RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14D

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14D

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D.

Other Exercises

Question 1.
Solution:
(i) First eight natural numbers are 1, 2, 3, 4, 5, 6, 7, 8
∴ Mean = \(\frac { 1+2+3+4+5+6+7+8 }{ 8 } \) = \(\frac { 36 }{ 8 } \) = \(\frac { 9 }{ 2 } \) = 4.5
(ii) First ten odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
∴ Mean = \(\frac { 1+3+5+7+9+11+13+15+17+179 }{ 10 } \) = \(\frac { 100 }{ 10 } \) = 10
(iii) First five prime numbers are 2, 3, 5, 7, 11.
∴ Mean = \(\frac { 2+3+5+7+11 }{ 5 } \) = \(\frac { 28 }{ 5 } \) = 5.6
(iv) First six even numbers are 2, 4, 6, 8, 10, 12
∴ Mean = \(\frac { 2+4+6+8+10+12 }{ 10 } \) = \(\frac { 42 }{ 6 } \) = 7
(v) First seven multiples of 5 are 5, 10, 15, 20, 25, 30, 35
∴ Mean = \(\frac { 5+10+15+20+25+30+35 }{ 7 } \) = \(\frac { 140 }{ 7 } \) = 20
(vi) All the factors of 20 are, 1, 2, 4, 5, 10, 20
∴ Mean = \(\frac { 1+2+4+5+10+20 }{ 6 } \) = \(\frac { 42 }{ 6 } \) = 7

Question 2.
Solution:
No. of families (n) = 10
Sum of children (∑x1) = 2 + 4 + 3 + 4 + 2 + 0 + 3 + 5 + 1 + 6 = 30
∴ Mean \(\left( \overline { x } \right) =\frac { \sum { x1 } }{ n } =\frac { 30 }{ 10 } =3\)

Question 3.
Solution:
Here number of days (n) = 7
Number of books (∑x1) = 105 + 216 + 322 + 167 + 273 + 405 + 346 = 1834
∴ Mean \(\left( \overline { x } \right) =\frac { \sum { x1 } }{ n } =\frac { 1864 }{ 7 } =262 books\)

Question 4.
Solution:
Number of days (n) = 6
Sum of temperature (∑x) = 35.5 + 30.8 + 27.3 + 32.1 + 23.8 + 29.9 = 179.4°F
∴ Mean temperature = \(\frac { \sum { x } }{ n } =\frac { 179.4 }{ 6 } \)=29.9°F

Question 5.
Solution:
Number of students (n) = 12
Sum of marks (∑x) = 64 + 36 + 47 + 23 + 0 + 19 + 81 + 93 + 72 + 35 + 3 + 1 = 474
= \(\frac { \sum { x1 } }{ n } =\frac { 474 }{ 12 } \) = 39.5

Question 6.
Solution:
Here, n = 6
and arithmetic mean =13
Total sum = 13 x 6 = 78.
But sum of 7 + 9+ 11 + 13 + 21=61
Value of x = 78 – 61 = 17
Hence x = 17 Ans.

Question 7.
Solution:
Let x1, x2, x3, … x24 be the 24 numbers
\(\frac { x1+x2+x3+…..+x24 }{ 24 } =35\)
=> x1 + x2 + x3 +….+ x24 = 35 x 24
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14D Q7.1

Question 8.
Solution:
Let x1 + x2 + x3……..x20 be the 20 numbers
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14D Q8.1

Question 9.
Solution:
Let x1, x2, x3 … x15 be the numbers
Mean = \(\frac { x1+x2+x3+…..+x15 }{ 15 } = 27\)
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14D Q9.1

Question 10.
Solution:
Let x1, x2, x3 … x12 be the numbers
Mean = \(\frac { x1+x2+x3+…..+x12 }{ 12 }\) = 40
=> x1+x2+x3+….+x12 = 40 X 12 =480
Now,new numbers are \(\frac { x1 }{ 8 } ,\frac { x2 }{ 8 } ,\frac { x3 }{ 8 } ,..\frac { x12 }{ 8 } \)
Mean = \(\frac { \frac { x1 }{ 8 } +\frac { x2 }{ 8 } +\frac { x3 }{ 8 } +…..+\frac { x12 }{ 8 } }{ 12 } \)
= \(\frac { 1 }{ 8 } \frac { \left( x1+x2+x3+….x12 \right) }{ 12 } \)
= \(\frac { 480 }{ 8X12 } \) = 5
Mean of new numbers =5

Question 11.
Solution:
Let x1, x2, x3,…..x20 are the numbers
Mean = \(\frac { x1+x2+x3+…x20 }{ 20 }\) = 18
=> x1 + x2 + x3 +….+ x20 = 18 X 20 =360
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14D Q11.1

Question 12.
Solution:
Mean weight of 6 boys = 48 kg
Their total weight = 48 x 6 = 288 kg
Weights of 5 boys among them, are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg
Sum of weight of 5 boys = (51 + 45 + 49 + 46 + 44) kg = 235 kg
Weight of 6th boy = 288 kg – 235 kg = 53 kg Ans.

Question 13.
Solution:
Mean of 50 students = 39
Total score = 39 x 50 = 1950
Now correct sum of scores = 1950 – Wrong item + Correct item= 1950 – 23 + 43
= 1950 + 20 = 1970
Correct mean = \(\frac { 1970 }{ 50 } \) = 39.4 Ans.

Question 14.
Solution:
Mean of 100 items = 64
The sum of 100 items = 64 x 100 = 6400
New sum of 100 items = 6400 + 36 + 90 – 26 – 9 = 6526 – 35 = 6491,
Correct mean = \(\frac { 6491 }{ 100 }\) =64.91 = 64.91 Ans.

Question 15.
Solution:
Mean of 6 numbers = 23
Sum of 6 numbers = 23 x 6 = 138
Excluding one number, the mean of remaining 5 numbers = 20
Total of 5 numbers = 20 x 5 = 100
Excluded number = 138 – 100 = 38 Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A

RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1A.

Other Exercises

Question 1.
Solution:
(i) 20 = 5 x 4
\(\\ \frac { -3 }{ 5 } \) = \(\\ \frac { -3X4 }{ 5X4 } \)
(multiplying numerator and denominator by 4)
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 1
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 2

You can also Download NCERT Solutions for Class 8 Maths to help you to revise complete Syllabus and score more marks in your examinations.

Question 2.
Solution:
98 = 7 x 14
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 3

Question 3.
Solution:
60 = 5 x 12
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 4

Question 4.
Solution:
(i) \(\\ \frac { -12 }{ 30 } \)
H.C.F of 12 and 30 = 6
Dividing the numerator and denominator
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 5
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 6

Question 5.
Solution:
(i) \(\\ \frac { 3 }{ 8 } \) or 0
We know that every positive rational number is greater than 0.
∴\(\\ \frac { 3 }{ 8 } \) is greater
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 7
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 8
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 9
and clearly \(\frac{-1}{2} \) is greater. Ans.

Question 6.
Solution:
(i) \(\\ \frac { -4 }{ 3 } \) or \(\\ \frac { 8 }{ 7 } \)
Her,denominator are not same
LCM of 3 and 7 = 21
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 10
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 11
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 12
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 13
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 14

Question 7.
Solution:
(i) Between \(\\ \frac { -3 }{ 7 } \) and \(\\ \frac { 6 }{ -13 } \)
or \(\\ \frac { -3 }{ 7 } \) and \(\\ \frac { -6 }{ 13 } \)
LCM of 7 and 13 = 91
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 15
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 16
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 17
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 18
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 19

Question 8.
Solution:
Among, \(\\ \frac { 4 }{ -9 } \),\(\\ \frac { -5 }{ 12 } \),\(\\ \frac { 7 }{ -18 } \),\(\\ \frac { -2 }{ 3 } \) making their denominator positive,we get:
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 20
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 21
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 22
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 23
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 24
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 25

Question 9.
Solution:
(i)Among – 2,\(\\ \frac { -13 }{ 6 } \),\(\\ \frac { 8 }{ -3 } \),\(\\ \frac { 1 }{ 3 } \)
Making the denominator of \(\\ \frac { 8 }{ -3 } \) as
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 26
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 27
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 28
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 29
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 30
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1A 31

Question 10.
Solution:
(i) True, as the set of whole numbers is a subset of the set of rational numbers
(ii) True, as the set of integers is a subset of the set of rational numbers.
(iii) False, as 0 is a whole number and set of the whole number is a subset of rational numbers
∴ 0 is also a rational number.

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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.

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

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

These Solutions are part of NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Biology) Chapter 5 – The Fundamental Unit of Life solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 5 – The Fundamental Unit of Life Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

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NCERT TEXT BOOK QUESTIONS

IN TEXT QUESTIONS

Question 1.
Who discovered cell, and how ?
Answer:
Robert Hooke (1665). He thinly pared a piece of cork and observed it under his self made primitive microscope. The scientist found that cork possesses a number of small box-like structures which he named cells (cellulae which later’abbreviated to cells). His work was published in the form of a book called Micrographia.

Question 2.
Why is cell called structural and functional unit of life ?
Answer:
Structural Unit: A living organism is made up of one or more cells. Therefore, cell is structural unit of life. Functional Unit. All life functions of an organism reside in its cells. Cells may also become specialised to perform specific functions like contraction in muscle cell or impulse transmission in nerve cell. Therefore, cells are functional units of life.

Question 3.
How do substances like CO2 and water move into and out of the cell ? Discuss. (CCE 2011)
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,
(i) 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,
(ii) 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. Osmosis. It is movement of water from the region of its higher concentration (pure water or dilute solution) to the region of its lower concentration (strong solution) when the two are separated by a semipermeable membrane. Plasma membrane functions as a semipermeable membrane. Cell sap functions as strong solution. Therefore, external water enters the cell (endosmosis) till wall pressure counter-balances this tendency. If the external medium has a very strong solution, water would pass out from the cell into the external medium. The phenomenon is called exosmosis.

Question 4.
Why is plasma membrane called selectively permeable membrane ? (CCE 2012)
Answer:
Cell membrane is semipermeable membrane for water. It permits the entry of gases through diffusion. Ions, sugar, amino acids, etc. pass through the plasma membrane by an active process. Plasma membrane is impermeable to certain other materials. Therefore, it is selectively permeable.

Question 5.
Fill in the gaps in the following table illustrating differences between prokaryotic and eukaryotic cells.
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 1
Answer:
2. (Left side), (i) Poorly defined due to absence of nuclear envelope (ii) Nucleoid.
4. (Right side). Membrane bound cell organelles are present.

Question 6.
Can you name the two organelles, we have studied that contain their own genetic material ?
Answer:
Yes. Mitochondria and plasdds.

Question 7.
If the organisation of a cell is destroyed due to some physical or chemical influence, what will happen ?
Answer:
Lysosome will burst to release digestive enzymes. Digestive enzymes will cause breakdown of various cellular components causing destruction of the cell.

Question 8.
Why are lysosomes known as suicide bags ? (CCE 2011, 2012, 2013)
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.

Question 9.
Where are proteins synthesised inside the cell ?
Answer:
Proteins are synthesized over the ribosomes.

NCERT Exercise

Question 1.
Make a comparison to write down ways in which plant cells are different from animal cells.
Answer:
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 2

Question 2.
How is prokaryotic cell different from eukaryotic cell ?
Answer:
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 3

Question 3.
What would happen if the plasma membrane ruptures or breaks down ?
Answer:
There will be spilling of cytoplasm and cell organelles, bursting of lysosomes and digestion of cellular contents.

Question 4.
What would happen to the life of a cell if there was no Golgi apparatus ? (CCE 2011)
Answer:
There would not be any lysosome for intracellular digestion and cleansing, no complexing of molecules, no excretion and no formation of new plasma membrane.

Question 5.
Which organelle is known as power house of the cell ? Why ? (CCE 2011, 2013)
Answer:
Mitochondrion is known as power house of the cell because it produces most of the molecules of ATP (adenosine triphosphate) which are required for providing energy for synthesis of new chemicals, mechanical and other cellular functions.

Question 6.
Where do the lipids and proteins constituting the cell membrane get synthesised ?
Answer:
Proteins are synthesised over ribosomes of RER while lipids are synthesised over SER.

Question 7.
How does Amoeba obtain 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 8.
What is osmosis ?
Answer:
Osmosis is diffusion of water from the region of its higher concentration (pure water or dilute solution) to the region of its lower concentration (strong solution) through a semipermeable membrane.

Question 9.
Carry out the following osmosis experiment.
Take four peeled potato halves and hollow each one out to make potato cups. One of these potato cups should be made from the boiled potato. Put each potato cup in the trough containing water. Now
(a) Keep cup A empty,
(b) Put one tea spoon sugar in cup B.
(c) Put one tea spoon of salt in cup C.
(d) Put one tea spoon sugar in boiled cup D.
Keep this set up for two hours. Then observe the four potato cups and answer the following :
(i) Explain why water gathers in the hollowed portion of B and C.
(ii) Why is potato A necessary for this experiment ?
(iii) Explain why water does not gather in the hollowed out portion of A and D.
Answer:
(i) Sugar and salt increases osmotic concentration which results in passage of water osmotically from the trough through the cells of potato B and C into its cavity.
(ii) Potato A functions as control experiment which indicates that the cavity of potato does not induce movement of water. Water does not gather in the hollowed out portion of A because it does not have a higher osmotic concentration than the cells of potato tuber.
(iii) Potato tuber D does not have living cells. Osmosis does not occur in dead cells. Therefore, despite presence of sugar in the cavity of D, no water passes from trough through dead potato cells into cavity of the tuber.

PRACTICAL BASED TWO MARKS QUESTIONS

Question 1.
Three students ‘A’, ‘B’ and ‘C were given five raisins each of equal mass. The raisins were soaked in distilled water at room temperature. ‘A’ soaked the raisins for 10 minutes, ‘B’ for overnight and ‘C for 60 minutes. They calculated the percentage of water absorbed by raisins. Now answer the following question :
(a) Name the student whose raisins will show the maximum percentage of water absorbed.
(b) Name the student whose raisins will show the minimum percentage of water absorbed. (CCE 2013)
Answer:
(a) Maximum Percentage ofWater Absorbed. Raisins of ‘B’ student.
(b) Minimum Percentage of Water Absorbed. Raisins of ‘A’ student.

Question 2.
A teacher soaked 10 g raisins in 35 ml of distilled water in a beaker A and a similar amount in beaker B. She maintained the temperature of beaker A at 20°C and beaker B at 40°C. After an hour, compare the percentage of water absorbed by raisins in beakers A and B. (CCE 2013, 2014)
Answer:
More water was absorbed in beaker B where temperature was 40°C. Permeability of cell membrane increases from 0°C to 40°C beyond which dénaturation sets in.

Question 3.
5g of raisins were placed in distilled water for 24 hours. The mass of water soaked raisins was found to be 7g. Calculate the percentage of water absorbed by raisins. (CCE 2013)
Answer:
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 4

Question 4.
A student puts five raisins in two beakers A and B. Beaker A contained 50 ml of distilled water at room temperature and beaker B had 50 ml of ice cold water. After some time what will be the observation of the student ? State reason for this observation.
(CCE 2013, 2015)
Answer:
Raisins in beaker A swell up. Those of beaker B do not. Low temperature of B water reduced membrane permeability as well as kinetic energy of water for osmosis.

Question 5.
A student recorded a mass of dry raisins as 6.0 g and mass of raisins after soaking them in water for about four hours as 10.5 g. Calculate the percentage of water absorbed by raisins. (CCE 2013)
Answer:
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 5

Question 6.
Write the mathematic equation used to determine the mass percentage of water absorbed by raisins.
Answer:
If initial weight is W1 and final weight after soaking is W2
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 6

Question 7.
Ravi took weight of five dry raisins and five swollen raisins of approximately equal size. If the weight of dry raisins was 7 g and weight of swollen raisins is ‘X’ g, then
(i) Write the formula to calculate the percentage of water absorbed by the raisins and
(ii) If the value of ‘X’ is 10.5 g, then what will be the percentage of water absorbed by the raisins ? (CCE 2014)
Answer:
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 7

Question 8.
Write four main steps of the method involved in an experiment “On determination of the percentage of water absorbed by raisins in the laboratory”. (CCE 2014)
Answer:
(i) Selection of raisins with intact stalks and their weight as W1
(ii) Soaking the raisins in water at room temperature for at least 1-2 hours
(iii) Taking out wet raisins, wiping out water with blotting paper and their final weight as W2
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 8

Question 9.
In the experiment of determining the percentage of water absorbed by raisins, two students Samya and Zahira used the following formula respectively
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 8
Which student used wrong formula ? What are W1 and W2 in the correct formula ? (CCE 2014)
Answer:
Student (i) used the wrong formula.
W1 is initial weight of dry raisins. W2 is final weight of swollen raisins.

Question 10.
(a) A student recorded the mass of dry raisins as 4.0 g and mass of raisins after soaking as 7.0 g. Calculate the percentage of water absorbed by raisins. (CCE 2014, 2016)
(b) Mention one application of the phenomenon of osmosis in plants. (CCE 2016)
Answer:
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 10
(b) Application of Osmosis. Turgidity of cells, root absorption.

Question 11.
(a) Ram while doing an experiment to find out the percentage of water absorbed by raisins measured the mass of dry raisins as 50 g. He soaked the raisins in water for four hours and again measured the mass as 80 g. Calculate the percentage of water absorbed by the raisins.
(b) He then placed raisins in concentrated salt solution. What will he observe ? (CCE 2015)
Answer:
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 11
(b) After Placing in Concentrated Salt Solution. Raisins will lose water and shrink to the maximum.

Question 12.
A student recorded the mass of dry raisins as 6.0 g and mass of raisins after soaking them in water for about four hours as 10.5 g. Calculate the percentage of water absorbed by raisins. Why do raisins get swelled up ? (CCE 2015)
Answer:
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 16
The raisins swell up due to absorption of water through osmosis.

Question 13.
A student recorded the following observations in an experiment for finding the percentage of water absorbed by raisins.
(i) Mass of water taken in beaker – 50 g
(ii) Mass of dry raisins before soaking water – 20 g
(iii) Mass of raisins after soaking water – 30 g.
(iv) Mass of remaining water in beaker after experiment – 40 g
Calculate the percentage of water absorbed by raisins. ( CCE 2015)
Answer:
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 12

Question 14.
If ‘X’ is the initial mass of the raisins and ‘Y’ is the final mass of raisins after soaking in water, calculate the percentage of water absorbed by raisins. Name the process due to which raisins absorb water. (CBSE 2015, 2016)
Answer:
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 13

Question 15.
A group of students selected 10 raisins with stalks and weighed them using digital balance. Then, they soaked them for a few hours. The weight of the swollen raisins was 9.2 g which was 4.6 g more than weight of dry raisins. Calculate the water imbibed by the raisins. ( CCE 2015)
Answer:
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 14

Question 16.
In the experiment “To determine the mass percentage of water imbibed by raisins”, the raisins absorb water when kept in water for 5-6 hours. Why does water absorption take place ? What is the phenomenon called ? (CCE 2015, 2016)
Answer:
Water moves into raisins due to endosmosis. Endosmosis is entry of water into a system, cell or organ due to presence of hypertonic solution in it and its separation from pure water or dilute solution by a semipermeable membrance. Skin of raisins function as semipermeable membrane. There is high concentration of sugar inside them. Therefore, external water passes into raisins and cause their swelling.

Question 17.
Before placing the raisins in water, the raisins weighed 10 g. The raisins were taken out of water, wiped well and then the weight was found to be 12.5 g. Determine the percentage of water absorbed by raisins. Define the process due to which raisins absorb water. (CCE 2016)
Answer:
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 15
Osmosis: Osmosis is the diffusion of water or solvent across a semipermeable membrane (which does not allow passage of solutes) from a region of its higher concentration to the region of its lower concentration. Plasma membrane functions as semipermeable membrane. There will be osmotic entry of water into cell if the external solution is pure water or dilute as compared to cell sap.

Question 18.
(a) In the experiment to determine the percentage of water absorbed by raisins, the raisins are wiped before weighing. Why ?
(b) While preparing a temporary stained mount of onion peel, Veena added a drop of glycerine. Why ? ( CCE 2016)
Answer:
(a) To remove unabsorbed water sticking to the surface of raisins.
(b) To prevent drying of onion peel.

Question 19.
A student took x gram water in a beaker and dipped p gram raisins in it. After keeping raisins in water for 2 hours, he measured the mass of soaked raisins as q grams. He also measured the mass of water left in the beaker which was y grams. On the
Answer:
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 17

SELECTION TYPE QUESTIONS

basis of his observations write correct formula to find the percentage of water absorbed by raisins. Mention the process due to which weight of raisins increased. (CCE 2016)

Alternate Response Type Questions
(True/False, Right/Wrong, Yes/No)

Question 1.
Cork comes from bark.
Question 2.
Robert Brown discovered protoplasm in 1831.
Question 3.
Amoeba has an everchanging shape.
Question 4.
Movement of a substance from the area of low concentration to an area of high concentration is called diffusion.
Question 5.
A dilute soludon is called hypertonic solution.
Question 6.
Lysosomes keep the cells clean by digesting foreign materials and worn out cell organelles.
Question 7.
SER detoxifies many poisons and drugs.
Question 8.
Central vacuole occupies 10-20% of cell volume.

Matching Type Questions :

Question 9.
Match the contents of the columns I and II (single matching)
NCERT Solutions for Class 9 Science C

Question 10.
Match the contents of columns I, II and III (double matching)
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 19

Question 11.
Which type of metabolism, anabolism (A) and catabolism (C) are performed by the following organelles (key or check list items) :
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 20

Question 12.
Match the stimulus with Appropriate Response.
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 21

Fill In the Blanks

Question 13. Cristae create a large surface area for …………… generating reactions.
Question 14. Plant cell wall is mainly composed of ……………..
Question 15. Same organelles perform …………… function in all organisms.
Question 16. Cell theory was proposed by …………. and ……………. .
Question 17. Cells were discovered by Robert Hooke in ………….. .

Answers:
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 22

SOME TYPICAL QUESTIONS

Question 1.
What is the junctional unit of life ? Define it.
Answer:
Cell is the functional unit of life. It can be defined as a tiny mass of protoplasm covered by plasma membrane which is capable of performing all functions of life.

Question 2.
What is the difference between plasma membrane and cell wall i Give the functions of each one. (CCE 2011)
Answer:
Plasma membrane is an elastic living membrane made up of lipids and proteins, whereas cell wall is a rigid non-living covering made up of cellulose.
Function of Plasma membrane. It acts as a selectively semipermeable membrane which allows only selective substances to pass through it.
Function of Cell Wall. It provides rigidity and protection to cell.

Question 3.
Differentiate between chromatin and chromosome.
Answer:
Chromatin. It is the nucleoprotein fine fibrous mass which stains strongly with basic dyes and is present as a network inside the nucleus.
Chromosome. Rod-like, stainable, condensed chromatin unit, visible at cell division and containing hereditary information in the form of genes.

Question 4.
Differentiate between RER and SER.
Answer:
NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life image - 23

Question 5.
Which type of ribosomes are found in prokaryotes and eukaryotes ?
Answer:
Prokaryotes have 70 S ribosomes and eukaryotes have 80 S ribosomes.

Question 6.
Which structure is called little nucleus ?
Answer:
Nucleolus.

Question 7.
Why is nucleus called director of the cell ?
Answer:
Nucleus controls and coordinates all the metabolic functions of the cell.

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

Hope given NCERT Solutions for Class 9 Science Chapter 5 are helpful to complete your science homework.

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RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1

RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1

Other Exercises

Question 1.
If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively. Find the mean height.
Solution:
Heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1 1.1

Question 2.
Find the mean of 994, 996, 998, 1002 and 1000.
Solution:
Mean of 994, 996, 998, 1002 and 1000
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1 2.1

Question 3.
Find the mean of first five natural numbers.
Solution:
First five natural numbers are 1, 2, 3, 4, 5
∴ Mean = \(\overline { x } =\frac { 1+2+3+4+5 }{ 5 } =\frac { 15 }{ 5 } \) = 3

Question 4.
Find the mean all factors of 10.
Solution:
Factors of 10 = 1, 2, 5, 10
∴ Mean = \(\overline { x } =\frac { 1+2+5+10 }{ 4 } =\frac { 18 }{ 4 } \) = 4.5

Question 5.
Find the mean of first 10 even natural numbers.
Solution:
First 10 even natural numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
∴ Mean = \(\overline { x } =\frac { 2+4+6+8+10+12+14+16+18+20 }{ 10 } =\frac { 110 }{ 10 } \) = 11

Question 6.
Find the mean of x, x + 2, x + 4, x + 6, x + 8.
Solution:
Sum = x + x + 2+ x + 4 + x + 6 + x + 8 = 5x + 20
∴ Mean = \(\overline { x } =\frac { \sum { { x }_{ i } } }{ n } \frac { 5x+20 }{ 5 } =x+4 \)

Question 7.
Find the mean of first five multiples of 3.
Solution:
First 5 multiples of 3 are = 3, 6, 9, 12, 15
∴ Mean = \(\overline { x } =\frac { 3+6+9+12+15 }{ 5 } =\frac { 45 }{ 5 } \) = 9

Question 8.
Following are the weights (in kg) or 10 new born babies in a hospital on a particular day:
3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6. Find the mean \(\overline { X } \).
Solution:
Weights of 10 new bom babies are 3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6
∴ Mean \(\overline { X } \) = \(\frac { 3.4+3.6+4.2+4.5+3.9+4.1+3.8+4.5+4.4+3.6 }{ 10 } \)
= \(\frac { 40.0 }{ 10 } \) = 4kg

Question 9.
The percentage of marks obtained by students of a class in mathematics are : 64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1. Find their mean.
Solution:
Percentage of 12 students are 64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1
∴ Mean \(\overline { X } \) = \(\frac { 64+36+47+23+0+19+81+93+72+35+3+1 }{ 12 } \)
= \(\frac { 474 }{ 12 } \) = 39.5

Question 10.
The numbers of children in 10 families of a locality are : 2, 4, 3, 4, 2, 0, 3, 5, 1, 1, 5. Find the mean number of children per family.
Solution:
Number of children in 10 families are 2, 4, 3, 4, 2, 0, 3, 5, 1, 1, 5
∴ Mean \(\overline { X } \) = \(\frac { 2+4+3+4+2+0+3+5+1+1+5 }{ 10 } \)
= \(\frac { 30 }{ 10 } \) = 3

Question 11.
Explain, by taking a suitable example, how the arithmetic mean alters by (i) adding a constant k to each term, (ii) subtracting a constant k from each them, (iii) multiplying each term by a constant k and (iv) dividing each term by a non-zero constant k.
Solution:
Let x1, x2, x3, x4, x5 are five numbers whose mean is \(\overline { x } \) i.e. = \(\frac { x1+x2+x3+x4+x5 }{ 5 } \) = \(\overline { x } \)
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1 11.2
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1 11.3
Hence we see that in each case, the mean is changed.

Question 12.
The mean of marks scored by 100 students was found to be 40. Later on its was discovered that a score of 53 was misread as 83. Find the correct mean.
Solution:
Mean score of 100 students = 40
∴Total = 100 x 40 = 4000
Difference in one score by mistake = 83 – 53 = 30
Actual total scores = 4000 – 300 = 3970
Actual mean = \(\frac { 3970 }{ 100 } \) = 39.70 = 39.7

Question 13.
The traffic police recorded the speed (in km/hr) of 10 motorists as 47, 53, 49, 60, 39, 42, 55, 57, 52, 48. Later on an error in recording instrument was found. Find the correct average speed of the motorists if the instrument recorded 5 km/hr less in each case.
Solution:
Speed of 10 motorist as recorded = 47, 53, 49, 60, 39, 42, 55, 57, 52, 48
Total of speed of 10 motorists = 47 + 53 + 49 + 60 + 39 + 42 + 55 +57 + 52 + 48 = 502
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1 13.1

Question 14.
The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.
Solution:
Mean of 5 numbers = 27
Total = 27 x 5 = 135
By excluded one number, then mean of remaining 4 numbers = 25
Total = 4 x 25 = 100
Excluded number = 135 – 100 = 35

Question 15.
The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.
Solution:
Mean weight of 7 students = 55 kg
Total weight of 7 students = 55 x 7 kg = 385 kg
Total weights of 6 students among them = 52 + 54 + 55 + 53 + 56 + 54 = 324 kg
Weight of 7th student = 385 – 324 = 61 kg

Question 16.
The mean weight of 8 numbers is 15. If each number is multiplied by 2, what will be the new mean?
Solution:
Weight of 8 numbers =15
By multiplying each number by 2, then the average will be = 15 x 2 = 30
New average = 30

Question 17.
The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.
Solution:
Mean of 5 numbers = 18
Total = 18 x 5 = 90
By excluding one number, the mean of remaining 5 – 1=4 numbers = 16
Total = 16 x 4 = 64
Excluded number = 90 – 64 = 26

Question 18.
The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.
Solution:
Mean of 200 items = 50
Total = 50 x 200 = 10000
The number were misread as 92 instead of 192 and 8 instead of 88
Difference = 192 – 92 + 88 – 8 = 180
New total = 10000 + 180 = 10180
and new mean = \(\frac { 10180 }{ 200 } \) = 50.9

Question 19.
If M is the mean of x1, x2, xr3, x4, x5 and x6, prove that
(x1 – M) + (x2 – M) + (x3 – M) + (x4 – M) + (x5 – M) + (x6 – M) = 0.
Solution:
∵ M is the mean of x,, x2, x3, x4, x5, x6
Then M = \(\frac { x1+x2+x3+x4+x5+x6 }{ 6 } \)
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1 19.1

Question 20.
Durations of sunshine (in hours) in Amritsar for first 10 days of August 1997 as reported by the Meteorological Department are given below:
9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, 10.9
(i) Find the mean \(\overline { X }\)
(ii) Verify that \( \sum _{ i=1 }^{ 10 }{ \left( { x }_{ i }-\overline { X } \right) } \) = 0
Solution:
Duration of sun shine for 10 days (in hours)
= 9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, 10.9
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1 20.1
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1 20.2

Question 21.
Find the values of n and X in each of the following cases:
(i) \(\sum _{ i=1 }^{ n }{ \left( { x }_{ i }-12 \right) } =-10\quad and\sum _{ i=1 }^{ n }{ \left( { x }_{ i }-3 \right) } =62\)
(ii) \(\sum _{ i=1 }^{ n }{ \left( { x }_{ i }-10 \right) } =30\quad and\sum _{ i=1 }^{ n }{ \left( { x }_{ i }-6 \right) } =150\)
Solution:
(i) \(\sum _{ i=1 }^{ n }{ \left( { x }_{ i }-12 \right) } =-10\)…(i)
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1 21.1
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1 21.2
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1 21.3

Question 22.
The sums of the deviations of a set of n values x1, x2,… xn measured from 15 and -3 are -90 and 54 respectively. Find the value of n and mean.
Solution:
In first case,
(x1 – 15) + (x2 – 15) + (x3 – 15) + … + (xn – 15) = – 90
=> x1 + x2 + x3 + … + xn – 15 x n = – 90
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1 22.1

Question 23.
Find the sum of the deviations of the variate values 3, 4, 6, 7, 8, 14 from their mean.
Solution:
Mean of 3, 4, 6, 7, 8, 14 = \(\frac { 42 }{ 6 } \) = 7
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1 23.1

Question 24.
If \(\overline { X } \) is the mean of the ten natural numbers x1, x2, x3, …, x10, show that (x1 – \(\overline { X } \)) + (x2 – \(\overline { X } \)) + … + (x10 – \(\overline { X } \)) = 0.
Solution:
\(\overline { X } \) is the mean of 10 natural numbers
x1, x2, x3, …, x10
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1 24.1

 

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RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E

RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10E.

Other Exercises

PROBLEMS ON NUMBERS
Question 1.
Solution:
Let natural number be = x
According to the condition,
x² + x = 156
⇒ x² + x – 156 = 0
⇒ x² + 13x – 12x – 156 = 0
⇒ x (x + 13) – 12 (x + 13) = 0
⇒ (x + 13) (x – 12) = 0
Either, x + 13 = 0, then x = -13 but it is not a natural number
or x – 12 = 0, then x = 12
Hence, required number = 12

Question 2.
Solution:
Let the natural number be x,
Then according to the condition,
x + √x = 132
Let √x = y, then
y² + y = 132
⇒ y² + y – 132 = 0
⇒ y² + 12y – 11y – 132 = 0
⇒ y (y + 12) – 11 (y + 12) = 0
⇒ (y + 12) (y – 11) = 0
Either, y + 12 = 0, then y = -12 but it is not a natural number.
or y – 11 = 0 ⇒ y = 11
√x = 11 ⇒ x = 121
Hence, required number =121

Question 3.
Solution:
Sum of two natural numbers = 28
Let first number = x
Then second number = 28 – x
According to the condition,
x (28 – x) = 192
⇒ 28x – x² = 192
⇒ x² – 28x + 192 = 0
⇒ x² – 16x – 12x + 192 = 0
⇒ x (x – 16) – 12 (x – 16) = 0
⇒ (x – 16)(x – 12) = 0
Either, x – 16 = 0, then x = 16
or x – 12 = 0, then x = 12
Required natural numbers are 16 and 12

Question 4.
Solution:
Let first integer = x
Then second = x + 1
According to the condition,
(x)² + (x + 1)² = 365
⇒ x² + x² + 2x + 1 = 365
⇒ 2x² + 2x + 1 – 365 = 0
⇒ 2x² + 2x – 364 = 0
⇒ x² + x – 182 = 0
⇒ x² + 14x – 13x – 182 = 0
⇒ x(x + 14) – 13(x + 14) = 0
⇒ (x + 14)(x – 13) = 0
Either, x + 14 = 0, then x = -14 but it is negative.
or x – 13 = 0, then x = 13
First number = 13
and second number = 13 + 1 = 14
Hence, the numbers are 13, 14

Question 5.
Solution:
Let first odd number = x
Then second consecutive odd number = x + 2
According to the condition,
x² + (x + 2)² = 514
⇒x² + x² + 4x + 4 = 514
⇒ 2x² + 4x + 4 – 514 = 0
⇒ 2x² + 4x – 510 = 0
⇒ x² + 2x – 255 = 0
⇒ x² + 17x – 15x – 255 = 0
⇒ x (x + 17) – 15 (x + 17) = 0
⇒ (x + 17) (x – 15) = 0
Either, x + 17 = 0, then x = -17 but it is not positive.
or x – 13 = 0, then x = 13
First odd number = 13
and second number = 13 + 2 = 15
Hence, the number are 15, 17

Question 6.
Solution:
Let first positive even number = x
Then second even number = x + 2
According to the condition, x² + (x + 2)² = 452
⇒ x² + x² + 4x + 4 = 452
⇒ 2x² + 4x + 4 – 452 = 0
⇒ 2x² + 4x – 448 = 0
⇒ x² + 2x – 224 = 0
⇒ x² + 16x – 14x – 224 = 0
⇒ x (x + 16) – 14 (x + 16) = 0
⇒ (x + 16) (x – 14) = 0
Either, x + 16 = 0, then x = 16 which is not positive.
or x – 14 = 0, then x = 14
First even number = 14
and second number = 14 + 2 = 16
Hence, the numbers are 14, 16

Question 7.
Solution:
Let first positive integer = x
Then second integer = x + 1
According to the condition,
x (x + 1) = 306
⇒ x² + x – 306 = 0
⇒ x² + 18x – 17x – 306 = 0
⇒ x (x + 18) – 17 (x + 18) = 0
⇒ (x + 18)(x – 17) = 0
Either, x + 18 = 0, then x = -18 which is not positive.
or x – 17 = 0, then x = 17
First number =17
and second integer = 17 + 1 = 18
Hence, the numbers are 17, 18

Question 8.
Solution:
Let first natural number = x
Then second number = x + 3
According to the condition,
x (x + 3) = 504
⇒ x² + 3x – 504 = 0
⇒ x² + 24x – 21x – 504 = 0
⇒ x(x + 24) – 21(x + 24) = 0
⇒ (x + 24)(x – 21) = 0
Either x + 24 = 0, then x = -24 but it is not positive.
or x – 21 =0, then x = 21
First natural number = 21
and second number = 21 + 3 = 24
Hence, two required numbers are 21, 24

Question 9.
Solution:
Let first multiple of 3 = 3x
and second = 3x + 3
According to the condition,
3x (3x + 3) = 648
⇒ 9x² + 9x – 648 = 0
⇒ x² + x – 72 = 0
⇒ x² + 9x – 8x – 72 = 0
⇒ x (x + 9) – 8 (x + 9) = 0
⇒ (x + 9)(x – 8) = 0
Either, x + 9 = 0, then x = -9 which is negative.
or x – 8 = 0, then x = 8
First multiple of 3 = 3 x 8 = 24
and second = 24 + 3 = 27
Hence, numbers are 24, 27

Question 10.
Solution:
Let first positive odd integer = x
Then second = x + 2
According to the condition,
x (x + 2) = 483
⇒ x² + 2x – 483 = 0
⇒ x² + 23x – 21x – 483 = 0
⇒ x (x + 23) – 21(x + 23) = 0
⇒ (x + 23)(x – 21) = 0
Either, x + 23 = 0, then x = -23 which is negative.
or x – 21 =0, then x = 21
First odd number = 21
and second = 21 + 2 = 23
Hence, numbers are 21, 23

Question 11.
Solution:
Let first positive even integer = 2x
Then second even integer = 2x + 2
According to the condition,
2x (2x + 2) = 288
⇒ 4x² + 4x – 288 = 0
⇒ x² + x – 12 = 0
⇒ x² + 9x – 8x – 72 = 0
⇒ x (x + 9) – 8 (x + 9) = 0
⇒ (x + 9) (x – 8) = 0
Either, x + 9 = 0, then x = -9 which is negative.
or x – 8 = 0, then x = 8
First even integer = 2 x 8 = 16
and second even integer = 16 + 2 = 18
Hence, required integers are 16, 18

Question 12.
Solution:
Sum of two natural numbers = 9
Let first number = x
Then second number = 9 – x
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 1

Question 13.
Solution:
Sum of two natural numbers = 15
Let first number = x
Then second number = 15 – x
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 2

Question 14.
Solution:
Difference of two natural numbers = 3
Let first number = x
Then second number = x + 3
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 3
Either, x + 7 = 0, then x = -7 which is negative.
or x – 4 = 0, then x = 4.
First number = 4
and second number = 4 + 3 = 7
Hence, the numbers are 4, 7

Question 15.
Solution:
Difference of two natural numbers = 5
Let first number = x
Then second number = x – 5
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 4
⇒ x (x – 7) + 2 (x – 7) = 0
⇒ (x – 7) (x – 2) = 0
Either, x – 7 = 0, then x = 7
or x + 2 = 0, then x = -2 which is negative
First number = 7
and second number = 7 – 5 = 2
Hence, numbers are 7, 2

Question 16.
Solution:
Let first number = 7x
Then second number = 7x + 7
According to the condition,
(7x)² + (7x + 7)² = 1225
⇒ 49x² + 49x² + 49 + 98x = 1225
⇒ 98x² + 98x + 49 – 1225 = 0
⇒ 98x² + 98x – 1176 = 0
⇒ x² + x – 12 = 0 (Dividing by 98)
⇒ x² + 4x – 3x – 12 = 0
⇒ x (x + 4) – 3 (x + 4) = 0
⇒ (x + 4) (x – 3) = 0
Either, x + 4 = 0, then x = -4 which is negative.
or x – 3 = 0, then x = 3
First number 3 x 7 = 21
and second number = 21 + 7 = 28
Hence, numbers are 21, 28

Question 17.
Solution:
Let the natural number = x
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 5
⇒ 8x² – 64x – x + 8 = 0
⇒ 8x (x – 8) – 1 (x – 8) = 0
⇒ (x – 8) (8x – 1) = 0
Either, x – 8 = 0, then x = 8
or 8x – 1 = 0, then 8x = 1 ⇒ x = \(\frac { 1 }{ 8 }\) which is not a natural number.
Required natural number = 8

Question 18.
Solution:
Let first part = x
Then second part = 57 – x
According to the condition,
x (57 – x) = 680
⇒ 57x – x² = 680
⇒ x² – 57x + 680 = 0
⇒ x² – 40x – 17x + 680 = 0
⇒ x (x – 40) – 17 (x – 40) = 0
⇒ (x – 40) (x – 17) = 0
Either, x – 40 = 0, then x = 40
or x – 17 = 0, then x = 17
If first part is 40, then second part = 57 – 40 = 17
Parts are 40, 17 or 17, 40

Question 19.
Solution:
Let first part = x
Then second part = 27 – x
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 6
⇒ 81x – 3x² = 540
⇒ 3x² – 81x + 540 = 0
⇒ x² – 27x + 180 = 0
⇒ x² – 12x – 15x + 180 = 0
⇒ x(x – 12) – 15(x – 12) = 0
⇒ (x – 12)(x – 15) = 0
Either, x – 12 = 0, then x = 12
or x – 15 = 0, then x = 15
If first part is 12, then
second part = 27 – 12 = 15
Parts are 12, 15 or 15, 12

Question 20.
Solution:
Let first part (larger) = x
Then second part = 16 – x
According to the condition,
2(x²) – (16 – x)² = 164
⇒ 2x² – (256 + x² – 32x) = 164
⇒ 2x² – 256 – x² + 32x – 164 = 0
⇒ x² + 32x – 420 = 0
⇒ x² + 42x – 10x – 420 = 0
⇒ x (x + 42) – 10 (x + 42) = 0
⇒ (x + 42) (x – 10) = 0
either x + 42 = 0, then x = -42 which is negative.
or x – 10 = 0, then x = 10
First part =10
and second part =16 – 10 = 6
Hence, parts are 10 and 6

Question 21.
Solution:
Let a and b be the two natural numbers.
According to the condition,
a² + b² = 25(a + b) and a² + b² = 50(a – b)
25(a + b) = 50(a – b)
⇒ a + b = 2(a – b)
⇒ a + b = 2a – 2b
⇒ 2a – a = 2b + b
⇒ a = 3b
Now, a² + b² = 25(a + b)
⇒ (3b)² + b² = 25(3b + b)
⇒ 9b² + b² = 25 x 4b
⇒ 10b² = 100b
⇒ b² = 10b
⇒ b² – 10b = 0
⇒ b(b – 10) = 0
Either, b = 0 which is not possible
or b – 10 = 0, then b = 10
a = 3b = 3 x 10 = 30
Numbers are 30 and 10

Question 22.
Solution:
Let the two natural numbers be a and b.
According to the condition,
a² – b² = 45 and b² = 4a
⇒ a² – 4a = 45
⇒ a² – 4a – 45 = 0
⇒ a² – 9a + 5a – 45 = 0
⇒ a(a – 9) + 5 (a – 9) = 0
⇒ (a – 9) (a + 5) = 0
Either, a – 9 = 0, then a = 9
or a + 5 = 0, then a = -5 which is not a natural number.
a = 9
and b² = 4a = 4 x 9 = 36 = (6)²
b = 6
Hence, numbers are 9, 6

Question 23.
Solution:
Let three consecutive positive integers be x, x + 1 and x + 2
According to the condition,
x² + (x + 1) (x + 2) = 46
⇒ x² + x² + 3x + 2 – 46 = 0
⇒ 2x² + 3x – 44 = 0
⇒ 2x² – 8x + 11x – 44 = 0
⇒ 2x (x – 4) + 11 (x – 4) = 0
⇒ (x – 4) (2x + 11) = 0
Either, x – 4 = 0, then x = 4
or 2x + 11 = 0, then 2x = -11 ⇒ x = \(\frac { -11 }{ 2 }\)
But it is not a natural number.
x = 4
Hence, three consecutive numbers are 4, 5, 6.

Question 24.
Solution:
Let the digits of a two digit number be x and y
Let one’s digit = x
and ten’s digit = y
Number = x + 10y
According to the condition,
x + 10y = 4(x + y)
and x + 10y = 2xy
Now, x + 10y = 4x + 4y
⇒ 10y – 4y = 4x – x
⇒ 3x = 6y
⇒ x = 2y …(i)
x + 10y = 2xy
⇒ 2y + 10y = 2 x 2y x y
⇒ 12y = 4y²
⇒ 3y = y²
⇒ y² – 3y = 0
⇒ y (y – 3) = 0
Either, y = 0, which is not possible,
or y – 3 = 0, then y = 3
x = 2y = 2 x 3 = 6
and number = x + 10y = 6 + 10 x 3 = 6 + 30 = 36

Question 25.
Solution:
Let the ones digit of a two digit number = x
and ten’s digit = y
Number = x + 10y
By reversing the digits,
the One’s digit = y
and ten’s digit = x
Number = y + 10x
According to the condition,
x + 10y + 45 = y + 10x
⇒ 10y – y + x – 10x = -45
9y – 9x = -45
⇒ x – y = 5
⇒ x = y + 5
and xy = 14
⇒ (y + 5) x y = 14
⇒ y² + 5y – 14 = 0
⇒ y² + 7y – 2y – 14 = 0
⇒ y (y + 7) – 2(y + 7) = 0
⇒ (y + 7)(y – 2) = 0
Either, y + 7 = 0, then x = -7 which is negative, so not possible.
or y – 2 = 0, then y = 2
x = y + 5 = 2 + 5 = 7
Number = x + 10y = 7 + 10 x 2 = 7 + 20 = 27

Question 26.
Solution:
Let numerator = x
Then denominator = x + 3
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 7
⇒ x² + 5x – 2x – 10 = 0
⇒ x (x + 5) – 2 (x + 5) = 0
⇒ (x + 5)(x – 2) = 0
Either x + 5 = 0, then x = -5 which not possible being negative,
or x – 2 = 0, then x = 2
Fraction = \(\frac { x }{ x + 3 }\) = \(\frac { 2 }{ 2 + 3 }\) = \(\frac { 2 }{ 5 }\)

Question 27.
Solution:
Let the denominator of a fraction = x
Then its numerator = x – 3
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 8

Question 28.
Solution:
Let the number = x
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 9

SOME GENERAL PROBLEMS
Question 29.
Solution:
Let number of students = x
According to the given condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 10
x² = (x + 25)(x – 24)
x² = x² + (25 – 24) x + 25 x (-24)
x² = x² + x – 600
x = 600
Number of students = 600

Question 30.
Solution:
Total number of apples = 300
Let number of students = x
Then according to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 11

Question 31.
Solution:
Total number of marks = 40
Let marks in Mathematics = x
Then marks in English = 40 – x
According to the given condition,
(x + 3)(40 – x – 4) = 360
(x + 3)(36 – x) = 360
⇒ 36x – x² + 108 – 3x = 360
⇒ -x² + 33x + 108 – 360 = 0
⇒ -x² + 33x – 252 = 0
⇒ x² – 33x + 252 = 0
⇒ x² – 21x – 12x + 252 = 0
⇒ x(x – 21) – 12(x – 21) = 0
⇒ (x – 21)(x – 12) = 0
Either, x – 21 =0, then x = 21
or x – 11 = 0, then x = 11
Marks obtain in Mathematic and English = (21 and 19) or (12 or 28)

Question 32.
Solution:
Total budget = ₹ 2000
Let number of students = x
5 students failed to attend
Remaining students = x – 5
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 12
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 13

Question 33.
Solution:
Let original price of each book = ₹ x
and total amount = ₹ 600
Commission in price = ₹ 5 per book
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 14
⇒ x² – 30x + 25x – 750 = 0
⇒ x(x – 30) + 25(x – 30) = 0
⇒ (x – 30)(x + 25) = 0
Either, x – 30 = 0, then x = 30
or x + 25 = 0, then x = -25 which is not possible being negative.
Original price of each book = ₹ 30
and number of books = \(\frac { 600 }{ 30 }\) = 20

Question 34.
Solution:
Total expenses on tour = ₹ 10800
Let number of days = x
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 15
⇒ x(x + 24) – 20(x + 24) = 0
⇒ (x + 24)(x – 20) = 0
Either, x + 24 = 0, then x = -24 which is not possible being negative,
or x – 20 = 0, then x = 20
Number of days of tour = 20 days

Question 35.
Solution:
Total marks obtained = 28
Let marks in Mathematics = x
Then marks in Science = 28 – x
According to the condition,
(x + 3)(28 – x – 4) = 180
⇒ (x + 3)(24 – x) = 180
⇒ 24x – x² + 72 – 3x = 180
⇒ -x² + 21x – 180 + 72 = 0
⇒ x² – 21x + 108 = 0
⇒ x² – 12x – 9x + 108 = 0
⇒ x(x – 12) – 9(x – 12) = 0
⇒ (x – 12)(x – 9) = 0
Either, x – 12 = 0, then x = 12
or x – 9 = 0, then x = 9
Marks obtained in Mathematics and Science = (12, 16) or (9, 19)

Question 36.
Solution:
Amount = ₹ 180
Let number of pens purchased = x
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 16
⇒ x(x + 15) – 12(x + 15) = 0
⇒ (x + 15) (x – 12) = 0
Either, x + 15 = 0, then x = -15 which is not possible being negative,
or x – 12 = 0, then x = 12
Number of pens purchased = 12

Question 37.
Solution:
S.P. of an article = ₹ 75
Gain % = C.P. of the article
Let gain % = x %
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 17
⇒ x² + 150x – 50x – 7500 = 0
⇒ x(x + 150) – 50(x + 150) = 0
⇒ (x + 150) (x – 50) = 0
Either, x + 150 = 0, then x = -150 which is not possible being negative,
or x – 50 = 0, then x = 50
Cost price of the article = ₹ 50

PROBLEMS ON AGES
Question 38.
Solution:
(i) Let present age of son = x years
His age 1 year ago = (x – 1) years
His father’s age = (x – 1) x 8
According to the condition,
(x – 1) x 8 + 1 = x2
⇒ 8x – 8 + 1 = x2
⇒ x² – 8x + 7 = 0
⇒ x² – x – 7x + 7 = 0
⇒ (x – 1)(x – 7) = 0
Either, x – 1 = 0, then x = 1 which is not possible.
or x – 7 = 0, then x = 7
Son’s age = 7 years
and his father’s age = (7)² = 49 years
(ii) Let the age of the son = x years
and the age of father = 3.5x years
According to the condition,
(x)² + (3.5x)² = 1325
⇒ x² + 12.25x² = 1325
⇒ 13.25x² = 1325
⇒ x² = 100
⇒ x = 10
Hence, the age of the son = 10 years
and the age of father 3.5 x 10 = 35 years

Question 39.
Solution:
Present age of Meena = x years
3 years ago her age = (x – 3) years
and 5 years hence, her age = (x + 5) years
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 18
(x – 7) (x + 3) = 0
Either, x – 7 = 0, then x = 7
or x + 3 = 0, then x = -3 which is not possible being negative.
Her age = 7 years

Question 40.
Solution:
Sum of ages of a boy and his brother = 25 years
Let age of the boy = x years
Then age of his brother = (25 – x) years
According to the condition,
x (25 – x) = 126
⇒ 25x – x² – 126 = 0
⇒ x² – 25x + 126 = 0
⇒ x² – 18x – 7x + 126 = 0
⇒ x (x – 18) – 7 (x – 18) = 0
⇒ (x – 18) (x – 7) = 0
Either, x – 18 = 0, then x = 18
or x – 7 = 0, then x = 7
Age of boy = 7 years or 18 years
and age of his brother = 18 years or 7 years.

Question 41.
Solution:
Let present age of Tanvy = x years
Then 5 years ago her age = x – 5 years
8 years hence, her age = x + 8 years
According to the condition,
(x – 5) (x + 8) = 30
⇒ x² + 3x – 40 = 30
⇒ x² + 3x – 40 – 30 = 0
⇒ x² – 10x + 7x – 70 = 0
⇒ x (x – 10) + 7 (x – 10) = 0
⇒ (x – 10) (x + 7) = 0
Either, x – 10 = 0, then x = 10
or x + 7 = 0, then x = -7 which is not possible being negative.
Her present age = 10 years

Question 42.
Solution:
2 years ago,
Let son’s age = x years
and his father’s age = 3x²
Present age of son = (x + 2) years
and his father’s age = 3x² + 2
3 years hence, son’s age will be = x + 2 + 3 = x + 5 years
and age of his father = 3x² + 2 + 3 = (3x² + 5) years
According to the condition,
3x² + 5 = 4(x + 5)
⇒ 3x² + 5 = 4x + 20
⇒ 3x² – 4x + 5 – 20 = 0
⇒ 3x² – 4x – 15 = 0
⇒ 3x² – 9x + 5x – 15 = 0
⇒ 3x (x – 37) + 5 (x – 3) = 0
⇒ (x – 3)(3x + 5) = 0
Either, x – 3 = 0, then x = 3
or 3x + 5 = 0, then 3x = -5 ⇒ x = \(\frac { -5 }{ 3 }\) which is not possible being negative.
x = 3
Son’s present age = 3 + 2 = 5 years
and his father’s age = 3x² + 2
= 3 x (3)² + 2
= 27 + 2 = 29 years

PROBLEMS ON TIME AND DISTANCE
Question 43.
Solution:
Distance travelled = 150 km
Let the average speed of the truck = x km/hr
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 19
⇒ x (x – 60) + 10 (x – 60) 0
⇒ (x – 60) (x + 10) = 0
Either, x – 60 = 0, then x = 60
or x + 10 = 0, then x = -10 which is not possible being negative.
Original speed of the truck = 60 km/h

Question 44.
Solution:
Distance covered = 1500 km
Let original speed of the plane = x km/h
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 20
⇒ x² + 600x – 500x – 300000 = 0
⇒ x(x + 600) – 500(x + 600) = 0
⇒ (x + 600) (x – 500) = 0
Either, x + 600 = 0, then x = -600 which is not possible being negative.
or x – 500 = 0, then x = 500
Original speed of the plane = 500 km/hr
Yes, his promptness was appreciable.

Question 45.
Solution:
Distance covered by a train = 480 km
Let uniform speed = x km/h
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 21
⇒ x² – 40x + 32x – 1280 = 0
⇒ x(x – 40) + 32(x – 40) = 0
⇒ (x – 40) (x + 32) = 0
Either, x – 40 = 0, then x = 40
or x + 32 = 0, then x = -32 which is not possible being negative.
Speed of the train = 400 km/hr

Question 46.
Solution:
Distance covered = 54 km and then 63 km
Let first average speed = x km/hr
and next speed = (x + 6) km/hr
Total time taken = 3 hours
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 22
⇒ x² – 33x – 108 = 0
⇒ x² – 36x – 3x – 108 = 0
⇒ x (x – 36) + 3 (x – 36) = 0
⇒ (x – 36) (x + 3) = 0
Either, x – 36 = 0, then x = 36
or x + 3 = 0, then x – 3 which is not possible being negative.
Speed of the train = 36 km/hr.

Question 47.
Solution:
Distance travelled by a train =180 km
Let the uniform speed = x km/hr
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 23
⇒ x² + 45x – 36x – 1620 = 0
⇒ x (x + 45) – 36 (x + 45) = 0
⇒ (x + 45) (x – 36) = 0
Either, x + 45 = 0, then x = -45 which is not possible being negative,
or (x – 36) = 0, then x = 36
Uniform speed of the train = 36 km/hr.

Question 48.
Solution:
Distance travelled = 300 km
Let the original speed of train be x km
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 24
⇒ 300x + 1500 – 300x = 2x² + 10x
⇒ 2x² + 10x – 1500 = 0
⇒ x² + 5x – 750 = 0 (Dividing by 2)
⇒ x² + 30x – 25x – 750 = 0
⇒ x (x + 30) – 25 (x + 30) = 0
⇒ (x + 30) (x – 25) = 0
Either, x + 30 = 0 ⇒ x = -30 (Rejected ∵ speed cannot be negative)
or x – 25 = 0 ⇒ x = 25
Speed of the train = 25 km/hr

Question 49.
Solution:
Distance = 300 km
Let the speed of the passenger = x km/hr
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 25
⇒ x (x + 30) – 25 (x + 30) = 0
⇒ (x + 30) (x – 25) = 0
Either, x + 30 = 0, then x = -30 which is not possible being negative,
or x – 25 = 0, then x = 25
Speed of the train = 25 km/hr.

Question 50.
Solution:
Distance between Mumbai and Pune = 192 km
Let the speed of Deccan Queen = x km/hr
and speed of another train = (x – 20) km/hr
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 26
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 27

Question 51.
Solution:
Distance travelled = 24 km up and down
Speed of motor boat in still water = 18 km/hr
Let speed of stream = x km/hr
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 28
⇒ x (x + 54) – 6 (x + 54) = 0
⇒ (x + 54) (x – 6) = 0
Either, x + 54 = 0, then x = -54 which is not possible being negative,
or x – 6 = 0, then x = 6
Speed of stream = 6 km/hr

Question 52.
Solution:
Speed of a boat in still water = 8 km/hr
Let speed of water in stream = x km/hr
Total time taken = 5 hours
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 29

Question 53.
Solution:
Speed of motorboat in still water = 9 km/hr
Let speed of water = x km/hr
Total time taken = 3 hours
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 30
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 31

PROBLEMS ON TIME AND WORK AND PIPES AND CISTERN
Question 54.
Solution:
Let B finishes the work in = x days
Then A will finish it in = (x – 10) days
According to the condition,
(A + B)’s one day’s work
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 32
⇒ x² – 30x – 4x + 120 = 0
⇒ x (x – 30) – 4 (x – 30) = 0
⇒ (x – 30)(x – 4) = 0
Either, x – 30 = 0, then x = 30
or x – 4 = 0, then x = 4
But x = 4 is not possible
x = 30
B can finish the work in 30 days
and A can in 30 – 10 = 20 days.

Question 55.
Solution:
2 pipes can fill a cistern in 3\(\frac { 1 }{ 13 }\) = \(\frac { 40 }{ 13 }\) minutes
Let one pipe can fill it in = x minute
Then second will fill it in = (x + 3) minutes
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 33
⇒ 13x (x – 5) + 24 (x – 5) = 0
⇒ (x – 5) (13x + 24) = 0
Either, x – 5 = 0, then x = 5
or 13x + 24 = 0
⇒ 13x = -24
⇒ x = \(\frac { -24 }{ 13 }\) which is not possible being negative.
x = 5
First pipe can fill the tank in 5 minutes
and second pipe will fill it in 5 + 3 = 8 minutes.

Question 56.
Solution:
Two pipes can fill a tank together in = 11\(\frac { 1 }{ 9 }\) = \(\frac { 100 }{ 9 }\) minutes
Let first pipe can fill it in = x minute
Then second pipe will fill it in = (x + 5) min
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 34
⇒ 9x² – 155x – 500 = 0
⇒ 9x² – 180x + 25x – 500 = 0
⇒ 9x (x – 20) + 25(x – 20) = 0
⇒ (x – 20) (9x + 25) = 0
Either, x – 20 = 0, then x = 20
or 9x + 25 = 0, then 9x = -25 ⇒ x = \(\frac { -25 }{ 9 }\)
But it is not possible being negative.
First pipe will fill it in 20 minutes
and second will fill in = 20 + 5 = 25 minutes-

Question 57.
Solution:
Two water taps together can fill a tank in = 6 hours
Let first tap of larger diameter can fill it in x hour
Then second tap will fill it in (x – 9) hour
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 35

PROBLEMS ON AREA AND GEOMETRY
Question 58.
Solution:
Let breadth of a rectangle (b) = x cm
Then length (l) = 2x cm
Area = l x b = 2x x x
2x x x = 288
⇒ 2x² = 288
⇒ x² = 144 = (12)²
x = 12
Length = 2x = 2 x 12 = 24 cm
and breadth = x = 12 cm

Question 59.
Solution:
Let breadth of a rectangular field (b) = x m
Then its length (l) = 3x m
Area = l x b = 3x x x = 3x²
3x² = 147
⇒ x² = 49 = (7)²
x = 7
Length = 3x = 3 x 7 = 21 m
and breadth = x = 7 m

Question 60.
Solution:
Let breadth of a hall (b) = x m
Then its length (l) = (x + 3) m
Area = l x b = (x + 3) x x = x² + 3x
x² + 3x = 238
⇒ x² + 3x – 238 = 0
⇒ x² + 11x – 14x – 238 = 0
⇒ x (x + 17) – 14 (x – 17) = 0
⇒ (x + 17) (x – 14) = 0
Either, x + 17 = 0, then x = -17 which is not possible being negative.
or x – 14 = 0, then x = 14
Breadth = 14 m
and length = 14 + 3 = 17 m

Question 61.
Solution:
Perimeter of a rectangular plot = 62 m
and area = 228 m²
Let length of plot = x m
Then width = \(\frac { 62 }{ 2 }\) – x = 31 – x
Area = l x b = x (31 – x)
x (31 – x) = 228
⇒ 31x – x² = 228
⇒ x² – 31x + 228 = 0
⇒ x² – 19x – 12x + 228 = 0
⇒ x (x – 19) – 12 (x – 19) = 0
⇒ (x – 19) (x – 12) = 0
Either, x – 19 = 0, then x = 19
or x – 12 = 0, then x = 12
If length is 19 m, then breadth = 31 – 19 = 12 m
and if length is 12 m then breadth = 31 – 12 = 19 m
which is not possible as length > breadth
Dimensions are (19, 12) m

Question 62.
Solution:
Area of path around it = 120 m²
Length of a rectangular field (l) = 16 m
and breadth (b) = 10 m
Let a path of x m wide be made around it, then its
Outer length (L) = (16 + 2x) m
and breadth (B) = (10 + 2x) m
Area of path = L x B – l x b
= (16 + 2x) (10 + 2x) – 16 x 10
⇒ 160 + 52x + 4x² – 160 = 120
⇒ 4x² + 52x – 120 = 0
⇒ x² + 13x – 30 = 0 (Dividing by 4)
⇒ x² + 15x – 2x – 30 = 0
⇒ x (x + 15) – 2 (x + 15) = 0
⇒ (x + 15) (x – 2) = 0
Either, x + 15 = 0, then x = -15 which is not possible being negative,
or x – 2 = 0, then x = 2
Width of path = 2 m

Question 63.
Solution:
The sum of areas of two squares = 640 m²
and difference between their perimeters = 64 m
Let 4 sides of first square = x m
Then perimeter = 4x
Perimeter of second square = 4x – 64
and side = \(\frac { 4x – 64 }{ 4 }\) = (x – 16) m
Now sum of their areas,
x² + (x – 16)² = 640
⇒ x² + x² – 32x + 256 = 640
⇒ 2x² – 32x + 256 – 640 = 0
⇒ 2x² – 32x – 384 = 0
⇒ x² – 16x – 192 = 0
⇒ x² – 24x + 8x – 192 = 0
⇒ x (x – 24) + 8 (x – 24) = 0
⇒ (x – 24) (x + 8) = 0
Either, x – 24 = 0, then x = 24
or x + 8 = 0, then x = -8 which not possible being negative,
x = 24
Side of first square = 24 m
and side of second square = 24 – 16 = 8 m

Question 64.
Solution:
Let side of a square = x m
Then length of rectangle = 3x m
and width = (x – 4) m
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 36
Now area of square = x² m²
and area of rectangle = l x b = (3x) x (x – 4) m²
According to the condition,
x² = 3x (x – 4)
⇒ x² = 3x² – 12x
⇒ 3x² – 12x – x2 = 0
⇒ 2x² – 12x = 0
⇒ 2x (x – 6) = 0
⇒ x – 6 = 0
⇒ x = 6m
Side of square = 6 m
and length of rectangle = 3 x 6 m = 18m
and width = x – 4 = 6 – 4 = 2m

Question 65.
Solution:
Area of a rectangular garden = 180 sq m
Let length of garden = x m
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 37
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 38

Question 66.
Solution:
Area of a right triangle = 600 cm²
Let altitude = x cm
Then base = (x + 10) cm
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 39
Area = \(\frac { 1 }{ 2 }\) base x altitude
600 = \(\frac { 1 }{ 2 }\) x (x + 10) x
x² + 10x = 1200
⇒ x² + 10x – 1200 = 0
⇒ x² + 40x – 30x – 1200 = 0
⇒ x (x + 40) – 30 (x + 40) = 0
⇒ (x + 40) (x – 30) = 0
Either, x + 40 = 0, then x = -40 which is not possible being negative.
or x – 30 = 0, then x = 30
Altitude = 30 cm
Base = 30 + 10 = 40 cm
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 40

Question 67.
Solution:
Area of right triangle = 96 sq. m
Let altitude of the triangle = x m
Then base = 3x m
Area = \(\frac { 1 }{ 2 }\) x base x altitude
96 = \(\frac { 1 }{ 2 }\) x 3x x x
⇒ 3x² = 96 x 2
x² = 64 = (8)²
x = 8 m
Base = 3x = 3 x 8 = 24m
and altitude = x = 8 m

Question 68.
Solution:
Area of right triangle = 165 sq m
Let altitude = x m
Then base = – (x + 7) m
Area = \(\frac { 1 }{ 2 }\) x base x altitude
165 = \(\frac { 1 }{ 2 }\) x x x (x + 7)
330 = x² + 7x
⇒ x² + 7x – 330 = 0
⇒ x² + 22x – 15x – 330 = 0
⇒ x (x + 22) – 15 (x + 22) = 0
⇒ (x + 22) (x – 15) = 0
Either, x + 22 = 0, then x = -22 which is not possible being negative,
or x – 15 = 0, then x = 15
Altitude = x + 7 = 15 + 7 = 22m
and base = x = 15 m

Question 69.
Solution:
In a right triangle,
Hypotenuse = 20 m
Difference of other two sides = 4 m
Let first side = x m
Then second side = (x + 4) m
According to the condition,
(x)² + (x + 4)² = (20)²
⇒ x² + x² + 8x + 16 = 400
⇒ 2x² + 8x + 16 – 400 = 0
⇒ 2x² + 8x – 384 = 0
⇒ x² + 4x – 192 = 0
⇒ x² + 16x – 12x – 192 = 0
⇒ x (x + 16) – 12 (x + 16) = 0
⇒ (x + 16) (x – 12) = 0
Either, x + 16 = 0, then x = -16 which is not possible being negative,
or x – 12 = 0, then x = 12
First side = 12 m
Second side = x + 4= 12 + 4 = 16 m

Question 70.
Solution:
In a right triangle, let base = x m
Length of hypotenuse = (x + 2) m
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 41

Question 71.
Solution:
Let shorter side of a right triangle = x m
Then hypotenuse = (2x – 1) m
and longer side = (x + 1) m
According to the condition,
(2x – 1)² = x² + (x + 1)² (Use of Pythagoras Theorem)
⇒ 4x² – 4x + 1 = x² + x² + 2x + 1
⇒ 4x² – 4x + 1 – x² – x² – 2x – 1 = 0
⇒ 2x² – 6x = 0
⇒ 2x (x – 3) = 0
⇒ x – 3 = 0
⇒ x = 3
Shorter side = 3 m
Hypotenuse = 2x – 1 = 3 x 2 – 1 = 6 – 1 = 5 m
and longer side = x + 1 = 3 + 1 = 4m
Sides 3 m, 4 m, 5 m

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NCERT Solutions for Class 9 Science Chapter 6 Tissues

NCERT Solutions for Class 9 Science Chapter 6 Tissues

These Solutions are part of NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 6 Tissues. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Biology) Chapter 6 – Tissues solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 6 – Tissues Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

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NCERT TEXT BOOK QUESTIONS

IN TEXT QUESTIONS

Question 1.
What is a tissue ?
Answer:
Tissue is a group of related cells that have a common origin and perform a common function.

Question 2.
What is the utility of tissues in multicellular organisms ?
Answer:

  1. Division of Labour: Tissues bring about division of labour in multicellular organisms. It increases efficiency.
  2. Higher Organisation: Tissues become organised to form organs and organ systems.
  3. Individual Cells: Work load of individual cells has decreased.
  4. Higher Survival: Because of division of labour, higher efficiency and organisation, the multicellular organisms have high survival.

Question 3.
Name types of simple tissues.
Answer:
Three – parenchyma, collenchyma and sclerenchyma. (Meristematic tissue is also a simple tissue).

Question 4.
Where is apical meristem found ?
Answer:
Apical meristem occurs at root and stem tips.

Question 5.
Which tissue makes up the husk of coconut ?
Answer:
Sclerenchyma.

Question 6.
What are the constituents of phloem ?
Answer:
Sieve tubes, companion cells, phloem parenchyma and phloem fibres.

Question 7.
Name the tissue responsible for movement of our body.
Answer:
Muscular tissue.

Question 8.
What does a neuron look-like ?
Answer:
A miniature tree with thin hair like parts arising from its ends.

Question 9.
Give three features of cardiac muscles.
Answer:

  1. Cells/Fibres: They are small, cylindrical, uninucleate striated with short lateral branches.
  2. Intercalated Discs: In the area of union becween the two adjacent cardiac muscle fibres, zig-zag junctions called intercalated discs develop. The intercalated discs function as impulse boosters.
  3. Rhythmic Contractions: The muscles are involuntary and nonfatigued which continue to contract and relax tirelessly throughout life.

Question 10.
What are the functions of areolar tissue ?
Answer:

  1. Packing: Areolar tissue provides packing material in various organs.
  2. Binding: It binds various structures with one another in such a way as to prevent their dislocation while allowing Macrophage limited movement.
  3. Covering: It provides covering over nerves, muscles and blood vessels.
  4. Repair: The tissue provides materials for repair of injury.

NCERT CHAPTER END EXERCISES

Question 1.
Define the term “tissue”.
Answer:
Tissue is a group of related cells that have a common origin and perform a common function.

Question 2.
How many types of elements together make up the xylem tissue ? Name them.
Answer:
Xylem tissue is formed of four types of elements. They are tracheids, vessels, xylem parenchyma and xylem fibres.

Question 3.
How are simple tissues different from complex tissues in plants ? (CCE 2014)
Answer:
Differences between Simple and Complex Tissues

Simple Tissues Complex Tissues
1. Cells: A simple tissue is formed of only one type of cells. A complex tissue is made of more than one type of cells.
2. Activity: All the cells perform the same function. The different cells perform different fractions of a function.
3. Types: There are three types of simple plant tissues— . parenchyma, collenchyma and sclerenchyma. There are two types of complex plant tissues— xylem and phloem.
4. Function: They form primary structure of the plant. They form transport system of the plant.

Question 4.
Differentiate amongst parenchyma, collenchyma and sclerenchyma on the basis of the cell wall.
Answer:

Parenchyma Collenchyma Sclerenchyma

1. Thickness: The cell wall is thin.

2. Smoothness: It is smooth.

3. Nature: Wall is formed of cellulose.

 It is thickened.

It is unevenly thickened.

The thickening is pectocellulosic.

 It is thickened.

The wall is uniformly thickened.

The thickening is generally of lignin.

Question 5.
What are the functions of stomata ?
Answer:
Functions of Stomata:

  1. Gaseous Exchange: Stomata are sites where exchange of gases (carbon dioxide and oxygen) occurs between the plant interior and external environment.
  2. Transpiration: Major part of transpiration occurs through stomata. Transpiration removes excess water and keeps plant surfaces cool even in bright sun.
  3. Regulation: They regulate both gaseous exchange and transpiration.

Question 6.
Diagramatically show the difference amongst three types of muscle fibres. (CCE 2014)
Answer:
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 1
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 2
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 3

Question 7.
What is the specific function of cardiac muscle ?
Answer:
Rhythmic contraction and relaxation continuously throughout life without getting fatigued.

Question 8.
Differentiate amongst striated, unstriated and cardiac muscles on the basis of their structure and site/ location in the body.
Answer:

Striated Muscle Fibres

Smooth Muscle Fibres

Cardiac Muscle Fibres
1. Cells. They are long cylindrical cells. The fibres are elongated and spindle­shaped. The cells are small and cylindrical.
2. Ends. The fibres have blunt ends. The fibres have pointed ends. The fibres have broad ends.
3. Striations. They possess striations Striations or light and dark bands Striations are present but they are
or alternate light and dark bands. are absent. fainter than those of striated muscle fibres.
4. Intercalated Discs and Cross Intercalated discs and cross- Intercalated discs and cross-
Connections. They are absent. connections are absent. connections are present.
5. Nucleus. The muscle fibre is Smooth muscle fibre is uninucleate. The cells are uninucleate. Nucleus
multinucleate. Nuclei are oval in Nucleus is centrally placed, oval or in oval-rounded. It is centrally
outline. They occur peripherally below the sarcolemma. elongated. placed.
6. Arrangement. They occur in bundles. They generally form sheets. They form a network.

Question 9.
Draw a labelled diagram of a neuron. (CCE 2013)
Answer:
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 4

Question 10.
Name the following :
(a) Tissue that forms the inner lining of our mouth,
(b) Tissue that connects muscle to bone in humans.
(c) Tissue that transports food in plants.
(d) Tissue that stores fat in our body.
(e) Connective tissue with a fluid matrix.
(f) Tissue present in the brain.
Answer:
(a) Epithelial tissue
(b) Tendon
(c) Phloem
(d) Adipose
(e) Blood
(f) Nervous tissue.

Question 11.
Identify the types of tissue in the following : skin, bark of tree, bone, lining of kidney tubule, vascular bundle.
Answer:
(a) Skin— Epithelial tissue
(b) Bark of Tree— Cork (Protective tissue)
(c) Bone— Connective tissue with solid matrix
(d) Lining of Kidney Tubule— Epithelial tissue
(e) Vascular Bundle— Complex or vascular tissues, xylem and phloem.

Question 12.
Name the regions in which parenchyma tissue is present.
Answer:
It occurs in almost all nonwoody parts of the plants— cortex, pith, medullary rays of stem, cortex and pith of root, chlorenchyma of leaves, flowers, pith of fruits, etc. Epidermis is special type of parenchyma.

Question 13.
What is the role of epidermis in plants ?
Answer:

  1. Protection,
  2. Regulation of transpiration,
  3. Formation of insulating stationary air layer with the help of hair,
  4. Exchange of gases.

Question 14.
Flow does the cork act as a protective tissue ?
Answer:
Cork acts as a protective layer because its cells are dead, filled with tannins, resin and air, impermeable due to deposition of suberin over the cell walls and absence of intercellular spaces. It is insulating (heat proof), fire proof, shock proof, water proof and repellent to microbes and animals.

Question 15.
Complete the table :
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 5
Answer:
(i) Parenchyma
(ii) Sclerenchyma
(iii) Phloem.

SELECTION TYPE QUESTIONS

Alternate Response Type Questions
(True/False, Right/Wrong, Yes/No)

Question 1.
Vacuoles are absent in meristematic plant cells.
Question 2.
Sderenchyma has irregularly thickened cells.
Question 3.
Absorptive surface areas of roots are increased by the presence of root hair.
Question 4.
Cells of connective tissue are compactly packed with no intercellular spaces.
Question 5.
Cardiac muscles undergo rhythmic contraction and relaxation throughout life.
Question 6.
Areolar connective tissue binds muscles with bones.
Question 7.
Cells of cork are dead, suberised and compacdy arranged.
Question 8.
Voluntary muscles control the movement of iris of eye.

Matching Type Questions

Question 9.
Match the contents of the columns A and B (single matching)
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 6

Question 10.
Match the contents of columns I, II and III (Double matching)
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 7

Question 11.
Which one of the following tissues are involved in growth (G), absorption (A), transportation (T) (Key or check list items)
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 8

Question 12.
Match the Stimulus with Appropriate Response.
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 9

Fill in the Blanks

Question 13. Animals move around in search of ………….. mate and shelter.
Question 14. A thick waterproof coating of ………… occurs over the epidermis in desert plants.
Question 15. ……………. consists of tracheids, vessels, parenchyma and fibres.
Question 16. ………….. epithelium occurs in the lining of renal tubules and ducts of salivary glands.
Question 17. Tendons connect …………….. with bones.

Answers:
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 10

SOME ACTIVITY BASED QUESTIONS

Question 1.
Why is blood considered to be connective tissue ?
Answer:

  1. Like other connective tissues, blood consists of living cells scattered in an abundant matrix. The matrix is liquid or plasma in blood.
  2. Blood circulates throughout the body, receiving and providing materials to all tissues and organs of the body. It thus connects all parts of the body.

Question 2.
What is aerenchyma ? Give its functions.
Answer:
Definition: It is a specialised parenchyma found in aquatic plants which consists of network of small cells that enclose large air cavities.
Functions:

  1. Storage of Gases: It stores metabolic gases, O2 and CO2, for use inside the plants in respiration and photosynthesis.
  2. Buoyancy: It makes the aquatic plant buoyant both inside and over the surface of water.

Question 3.
What is skeletal connective tissue ? Give its functions.
Answer:
Definition: Skeletal connective tissue is that connective tissue in which the matrix is solid and the living cells occur inside fluid filled spaces called lacunae. It is of two types, cartilage and bone.
Functions:

  1. Endoskeleton: It forms the internal supporting framework of the animal body.
  2. Protection: The tissue protects the vital organs like brain, spinal cord, heart, lungs, etc.
  3. Joints: The tissue forms joints which allow for growth and movement of body parts.
  4. Muscles: It provides surface for attachement to muscles.
  5. Blood Cells: They develop inside red marrow of bones.
  6. Minerals: Bony skeleton stores minerals, some of which are withdrawn by the body in case of emergency.

Question 4.
What is synapse ? Explain.
Answer:
It is a junction between two neurons without developing an organic union. The terminal knobbed branch end of an axon of one neuron comes in near contact with a dendrite terminal of the next neuron. A narrow fluid filled space occurs between the two. An activated axon end passes out a neurotransmitter like acetylcholine which provides sensation to dendrite terminal. This helps in transfer of impulse from one neuron to the next.

NCERT Solutions for Class 9 Science Chapter 6 Tissues

Hope given NCERT Solutions for Class 9 Science Chapter 6 are helpful to complete your science homework.

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RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1

RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1

Other Exercises

Question 1.
In a ∆ABC, if ∠A = 55°, ∠B = 40°, find ∠C.
Solution:
∵ Sum of three angles of a triangle is 180°
∴ In ∆ABC, ∠A = 55°, ∠B = 40°
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 Q1.1
⇒ 55° + 40° + ∠C = 180°
⇒ 95° + ∠C = 180°
∴ ∠C= 180° -95° = 85°

Question 2.
If the angles of a triangle are in the ratio 1:2:3, determine three angles.
Solution:
Ratio in three angles of a triangle =1:2:3
Let first angle = x
Then second angle = 2x
and third angle = 3x
∴ x + 2x + 3x = 180° (Sum of angles of a triangle)
⇒6x = 180°
⇒x = \(\frac { { 180 }^{ \circ } }{ 6 }\)  = 30°
∴ First angle = x = 30°
Second angle = 2x = 2 x 30° = 60°
and third angle = 3x = 3 x 30° = 90°
∴ Angles are 30°, 60°, 90°

Question 3.
The angles of a triangle are (x – 40)°, (x – 20)° and (\(\frac { 1 }{ 2 }\) x – 10)°. Find the value of x.
Solution:
∵ Sum of three angles of a triangle = 180°
∴ (x – 40)° + (x – 20)° + (\(\frac { 1 }{ 2 }\)x-10)0 = 180°
⇒ x – 40° + x – 20° + \(\frac { 1 }{ 2 }\)x – 10° = 180°
⇒ x + x+ \(\frac { 1 }{ 2 }\)x – 70° = 180°
⇒ \(\frac { 5 }{ 2 }\)x = 180° + 70° = 250°
⇒ x = \(\frac { { 250 }^{ \circ }x 2 }{ 5 }\)  = 100°
∴ x = 100°

Question 4.
Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle.
Solution:
Let each of the two equal angles = x
Then third angle = x + 30°
But sum of the three angles of a triangle is 180°
∴ x + x + x + 30° = 180°
⇒ 3x + 30° = 180°
⇒3x = 150° ⇒x = \(\frac { { 150 }^{ \circ } }{ 3 }\) = 50°
∴ Each equal angle = 50°
and third angle = 50° + 30° = 80°
∴ Angles are 50°, 50° and 80°

Question 5.
If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.
Solution:
In the triangle ABC,
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 Q5.1
∠B = ∠A + ∠C
But ∠A + ∠B + ∠C = 180°
⇒∠B + ∠A + ∠C = 180°
⇒∠B + ∠B = 180°
⇒2∠B = 180°
∴ ∠B = \(\frac { { 180 }^{ \circ } }{ 2 }\) = 90°
∵ One angle of the triangle is 90°
∴ ∆ABC is a right triangle.

Question 6.
Can a triangle have:
(i) Two right angles?
(ii) Two obtuse angles?
(iii) Two acute angles?
(iv) All angles more than 60°?
(v) All angles less than 60°?
(vi) All angles equal to 60°?
Justify your answer in each case.
Solution:
(i) In a triangle, two right-angles cannot be possible. We know that sum of three angles is 180° and if there are two right-angles, then the third angle will be zero which is not possible.
(ii) In a triangle, two obtuse angle cannot be possible. We know that the sum of the three angles of a triangle is 180° and if there are
two obtuse angle, then the third angle will be negative which is not possible.
(iii) In a triangle, two acute angles are possible as sum of three angles of a trianlge is 180°.
(iv) All angles more than 60°, they are also not possible as the sum will be more than 180°.
(v) All angles less than 60°. They are also not possible as the sum will be less than 180°.
(vi) All angles equal to 60°. This is possible as the sum will be 60° x 3 = 180°.

Question 7.
The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angle is 10°, find the three angles.
Solution:
Let three angles of a triangle be x°, (x + 10)°, (x + 20)°
But sum of three angles of a triangle is 180°
∴ x + (x+ 10)° + (x + 20) = 180°
⇒ x + x+10°+ x + 20 = 180°
⇒ 3x + 30° = 180°
⇒ 3x = 180° – 30° = 150°
∴ x = \(\frac { { 180 }^{ \circ } }{ 2 }\) = 50°
∴ Angle are 50°, 50 + 10, 50 + 20
i.e. 50°, 60°, 70°

Question 8.
ABC is a triangle is which ∠A = 72°, the internal bisectors of angles B and C meet in O. Find the magnitude of ∠BOC.
Solution:
In ∆ABC, ∠A = 12° and bisectors of ∠B and ∠C meet at O
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 Q8.1
Now ∠B + ∠C = 180° – 12° = 108°
∵ OB and OC are the bisectors of ∠B and ∠C respectively
∴ ∠OBC + ∠OCB = \(\frac { 1 }{ 2 }\) (B + C)
= \(\frac { 1 }{ 2 }\) x 108° = 54°
But in ∆OBC,
∴ ∠OBC + ∠OCB + ∠BOC = 180°
⇒ 54° + ∠BOC = 180°
∠BOC = 180°-54°= 126°
OR
According to corollary,
∠BOC = 90°+ \(\frac { 1 }{ 2 }\) ∠A
= 90+ \(\frac { 1 }{ 2 }\) x 72° = 90° + 36° = 126°

Question 9.
The bisectors of base angles of a triangle cannot enclose a right angle in any case.
Solution:
In right ∆ABC, ∠A is the vertex angle and OB and OC are the bisectors of ∠B and ∠C respectively
To prove : ∠BOC cannot be a right angle
Proof: ∵ OB and OC are the bisectors of ∠B and ∠C respectively
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 Q9.1
∴ ∠BOC = 90° x \(\frac { 1 }{ 2 }\) ∠A
Let ∠BOC = 90°, then
\(\frac { 1 }{ 2 }\) ∠A = O
⇒∠A = O
Which is not possible because the points A, B and C will be on the same line Hence, ∠BOC cannot be a right angle.

Question 10.
If the bisectors of the base angles of a triangle enclose an angle of 135°. Prove that the triangle is a right triangle.
Solution:
Given : In ∆ABC, OB and OC are the bisectors of ∠B and ∠C and ∠BOC = 135°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 Q10.1
To prove : ∆ABC is a right angled triangle
Proof: ∵ Bisectors of base angles ∠B and ∠C of the ∆ABC meet at O
∴ ∠BOC = 90°+ \(\frac { 1 }{ 2 }\)∠A
But ∠BOC =135°
∴ 90°+ \(\frac { 1 }{ 2 }\) ∠A = 135°
⇒ \(\frac { 1 }{ 2 }\)∠A= 135° -90° = 45°
∴ ∠A = 45° x 2 = 90°
∴ ∆ABC is a right angled triangle

Question 11.
In a ∆ABC, ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that ∠BOC = 120°. Show that ∠A = ∠B = ∠C = 60°.
Solution:
Given : In ∠ABC, BO and CO are the bisectors of ∠B and ∠C respectively and ∠BOC = 120° and ∠ABC = ∠ACB
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 Q11.1
To prove : ∠A = ∠B = ∠C = 60°
Proof : ∵ BO and CO are the bisectors of ∠B and ∠C
∴ ∠BOC = 90° + \(\frac { 1 }{ 2 }\)∠A
But ∠BOC = 120°
∴ 90°+ \(\frac { 1 }{ 2 }\) ∠A = 120°
∴ \(\frac { 1 }{ 2 }\) ∠A = 120° – 90° = 30°
∴ ∠A = 60°
∵ ∠A + ∠B + ∠C = 180° (Angles of a triangle)
∠B + ∠C = 180° – 60° = 120° and ∠B = ∠C
∵ ∠B = ∠C = \(\frac { { 120 }^{ \circ } }{ 2 }\) = 60°
Hence ∠A = ∠B = ∠C = 60°

Question 12.
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.
Solution:
In a ∆ABC,
Let ∠A < ∠B + ∠C
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 Q12.1
⇒∠A + ∠A < ∠A + ∠B + ∠C
⇒ 2∠A < 180°
⇒ ∠A < 90° (∵ Sum of angles of a triangle is 180°)
Similarly, we can prove that
∠B < 90° and ∠C < 90°
∴ Each angle of the triangle are acute angle.

Hope given RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C

RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1C.

Other Exercises

Question 1.
Solution:
(i) \(\\ \frac { -2 }{ 5 } \) + \(\\ \frac { 4 }{ 5 } \)
= \(\\ \frac { -2+4 }{ 5 } \) = \(\\ \frac { 2 }{ 5 } \)
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 1

Question 2.
Solution:
(i) \(\frac { 3 }{ 4 } +\left( \frac { -3 }{ 5 } \right) \)
\(\frac { 15+\left( -12 \right) }{ 20 } =\frac { 15-12 }{ 20 } =\frac { 3 }{ 20 } \)
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 2
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 3
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 4

Question 3.
Solution:
(i) L.H.S = \(\\ \frac { -12 }{ 5 } \) + \(\\ \frac { 2 }{ 7 } \)
= \(\\ \frac { -84+10 }{ 35 } \) = \(\\ \frac { -74 }{ 35 } \)
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 5
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 6
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 7

Question 4.
Solution:
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 8
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 9
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 10
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 11
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 12

Question 5.
Solution:
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 13
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 14

Question 6.
Solution:
We know that additive inverse of \(\\ \frac { a }{ b } \) is \(\\ \frac { -a }{ b } \) and of \(\\ \frac { -a }{ b } \) is \(\\ \frac { a }{ b } \).
Therefore Additive inverse of
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 15
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 16

Question 7.
Solution:
(i) \(\\ \frac { 3 }{ 4 } \) from \(\\ \frac { 1 }{ 3 } \) or \(\\ \frac { 1 }{ 3 } \) – \(\\ \frac { 3 }{ 4 } \)
= \(\\ \frac { 4-9 }{ 12 } \) = \(\\ \frac { -5 }{ 12 } \)
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 17
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 18
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 19
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 20

Question 8.
Solution:
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 21
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 22
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 23

Question 9.
Solution:
Sum of two numbers = – 2
one number = \(\\ \frac { -14 }{ 5 } \)
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 24

Question 10.
Solution:
Sum of two numbers = \(\\ \frac { -1 }{ 2 } \)
One number = \(\\ \frac { 5 }{ 6 } \)
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 25

Question 11.
Solution:
Sum of two numbers = \(\\ \frac { -3 }{ 2 } \)
one number = \(\\ \frac { -5 }{ 8 } \)
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 26

Question 12.
Solution:
Sum of two numbers = \(\\ \frac { 5 }{ 7 } \)
One number = – 1
Second number = \(\\ \frac { 5 }{ 7 } \) – ( – 1)
= \(\\ \frac { 5 }{ 7 } \) + \(\\ \frac { 1 }{ 1 } \)
\(\\ \frac { 5+7 }{ 7 } \) = \(\\ \frac { 12 }{ 7 } \)

Question 13.
Solution:
Difference of two numbers = \(\\ \frac { -1 }{ 6 } \)
One number = \(\\ \frac { -2 }{ 3 } \)
RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1C 27

Question 14.
Solution:
(i) 0 is the rational number.
(ii) Yes, the difference of two rational numbers is also rational.
(iii) Yes, addition is commutative.
(iv) Yes, addition associative.
(v) No, subtraction is not commutative.
(vi) No, subtraction is not associative.
(vii) The number itself.

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NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom

NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom

Multiple Choice Questions

Question 1.
Which of the following correctly represents the electonic distribution in the Mg atom ?
(a) 3, 8, 1         (b) 2, 8, 2
(c) 1, 8, 3         (d) 8, 2, 2.
Correct Answer:
(b).

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Question 2.
Rutherfords alpha (a) particles scattering experiment resulted in to discovery of :
(a) Electron
(b) Proton
(c) Nucleus in the atom
(d) Atomic mass.
Correct Answer:
(c).

Question 3.
The number of electrons in an element X is 15 and the number of neutrons is 16. Which of the following is the correct representation of the element ?
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 1
Correct Answer:
(a) Mass no. = no. of electrons (protons) + no. of neutrons = 15 + 16 = 31.

Question 4.
Daltons atomic theory successfully explained :
(i) Law of conservation of mass
(ii) Law of constant composition
(iii)Law of radioactivity
(iv) Law of multiple proportion
(a) (i), (ii) and (iii)         (b) (i), (ii) and (iv)
(c) (ii), (iii) and (iv)       (d) (i), (ii) and (iv).
Correct Answer:
(d) Except for the law of radioactivity, Dalton’s atomic theory explained all other laws which have been listed.

Question 5.
Which of the following statements about Rutherford’s model of atom are correct ?
(i) Considered the nucleus as positively charged
(ii) Established that the α-particles are four times as heavy as a hydrogen atom
(iii) Can be compared to solar system
(iv) Was in agreement with Thomson’s model
(a) (i) and (iii)         (b) (ii) and (iii)
(c) (i) and (iv)         (d) only (i).
Correct Answer:
(a) The statements (i) and (iii) are both correct.

Question 6.
Which of the following are true for an element ?
(i) Atomic number = number of protons + number of electrons
(ii) Mass number = number of protons + number of neutrons
(iii) Atomic Mass = number of protons + number of neutrons
(iv) Atomic number = number of protons = number of electrons
(a) (i) and (ii)           (b) (i) and (iii)
(c) (ii) and (iii)         (d) (ii) and (iv).
Correct Answer:
(d) The statements (it) and (iv) are both correct.

Question 7.
In the Thomson’s model of atom, which of the following statements are correct ?
(i) The mass of the atom is assumed to be uniformly distributed over the atom
(ii) The positive charge is assumed to be uniformly distributed over the atom
(iii) The electrons are uniformly distributed in the positively charged sphere
(iv) The electrons attract each other to stabilise the atom
(a) (i), (ii) and (iii)             (b) (i) and (iii)
(c) (i) and (iv)                   (d) (i), (iii) and (iv).
Correct Answer:
(a) Except for the statements (iv), all other statements are correct.

Question 8.
Rutherford’s α-particle scattering experiment showed that
(i) electrons have negative charge
(ii) the mass and positive charge of the atom is concentrated in the nucleus
(iii) neutron exists in the nucleus
(iv) most of the space in atom is empty Which of the above statements are correct ?
(a) (i) and (iii)        (b) (ii) and (iv)
(c) (i) and (iv)        (d) (iii) and (iv).
Correct Answer:
(b) The statements (ii) and (iv) are correct.

Question 9.
The ion of an element has 3 positive charges. Mass , number of the atom is 27 and the number of neutrons is 14. What is the number of electrons in the ion ?
(a) 13      (b) 10
(c) 14      (d) 16.
Correct Answer:
(b) The ion has 13 protons (27-14) and 10 electrons. It is Al3+ ion.

Question 10.
Identify the Mg2+ ion from the given figure where n and p represent the number of neutrons and protons respectively
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 2
Correct Answer:
(d) Since the ion has 10 electrons, it is Mg2+ ion.

Question 11.
In a sample of methyl ethanoate (CH3COOCH3) the two oxygen atoms have the same number of electrons but different number of neutrons. Which of the following is the correct reason for it ?
(a) One of the oxygen atoms has gained electrons
(b) One of the oxygen atoms has gained two neutrons
(c) The two oxygen atoms are isotopes
(d) The two oxygen atoms are isobars.
Correct Answer:
(c) The two oxygen atoms are the isotopes. These are the different atoms of the same element which have same number of electrons (or atomic number) but different number of neutrons.

Question 12.
Elements with valency 1 are :
(a) always metals
(b) always metalloids
(c) either metals or non-metals
(d) always non-metals.
Correct Answer:
(c) Elements with valency 1 may be metals (+1 valency) or non-metals (-1 valency)

Question 13.
The first model of an atom was given by :
(a) N. Bohr
(b) E. Goldstein
(c) Rutherford
(d) J.J. Thomson.
Correct Answer:
(d).

Question 14.
An atom with 3 protons and 4 neutrons will have a valency of :
(a) 3         (b) 7
(c) 1         (d) 4.
Correct Answer:
(c) The atom has also 3 electrons with electronic configuration 2, 1. It has valency of 1.

Question 15.
The electronic distribution in an aluminium atom is :
(a) 2, 8, 3         (b) 2, 8, 2
(c) 8, 2, 3         (d) 2, 3, 8.
Correct Answer:
(d).

Question 16.
Which of the following do not represent Bohr’s model of an atom correctly ?
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 3
(a) (i) and (ii)           (b) (ii) and (iii)
(c) (ii) and (iv)          (d) (i) and (iv).
Correct Answer:
(c) The first shell (K) cannot have more than two electrons in it.

Question 17.
Which of the following statements is always correct ?
(a) An atom has equal number of electrons and protons.
(b) An atom has equal number of electrons and neutrons.
(c) An atom has equal number of protons and neutrons.
(d) An atom has equal number of electrons, protons and neutrons.
Correct Answer:
(a) The number of electrons or protons in an atom are always equal.

Question 18.
Atomic models have been improved over the years. Arrange the following atomic models in the order of their chronological order :
(i) Rutherford’s atomic model
(ii) Thomson’s atomic model
(iii) Bohr’s atomic model
(a) (i), (ii) and (iii)
(b) (ii), (iii) and (i)
(c) (ii), (i) and (iii)
(d) (iii), (ii) and (i).
Correct Answer:
(c). It is the correct order.

Short Answer Questions

Question 19.
An element has one proton, one electron and no neutron. Name the element. How will you represent it ?
Answer:
The element is called protium or hydrogen (H). It can be represented as 11H.

Question 20.
Write any two observations to show that atoms are divisible.
Answer:
The discovery of electrons and protons led to the opinion that the atoms are divisible. Even neutrons were also discovered at a later stage.

Question 21.
Will 35Cl and 37Cl have different valencies ? Justify your answer.
Answer:
The atomic number (Z) of the element (Cl) is 17. Its electronic distribution is 2, 8, 7. It has valency (8 – 7) = 1 which is the same for both the species, which act as isotopes.

Question 22.
Why did Rutherford select a gold foil in his α-scattering experiment ?
Answer:
Gold is a highly malleable. It could be converted into very fine foils or leaves. Moreover, the nucleus of the element is very heavy. It could not be displaced by the impact of fast moving α-particles.

Question 23.
Find out the valency of the atoms represented by the Fig. (a) and (b).
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 4Answer:
(a) The electronic configuration of the atom is 2, 8, 8. It has completely filled K, L, M shells. Its valency is zero. The atom belongs to the element Argon (Ar).
(b) The electronic configuration of the atom is 2, 7. It has seven electrons in the valence shell. Its valency is (8 – 7) equal to one. The atom belongs to the element fluorine (F).

Question 24.
One electron is present in the outermost shell of the atom of an element X. What would be the nature and value of charge on the ion formed if this electron is removed from the outermost shell ?
Answer:
Since the atom of the element X has lost one electron from the outermost shell, it has +1 unit charge.

Question 25.
Write down the electron distribution of chlorine atom. How many electrons are there in the L shell ? (Atomic number of chlorine is 17).
Answer:
The electronic distribution in chlorine (Z = 17) is K(2), L(8) and M(7). The L-shell has eight electrons.

Question 26.
In the atom of an element X, 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed ?
Answer:
With six (6) electrons in the outermost shell, the atom of the element needs two (2) electrons more to have eight (8) electrons in the outermost shell. The ion (x2-) has -2 units charge.

Question 27.
What information do you get from the following electronic distribution about the atomic number, mass number and valency of atoms X, Y and Z ? Give your answer in a tabular form.
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 5
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 6

  • In atom X, valency is equal to number of valence electrons (2, 3) = 3
  • Atom Y has electronic configuration 2, 6. Valency is eight – No. of valence electrons (8-6)
  • In atom Z, valency is equal to number of valence electrons (2, 8, 5), or eight – no. of valence electrons
    (8 – 5) = 3.

Question 28.
In response to a question, a student stated that in an atom, the number of protons is greater than the number of neutrons, which in turn is greater than the number of electrons. Do you agree with the statement ? Justify your answer.
Answer:
The statement is wrong. In an atom, number of protons and electrons are always equal. These can be different only in an ion (positive or negative).

Question 29.
Calculate the number of neutrons present in the nucleus of an element 3115X which is represented as .
Answer:
Mass no. of the element (A) = 31
Atomic no. of the element (Z) =15
No. of neutrons (n) = A – Z = 31 – 15 = 16.

Question 30.
Match the names of the Scientists given in column A with their contributions towards the understanding of the atomic structure
as given in column B.
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 7
Answer:
(a) — (iii) ;
(b) — (iv);
(c) — (i) ;
(d) — (ii)
(e) — (vi) ;
(f) — (vii) ;
(g) — (v).

Question 31.
The atomic numbers of calcium and argon are 20 and 18 respectively, but the mass number of both these elements is 40. What is the name given to such a pair of elements ?
Answer:
The elements having same mass number but different atomic numbers are known as isobars.

Question 32.
Complete the following table on the basis of information available in the symbols given below :
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 8
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 9

Question 33.
Helium atom has 2 electrons in its valence shell but its valency is not 2. Explain.
Answer:
Helium (He) atom with atomic number (Z) equal to two has only one shell (K-shell). It can have a maximum of two electrons which it has. Therefore, the valence shell is complete and valency of the atom is zero.

Question 34.
Fill in the blanks in the following statements :
(a) Rutherfords α-particle scattering experiment led to the discovery of the .
(b) Isotopes have same but different
(c) Neon and chlorine have atomic numbers 10 and 17 respectively. Their valencies will be
and respectively.
(d) The electronic configuration of silicon is and that of sulphur is
Answer:
(a) nucleus
(b) atomic numbers, mass numbers
(c) zero and one
(d) 2, 8, 4 and 2, 8, 6.

Question 35.
An element X has a mass number 4 and atomic number 2. Write the valency of this element.
Answer:
The element X with mass number 4 and atomic number 2 is helium (He). It has 2 electrons (maximum possible) in its only shell which is K-shell. Therefore, the valency of the element is zero.

Long Answer Questions

Question 36.
Why do Helium, Neon and Argon have zero valency ?
Answer:
All the three elements have completely filled valence shells, according to Bohr Bury scheme. Therefore, their valencies are zero.

Question 37.
The ratio of the radii of hydrogen atom and its nucleus is 105. Assuming the atom and the nucleus to be spherical,
(i) what will be the ratio of their sizes ?
(ii) If atom is represented by planet earth ‘Re’ = 6.4 x 106 m, estimate the radius of the nucleus.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 10

Question 38.
Enlist the conclusions drawn by Rutherford from his α-ray scattering experiment.
Answer:
He made the following conclusions from the experiment.

  1. As most of the alpha particles passed through undeflected, this means that they did not come across any obstruction in their path. Thus, most of the space in an atom is expected to be empty.
  2. As a few alpha particles suffered minor deflections and a very few major deflections, this means that these must have met with some obstructions in their path.
  3. This obstruction must be :
    1. Very small : Only a few particles were obstructed by it.
    2. Massive : Each alpha particle has 4 u mass and is quite heavy. It could easily pass through a light obstruction by pushing it aside.

Question 39.
In what way is the Rutherfords atomic model different from Thomsons atomic model ?
Answer:
According to Thomsons model, an atom may be regarded as a positively charged sphere containing protons in which the negatively charged protons are supposed to be studied or embedded. He gave no clue about the nucleus and extranuclear portion. This was given for the first time by Rutherfords model atom with the help of α-ray scattering experiment.

Question 40.
What were the drawbacks of Rutherfords model of an atom ?
Answer:
There are infact, two main drawbacks of Rutherford Model atom

  1. Rutherford model of atom could not explain the stability of the atom.
  2. Rutherford model of atom could not explain as to how the electrons are distributed in the extra nuclear portion in an atom.

Question 41.
What are the postulates of Bohr’s model of an atom ?
Answer:
The main postulates of the theory are listed :

  1. In the extra nuclear portion of an atom, the electrons revolve in well defined circular paths known as orbits.
  2. These circular orbits are also known as energy levels or energy shells.
  3. These have been designated as K, L, M, N, O, … (or as 1, 2, 3, 4, 5, …) based on the energy present.
  4. The order of the energy of these energy shells is :
    K<L<M<N<0 <…. or 1< 2< 3 < 4<5 <….
  5. While revolving in an orbit, the electron is not in a position to either lose or gain energy. In other words, its energy remains stationary. Therefore, these energy states for the electrons are also known as stationary states.

Question 42.
Show diagramatically the electronic distribution in a sodium atom and a sodium ion and also give their atomic number.
Answer:
The atomic number of sodium (Na) is 11. The electronic distribution in the atom is K(2) L(8) and L(1). Sodium
ion (Na+) is formed by removal of one electron from the atom. Its electronic distribution is K(2) and L(8). This means that Na+ ion has the elecronic configuration of Ne atom which is inert gas atom. The electronic distribution of the atom and ion are shown as follows :
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 11

Question 43.
In the Gold foil experiment of Geiger and Marsden, that paved the way for Rutherfords model of an atom, 1.00% of the α-particles were found to deflect at angles > 50°. If one mole of α-particles were bombarded on the gold foil, compute the number of α-particles that would deflect at angles less than 50°.
Answer:
Percentage (%) of particles deflected at an angle more than 50° = 1%
Percentage (%) of α-particles deflected at an angle less than 50° = 100 – 1 = 99%
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 12

Hope given NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom 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 9 Science Chapter 6 Tissues

NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues

Question 1.
Animals of colder region and fishes of cold water have thicker layer of subcutaneous fat. Describe why ?
Answer:
Subcutaneous fat functions as insulating layer that prevents heat loss from the body in colder environment. Therefore, animals of colder areas and fishes of colder water possess thicker layer of subcutaneous fat. Fat also functions as reserve food during periods of scarcity.

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Question 2.
Match the items of columns A and B.
NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues image - 1
Answer:
a—v,
b—iv,
c—iii,
d—i,
e—ii,
f—vi.

Question 3.
Malch lhe column A witli column B.
NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues image - 2
Answer:
a—i,
b—ii,
c—iv,
d—iii,
e—v.

Question 4.
If a potted plant is covered with a glass jar, water vapours appear on the wall of galss jar. Explan why ?
Answer:
Plants lose water vapours in transpiration from their leaves and young stems. When a potted plant is placed inside a glass jar, the water vapours cannot escape but remain inside. The wall of the glass jar being cool, water vaours condense there.

Question 5.
Name the different components of xylem and draw a living component.
Answer:
Xylem is made of four components — tracheids, vessels (tracheae), xylem fibres and xylem parenchyma. Only xylem parenchyma is the living component.
NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues image - 3

Question 6.
Draw and identify different elements of phloem.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues image - 4
The various components are phloem fibres, phloem parenchyma, sieve tubes and companion cells.

Question 7.
Write true (T) and false (F).
(a) Epithelial tissue is protective tissue in animal body.
(b) The lining of blood vessels, lung alveoli and kidney tubules are all made up of epithelial tissue.
(c) Epithelial cells have a lot of intercellular spaces.
(d) Epithelial layer is permeable layer.
(e) Epithelial layer does not allow regulation of materials between body and external environment.
Answer:
a—T,
b—T,
c—F,
d—T,
e—F.

Question 8.
Differentiate between voluntary and involuntary muscles. Given an example of each type.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues image - 5

Question 9.
Differentiate the following activities on the basis of voluntary (V) or involuntary (IV) muscles.
(a) Jumping of frog,
(b) Pumping of heart
(c) Writing with hand
(d) Movement of chocolate in your intestine.
Answer:
a—V,
b—IV,
c—V,
d—IV.

Question 10.
Fill in the blanks :

  1. Lining of blood vessels is made up of ……………… .
  2. Lining of small intestine is made up of ………….. .
  3. Lining of kidney tubules is made up of ……………. .
  4. Epithelial cells with cilia are found in ……………… of our body.

Answer:

  1. squamous epithelium
  2. columuar epithelium
  3. cuboidal epithelium
  4. respiratory tract.

Question 11.
Water Hyacinth floats on water surface. Explain. (CCE 2012)
Answer:
Water Hyacinth has spongy petioles which enclose a lot of air in its aerenchyma. Air makes the plant lighter than water so that it is able to float on surface of water.

Question 12.
Which structure protects the plant body against the invasion of parasites ?
Answer:
Epidermis is the outer protective layer of plant body which does not allow the parasites to gain entry into the internal tissues due to

  1. Absence of intercellular spaces
  2. Thick outer walls
  3. Deposition of cutin and wax in the cuticle covering the epidermis
  4. Silica and other deposidons.

Question 13.
Fill in the blanks :
(a) Cork cells possess on …………….. their walls that makes them impervious to gases and water.
(b) ……………. have tubular cells with perforated walls and are living in nature.
(c) Bone possesses a hard matrix composed of …………….. and ……………
Answer:
a —suberin,
b —sieve tubes,
c—calcium, phosphorus.

Question 14.
Why is epidermis important for the plants ?
Answer:

  1. Strength: Collenchyma is a living mechanical tissue which provides both mechanical strength as well as flexibility. Because of collenchyma, plant organs can bend without breaking.
  2. Growth: Collenchyma allows growth and elongation of organs.
  3. Storage: Being living tissue, collenchyma stores food.
  4. Photosynthesis: Collenchyma cells may contain chloroplasts and take part in photosynthesis.

Question 15.
Fill in the blanks :

  1. …………….. are forms of complex tissue.
  2. …………….. have guard cells.
  3. Cells of cork contain a chemical called ……………..
  4. Husk of coconut is made of ……………. tissue.
  5. …………… and …………….. are both conducting tissues.
  6. …………….. gives flexibility in plants.
  7. Xylem transports …………….. and ……………. from soil.
  8. Phloem transports ……………… from …………….. to other parts of the plant.

Answer:

  1. Xylem and phloem
  2. Stomata
  3. Suberin
  4. sclerenchyma
  5. Xylem, phloem,
  6. Collenchyma
  7. water, minerals
  8. food, leaves.

Question 16.
Differentiate between sclerenchyma and parenchyma tissue. Draw diagram as well.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues image - 6
NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues image - 7
Differences between Sclerenchyma and Parenchyma
NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues image - 8

Question 17.
Describe the structure and function of different types of epithelial tissues. Draw well labelled diagrams.
Answer:

  1. Protection: Surface epithelium lying over the skin protects the body from drying up, microbes, chemicals and injury. A similar function of protection is carried out by epithelium lining mouth, nasal tract and alimentary canal.
  2. Absorption: Some epithelia have become specialised for absorption, e.g., intestinal mucosa.
  3. Epithelium lining the different parts of uriniferous tubules takes part in ultrafiltration, secretion and reabsorption to produce urine.

NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues image - 9

Question 18.
Draw well albelled diagrams of various types of muscles found in human body.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues image - 10
NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues image - 11
NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues image - 12

Question 19.
Give reasons for
(a) Meristematic cells have a prominent nucleus and dense cytoplasm but they lack vacuole.
(b) Intercellular spaces are absent in sclerencymatous tissue.
(c) We get a crunchy and granular feeling when we chew pear fruit.
(d) Branches of a tree move and bend freely in high wind velocity.
(e) It is difficult to pull out the husk of coconut.
Answer:
{a) They are undergoing divisions and do not store food.
(b) They have lignified walls and form bundles for mechanical function.
(c) It has stone cells or sclereids.
(d) Collenchyma provides them flexibility.
(e) Sderenchyma fibres are closely packed.

Question 20.
List the characteristics of cork. How is it formed ? (CCE 2014) Mention its role.
Answer:
(a) Characteristics:

  1. Outer Tissue: Cork is outer protective tissue of older stems and roots.
  2. Dead cells: The mature cork cells become dead and filled with tannins, resins and air.
  3. Compact Tissue: Intercellular spaces are absent. Cork is a compact tissue.
  4. Suberin: Cork cells are impermeable due to deposition of suberin in their walls.
  5. Multilayered: Cork consists of several layers of cells.
  6. Shape: Cork cells are rectangular in outline.
  7. Lenticels: At places cork bears aerating pores called lenticels.

(b) Formation: Cork is formed from a secondary lateral meristem called phellogen or cork cambium. It develops subepidermally in older stems and roots. Cells cut out on the outer side by cork cambium form cork or phellem while cells cut out on the inner side give rise to secondary cortex or phelloderm. The whole tissue (cork, cork cambium and secondary cortex) is called periderm.
(c) Functions:

  1. Protection: Epidermis is the outer protective layer of the plant that prevents entry of pathogens and pests,
  2. Water Loss: By presence of cuticle, it checks the rate of water loss from aerial parts,
  3. Hair: Occurrence of epidermal hair produces an insulating stationary layer of air. (iv) Stomata. They regulate exchange of gases. Stomata are also the seat of major loss of water in transpiration. Transpiration keeps the aerial parts cool.
  4. Epiblema or epidermis of the root alongwith its root hairs takes part in absorption of water and minerals.

Question 21.
Why are xylem and phloem called complex tissues ? How are they different from one another ?
Answer:
Xylem and phloem are called complex tissues because each of them is made of more than one type of cells which coordinate their activities to perform a common function.
Differences between Xylem and Phloem

Xylem Phloem
I. Conduction. It conducts water and minerals. Phloem conducts organic solutes or food materials.
2. Direction. Conduction is mostly unidirectional. Conduction can be bidirectional.
3. Channels. Conducting channels or tracheary elements are tracheids and vessels. Conducting channels are sieve tubes.
4. Components. Xylem consists of tracheids, vessels, xylem parenchyma and xylem fibres. Phloem consists of sieve tubes, companion cells, phloem parenchyma and phloem fibres.
5. Dead/Living Parts. Three of the four elements of xylem are dead (viz., tracheids, vessels and fibres). Only xylem parenchyma is living. Three of the four elements are living (viz., sieve tubes, companion cells and phloem parenchyma). Only phloem/fibres are dead.
6. Mechanical Strength. In addition to conduction, xylem provides mechanical strength to the plant. There is little mechanical function of phloem.

Question 22.
(a) Differentiate between meristematic and permanent tissues in plants
(b) Define the process of differentiation.
(c) Name any two simple and two complex permanent tissues in plants.
Answer:
(a) Differences between Meristematic and Permanent Tissues

Meristematic Tissue Permanent Tissue
1. Nature. The cells are small, isodiametric and undifferentiated. The cells are large, differentiated with different shapes.
2. Spaces. Intercellular spaces are absent. Intercellular spaces are often present.
3. Vacuoles. They are nearly absent. Large central vacuole occurs in living permanent cells.
4. Nucleus. It is large and prominent. Nucleus is less conspicuous.
5. Wall. Cell wall is thin. Cell wall is thin or thick.
6. Divisions. The cells undergo regular divisions. The cells do not normally divide.
7. Metabolism. Rate of metabolism is high. Metabolic rate is comparatively slower.
8. Life. The cells are living. The cells may be living or dead.
9. Function. It takes part in growth. It provides protection, support, conduction,photosynthesis, storage, etc.

(b) Differentiation : The process of loss of ability to divide and taking up a permanent shape, size, structure and fucntion by newly formed cells is called differentiation.
(c)

  1. Simple Permanent Tissues : Parenchyma, Collenchyma, Sderenchyma.
  2. Complex Permanent Tissues : Xylem, phloem.

Hope given NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues 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.