Engineering
ISBT Reasoning for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q1. |
|
1) | 47% | 2) | 57% |
3) | 67% | 4) | 77% |
5) | None of these |
Answer : 57%
Explanation :
Explanation :
Voltage regulation =(VNL - VFL) / VFL
= (600 - 420)/420 = 0.57 or 57%

Q2. |
|
1) | 68.80 ×10^6 bytes/s | 2) | 24.58 ×10^6 bytes/s |
3) | 68.80 ×10^3 bytes/s | 4) | 24.58 ×10^3 bytes/s |
5) | None of these |
Answer : 24.58 ×10^6 bytes/s
Explanation :
Explanation :
The data transfer rate = (No. of sector/track)×(Bytes in each sector)×(Rotational speed (revolution per sec.))
Rotational Speed = 7200 rpm × (1/60) = 120 rev. per sec.
= 400×512×120 = 24.58×106

Q3. |
|
1) | 17 ms | 2) | 19 ms |
3) | 21 ms | 4) | 23 ms |
5) | None of these |
Answer : 17 ms

Q4. |
|
1) | 1, 2 and 3 only | 2) | 1, 2 and 4 only |
3) | 3 and 4 only (d) | 4) | 1, 2, 3 and 4 |
5) | None of these |
Answer : 1, 2, 3 and 4

Q5. |
|
1) | 0.25 | 2) | 0.5 |
3) | 2 | 4) | 4 |
5) | None of these |
Answer : 4
Explanation :
Explanation :
Insertion loss= 6dB
Insertion loss (IL) is given by
I.L. = 10 log (PT/PR), where PT is the transmitted power PR is the power received by the load after insertion 10 log (PT/PR) = 6 dB
So, (PT/PR) = 100.6 = 4
Noise factor = (PT/PR) = 4

Q6. |
|
1) | n2/2 | 2) | n/2 |
3) | 2n – 2 | 4) | n – 1 |
5) | None of these |
Answer : 2n – 2

Q7. |
|
1) | 480 bytes/s | 2) | 488 bytes/s |
3) | 4800 bytes/s | 4) | 4880 bytes/s |
5) | None of these |
Answer : 4880 bytes/s
Explanation :
Explanation :
In 8 bit of a single character total bit = 8 + start bit + stop bit = 8 + 1 + 1 = 10 bit
For every 10 bit wastage of bits = 2
For 48.8 × 103 bit wastage of bits.
(2 /10) x 48.8 x 103 = 9760 bit
Effective Transmission rate = 48.8 x 103 - 9760 = 39040 bits/s
= 39040/8 bytes/ s = 4880 bytes/s

Q8. |
|
1) | 32 ns | 2) | 64 ns |
3) | 96 s | 4) | 192 ns |
5) | None of these |
Answer : 192 ns
Explanation :
Explanation :
Bandwidth 1GB/s means 109 bytes can be loaded online in 1 second. To load 128 bytes
= 128 x (1/109 ) = 128 nsec.
Main memory latency = 64 nsec.
Time required to fetch cacheline = 128 + 64 = 192 nsec.

Q9. |
|
1) | 4 kbits | 2) | 8 kbits |
3) | 16 kbits | 4) | 32 kbits |
5) | None of these |
Answer : 16 kbits
Explanation :
Explanation :
Given, total capacity = 32 kB = 32k × 8 bits = (256 /16) k bits
Since, IC’s are used.
So each IC provides = 256 kbits 16
14 bits address gives 16 k location. Each chip has 16 k location.
Capacity of each IC = 16 k bits.

Q10. |
|
1) | derivative of the error signals | 2) | integral of the error signals |
3) | steady-state error | 4) | a constant which is a function of the system type |
5) | None of these |
Answer : derivative of the error signals
