solution :
Neither of the shaded portions represents the corresponding given fractions. 4.

Question 4.
What fraction of a day is 8 hours?
Solution :
1 day = 24 hours
∴ Required fraction = \(\frac { 8 }{ 24 }\)

Question 5.
What fraction of an hour is 40 minutes?
Solution :
1 hour = 60 minutes
∴ Required fraction = \(\frac { 40 }{ 60 }\)

Question 6.
Arya, Abhimanyu, and Vivek shared lunch. Arya brings two sandwiches, one made of vegetable and one of jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that each person will have an equal share of each sandwich.
(a) How can Arya divide his sandwiches so that each person has an equal share?
(b) What part of a sandwich will each boy receive?
Solution :
(a) Arya will divide each sandwich into three equal parts, and give one part of each sandwich to each one of them.
(b) Each boy will receive \(\frac { 1 }{ 3 }\) part of a sandwich.

Question 7.
Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?
Solution :
She has finished \(\frac { 2 }{ 3 }\) fraction of the dresses.

Question 8.
Write the natural numbers from 2 to 12. What fraction of them are prime numbers?
Solution :
The natural numbers from 2 to 12 are 2, 3, 4,5,6, 7, 8, 9, 10, 11 and 12
Total number of natural numbers = 11
Out of these, the prime numbers are 2, 3, 5, 7, 11

Total number of these prime numbers = 5 5
∴ Required fraction = \(\frac { 5 }{ 11 }\).

Question 9.
Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Solution :
The natural numbers from 102 to 113 are
102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112 and 113
Total number of natural numbers =12
Out of these, the prime numbers are 103, 107, 109, 113.
Total number of these prime numbers = 4 . 4
∴ Required fraction = \(\frac { 4 }{ 12 }\).

Question 10.
What fraction of these circles have X’s in them?
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 8
Solution :
Total number of circles = 8
Number of circles which have X’s in them = 4
∴ Required fraction = \(\frac { 4 }{ 8 }\).

Question 11.
Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?
Solution :
Number of CDs bought = 3
Number of CDs received as gifts = 5
∴ Total number of CDs = 3 + 5 = 8
Fraction of her total CDs that she bought \(\frac { 3 }{ 8 }\) and, fraction of her total CDs that she received as gifts \(\frac { 5 }{ 8 }\).

 

We hope the NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.1 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 8
Chapter Name  Decimals
Exercise  Ex 8.1
Number of Questions Solved 10
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1

Question 1.
Write the following as numbers in the given table:
(a)
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 1
(b)
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 2

Hundreds
(100)
Tens
(10)
Ones
(1)

Tenths
(\(\frac { 1 }{ 10 } \))

Solution.

Hundreds
(100)
Tens
(10)
Ones
(1)
Tenths
(\(\frac { 1 }{ 10 } \))
(a) 0 3 1 2
(b)  1 1 0 4

Question 2.
Write the following decimals in the place value table :
(a)
19.4
(b)
0.3
(c)
10.6
(d)
205.9
Solution.

Hundreds Tens
Ones Tenths
(a) 0 1 9 4
(b)  0 0 0 3
(c)  0 1 0 6
(d)  2 0 5 9

Question 3.
Write each of the following as decimals:
(a)
Seven-tenths
(b) Two tens and nine-tenths
(c) Fourteen point six
(d) One hundred and two ones
(e) Six hundred point eight
Solution.
(a) 7 tenths : \(\frac { 7 }{ 10 } \) =0.7.
(b) Two tens and nine-tenths = 20 + \(\frac { 9 }{ 10 } \)  =20.9
(c) Fourteen point six = 14.6
(d) One hundred and two-ones =100 + 2=102
(e) Six hundred point eight = 600.8

Question 4.
Write each of the following as decimals:
(a) 
\(\frac { 5 }{ 10 } \)
(b) 3 +\(\frac { 7 }{ 10 } \)
(c) 200+60+5+\(\frac { 1 }{ 10 } \)
(d) 70+\(\frac { 8 }{ 10 } \)
(e) \(\frac { 2 }{ 10 } \)
(g) \(\frac { 3 }{ 2 } \)
(h) \(\frac { 2 }{ 5 } \)
(i) \(\frac { 12 }{ 5 } \)
(j) \(\frac { 3 }{ 5 } \)
(k) 4\(\frac { 1 }{ 2 } \)
Solution.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 3
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 4

Question 5.
Write the following decimals as fractions. Reduce the fractions to lowest form :
(a) 0.6
(b) 205
(c) 1.0
(d) 3.8
(e) 13.7
(f) 21.2
(g) 6.4
Solution.
(a)
 \(0.6=\frac { 6 }{ 10 } =\frac { 6\div 2 }{ 10\div 5 } \)
∵  H.C.F.(6,10) = 2
= \(\frac { 3 }{ 5 } \)

(b)
\(2.5=\frac { 25 }{ 10 } =\frac { 25\div 5 }{ 10\div 5 } \)
∵  H.C.F.(25,10) = 5
= \(\frac { 5 }{ 2 } \)

(c)
\(1.5=\frac { 10 }{ 10 } =\frac { 10\div 10 }{ 10\div 10 } \)
∵  H.C.F.(10,10) = 10
= \(\frac { 1 }{ 1 } \) =1

(d)
\(3.8 =\frac { 38 }{ 10 } =\frac { 38\div 2 }{ 10\div 2 } \)
∵  H.C.F.(38,10) = 2
= \(\frac { 19 }{ 5 } \)

(e)
\(13.7=\frac { 137 }{ 10 } \)

(f)
\( 21.2=\frac { 212 }{ 10 } =\frac { 212\div 2 }{ 10\div 5 } \)
∵  H.C.F.(212,10) = 5
= \(\frac { 106 }{ 5 } \)

(g)
\(6.4=\frac { 64 }{ 10 } =\frac { 64\div 2 }{ 10\div 5 } \)
∵  H.C.F.(64,10) = 2
= \(\frac { 32 }{ 5 } \)

Question 6.
Express the following as cm using decimals:
(a)
2 mm
(b)
30 mm
(c)
116 mm
(d)
4cm 2mm
(e)
11cm 52mm
(f)
83 mm
Solution.
(a)
\(2mm =\frac { 2 }{ 10 } \)
∵  \( 1mm =\frac { 1 }{ 10 } \)
= 0.2cm

(b)
\(30mm =\frac { 30 }{ 10 } \)
∵  \( 1mm =\frac { 1 }{ 10 } \)
= 3cm = 3.0cm
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 5
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 6
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 7

Question 7.
Between which two whole numbers on the number line are the given numbers lie? Which of these whole numbers is nearer the number?
(a) 0.8
(b) 5.1
(c) 2.6
(d) 6.4
(e) 9.0
(f) 4.9
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 8
Solution.
(a) 0.8. The number 0.8 is between the two whole numbers 0 and 1. The whole number 1 is nearer the number 0.8.

(b) 5.1. The number 5.1 is between the two whole numbers 5 and 6. The whole number 5 is nearer the number 5.1.

(c) 2.6. The number 2,6 is between the two whole numbers 2 and 3. The whole number 3 is nearer the number 2.6.

(d) 6.4 The number 6.4 is between the two whole numbers 6 and 7. The whole number 6 is nearer the number 6.4.

(e) 0. 9.0 itself is a whole number.

(f) 4.9. The number 4.9 is between the two whole numbers 4 and 5. The whole number 5 is nearer the number 4.9.

Question 8.
Show the following numbers on the number line:
(a)
2
(b) 1.9
(c) 1
(d) 2.5.
Solution.
(a) 0.2. We know that 0.2 is more than zero but less than one. There are 2-tenths in it. Divide the unit length between 0 and 1 into 10 equal parts and take 2 parts as shown below:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 9

(b) 1.9. We know that 1.9 is more than one but less than two. There are 9-tenths in’it. Divide the unit length between 1 and 2 into 10 equal parts and take 9 parts as shown below:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 10

(c) 1.1. We know that 1.1 is more than one but less than two. There is one-tenth in it. Divide the unit length between 1 and 2 into 10 equal parts and take 1 part as shown below:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 11

(d) 2.5. We know that 2.5 is more than two but less than three. There are 5-tenths in it. Divide the unit length between 2 and 3 into 10 equal parts and take 5 parts as shown below :
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 12

Question 9.
Write the decimal number represented by the points A, B, C, D on the given number line :
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 13
Solution.
(i) A. The decimal number represented by the point A is 0.8 as the unit length between 0 and 1 has been divided into 10 equal parts and 8 parts have been taken.

(ii) B. The decimal number represented by the point B is 1.3 as the unit length between 1 and 2 has been divided into 10 equal parts and 3 parts have been taken.

(iii) C. The decimal number represented by the point C is 2.2 as the unit length between 2 and 3 has been divided into 10 equal parts and 2 parts have been taken.

(iv) D. The decimal number represented by the point D is 2.9 as the unit length between 2 and 3 has been divided into 10 equal parts and 9 parts have been taken.

Question 10.
(a) The length of Ramesh’s notebook is 9 cm 5 mm. What will be its length in cm?
(b) The length of a young gram plant is 65 mm. Express its length in cm.
Solution.
(a) Length of Ramesh’s notebook = 9 cm 5mm
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 14

NCERT Solutions for Class 6 Maths Chapter 8 Decimals 15
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 16

 

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NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.1

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.1.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 13
Chapter Name Symmetry
Exercise  Ex 13.1
Number of Questions Solved 6
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.1

Question 1.
List any four symmetrical objects from your home or school.
Solution :
The blackboard, the table top, a pair of scissors, the computer disc.

Question 2.
For the given figure, which one is the mirror line, l1 or l2?
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 1
Solution :
I2 is the mirror line.

Question 3.
Identify the shapes given below. Check whether they are symmetric or not. Draw the line of symmetry as well.
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 2
Solution :
(a) Symmetric
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 3
(b) Symmetric
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 4
(c) Not symmetric
(d) Symmetric
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 5
(e) Symmetric
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 6
(f) Symmetric
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 7

Question 4.
Copy the following on a squared paper. A square paper is what you would have used in your arithmetic notebook in earlier classes. Then complete them such that the dotted line is the line of symmetry.
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 8
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 9
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 10
Solution :
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 11
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 12
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 13

Question 5.
In the figure, l is the line of symmetry. Complete the diagram to make it symmetric.
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 14
Solution :
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 15
Question 6.
In the figure, l is the line of symmetry. Draw the image of the triangle and complete the diagram so that it becomes symmetric.
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 16
Solution :
NCERT Solutions for Class 6 Maths Chapter 13 Symmetry 17

 

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NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1 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 5.1.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 5
Chapter Name Understanding Elementary Shapes
Exercise  Ex 5.1
Number of Questions Solved 7
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1

Question 1.
What is the disadvantage in comparing line segments by mere observation?
Solution :
Sometimes the difference in lengths betweenthe two line segments is not obvious. So, we are not always sure about our usual judgment.

Question 2.
Why is it better to use a divider than with a ruler, while measuring the length of a line segment?
Solution :
There may be errors due to the thickness of the ruler and angular viewing by using a ruler. These errors are eradicated by using a divider. So, it is better to use a divider, than a ruler, while measuring the length of a line segment.

Question 3.
Draw any line segment, say \(\overline { AB }\). Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AB = AC + CB? [Note : If A, B, C are any three points on a line such that AC + CB = AB, then we can be sure that C lies between A and B.]
Solution :
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 1
Length of AB = 7 cm
Length of BC = 3 cm
Length of AC = 4 cm
AC + CB = 4 cm + 3 cm = 7 cm
But AB = 7 cm
So, AB = AC A- CB.

Question 4.
If A, B, C are three points on a line such that AB = 5 cm, BC – 3 cm and AC – 8 cm, which one of them lies between the other two?
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 2
Solution :
AB + BC = AC, so, the point B lies between the point A and point C.

Question 5.
Verify, whether D is the mid-point of AG.
Solution :
AD = AB + BC + CD = 3 units
DG = OE + EF + FG = 3 units
∴ Yes ! D is the mid-point of AG.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 3
Question 6.
If B is the mid-point of \(\overline { AC }\) and C is the mid-point of \(\overline { BD }\), where A, B, C, D lie on a straight line, say why AB = CD ?
Solution :
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 4
∴ B is the mid-point of \(\overline { AC }\)
∴ AB = BC …(1)
∴ C is the mid-point of \(\overline { BD }\)
∴ BC = CD … (2)
In view of (1) and (2), we get AB = CD.

Question 7.
Draw five triangles and measure their sides. Check in each case, if the sum of the lengths of any two sides is always less than the third side.
Solution :
(i) AB = 3.7 cm
BC = 3 cm
AC = 3.8 cm
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 5
Clearly, AB + BC > AC
BC + AC > AB
AC + AB > BC

(ii) AB = 3 cm
BC = 3 cm
CA = 3 cm
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 6
Clearly, AB + BC > AC
BC + AO AB
AC + AB > B C

(iii) AB = 4 cm
BC = 3 cm
AC = 5 cm
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 7
Clearly, AB + BC > AC
BC+ AC > AB
AC + AB > BC.

(iv) AB = 2 cm
BC = 2 cm
AC = 2.8 cm
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 8
Clearly, AB + BC > AC
BC + AC > AB
AC +AB> BC

(v) AB = 3 cm
BC = 4 cm
CA = 3 cm
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 9
Clearly, AB + BC > AC
BC + AC> AB
AC + AB> BC
In each case, we observe that the sum of the lengths of any two sides is always greater than the third side.

 

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NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.1

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.1 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 3.1.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 3
Chapter Name Playing With Numbers
Exercise  Ex 3.1
Number of Questions Solved 4
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.1

Question 1.
Write all the factors of the following numbers:
(i) 24
(ii) 15
(iii) 21
(iv) 27
(v) 12
(vi) 20
(vii) 18
(viii) 23
(ix) 36.
Solution :
(i) 24
24 = 1 × 24
24 = 2 × 12
24 = 3 × 8
24 = 4 × 6
24 = 6 × 4
Stop here, because 4 and 6 have occurred earlier. Thus, all the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.

(ii) 15
15 = 1 × 15
15 = 3 × 5
15 = 5 × 3
Stop here, because 3 and 5 have occurred earlier. Thus, all the factors of 15 are 1, 3, 5 and 15.

(iii) 21
21 = 1 × 21
21 = 3 × 7
21 = 7 × 3
Stop here, because 3 and 7 have occurred earlier. Thus, all the factors of 21 are 1, 3, 7 and 21.

(iv) 27
27 = 1 × 27
27 = 3 × 9
27 = 9 × 3
Stop here, because 3 and 9 have occurred earlier. Thus, all the factors of 27 are 1, 3, 9 and 27.

(v) 12
12 = 1 × 12
12 = 2 × 6
12 = 3 × 4
12 = 4 × 3
Stop here, because 3 and 4 have occurred earlier. Thus, all the factors of 12 are 1,2,3,4,6 and 12.

(vi) 20
20 = 1 × 20
20 = 2 × 10
20 = 4 × 5
20 = 5 × 4
Stop here, because 4 and 5 have occurred earlier. Thus, all the factors of 20 are 1, 2, 4, 5, 10 and 20.

(Vii) 18
18 = 1 × 18
18 = 2 × 9
18 = 3 × 6
18 = 6 × 3
Stop here, because 3 and 6 have occurred earlier. Thus, all the factors of 18 are 1,2,3,6,9 and 18.

(viii) 23
23 = 1 × 23
Thus, all the factors of 23 are 1 and 23.

(ix) 36
36 = 1 × 36
36 = 2 × 18
36 = 3 × 12
36 = 4 × 9
36 = 6 × 6
36 = 6 × 6
Stop here, because both the factors 6 and 6 have occurred earlier. Thus, all the factors of 36 are 1,2,3,4,6,9,12, 18 and 36.

Question 2.
Write the first five multiples of:
(i) 5
(ii) 8
(iii) 9.
Solution :
(i) 5
First, five multiples of 5 are 5 × 1, 5 × 2, 5 × 3, 5 × 4 and 5 × 5
i. e.,5, 10, 15, 20 and 25.

(ii) 8
First, five multiples of 8 are 8×1,8×2,8×3,8 × 4 and 8×5
i.e., 8, 16, 24, 32 and 40.

(iii) 9
First, five multiples of 9 are 9 × 1,9 × 2,9 × 3,9 × 4 and 9×5
i.e., 9, 18,27, 36 and 45.

Question 3.
Match the items in column 1 with the items in column 2 :
Solution :

Question 4.
Find all the multiples of 9 up to 100.
Solution :
The multiples of 9 are
9 × 1, 9 × 2, 9 × 3. 9 × 4, 9 × 5, 9 × 6, 9 × 7, 9 × 8, 9 × 9, 9 × 10, 9 × 11, 9 × 12……….
i.e.. 9, 18, 27, 36, 45, 54. 63, 72, 81, 90, 99, 108,
Thus, all the multiples of 9 up to 100 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.

 

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NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 10
Chapter Name Mensuration
Exercise  Ex 10.1
Number of Questions Solved 17
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1

Question 1.
Find the perimeter of each of the following figures:
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 1
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 2
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 3
Solution :
(a) Perimeter
= 5 cm + I cm + 2 cm + 4 cm
= 12 cm
(b) Perimeter
= 40 cm + 35 cm + 23 cm + 35 cm
= 133 cm
(c) Perimeter
= 15 cm + 15 cm + 15 cm+ 15 cm
= 60 cm
(d) Perimeter
= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm
= 20 cm
(e) Perimeter
= 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm
= 15 cm
(f) Perimeter = 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm
= 52 cm

Question 2.
The lid of a rectangular box of sides 40 cm by 10 cm is sealed all around with tape. What is the length of the tape required?
Solution :
Length of the tape required
= Perimeter of the rectangular box = 2 × (Length + Breadth)
= 2 × (40 cm + 10 cm)
= 2 × (50 cm)
= 100 cm = l m.

Question 3.
A table-top measures 2 m 25 cm by l m 50 cm. What is the perimeter of the table-top?
Solution :
Perimeter of the table-top
= 2 × (Length + Breadth)
= 2 × (2 m 25 cm + 1 m 50 cm)
= 2 × (2.25 m+1.50 m)
= 2 × (3.75 m)
= 7.5 m

Question 4.
What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Solution :
Length of the wooden strip required = 2 × (Length + Breadth)
= 2 × (32 cm + 21 cm)
= 2 × (53 cm)
= 106 cm = 1.06 m.

Question 5.
A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Solution :
Perimeter of the rectangle
= 2 × (Length + Breadth )
= 2 × (0.7 km + 0.5 km)
= 2 × (1.2km)
= 2.4 km
Length of the wire needed
= 4 × Perimeter of the rectangle = 4 × (2.4 km)
= 9.6 km.

Question 6.
Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm, and 5 cm
(b) An equilateral triangle of side 9 cm
(c) An isosceles triangle with equal sides 8 cm each a mi third side 6 cm.
Solution :
(a) Perimeter of the triangle = 3 cm + 4 cm + 5 cm = 12 cm
(b) Perimeter of the equilateral triangle = 3 × Length of a side = 3 × (9 cm) = 27 cm
(c) Perimeter of the isosceles triangle = 8 cm + 8 cm + 6 cm = 22 cm.

Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm, and 15 cm.
Solution :
Perimeter of the triangle
= Sum of the lengths of its three sides
= 10 cm + 14 cm + 15 cm
= 39 cm.

Question 8.
Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution :
Perimeter of the regular hexagon
= 6 × Length of a side
= 6 × (8m)
= 48 m.

Question 9.
Find the side of the square whose perimeter is 20 m.
Solution :
Perimeter of the square
= 4 × Length of a side
⇒ Length of one side
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 4

Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its every side?
Solution :
Perimeter of the regular pentagon = 5 × Length of a side
⇒ Length of one (each) side
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 5

Question 11.
A piece of string is 30 cm long. What will be the length of each side if the string is used to form:
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
Solution :
(a) Perimeter of the square = 4 × Length of a side
⇒ Length of a side
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 6

(b) Perimeter of the equilateral triangle
= 3 × Length of a side
⇒ Length of a side
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 7

(c)
Perimeter of the regular hexagon = 6 × Length of a side.
⇒ Length of a side
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 8

Question 12.
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is the third side?
Solution :
Perimeter of a triangle = Sum of the lengths of its three sides
⇒ 36 cm = 12 cm + 14 cm + Length of the third side
⇒ 36 cm = 26 cm + Length of the third side
⇒ Length of the third side = 36 cm – 26 cm = 10 cm.

Question 13.
Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per meter.
Solution :
Perimeter of the square park = 4 × Length of a side = 4 × (250m)
= 1000 m
∴ Cost of fencing the square park at the rate of?
20 per metre = ₹ 1000 × 20
= ₹ 20,000

Question 14.
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of? 12 per meter.
Solution :
Perimeter of the rectangular park = 2 × (Length + Breadth)
= 2 × (175m + 125 m)
= 2 × (300 m)
= 600 m
Cost of fencing the rectangular park at the rate of ?
12 per metre = ₹ 600 × 12
= ₹ 7200.

Question 15.
Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
Solution :
Perimeter of the square park = 4 × Length of a side = 4 × (75 m)
= 300 m
Perimeter of the rectangular park
= 2 × (Length + Breadth)
= 2 × (60 m + 45 m)
= 2 × (105 m)
= 210 m.
Since, the perimeter of the rectangular park is less than the perimeter of the square park, therefore. Bulbul covers less distance.

Question 16.
What is the perimeter of each of the following figures? What do you infer from the answers?
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 9
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 10
Solution :
(a) Perimeter
= Sum of the lengths of all the sides
= 25 cm + 25 cm + 25 cm + 25 cm
= 100 cm.

(b) Perimeter
= Sum of the lengths of all the sides
= 40 cm + 10 cm + 40 cm + 10 cm
= 100 cm.

(c) Perimeter
= Sum of the lengths of all the sides
= 30 cm + 20 cm + 30 cm + 20 cm
= 100 cm.

(d) Perimeter
= Sum of the lengths of all the sides
= 30 cm + 40 cm + 30 cm
= 100 cm.
The inference from the answers. All the figures have the same perimeter.

Question 17.
Avneet buys 9 square paving slabs, each with a side of \(\frac { 1 }{ 2 }\) m. He lays them in the form of a square.
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 11
(a) What is the perimeter of his arrangement (Fig. i)?
(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement (Fig. ii)?
(c) Which has a greater perimeter?
(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be broken.)
Solution :
(a) Perimeter of his arrangement = 4 × Length of one side
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 12
(b) Perimeter of her arrangement = Sum of the lengths of all the sides
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 13
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 14
(c) Cross has greater perimeter.
(d) Yes ! there is a way of getting an even greater perimeter. It is shown below:
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 15

 

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NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 9
Chapter Name Data Handling
Exercise  Ex 9.1
Number of Questions Solved 7
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1

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

8 I 3 7 6 5 5 4 4 2
4 9 5 3 7 I 6 5 2 7
7 3 8 4 2 8 9 5 8 6
7 4 5 6 9 6 4 4 6 6

(a) Find how many students obtained marks equal to or more than 7.
(b) How many students obtained marks below 4?
Solution.
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 1
(a) 5 + 4 + 3=12 students obtained marks equal to or more than 7.
(b) 3 + 3 + 2 = 8 students obtained marks below 4.

Question 2.
Following is the choice of sweets of 30 students of Class VI
Ladoo, Barfi, Ladoo, Jalebi, Ladoo, Rasgulla, Jalebi, Ladoo, Barfi, Rasgulla, Ladoo, Jalebi Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Ladoo, Ladoo, Barfi, Rasgulla, Rasgulla, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo.
(a) Arrange the names of sweets in a table using tally marks.
(b) Which sweet is preferred by most of the students?
Solution.
(a)
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 2

(b) Ladoo is preferred by most of the students.

1 3 5 6 6 3 5 4 1 6
2 5 3 4 6 1 5 5 6 1
1 2 2 3 5 2 4 5 5 6
5 1 6 2 3 5 2 4 1 5

Question 3.
Make a table and enter the data using tally- marks. Find the number that appeared.
(a) The minimum number of times.
(b) The maximum number of times.
(c) Find those numbers that appear an equal number of limes.
Solution.
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 3
(a) The number that appeared the minimum number of times is 4.
(b) The number that appeared the maximum number of times is 5.
(c) The numbers that appeared an equal number of times are 1 and 6.

Question 4.
Following pictograph shows the number of tractors in five villages:
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 4
Observe the pictograph and answers the following questions:
(i)
Which village has the minimum number of tractors?
(ii) Which village has the maximum number of tractors?
(iii) How many more tractors village C has as compared to village B?
(iv) What is the total number of tractors in all the five villages?
Solution.
(i) Village D has the minimum number of tractors.
(ii) Village C has the maximum number of tractors.
(iii) Village C has 8-5 = 3 more tractors as compared to village B.
(iv) Total number of tractors in all the five villages = 6 +5+ 8 + 3 + 6 = 28.

Question 5.
The number of girl students in each class of a co-educational middle school is depicted by the pictograph:
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 5
Observe this pictograph and answer the following questions:
(a) Which class has the minimum number of girl students?
(b) Is the number of girls in class VI less than the number of girls in class V?
(c) How many girls are there in VII class?
Solution.
(a) Class VIII has the minimum no. of girl students.
(b) No! the number of girls in class VI is not less than the number of girls in class V.
(c) Number of girls in class VII – 3 x 4 = 12.

Question 6.
The sale of electric bulbs on different days of a week is shown below:
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 6
What can we conclude from the said pictograph?
Observe the pictograph and answer the following questions:
(a) How many bulbs were sold on Friday?
(b) On which day the maximum number of bulbs were sold?
(c) If one bulb was sold at the rate off 10, what was the total sale on Sunday?
(d) Can you find out the total sale of the week?
(e) If one big carton can hold 9 bulbs, how many cartons were needed in the given week?
Solution.
(a) Number of bulbs sold on Friday = 7×2 = 14.
(b) The maximum number of bulbs were sold on Sunday.
(c) Number of bulbs sold on Sunday
= 9 x 2=18.
∴ Total sale on Sunday
= Rs.18 x 10 = Rs. 180.
(d) Total number of bulbs sold in the week
= (6 + 8 + 4 + 5 + 7 + 4 + 9) x 2
= 43 x 2 = 86.
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 7
Hence, 10 cartons were needed in the given week.

Question 7.
In a village six fruit merchants sold the following number of fruit baskets in a particular season:
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 8
Observe this pictograph and answer the following questions:
(a) Which merchant sold the maximum number of baskets?
(b) How many fruit baskets were sold by Anwar?
(c) The merchants who have sold 600 or number of baskets are planning to buy a godown for the next season. Can you name them?
Solution.
(a) Martin sold the maximum number of baskets.
(b) 7 x 100 = 700 fruit baskets were sold by Anwar.
(c) Yes! Anwar. Martin and Ranjit Singh are planning to buy a godown for the next season.

 

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NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 6
Chapter Name Integers
Exercise  Ex 6.1
Number of Questions Solved 10
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1

Question 1.
Write opposite of the following:
(a) Increase in weight
(b) 30 km north
(c) 326 BC
(d) Loss of Rs. 700
(e) 100 m above sea level.
Solution.
(a) Decrease in weight
(b) 30 km south
(c) 326 AD
(d) Gain of Rs. 700
(e) 100 m below sea level

Question 2.
Represent the following numbers as integers with appropriate signs.
(a) An airplane is flying at a height two thousand meters above the ground.
(b) A submarine is moving at depth, eight hundred meters below the sea level.
(c) A deposit of rupees two hundred.
(d) Withdrawal of rupees seven hundred.
Solution.
(a) + 2000
(b) – 800
(c) + 200
(d) – 700.

Question 3.
Represent the following numbers on a number line:

(a) + 5
(b) -10
(c) + 8
(d) -7
Solution.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 1

Question 4.
Adjacent figure is a vertical number line, representing integers. Observe it and locate the following points :
(a) If point D is + 8, then which point is – 8?
(b) Is point G a negative integer or a positive integer?
(c) Write integers for points B and E.
(d) Which point marked on this number line has the least value?
(e) Arrange all the points in p-decreasing order of value.
Solution.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 2
(a) F
(b) negative integer
(c) B → + 4, E → 10
(d) E
(e) D, C, B, A, O, H, G, F, E.

Question 5.
Following is the list of temperatures of five places in India, on a particular day of the year.
Solution.

Place Temperature
Siachin 10°C below 0°C ………….
Shimla 2°C below 0°C …………
Ahmedabad 30°C above 0°C ………….
Delhi 20°C above 0°C ………..
Srinagar 5°C below 0°C ………….

NCERT Solutions for Class 6 Maths Chapter 6 Integers 3
(a)
Write the temperature of these places in the form of integers in the blank column.
(b) Following is the number line representing the temperature in degree Celsius.
Plot the name of the city against its temperature.
(c) Which is the coolest place?
(d) Write the names of the places whose temperature are above 10°C.
Solution.
(a)

Place Temperature
Siachin 10°C below 0°C – 10°C
Shimla 2°C below 0°C – 2°C
Ahmedabad 30°C above 0°C + 30°C
Delhi 20°C above 0°C + 20°C
Srinagar 5°C below 0°C – 5°C

(b)
NCERT Solutions for Class 6 Maths Chapter 6 Integers 4
(c) Siachin is the coolest place.
(d) Ahmedabad and Delhi.

Question 6.
In each of the following pairs, which number is to the right of the other on the number line?
(a) 2, 9
(b) -3,-8
(c) 0,-1
(d) – 11, 10
(e) -6,6
(f) 1,- 100
Solution.
(a) 2, 9
The number 9 is to the right of the number 2.
(b) -3,-8
The number – 3 is to the right of the number -8.
(c) 0,-1
The number 0 is to the right of the number – 1.
(d) – 11,10
The number 10 is to the right of the number -11.
(e) -6, 6
The number 6 is to the right of the number – 6.
(f) 1, -100
The number 1 is to the right of the number – 100.

Question 7.
Write all the integers between the given pairs (write them in the increasing order)
(a) 0 and – 7
(b) -4 and 4
(c) – 8 and -15
(d) – 30 and – 23.
Solution.
(a) 0 and – 7
The integers between 0 and – 7 in increasing order are – 6, – 5, – 4, – 3, – 2 and – 1.
(b) – 4 and 4
The integers between – 4 and 4 in increasing order are – 3, – 2, – 1, 0, 1, 2 and 3.
(c) – 8 and – 15
The integers between – 8 and – 15 in increasing order are – 14, – 13, – 12, – 11, – 10 and – 9.
(d) – 30 and – 23
The integers between – 30 and – 23 in increasing order are – 29, – 28, – 27, – 26, – 25 and – 24.

Question 8.
(a) Write four negative integers greater than – 20.
(b) Write four negative integers less than -10.
Solution.
(a) Four negative integers greater than 20 are – 19, – 18, – 17 and – 16.
(b) Four negative integers less than – 10 and 11,- 12,- 13 and -14

Question 9.
For the following statements write True (T) or False (F). If the statement is false, correct the state­ment.
(a)
-8 is to the right of- 50 on a number line.
(b) – 100 is to the right of – 50 on a number line.
(c) A smallest negative integer is – 1.
(d) – 26 is larger than -25
Solution.
(a) True (T)
(b) False (F); – 100 is to the left of – 50 on a number line.
(c) False (F); Greatest negative integer is – 1.
(d) False (F); – 26 is smaller than – 25.

Question 10.
Draw a number line and answer the following:
(a) Which number will we reach if we move 4 numbers to the right of-2?
(b) Which number will we reach if we move 5 numbers to the left of 1?
(c) If we are at -8 on the number line, in which direction should we move to reach -13?
(d) If we are at-6 on the number line, in which direction should we move to reach -l?
Solution.
(a) We will reach number 2.
(b) We will reach the number – 4.
(c) We should move in the left direction.
(d) We should move in the right direction

 

We hope the NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1, 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 4.1

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.1 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 4.1.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 4
Chapter Name Basic Geometrical Ideas
Exercise  Ex 4.1
Number of Questions Solved 6
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.1

Question 1.
Use the figure to name:
(a) Five points
(b) A line
(c) Four rays
(d) Five line segments.
Solution :
(a) O, B, C. D, E
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 1
Question 2.
Name the line given in all possible (twelve) ways, choosing only two letters at a time from the four given.
Solution :
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 2
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 3

Question 3.
Use the figure to name:
(a) The line containing point E.
(b) The line passing through A.
(c) The line on which O lies
(d) Two pairs of intersecting lines.
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 4
Solution :
(a) \(\overleftrightarrow { AE }\), etc.
(b) \(\overleftrightarrow { AE }\), etc.
(c) \(\overleftrightarrow { CO } or\overleftrightarrow { OC }\)
(d) \(\overleftrightarrow { CO, } \overleftrightarrow { AE } ;\overleftrightarrow { AE } ,\overleftrightarrow { EF }.\)

Question 4.
How many lines can pass through
(a) one given point?
(b) two given points?
Solution :
(a) Countless lines can pass through one given point.
(b) One and only one line can pass through two given points.

Question 5.
Draw a rough figure and label suitably in each of the following cases :
(a) Point P lies on \(\bar { AB } \)
(b) \(\overleftrightarrow { XY }\) and \(\overleftrightarrow { PQ }\) intersect at M.
(c) Line contains E and F but not D.
(d) \(\bar { Op } \) and \(\bar { OQ } \) meet at O.
Solution :
(a)
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 5
(b)
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 6
(c)
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 7
(d)
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 8
Question 6.
Consider the following figure of line \(\bar { MN } \) Say whether following statements are true or false in context of the given figure.
(a) Q, M, O, N, P are points on the line \(\bar { MN } \)
(b) M, O, N are points on a line segment \(\bar { MN } \).
(c) M and N are end points of line segment \(\bar { MN } \).
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 9
(d) O and N are end points of line segment \(\bar { OP } \).
(e) M is one of the end points of line segment \(\bar { QO } \).
(f) M is point on ray \(\overrightarrow { OP }\).
(g) Ray \(\overrightarrow { OP }\) is different from ray \(\overrightarrow { QP }\).
(h) Ray \(\overrightarrow { OP }\) is same as ray \(\overrightarrow { OM }\).
(i) Ray \(\overrightarrow { OM }\) is not opposite to ray \(\overrightarrow { OP }\).
(j) O is not an initial point of \(\overrightarrow { OP }\).
(k) N is the initial point of \(\overrightarrow { NP }\) and \(\overrightarrow { NM }\).
Solution :
(a) True
(b) True
(c) True
(d) False
(e) False
(f) False
(g) True
(h) False
(i) False
(j) False
(k) True.

 

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