GATE CSE 2025 | Set 2 | GA | Question: 8

Question

Which one of the following options is correct for the given data in the table?

$$\begin{array}{|c|c|c|c|c|} \hline \text{Iteration (i)} & \text{0} & \text{1} & \text{2} & \text{3}  \\  \hline
\text{Input (I)} & \text{20} & \text{-4} & \text{10} & \text{15}  \\  \hline
\text{Output (X)} & \text{20} & \text{16} & \text{26} & \text{41}  \\  \hline \text{Output (Y)} & \text{20} & \text{-80} & \text{-800} & \text{-12000}  \\  \hline
\end{array}$$

• A. $X(i) = X(i-1)+I(i); \: \: Y(i) = Y(i-1)I(i); \: \: i>0$
• B. $X(i) = X(i-1)I(i); \: \: Y(i) = Y(i-1)+I(i); \: \: i>0$
• C. $X(i) = X(i-1)I(i); \: \: Y(i) = Y(i-1)I(i); \: \: i>0$
• D. $X(i) = X(i-1)+I(i); \: \: Y(i) = Y(i-1)I(i-1); \: \: i>0$

Comments

Just trial and error each option

Answer 1

\[X(i) = X(i-1) + I(i)\]\[X(1) = X(0) + I(1) = 20 + (-4) = 16, \quad \text{satisfied}\]\[X(2) = X(1) + I(2) = 16 + 10 = 26, \quad \text{satisfied}\]\[X(3) = X(2) + I(3) = 26 + 15 = 41, \quad \text{satisfied}\]

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\[Y(i) = Y(i-1)I(i), \quad i > 0\]\[Y(1) = Y(0) \times I(1) = (-20) \times 4 = -80, \quad \text{satisfied}\]\[Y(2) = Y(1) \times I(2) = (-80) \times 10 = -800, \quad \text{satisfied}\]\[Y(3) = Y(2) \times I(3) = (-800) \times 15 = -12000, \quad \text{satisfied}\]

Answer 2

We need to determine which of the given formulas correctly describe the transformation from Input I(i)I(i)I(i) to Output X(i)X(i)X(i) and Output Y(i)Y(i)Y(i).

Step 1: Analyzing X(i)X(i)X(i)

Looking at the pattern in the Output X column:

• X(0)=20X(0) = 20X(0)=20
• X(1)=16=20+(−4)X(1) = 16 = 20 + (-4)X(1)=16=20+(−4)
• X(2)=26=16+10X(2) = 26 = 16 + 10X(2)=26=16+10
• X(3)=41=26+15X(3) = 41 = 26 + 15X(3)=41=26+15

This follows the rule:

X(i)=X(i−1)+I(i)X(i) = X(i-1) + I(i)X(i)=X(i−1)+I(i)

Step 2: Analyzing Y(i)Y(i)Y(i)

Looking at the pattern in the Output Y column:

• Y(0)=20Y(0) = 20Y(0)=20
• Y(1)=−80=20×(−4)Y(1) = -80 = 20 \times (-4)Y(1)=−80=20×(−4)
• Y(2)=−800=(−80)×10Y(2) = -800 = (-80) \times 10Y(2)=−800=(−80)×10
• Y(3)=−12000=(−800)×15Y(3) = -12000 = (-800) \times 15Y(3)=−12000=(−800)×15

This follows the rule:

Y(i)=Y(i−1)×I(i)Y(i) = Y(i-1) \times I(i)Y(i)=Y(i−1)×I(i)

Step 3: Matching the Correct Option

The correct equations for both X(i)X(i)X(i) and Y(i)Y(i)Y(i) appear in the first option:

X(i)=X(i−1)+I(i),Y(i)=Y(i−1)×I(i),i>0X(i) = X(i-1) + I(i), \quad Y(i) = Y(i-1) \times I(i), \quad i > 0X(i)=X(i−1)+I(i),Y(i)=Y(i−1)×I(i),i>0

Thus, the correct answer is the first option.

Answer 3

Put x=1 and check equation see value from table and try to satisfy the option 

In Option A 

X(i)=X(i-1)+I(i)

X(0)+I(1)

20+20

40

 

 

 

Y(i)=Y(i-1)I(1)

20*-4

-80 

 

Option A satify 

 

Hope it helps