1. Why we use Treeview Control? To list the hierarchial list of the node objects. Such of files and Directories. 2. Why we need OLE-Automation? Advantages? Enables an application to exposes objects and methods to other Applications. No need to reserve memory. No need to write functions. Object library that simplify programming tasks. i.e., No need to Object library. (OLB, TLB). 3. What is the diff between the Create Object and Get object? Create Object - To create an instance of an object. Get Object ? To get the reference to an existing object. 4. Have you create Properties and Methods for your own Controls? Properties ? Public variable of a Class Method ? Public procedure of a classWhat is Collection Objects? Similarly to arrays but is preferred over an array because of the following reasons. 1. A collection objects uses less Memory than an array. 2. It provides methods to add and delete members. 3. It does not required reason statement when objects are added or deleted. 4. It does not have boundary limitations. 5. What is Static Variable? Its Scope will be available through out the life time. Private Dim x as integer. Is it true? No. Private cannot be used in front of DIM. 6. What is Implicit? Instance of specific copy of a class with its own settings for the properties defined in that class. Note: The implicity defined variable is never equal to nothing. 7. What are the scope of the class? Public , private, Friend Can we able to set Instancing properties like Singleuse, GlobalSingleuse to ActiveXDll? No. In project properties if we set Unattended what is it mean? This cannot have user interface. This can be used for the COM creation.What are the Style Properties of Combo Box? Simple, Dropdown list ? We can type and select. Dropdown Combo ? Only Drop Down. 8. What are the Style properties of List Box? Simple ?Single Select , Extended. ? Multiple Select. 9. What are the different types of Dialog Box? Predefined, Custom, User Defined. 10. What is Parser Bug? It is difficult to use database objects declared in a module from within a form. 11. What is the Dll required for running the VB? Vbrun300.dllCan We create CGI scripts in VB? Yes. 12. How to change the Mouse Pointer? Screen. Mouse Pointer = VBHourGlass/VBNormal. 13. How to check the condition in Msgbox? If (Msgbox("Do you want to delete this Record",VbYesNo)=VbYes)Then End if 14. What is difference between datagrid and flexgrid? Datagrid ? Editable. Flexigrid ? Non-Editable. (Generally used for Read only purpose.) 15. What is ADO? What are its objects ? ActiveX Data Object. ADO can access data from both flat files as well as the databases. I.e., It is encapsulation of DAO, RDO, and OLE that is why we call it as OLE-DB Technology. Objects are Connection, Record Set, Command, Parameter, field, Error, Property. 16. What is Dataware Control? Any control bound to Data Control. Ex:- Textbox, Check Box, Picture Box, Image Control, Label, List box, Combo Box, DB Combo, What are two validate with Data Control? Data_Validate, Data_Error. Record set types and Number available in VB? 3. 1- Dynaset, 0 ? Table, 2 ? Snap Shot. Referential Integrity (Take care By jet database Engine). Cascade Delete, Cascade Update ? is done setting property of Attributes. Db Relation Delete Cascade, Db Relation Update Cascade. 17. What are the locks available in Visual Basic? Locking is the process by which a DBMS restricts access to a row in a multi-user environment 4 types of locks. They are 1.Batch Optimistic 2.Optimistic 3.Pessimistic 4.ReadOnly 18. What is the diff between RDO and ADO? RDO is Hierarchy model where as ADO is Object model. ADO can access data from both flat files as well as the data bases. I.e., It is encapsulation of DAO, RDO , OLE that is why we call it as OLE-DB Technology. 19. How can we call Stored procedure of Back End in RDO and ADO ? In RDO ? We can call using RDO Query Objects. In ADO ? We can call using Command Objects. 20. What is the different between Microsoft ODBC Driver and Oracle OBDC Driver? Microsoft ODBC driver will support all the methods and properties of Visual Basic. Where as the Oracle not. 21. What are the Technologies for Accessing Database from Visual Basic? DAO, Data Control, RDO, ODBCDIRECT, ADO, ODBC API. Calling Stored Procedures in VB? 1. Calling Simply the Procedure with out Arguments "Call Procedure Name}" If it is with Arguments Means then Declare the Query Def qy Set Qy as New Query def Qy.SQL = "{Call ProcedureName(?,?,?)}" qy(0)=val(Txt1.Text) qy(1)=val(Txt2.Text) qy(2)=val(Txt3.Text) Set Rs = Qy.OpenresultSet Txt(1)=Rs.RdoColumns(0)

Aptitude Questions 1. A sum of Rs.312 was divided among 100 boys and girls in such a way that the boy gets Rs.3.60 and each girl Rs.2.40 the number of girls is A.35 B.40 C.45 D.50 Ans: B Ex:Step (i) Let x be the number of boys and y be the number of girls. Given total number of boys and girls = 100 x + y = 100 -------------- (i) Step (ii) A boy gets Rs. 3.60 and a girl gets Rs. 2.40 The amount given to 100 boys and girls = Rs. 312 3.60x +2.40y = 312 -------------- (ii) Step (iii) Solving (i) and (ii) 3.60 x + 3.60y = 360 ------- Multiply (i) by 3.60 => 3.60 x + 2.40y = 312 --------- (ii) 1.20y = 48 y = 48/1.20 = 40 => Number of girls = 40 2. If 20 men or 24 women or 40 boys can do a job in 12 days working for 8 hours a day, how many men working with 6 women and 2 boys take to do a job four times as big working for 5 hours a day for 12 days? A.8 men B.12 men C.2 men D.24 men Ans: C Ex:Let?s try solving this Problem using ratio approach. "Amount of work done by 20 men = 24 women = 40 boys or 1 man = 1.2 woman = 2 boys." Let us therefore, find out the amount of men required, if only men were working on the job, to complete the new job under the new conditions and then make adjustments for the women and children working with the men. The man hours required to complete the new job = 4 times the man hours required to complete the old job.(As the new job is 4 times as big as the old job) Let 'n' be the number of men required. 20 * 12 * 8 = n * 5 * 12 * 4 => n = 8 8 men working will be able to complete the given job. However, the problem states that 6 women and 2 boys are working on the job. 6 women = 6/12= 5 men and 2 boys = 1 man. The equivalent of 5 + 1 = 6 men are already working. 3. When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200g. What is the average weight of the remaining 59 students? A.57 B.56.8 C.58.2 D.52.2 Ans:A Ex:Let the average weight of the 59 students be A. Therefore, the total weight of the 59 of them will be 59A. The questions states that when the weight of this student who left is added, the total weight of the class = 59A + 45 When this student is also included, the average weight decreases by 0.2 kgs. 59A + 45/ 60 = A - 0.2 => 59A + 45 = 60A ? 12 => 45 + 12 = 60A - 59A => A = 57 4. The difference between two angles of a triangle is 240. The average of the same two angles is 540.Which one of the following is the value of the greatest angle of the triangle? A.450 B.600 C.660 D.720 Ans Ex:Let a and b be the two angles in the question, with a > b. We are given that the difference between the angles is 240. => a - b = 24. Since the average of the two angles is 540, we have (a + b)/2 = 54. Solving for b in the first equation yields b = a - 24, and substituting this into the second equation yields ({a + (a - 24)}/2) = 54 ({2a ? 24}/2) = 54 2a - 24 = 54*2 2a - 24 = 108 2a = 108 + 24 2a = 132 a = 66 Also, b = a - 24 = 66 - 24 = 42 Now, let c be the third angle of the triangle. Since the sum of the angles in the triangle is 1800, a + b + c =180. Plugging the previous results into the equation yields 66 + 42 + c = 180. Solving for c yields c = 72 Hence, the greatest of the three angles a, b and c is c, which equals 720. 5. If we write down all the natural numbers from 259 to 492 side by side get a very large natural number 259260261....491492. How many 8's will be used to write this large natural number? A.32 B.43 C.52 D.53 Ans Ex:From 259 to 458, there are 200 natural numbers so there will be 2*20 =40 8's From 459 to 492 we have 13 more 8's and so answer is 40+13 =53 6. If each of the three nonzero numbers a , b , and c is divisible by 3, then abc must be divisible by which one of the following the numbers? A.8 B.27 C.81 D.121 Ans:B Ex:Since each one of the three numbers a, b, and c is divisible by 3, the numbers can be represented as 3p,3q, and 3r, respectively, where p, q, and r are integers. The product of the three numbers is 3p*3q*3r =27(pqr). Since p, q, and r are integers, pqr is an integer and therefore abc is divisible by 27. 7. A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets? A.45% B.45{5/11}% C.54{6/11}% D.55% Ans:B Ex:Number of runs made by running = 110 - (3 * 4 + 8 * 6) = 110 - (60) = 50. Required percentage = ({50/110}*100)% = 45{5/11}% 8. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are: A.39, 30 B.41, 32 C.42, 33 D.43, 34 Ans:C Ex:Let their marks be (x + 9) and x. Then, x + 9 = (56/100)(x + 9 + x) 25(x + 9) = 14(2x + 9) 3x = 99 x = 33 So, their marks are 42 and 33. 9. The letter of the word LABOUR are permuted in all possible ways and the words thus formed are arranged as in a dictionary. What is the rank of the word LABOUR? A.275 B.251 C.240 D.242 Ans Ex:The order of each letter in the dictionary is ABLORU. Now, with A in the beginning, the remaining letters can be permuted in 5! ways. Similarly, with B in the beginning, the remaining letters can be permuted in 5! ways. With L in the beginning, the first word will be LABORU, the second will be LABOUR. Hence, the rank of the word LABOUR is 5!+5!+2 =242 10. A class photograph has to be taken. The front row consists of 6 girls who are sitting. 20 boys are standing behind. The two corner positions are reserved for the 2 tallest boys. In how many ways can the students be arranged? A.6! * 1440 B.18! * 1440 C.18! *2! * 1440 D.None of these Ans:B EX:Two tallest boys can be arranged in 2! ways, rest 18 can be arranged in 18! ways. Girls can be arranged in 6! ways. Total number of ways in which all the students can be arranged = 2! * 18! * 6! = 18! *1440 11. Ten different letters of alphabet are given, words with 5 letters are formed from these given letters. Then, the number of words which have at least one letter repeated is: A.69760 B.30240 C.99748 D.42386 Ans:A Number of words which have at least one letter replaced: = Total number of words - total number of words in which no letter is repeated" =>105 - {16P5} =>100000 - 30240 =69760 12. In a charity show tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random (blindly) will have a number with a hundredth digit of 2? A.0.285 B.0.40 C.100/249 D.99/250 Ans:B Ex:250 numbers between 101 and 350 i.e. n(S) = 250 n(E) = 100th digits of 2 = 299-199 = 100 P(E) = {n(E)}/{n(S)} = 100/250 =0.40 13. The sum of three numbers is 98. If the ratio of the first to second is 2 :3 and that of the second to the third is 5 : 8, then the second number is: A.20 B.30 C.48 D.58 Ans:B Ex:Let the three parts be A, B, C. Then, A : B = 2 : 3 and B : C = 5 : 8 = 5*{3/5} : 8*{3/5} = 3 : 24/5 => A : B : C = 2 : 3 : 24/5 = 10 : 15 : 24 => B = (98*{15/49}) =30 14. Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is Sumit's salary? A.Rs. 17,000 B.Rs. 20,000 C.Rs. 34,000 D.Rs. 38,000 Ans Ex:Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively. Then, {2x + 4000}/{3x + 4000} = 40/57 => 57*(2x + 4000) = 40*(3x + 4000) => 6x = 68,000 => 3x = 34,000 Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) =Rs. 38,000 15. Sachin can cover a distance in 1hr 24min by covering 2/3 of the distance at 4 kmph and the rest at 5kmph.the total distance is? A.5 km B.6 km C.7 km D.8 km Ans:B EX:Let total distance = D Distance travelled at 4 kmph speed = (2/3)D Distance travelled at 5 kmph speed = (1 - 2/3)D = (1/3)D Total time = 1hr 24 min = (60+24)min = (84/60)hr = (21/15)hr We know, time = distance/speed Total time = (21/15)={(2/3)D}/4 + {(1/3)D} /5 21/15 = {2D}/12 + {D}/15 21/15={14D}/60 84 = 14D D=6km