RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3

RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3

Question 1.
Find the cube of each of the following binomial expressions:
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q1.1
Solution:
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q1.2
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q1.3
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q1.4
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q1.5

Question 2.
If a + b = 10 and ab = 21, find the value of a3 + b3.
Solution:
a + b = 10, ab = 21
Cubing both sides,
(a + b)3 = (10)3
⇒ a3 + 63 + 3ab (a + b) = 1000
⇒  a3 + b3 + 3 x 21 x 10 = 1000
⇒  a3 + b3 + 630 = 1000
⇒  a3 + b3 = 1000 – 630 = 370
∴ a3 + b3 = 370

Question 3.
If a – b = 4 and ab = 21, find the value of a3-b3.
Solution:
a – b = 4, ab= 21
Cubing both sides,
⇒ (a – A)3 = (4)3
⇒ a3 – b3 – 3ab (a – b) = 64
⇒ a3-i3-3×21 x4 = 64
⇒  a3 – 63 – 252 = 64
⇒  a3 – 63 = 64 + 252 =316
∴ a3 – b3 = 316

Question 4.
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q4.1
Solution:
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q4.2

Question 5.
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q5.1
Solution:
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q5.2

Question 6.
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q6.1
Solution:
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q6.2

Question 7.
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q7.1
Solution:
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q7.2
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q7.3

Question 8.
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q8.1
Solution:
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q8.2
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q8.3

Question 9.
If 2x + 3y = 13 and xy = 6, find the value of 8x3 + 21y3.
Solution:
2x + 3y = 13, xy = 6
Cubing both sides,
(2x + 3y)3 = (13)3
⇒ (2x)3 + (3y)3 + 3 x 2x x 3X2x + 3y) = 2197
⇒ 8x3 + 27y3 + 18xy(2x + 3y) = 2197
⇒ 8x3 + 27y3 + 18 x 6 x 13 = 2197
⇒ 8X3 + 27y3 + 1404 = 2197
⇒  8x3 + 27y3 = 2197 – 1404 = 793
∴ 8x3 + 27y3 = 793

Question 10.
If 3x – 2y= 11 and xy = 12, find the value of 27x3 – 8y3.
Solution:
3x – 2y = 11 and xy = 12 Cubing both sides,
(3x – 2y)3 = (11)3
⇒  (3x)3 – (2y)3 – 3 x 3x x 2y(3x – 2y) =1331
⇒  27x3 – 8y3 – 18xy(3x -2y) =1331
⇒   27x3 – 8y3 – 18 x 12 x 11 = 1331
⇒  27x3 – 8y3 – 2376 = 1331
⇒  27X3 – 8y3 = 1331 + 2376 = 3707
∴ 2x3 – 8y3 = 3707

Question 11.
Evaluate each of the following:
(i)  (103)3
(ii) (98)3
(iii) (9.9)3
(iv) (10.4)3
(v) (598)3
(vi) (99)3
Solution:
We know that (a + bf = a3 + b3 + 3ab(a + b) and (a – b)3= a3 – b3 – 3 ab(a – b)
Therefore,
(i)  (103)3 = (100 + 3)3
= (100)3 + (3)3 + 3 x 100 x 3(100 + 3)    {∵ (a + b)3 = a3 + b3 + 3ab(a + b)}
= 1000000 + 27 + 900 x 103
= 1000000 + 27 + 92700
= 1092727
(ii) (98)3 = (100 – 2)3
= (100)3 – (2)3 – 3 x 100 x 2(100 – 2)
= 1000000 – 8 – 600 x 98
= 1000000 – 8 – 58800
= 1000000-58808
= 941192
(iii) (9.9)3 = (10 – 0.1)3
= (10)3 – (0.1)3 – 3 X 10 X 0.1(10 – 0.1)
= 1000 – 0.001 – 3 x 9.9
= 1000 – 0.001 – 29.7
= 1000 – 29.701
= 970.299
(iv) (10.4)3 = (10 + 0.4)3
= (10)3 + (0.4)3 + 3 x 10 x 0.4(10 + 0.4)
= 1000 + 0.064 + 12(10.4)
= 1000 + 0.064 + 124.8 = 1124.864
(v) (598)3 = (600 – 2)3
= (600)3 – (2)3 – 3 x 600 x 2 x (600 – 2)
= 216000000 – 8 – 3600 x 598
= 216000000 – 8 – 2152800
= 216000000 – 2152808
= 213847192
(vi) (99)3 = (100 – 1)3
= (100)3 – (1)3 – 3 x 100 x 1 x (100 – 1)
= 1000000 – 1 – 300 x 99
= 1000000 – 1 – 29700
= 1000000 – 29701
= 970299

Question 12.
Evaluate each of the following:
(i)  1113 – 893
(ii) 463 + 343
(iii) 1043 + 963
(iv) 933 – 1073
Solution:
We know that a3 + b3 = (a + bf – 3ab(a + b) and a3 – b3 = (a – bf + 3 ab(a – b)
(i) 1113 – 893
= (111 – 89)3 + 3 x ill x 89(111 – 89)
= (22)3 + 3 x 111 x 89 x 22
= 10648 + 652014 = 662662
(OR)
(a + b)3 – (a – b)3 = 2(b3 + 3a2b)
= 1113 – 893 = (100 + 11)3 – (100 – 11)3
= 2(113 + 3 x 1002 x 11]
= 2(1331 + 330000]
= 331331 x 2 = 662662
(a + b)3 + (a- b)3 = 2(b3 + 3ab2)
(ii) 463 + 343 = (40 + 6)3 + (40 – 6)3
= 2[(40)3 + 3 x 40 x 62]
= 2[64000 + 3 x 40 x 36]
= 2[64000 + 4320]
= 2 x 68320 = 136640
(iii) 1043 + 963 = (100 + 4)3 + (100 – 96)3
= 2 [a3 + 3 ab2]
= 2[(100)3 + 3 x 100 x (4)2]
= 2[ 1000000 + 300 x 16]
= 2[ 1000000 + 4800]
= 1004800 x 2 = 2009600
(iv) 933 – 1073 = -[(107)3 – (93)3]
= -[(100 + If – (100 – 7)3]
= -2[b3 + 3a2b)]
= -2[(7)3 + 3(100)2 x 7]
= -2(343 + 3 x 10000 x 7]
= -2[343 + 210000]
= -2[210343] = -420686

Question 13.
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q13.1
Solution:
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q13.2
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q13.3
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q13.4

Question 14.
Find the value of 27X3 + 8y3 if
(i) 3x + 2y = 14 and xy = 8
(ii) 3x + 2y = 20 and xy = \(\frac { 14 }{ 9 }\)
Solution:
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q14.1
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q14.2

Question 15.
Find the value of 64x3 – 125z3, if 4x – 5z = 16 and xz = 12.
Solution:
4x – 5z = 16, xz = 12
Cubing both sides,
(4x – 5z)3 = (16)3
⇒ (4x)3 – (5y)3 – 3 x 4x x 5z(4x – 5z) = 4096
⇒ 64x3 – 125z3 – 3 x 4 x 5 x xz(4x – 5z) = 4096
⇒  64x3 – 125z3 – 60 x 12 x 16 = 4096
⇒ 64x3 – 125z3 – 11520 = 4096
⇒  64x3 – 125z3 = 4096 + 11520 = 15616

Question 16.
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q16.1
Solution:
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q16.2
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q16.3

Question 17.
Simplify each of the following:
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q17.1
Solution:
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q17.2
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q17.3
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q17.4
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q17.5

Question 18.
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q18.1
Solution:
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q18.2
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q18.3
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q18.4

Question 19.
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q19.1
Solution:
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q19.2
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 Q19.3

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RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A

RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A.

Question 1.
Solution:
A theorem is a statement that requires a proof while an axiom is the basic fact which is taken for granted without proof.

Question 2.
Solution:
(i) Line segment: The straight line between two points A and B is a called a line segment \(\overline { AB } \)
RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A 1
(ii) Ray : A line segment \(\overline { AB } \) when extended indefinitely is one direction is called a ray \(\overrightarrow { AB } \) It has no definitely length.
RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A 1.1
(iii) Intersecting lines : Two lines having one common point are called intersecting lines and the common point is called the point of intersection.
RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A 1.2
(iv) Parallel Lines : If two lines lying in the same plane do not intersect each other when produced on either side, then these two lines are called parallel lines. The distance between two parallel hues always remains the same.
RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A 1.3
(v) Half line : If we take a point P on a line \(\overleftrightarrow { AB } \), its divides the line into two parts. Each part is called half line or two ray i.e. \(\overrightarrow { PA } \) and \(\overrightarrow { PB } \) .
RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A 1.1.4
(vi) Concurrent lines : Three or more lines intersecting at the same point are called concurrent lines.
RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A 1.1.5
(vii) Collinear points : Three or more points lying on the same line are called collinear points.
RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A 1.1.6
(viii) Plane : A plane is a surface such that every point of the line joining any two points on it, lies on it.

Question 3.
Solution:
(i) Six points are : A, B, C, D, E and F
(ii) Five line segments are : \(\overline { EG }\), \(\overline { FH }\), \(\overline { EF }\), \(\overline { GH }\) and \(\overline { MN }\)
(iii) Four rays are : \(\overrightarrow { EP } \) , \(\overrightarrow { GR } \),\(\overrightarrow { GB } \) and \(\overrightarrow { HD } \)
(iv) Four lines are : \(\overrightarrow { AB } \),\(\overrightarrow { CD } \),\(\overrightarrow { PQ } \) and \(\overrightarrow { RS } \)
(v) Four collinear points are M, E, G, B. Ans

Question 4.
Solution:
(i) \(\overleftrightarrow { EF } \) and \(\overleftrightarrow { GH } \) is a pair of intersecting line whose point of intersection is R
and second pair of intersecting lines is \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { CD } \) and point of intersection is P.
(ii) Three concurrent lines are \(\overleftrightarrow { AB } \), \(\overleftrightarrow { EF } \) and \(\overleftrightarrow { GH } \) and the point of intersection is R.
(iii) Three rays are \(\overleftrightarrow { RB } \),\(\overleftrightarrow { RH } \) and \(\overleftrightarrow { RG } \)
(iv) Two line segments are \(\overleftrightarrow { RQ } \) and \(\overleftrightarrow { RP } \)

Question 5.
Solution:
(i) Through a given point, infinitely many lines can be drawn.
(ii) Only one line can be drawn to pass through two given points.
(iii) Two lines can intersect each other at the most one point
(iv) A, B and C are three collinear points. Then the line segments will be \(\overline { AB } \), \(\overline { BC } \) and \(\overline { AC } \).

Question 6.
Solution:
(iv), (vi), (vii), (viii) and (ix) are true and others are not true.

 

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RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS

RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS

Question 1.
Write the value of (2 + \(\sqrt { 3 } \) ) (2 – \(\sqrt { 3 } \)).
Solution:
(2+ \(\sqrt { 3 } \) )(2- \(\sqrt { 3 } \) ) = (2)2-(\(\sqrt { 3 } \) )2
{∵ (a + b) (a – b) = a2 – b2}
= 4-3=1

Question 2.
Write the reciprocal of 5 + \(\sqrt { 2 } \).
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q2.1

Question 3.
Write the rationalisation factor of 7 – 3\(\sqrt { 5 } \) .
Solution:
Rationalising factor of 7 – 3\(\sqrt { 5 } \) is 7 + 3\(\sqrt { 5 } \)
{∵ (\(\sqrt { a } \) + \(\sqrt { b } \)  ) (\(\sqrt { a } \) – \(\sqrt { b } \)) = a-b}

Question 4.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q4.1
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q4.2

Question 5.
If x =\(\sqrt { 2 } \) – 1 then write the value of \(\frac { 1 }{ x }\).
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q5.1

Question 6.
If a = \(\sqrt { 2 } \) + h then find the value of a –\(\frac { 1 }{ a }\)
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q6.1

Question 7.
If x = 2 + \(\sqrt { 3 } \), find the value of x + \(\frac { 1 }{ x }\).
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q7.1

Question 8.
Write the rationalisation factor of \(\sqrt { 5 } \) – 2.
Solution:
Rationalisation factor of \(\sqrt { 5 } \) – 2 is \(\sqrt { 5 } \) + 2 as
(\(\sqrt { a } \) + \(\sqrt { b } \))(\(\sqrt { a } \) – \(\sqrt { b } \)) = a – b

Question 9.
Simplify : \(\sqrt { 3+2\sqrt { 2 } }\).
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q9.1

Question 10.
Simplify : \(\sqrt { 3-2\sqrt { 2 } }\).
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q10.1

Question 11.
If x = 3 + 2 \(\sqrt { 2 } \), then find the value of  \(\sqrt { x } \) – \(\frac { 1 }{ \sqrt { x } }\).
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q11.1

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HOTS Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure

HOTS Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure

These Solutions are part of HOTS Questions for Class 9 Science. Here we have given HOTS Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure.

Question 1.
A house wife churned full cream with a milk churner

  1. What will she observe after churning the milk ?
  2. What could be the possible reason for the observation ?

Answer:

  1. Churning of milk is a centrifugation process. As a result, lighter particles of cream or butter will move upwards and collect at the top. Heavier residual particles will remain at the bottom. This is a very common process used to separate cream from milk or butter from yogurt.
  2. The separation is based on the principle that lighter particles move upwards while denser particle downwards upon centrifugation.

More Resources

Question 2.
Based on separation techniques, complete the following. The first one is done for you.
HOTS Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 1
Answer:
2. Homogeneous, Evaporation, Difference in nature.
3. Hetereogeneous, Filtration, Difference in solubility in water
4. Homogeneous, Chromatography, Difference in adsorption of different components.

Question 3.
The table given below shows number of grams of five different solids dissolving in 100 g of the solvents : water, alcohol and chloroform (all at 20°C).
HOTS Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 2

  1. Which solid dissolves best in water at 20°C ?
  2. Which solid is maximum soluble in alcohol ?
  3. Which solid is insoluble in all the three solvents ?

Answer:

  1. Sugar is best soluble in water at 20°C
  2. Iodine is maximum soluble in alcohol.
  3. Chalk is insoluble in all the three solvents.

Question 4.
Some solids dissolve easily in liquids while the others donot

  1. What is the name given to the liquids which dissolve solids ?
  2. What is the name given to the clear liquid formed when a solid dissolves in a liquid ?
  3. What is the name given to the liquid which contains in it some suspended particles ?

Answer:

  1. The liquids are known as solvents
  2. The clear liquid is called solution or true solution. .
  3. The liquid is known as dispersion medium or dispersing medium.

Question 5.
Butter is an example of one type of colloidal solution. Name it. Give a reason for your choice.
Answer:
The colloidal solution is an example in which solid acts as the dispersion medium while liquid as the dispersed phase. It is also called gel.
Reason for the choice. On pressing butter, liquid drops come out of it leaving behind a solid. This clearly shows that butter is a gel.

Question 6.

  1. The solubility of sodium chloride in water increases with rise in temperature while that of lithium carbonate decreases. Assign reason.
  2. Water containing 88-8% oxygen and 11-2% hydrogen is often used as a fire extinguisher. Can a mixture containing the two gases in the same ratio by mass be used for extinguishing fire ?
  3. The melting point of a solid when determined experimentally comes out to be 160°C. But its actual melting point as given in standard books is 150°C. Predict the nature of the solid.

Answer:

  1. When sodium chloride is dissolved in water, the process is endothermic in nature. This means that heat energy is absorbed in the process. Therefore, solubility increases with rise in the temperature. In case of lithium carbonate, the process of dissolution is exothermic. This means that heat is evolved in the process. Therefore, its solubility in water decreases with rise in temperature.
  2. No, it cannot be used. Actually, in water the two elements are chemically combined with each other. They therefore, lose their identity. But in the mixture, no chemical combination between hydrogen and oxygen has taken place. Even water cannot be formed on mixing the gases. Therefore, the mixture does not extinguish any fire.
  3. Since the experimentally determined melting point of the solid is more than the standard value of the melting point, this means that the solid is not in pure state. It has some impurities present. Please note that the purity of a solid can be determined by finding its melting point and comparing it with the standard value.

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HOTS Questions for Class 9 Science Chapter 1 Matter in Our Surroundings

HOTS Questions for Class 9 Science Chapter 1 Matter in Our Surroundings

These Solutions are part of HOTS Questions for Class 9 Science. Here we have given HOTS Questions for Class 9 Science Chapter 1 Matter in Our Surroundings

Question 1.
The diagram shows an experiment in which gases hydrogen and carbon dioxide are placed in two jars as shown in the figure. If the lid separating the two jars be removed, what will the constituents in the gas jar A after a few minutes ?
HOTS Questions for Class 9 Science Chapter 1 Matter in Our Surroundings image - 1
(a) carbon dioxide only
(b) hydrogen only
(c) mixture of carbon dioxide and hydrogen.
Answer:
(c) The gas jar A will contain both the gases carbon dioxide and hydrogen. Actually, the molecules of the gas present in one jar will move into the empty spaces present in the other jar and vice versa.

More Resources

Question 2.
The given graph shows the heating curve for a pure substance. The temperature rises with time as the substance is heated :
HOTS Questions for Class 9 Science Chapter 1 Matter in Our Surroundings image - 2

  1. What is the physical state of the substance at the points A, B, C and D ?
  2. What is the melting point of the substance ?
  3. What is its boiling point ?
  4. What happens to the temperature while the substance is changing state ?
  5. The substance is not water. How can you judge from the graph ?

Answer:

  1. At point A : The substance is in the solid state.
    At point B : The substance has started melting. It exists both in the solid and liquid states.
    At point C : The substance is in the liquid state.
    At point D : The substance has started boiling. It exists both in the liquid and gaseous states.
  2. The melting point of the substance is 15°C.
  3. The boiling point of the substance is 110°C.
  4. The temperature remains the same during the change of state.
  5. Had substance been water, its melting point should have been 0°C and boiling point 100°C. It is therefore, not water.

Question 3.
In severe cold weather, a family burnt wood in the room during the night by keeping the windows close. After sometime, they felt suffocated. They immediately opened the windows and got relief. What did actually happen ?
Answer:
When wood burns, the carbon present is oxidised to carbon dioxide (CO2) which is non-poisonous. When the windows are close, the air or oxygen cannot enter the room. In the incomplete supply of oxygen, carbon is oxidised to carbon monoxide (CO) which is a highly poisonous gas. It caused suffocation in a close room. On opening the windows, the poisonous gas slowly diffused out the room and fresh air came inside. That is how the poisonous effect of carbon monoxide gas was neutralised and the family got relief.

Question 4.
Small quantities of water and ether are placed on the palms of both the hands. Which will experience more cooling ?
Answer:
The palm containing ether will experience extra cooling. Actually ether is more volatile than water. Therefore, it will evaporate at a faster rate than water. Since cooling is always caused during evaporation, the palm containing ether will become cooler.

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RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers MCQS

RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers MCQS.

Other Exercises

Choose the correct answer in each of the following questions.
Question 1.
Solution:
(b) We know that HCF of two co-prime number is 1
HCF of 14, 35 is 7
HCF of 18, 25 is 1
HCF of 31, 93 is 31
HCF of 32, 60 is 4
Required co-prime number is (18, 25)

Question 2.
Solution:
(b) a = (22 x 33 x 54), b = (23 x 32 x 5)
HCF = 22 x 32 x 5 = 2 x 2 x 3 x 3 x 5 = 180

Question 3.
Solution:
(c) HCF of 23 x 32 x 5, 22 x 33 x 52, 24 x 3 x 53 x 7
HCF = 22 x 3 x 5 = 2 x 2 x 3 x 5 = 60

Question 4.
Solution:
(d) LCM of 23 x 3 x 5, 24 x 5 x 7 = 24 x 3 x 5 x 7
=2 x 2 x 2 x 2 x 3 x 5 x 7
= 1680

Question 5.
Solution:
(d) HCF of two numbers = 27
LCM = 162
One number = 54
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers MCQS 1

Question 6.
Solution:
(c) Product of two numbers = 1600
HCF = 5
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers MCQS 2

Question 7.
Solution:
(c) Largest number that divides each one of 1152 and 1664
HCF of 1152 and 1664 =128
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers MCQS 3

Question 8.
Solution:
(a) Largest number that divides 70 and 125 leaving remainders as 5 and 8 respectively.
Required number = 70 – 5 = 65
and 125 – 8= 117
HCF of 65, 117 = 13
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers MCQS 4

Question 9.
Solution:
(b) Largest number that divides 245 and 1029 leaving remainder as 5 in each case. .
Required number = 245 – 5 = 240 and 1029 – 5 = 1024
Now, HCF of 240 and 1020 = 16
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers MCQS 5

Question 10.
Solution:
(d)
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers MCQS 6

Question 11.
Solution:
(c) In a = bq + r
r must satisfy i.e. 0 ≤ r < b

Question 12.
Solution:
(d) Let the given number when divided by 143 gives q as quotient and 31 as remainder.
Number = 143q + 31
= (13 x 11) q + 31
= 13 x 11 q+ 13 x 2 + 5
= 13 (110 + 2) + 5
The number where divided by 73, gives 5 as remainder.

Question 13.
Solution:
(d) 3.141141114… is irrational because it is non terminating non-repeating.

Question 14.
Solution:
(c) π is an irrational number.

Question 15.
Solution:
(b) \(2.\bar { 35 }\) is a rational number as it is non-terminating repeating decimal.

Question 16.
Solution:
(c) 2.13113111311113… is an irrational number.
It is non-terminating non-repeating decimal.

Question 17.
Solution:
(b) 3.24636363…
= \(3.24\bar { 63 }\)
It is non-terminating repeating decimal.
It is a rational number.

Question 18.
Solution:
(c) \(\frac { 2027 }{ 625 }\) = \(\frac { 2027 }{ { 5 }^{ 4 } }\) is a rational because it has terminating decimal as q = 54 which is in form of 2m x 5n.

Question 19.
Solution:
(b)
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers MCQS 7

Question 20.
Solution:
(d)
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers MCQS 8
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers MCQS 9

Question 21.
Solution:
(b) 1.732 is a rational number.
As it is terminating decimal.

Question 22.
Solution:
(a) Least prime factor of a positive integer a is 3 and b is 5
2 is neither a factor of a nor of b
a and b are odd
Then (a + b) = even
(Sum of two odd numbers is even)
(a + b) is divisible by 2
Which is the least prime factor.

Question 23.
Solution:
(b) √2 is an irrational number.

Question 24.
Solution:
(c)
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers MCQS 10

Question 25.
Solution:
(c) 2 + √2 is an irrational number as sum of a rational and an irrational is an irrational

Question 26.
Solution:
(c) LCM of 1 to 10 = 2 x 2 x 2 x 3 x 3 x 5 x 7 = 2520
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers MCQS 11

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RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.1

RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.1

Question 1.
Simplify each of the following:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.1 Q1.1
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.1 Q1.2

Question 2.
Simplify the following expressions:
(i)  (4 + \(\sqrt { 7 } \)) (3 + \(\sqrt { 2 } \))
(ii) (3 + \(\sqrt { 3 } \) )(5- \(\sqrt { 2 } \))
(iii) (\(\sqrt { 5 } \) -2)(\(\sqrt { 3 } \) – \(\sqrt { 5 } \))
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.1 Q2.1

Question 3.
Simplify the following expressions:
(i)  (11 + \(\sqrt { 11 } \)) (11 – \(\sqrt { 11 } \))
(ii) (5 + \(\sqrt { 7 } \) ) (5 – \(\sqrt { 7 } \) )
(iii) (\(\sqrt { 8 } \) – \(\sqrt { 2 } \)) (\(\sqrt { 8 } \)+ \(\sqrt { 2 } \))
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.1 Q3.1
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.1 Q3.2

Question 4.
Simplify the following expressions:
(i) (\(\sqrt { 3 } \)+\(\sqrt { 7 } \))2
(ii) (\(\sqrt { 5 } \) – \(\sqrt { 3 } \) )2
(iii) (2\(\sqrt { 5 } \) + 3 \(\sqrt { 2} \))2
Solution:
(i)
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.1 Q4.1
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.1 Q4.2
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RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5

RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5

Other Exercises

Question 1.
Multiply:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 1
Solution:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 2
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 3

Question 2.
Multiply:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 4
Solution:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 5
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 6
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 7
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 8

Question 3.
Simplify each of the following and express the result as a rational number in standard form :
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 9
Solution:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 10
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 11

Question 4.
Simplify :
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 12
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 13
Solution:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 14
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 15
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 16
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 17

Question 5.
Simplify :
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 18
Solution:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 19
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 20
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5 21

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RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B

RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2B.

Other Exercises

Question 1.
Solution:
p(x) = 5 – 4x + 2x2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q1.1

Question 2.
Solution:
p(y) = 4 + 3y – y2 + 5y3
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q2.1

Question 3.
Solution:
f(t)=4t2-3t+6
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q3.1

Question 4.
Solution:
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q4.1
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q4.2

Question 5.
Solution:
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q5.1
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q5.2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q5.3

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RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2A

RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2A.

Other Exercises

Question 1.
Solution:
(i) x5-2x3+x+7 is a polynomial and its degree is 5
(ii) y3-√3y is a polynomial and its degree is 3
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2A 1

Question 2.
Solution:
(i) Degree is 1
(ii) degree is 3
(iii) degree is zero
(iv) degree is 7
(v) degree is 10
(vi) degree is 2

Question 3.
Solution:
(i) Coefficient of x3 is -5.
(ii) Coefficient of x is -2√2
(iii) Coefficient of x2 is \(\frac { \pi }{ 3 } \)
(iv) Coefficient of x2 is 0

Question 4.
Solution:
(i) Example of a binomial of degree 27 is 4x27 – 5
(ii) Example of a monomial of degree 16 is x16
(iii) Example of trinomial of degree 3 is 2x3 + 7x + 4

Question 5.
Solution:
(i) 2x2 + 4x is a quadratic polynomial. (Degree 2)
(ii) x – x3 is a cubic polynomial (Degree 3)
(iii) 2 – y – y2 is a quadratic polynomial (Degree 2)
(iv) – 7 + z is a linear polynomial (Degree 1)
(v) 5t is a linear polynomial (Degree 1)
(vi) p3 is a cubic polynomial (Degree 3)

 

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RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4

RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4

Other Exercises

Question 1.
Simplify each of the following and write as a rational number of the form :
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4 1
Solution:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4 2
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4 3
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4 4

Question 2.
Express each of the following as a rational number of the form \(\frac { p }{ q }\):
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4 5
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4 6
Solution:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4 7
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4 8
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4 9

Question 3.
Simplify :
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4 10
Solution:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4 11
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4 12
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4 13

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