Question 15 Marks
Prove that the following inequality holds true:
$
\sqrt{5}+\sqrt{3}>\sqrt{6}+\sqrt{2}
$
$
\sqrt{5}+\sqrt{3}>\sqrt{6}+\sqrt{2}
$
Answer
View full question & answer→self
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Question 25 Marks
Solve the following inequality:
(i) $(-2 z-6)<10$
(ii) $2 a < a -4 \leq 3 a +8$
(iii) $\frac{(y-1)}{3}+4<\frac{(y-5)}{5}-2$
(i) $(-2 z-6)<10$
(ii) $2 a < a -4 \leq 3 a +8$
(iii) $\frac{(y-1)}{3}+4<\frac{(y-5)}{5}-2$
Answer
View full question & answer→(i) $z>-8$
(ii) $-6 \leq x<-4$
(iii) $(-\infty,-50)$
(ii) $-6 \leq x<-4$
(iii) $(-\infty,-50)$
Question 35 Marks
Nikhil, Priyesh and Ritik took a house on rent for ₹ 13824. They remained together for 4 months and then Ritik left the house. After 5 more months, Priyesh also left the house. How much rent should each pay?
Answer
View full question & answer→Nikhil, Priyesh and Ritik should pay ₹ 7872 , ₹ 4416 and ₹ 1536 respectively.
Question 45 Marks
If n = 2p and p is an odd prime number less than 14, then:
i) Evaluate $\varphi(n), \tau(n), \sigma(n)$ and represent the results in tabular form.
ii) Verify the relation $n+\varphi(n)+\tau(n)=\sigma(n)$.
View full question & answer→i) Evaluate $\varphi(n), \tau(n), \sigma(n)$ and represent the results in tabular form.
ii) Verify the relation $n+\varphi(n)+\tau(n)=\sigma(n)$.
Question 55 Marks
If p is a prime then $\varphi(p)+\tau(p)=\sigma(p)$, Using the information complete the following table:
View full question & answer→Question 65 Marks
Calculate $\varphi(90)+\tau(42)+\sigma(72)+\mu(70)$
Answer
View full question & answer→226
Question 75 Marks
Consider the below processes available in the ready queue for execution and with given burst time.
a) What is the time at which all the processes get executed?
b) Find the average waiting time and average turnaround time using the non- pre-emptive SJF
scheduling algorithm
| Process No | Arrival Time | Burst time |
| P1 | 1 | 3 |
| P2 | 2 | 4 |
| P3 | 3 | 2 |
| P4 | 4 | 4 |
b) Find the average waiting time and average turnaround time using the non- pre-emptive SJF
scheduling algorithm
Answer
Arrival time of both P1 and P3 is 1 unit of time. Thus, the processor remains idle for 1 unit of
time. Since both P1 and P3 arrive at the same time, priority will be given to the one with lesser burst
time therefore P3 gets executed first for 2 units of time.P3 is over after 3 units of time and now P1
gets executed for 3 units of time. P2 has already arrived in ready queue after 2 units of time and
the completion time for P2 is 10 units. When time is 4 units P4 arrives and gets executed for 4 units
of time. Thus, its completion time is 14 units.
a) All the processes got executed at 14 units of time.
b) Average waiting time $=\frac{2+4+0+6}{4}=3$ units
Average turnaround time $=\frac{5+8+2+10}{4}=6.25$ units
View full question & answer→Arrival time of both P1 and P3 is 1 unit of time. Thus, the processor remains idle for 1 unit of
time. Since both P1 and P3 arrive at the same time, priority will be given to the one with lesser burst
time therefore P3 gets executed first for 2 units of time.P3 is over after 3 units of time and now P1
gets executed for 3 units of time. P2 has already arrived in ready queue after 2 units of time and
the completion time for P2 is 10 units. When time is 4 units P4 arrives and gets executed for 4 units
of time. Thus, its completion time is 14 units.
a) All the processes got executed at 14 units of time.
b) Average waiting time $=\frac{2+4+0+6}{4}=3$ units
Average turnaround time $=\frac{5+8+2+10}{4}=6.25$ units
Question 85 Marks
Consider the available processes given below in the ready queue for execution and with given burst
time.
a) What is the time at which all the processes get executed?
b) Find the average waiting time and average turnaround time using thenon - pre-emptive FCFS
scheduling algorithm.
time.
| Process No | Arrival Time | Burst time |
| P1 | 0 | 2 |
| P2 | 1 | 2 |
| P3 | 5 | 3 |
| P4 | 6 | 4 |
b) Find the average waiting time and average turnaround time using thenon - pre-emptive FCFS
scheduling algorithm.
Answer
Arrival time of P1 is 0 and it gets executed for 2 units of time. While P1 is being executed P2
has already arrived in ready queue after 1 unit of time. Completion time for P2 is 4 units. Since P3
arrives after 5 units of time, the processor remains idle for 1 unit of time. When time is 5 units P3
arrives and get executed for 3 units of time. After 6 units of time P4 is in ready queue. Thus, its
completion time is 12 units.
a) All the processes got executed at 12 units of time
b) Average waiting time $=\frac{0+1+0+2}{4}=0.75$ units
Average turnaround time $=\frac{2+3+3+6}{4}=3.5$ units
View full question & answer→Arrival time of P1 is 0 and it gets executed for 2 units of time. While P1 is being executed P2
has already arrived in ready queue after 1 unit of time. Completion time for P2 is 4 units. Since P3
arrives after 5 units of time, the processor remains idle for 1 unit of time. When time is 5 units P3
arrives and get executed for 3 units of time. After 6 units of time P4 is in ready queue. Thus, its
completion time is 12 units.
a) All the processes got executed at 12 units of time
b) Average waiting time $=\frac{0+1+0+2}{4}=0.75$ units
Average turnaround time $=\frac{2+3+3+6}{4}=3.5$ units