# Finding number of memory locations in chip

I was solving some old MCQs and found this question:

A RAM chip has 7 address lines, 8 data lines and 2 chip select lines. Then the number of memory locations is _____.

(A) $$2^{12}$$ (B) $$2^{10}$$ (C) $$2^{19}$$ (D) $$2^{13}$$

Assuming a word consisting of a byte, this should have

2 chip select lines, meaning total $$2^2$$ chips.

With 7 address lines, we can address $$2^7$$ memory locations in a chip.

8 data lines should be used to access only the data in the memory location, and not to specify any location.

That'll make for a total of $$2^2\times2^7=2^9$$ memory locations. But none of the option matches my answer. What should be the correct answer?

• My guess would be $2^2 \times 2^7 \times 8 = 2^{12}$. – Yuval Filmus Apr 10 at 12:57
• There's always a correct answer and an expected answer, and they are often not the same. With the information given, Yuval's is a good guess that the question considers each bit to be stored in its own memory location, making the answer $2^{12}$ - 512 addresses, and 8 bits per address. If $2^9$ was in the list of possible answers, or "none of the above", you'd pick that instead. – gnasher729 Apr 10 at 14:44
• I'm nitpicking, but the fact that there are 2 chip select lines doesn't mean that all 4 chips exist. The same goes for the 7 address lines.. maybe not all addresses are valid. I could even think of a RAM chip with unaddressable locations... A better question would have been "What is the maximum number of addressable locations?" – Steven Apr 11 at 1:13
• What I don't understand is why do we have to multiply the 8 data lines in it? any reason? – Adnan Apr 12 at 19:16

Here is an actual chip! From the '70s :

Motorola MC6810

128 bytes : 128 * 8bits = 1024 bits = $$2^{10}$$ "memory cells".