RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1

RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1

Other Exercises

Question 1.
A coin is tossed 1000 times with the following frequencies
Head : 455, Tail : 545.
Compute the probability for each event.
Solution:
Total number of events (m) 1000
(i) Possible events (m) 455
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 455 }{ 1000 } \)
= \(\frac { 91 }{ 200 } \) = 0.455
(ii) Possible events (m) = 545
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 545 }{ 1000 } \) = \(\frac { 109 }{ 200 } \) = 0.545

Question 2.
Two coins are tossed simultaneously 500 times with the following frequencies of different
outcomes:
Two heads : 95 times
One tail : 290 times
No head: 115 times
Find the probability of occurrence of each of these events.
Solution:
Two coins are tossed together simultaneously 500 times
∴ Total outcomes (n) 500
(i) 2 heads coming (m) = 95 times
∴Probability P(A) = \(\frac { m }{ n } \)
= \(\frac { No. of possible events }{ Total number of events } \)
= \(\frac { 95 }{ 500 } \) = \(\frac { 19 }{ 100 } \) = 0.19
(ii) One tail (m) = 290 times
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 290 }{ 500 } \) = \(\frac { 580 }{ 1000 } \) = \(\frac { 58 }{ 100 } \) = 0.58
(iii) No head (m) = 115 times
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 115 }{ 500 } \) = \(\frac { 23 }{ 100 } \) = 0.23

Question 3.
Three coins are tossed simultaneously 1oo times with the following frequencies of different outcomes:
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 3.1
If the three coins are simultaneously tossed again, compute the probability of:
(i) 2 heads coming up.
(ii) 3 heads coming up.
(iii) at least one head coming up.
(iv) getting more heads than tails.
(v) getting more tails than heads.
Solution:
Three coins are tossed simultaneously 100 times
Total out comes (n) = 100
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 3.2
(i) Probability of 2 heads coming up (m) = 36
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 36 }{ 100 } \) = 0.36
(ii) Probability of 3 heads (m) = 12
ProbabilityP(A)= \(\frac { m }{ n } \) = \(\frac { 12 }{ 100 } \) = 0.12
(iii) Probability of at least one head coming up (m) = 38 + 36 + 12 = 86
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 86 }{ 100 } \) = 0.86
(iv) Probability of getting more heads than tails (m) = 36 + 12 = 48
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 48 }{ 100 } \) = 0.48
(v) Getting more tails than heads (m) = 14 + 38 = 52
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 52 }{ 100 } \) = 0.52

Question 4.
1500 families with 2 children were selected randomly and the following data were recorded:
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 4.1
If a family is chosen at random, compute the probability that it has:
(i) No girl
(ii) 1 girl
(iii) 2 girls
(iv) at most one girl
(v) more girls than boys
Solution:
Total number of families (n) = 1500
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 4.2
(i) Probability of a family having no girls (m) = 211
∴Probability P(A)= \(\frac { m }{ n } \) = \(\frac { 211 }{ 1500 } \) = 0.1406
(ii) Probability of a family having one girl (in) = 814
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 814 }{ 1500 } \) = 0.5426
(iii) Probability of a family having 2 girls (m) = 475
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 475 }{ 1500 } \) = 0.3166
(iv) Probability of a family having at the most one girls
∴m = 814 + 211 = 1025
∴Probability P(A) =\(\frac { m }{ n } \) = \(\frac { 1025 }{ 1500 } \) = 0.6833
(v) Probability of a family having more girls than boys (m) = 475
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 475 }{ 1500 } \) = 0.3166

Question 5.
In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that on a ball played:
(i) he hits boundary
(ii) he does not hit a boundary.
Solution:
Total balls played (n) 30
No. of boundaries = 6
(i) When the batsman hits the boundary = 6
∴m = 6
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 6 }{ 30 } \) = \(\frac { 1 }{ 5 } \) = 0.2
(ii) When the batsman does not hit the boundary (m) = 30 – 6 = 24
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 24 }{ 30 } \) = \(\frac { 4 }{ 5 } \) = 0.8

Question 6.
The percentage of marks obtained by a student in monthly unit tests are given below:
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 6.1
Find the probability that the student gets:
(i) more than 70% marks
(ii) less than 70% marks
(iii) a distinction.
Solution:
Percentage of marks obtain in
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 6.2
(i) Probability of getting more than 70% marks (m) = In unit test II, III, V = 3
Total unit test (n) = 5
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 3 }{ 5 } \) = 0.6
(ii) Getting less then 70% marks = units test I and IV
∴m = 2
Total unit test (n) = 5
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 2 }{ 5 } \) = 0.4
(iii) Getting a distinction = In test V (76 of marks)
∴m = 1
Total unit test (n) = 5
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 1 }{ 5 } \) = 0.2

Question 7.
To know the opinion of the students about Mathematics, a survey of 200 students was conducted. The data is recorded in the following table:
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 7.1
Find the probability that a student chosen at random
(i) likes Mathematics
(ii) does not like it.
Solution:
Total number of students (n) = 200
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 7.2
(i) Probability of students who like mathematics (m) = 135
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 135 }{ 200 } \) = 0.675
(ii) Probability of students who dislike mathematics (m) = 65
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 65 }{ 200 } \) = 0.325

Question 8.
The blood groups of 30 students of class IX are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O,
A student is selected at random from the class from blood donation. Find the probability that the blood group of the student chosen is:
(i) A (ii) B (iii) AB (iv) O
Solution:
Total number of students of IX class = 30
No. of students of different blood groups
A AB B O
9 3 6 12
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 8.1

Question 9.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour
(in kg):
4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Solution:
Number of total bags (n) = 11
No. of bags having weight more than 5 kg (m) = 7
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 7 }{ 11 } \)

Question 10.
Following table shows the birth month of 40 students of class IX.
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 10.1
Find the probability that a student was born in August.
Solution:
Total number of students (n) = 40
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 10.2
Number of students who born in Aug. (m) = 6
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 6 }{ 40 } \) = \(\frac { 3 }{ 20 } \)

Question 11.
Given below is the frequency distribution table regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days.
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 11.1
Find the probability of concentration of sulphur dioxide in the interval 0.12 – 0.16 on any of these days.
Solution:
Total number of days (n) = 30
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 11.2
Probability of cone, of S02 of the interval 0.12-0.16 (m) = 2
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 2 }{ 30 } \) = \(\frac { 1 }{ 15 } \)

Question 12.
A company selected 2400 families at random and survey them to determine a relationship between income level and the number of vehicles in a home. The information gathered is listed in the table below:
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 12.1
If a family is chosen, find the probability that the family is:
(i)earning Rs 10000-13000 per month and owning exactly 2 vehicles.
(ii)earning Rs 16000 or more per month and owning exactly I vehicle.
(iii)earning less than Rs 7000 per month and does not own any vehicle.
(iv)earning Rs 13000-16000 per month and owning more than 2 vehicle.
(v)owning not more than 1 vehicle.
(vi)owning at least one vehicle.
Solution:
Total number of families (n) = 2400
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 12.2
(i) Number of families earning income Rs 10000-13000 and owning exactly 2 vehicles (m) = 29
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 29 }{ 2400 } \)
(ii) Number of families earning income Rs 16000 or more having one vehicle (m) = 579
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 579 }{ 2400 } \)
(iii) Number of families earning income less than Rs 7000 having no own vehicle (m) = 10
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 10 }{ 2400 } \) = \(\frac { 1 }{ 240 } \)
(iv) Number of families having X13000 to X16000 having more than two vehicles (m) = 25
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 25 }{ 2400 } \) = \(\frac { 1 }{ 96 } \)
(v) Number of families owning not more than one vehicle (m)
= 10 + 1 + 2 + 1 + 160 + 305 + 533 + 469 + 579 = 2062
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 2062 }{ 2400 } \) = \(\frac { 1031 }{ 1200 } \)
(vi) Number of families owning at least one vechile (m) = 2048 + 192 + 110 = 2356
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 2356 }{ 2400 } \) = \(\frac { 589 }{ 600 } \)

Question 13.
The following table gives the life time of 400 neon lamps:
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 13.1
A bulb is selected at random. Find the probability that the life time of the selected bulb is: (i) less than 400 (ii) between 300 to 800 hours (iii) at least 700 hours.
Solution:
Total number of neon lamps (n) = 400
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 13.2
A bulb is chosen:
(i)No. of bulbs having life time less than 400 hours (m) = 14
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 14 }{ 400 } \) = \(\frac { 7 }{ 200 } \)
(ii)No. of bulbs having life time between 300 to 800 hours (m) = 14 + 56 + 60 + 86 + 74 = 290
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 290 }{ 400 } \) = \(\frac { 29 }{ 40 } \)
(iii)No. of bulbs having life time at least 700 hours (m) = 74 + 62 + 48 = 184
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 184 }{ 400 } \) = \(\frac { 23 }{ 50 } \)

Question 14.
Given below is the frequency distribution of wages (in Rs) of 30 workers in a certain factory:
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 14.1
A worker is selected at random. Find the probability that his wages are:
(i) less than Rs 150
(ii) at least Rs 210
(iii) more than or equal to 150 but less than Rs 210.
Solution:
Number of total workers (n) = 30
RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 14.2
A worker is selected.
(i)No. of workers having less than Rs 150 (m) = 3 + 4 = 7
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 7 }{ 30 } \)
(ii)No. of workers having at least Rs 210 (m) = 4 + 3 = 7
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 7 }{ 30 } \)
(iii)No. of workers having more than or equal to Rs 150 but less than Rs 210 = 5 + 6 + 5 = 16
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 16 }{ 30 } \) = \(\frac { 8 }{ 15 } \)

 

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NCERT Exemplar Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life

NCERT Exemplar Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life

Question 1.
Why are lysosomes known as “suicide bags” of a cell ?
Answer:
Lysosomes contain digestive enzymes against all types of organic materials. If their covering membrane breaks as it happens during injury to cell, the digestive enzymes will spill over the cell contents and digest the same. As lysosomes are organelles which on bursting can kill the cells possessing them, they are called suicide bags.

More Resources

Question 2.
Do you agree that “A cell is a building unit of an organism” ?
Answer:
Yes. Cell is a building unit of every living organism as every living being is made up of one or more cells. In unicellular or acellular organisms, the single cell performs all the functions of life. In multicellular organisms all the cells have a similar basic structure and perform similar basic life activities. However, they become specialised to form components of different structures that perform different functions. Cells are first organised into tissues, each with a specific function, e.g., contraction by muscular
tissue. Tissues are organised to form organs with each organ performing a specific fonction, e.g., heart, stomach, kidney. Organs are grouped into organ systems, each with a major fonction, e.g., circulatory system, excretory system, respiratory system. A living being has a number of organ systems. However, in all such organisational complexity, cell remains the basic building unit of the organism.

Question 3.
Why does the skin of your fingers shrink when you wash clothes for a long time ?
Answer:
Clothes are washed with soap or detergent solution. This solution is hypertonic as composed to osmotic concentration of our skin cells. The washing solution, therefore, causes exosmosis in the skin cells that come in contact with it for some time. Because of it, the skin over the fingers shrinks while washing clothes for a long time.

Question 4.
Why is endocytosis found in animals only ? (CCE 2012)
Answer:
Endocytosis is engulfment of food and other substances from external medium by plasma membrane. This is possible only when plasma membrane is in direct contact with external medium. It occurs only in animal cells. In plant cells, a cell wall is present over the plasma membrane. Therefore, their plasma membrane cannot perform endocytosis.

Question 5.
A person takes concentrated solution of salt. After some time he starts vomiting. What is the phenomenon responsible for such a situation ? Explain.
Answer:
Concentrated salt solution causes irritation and excessive dehydration in the wall of alimentary canal due to exosmosis. There is uncomfortable stretching which causes reverse movements and hence vomiting.

Question 6.
Name any cell organelle which is non-membranous.
Answer:
Ribosome.

Question 7.
We eat food composed of all the nutrients like carbohydrates, proteins, fats, vitamins, minerals and water. After digestion, they are absorbed in the form of glucose, amino acids, fatty acids, glycerol, etc. What mechanisms are involved in the absorption of digested food and water ?
Answer:
Digested Food :

  1. Glucose, amino acids and some ions — Active transport.
  2. Fatty acids, Glycerol — Diffusion (Passive transport).
    Water : Osmosis.

Question 8.
If you are provided with some vegetables to cook, you genenrally add salt into vegetables during cooking process. After adding salt, vegetables release water. What mechanism is responsible for this ? (CCE 2014)
Answer:
On adding salt, vegetables release water due to exosmosis. Exosmosis occurs whenever the external medium is hypertonic as compared to the osmotic concentration inside living cells.

Question 9.
If cells of Onion peel and RBC are separately kept in hypotonic solution what among the following will take place ? Explain the reason for your answer,
(a) Both the cells will swell
(b) RBC will burst easily while cells of Onion peel will resist the bursting to some extent
(c) a and b both are correct
(d) RBC and Onion peel cells will behave similarly.
Answer:
(b) RBC will burst as there is no mechanism to restrict entry of water into them. Onion peel cells do not burst. Endosmosis causes some initial swelling but cell wall puts a mechanical barrier to further swelling and entry of water. Therefore, they do not burst.

Question 10.
Bacteria do not have chloroplast but some bacteria are photoautotrophic in nature and perform photosynthesis. Which part of bacterial cell performs this ?
Answer:
Photoautotrophic bacteria posses photosynthetic pigments inside small vesicles which may be attached to the plasma membrane.

Question 11.
Match the items of A and B.
NCERT Exemplar Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image -1
Answer:
a —iv, b —v, c —iii, d —i, e —ii

Question 12.
Write the name of different plant parts in which chloroplast, chromoplast and leucoplast are present.
Answer:

  1. Chromoplast : Flower (petals) and fruits
  2. Chloroplast : Green leaves and green parts of young stems.
  3. Leucoplast : Root and underground stem .

Question 13.
Name the organelles which show analogy written as under
(a) Transporting channels of the cell
(b) Power house of the cell
(c) Packaging and despatching unit of the cell
(d) Digestive bag of the cell
(e) Storage sac of the cell
(f) Kitchen of the cell
(g) Control room of the cell
Answer:
(a) Endoplasmic reticulum
(b) Mitochondria
(c) Golgi apparatus
(d) Lysosome
(e) Vacuole
(f) Chloroplast
(g) Nucleus.

Question 14.
How is bacterial cell different from an Onion peel cell ?
Answer:
See differences between prokaryotic (bacterial cell) and eukaryotic (Onion peel cell) cells.
NCERT Exemplar Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image -2

Question 15.
How do substance like carbon dioxide and water move in and out of the cell ?
Answer:
CO2 moves into and out of cells by diffusion while water does it through osmosis.
Diffusion. It is movement of particles of various substances from the region of their higher concentration to the region of their lower concentration,

  1. In a respiring cell, more CO2 is produced internally. As a result its internal concentration rises. As concentration of CO2 is lower in the outside medium, CO2 passes out from cell into external medium,
  2. In photosynthetic cell, CO2 is being consumed in photosynthesis. Its intracellular concentration is lower than outside medium. Therefore, CO2 diffuses from outside to inside of the cell.

Question 16.
How does Amoeba obtain its food ?
Answer:
Plasma membrane of Amoeba is flexible. With its help, Amoeba engulfs food particle. The engulfed food particle passes into the body of Amoeba as a phagosome. Phagosome combines with lysosome to produce digestive or food vacuole. Digestion occurs in food vacuole. The digested food passes into surrounding cytoplasm. The undigested matter is thrown out of the cell in exocytosis.

Question 17.
Name two organelles in the plant cell that contain their own genetic materials and ribosomes.
Answer:

  1. Plastids
  2. Mitochondria.

Question 18.
Why are lysosomes also known as “scavengers—waste disposal system of the cell ?” (CCE 2011)
Answer:
Lysosomes are called scavengers of the cell because they remove cell debris consisting of dead and worn out cell organelles by digesting the same. For this, they possess powerful digestive enzymes capable of breaking down all organic nutrients. Lysosomes also.nourish the cells by sending out digested nutrients into the cytoplasm.

Question 19.
Which cell organelle controls most of the activities of the cell ?
Answer:
Nucleus, by controlling metabolism and cell activities. Genes express their effect through RNAs. RNAs control synthesis of proteins and enzymes.

Question 20.
Which kind of plastid is more common in
(a) Root of the plant
(b) Leaves of the plant
(c) Flowers and fruits ?
Answer:
(a) Roots : Leucoplasts
(b) Leaves : Chloroplasts
(c) Flowers and Fruits : Chromoplasts.

Question 21.
Why do plant cells possess large sized vacuole ? (CCE 2011)
Answer:
Vacuole of plant cells has to be large because it takes part in

  1. Storage : It stores salts, sugar, amino acids, organic acids and some proteins.
  2. Cellular Wastes : They are dumped in the vacuole.
  3. Lysosomal Enzymes : They occur in the vacuole of plant cells.
  4. Turgidity : The vacuole contains cell sap which provides turgidity to the cells.
  5. Absorption of Water : Plant cell vacuole contains an osmotic concentration required for absorption of water.

Question 22.
How are chromatin, chromatid and chromosomes related to each other ? (CCE 2012)
Answer:
Chromatin is intertwined mass of fine thread like structures made of DNA and protein. During cell division, chromatin condenses to form thicker rod like structures called chromosomes. Each chromosome consists of two similar halves called chromatids. Formation of chromosomes having two similar halves or chromatids is meant for equitable distribution of chromatin which is hereditary material.

Question 23.
What are consequences of the following conditions ?

  1. A cell containing higher water concentration than the surrounding medium.
  2. cell having low water concentration as compared to its surrounding medium.
  3. A cell having equal water concentration to its surrounding medium.

Answer:
High water concentration occurs in hypotonic solution, low water concentration in hypertonic solution while equal water concentration occurs in isotonic solution.

  1. A cell having higher water content or hypotonic cell sap will undergo exosmosis and therefore, lose water. It may undergo plasmolysis.
  2. A cell having low water concentration or hypertonic cell sap will undergo endosmosis and absorb water from outside. It would become turgid.
  3. A cell having isotonic cell sap will neither gain or lose water to the external medium.

Question 24.
Draw a plant cell and label the parts which

  1. Determine function and development of the cell.
  2. Provides resistance to microbes and to withstand hypotonic external medium without injury.
  3. Packages materials coming from the endoplasmic reticulum.
  4. Is a fluid contained inside the nucleus.
  5. Is site for many biochemical reactions necessary to sustain life.

Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image -3

  1. Nucleus
  2. Cell wall
  3. Golgi apparatus
  4. Nucleoplasm
  5. Cytoplasm.

Question 25.
Illustrate only a plant cell as seen under electron microscope. How is it different from animal cell ?
Answer:
(a)
NCERT Exemplar Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image -4
(b) Differences between Plant and Animal Cells.
NCERT Exemplar Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image -5

Question 26.
Draw a neat labelled diagram of an animal cell.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image -6

Question 27.
Draw a well labelled diagram of eukaryotic nucleus. How is it different from nucloid ?
Answer:
(a)
NCERT Exemplar Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image -7
(b) Differences between Nucleus and Nucleoid
NCERT Exemplar Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image -8

Question 28.
Differentiate between rough and smooth endoplasmic recticulum. How is endoplasmic reticulum important for membrane biosynthesis ?
Answer:
(a) Differences.
NCERT Exemplar Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image -9(b) Membrane Biosynthesis : Both the types of E.R. take part in membrane biosynthesis. Rough endoplasmic reticulum synthesises proteins which are passed over to Golgi apparatus for glycolisation. Smooth endoplasmic reticulum forms lipids over its surface. They are also transported to Golgi apparatus. Golgi apparatus builds membrane with the help of lipids and proteins.

Question 29.
In brief, state what happens when

  1. Dry apricots are left for some time in pure water and later transferred to sugar solution.
  2. A red blood cell is kept in concentrated salt solution.
  3. The plasma membrane of a cell breaks down.
  4. Rheo leaves are boiled in water first and then drop of sugar syrup is put on it.
  5. Golgi apparatus is removed from the cell.

Answer:

  1. When placed in pure water, dry apricots swell up the due to endosmosis. On being transferred to sugar solution, they shrink due to exosmosis.
  2. In concentrated salt solution, red blood cell will shrink and give a shrivelled appearance (crenation).
  3. Breakdown of plasma membrane will result in death of the cell as protoplasmic structures will get dispersed.
  4. Boiling shall kill the leaves. The dead leaves and their cells do not undergo plasmolysis.
  5. Formation of lysosomes, secretory and excretory vesicles will stop.

Question 30.
Draw a neat diagram of plant cell and label any three parts which differentiate it from animal cell.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image -10
Show chloroplast, large vacuole and cell wall. The three do not occur in the animal cell.

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RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A

RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A.

Question 1.
Solution:
Number of trials = 500 times
Let E be the no. of events in each case, then
∴No. of heads (E1) = 285 times
and no. of tails (E2) = 215 times
∴ Probability in each case will be
∴(i)P(E1) = \(\frac { 285 }{ 500 } \) = \(\frac { 57 }{ 100 } \) = 0.57
(ii) P(E2) = \(\frac { 215 }{ 500 } \) = \(\frac { 43 }{ 100 } \) = 0.43

Question 2.
Solution:
No. of trials = 400
Let E be the no. of events in each case, then
No. of 2 heads (E1) = 112
No. of one head (E2) = 160 times
and no. of O. head (E3) = 128 times
∴ Probability in each case will be:
∴ (i)P(E1) = \(\frac { 112 }{ 400 } \) = \(\frac { 28 }{ 100 } \) = 0.28
(ii)P(E2) = \(\frac { 160 }{ 400 } \) = \(\frac { 40 }{ 100 } \)= 0.40
(iii) P(E3) = \(\frac { 128 }{ 400 } \) = \(\frac { 32 }{ 100 } \) = 0.32 Ans.

Question 3.
Solution:
Number of total trials = 200
Let E be the no. of events in each case, then
No. of three heads (E1) = 39 times
No. of two heads (E2) = 58 times
No. of one head (E3) = 67 times
and no. of no head (E4) = 36 times
∴ Probability in each case will be .
(i) P(E1) = \(\frac { 39 }{ 200 } \) = 0.195
(ii) P(E3) = \(\frac { 67 }{ 200 } \) = 0.335
(iii) P(E4) = \(\frac { 36 }{ 200 } \) = \(\frac { 18 }{ 100 } \) = 0.18
(iv) P(E2) = \(\frac { 58 }{ 200 } \) = \(\frac { 29 }{ 100 } \) = 0.29

Question 4.
Solution:
Solution No. of trials = 300 times
Let E be the no. of events in each case, then
No. of outcome of 1(E1) = 60
No. of outcome of 2(E2) = 72
No. of outcome of 3(E3) = 54
No. of outcome of 4(E4) 42
No. of outcome of 5(E5) = 39
No. of outcome of 6(E6) = 33
The probability of
(i) P(E3) = \(\frac { 54 }{ 300 } \) = \(\frac { 18 }{ 100 } \) = 0.18
(ii) P(E6) = \(\frac { 33 }{ 100 } \) = \(\frac { 11 }{ 100 } \)= 0.11
(iii) P(E5) = \(\frac { 39 }{ 300 } \) = \(\frac { 13 }{ 100 } \) = 0.13
(iv) P(E1) = \(\frac { 60 }{ 300 } \) = \(\frac { 20 }{ 100 } \)= 0.20 Ans.

Question 5.
Solution:
No. of ladies on whom survey was made = 200.
Let E be the no. of events in each case.
No. of ladies who like coffee (E1) = 142
No. of ladies who like coffee (E2) = 58
Probability of
(1) P(E1) = \(\frac { 142 }{ 200 } \) = \(\frac { 71 }{ 100 } \) = 0.71
(ii) P(E2) = \(\frac { 58 }{ 200 } \) = \(\frac { 29 }{ 100 } \) = 0.29 Ans.

Question 6.
Solution:
Total number of tests = 6
No. of test in which the students get more than 60% mark = 2
Probability will he
P(E) = \(\frac { 2 }{ 6 } \) = \(\frac { 1 }{ 3 } \)Ans.

Question 7.
Solution:
No. of vehicles of various types = 240
No. of vehicles of two wheelers = 64.
Probability will be P(E) = \(\frac { 84 }{ 240 } \) = \(\frac { 7 }{ 20 } \) = 0.35 Ans.

Question 8.
Solution:
No. of phone numbers are one page = 200
Let E be the number of events in each case,
Then (i) P(E5) = \(\frac { 24 }{ 200 } \) = \(\frac { 12 }{ 100 } \) = 0.12
(ii) P(E8) = \(\frac { 16 }{ 200 } \) = \(\frac { 8 }{ 100 } \) = 0.08 Ans.

Question 9.
Solution:
No. of students whose blood group is checked = 40
Let E be the no. of events in each case,
Then (i) P(E0) = \(\frac { 14 }{ 40 } \) = \(\frac { 7 }{ 20 } \) = 0.35
(ii) P(EAB) = \(\frac { 6 }{ 40 } \) = \(\frac { 3 }{ 20 } \) = 0.15 Ans.

Question 10.
Solution:
No. of total students = 30.
Let E be the number of elements, this probability will be of interval 21 – 30
P(E) = \(\frac { 6 }{ 30 } \) = \(\frac { 1 }{ 5 } \) = 0.2 Ans.

Question 11.
Solution:
Total number of patients of various age group getting medical treatment = 360
Let E be the number of events, then
(i) No. of patient which are 30 years or more but less than 40 years = 60.
P(E) = \(\frac { 60 }{ 360 } \) = \(\frac { 1 }{ 6 } \)
(ii) 50 years or more but less than 70 years = 50 + 30 = 80
P(E) = \(\frac { 80 }{ 360 } \) = \(\frac { 2 }{ 9 } \)
(iii) Less than 10 years = zero
P(E) = \(\frac { 0 }{ 360 } \) = 0
(iv) 10 years or more 90 + 50 + 60 + 80 + 50 + 30 = 360

Hope given RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A 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.

Value Based Questions in Science for Class 9 Chapter 5 The Fundamental Unit of Life

Value Based Questions in Science for Class 9 Chapter 5 The Fundamental Unit of Life

These Solutions are part of Value Based Questions in Science for Class 9. Here we have given Value Based Questions in Science for Class 9 Chapter 5 The Fundamental Unit of Life

Question 1.

  1. A fruit is green when unripe but become beautifully coloured when ripe. How does this change occur ?
  2. What is the importance of this change ?
  3. What is mutualism involved ?
  4. Give an example of such a mutualism in our society.

Answer:

  1. An unripe fruit is green because it contains chloroplasts in its skin. Towards ripening of the fruit, the chloroplasts are changed into chromoplasts which give the fruit an attractive non-green (reddish, pink, purple, brownish) colour.
  2. The colour of ripe fruits attracts animals. The coloured fruits are often sweet and fleshy. Animals, like birds, come to feed over the fruits.
  3. While eating the flesh of fruits, the animals pick up the seeds and take them to different places visited by them. This helps in the dispersal of seeds. Therefore, both the animals and the plants are benefitted. It is mutualism.
  4. Mutualism or mutual dependence is the law of human society. No body can live and work alone. Every body is dependent on somebody else for most of one’s activities. For example, household helper, municipal worker, transport staff etc. work for you for which you pay so that the worker is able to run his/her family.

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Question 2.
How does a cell show division of labour ? Is there any parallelism between the working of the cell and our society ?
Answer:
A cell has a number of organelles, viz., chloroplasts, mitochondria, Golgi apparatus, lysosomes, ribosomes, endoplasmic reticulum, nucleus, etc. Every organelle performs its specific function like food manufacture by chloroplasts, energy liberation by mitochondria, protein synthesis by ribosomes, transport by endoplasmic reticulum, control by nucleus. All the organelles coordinate their activities for the smooth functioning of the cell.
Human society has also a number of specialised activities being performed by different groups of persons, e.g., electricians, carpenters, plumbers, masons, transporters, scavengers, traffic controllers, teachers, doctors, agriculturists, engineers, manufactures, shopkeepers, managers, security persons, etc. All of them are however, coordinating with one another towards the smooth running of the society.

Question 3.
What are the functions of cell membrane ? How is the cell membrane able to perform diverse functions ? Give an example of diversity in functioning in any segment of human society.
Answer:
Cell membrane has a number of functions like providing shape to cell, acting as mechanical barrier between cell contents and environment, selective permeability, endocytosis, recognition, flow of information, passage of water, flow of external fluids by cilia or absorption of nutrients by its microvilli.
The diverse functions of cell membrane are possible due to adaptation of proteins to form channels, enzymes, carriers, receptors, etc. and their attachment to small carbohydrates.
Every human being is performing diverse functions. Seema is daughter of her parents, sister of her brother, student of her school, a badminton player, a friend to several classfellows, a companion of her pet, a gardener in home garden, a helper to her mother, a caretaker of her grand parents, an active social worker, etc.

Question 4.
What is the basis of long life of pickles and jams ? What lesson one gets from this fact ?
Answer:
Pickles and jams possess high concentration of osmotically active substances (salt in pickles, sugar in jams). They do not allow the microbial spores to germinate over them. Even on contamination, a microbe cannot survive in them as it will undergo exosmosis and die. Because of being microbe/germ free, pickles and jams do not get easily spoiled.
An active and long life can also occur in humans if they live in hygienic environment, avoid coming in contact with contaminated articles and visiting crowded places.

Hope given Value Based Questions in Science for Class 9 Chapter 5 The Fundamental Unit of Life 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.

HOTS Questions for Class 9 Science Chapter 5 The Fundamental Unit of Life

HOTS Questions for Class 9 Science Chapter 5 The Fundamental Unit of Life

These Solutions are part of HOTS Questions for Class 9 Science. Here we have given HOTS Questions for Class 9 Science Chapter 5 The Fundamental Unit of Life

Question 1.
What will happen if chloroplast is taken out of the cell and illuminated ?
Answer:
Chloroplast is a semiautonomous cell organelle which on illumination can perform its function of photosynthesis and release oxygen even outside the cell provided it is kept in isotonic medium and receives raw material of carbon dioxide.

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Question 2.

  1. Identify the above figure.
    HOTS Questions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 1
  2. Label X and Y
  3. What is the function of X?

Answer:

  1. It is figure of Nucleus.
  2. X- Nucleous.Y-Chromatin.
  3. Function of X. Synthesis of ribosomes.

Question 3.

  1. Identify A and B cells.
    HOTS Questions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 2
  2. What will happen if B cells are kept in hypotonic solution ?
  3. What will happen if A cells are kept in hypertonic solution ?

Answer:

  1. A cells- Turgid cells. B cells- Plasmolysed cells.
  2. B cells kept in hypotonic solution will become deplasmolysed if done so immediately after plasmolysis.
  3. A cells kept in hypertonic solution will become plasmolysed.

Question 4.
What will happen if

  1. Excess amount of fertiliser is added to green grass lawn.
  2. Salt is added to cut pieces of raw mango ?

Answer:

  1. Excess Fertiliser. It will kill grass plants due to exosmosis and plasmolysis.
  2. Salt to mango pieces. It will cause exosmosis. Salt also protects the cut pieces from bacterial and fungal attack.

Question 5.

  1. Label X and Y
    HOTS Questions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 3
  2. What is the function of X ?
    What is the composition of Y?
  3. Identify the above diagram and what is its common name ?

Answer:

  1. X-crista. Y-matrix. iM
  2. Function of X. Increases surface area of inner membrane for ATP synthesis.
  3. Mitochondrion (cut lengthwise). Common Name. Power house of cell.

Question 6.

  1. Label W, X, Y and Z.
    HOTS Questions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 4
  2. What is the covering membrane of X known as ?
  3. Which of them contains hydrolytic enzymes ?
  4. Which one of them takes part in storage, modification and packaging of various chemicals.

Answer:

  1. W-Golgi apparatatus. X-Vacuole. Y-Ribosomes. Z-Lysosome.
  2. Covering membrane of X. Tonoplast.
  3. Hydrolytic Enzymes. Lysosome.
  4. Storage, Modification and Packaging. Golgi apparatus.

Hope given HOTS Questions for Class 9 Science Chapter 5 The Fundamental Unit of Life 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.