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The mode of 4,2,5,2,0,3,2,3 is  2

When a die is thrown, there are total six outcomes, i.e., 1, 2, 3, 4, 5 and 6.

(i) (a) 2, 3, 5 are prime numbers. So there are 3 outcomes out of 6. Therefore, probability of getting a prime number = 3/6 = 1/2

(b) 1, 4, 6 are not the prime numbers. So there are 3 outcomes out of 6. Therefore, probability of getting a prime number = 3/6 = 1/2

(ii) (a) Only 6 is greater than 5. So there is one outcome out of 6. Therefore, probability of getting a number greater than 5 = 1/6

(b) Numbers not greater than 5 are 1, 2, 3, 4 and 5. So there are 5 outcomes out of 6. Therefore, probability of not getting a number greater than 5 = 5/6

In the given case, the person is able to see vertical lines more distinctly than horizontal lines. This means that the refracting system (cornea and eye-lens) of the eye is not working in the same way in different planes. This defect is called astigmatism. The person’s eye has enough curvature in the vertical plane. However, the curvature in the horizontal plane is insufficient. Hence, sharp images of the vertical lines are formed on the retina, but horizontal lines appear blurred. This defect can be corrected by using cylindrical lenses

On Indian railways standard length of BG rail = 13 m 

So, total no. of rail required in 1.024 km or 1024 m

= 1024/13 = 78.78 = 80 (in round figure)

Sleeper density = M + 7

M = length of rail

So, sleeper density = 13 + 7 = 20

No. of sleepers = 20 × 80 = 1600

Given: Fundamental frequency for closed pipe = Third overtone of open pipe. 

To find: Ratio of lengths of air columns in both the pipes \((\frac{l_c}{l_o})\) 

Formulae: a. Fundamental frequency of pipe closed at one end: 

n_c = \(\frac{v}{4l_c}\) 

b. Fundamental frequency of pipe open at both ends: 

n_o = \(\frac{v}{2l_o}\) 

c. Third overtone of open pipe: 

n₃ = 4n₀

Calculation: Fundamental frequency of closed pipe (n_c) is same as third overtone of open pipe.

∴ n_c = n₃ 

∴ n_c = 4n₀ 

∴ \(\frac{v}{4l_c}\) = 4[\(\frac{v}{2l_o}\)]

∴ \(\frac{v}{4l_c}\) = \(\frac{2}{l_o}\)

∴ \(\frac{l_c}{l_o}\)= \(\frac18\)

The ratio of lengths of air columns in closed pipe and open pipe is 1:8.

Infinity is the bulk  modulus of a perfectly rigid body.

(4)

n = PV/RT = 

= 128 x 10^-5 moles

= (3770 x 10^-5atm x 1L/0.0821 L atm k^-1mol^-1 x 300K) = 1.27 x 10^-3 mol

The dimensions of the cuboid are 100 cm × 80 cm × 64 cm. 

Therefore, the volume of the cuboid is V = l × b × h 

= 100 × 80 × 64 = 512000 cm³ . 

Given that the cuboid is melted and recast into a cube. 

∴ The volume of the cube = the volume of the cuboid. 

∴ The volume of the cube = 512000 cm³ . 

(a) If the edge of the cube is a then, the volume of the cube is given by = a³ . 

Hence, option (ii) is correct. 

(b) Formula of finding volume of cuboid = l × b × ℎ. 

Hence, option (i) is correct. 

(c) Volume of cuboid is V = l ×b × ℎ. 

= 100 × 80 × 64 = 512000 cm³ . 

Hence, option (i) is correct. 

(d) Volume of the cube = 512000 cm³ . 

Hence, option (i) is correct. 

(e) Since, volume of the cube whose side length is a is given by a³ . 

And volume of the cube is 512000 cm³ . 

∴ a³ = 512000 = 8³×10³ = 80³ 

⇒ a = 80 cm. 

Hence, the side length of the cube is a = 80 cm. 

∴ Surface area of the cube = 6a² = 6×80² = 38400 cm² . 

Hence, option (iii) is correct.

(a) (i) a³

(b) (i) l × b × h

(c) (i) 512000 cm³

(d) (i) 512000 cm³

(e) (iii) 38400 cm²

Correct option (d) None of these   

Explanation: 

For n = 1 sum = 2/3

For n = 2 sum = 20/27

For n = 3 sum = 1640/2187

None of the options matches these numbers and hence option (d) is correct.

Correct option (a) n( 2n² + 2n + 1) 

Explanation: 

S = n² + 1 + n² + 2 +… + n² + 2n = 2n³ + n(2n + 1) = n(2n² + 2n + 1) 

Alternative Method: Suppose n = 2 then 1^st term of the series will be 22 + 1 = 5. Now we want to find the sum of first 2n = 4 terms. First 4 terms of the series will be 5, 6, 7, 8. Sum = 26. If we put n = 2 in the above options then only option (a) satisfies.

\(Z_{total}=Z_{1}+Z_{2}+Z_{3}=9.487∟ -18.434°\Omega\)

\(I=15.8∟18.45^\circ A\)

\(V=150∟0^\circ V\)

2x³ + x² – 3x + 4.

[2,1, – 3, 4]

Ratio ⇒ 120/5 dozen = 120/5 x 12 = 2/1 = 2:1

5√7 x 2√7 = 10 x 7 = 70

√ 2450 = 35√ 2

Now, √ 2450/5√ 2

= 35√ 2/5√ 2

= 7

(11√ 2)² = 242 and (9√ 3)² = 243

Now (11√ 2)² - (9√ 3)² = 242 - 243 = - 1

(5√7)² = 175 and √49 = 7

Now, (5√ 7)² ÷ √ 49 

⇒ 175 ÷ 7

∴ 25

Correct option is: (d) 625

Solution:
Since, 1, 2 and 3 are the remainders of 1251, 9377 and 15628 respectively.
So, 1251 – 1 = 1250 is exactly divisible by the required number,
9377 – 2 = 9375 is exactly divisible by the required number,
15628 – 3 = 15625 is exactly divisible by the required number.
So, required number = HCF of 1250, 9375 and 15625.
By Euclid’s division algorithm,
15625 = 9375 x 1 + 6250
9375 = 6250 x 1 + 3125
6250 = 3125 x 2 + 0
=> HCF (15625, 9375) = 3125
3125 = 1250 x 2 + 625
1250 = 625 x 2 + 0

HCF(3125, 1250) = 625
So, HCF (1250, 9375, 15625) = 625
Hence, the largest number is 625.

Metals that are low in the activity series are very un-reactive. The oxides of these metals can be reduced to metals by heating alone. 

For example, mercury is obtained from its ore, cinnabar (HgS), by the process of heating. 

2HgS + 3O₂ + ∆ → 2HgO + 2SO₂ (Roasting)

 2HgO + ∆ → 2Hg +O₂ (Reduction) 

Another instance is reduction of Cu₂S (ore of copper) to copper by heating. 

2Cu₂S + 3O₂ + ∆ → 2Cu₂O + 2SO₂ (Roasting) 

2Cu₂O + Cu₂S + ∆ → 6Cu + SO₂ (Reduction) 

Metals falling low in activity series are found in free states or in the form of sulphides ores in nature.

According to National Curriculum Framework-2005, the main goal of Mathematics education in school is the mathematization of the child’s thought process. Mathematics relies on logic, reasoning, problem-solving, creativity, and mathematical way of thinking. These skills can be useful in many other subjects.

• Primary students need a solid foundation in geometry in order to be successful later in middle school and high school geometry.
• Classroom activities should encourage students to observe, compare, make hypotheses, and then prove or disprove their hypotheses.
• The geometry presented in primary classrooms focuses mainly on the identification and observation of basic shapes. 
  • At the primary level, the teacher mainly expects their students to observe and identify the different shapes such as square, rectangle, triangle, etc.
  • When students are able to observe and identify the geometrical shapes only then they will be able to define them and draw them.
  • The students at the primary level are not able to give logical proof of the geometrical figures as this skill is needed to be acquired by them at higher levels of classes.
• The effective teaching of geometry at the primary level lays the foundation for the understanding of basic geometrical shapes that will further help the students in conceptual understanding of geometry at higher levels of classes.

Hence, it could be concluded that the objective of teaching geometry at the primary stage is the observation of geometrical figures.

National Curriculum Framework (NCF), 2005 provides a guideline with which teachers and schools can choose and plan experiences that they think children should have.

• It seeks to reform the curriculum and to bring learning experiences in and outside the classroom.
• According to the NCF 2005, “Developing children’s abilities for mathematization is the main goal of mathematics education.”

There are two kinds of aims in mathematics for school education such as broader and narrower aims.

Narrower aim-

• to develop numeracy related skills
• to develop ‘useful’ capabilities particularly those relating to numeracy- numbers, number operations, measurements, decimals and percentages

Broader aim-

• Problem- solving
• Use of heuristics
• Estimation and approximation
• Optimisation
• Use of patterns
• Visualization
• Representation
• Reasoning and proof
• Making connections
• Mathematical communication

Hence, we conclude that to make students proficient in handling numbers and number operations is one of the narrow aims of NCF.

Learning strategy is 'the sum of an individual's preferences for physical, social, emotional, and environmental elements in the course of learning'. Each learner develops his/her strategy to learning which is rooted in his/her interests and habits. Note that:

• While teaching decimals to students, students should be taught using visual method or aids so that they can understand the concept in a better way.
• 'Multiplication as repeated addition' can be emphasised when teaching multiplication of non-decimal numbers because repeated addition of decimals can be complicated for students.
• Algorithm shows how to solve the problem but it does not connect with the real-life meaning of multiplication.
• Teaching multiplication as 'inverse of division' will not concretely explaining the concept of multiplication.

Hence, the most appropriate strategy is to teach visually.

Concept:

A relation R in a set A is called

• Reflexive, if (a, a) ∈ R, for every a ∈ A.
• Symmetric, if (a, b) ∈ R implies that (b, a) ∈ R, for all a, b ∈ A.
• Transitive, if (a, b) ∈ R and (b, c) ∈ R  implies that (a, c) ∈ R, for all a, b, c ∈ A.

A relation R in a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive.

Calculation:

Given: R = {(T1,T2): T1 is similar to T₂ where T₁, T₂ ∈ T} and T is the set of all triangles in a plane

Reflexive:

As we know that, every triangle is similar to itself, so (T₁, T₁) ∈ R ∀ T₁ ∈ T

Hence, relation R is reflexive.

Symmetric:

Suppose if (T₁, T₂) ∈ R ⇒T₁ is similar to T₂ ⇒T₂ is also similar to T₁ ⇒ (T₂, T₁) ∈ R.

Hence, relation R is symmetric.

Transitive:

Now suppose, (T₁, T₂), (T₂, T₃) ∈ R ⇒T₁ is similar to T₂ and T₂ is similar to T₃ ⇒T₁ is similar to T₃ . So (T₁, T₃) ∈ R.

Hence, relation R is transitive.

Hence, relation R is an equivalence relation.

Correct option is (B) 6/11 

Total number of solutions = ^{10 + 3 – 1}C_{3 – 1} = 66 

Favourable number of solutions = ¹¹C₁ + ⁹C₁ + ⁷C₁ + ⁵C₁ + ³C₁ + ¹C₁ = 36

P(req) = \(\frac{36}{66}=\frac{6}{11}\)