question of type where propagation delay is given

Question

Answer 1

For single addition, we need output at the gate 2 in the above figure.
Now, it's given that 1 NAND gate takes 15 ns,
1 EXOR gate will take 15+15+15 = 45 ns (as shown in the figure).

So time required for single addition = 45 + 45 = 90 ns. 

B is the answer. 

Comments

Thank you...!!!

can you give detail about how to solve such questions.....!!

not particularly this one. @Soumya29

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Yes..just calculate, after how much time input is available to any gate and how much time does it take to produce the output. 
For example - in the above question- we want to calculate s_n.

First, we need to calculate time taken by EXOR gate -
In the equivalent circuit of EXOR gate. 1^st NAND gate will take 15 ns. After that input is available to both 2^nd and 3^rd NAND gate. So they can operate in parallel and take 15 ns to produce their output. After that input is available to 4^th NAND gate and it will take 15 ns to produce the final output C. 
So total time take by EXOR gate - 15 +15+15 = 45 ns .

Now see. Input is available to EXOR gate 1 directly. It will take 45 ns to produce the output. Its output goes to EXOR gate 2 as input. So input is available for EXOR gate 2 after 45 ns and it will take 45 ns to produce the output which is s_n. 
So total time circuit takes to product s_n = Time take by EXOR gate 1 + Time take by EXOR gate 2 = 45 ns.
 

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yes...
i got it
thanks...!!!