Sunday, April 5, 2020

Decision Making

It is a process in which a final outcome is derived by analysing the given data. A set of rules is specified as selection criteria. It is followed by qualification of the candidate. In selection procedure, two conditions are required these are;
  • Primary Condition
  • Additional condition
In decision making, questions contain following −
  • Information about a candidate is provided.
  • Some conditions are there which a candidate has to fulfil in order to get selected for a particular job or post.
  • When a candidate fulfils many criteria except some, some different course of action has to be taken for him.

Primary Conditions and Additional Conditions

There are some conditions in this type of questions which a candidate has to fulfil in order to get selected for a particular job. These conditions are called Original or Primary conditions. Similarly, in such problems, there are certain more conditions. We shall call them Additional conditions.
To solve this type of questions, little logic is required. All you have to do is to see the candidate fulfils some of the given conditions to decide the correct choice. Taking into account various aspects of this problem, a method is suggested below.
These types of problems can be solved easily by constructing a table using the given information.
Step I − First of all write the symbols of primary conditions (Say a, b, c, d etc.) and additional conditions (Say i, ii, iii etc.) to the top row of the table. Then write the question number and the name of candidates to the extreme left column of table. Put the primary conditions without brackets and the additional conditions within brackets.
Again, write the symbols of additional conditions with an oblique (/) mark placed before them. Then next to the symbol of that primary condition with which these might be related. Suppose we have 5 questions and 4 primary conditions and two additional conditions, then we have to construct a table as following,
Q. No.ab/(i)c/(ii)f
1.
2.
3.
4.
5.
Step II − In this step, read each question carefully and compare the facts given in it with various conditions one by one. Put the appropriate mark {‘✓’, or ‘×’ or ‘(✓)’ or ‘(×)’, as suitable}.
In the process of comparing and putting appropriate marks below the conditions indicates the following;
  • If a primary condition is satisfied, put a ‘✓’ mark below it.
  • If a primary condition is violated, put a ‘×’ mark below it.
  • If a primary condition is violated but there exist additional conditions attached to it, then if additional condition is also violated, put a ‘(×)’ mark below it, and if additional condition is satisfied put a ‘(✓)’ mark below it.
  • When no information is given then put a ‘?’ or ‘-‘ mark below it.
The table given below is an example.
Q. No.ab/(i)c/(ii)f
1.×
2.
3.(✓)(✓)
4.(×)
5.-
The above example illustrates the following;
  • a, b, c and d are four primary conditions whereas, (i) and (ii) are two additional conditions. Condition (i) is attached to b while condition (ii) is attached to c.
  • In question (1), a, b, c are satisfied where d is violated.
  • In question (2), all primary conditions a, b, c and d are satisfied.
  • In question (3), the primary conditions a and d are satisfied and though the primary conditions b and c are violated, the additional conditions (i) and (ii) are satisfied.
  • In question (4), the primary conditions a, b and d are satisfied while both c and (ii) are violated.
  • In question (5), the primary conditions a, c and d are satisfied but no information is given about b or (ii).
Step III − Now select your answer according to following rule.
  • Whenever you obtain a ‘×’ and a ‘(×)’ sign do not bother to examine the remaining conditions. Select the choice candidate not selected and move over to next question.
  • If there is no cross but a ‘?’ mark or ‘-’ mark below any condition, select the choice data is inadequate”. This is called inadequacy of data.
  • If there are no marks present, then look carefully at the answer and decide which combination of conditions lead to which conclusion and prepare a combination chart as discussed in the illustrative example given below.

Illustrative Examples

Study the following information carefully and answer the questions based on it. The following are the conditions for the requirement of person as faculty members in a computer education centre. The candidate must
a. Be in the range of 23 years to 28 years as on 1 November, 1993.
b. Have work experience- or programming experience.
c. Have a postgraduate degree in computer application. I.e. MCA, M.Tech etc. with at least 60% of marks.
d. Obtain at least 25 marks in the interview out of a total of 50 marks.
In case of a candidate who
e. Fulfils the above criteria, he or she shall be appointed as senior faculty member.
f. Has a post graduate degree in computer application with less than 60% marks but 50% marks, he/she should be appointed as junior faculty member.
g. Is of age more than 28 years but less than 32 years as on 1 Nov., 1993, in case may be referred to director of centre.
On the above condition and the information provided, we have to decide which of the following courses of action should be taken against each candidate. Don’t assume anything. The decisions will be made only on the information provided.
Mark answer
1) If the candidate is to be selected as junior faculty member.
2) If the candidate is to be selected as senior faculty member.
3) If the case is to be referred to the director of the centre.
4) If data is inadequate.
5) If candidate is not to be selected.
Questions −
1. Manisha Punjabi is M.Sc. in computer science with 65% of marks. Her date of birth is 31st July, 1965. She has been working in an institution as a programmer for last 4 years.
2. Kishore Garg is MCA with 61% marks. His date of birth is 14 August, 1970. He has worked as a computer teacher for 3 years. He has obtained 60% marks in the interview.
3. Babli Sarkar is M.Tech with 58% marks. Her date of birth is 31st December, 1971. She has been working as a programmer for last two-and-half years. She has obtained 40 marks in an interview.
4. Manish Kumar is M.Sc. with 52% marks. His date of birth is 10 July, 1968. He has been working as computer teacher for 4 years. He has obtained 40% mark in interview.
5. Sudha Ranjan is MCA with 56% marks. She has been working as a programmer for 3 years. Her date of birth is 12 February, 1969. She has obtained 55% mark in interview.
Solution −
Step I − There are 4 primary conditions ‘a’, ‘b’, ‘c’ and ‘d’, and two additional conditions are ‘f’ and ‘g’.
Q. No.a/(g)bc/(f)d
1.
2.
3.
4.
5.
Step II − Read each question carefully and compare facts given in it with the various conditions and put appropriate mark below.
Q. No.a/(g)bc/(f)d
1.(✓)-
2.
3.(×)(✓)
4.(✓)(✓)×
5.(✓)(✓)
Question 1 − Manisha is a M.Sc. student with 65% marks. This fulfils condition c. Condition a is violated though on 31st July, 1993, she is more than 28 years. Additional condition of a is g. g is satisfied, that’s why we put ‘(✓)’ mark here. Manisha has more than 2 years of experience. So we put ✓ mark below condition b. Below condition d we can see the vacant space. So we refer to d in the question. Information about Manisha’s mark in interview is missing so we put ‘-‘ mark there.
Question 2 − Kishore Garg is an MCA with 61% marks. It fulfils condition c, so we put a ✓ mark below c. According to his date of birth, he will be slightly more than 23 years. So it fulfils a. We put a ✓ mark below a. He has a work experience of more than 3 years. So it fulfils b. He acquired 60% marks in interview that fulfils d.
Question 3 − Babli Sarkar is an M.Tech with 58% marks so it violates c. An additional condition is attached to c, is f. Referring to f, we see that f is satisfied. So, we put ‘(✓)’ mark over there. She is certainly below 23 as her date of birth is in 1971. This violates condition a. The additional condition attached to a is g. so g is also violated. We put a (×) mark there.
Question 4 − Manish is M.Sc. with 52% marks. So it violates c but fulfils the additional condition f attached to c. We put (✓) mark there. He is born in 1968, so he is nearly 25 years. Hence a is satisfied. We put ✓ mark below a. He worked as computer teacher for 4 years. So it fulfils condition b. Marks obtained in the interview is 40%. It violates d. so we put × mark below d.
Question 5 − Sudha is an MCA with 56% marks. This violates condition c but fulfils f. We put a (✓) mark there. She has an experience of 3 years. It fulfils condition b. She was born in 1969, so she is 24 years old. So, condition a is satisfied. She has obtained more than 50% mark. So, condition d is satisfied.
Step IV − Now we have to decide which combination of conditions leads to which conclusion and the following are the conditions.
a + b + c + d ⇒ (2) (senior faculty member)
g + b + c + d ⇒ (3) (to be referred to director)
a + b + f + d ⇒ (1) (junior faculty member)
Question 1 − No cross mark and question mark; so, data inadequate.
Question 2 − a + b + c + d ⇒ 2
Question 3 − A cross-mark is there. So not selected.
Question 4 − Answer choice: 5.
Question 5 − a + b + f + d ⇒ 1
Following are the criteria for selection of chemistry lectures in an organization. The candidate must −
a. Have passed +2 examination in first class with at least 60% marks.
b. Have passed graduation with chemistry honours or pass degree with at least 55% marks.
c. Have experience in chemistry teaching for a minimum of 1 year.
d. Be not less than 22 years and not more than 30 years of age as on 1.1.2014.
If a candidate satisfies all the above mentioned criteria except
I. (b) Above but has a post-graduate degree in chemistry, case may be assigned to the executive director(ED).
II. (c) Above but can has experience and can do marketing, the case may be assigned to the vice president (VP).
Q 1 − Sambhrant has completed +2 from BJB junior college with 65% marks and graduated in chemistry from Utkal University with 70%. He has teaching experience in chemistry for more than 3 years and his age is 27 years.
Give answer:
Answer - A
Explanation
Sambhrant will be selected as he has all the required qualifications.
Following are the criteria for selection of chemistry lectures in an organization. The candidate must −
a. Have passed +2 examination in first class with at least 60% marks.
b. Have passed graduation with chemistry honours or pass degree with at least 55% marks.
c. Have experience in chemistry teaching for a minimum of 1 year.
d. Be not less than 22 years and not more than 30 years of age as on 1.1.2014.
If a candidate satisfies all the above mentioned criteria except
I. (b) Above but has a post-graduate degree in chemistry, case may be assigned to the executive director(ED).
II. (c) Above but can has experience and can do marketing, the case may be assigned to the vice president (VP).
Q 2 − Laxmi Rattan Shukla has passed +2 with 70% marks and is a chemistry graduate with 50% and he has also done M SC in chemistry. He has experience of teaching for more than 1 year and is 29 years of age.
Give answer:
Answer - B
Explanation
This case will be assigned to executive director of the institute because some points related to Mr. Shukla is confusing.
Following are the criteria for selection of chemistry lectures in an organization. The candidate must −
a. Have passed +2 examination in first class with at least 60% marks.
b. Have passed graduation with chemistry honours or pass degree with at least 55% marks.
c. Have experience in chemistry teaching for a minimum of 1 year.
d. Be not less than 22 years and not more than 30 years of age as on 1.1.2014.
If a candidate satisfies all the above mentioned criteria except
I. (b) Above but has a post-graduate degree in chemistry, case may be assigned to the executive director(ED).
II. (c) Above but can has experience and can do marketing, the case may be assigned to the vice president (VP).
Q 3 − Mamta Kulkarni is a BSc chemistry student with 53% marks. She has secured 58% in the +2 but has teaching experience of 2 years. She is 28 years of age.
Give answer:
Answer - E
Explanation
The candidate will not be selected because Miss Mamta doesn’t satisfy the criteria.
Following are the criteria for selection of chemistry lectures in an organization. The candidate must −
a. Have passed +2 examination in first class with at least 60% marks.
b. Have passed graduation with chemistry honours or pass degree with at least 55% marks.
c. Have experience in chemistry teaching for a minimum of 1 year.
d. Be not less than 22 years and not more than 30 years of age as on 1.1.2014.
If a candidate satisfies all the above mentioned criteria except
I. (b) Above but has a post-graduate degree in chemistry, case may be assigned to the executive director(ED).
II. (c) Above but can has experience and can do marketing, the case may be assigned to the vice president (VP).
Q 4 − Ritu Chandra is a chemistry graduate. He has no experience in teaching but he can do marketing for the institute. He has experience in marketing. His age is around 26 years.
Give answer:
Answer - C
Explanation
This case will be assigned to the vice president of the institute as the candidate is able to do marketing. So option C is correct.
Following are the criteria for selection of chemistry lectures in an organization. The candidate must −
a. Have passed +2 examination in first class with at least 60% marks.
b. Have passed graduation with chemistry honours or pass degree with at least 55% marks.
c. Have experience in chemistry teaching for a minimum of 1 year.
d. Be not less than 22 years and not more than 30 years of age as on 1.1.2014.
If a candidate satisfies all the above mentioned criteria except
I. (b) Above but has a post-graduate degree in chemistry, case may be assigned to the executive director(ED).
II. (c) Above but can has experience and can do marketing, the case may be assigned to the vice president (VP).
Q 5 − Mannishree Malhotra is a B.Sc. graduate in chemistry and has scored 65% in his graduation. He has teaching experience of 3 years and his age is not more than 27 years.
Give answer:
Answer - D
Explanation
The data is not sufficient as nothing is mentioned about +2 of Mannishree Malhotra. So option D is correct.
Following are the criteria for selection of chemistry lectures in an organization. The candidate must −
a. Have passed +2 examination in first class with at least 60% marks.
b. Have passed graduation with chemistry honours or pass degree with at least 55% marks.
c. Have experience in chemistry teaching for a minimum of 1 year.
d. Be not less than 22 years and not more than 30 years of age as on 1.1.2014.
If a candidate satisfies all the above mentioned criteria except
I. (b) Above but has a post-graduate degree in chemistry, case may be assigned to the executive director(ED).
II. (c) Above but can has experience and can do marketing, the case may be assigned to the vice president (VP).
Q 6 − Rani Mukherjee is a chemistry graduate. She has no experience in teaching but she can do marketing for the institute. She has experience in marketing. Her age is around 26 years.
Give answer:
Answer - C
Explanation
This case will be assigned to the vice president of the institute as the candidate is able to do marketing. Hence option C is correct.
Following are the criteria for selection of chemistry lectures in an organization. The candidate must −
a. Have passed +2 examination in first class with at least 60% marks.
b. Have passed graduation with chemistry honours or pass degree with at least 55% marks.
c. Have experience in chemistry teaching for a minimum of 1 year.
d. Be not less than 22 years and not more than 30 years of age as on 1.1.2014.
If a candidate satisfies all the above mentioned criteria except
I. (b) Above but has a post-graduate degree in chemistry, case may be assigned to the executive director(ED).
II. (c) Above but can has experience and can do marketing, the case may be assigned to the vice president (VP).
Q 7 − Rajendra Sharma has passed +2 with 75% marks and is a chemistry honours graduate with 52% and he has also done M Sc. in chemistry. He has experience of teaching for more than 2 years and is 29 years of age.
Give answer:
Answer - B
Explanation
This case will be assigned to executive director of the institute because he don’t have 55% in graduation.
Following are the criteria for selection of chemistry lectures in an organization. The candidate must −
a. Have passed +2 examination in first class with at least 60% marks.
b. Have passed graduation with chemistry honours or pass degree with at least 55% marks.
c. Have experience in chemistry teaching for a minimum of 1 year.
d. Be not less than 22 years and not more than 30 years of age as on 1.1.2014.
If a candidate satisfies all the above mentioned criteria except
I. (b) Above but has a post-graduate degree in chemistry, case may be assigned to the executive director(ED).
II. (c) Above but can has experience and can do marketing, the case may be assigned to the vice president (VP).
Q 8 − Jatindranath is a BSc Hons. Graduate in chemistry and has scored 65% in his graduation. He has teaching experience of 3 years and his age is not more than 28 years.
Give answer:
Answer - D
Explanation
The data is not sufficient as nothing is mentioned about his +2. Hence option D is correct.
Following are the criteria for selection of chemistry lectures in an organization. The candidate must −
a. Have passed +2 examination in first class with at least 60% marks.
b. Have passed graduation with chemistry honours or pass degree with at least 55% marks.
c. Have experience in chemistry teaching for a minimum of 1 year.
d. Be not less than 22 years and not more than 30 years of age as on 1.1.2014.
If a candidate satisfies all the above mentioned criteria except
I. (b) Above but has a post-graduate degree in chemistry, case may be assigned to the executive director(ED).
II. (c) Above but can has experience and can do marketing, the case may be assigned to the vice president (VP).
Q 9 − Madanjeet Singh is a BSc pass chemistry student with 63% marks. He has secured 48% in the +2 but has teaching experience of 2 years. He is 27 years of age.
Give answer:
Answer - E
Explanation
The candidate will not be selected because he does not fulfil the required criteria.
Following are the criteria for selection of chemistry lectures in an organization. The candidate must −
a. Have passed +2 examination in first class with at least 60% marks.
b. Have passed graduation with chemistry honours or pass degree with at least 55% marks.
c. Have experience in chemistry teaching for a minimum of 1 year.
d. Be not less than 22 years and not more than 30 years of age as on 1.1.2014.
If a candidate satisfies all the above mentioned criteria except
I. (b) Above but has a post-graduate degree in chemistry, case may be assigned to the executive director(ED).
II. (c) Above but can has experience and can do marketing, the case may be assigned to the vice president (VP).
Q 10 − Sarabjit Singh has completed +2 from Sitaram Junior College with 75% marks and graduated with chemistry hons. from BJB College with 70%. He has teaching experience of more than 4 years and his age is 28 years.
Give answer:
Answer - A
Explanation
The candidate will be selected as he has all the required qualifications. Hence option A is correct

Cube and Cuboid

Nowadays, questions based on ‘Cubes and Cuboids’ are being asked in almost every competitive exam. These problems are very frequent in various competitive exams.
The methods described below are simple yet elegant. They should be very easy to understand and with a little practice you should master them. A cube is a three dimensional solid having 6 faces, 12 edges, and corners. All edges of a cube are equal and all the faces are square in shape. That is a solid figure which has 6 faces; each face being a square is called a cube. If each of the six faces be a rectangle, it is called cuboid. A cuboid is also called a rectangular parallelepiped.
The questions asked on cube and cuboids may be of the following types.

Type I

Several views of a complete cube are given and you have to find which part of the cube lies exactly below a particular part.

Type II

An opened-up cube is given and you have to predict what it will look like when it is closed into a cube.

Type III

A cube could not be varnished on or some of its faces with the same colour or different colours and then cut into a certain specified number of identical pieces. Then question of the form- “how many small cubes have 2 faces varnished?” “How many small cubes have only one face varnished?” etc.
There are two types of problems that appear in exam. At first, you are given several views of a complete cube, and you have to state which part of the cube lies exactly below a particular part. In another type, you are given an opened-up cube, and you have to predict what it will look like when it is closed into a cube.
For Example −
Several faces of a cube are shown below −
Several faces of Cuboid
Which number would lie opposite to 2?
A − 1
B − 6
C − 5
D − 4
The fundamental approach is as follows −

Type I

A fundamental rule: Opposite cannot be together;
Whenever we see a cube, with only three of its faces visible to us, we can never see two opposite faces together. With all this rules, we can easily solve the type of problem discussed above. In the above question where we have to find opposite face of a particular face, we can eliminate those faces which have occurred together with X in any view. Thus, we can eliminate all other choice and remaining will be our answer.
At this point, you should go through the previous paragraph once more and see that you understand the concept. After this, you should try to solve the above example and see if you can apply the concept discussed above. However, you find that you have not understood the concept fully, no problem. Continue reading this section. Things will become clear once you finish the section. With the foregoing fundamental rule at the back of your mind, you can solve the above type of question.
For the question, the rule is sufficient in itself. After that, you can solve it more quickly by three secondary rules.
Solution for above example −
In the given example, we have to find the face opposite 2. Now in the first figure, 2 is appearing along with 1 and 3. It means that neither 1 nor 3 can be opposite to 2. It means that opposite of 1 we can have either 4 or 5 or 6. Similarly, opposite 3 we can have either 4 or 5 or 6. Now, look at the second figure. Here, 3 and 1 occur together with 5. It means that 5 is opposite to neither 3 nor 1. So, it means that either 4 or 6 is opposite 1 and other is opposite 3 so 5 must be opposite 2. Hence 5 is correct answer.

Some Quicker Rules

Now you must have understood the basic trick of solving such questions. The trick is that you should eliminate those choices which are not possible. For this, you take help of fundamental rule which says that if two faces are opposite to each other, their simultaneous occurrence in one view of the cube is not possible. However, in today’s time-precious competitions, merely the concept will not do. You must be able to solve a question quickly. There are some secondary rules for solving questions.
Rule I − Let us call that figure X, the opposite of which you have to find. Suppose in any one view of the cube, X appears with Y and Z. Along with a third figure (say A), then X will be opposite A.
So as for example, you have to find the face opposite 2. (This is our X). Now, 2 appears in one figure along with 1 and 3. (Y and Z). Also 1 and 3 appear together in one more figure, along with 5. (That is A). Hence 2 must be opposite 5.
Rule II − We have to find opposite face of ‘X’. Suppose that in any one view of the cube, X appears with Y and Z. Now, suppose Y and Z don’t appear together in any more views, but they appear separately in two or more different views. Then the common figure between the two more views in which Y and Z appear separately, will be the figure opposite X.
Rule III − Let’s call the figure X, the opposite of which you have to find. Now, suppose X appears in two views and in these two views four different figures are seen with X. Then the only figure not seen with X in these two views must be opposite to X.
Summary − You have to keep the fundamental rule at the back of your mind and then apply three secondary rules for quick answers. The entire approach can be summarised by following diagram.
First Type

Type II

In this type, we use fundamental rule. This rule helps us to eliminate those combinations where opposite faces are shown in a single view. So it will lead to the elimination of a choice provided we know how to determine which face will be opposite to each other, by looking at the “opened-up cube”. For this purpose, there is a very simple rule using which you can tell by looking at the opened-up cube, which faces will be opposite to each other by just looking at it.
The rule is given below;
Third is opposite rule −
When you want to find out the opposite face of a face (say X), in figure I, II, III; an opened up cube is given. We have to find which faces opposite each other when the cube is closed.
Second Type
Explanation −
In figure (I), the third figure to A is C. So A is opposite of C. So, D and F will be opposite. B and E will be opposite.
In figure (II), B is third to D, so B will be opposite to D. Similarly, C will be opposite to E and A will be opposite to F.
In figure (III), A is opposite to E, B is opposite F. Hence, C is opposite D.
Steps to solve problem
We can now solve questions of this type. We know how to find the opposite face by looking at an Opened-up cube. We also know that in any view of the cube, opposite faces can’t be together. Hence, combining two rules, we can easily solve problems.

Summary

With this, the discussion on how to solve questions of type II is completed. You have to use the rule third to determine which faces are opposite to each other. The following diagram will give the complete information about this approach.
Summary

Type III

Counting of Cubes (when a varnished solid cube is cut);
In the previous section, we have discussed the problem of finding the opposite face of a cube. There is another type of question related to cubes wherein a larger cube with different colours varnished on different sides, is broken into several smaller cubes and you have to find the number of cubes having only one side varnished or two sides varnished.
Format of this problem −
Example −
A cube is varnished with three colours green, blue and red on its sides, with every colour varnished on two opposite faces of the cube. Now the cube is broken into 64 cubes of equal size. Based on this information, answer the following questions −
1. How many cubes have two sides varnished and remaining sides unvarnished?
A − 18
B − 20
C − 22
D − 24
2. How many cubes have only one side varnished (with either green or blue colour only)
A − 4
B − 24
C − 16
D − 12
3. How many cubes have no sides varnished?
A − 0
B − 8
C − 12
D − 64