Solution

To arrange 1369 students in an equal number of rows and columns, we need to find a square number that is closest to 1369 because the square root of that number will give us the number of students in each row and column.

Finding the Square Root

The square root of 1369 gives us:

\[
\sqrt{1369} = 37
\]

This means that the teacher can arrange the students in 37 rows and 37 columns, with each row and column having exactly 37 students. Therefore, the number of students in the last row is 37.

The correct answer is (a) 37.

Given the equation \(a^{2}+b^{2}+c^{2}=2(a-b-c)-3\), we can rewrite it as:

\[
a^2 + b^2 + c^2 – 2a + 2b + 2c + 3 = 0
\]

This equation can be rearranged to form perfect squares:

\[
(a^2 – 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0
\]

The expressions on the left-hand side can be rewritten as the squares of binomials:

\[
(a – 1)^2 + (b + 1)^2 + (c + 1)^2 = 0
\]

For the sum of squares to be zero, each square must individually be zero. Thus, we have:

\[
(a – 1)^2 = 0, \quad (b + 1)^2 = 0, \quad (c + 1)^2 = 0
\]

Solving these equations gives:

\[
a = 1, \quad b = -1, \quad c = -1
\]

Substituting these values into the expression \(2a – 3b + 4c\):

\[
2(1) – 3(-1) + 4(-1) = 2 + 3 – 4 = 1
\]

Therefore, the value of \(2a – 3b + 4c\) is \(1\).

The correct answer is (a) 1.

To simplify the given expression \(\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\), let’s start by simplifying each square root by expressing them in terms of their prime factors and then simplify:

\[
\sqrt{32} = \sqrt{2^5} = \sqrt{16 \cdot 2} = 4\sqrt{2}
\]
\[
\sqrt{48} = \sqrt{2^4 \cdot 3} = 4\sqrt{3}
\]
\[
\sqrt{8} = \sqrt{2^3} = \sqrt{4 \cdot 2} = 2\sqrt{2}
\]
\[
\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}
\]

Substituting these simplified forms back into the original expression gives:

\[
\frac{4\sqrt{2} + 4\sqrt{3}}{2\sqrt{2} + 2\sqrt{3}}
\]

We can take a 2 out of both the numerator and the denominator to simplify further:

\[
= \frac{2(2\sqrt{2} + 2\sqrt{3})}{2(\sqrt{2} + \sqrt{3})}
\]

Upon simplification, we find:

\[
= \frac{2\sqrt{2} + 2\sqrt{3}}{\sqrt{2} + \sqrt{3}}
\]

Since the terms in the numerator are double those in the denominator, simplifying the expression incorrectly suggested the terms could cancel out directly. The correct simplification should consider factoring correctly and assessing the common terms. The step where the 2 is factored out both in the numerator and denominator simplifies the expression directly, leading to:

\[
= 2
\]

Thus, the simplified value of the expression \(\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\) is indeed \(2\).

The correct answer is (c) 2.

Calculation of the Given Expression

Given the equation \(a^{2}+b^{2}=5ab\), we are tasked with determining the value of the expression \(\left(\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}\right)\).

Step 1: Simplify the Given Relation

Starting with the given equation, we divide both sides by \(ab\) to simplify:

\[
\frac{a^2 + b^2}{ab} = 5
\]

This leads to:

\[
\frac{a}{b} + \frac{b}{a} = 5
\]

Step 2: Square Both Sides

To find the value of \(\left(\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}\right)\), we square both sides of the simplified equation:

\[
\left(\frac{a}{b} + \frac{b}{a}\right)^2 = 5^2
\]

This yields:

\[
\frac{a^2}{b^2} + 2\left(\frac{a}{b}\cdot\frac{b}{a}\right) + \frac{b^2}{a^2} = 25
\]

Given that \(\frac{a}{b}\cdot\frac{b}{a} = 1\), we simplify further:

\[
\frac{a^2}{b^2} + \frac{b^2}{a^2} + 2 = 25
\]

Step 3: Isolate the Target Expression

Subtracting 2 from both sides to isolate the expression gives us:

\[
\frac{a^2}{b^2} + \frac{b^2}{a^2} = 25 – 2 = 23
\]

Conclusion

Therefore, the value of the expression \(\left(\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}\right)\) is \(\textbf{23}\), making the correct answer:

(c) \(\textbf{23}\)

This solution methodically derives the value of the given expression by leveraging the initial condition and algebraic manipulation, leading to a clear and logical conclusion.

Calculation of Water Levels in Jars with Different Capacities

This problem presents a scenario where equal volumes of water are introduced into two jars of distinct capacities. Here’s a breakdown of the situation and the outcome when the water from one jar is transferred to the other.

Initial Setup

– Larger Jar: When filled with a certain volume of water, it reaches \(\frac{1}{4}\) of its full capacity.
– Smaller Jar: The same volume of water fills this jar to \(\frac{1}{3}\) of its capacity, indicating its smaller size compared to the larger jar.

Transfer Process and Outcome

Upon transferring the water from the smaller jar (which is \(\frac{1}{3}\) full) into the larger jar (\(\frac{1}{4}\) full), we aim to understand how the water level changes in the larger jar.

– Observation: Since the initial amounts of water in both jars are equal, transferring the water from the smaller jar to the larger one effectively doubles the amount of water in the larger jar.

– Mathematical Representation: The act of pouring water from the smaller jar doubles the water volume in the larger jar, leading to the equation \(2 \times \frac{1}{4} = \frac{1}{2}\).

Conclusion

Therefore, after the water from the smaller jar is poured into it, the larger jar becomes \(\frac{1}{2}\) full.