GATE CSE 2009 | Question: 7, ISRO2015-3

Question

How many $32K \times 1$ RAM chips are needed to provide a memory capacity of $ 256K$ bytes?

• A. $8$
• B. $32$
• C. $64$
• D. $128$

Comments

64

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How did you guess that the 32K x 1 chips has size in bits and not bytes??

Answer 1

Answer is : (c) 64

Total number of RAM chips

$$\begin{align*}&= \frac{\text{Total size}}{\text{1 RAM size}} \\
&=\frac{256 \text{ K-bytes}}{ 32 \times 1} \\
&= \frac{2^8 \times 2^{10} \times 2^3}{ 2^5\times 2^{10}}\\
&=  2^6 = 64 \end{align*}$$

Comments

C) 64

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$32K \times 1$ RAM

Is specifying no unit a new standard? or "$\times 1$" is a new notation for "bits"? Any link / wiki explaining this new notation?
To me "$32K\times 1$ RAM" looks like a printing mistake. How one is supposed to decode it straight that it means 32Kbits without having any doubt?

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@GateAspirant999 even I would like to know, whether this is a new standard?

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GateAspirant999 the part before X means no. of row in the ram (ie addrasable memory locations) and and after X part means the no. of memory bits accessed at each address.

Answer 2

answer - C

(256K x 8) / (32K x 1) = 64

Answer 3

total number of RAM chip = Total size /1 RAM size
                                       =256K bytes/ 32*1
                                       = 2⁸ * 2¹⁰ *2³/ 2^5*2¹⁰

                                       =  2⁶ = 64
Option C

Comments

Is there any reason, why should we convert byte into bit?

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I am also confused as 32k * 1 should mean 32k addresses each 1 byte, as a system is byte addresable?

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Did u get the answer of multiplying by 8

Answer 4

*1 refers bit so each row requires 8bits i.e that becomes byte for row 8 chips are needed..like that 256/32=8 columns are there

totally 8columns*8=64