Operating Systems Preparation Resources for GATE CSE

July 29, 2015 by Arjun Suresh 13 Comments

Operating Systems Syllabus

System calls, Processes, threads, inter‐process communication, concurrency and synchronization.
Deadlock.
CPU and I/O scheduling.
Memory management and virtual memory.
File systems. Disks is also under this

” System Calls is newly added. CPU scheduling renamed to CPU and I/O scheduling.

Book

Galvin

Video

GO Classroom

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Important Questions

Important Questions in Operating Systems

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2016

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Let the cost price per kg of type $A,$ and $B$ be $\text{Rs.} x$ and $\text{Rs.} y$ respectively. When, $A$ and $B$ are mixed in the ratio of $3:2,$ then the profit is $10\%,$ So, selling price $:$ $ \left( \frac{3x+2y}{5} \right) \times \frac{110}{100} = 40 $ $ \Rightarrow \left( \frac{3x+2y}{5} \right) \times \frac{11}{10} = 40 \quad \longrightarrow (1) $ When, $A$ and $B$ are mixed in the ratio of $2:3,$ then the profit is $5\%.$ So, selling price $:$ $ \left( \frac{2x+3y}{5} \right) \times \frac{105}{100} = 40 $ $ \Rightarrow \left( \frac{2x+3y}{5} \right) \times \frac{21}{20} = 40 \quad \longrightarrow… [...]

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Let the cost price of peanuts per kg be $\text{Rs}. x.$ Then, the cost price of walnuts per kg be $\text{Rs}. 3x.$ A wholesaler sold $8 \; \text{kg}$ of peanuts at a profit of $10\%.$ So, the cost price of $8 \; \text{kg}$ peanuts $ = \text{Rs}. 8x.$ Therefore, selling price of $8 \; \text{kg}$ peanuts $ = 8x \times \frac{110}{100} = \text{Rs.} \frac{88x}{10} $ And, he also sold $16 \; \text{kg}$ walnuts at a profit of $20 \%.$ So, the cost price of $16 \; \text{kg}$ walnuts $ = 16 \times 3x = \text{Rs.} 48x $ Therefore, selling price… [...]

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Given that, points $A$ and $B$ are $150 \; \text{km}$ apart. Each car travel at a speed of $100 \; \text{km/hr}$ for the first $50 \; \text{km}.$ $ \boxed{\text{Time} = \frac{\text{Distance}}{\text{Speed}}} $ So, $ t_{1} = \frac{50}{100} $ $ \Rightarrow \boxed{t_{1} = \frac{1}{2} \; \text{hour}} $ And, each car travel at a speed of $ 50 \; \text{kmph}$ for the $50 \; \text{km}.$ So, $ t_{2} = \frac{50}{50} $ $ \Rightarrow \boxed{t_{2} = 1 \; \text{hour}} $ And, each travel at a speed of $25 \; \text{kmph}$ for the last $50 \; \text{km}. $ So, $t_{3} = \frac{50}{25} $ $… [...]

Answered: CAT2018-2: 71

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