Prasanna – Page 197

RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.1

March 26, 2023

RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.1

Other Exercises

Question 1.
Without performing actual addition and division, write the quotient when the sum of 69 and 96. is divided by
(i) 11
(ii) 15
Solution:
Two numbers are 69 and 96 whose digits are reversed Here a = 6,= 9
(i) Sum if 69 + 96 is divisible by 11, then quotient = a + 6 = 6 + 9 = 15
(ii) If it is divided by a + b i.e., 6 + 9 = 15, then quotient = 11

Question 2.
Without performing actual computations, find the quotient when 94 – 49 is divided by
(i) 9
(ii) 5
Solution:
Two given numbers are 94 and 49. Whose digits are reversed.
(i) If 94 – 49 is divided by 9, then the quotient = a-b = 9-4 = 5
(ii) and when it is divided by a – b i.e. 9-4 = 5, then quotient will be = 9

Question 3.
If sum of the number 985 and two other numbers obtained by arranging the digits of 985 in cyclic order is divided by 111, 22 and 37 respectively. Find the quotient in each case.
Solution:
The given number is 985
The other two numbers by arranging its digits
in cyclic order, will be 859, 598 of the form
\(\overline{ abc } ,\overline{ bca } ,\overline{ cba }\)
Therefore,
If 985 + 859 + 598 is divided by 111, then quotient will bea + 6 + c = 9 + 8 + 5 = 22
If this sum is divided by 22, then the quotient = 111
and if it is divided by 37, then quotient = 3 (a + b + c) = 3 (22) = 66

Question 4.
Find the quotient when difference of 985 and 958 is divided by 9.
Solution:
The numbers of three digits are
985 and 958 in which tens and ones digits are reversed, then
\(\overline{ abc } -\overline{ acb }\) = 9 (b – c)
985 – 958 = 9 (8 – 5) = 9 x 3
i. e., it is divisible by 9, then quotient = b-c =8-5=3

Hope given RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.1 are helpful to complete your math homework.

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

CA Foundation Business Economics Study Material – Concepts of Product

March 26, 2023

CA Foundation Business Economics Study Material Chapter 3 Theory of Production and Cost – Concepts of Product

Product i.e. output refers to the volume of goods produced by a firm in a particular period of time.
There are three concepts relating to the physical production by factors namely-

Total Product (TP),
Average Product (AP), and
Marginal Product (MP).

1. Total Product (TP):

The total output produced by all the factors per unit of time is called total product.
Total product increases with an increase in the variable factor input.
Column Nos. (1) and (2) of the following table shows a total product schedule.

2. Average Product (AP):

The. average product means the total product per unit of a variable factor.
In other words, it is the total product divided by the number of units of a variable factor.<
Column No. (3) of the following table shows the average product of variable factor.

3. Marginal Product (MP):

The marginal product means addition made to total product by the use of an extra unit of variable factor.
It may be stated as-
MP_n = TP_n – TP_n-1
where,
MP_n = Marginal product when ‘n ’ units of variable factors are used
TP = Total Product
n = number of units of variable factors used.
Marginal Product may also be defined as the change in total output due to use of additional unit of variable factor

Where –
Δ = a small change Column No. (4) of the following table shows the marginal product schedule.

Table: Product Schedule

Units of Variable	Total Product (TP) factor E.g. LABOUR	Average Product (AP)	Marginal Product (MP)
1	10	10	10
2	30	15	20
3	60	20	30
4	80	20	20
5	90	18	10
6	90	15	0
7	85	12.1	-5

Average product and Marginal product are related to one another.

(i) – When average product of the variable factor is rising, marginal product of the variable factor is more than its average product.
– So when average product curve is rising, the marginal product curve will lie somewhere above it.

(ii) – When average product of the variable factor is falling, marginal product of the variable factor is less than its average product.
– So when average product curve is falling, the marginal product curve will lie somewhere below it.

(iii) – When average product of the variable factor is maximum and constant, marginal product is equal to average product.
– In other words, the marginal product curve cuts the average product curve at its maximum point.

NCERT Exemplar Solutions for Class 10 Science Chapter 7 Control and Coordination

March 26, 2023

NCERT Exemplar Solutions for Class 10 Science Chapter 7 Control and Coordination

These Solutions are part of NCERT Exemplar Solutions for Class 10 Science. Here we have given NCERT Exemplar Solutions for Class 10 Science Chapter 7 Control and Coordination

NCERT Exemplar Solutions for Class 10 Science Chapter 7 Short Answer Questions

Question 1.

Label the parts a, b, c and d and show the direction of electrical signals in the figure. (CBSE 2010)
Draw the figure shown here and label motor neuron, relay neuron and spinal cord. What is the name of this connection ?
(CCE 2011)

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Answer:

a – Sensory neuron.
b – Spinal cord (CNS).
c – Mortor neuron.
d – Effector (muscle in arm)
Name — Reflex arc.
Direction :
1. Receptor (hand) to spinal cord through sensory neuron,
2. Sensory neuron to motor neuron through Spinal cord to effector through motor neuron.

Question 2.
Name the plant hormone responsible for the following :
(a) Elongation of cells
(b) Growth of stem
(c) Promotion of cell division
(d) Falling of senescent leaves.
Answer:
(a) Elongation of Cells. Auxin.
(b) Growth of Stem. Gibberellin.
(c) Promotion of Cell Division. Cytokinin.
(d) Falling of Senescent Leaves. Abscisic acid.

Question 3.
Label the endocrine glands in the figure
NCERT Exemplar Solutions for Class 10 Science Chapter 7 Control and Coordination image - 2
Answer:
a – Pineal gland.
b – Pituitary gland.
c – Thyroid
d – Thymus.

Question 4.
In the figures (a), (b) and (c) which appears more accurate and why ?
NCERT Exemplar Solutions for Class 10 Science Chapter 7 Control and Coordination image - 3
Answer:
Figure (a) is more accurate as it shows appropriate response of both shoot and root to the vector of gravity. Shoot is negatively geotropic. It, therefore, bends upwardly. Root is positively geotropic. It, therefore, bends downwardly.

Question 5.
Label the parts of a neuron shown in the figure.
NCERT Exemplar Solutions for Class 10 Science Chapter 7 Control and Coordination image - 4
Answer:
a – Dendrite.
b – Cyton (Cell body).
c – Axon.
d – Axon terminal (bouton).

Question 6.
Match the terms of column A with those of column B.

(a) Olfactory receptors

(b) Thermoreceptors (temperature receptors)

(d) Photoreceptors

(I) Tongue

(ii) Eye

(iiî) Nose

(iv) Skin

Answer:
a – IIi,
b — iv,
c — i,
d — ii.

Question 7.
What is a tropic movement ? Explain with an example. (CCE 2011)
Answer:
Tropic Movement: It is directional paratonie growth movement of curvature in which direction of movement is determined by the direction of the stimulus. They mostly occur in cylindrical organs. Growth response occurs due to differences in the distribution of auxin.
Example. On unidirectional exposure to light, shoots bend towards the source of light (positive phototropism) while roots may bend away from the source of light (negative phototropism).

Question 8.
What will happen if intake of iodine in our diet is low ?
Answer:

Low intake of iodine will affect production of thyroxine. Reduced thyroxine reduces metabolism of protein, carbohydrate and fat. Physical activity and consumption of energy are impaired.
With continued reduced intake of iodine, thyroid gland enlarges (in order to produce more thyroxine). It results in swelling of neck called goitre.

Question 9.
What happens at the synapse between the neurons ?

State the events in sequence that take place when an electrical impulse travels from a dendritic tip of a nerve cell to another nerve cell. (CCE 2011, 2012, 2013)
Answer:
At synapse, axon terminal is expanded to form presynaptic knob. The dendrite terminal that lies close to it is slightly broadened and depreseed to form post-synaptic depression. A fluid filled narrow space, called synaptic cleft, occurs between the two. When an impulse reaches the presynaptic knob, it stimulates the release of neurotransmitter into synaptic cleft. Neurotransmitter comes in contact with chemoreceptor sites of the membrane of postsynaptic depression. This generates an electrochemical signal or impulse in the dendrite part of second neuron.

Question 10.
Answer the following :
(a) Which hormone is responsible for the changes noticed in female at puberty ?
(b) Dwarfism results due to deficiency of which hormone ?
(c) Blood sugar level rises due to deficiency of which hormone ?
(d) Iodine is necessary for synthesis of which hormone ?
Answer:
(a) Puberty Changes in Females. Oestrogen.
(b) Dwarfism. Growth hormone.
(c) Blood Sugar. Insulin.
(d) Iodine Hormone. Thyroxine

Question 11.
Answer the following :
(a) Name the endocrine gland associated with brain.
(b) Which gland secrets digestive enzymes as well as hormones ?
(c) Name the endocrine gland associated with kidney.
(d) Which endocrine gland is present in males but not in females ?
Answer:
(a) Endocrine Gland Associated with Brain. Pituitary.
(b) Gland with Digestive Enzymes and Hormones. Pancreas.
(c) Endocrine Gland Associated with Kidneys. Adrenal glands.
(d) Endocrine Gland in Males only. Testis.

NCERT Exemplar Solutions for Class 10 Science Chapter 7 Long Answer Questions

Question 12.
Draw the structure of a neuron and explain its function. (CCE 2011)
Answer:
NCERT Exemplar Solutions for Class 10 Science Chapter 7 Control and Coordination image - 5
Functions:

Dendrites: Picking up sensations and transmitting the same to cell body.
Cell Body:
1. Sustaining structure and function of dendrites and axon,
2. Functioning as passage¬way for transmission of sensation or impulse to axon.
Axon: Carrying impulse to another neuron, muscle, gland and organ. A single impulse can be transmitted to several structures with the help of axon terminals.

Question 13.
What are the major parts of the brain ? Mention the function of different parts.
Answer:
Major Parts of Brain
Fore-Brain: Olfactory lobes (2), Cerebral hemispheres (2) or cerebrum, Diencephalon.
Mid-Brain: Cerebral peduncles, Corpora quadrigemina.
Hind Brain: Pons, Cerebellum, Medulla oblongata.
Functions

Olfactory Lobes: Receive sensation of smell from olfactory epithelium and relay the same to temporal lobes of cerebral hemispheres.
Cerebral Hemispheres:
1. Frontal Lobes: Centres of intelligence, control of movements and facial muscles, speech.
2. Parietal Lobes: Taste and cutaneous sensations with some components of speech.
3. Temporal Lobes: Hearing, smell, recall of audiovisual memory, some components of speech.
4. Occipital Lobes: Sight.
Diencephalon: Hypothalamus part regulates activity of smooth muscles and functions as control centre for hunger, thirst, fatigue, sleep, sweating, temperature and emotions. Pituitary gland and pineal gland are components of diencephalon.
Corpora Quadrigemina: Superior – sight reflexes. Inferior – auditory reflexes.
Cerebellum: Controls posture and equilibrium.
Pons: Part of respiratory centre.
Medulla Oblongata: Centre for reflexes, blood pressure, heart beat, breathing.

Question 14.
What constitutes the central and peripheral nervous systems ? How are the components of central nervous system protected?
(CCE 2012)
Answer:
Central Nervous System (CNS): It is hollowed part of nervous system that lies along the mid-dorsal part of the body. It has two parts, brain (located in the head) and spinal cord (located in the trunk).
Peripheral Nervous System (PNS): It is soild lateral part of nervous system that develops from CNS and connects different parts of the body with CNS. Peripheral nervous system has two components, voluntary and involuntary. Voluntary peripheral nerous system is under the control of will. It consists of cranial nerves from brain and spinal nerves from spinal cord. Involuntary peripheral nervous system works independent of will. It develops from some cranial and spinal nerves. Involuntary peripheral nervous system is also called autonomic nervous system. It has two parts, sympathetic (for emergency) and parasympathetic (for moderation). They control the functioning of various internal body parts.
Protection of Brain: Brain box or Cranium, meninges and cerebrospinal fluid.
Protection of Spinal Cord: Vertebral Column, meninges and cerebrospinal fluid.

Question 15.
Mention one function for each of these hormones.
(a) Thyroxine
(b) Insulin
(c) Adrenaline
(d) Growth hormone
(e) Testosterone. (CCE 2010, 2013, 2015)
Answer:
(a) Thyroxine: Regulates metabolism of carbohydrates, fats and proteins, release of energy and body activity.
(b) Insulin: Regulates blood glucose by its absorption in liver, muscles (for formation of glycogen) and individual cells (for metabolic activity).
(c) Adrenaline: For meeting an emergency by increasing heart rate and supply of blood to various concerned organs.
(d) Growth Hormone (GH) or Somatotrophic Hormone: Regulates growth and development of the body through anabolic activity for growth of bones, muscles and visceral organs.
(e) Testosterone (Male Sex Hormone): Controls changes in body associated with puberty in males (beard, moustache, low pitch voice, increased growth of bones and muscles).

Question 16.
Name the various plant hormones. Also give their physiological effects on plant growth and development. (CCE 2016)
Answer:
Five types — auxin, gibberelin, cytokinin, ethylene and abscisic acid.
Auxin: Cell enlargement, root formation, apical dominance, inhibition of abscission, fruit growth.
Gibberellin: Growth in stem and leaves, higher fruit yield, overcoming dormancy.
Cytokinin: Essential for cell division, differentiation, prevention of senescence and overcoming apical dominance.
Ethylene: Promotes transverse growth, fruit ripening and overcoming dormancy of some parts.
Abscisic Acid (ABA): Induces dormancy, senescence and abscission, checking excessive activity of growth promoting hormones, closure of stomata under water stress.

Question 17.
What are reflex actions ? Give two examples. Explain a reflex arc.
Answer:
Definition: Reflex actions seem to have evolved quite early in the physiology of animals when complex neuron network for processing had not been evolved. Even after evolution of processing centres, reflex actions have continued to persist because of their more efficiency for quick responses.
Two Examples:

On being pricked or coming in contact with hot surface, hand is withdrawn even before pain is perceived (by brain),
Wider openin0 of pupil in dim light and its narrowing in strong light.

Question 18.
“Nervous and hormonal systems together perform the function of control and coordination in human being.” Justify the statement.
Answer:
Both nervous system and hormonal (or endocrine) system are involved in control, regulation and coordination of body parts. Nervous system is connected to receptors of all senses. Information obtained from sensory organs is passed rapidly to CNS for interpretation.
On the basis of interpretation, a message is sent to effector organ or organs (muscles, glands, etc.). The rate of information or impulse transfer is very high, some 100m/sec. Every action and activity is well coordinated as the information is sent to all the regions required for that action. The action can be voluntary (under the will) or involuntary. Passage of food in the alimentary canal is due to an involuntary movement of alimentary canal called peristalsis. Picking up food and placing it inside the mouth is a voluntary movement. Both these movements are highly coordinated by nervous system.
Nervous system is also connected with the functioning of endocrine or hormonal system. The endocrine system functions at the biochemical level while nervous system functions at the physical and physiological level. For their functioning endocrine glands pour their secretion into blood which transports them to all parts of the body. Target cells have receptors for picking up the hormones and working as per hormonal stimulus. There is a feedback system which determines the requirement of hormones and the activity of endocrine glands. Other stimuli are also involved for coordinated functioning. Presence of food in the stomach stimulates its wall to secrete hormone gastrin. Gastrin stimulates gastric glands to pour gastric juice over the food. As the partially digested food passes into duodenum, the latter forms hormone secretin which induces passage of bile and pancreatic juice into duodenum.
Joint working of both the systems is observed during an emergency. Both sympathetic nervous system and adrenal glands prepare the body for meeting the emergency. There is higher rate of heart beat, more blood supply to cardiac and skeletal muscles and higher rate of breathing for quicker oxygenation of blood.

Question 19.
How does chemical co-ordination take place in animals ?
Answer:
In animals, chemical coordination is achieved through the agency of hormones which function as chemical messengers or informational molecules. Hormones are secreted by ductless glands in response to specific conditions or nervous stimulation. Timing and amount of a hormone released are regulated by feed-back mechanism. After a meal, sugar level of blood rises. It is detected by pancreas. Pancreas responds by producing hormone insulin from (3-cells of islets of Langerhans. Insulin causes glucose to be absorbed by all cells as well as get stored in liver and muscles in the form of glycogen. As the level of glucose falls in blood, insulin secretion is reduced.

Question 20.
Why the flow of signals in a synapse is from axonal end of one neuron to dendrite end of another neuron but not the reverse ?
Answer:
As the electrical impulse reaches the axon terminal in the region of axon-dendrite synapse, it stimulates the exocytosis of vesicles containing neurotransmitter (e.g. acetylcholine). Neurotransmitter attaches to the chemoreceptor sites of the membrane covering the dendrite end of synapse. It creates a new impulse that travels through cell body and axon of the second neuron. At synapse, the axon end does not contain any chemoreceptor sites so that reverse flow of electrochemical impulses is not possible.

Hope given NCERT Exemplar Solutions for Class 10 Science Chapter 7 Control and Coordination are helpful to complete your science homework.

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

NCERT Exemplar Solutions for Class 10 Science Chapter 8 How do Organisms Reproduce?

March 26, 2023

NCERT Exemplar Solutions for Class 10 Science Chapter 8 How do Organisms Reproduce?

These Solutions are part of NCERT Exemplar Solutions for Class 10 Science. Here we have given NCERT Exemplar Solutions for Class 10 Science Chapter 8 How do Organisms Reproduce?

NCERT Exemplar Solutions for Class 10 Science Chapter 8 Short Answer Questions

Question 1.
In a bisexual flower, inspite of the young stamens being removed artificially, the flower produces fruit. Provide a suitable explanation for the above situation.
Answer:
The open flowers are generally cross pollinated. Removal of stamens of a bisexual flower will not affect pollination of its intact pistil and formation of fruit.

More Resources

Question 2.
Can you consider cell division as a type of reproduction in unicellular organisms ? Give one reason.
Answer:
Yes. Cell division in a unicellular organism which produces two d&ughter cells, actually forms two daughter individuals.

Question 3.
What is a clone ? Why do offspring formed by asexual reproduction exhibit remarkable similarity ?
Answer:
Clone is an exact genetic replica of another individual. All the offspring formed from a parent through asexual reproduction are clones of one another as well as their parent. The remarkable similarity of asexually produced daughter individuals is due to genetic similarity as they possess exact copies of DNA of their parent.

Question 4.
Explain how, offspring and parents of organisms reproducing sexually have the same number of chromosomes/DNAs.
(CCE 2014, CBSE Delhi 2017)
Answer:
The parents are diploid (2n) as each of them has two sets of chromosomes (DNAs). They form haploid (In) male and female gametes through the process of meiosis. The haploid gametes have one set of chromosomes (DNAs). During fertilization, one male gamete fuses with one female gamete. It restores the diploid (2n) chromosome/DNA number in the offspring that is formed from fusion product or zygote (2n).

Question 5.
Colonies of yeast fail to multiply in water but multiply in sugar solution. Give one reason for this.
Answer:
Yeast is heterotrophic. It obtains its nourishment from outside. Plain water cannot provide nourishment to yeast while sugar solution can do so. Therefore, Yeast multiplies in sugar solution and not in plain water.

Question 6.
Why does Bread Mould grow profusely on a moist slice of bread rather than on a dry slice of bread ?
Answer:
Growth of Bread Mould requires both water and nutrients. Dry slice of bread contains nutrients but no water. Therefore, Bread Mould does not grow over it. Moist slice possesses both water and nutrients. Hence, Bread Mould grows over it.

Question 7.
Give two reasons for the appearance of variations among the progeny formed by sexual reproduction.
Answer:
Variations appear in the progeny of sexually reproducing organisms due to

Random separation and coming together of chromosomes during gamete formation and gamete fusion
Crossing over and mutations.
Coming together of chromosomes of the parents.

Question 8.
Would a Planaria cut vertically into two halves regenerate into two individuals ?
Answer:
Yes. Each piece of Planaria is able to grow the missing parts and form the complete organism.

Question 9.
Correlate the number of chromosomes with the size of the organism and answer the following questions :
(a) Do larger organisms have more number of chromosomes/cell ?
(b) Can organism with fewer chromosomes reproduce more easily than organism with more number of chromosomes ?
(c) More the number of chromosomes/cell, greater is the DNA content. Justify.
Answer:
(a) No. There is no correlation between chromosome number and size of the individual.
(b) No. Chromosome number does not affect reproduction.
(c) Yes. Each chromosome represents a single duplex of DNA. More chromosomes means more DNA.

Question 10.
In Tobacco, male gametes have 24 chromosomes.
(a) What is the number of chromosomes in the female gamete ?
(b) What is the number of chromosomes in the zygote ?
Answer:
(a) 24
(b) 48.

Question 11.
Why cannot fertilization take place in flowers if pollination does not occur ?
Answer:
Pollination is essential for bringing the male gametes. In the absence of pollination, there will be no male gamete to perform fertilization.

Question 12.
Is the chromosome number of zygote, emboryonal cells and adult of a particular organism always constant ? How is the constancy maintained in these three stages ?
Answer:
Zygote is the first diploid structure. It undergoes mitotic divisions to form embryonal cells. Embryonal cells further divide by mitosis to form the adult. Mitosis maintains the same chromosome number in the daughter cells.

Question 13.
Where is the zygote located in the flower after fertilization ?
Answer:
Zygote is the fertilized oosphere which occurs in the embryo sac present inside an ovule located in the ovary part of the pistil.

Question 14.
Reproduction is linked to stability of population of a species. Justify the statement.
Answer:
Reproduction maintains the size and composition of population of a species by regularly adding new individuals for replacing the ones getting killed due to ageing, disease or predation. It also adds variations that allow the individuals to adapt to changing environment.

Question 15.
How are general growth and sexual maturation different from each other ?
Answer:
General growth is the growth of an individual in size, height, shape, weight, etc. Sexual maturation is a set of changes in body of an individual that enable one to take part in reproduction, e.g., maturation of gonads, genitalia and other accessory structures. In human beings, beginning of sexual maturation is indicated by puberty changes like cracking of voice, new hair pattern, development of breast in female, etc.

Question 16.
Trace the path of sperm during ejaculation and mention the glands and their function associated with the male reproductive system.
Answer:
Ejaculated sperms are the ones which are stored in epididymes. They are formed regularly in seminiferous tubules from where they pass through vasa efferentia into epididymes. At the time of ejaculation, the sperms are first pushed through vasa deferentia, enter ejaculatory duct where they receive secretion of seminal vesicles and then urinogenital duct where the secretion of prostate gland is poured to form semen. The urinogenital duct is lubricated by secretion of a pair of Cowper’s glands (bulbourethral glands).
Secretion of Seminal Vesicles. Nourishment, activation and providing fluid medium for sperm transport.
Secretion of Prostates Gland. Motility of Sperms.

Question 17.
What changes are observed in the uterus if fertilization does not occur ? (CBSE Delhi 2017)
Answer:
Glands present in the mucosa of uterine endometrium begin to degenerate. This sloughs off the endometrial lining, releasing a lot of mucus and blood. It is called menstruation. Menstruation lasts for 3-5 days.

Question 18.
What changes are observed in the uterus subsequent to implantation of young embryo ? (CBSE Delhi 2017)
Answer:
Endometrium thickens, becomes glandular and highly vascular. The contact region between embryo and uterine wall grows into placenta. Placenta helps the embryo in obtaining nourishment and oxygen from blood sinuses of the uterus. It also helps in transfer of wastes from embryo to the blood of the mother.

Question 19.
What are the benefits of using mechanical barriers during sexual act ?
Answer:

Prevention of Pregnancy : Mechanical barriers (e.g., Condoms, diaphragms) prevent the passage of sperms into the genital tract of the female. Consequently, pregnancy does not occur.
Non-transmission of Infections: There is no transfer of venereal diseases from the infected partner to the non-infected partner.

Question 20.
In the female reproductive system, name the parts that are involved in
(a) Production of egg.
(b) Site of fertilization
(c) Site of implantation
(d) Entry of sperms.
Answer:
(a) Production of Egg: Ovary
(b) Site of Fertilization: Ampulla part of fallopian tube.
(c) Site of Implantation: Uterus.
(d) Entry of Sperms: Vagina.

Question 21.
What would be the ratio of chromosome number between an egg and its zygote ? How is the sperm genetically different from the egg ?
Answer:
Chromosome Number in Egg and Zygote. 1 : 2
Genetic Difference between Sperm and Egg. Sperms are genetically of two types, X-containing and Y-containing. Eggs are always of one type, X-containing.

NCERT Exemplar Solutions for Class 10 Science Chapter 8 Long Answer Questions

Question 22.
Why are budding, fragmentation and regeneration all considered as asexual types of reproduction ? With neat diagrams, explain the process of regeneration in Planaria.
Answer:
(a) Asexual Reproduction: Budding, fragmentation and regeneration are all considered to be modes of asexual reproduction as they involve formation of new individuals from single parents without fertilization or fusion of gametes.
(b) Regeneration in Planaria:
NCERT Exemplar Solutions for Class 10 Science Chapter 8 How do Organisms Reproduce image - 1

Question 23.
Write two points of differences between asexual and sexual types of reproduction. Describe why variations are observed in the offspring formed by sexual reproduction.
Answer:
(a) Differences between Asexual and Sexual Reproduction

Asexual Reproduction	Sexual Reproduction
1. Parents: It is monoparental.	It is generally biparental
2. Meiosis: Meiosis does not occur during asexual reproduction. There are no gametes and no fertilization.	Meiosis occurs. Gametes are formed and fertilization occurs in sexual reproduction.
Ex. Binary fission in Amoeba, Budding in Yeast	Ex. Offspring in Wheat, Human.

(b) Development of Variation in Sexual Reproduction:
Sexual reproduction produces a number of variations in the population due to

Chance separation of chromosomes at the time of meiosis.
Crossing over during meiosis.
Chance combination of chromosomes during fertilization. It produces a unique combination of variations already accumulated by each individual in its DNA obtained from its parents,
Mutations or defective DNA replication.

Question 24.
Distinguish between pollination and fertilization. Mention the site and product of fertilization in a flower. Draw a neat, labelled diagram of a pistil showing pollen tube growth and its entry into the ovule.
Answer:
(a) Differences between Pollination and Fertilization

Pollination

Fertilization

1. Definition: It is transfer of pollen grains from anther to the stigma of a flower.

2. Step: Pollination precedes fertilization.

3. Purpose: It carries the male gamete producing pollen grains to the female sex organ.

4. Process: Pollination is a physical process.

5. Occurrence: It occurs only in seed plants.

It is the fusion of male and female gametes.

Fertilization occurs only after pollination when the pollen grain has germinated and male gametes are carried into ovule.

It actually brings about fusion of gametes.

Fertilization is a physico-chemical (biological) process.

It occurs in both plants and animals of various types.

(b)

Site of Fertilization. Embryo sac in ovule.
Product of Fertilization. Zygote, Primary endosperm cell.

Question 25.
Distinguish between a gamete and zygote. Explain their roles in sexual reproduction.
Answer:
Differences between Gamete and Zygote

Gamete	Zygote
1. Nature: It is sex or germ cell that takes part in fertilization.	It is a product of fertilization.
2. Types: There are two types of gametes, male and female.	Zygote is of one type.
3. Chromosome Number: A gamete has haploid or In chromosome number.	Zygote has diploid or 2n chromosome number.
4. Characteristics: A gamete carries characteristics of only one parent.	It carries characteristics of both the parents.
5. Generation: Gamete is the last cell of its generation.	It is the first cell of new generation.

Role of Gamete. Gamete is sex or germ cell which is specialized to take part in sexual reproduction. There are two types of gametes, male and female. Each of them carries one set of chromosomes which are randomly obtained from its parent. Fusion of male gamete with a female gamete produces a zygote.
Role of Zygote. It is the first’cell of new generation. Zygote develops into embryo that later forms the new individual.

Question 26.
Draw the diagram of a flower and label the four whorls. Write the names of gamete producing organs in the flower.
Answer:
(a)
NCERT Exemplar Solutions for Class 10 Science Chapter 8 How do Organisms Reproduce image - 3
(b) Male Gamete Forming Part. Pollen grain, produced in anther of stamen.
Female Gamete Forming Part. Embryo sac developed inside ovule of ovary part of pistil.

Question 27.
What is placenta ? Mention its role during pregnancy. (CCE 2011, 2014)
Answer:
Placenta: It is a special double layered, spongy tissue connection between the foetus and uternine wall found in pregnant females. It has finger-like outgrowths or villi which are in contact with blood sinuses present in the uterine wall.
Role:

Attachment: Placenta attaches the foetus to uterine wall.
Villi: Placenta has finger-like outgrowths or villi which develop a large surface area for fixation and absorption.
Nutrients: Placenta picks up nutrients from mother’s blood and passes it to the blood of the foetus.
Waste Products: Waste products produced by the foetus passes out through the placenta into mother’s blood.
Gases: Foetus obtains oxygen supply from mother’s blood and eliminates carbon dioxide through placenta.

Question 28.
What are the various ways to avoid pregnancy ? Elaborate any one method.
Answer:
(a) Methods to Avoid Pregnancy: Prevention of pregnancy is called contraception. The techniques used in preventing the occurrence of pregnancy are called contraceptive devices.
(b) Mechanical Barrier Methods of Contraception:

Mechanical Barriers like condoms, cervical cap, diaphragm.
Oral Contraceptives or oral pills like Mala D, Saheli
Intrauterine Contraceptive Devices (IUCD) like loop, bow, Cu-T.
Surgical Methods like vasectomy in males and tubectomy in females.

Question 29.
Flow does fertilization take place ? Fertilization occurs once in a month. Comment.
Answer:
In human beings, fertilization is internal. It can occur within 2-3 days of ovulation (between 14th and 16th day of menstrual cycle). Fertilization generally takes place in ampulla (ampulla-isthmus border) part of fallopian tube where ovum rests for several hours after entry into oviduct. The sperms have to reach there. Sperms deposited in the vagina reach there partly by their own movement and partly by local movements of reproductive tract. Both sperms and eggs are viable for 24-48 hours. A sperm reaching the surface of ovum, dissolves its covering and passes its head and middle part into it. It results in fertilization and formation of zygote.
Fertilization only once in a Month. A single ovum is released by female roughly in the middle of menstrual cycle which has a span of 4 weeks. Therefore, fertilization can occur only once in a month.

Question 30.
Reproduction is essentially a phenomenon that is not for survival of individual but for the stability of a species. Justify. Reproduction is not essential for survival of the individual.
Answer:
Survival of the individual depends upon input of nutrients and energy and elimination of wastes. Reproduction has ho role in these. It is, however, essential for the stability and survival of the species. Reproduction takes part in
(a) Perpetuation of Species: Perpetuation or continuation of species.
(b) Replacement: Replacement of dead individuals and maintaining the organisation in population.
(c) Population Characteristics:

Education: There is an inverse ratio between education and population growth.
Population Education: It is imparting knowledge to public about the effects of excessive population, advantages of small families and means to achieve it.
Marriageable Age: Number of births is reduced if young persons marry late. Gainful employment and higher social status of women also reduce birth rate.

(d) Variations:
Variations are differences found in morphological, physiological and other traits of individuals belonging to the same organism, race or family. They develop due to

Faulty DNA replication
Crossing over
Chance separation of chromosomes during meiosis and chance combination during fertilization. Variations are highly important.

(e) Absence of Variations: Due to absence of variations, asexually reproduced organisms are unable to adapt themselves to changes in the environment.

Question 31.
Describe sexually transmitted diseases and mention the ways to prevent them.
Answer:
It is a group of infections caused by different types of pathogens that are transmitted by sexual contact between a healthy person and an infected person. The sexually transmitted diseases are also called venereal diseases (VDs). Some 30 different types of STDs are known.

It is caused by bacterium Neisseria gonorrhoeae. Gonorrhoea spreads through sexual contact, common toilets and under clothes. Incubation period is 2-5 days. Bacterium resides in genital tubes. It causes pus containing discharge, pain around genitalia and burning sensation during urination. Effective medicine is ampicillin.
The disease is caused by corkscrew like bacterium Treponema pallidum. It spreads through sexual contact and from mother to infants. Incubation periods is 3-5 weeks, Painless ulcer on genitalia and swelling of lymph glands occur in first stage. In second stage skin lesions, rashes and hair loss occur. Tertiary stage is characterised by chronic ulcers and damage to vital organs. Effective medicine is tetracycline.
Genital Warts: The disease is due to Human Papilloma virus (HPV). Hard benign outgrowths called warts appear over external genitalia and perianal area. In women infection may enter vagina and cervix causing acute pain. Cryosurgery and Podophyllum preparations are effective.
AIDS (Acquired Immune Deficiency Syndrome): It is caused by human immunodeficiency virus or HIV. The virus is transmitted through sexual contact, blood contact (as using common needles, syringes, razors, transfusion) and placenta (mother to foetus). Incubation period is generally 27-28 months but symptoms may appear early. Count of T-helper cells becomes low, 200/ml or less. There is headache, rashes, nausea, pharyngitis and fever. Immunity is drastically reduced so that many infections (opportunistic infections) begin to appear. A proper treatment has not yet been discovered. ART (antiretrovirus treatment) is given to patients to reduce the effect of infection.

Hope given NCERT Exemplar Solutions for Class 10 Science Chapter 8 How do Organisms Reproduce? are helpful to complete your science homework.

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RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3

March 26, 2023

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3

Other Exercises

Find each of the following products (1-8)
Question 1.
5x² x 4x³Solution:
5x² x 4x³ = 5 x 4 x x² x x³
= 20x^{2 + 3} = 20x^s

Question 2.
3a² x 4b⁴Solution:
-3a² x 4b⁴ = -3 x 4 x a²b⁴= -12a²b⁴

Question 3.
(-5xy) x (-3x²yz)
Solution:
(-5xy) x (-3x²yz)
= (-5) x (-3)xy x x²yz
= 15x^{1 +}²xy^{1+ 1}z= 15x³y²z

Question 4.

Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 2

Question 5.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 4
Solution:

Question 6.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 6
Solution:

Question 7.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 7
Solution:

Question 8.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 10
Solution:

Find each of the following products : (9-17)

Question 9.
(7ab) x (-5ab²c) x (6abc²)
Solution:
(7ab) x (-5ab²c) x (6abc²)
= 7 x (-5) x 6 x a x a x a x b x b² x b x c x c²=-210 x a¹⁺¹⁺¹x b¹⁺²⁺¹x c¹⁺²=-210 x a³b⁴c³

Question 10.
(-5a) x (-10a²) x (-2a³)
Solution:
(-5a) x (-10a²) x (-2a³)
= (-5) (-10) (-2) x a x a² x a³= -100a^{1 + 2 + 3} = -100a⁶

Question 11.
(-4x²) x (-6xy²) x (-3yz²)
Solution:
(-4x²) x (-6xy²) x (-3yz²)
= (-4) x (-6) x (-3) x² x x x y² x y xz²= -72x²⁺¹ x y²⁺¹ x _z²
= 72x³y³z³

Question 12.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 12
Solution:

Question 13.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 14
Solution:

Question 14.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 16
Solution:

Question 15.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 18
Solution:

Question 16.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 20
Solution:

Question 17.
(2.3xy) x (0.1x) x (0.16)
Solution:
(2.3xy) x (0.1x) x (0.16)
= 2.3 x 0.1 x 0.16 x x x x x y
= 0.0368x^{1 +1} x y = 0.0368x²y

Express each of the following products as a monomials and verify the result in each case for x = 1 : (18 -26)

Question 18.
(3x) x (4x) x (-5x)
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 23

Question 19.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 24
Solution:

Question 20.
(5x⁴) x (x²)³ x (2x)²Solution:
(5x⁴) x (x²)³ x (2x)²
= 5x⁴ x x^{2 x}³ x 2x x 2x
= 5x⁴ * x⁶ x 4x² = 5 x 4 x x^{4 + 6 + 2}= 20x¹²Verification:
L.H.S. = (5x⁴) x (x²)³ x (2x)²= 5 x (1)⁴ x [(1)²]³ x (2 x 1)²= 5 x 1 x (1)^{2 x}³_x (2)²= 5 x 1⁶ x 2² = 5 x 1 x 4 = 20
R.H.S. = 20x¹² = 20 (1)¹² = 20 x 1 = 20
∴ L.H.S. = R.H.S.

Question 21.
(x²)³ x (2x) x (-4x) x 5
Solution:
(x²)³ x (2x) x (-4x) x (5)
= x^{2 x} ³ X 2x X (-4x) X 5
= x⁶ x 2x x (-4x) x 5 = 2 x (-4) x 5x⁶^{+1 +1}
= -40x⁸Verification
L.H.S. = (x²)³ x (2x) x (-4x) x (5)
= (1²)³ x (2 x 1) x (-4 x 1) x 5
= 1^2 x³x 2 x (- 4) x 5 = 1⁶ x 2 x (-4) x 5
= 1 x 2 x (-4) x 5 = -40
R.H.S. = -40x⁸ = -40 x (1)⁸
= -40 x 1 = -40
∴ L.H.S. = R.H.S.

Question 22.
Write down the product of -8x²y⁶ and – 20xy Verify the product for x = 2.5, y = 1.
Solution:
Product of -8x²y⁶ and -20xy
= (-8x²y⁶) x (-20xy)
= -8 x (-20) x² x _x x y⁶ x y = 160x^{2 + 1} x y^{6 +}¹= 160x³y³
Verification.
L.H.S. = (-8x²y⁶) x (-20xy)
= -8 x (2.5)² x (1) x (-20 x 2.5 x 1)
= -8 x 6.25 x 1 x -20 x 2.5
= (-50) x (-50) = 2500
R.H.S. = 160 x = 160 (2.5)³ x (1)⁷= 160 x 15.625 x 1 =2500
∴ L.H.S. = R.H.S.

Question 23.
Evaluate : (3.2x⁶y³) x (2.1x²y²) when x = 1 and y = 0.5.
Solution:
3.2x⁶y³ x 2.1x²y²= 3.2 x 2.1 x x⁶⁺²x y³⁺²
= 6.72x⁸y⁵ = 6.72 x (1)⁸ x (0.5)⁵= 6.72 x 1 x 0.03125
= 0.21

Question 24.
Find the value of (5x⁶) x (-1.5x²y³) x (-12xy²) when x = 1, y = 0.5.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 27

Question 25.
Evaluate : (2.3a⁵b²) x (1.2a²b²) when a = 1, b = 0.5.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 28

Question 26.
Evaluate : (-8x²y⁶) x (-20xy) for x = 2.5 and y = 1.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 29

Express each of the following products as a monomials and verify the result for x = 1,y = 2: (27-31)

Question 27.
(-xy³) x (yx³) x (xy)
Solution:
(-xy³) x (yx³) x (xy)
= -x x x³x x x y³x y x y = -x¹⁺³⁺¹ x y^{3 +}¹⁺¹= -x⁵y⁵Verification:
L.H.S. = (-xy³) x (yx³) x (xy)
= (-1 x 2³) x [2 x (1)³] x (1 X 2)
= (-1 x 8) x (2 x 1) x (1 x 2)
= -8 x 2 x 2 = -32
R.H.S. =-x⁵y⁵ = -(1)⁵ (2)⁵= -1 x 32 =-32
∴ L.H.S. = R.H.S.

Question 28.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 30
Solution:

Question 29.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 33
Solution:

Question 30.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 35
Solution:

Question 31.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 37
Solution:

Question 32.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 40
Solution:

Question 33.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 43
Solution:

Hope given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5

March 26, 2023

RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5

Other Exercises

Making use of the cube root table, find the cubes root of the following (correct to three decimal places)
Question 1.
7
Solution:
\(\sqrt [ 3 ]{ 7 }\) =1.913 (From the table)

Question 2.
70
Solution:
\(\sqrt [ 3 ]{ 70 }\) =4.121 (From the table)

Question 3.
700
Solution:
\(\sqrt [ 3 ]{ 700 } =\sqrt [ 3 ]{ 7\times 100 } \)= 8.879 (from \(\sqrt [ 3 ]{ 10x }\))

Question 4.
7000
Solution:
\(\sqrt [ 3 ]{ 7000 } =\sqrt [ 3 ]{ 70\times 100 }\) = 19.13 (from \(\sqrt [ 3 ]{ 100x }\))

Question 5.
1100
Solution:
\(\sqrt [ 3 ]{ 1100 } =\sqrt [ 3 ]{ 11\times 100 }\) = 10.32 (from \(\sqrt [ 3 ]{ 100x }\))

Question 6.
780
Solution:
\(\sqrt [ 3 ]{ 780 } =\sqrt [ 3 ]{ 78\times 100 }\) = 9.205 (from \(\sqrt [ 3 ]{ 10x }\))

Question 7.
7800
Solution:
\(\sqrt [ 3 ]{ 7800 } =\sqrt [ 3 ]{ 78\times 100 }\) = 19.83 (from \(\sqrt [ 3 ]{ 100x }\))

Question 8.
1346
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 1

Question 9.
250
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 2

Question 10.
5112
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 4

Question 11.
9800
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 5

Question 12.
732
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 6

Question 13.
7342
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 7

Question 14.
133100
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 9

Question 15.
37800
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 10

Question 16.
0.27
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 11

Question 17.
8.6
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 12

Question 18.
0.86
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 14

Question 19.
8.65
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 15

Question 20.
7532
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 16

Question 21.
833
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 17

Question 22.
34.2
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 19

Question 23.
What is the length of the side of a cube whose volume is 275 cm³. Make use of the table for the cube root.
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 20

Hope given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G

March 26, 2023

RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 5 Fractions Ex 5G.

Other Exercises

Objective Questions :
Tick the correct answer in each of the following :

Question 1.
Solution:
(c) ∴ canceling the common factor 2, we get \(\\ \frac { 3 }{ 5 } \)

Question 2.
Solution:
(c) ∴ multiplying numerator and denominator by 4, we get \(\\ \frac { 8 }{ 12 } \)

Question 3.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 3.1

Question 4.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 4.1

Question 5.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 5.1

Question 6.
Solution:
(c) each of the fractions has the same denominator.

Question 7.
Solution:
(d) none of these has greater denominator than its numerator.

Question 8.
Solution:
(a) its denominator is greater than its numerator.

Question 9.
Solution:
(b) their numerators are same and 4 < 5 , \(\frac { 3 }{ 4 } >\frac { 3 }{ 5 } \)

Question 10.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 10.1

Question 11.
Solution:
(b) In \(\frac { 4 }{ 5 } ,\frac { 2 }{ 7 } ,\frac { 4 }{ 9 } ,\frac { 4 }{ 11 } \) numerator is same then the smallest denominator’s fraction is greater.

Question 12.
Solution:
(a) Denominators are same, then fraction of smallest numerator will be smallest.

Question 13.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 13.1

Question 14.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 14.1

Question 15.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 15.1

Question 16.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 16.1

Question 17.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 17.1

Question 18.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 18.1

Question 19.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 19.1

Question 20.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 20.1

Hope given RS Aggarwal Solutions Class 6 Chapter 5 Fractions Ex 5G are helpful to complete your math homework.

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CA Foundation Business Economics Study Material – Fixed Inputs (Fixed Factors) and Variable Inputs (Variable Factors)

March 26, 2023

CA Foundation Business Economics Study Material Chapter 3 Theory of Production and Cost – Fixed Inputs and Variable Inputs

Fixed Inputs (Fixed Factors) and Variable Inputs (Variable Factors)

Comparison	Fixed Inputs	Variable Inputs
(i) Meaning	The factors which cannot be easily and quickly changed and require long time to make adjustment in them with the changes in the level of output are called fixed inputs or fixed factors of production. In other words, factor inputs whose quantity does not vary from day-to-day are called as fixed inputs.	The factors which can be easily and quickly changed and readily adjusted with the changes in the level of output are called variable inputs or variable factors of production. In other words, factor inputs whose quantity may vary from day-to-day are called as variable inputs.
(ii) Examples	Examples of fixed inputs – buildings, machinery, plant, top management, etc. It requires long time to make variations in them. E.g. To construct a new factory building with a larger area and capacity.	Examples of variable inputs – ordinary labour, raw-material, power, fuel chemicals, etc. It can be readily changed.
(iii) Relation with Output	Fixed inputs do not vary with the level of output. Its quantity remains the same, whether the output is more or less or zero in SHORT RUN	Variable inputs vary directly with the level of output. Such factors are required more, when output is more; less, when output is less and zero, when output is zero in SHORT RUN.
(iv) Cost	The cost of the fixed inputs is called FIXED COST. In the short run the firm has to bear the fixed cost even if the output is zero. Since the quantity of fixed inputs remains the same, fixed cost remains the same whatever be the level of output.	The cost of the variable inputs is called VARIABLE COST. Since variable inputs vary directly with the level of output, variable costs are also positively related with output. If output is zero, variable cost is also zero. If output is increased variable cost also increases and vice-versa.

Short Run (Short Period) & Long Run (Long Period)

Comparison	Short Run	Long Run
(i) Meaning	The short run is defined as the period of time in which some factors of production or at least one factor is fixed i.e. does not vary with output. Thus, in the short period some factors are FIXED FACTORS E.g. Factory building, machinery, management, etc. and some are VARIABLE FACTORS E.g. Labour, raw-material, power, fuel, etc.	The long run is defined as the period of time in which all factors may vary. In the long run, all factors become variable and so there is no distinction between fixed and variable factors.
(ii) Scale of Production OR Size of the Firm	In the short run, the output is produced with a GIVEN SCALE OF PRODUCTION i.e. the size of plant or firm (and so the production capacity) remains unchanged. Hence, production can be increased or decreased only by changing the amount of variable factors.	In the long run, the output is produced with the CHANGE IN THE SCALE OF PRODUCTION i.e. the size of plant or firm can be increased (and so the production capacity). Hence, production can be increased by varying all factors i.e. fixed factors (of short period) as well as variable factors.
(iii) Production Law	The production function which is studied in the short run period is called as the Law of Variable Proportions.	The production function which is studied in the long run period is called as the Law of Returns to Scale.
(iv) Decisions about Change in factors	The decisions to change the amount of variable factors (like raw material, labour, etc.) are taken very frequently depending upon changes in demand of the commodity. Hence, short run is the ‘ACTUAL PRODUCTION PERIOD’ during which some factors are fixed while some are variable. Thus, firms operate in the short run period.	The decisions to change the amount of fixed factors i.e. scale of production or to close down the firm are taken only once in a while. Hence, long run is the ‘PLANNING PERIOD’. Thus, firms plan in the long run period.
(v) Nature of Supply	In the short run period, supply can be adjusted upto a limited extent as per changes in demand. In other words, supply is relatively inelastic.	In the long run period, supply can be fully adjusted as per changes in demand. In other words, supply is relatively elastic.
(vi) Nature of Cost	In short run period, cost is classified as FIXED COST and VARIABLE COST. Fixed cost is the cost of fixed inputs and Variable cost is the cost of variable inputs. Fixed cost is the main feature of short run period	In long run period ALL COSTS ARE VARIABLE. Variable cost is the main feature of long run period.
(vii) Effect on Price	In short-run, the price determination of a commodity is more influenced by – (a) The demand forces than supply forces because supply in short-run is relatively inelastic, and (b) The UTILITY of the commodity. The short-run price is called SUB-NORMAL PRICE	In long-run, the price determination of a commodity is more influenced by- (a) The supply forces than demand forces because supply in long-run is relatively elastic, and (b) The COST OF PRODUCTION of the commodity. The long-run price is called NORMAL PRICE.
(viii) Average Cost Curve	The short-run average cost curve is ‘U’ shaped. Its U-shape is explained with the Law of Variable Proportions.	The long-run average cost curve is also U shaped. But its U- shape is not as prominent as short-run average cost curve. Its U-shape is explained with the Law of Returns to Scale. Long-run average cost curve is also called ‘PLANNING CURVE’ and ‘ENVELOPE CURVE’.
(ix) Profit of Firms	In the short-run period – The firms under perfect competition on being at equilibrium may earn normal profits, super normal profits or incur losses; The monopoly firm on being at equilibrium may earn normal profits, super normal profits or incur losses; The firms under monopolistic competition on being at equilibrium may earn normal profits, super normal profits or incur losses.	In the long run period- The firms under perfect competition earn only NORMAL PROFITS and operate at optimum level. The monopoly firm can earn SUPER NORMAL PROFITS and operate at sub-optimum level. The firms under monopolistic competition earn only NORMAL PROFITS and operate at sub-optimum level.

The Comet 1 Summary in English by Jayant Narlikar

March 26, 2023

We have decided to create the most comprehensive English Summary that will help students with learning and understanding.

The Comet 1 Summary in English by Jayant Narlikar

The Comet 1 Summary in English

Duttada was not a scientist or astronomer. But star-gazing was his pastime. He used to spend long nights studying the stars. His one ambition in life was to discover a new comet. He longed for enough money to buy a good telescope and also free time to study the stars. He got them both after retirement. He bought an eight-inch telescope to fulfil his secret ambition. He knew that comets come from the solar system. Like other planets they too go round the sun. But their movement is unusual. They could disappear for ages.

Indrani Debi, his wife, cursed the telescope. She called it Dibya or Dibya Chakshu i.e. Divine Eye. She thought that the telescope was another woman who had trapped her husband. It made him careless about his own well-being and the practical problems of living. He did not take precautions against cold.

The professional astronomers used big telescopes. Duttada’s eight-inch Dibya stood very little chance of spotting a new comet. Still Duttada was hopeful. He knew that the great scientists looked chiefly at faint stars. They missed such an insignificant thing as comet.

One night Duttada noticed a new comet. Two days later the news appeared in the paper that a Calcutta man Manoj Dutta had seen a comet and it was reported at Indian Institute of Astrophysics. This comet was heading towards the earth, and it would be seen with naked eyes in the next few months. It was named ‘Comet Dutta.’ He became famous, a celebrity.

There were felicitations for Duttada. But he didn’t like functions organised in his honour. He wished he had not discovered the new comet. His wife agreed but for a different reason. She was highly educated, still supertitious. She feared that comets brought ill-luck and calamities on the earth.

At Cambridge University, Dr James Forsyth got a message to see John Macpherson, Defence Science Advisor to the British Government. Sir John handed him an article which James had written and sent for publication to Nature’. It was very important because James had forewarned that Comet Dutta would hit the Earth and destroy it.

The collision could be avoided if it got broken into pieces while coming near the sun or it might collide with some other comet. The tragedy could occur in just ten months.

It was decided to call an urgent meeting of experts to find out ways to save the Earth. They had to do something to push the deadly comet out of the way.

The Comet 1 Summary in Hindi

दत्तादा कोई वैज्ञानिक या खगोलविद् नहीं थे। पर तारों को देखना समझना उनका शौक था। वे लम्बी रातें तारों का अध्ययन करने में बिताया करते थे। उनके जीवन में एक महत्वाकांक्षा थी कि एक नये धूमकेतु को खोज ले। उनकी तम्मन्ना थी कि उनके पास इतना पैसा हो जाये कि वह एक अच्छी दूरबीन खरीद ले तथा सितारों का अध्ययन करने के लिये उनके पास अवकाश हो। ये दोनों चीजें उन्हें नौकरी से अवकाश के बाद मिल गई। उन्होंने एक आठ इंच वाली दूरबीन अपनी गुप्त इच्छा की पूर्ति के लिये खरीद ली। वह जानते थे कि धूमकेतू सौरमण्डल से आते हैं। अन्य ग्रहों की भाँति वे भी सूर्य की परिक्रमा करते हैं। पर उनकी परिक्रमा असामान्य तरीके की होती है। वे युगों तक दृष्टि से ओझल हो जाते हैं।

उनकी पत्नी इन्द्राणी देवी उस दूरबीन को कोसती रहती थी। वह उसे दिव्या या दिव्य चक्षु कहती थी। उनके विचार से दूरबीन उनकी सौतन थी जिसने उनके पति को फाँस रखा था। दूरबीन उन्हें अपनी ही खोजखबर नहीं रखने देती, न ही वह जीने की व्यावहारिक समस्याओं की ओर ध्यान देते थे। वह ठण्ड से अपना बचाव भी नहीं करते थे।

पेशेवर खगोल शास्त्री बड़ी दूरबीनों का इस्तेमाल करते थे। उनके सामने भला आठ इंच दूरबीन से किस प्रकार नये-नये पुच्छल तारों को देख पाना संभव था। फिर भी दत्तादा को आशा थी। वह जानते थे कि बड़े वैज्ञानिक तो प्रमुख रुप से मद्धिम तारों पर दृष्टि रखते हैं। वे धूमकेतु जैसी तुच्छ चीजों को अनदेखा कर देते हैं।

एक रात को एक नया धूमकेतू दिख गया। दो दिन बाद यह समाचार अखबार में छपा कि एक कलकत्तावासी मनोज दत्ता ने एक नया धूमकेतु खोज लिया है और उसकी सूचना बैंगलौर स्थित भारतीय खगोल भौतिकी संस्थान को दे दी है। उनकी खोज की पुष्टि हो गई। वह धूमकेतु पृथ्वी की ओर आ रहा है और कुछेक माह में नंगी आँखों से देखा जा सकेगा। इसका नाम रखा गया ‘दत्ता धूमकेतु’। वह विख्यात हस्ती हो गये।

दत्तादा को बधाईयाँ दी गईं। परन्तु अपने सम्मान में आयोजित समारोह उसे पसन्द नहीं थे। उनकी इच्छा थी कि उन्हें वह धमकेतु, नहीं खोजना चाहिए था। उनकी पत्नी भी उनसे सहमत हो गयी परन्तु उसका कारण अलग था। वह अत्यधिक शिक्षित थी, परन्तु अन्धविश्वासी थी। उसे भय था कि धूमकेतु पृथ्वी पर दुर्भाग्य तथा आपदाएँ लाते हैं।

केम्ब्रिज विश्वविद्यालय में Dr. James Forsyth को संदेश मिला कि वह अंग्रेजी सरकार के सुरक्षा सलाहकार John Macpherrson से आकर मिले। सर जॉन ने एक लेख उन्हें दिखाया जिसे James ने ही लिखकर ‘Nature’ पत्रिका में छपने के लिए भेजा था। यह बहुत महत्त्वपूर्ण था क्योंकि जेम्स ने भविष्यवाणी की थी कि दत्ता-कॉमेट पृथ्वी से टकरायेगा तथा पृथ्वी को नष्ट कर देगा।

इस टकराव को टाला जा सकता है यदि वह सूर्य के निकट आने पर चूर-चूर हो जाये अथवा वह किसी अन्य धूमकेतु से टकरा कर नष्ट हो जाये। यह त्रासदी दस माह में घट सकती है।

यह निर्णय लिया गया कि पृथ्वी को बचाने के लिये विशेषज्ञों की एक सभा बुलाई जाये। उन्हें घातक धूमकेतु को अपने रास्ते से हटाने के लिये कोई न कोई उपाय करना जरुरी था।

NCERT Solutions for Class 10 Hindi

March 26, 2023

NCERT Solutions for Class 10 Hindi are the part of NCERT Solutions for Class 10. Here we have given Class 10 Hindi NCERT Solutions.

NCERT Solutions for Class 10 Hindi Course A and B are designed by our experienced subject teachers according to the CBSE Latest Hindi Syllabus and guidelines. These solutions are prepared in a conceptual manner for easy understanding of the subject concepts. Students can learn all chapters from NCERT Class 10 Hindi Solutions and perform well in their final exams.

Students will understand how to answer a particular question in the exam after preparing the concepts from NCERT Solutions for Grade 10 Hindi. You can cover Hindi Course-A Kshitij Bhag 2 (17 chapters), Kritika Bhag 2 (5 chapters), and Hindi Course-B Sparsh Bhag 2(17 chapters), Sanchayan Bhag 2 (3 chapters) in the NCERT Solutions PDF links available here. So download them and ace up your preparation.

Download Chapterwise NCERT Solutions for Class 10 Hindi in PDF Free

Each and every chapter in the NCERT textbook for Class 10 Hindi are explained in a simple language for students to understand easily and prepare all answers properly. These NCERT 10th Hindi Textbook Solutions help you to practice all exercise questions and score well in the main examinations. Overall, Class 10 Hindi NCERT Solutions PDF for all chapters clear all your doubts and make you learn each and every point perfectly.

So, download the best study resources ie., NCERT Solutions for Class Hindi Course A and B in PDF format by tapping on the respective links provided here for future reference.

NCERT Solutions for Class 10 Hindi – A

NCERT Solutions for Class 10 Hindi Kshitij Bhag 2 क्षितिज भाग 2

काव्य – खंड

गद्य – खंड

NCERT Solutions for Class 10 Hindi Kritika Bhag 2 कृतिका भाग 2

NCERT Solutions for Class 10 Hindi – B

NCERT Solutions for Class 10 Hindi Sparsh Bhag 2 स्पर्श भाग 2

काव्य – खंड

गद्य – खंड

NCERT Solutions for Class 10 Hindi Sanchayan Bhag संचयन भाग 2

Benefits of Studying NCERT Class 10 Hindi Solutions Book PDF

Candidates of class 10 can find these solutions book very handy at the time of revision.
Students can identify and examine their performance by using the NCERT Solutions Study Book.
Also, they can spend the required amount of time for the weak areas and enhance their subject knowledge.
All main concepts and poetry lessons can easily understandable by the students while studying from NCERT Solutions PDF.
You can even complete your given Homework and assignments based on NCERT textbook solutions.

FAQs on NCERT Solutions for Class 10 Hindi PDF

1. How helpful are the NCERT Solutions for Class 10 Hindi?

NCERT Solution book is very helpful and essential for students’ exam preparation. With NCERT Class 10 Hindi Solutions, you can easily learn and understand all chapters quickly & easily. These solutions are written by subject experts as per the latest CBSE 10th Hindi syllabus. Also, you can find many benefits while preparing from NCERT Solutions Class 10 Hindi.

2. How many chapters are there in Class 10 Hindi?

There are 42 chapters total in both Class X Hindi Course A and B. In Hindi Course-A, Kshitij Bhag 2 (17 chapters), Kritika Bhag 2 (5 chapters). In Hindi Course-B, Sparsh Bhag 2(17 chapters), Sanchayan Bhag 2 (3 chapters).

3. How to download NCERT Solutions of grade 10 Hindi in PDF?

All students of class 10 can download 10th standard Hindi course A & B NCERT Solutions in PDF format by clicking on the available direct links over here. You can access these PDF links free of cost so download and prepare well.

Conclusion

We believe the details shared about the NCERT Solutions for Class 10 Hindi PDF will benefit you a lot. If you need any assistance to download NCERT 10th Hindi Solutions PDF, feel free to share it with us via the comment section. We’ll revert back to you soon with the best possible solutions meanwhile, visit our site learninsta.com and get updated info on NCERT Solutions for Class 10 subjects.

CA Foundation Business Economics Study Material – Production Function

March 26, 2023

CA Foundation Business Economics Study Material Chapter 3 Theory of Production and Cost – Production Function

Production Function

Output is a function of inputs i.e. factor services such as land, labour and capital which are used in production. In other words, production is a transformation of PHYSICAL INPUTS into PHYSICAL OUTPUT.
The functional relationship between physical inputs and physical output, per unit of time under a given state of technology is called production function.
It can also be expressed in the form of a mathematical equation in which output is the dependent variable and inputs are the independent variables.
Q = f (a, b, c ………… n)
Where –
Q denotes quantity of output of a commodity per unit of time
f stands for function of i.e. depends on a, b, c,… n denotes quantity of various inputs.

Assumptions of Production Function:
The production function is based on the following assumptions:

It is specified with reference to a specified period of time.
It is assumed that the state of technology remains the same, during the period of time.
It is assumed that the firm uses best and most efficient technique available in production.
It is assumed that the factors of production are divisible into viable units.

The production function can be explained under two heads:
1. The short run production function in which input – output relations are analysed where –

One input is variable, all other inputs are fixed, (described as the Law of Variable Proportions) OR
Two inputs are variable, all other factors are fixed (explained with the help of isoquants)

2. The long run production function in which input- output relations are analysed where all the inputs are variable (described as the Law of Returns to Scale).

Cobb-Douglas Production Function
Q = f (L, K).
Where –
Q = Output; L = Labour; K = Capital

Paul H. Douglas and C.W. Cobb of the U.S.A. studied the production function of the American manufacturing industries. This production function applies to the whole of manufacturing in U.S.A. rather than to an individual firm. In this case, output is manufacturing production and inputs used are labour and capital.

The conclusion of study is that labour contributed 3 /4th and capital about 1 /4th in the manufacturing production.