- Option : A
- Explanation :

Clearly 13 = 1 x 10 + 3 and 36 = 3 x 10 + 6 ⇒ base b = 10

The quadratic equation with solutions x = 5 and x = 6 is x^{2}- 11x + 30 = 0

According to the given condition, we have b + 3 = 11 and 3b + 6 = 30 ⇒ b = 8

Answer is 8

Alternative solution:

x^{2}- 13x + 36 = 0 (given quadratic equation)

In bae b, 13 = 1xb^{1}+ 3xb^{0}= b+3 and

36 = 3x^{1}+ 6x^{0}= 3b + 6

So the equation becomes x^{2}- (b + 3)x + (3b + 6) = 0

Since x = 5 is a solution

∴ 5^{2}- (b + 3)5 + (3b + 6) = 0 ⇒ b = 8

Similarly, by putting x = 6, we get b = 8

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- Option : C
- Explanation :

The given grammar with S as start symbol is

S → XY

X → aX∣a

Y → aYb∣ϵ

X generates atleast one 'a'. While Y generates equal no of a's and b's( including epsilon).

L = { a , ab, aab, aabb, aaabb ....}

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- Option : C
- Explanation :

void printxy (int x, int y) {

int *ptr;

x = 0; // x=0,y=1

ptr = &x; // x=0,y=1, and ptr points to x

y = * ptr; //x=0, y=0

* ptr = 1; //x=1, y=0

printf (“%d, %d,” x, y); // It prints 1 AND 0.

}

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- Option : B
- Explanation :

Record Route is an optional field used to record address.

In IPv4 header, 40 bytes are reserved for OPTIONS. For Record Route to stores, 1 byte is used to store type of option, 1 byte for length and 1 byte for pointer. Out of 40, 37 bytes are left.

An IP4 address takes 32 bits or 4 bytes.

We can store at most floor(37/4) = 9 router addresses.

Alternate Solution

In IPV4 frame format there are 40 bytes of option field, out of which only 38 bytes can be used, 2 bytes are reserved. Also one IPV4 address is of 4 bytes

Thus maximum IPV4 address that can be hold <= 38/4 = 9

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