The number of flip-flops required to design a modulo - 272 counter is :
| A. | 8 |
| B. | 9 |
| C. | 27 |
| D. | 11 |
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Option: B Explanation : for construction a Modulo - N counter we need Log2 number of flipflops.the value is rounded to next highest whole number. so log2271 = 9 Click on Discuss to view users comments. |
Let E1 and E2 be two entities in E-R diagram with simple single valued attributes. R1 and R2 are two relationships between E1 and E2 where R1 is one - many and R2 is many - many. R1 and R2 do not have any attribute of their own. How many minimum number of tables are required to represent this situation in the Relational Model?
| A. | 4 |
| B. | 3 |
| C. | 2 |
| D. | 1 |
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Option: B Explanation : Click on Discuss to view users comments. |
The STUDENT information in a university stored in the relation STUDENT (Name, SEX, Marks, DEPT_Name)
Consider the following SQL Query SELECT DEPT_Name from STUDENT where SEX = 'M' group by DEPT_Name having avg (Marks)>SELECT avg (Marks) from STUDENT. It Returns the Name of the Department for which:
| A. | The Average marks of Male students is more than the average marks of students in the same Department |
| B. | The average marks of male students is more than the average marks of students in the University |
| C. | The average marks of male students is more than the average marks of male students in the University |
| D. | The average marks of students is more than the average marks of male students in the University |
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Option: B Explanation : Click on Discuss to view users comments. |
Select the 'False' statement from the following statements about Normal Forms :
| A. | Lossless preserving decomposition into 3NF is always possible |
| B. | Lossless preserving decomposition into BCNF is always possible |
| C. | Any Relation with two attributes is in BCNF |
| D. | BCNF is stronger than 3NF |
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Option: B Explanation : Achieving Lossless and dependency-preserving decomposition property into BCNF is difficult. It is not always possible to decompose a table in BCNF and preserve dependencies. For example, a set of functional dependencies {AB �> C, C �> B} cannot be decomposed in BCNF. Click on Discuss to view users comments. |
The Relation Vendor Order (V_no, V_ord_no, V_name, Qty_sup, unit_price) is in 2NF because:
| A. | Non_key attribute V_name is dependent on V_no which is part of composite key |
| B. | Non_key attribute V_name is dependent on Qty_sup |
| C. | Key attribute Qty_sup is dependent on primary_key unit price |
| D. | Key attribute V_ord_no is dependent on primary_key unit price |
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Option: A Explanation : In the above question option C and Option d rejected as unit price cannot serve as primary key. option B is also irrelevant as qty_sup has no concern with v_name. so option A seems to be the best answer Click on Discuss to view users comments. |
