Question

The cost of 6 hard-discs equals the cost of 9 printers, the cost of 27 printers equals the cost of 30 scanners, and the cost of 300 scanners equals the cost of 9 computers. If the cost of 3 computers is Rs. 72,000, find the cost of a hard-disc.

• A. Rs. 1800
• B. Rs. 800
• C. Rs. 1500
• D. Rs. 1200

Hint:

Convert all three given relations into one chain connecting hard-discs, printers, scanners and computers through a common multiple.

Answer 1

The Correct Option is D

Step 1: Write the three given relations as equations.
Let the cost of one hard-disc, printer, scanner and computer be H, P, S and C respectively, in rupees.
\[ 6H = 9P \quad \text{...(i)} \]
\[ 27P = 30S \quad \text{...(ii)} \]
\[ 300S = 9C \quad \text{...(iii)} \]

Step 2: Combine all three into one chain using a common multiple.
Multiply (i) by 10: \( 60H = 90P \).
Multiply (ii) by \( \frac{10}{3} \): \( 90P = 100S \).
Divide (iii) by 3: \( 100S = 3C \).
Putting these together:
\[ 60H = 90P = 100S = 3C \]

Step 3: Use the given value of 3 computers.
We are told \( 3C = 72{,}000 \), and from the chain \( 60H = 3C \).
\[ 60H = 72{,}000 \Rightarrow H = \frac{72{,}000}{60} = 1{,}200 \]

Final Answer:
The cost of a hard-disc is Rs. 1,200. \[ \boxed{Rs.\ 1200} \]