Solution

Given the sum of the squares of two odd numbers is 11570, and the square of the smaller number is 5329, we can find the square of the other number as follows:

Step 1: Find the Square of the Other Number

The sum of the squares is given by:

\[
11570 = 5329 + x^2
\]

where \(x^2\) is the square of the other number. Solving for \(x^2\):

\[
x^2 = 11570 – 5329 = 6241
\]

Step 2: Determine the Other Number

To find the value of \(x\), we take the square root of 6241:

\[
x = \sqrt{6241} = 79
\]

Therefore, the other number is 79.

The correct answer is (d) 79.

To find the least number that can be added to 4800 to make it a perfect square, we observe that:

• \(4800\) is close to \(4900\), which is a perfect square.
• The square root of \(4900\) is \(70\), indicating \(4900\) is the nearest perfect square above \(4800\).

The calculation to find the required least number is:

\[
4900 – 4800 = 100
\]

Thus, the least number that needs to be added to 4800 to make it a perfect square is 100.

Since none of the provided options (a) through (d) match \(100\), the correct answer is indeed (e) None of these.

Solution

To determine which of the given statements is true, we evaluate each option:

Option (a): \(\sqrt{5}+\sqrt{3}>\sqrt{6}+\sqrt{2}\)

We calculate both sides of the inequality:

• Left side: \(\sqrt{5} + \sqrt{3}\)
• Right side: \(\sqrt{6} + \sqrt{2}\)

Approximating the square roots:

• \(\sqrt{5} \approx 2.236\)
• \(\sqrt{3} \approx 1.732\)
• \(\sqrt{6} \approx 2.449\)
• \(\sqrt{2} \approx 1.414\)

Summing up the approximations:

• Left side: \(2.236 + 1.732 = 3.968\)
• Right side: \(2.449 + 1.414 = 3.863\)

Since \(3.968 > 3.863\), option (a) \(\sqrt{5}+\sqrt{3}>\sqrt{6}+\sqrt{2}\) is true.

Option (b): \(\sqrt{5}+\sqrt{3}<\sqrt{6}+\sqrt{2}\)

From our calculation above, we know that the left side is greater than the right side, making this option false.

Option (c): \(\sqrt{5}+\sqrt{3}=\sqrt{6}+\sqrt{2}\)

As shown, the two sides are not equal, making this option false.

Option (d): \((\sqrt{5}+\sqrt{3})(\sqrt{6}+\sqrt{2})=1\)

This option can be quickly dismissed without calculation, as the product of these sums, given their approximate values, clearly does not equal 1.

Thus, the correct answer is (a) \(\sqrt{5}+\sqrt{3}>\sqrt{6}+\sqrt{2}\).

Given that ‘a’ divides 228 leaving a remainder of 18, we first calculate the value that ‘a’ divides exactly:

\[
228 – 18 = 210
\]

The question asks for the largest two-digit value of ‘a’. This means we are looking for the largest two-digit number that divides 210 without leaving a remainder.

Upon considering the mistake in the initial explanation, the largest two-digit divisor of 210 is indeed found to be 70. This is because 210 divided by 70 equals 3, without leaving any remainder, making 70 the largest two-digit number that fits the criteria.

Thus, the correct answer is (b) 70.

Solution

To find the minimum value of \(a^{2}+b^{2}+c^{2}\) given that \(a+b+c=9\), where \(a, b, c\) are real numbers, we can follow these steps:

Step 1: Express \(a^{2}+b^{2}+c^{2}\) in Terms of \(a+b+c\)

We start by expanding \((a+b+c)^2\), which gives us:

\[
\begin{aligned}
(a+b+c)^2 & = a^2 + b^2 + c^2 + 2(ab + bc + ca) \\
\end{aligned}
\]

Thus, we can express \(a^{2}+b^{2}+c^{2}\) as:

\[
\begin{aligned}
a^2+b^2+c^2 & =(a+b+c)^2-2(ab+bc+ca) \\
& =9^2-2(ab+bc+ca)
\end{aligned}
\]

Step 2: Maximizing \(ab + bc + ca\)

Since \(a^2+b^2+c^2\) will be minimum if \(ab + bc + ca\) is maximum, we consider the condition for maximizing \(ab + bc + ca\).

This condition is met when \(a = b = c\), due to the symmetry of the expression and given that their sum is fixed (\(a + b + c = 9\)). Thus, when \(a = b = c = 3\), \(ab + bc + ca\) is maximized.

Step 3: Calculating the Minimum Value

Substituting \(a = b = c = 3\) into our expression:

\[
\begin{aligned}
a^2+b^2+c^2 & = 81 – 2(3 \times 3 + 3 \times 3 + 3 \times 3) \\
& = 81 – 2(27) \\
& = 81 – 54 \\
& = 27
\end{aligned}
\]

Therefore, the minimum value of \(a^2+b^2+c^2\) is 27.

The correct answer is (d) 27.