Given

\[
\begin{aligned}
& a^2-4 a-1=0 \\
& a^2-4 a=1 \\
& a(a-4)=1 \\
& a-4=\frac{1}{a} \\
& a-\frac{1}{a}=4
\end{aligned}
\]

We have \(\mathrm{a}^2+3 \mathrm{a}+\frac{1}{\mathrm{a}^2}-\frac{3}{\mathrm{a}}\)
\[
\begin{aligned}
& \left(a^2+\frac{1}{a^2}\right)+3\left(a-\frac{1}{a}\right) \\
& \left(a-\frac{1}{a}\right)^2+3\left(a-\frac{1}{a}\right)+2 \\
& 4^2+3 \times 4+2=30
\end{aligned}
\]

Therefore, Correct option is (d).

Let the given number be \(x\). According to the problem:

– The correct operation was to multiply \(x\) by \(\frac{7}{19}\): \(x \times \frac{7}{19}\).
– By mistake, the boy divided \(x\) by \(\frac{7}{19}\), which is equivalent to multiplying \(x\) by \(\frac{19}{7}\): \(x \times \frac{19}{7}\).
– The mistake led to a result that was 624 more than the correct answer.

Thus, we have:

\[
x \times \frac{19}{7} = x \times \frac{7}{19} + 624
\]

Rearrange the equation to isolate \(x\):

\[
x \times \frac{19}{7} – x \times \frac{7}{19} = 624
\]

Factoring \(x\) out:

\[
x \left(\frac{19}{7} – \frac{7}{19}\right) = 624
\]

To solve for \(x\), simplify the expression in the parentheses:

\[
x \left(\frac{19^2 – 7^2}{7 \times 19}\right) = 624
\]

Calculate the difference of squares:

\[
19^2 – 7^2 = (19 + 7)(19 – 7) = 26 \times 12 = 312
\]

Substitute back into the equation:

\[
x \left(\frac{312}{7 \times 19}\right) = 624
\]

Simplify the fraction:

\[
x \left(\frac{312}{133}\right) = 624
\]

Solving for \(x\):

\[
x = 624 \times \frac{133}{312} = 2 \times 133 = 266
\]

The sum of the digits of the given number \(266\) is \(2 + 6 + 6 = 14\).

Answer: (d) 14.

To find the smallest positive integer that, when multiplied by 392, yields a perfect square, we first need to factorize 392 to understand its prime factorization.

Prime Factorization of 392

\[
392 = 2^3 \times 7^2
\]

For a number to be a perfect square, all the exponents in its prime factorization must be even. In the prime factorization of 392, the exponent of 2 is 3 (which is odd), and the exponent of 7 is 2 (which is even).

Finding the Smallest Positive Integer

To make the number a perfect square, we need to multiply it by a number that will make all the exponents even. Since the exponent of 2 is odd, we need to multiply 392 by another 2 to make the exponent of 2 become \(3 + 1 = 4\), which is even.

\[
392 \times 2 = 2^4 \times 7^2
\]

Now, the prime factorization \(2^4 \times 7^2\) indicates a perfect square because both exponents are even.

Conclusion

The smallest positive integer that, when multiplied by 392, yields a perfect square is \(2\).

Answer: (a) 2.

Solution

To solve this, let’s denote:

• \(I\) for ice-creams,
• \(C\) for cookies, and
• \(P\) for pastries.

Given conditions are:

• \(I \geq 9\),
• \(C \geq 9\),
• \(P \geq 9\),
• \(C > I\),
• \(P > C\),
• \(I + C + P = 32\).

Since Rachita buys a minimum of 9 units of each item, the smallest numbers she can buy, respecting the given conditions (\(C > I\) and \(P > C\)), are:

• \(I = 9\),
• \(C > 9\),
• \(P > C\).

Given \(I + C + P = 32\) and \(I = 9\), we have:

• \(9 + C + P = 32\),
• \(C + P = 23\).

To satisfy \(C > I\) and \(P > C\) with the least numbers:

• If \(C = 9\), then \(P\) cannot be equal to \(9\) since \(P > C\), contradicting \(C > I\).
• \(C\) must be more than 9, so the next possible value is \(C = 10\) or higher, and \(P\) must be higher than \(C\).

Let’s examine the possible combinations under these constraints:

• If \(C = 10\), then \(P = 13\) (since \(P > C\) and they must sum to 23), which satisfies all conditions.
• If \(C = 11\), then \(P = 12\) (still satisfies \(P > C\)), which also meets all requirements.

Therefore, based on the constraints and the need for \(C\) to be more than \(I\) and \(P\) to be more than \(C\), Rachita can buy either 10 or 11 cookies to keep the total count to 32 while adhering to all given conditions.

The correct answer is (c) Either 10 or 11.