Seating and Complex Arrangements

Arrangement problems are quite common in all entrance examinations. If you follow the instructions given

below, you can easily solve any problem given. First we learn different types of arrangements and tips to

crack. There are three types of arrangements: 1. Linear Arrangements : There are people sit in a row 2. Circular Arrangements : People sit in a row facing center. In most of the cases, problems with 6, 8

people given. 3. Complex Arrangements : In these questions, There are some persons or things which eat different foods,

wear different colored shirts, use different bikes, have some first and last names etc. We have to match

these persons and their interests according to the conditions given. Tips in solving linear and circular arrangement problems: 1. Always start filling in the details with Specific Statements. There are two types of statements given: 1. Specific 2. Non specific Specific statements always give only one type of arrangement. For example: D sits immediate left of F. So

we have to put the combination DF in the diagram. C sits opposite to A. This also a specific arrangement in a circular diagram as there is no other way to

represent this. Example for non-specific statements is B and E sits next to each other. This gives two types of

arrangement. BE and EB. A sits in between C and D. This gives CAD and DAC. So never solving question with statements like this. 2. Search for some continuation statement: If the first statement starts with D and F, search for another statement which has either D or F in that. So

that it will give you some continuation. 3. Draw the diagrams in the circular arrangement according to the shown below

Remember: Left side in a circular arrangement is always clock wise and right side means anti - clock

wise.

Set 1: Six persons A, B, C, D, E and F are sitting around a circular table facing the center. I. C is sitting in-between A and F. II. B is sitting two places to the left of E. III. D is sitting two places to the right of F. 1. Between which two persons does D is sitting? a. F B b. E B c. C B d. A B

2. Who is sitting diagonally opposite to A? a. F b. C c. E d. None of these Sol: From statement 1, ACF / FCA From statement 2, B _ E From statement 3, F _ D We can start with either statement 2 or 3, but starting with statement 3 gives us continuation with statement

1.

D sits to the right of F so When we fix D, we have to write F two places after CLOCK wise direction. Now

ACF or FCA possible. ACF is not possible as D occupied so FCA possible. There is only one position

available for B and E and the remaining place is occupied by C. So Option C and Option B are correct. Set 2: Eleven students A, B, C, D, E, F, G, H, I, J and K are sitting in a row of the class facing the teacher.

D, who is to the immediate left of F, is second to the right of C. A is second to the right of E, who is at one

of the ends. J is the immediate neighbor of A and B and third to the left of G. H is to the immediate left of

D and third to the right of I. 3. Who is sitting in the middle of the row? a. C b. I c. B d. G e. None 4. Which of the following groups of friends is sitting to the right of G? a. IBJA b. ICHDF c. CHDF d. CHDE Sol: _ _ _ _ _ _ _ _ _ _ _ _ Let us code all the given statements into some notation format so that it saves lot of time in going back and

forth to the question. 1. D, who is to the immediate left of F, is second to the right of C. This implies, D is sitting immediate left of F and D is setting second to the right of C. DF, C _ D 2. A is second to the right of E, who is at one of the ends. If E sits at one of the end he must sits at left end. Then only the following arrangement possible. E _ A 3. J is the immediate neighbor of A and B and third to the left of G. AJB / BJA possible and J _ _ G Therefore, A J B _ G / B J A _ G 4. H is to the immediate left of D and third to the right of I. HD and I _ _ H From 1, C _ D F From 4, I _ _ H D From 1 and 4, I _ C H D F ----(1) From 3, A J B _ G or B J A _ G possible

If we consider 2 also, above statement becomes, E _ A J B _ G - - - - - (2) Now from 1 and 2, we have three possibilities. 1. F sits to the left of E 2. I sits to the right of F. These two are not possible as total places are becoming more than 11. Now I should occupy the position

between B and G. So E _ A J B I G C H D F is the right arrangement. The remaining person K occupy the position between E and A. Now answers for the above questions are Option B and C Solving Complex Arrangement Questions: 1. Amit, Bharati, Cheryl, Deepak and Eric are five friends sitting in a restaurant. They are wearing caps of

five different colours yellow, blue, green, white and red. Also they are eating five different snacks burgers, sandwiches, ice cream, pastries and pizza. I. The person wearing a red cap is eating pastries. II. Amit does not eat ice cream and Cheryl is eating sandwiches. III. Bharati is wearing a yellow cap and Amit wearing a blue cap. IV. Eric is eating pizza and is not wearing a green cap. 8. What is Amit eating? a. Burgers b. Sandwiches c. Ice cream d. Pastries 9. Who among the following friends is wearing the green cap? a. Amit b. Bharati c. Cheryl d. Deepak 10. Who among the following friends is having ice cream? a. Amit b. Bharati c. Cheryl d. Deepak Sol: In this question, there are 3 variables. Name of the person, Color of the caps, and Snacks they take. Never try to write all the names and try to match them. This is a bad habit. Try this method. 1. Identify one variable and write all the names belong to it below it. Only write the variable names on

both sides of this column

Now try to fill in the details in the table according the conditions given.

After Filling in all the available details, table looks like above. Now we have to fit Pastries and Red some

where. Only one place left. It must be at D. Once you fit that one, C's color becomes Green and E's

become white. Similarly B takes Ice cream and A takes Burger. So final table looks like this

So answers for the above questions are A, C and B respectively. Complex arrangement problems are not this much straight forward. But the procedure to solve any

question is like this. If there are more variables, the complexity increases. But with adequate practice you can solve the

questions easily.

Syllogisms or Statements and Conclusions A syllogism question contains 2 or more Question statements, and 2 or more conclusions followed by 4

options. Statement 1: - - - - Statement 2: - - - - Conclusion 1: - - - - Conclusion 2: - - - - Answers options: Option 1: If only conclusion 1 can be drawn from the given statements. Option 2: If only conclusion 2 can be drawn from the given statements. Option 3: If both conclusions 1 and 2 can be drawn from the given statements. Option 4: None of the conclusions can be drawn from the given statements. Example Question: Statements: All MBAs are Graduates All graduates are Students Conclusions: 1: All MBAs are Students 2: Some students are MBAs There are two types of Conclusions: 1. Immediate Inferences 2. Logical Conclusions. Immediate inferences are the conclusion which are drawn from only one statement. For example, From the statement All A's are B's, we can draw some A's are B's, Some B's are A's. These are easy to draw. For Logical conclusions you have to follow complete theory. Understanding Syllogism question: Questions on syllogisms contains only the following 4 types of statements:

1. The universal positive : Eg: All X are Y

2. The universal negative : Eg: No X is Y

3. The particular positive: Eg: Some X are Y

4. The particular negative : Eg: Some X are not Ys

Here Checkmark () indicates "Distribution". If a term is distributed means It covers each and every

element of it. All X are Ys means X Y, But Y need not be a subset of X. So Y does not have check mark.

You should commit to memory, how to put marks and marks and to distinguish positive and negative statements, universal and particular statements.

Here two statements are universal (1 and 2), and two statements are particular (3 and 4). Two statements are positive (1 and 3) and two statements are negative (2 and 4).

I. The Universal Positive: All X are Y It states that every member of the first class is also a member of the second class. Take a statement "All

Tamilians are Indians". It does not necessarily follows All Indians are Tamilians. So Indians is not

distributed on Tamilians.

The general diagram for Universal Affirmative All X are Y is

Immediate inference:

1. Some X are Y

2. Some Y are X

II. The universal Negative: No X is Y It states that no member of the first class is a member of the second class. This proposition takes the form

- No X is Y. The general diagram for Universal Negative No X is Y is

Immediate Inferences:

1. No Y is X

2. Some X are not Ys,

3. Some Y are not Xs

III. The Particular Affirmative: Some X are Y

It states that at least one member, but never all, of the term designated by the class X is also a member of the class designated by the term Y. This proposition takes the form Some Xs are Ys. This possible diagrams as shown by the Eulers circles for this proposition are:

Immediate Inferences:

Some Y are X

IV. The particular Negative: Some X are not Ys. It states that at least one member of the class designated by the term X is excluded from the whole of the class designated by the term Y. This proposition takes the form Some Xs are not Ys. The Eulers circle diagrams for this proposition are as follows.

Immediate Inferences: None The shaded portion in each is that part of X that is not Y.

Most of the students get some doubt why "Y" is distributed here. Here is the explanation. Suppose take an example statement "Some students are not hardworking". This can be phrased as All the hardworking

students are not some students". Or let us say, Rama is one of the student who is not hardworking. So not

even a single hardworking student cannot be Rama. So All hardworking students cannot be Rama. Read this article for more information: Click Here

How to answer Syllogisms: There are two methods to answer syllogisms. 1. Euler venn diagram method 2. Aristotle's rules Method If all the statements are Universal you can easily draw venn diagrams and solve the questions. But If there

are more particular statements, then you better learn Aristotle method. But Aristotle's method initially

seems to be a bit difficult to understand, as one practices good number of questions, one can easily crack

these questions.

Aristotle's Rules to solve syllogisms: 1. If both the statements are particular, no conclusion possible (Explanation: Statements starting with "Some" are particular) 2. If both the statements are negative, no conclusion possible 3. If both the statements are positive, conclusion must be positive 4. If one statement is particular, conclusion must be particular 5. If one statement is negative, conclusion must be negative 6. Middle term must be distributed in atleast one of the premises (Explanation: Middle term is the common term between two given premises, and A terms is distributed

means it must have the "" mark above it)

7. If a term is distributed in the conclusion, the term must be distributed in atleast one of the premises. (Explanation: If any term is having star mark in the conclusion, it term must have star mark in the given

premises) 8. If a term is distributed in both the statements only particular conclusion possible.

Solved Example 1: Statements: All MBAs are Graduates All graduates are Students Conclusions: 1: All MBAs are Students 2: Some students are MBAs Explanation:

Statement 1: All MBAs are Graduates

Statement 2: All Graduates are Students

C1: All MBA are students

C2: Some Students are MBA Now Let us apply rules: 1. Both statements are positive, conclusion must be positive 2. Common term is Graduate and it has check mark in the second statement Conclusion 1: MBA in the conclusion has got a check mark so it must have check mark in atleast one of

the premises. MBA in S1 has got star mark. It satisfied all the rules. It is valid conclusion Conclusion 2: No term in the conclusion has got a check mark so no need to check anything. It followed

all the rules. This statement is a valid conclusion.

Solved Example 2: Statements: All Cats are Dogs No Dog is Fish Conclusions: 1. No Cat is Fish 2. Some Cats are Fish Explanation:

S1: All Cats are Dogs

S2: No Dog is Fish

C1: No Cat is Fish

C2: Some Cats are Fish Now Let us apply rules: 1. S2 is negative, so conclusion must be negative. So C2 is ruled out, as rule says that one statement

is negative conclusion must be negative. 2. Common terms is Dog and it has check mark in both the premises Conclusion 1: In the conclusion, both the terms Cat, Fish have check marks and These two terms have

check marks in at least one of the premises. So Conclusion 1 is valid Conclusion 2: As one of the premises is negative, conclusion must be negative. So this conclusion is not

valid Solved Example 3: Statements: Some books are toys. No toy is red. Conclusions: 1. Some toys are books. 2. Some books are not red. Explanation:

S1: Some Books are Toys

S2: No Toy is Red

C1: Some Toys are Books

C2: Some Books are not Red We can easily draw conclusion one as it is a direct inference from Statement 1. Now by applying the rules, Statement 2 is negative so conclusion must be negative. and middle term "Toy" should have

a mark. It has mark in statement 2. Now If a term has a mark in the conclusion it should have mark in atleast one of the statements. Here "red" has mark in the conclusion and also has mark in statement 2. So Statement 2 also following all the rules. Both conclusions are valid. Three Statement Types: We can apply all the rules we learnt above while solving 3 statement questions too. Look at the options and see from which two statements those words are derived. If Once word is from first

and another is from the third, we have to check that there are two middle terms with . Solved Example 4: Statements: All cats are dogs some pigs are cats. All dogs are tigers Conclusions: some tigers are cats some pigs are tigers all cats are tigers some cats are not tigers Answer options: 1. Only 1 and 2 2. Only 1, 2 and 3 3. All follow 4. None Follow Explanation: Let us rearrange for easy understanding.

S2: Some Pigs are Cats

S1: All Cats are Dogs

S3: All Dogs are Tigers

C1: Some Tigers are Cats

C2: Some Pigs are Tigers

C3: All Cats are Tigers

C4: Some Cats are not Tigers

From statements 1 and 3, All Cats are Tigers is correct as dogs has mark. So conclusion 3 is correct. Also from the above, Some tigers are cats is also true. So conclusion 1 is correct. Pigs is from 2nd statement and tigers is from third statements. If you observed our rearraged

statements, In the second and first statements Cats has markand first and third statements has dogs has mark. So we can conclude that Some Pigs are tigers. All the given statements are positive so conclusion should be positive. So option 4 is not correct. Correct Answer option is 2.

Cubes

A cube is a 3-dimensional diagram with all sides equal. If we divide it into the size (1n)

th part of

its side, we get n3smaller cubes.

Shown below is a cube which is painted on all the sides and the cut into (14)

th of its original side.

Some observations: A cube has 6 faces, 12 edges and 8 corners. We can see that the cubes which got all the three sides painting lies at the corners. So the number of cubes which got painted all the three sides is equal to 8. Cubes with 2 sides painting lie on the edges (see the diagram). But the cubes which are on the left and right side of the edge matches with the corners. So we have to substract these two cubes from the number of cubes lying on the edge to get the number of cubes with 2 sides painting. Cubes with 1 side painting lies on the surfaces. Since, the top row, bottom row, left column, and right column matches with the edges, We must exclude these cubes while calculating the single side painted cubes.

The following rules may be helpful: If a cube is divided into the size (1n)th of its original side after get painted all the sides, Then

Total number of cubes = n3 Cubes with 3 sides painting = 8

Cubes with 2 sides painting = 12(n2) Cubes with 1 sied painting = 6(n2)2 Cubes with no painting = (n2)3

Solved Examples (Level - 1) 1. A cube whose two adjacent faces are coloured is cut into 64 identical small cubes. How many of these small cubes are not coloured at all? Assume the top face of the cube and its right side are colored green and orange respectively.

Now If we remove the colored faces, we left with a cuboid, whose front face is indicated with dots.

So on the front face there are 9 cubes, and behind it lies 4 stacks. So total 9 x 4 = 36 2. A cube, painted yellow on all-faces is cut into 27 small cubes of equal size. How many small cubes got no painting?

Assume we have taken out the front 9 cubes. Then the cube looks like the one below.

Now the cube which is in the middle has not got any painting. The cubes on the Top row, bottom row, left column and right column all got painting on atleast one face. Alternative method:

Use formula: (n2)3 Here n = 3 So (32)3 = 1 3. All surfaces of a cube are coloured. If a number of smaller cubes are taken out from it, each side 1/4 the size of the original cube's side, Find the number of cubes with only one side painted.

The original (coloured) cube is divided into 64 smaller cubes as shown in the figure. The four central cubes on each face of the larger cube, have only one side painted. Since, there are six faces, therefore total number of such cubes = 4 x 6 = 24. Alternative Method:

Use formula : 6(n2)2 = 6(42)2 = 24 Level - 2

4. Directions: One hundred and twenty-five cubes of the same size are arranged in the form of a cube on a table. Then a column of five cubes is removed from each of the four corners. All the exposed faces of the rest of the solid (except the face touching the table) are coloured red. Now, answer these questions based on the above statement: (1) How many small cubes are there in the solid after the removal of the columns? (2) How many cubes do not have any coloured face? (3) How many cubes have only one red face each? (4) How many cubes have two coloured faces each? (5) How many cubes have more than 3 coloured faces each? The following figure shows the arrangement of 125 cubes to form a single cube followed by the removal of 4 columns of five cubes each.

When the corner columns of the original cube are removed , and the resulting block is coloured on all the exposed faces (except the base) then we get the right hand side diagram. We labelled the various columns from a to u as shown in the figure (1): Since out of 125 total number of cubes, we removed 4 columns of 5 cubes each, the remaining number of cubes = 125 - (4 x 5) = 125 - 20 = 105. (2): Cubes with no painting lie in the middle. So cubes which are blow the cubes named as s, t, u, p, q, r, m, n, o got no painting. Since there are 4 rown below the top layer, total cubes with no painting are (9 x 4) = 36. (3): There are 9 cubes namaed as m, n, o, p, q, r, s, t and u in layer 1, and 4 cubes (in columns b, e, h and k) in each of the layers 2, 3, 4 and 5 got one red face. Thus, there are 9 + (4 x 4) = 25 cuebs. (4) the columns (a, c, d, f, g, i, j, l) each got 4 cubes in the layers 2, 3, 4, 5. Also in the layer 1, h, k, b, e cubes got 2 faces coloured. so total cubes are 32 + 4 = 36 (5): There is no cube in the block having more than three coloured faces. There are 8 cubes (in the columns a, c, d, f, g, i, j and l) in layer 1 which have 3 coloured faces. Thus, there are 8 such cubes. Thus, there are 8 such cubes. 5. Directions: A cube of side 10 cm is coloured red with a 2 cm wide green strip along all the sides on all the faces. The cube is cut into 125 smaller cubes of equal size. Answer the following questions based on this statement: (1) How many cubes have three green faces each? (2) How many cubes have one face red and an adjacent face green? (3) How many cubes have at least one face coloured? (4) How many cubes have at least two green faces each? Clearly, upon colouring the cube as stated and then cutting it into 125 smaller cubes of equal size we get a stack of cubes as shown in the following figure.

The figure can be analysed by assuming the stack to be composed of 5 horizontal layers. (1): All the corner cubes are painted green. So there are 8 cubes with 3 sides painted green. (2): There is no cube having one face red and an adjacent face green as all the green painted cubes got paint on atleast 2 faces.

(3): Let us calculate the number of cubes with no painting. By formula, (n2)3 = (52)3 = 27 Therefore, there are 125 - 27 = 98 cubes having at least one face coloured. (4): From the total cubes, Let us substract the cubes with red painting, cubes with no painting. 125 - (9 x 6) - 27 = 44

Binary Logic Binary Logic questions are an important type which frequently appear in IT company entrances and MBA entrance exams. In these problems, you find people answer a questions in two or three different statements and some of them are true and some are false. Based on the clues given, we have to figure out the actual category of persons. Solved Example 1: Three persons give these statements. A says either Democratic or Liberal wins the elections. B says Democratic wins. C says neither Democratic nor Liberal wins the elections. Of these statements only one is wrong. Who wins the elections? As only one statement is wrong, other two statements will be true. Assume Democratic wins the election. Now Statements of A and B are true. Which satisfies our condition that 2 of them are truth tellers. So Democratic wins the election If you assume Liberal wins the election,Statements of B and C are becoming false which is against our condition. Note: Most of the binary logic questions can be solved easily if you start assuming like above. Solved Example 2: Consider the following statements: Albert: Dave did it. Dave : Tony did it. Gul: I did not do it. Tony: Dave lied when he said that I did it. (a) If only one out of all above statements is true, who did it? (b) if only one out of all above statements is false, who did it? We solve this question by assuming that Albert is thief. Then Dave, there after Gul. and put it in a table

From the table, it is clear that only one statement is false when we assume Dave is thief. So answer for (b) is Dave. And only one statement became true, when we assume Gul is thief. So answer for (a) is Gul Solved Example 3: The police rounded up Jim, Bud and Sam yesterday because one of them was suspected of having robbed the local bank. The three suspects made the following statements under intensive questioning. Jim : Im innocent. Bud: Im innocent. Sam: Bud is the guilty one. If only one of the statements turned out to be true, who robbed the bank? Assume Jim is the thief. Now Except Bud statement, remaining two statements became false which is given in the question. So Jim is the thief. Solved Example 4: Directions : Three criminals were arrested for shop lifting. However, when interrogated only one told the truth in both his statements, while the other two each told one true statement and one lie. The statement were: ALBERT: (a) Clive passed the goods. (b) Bruce created the diversion. BRUCE : (a) Albert passed the goods. (b) I created the diversion. CLIVE : (a) I took the goods out of the shop. (b) Bruce passed goods. Who created the diversion? (a) Albert (b) Clive (c) Bruce (d) either (a) or (c) (e) either (b) or (c) Solved Example 5: Using the data from the above question, which of these statements is correct? (a) Clive created the diversion. (b) Albert took the goods out of the shop. (c) Clive passed the goods. (d) Albert created the diversion. (e) Albert passed the goods. Let T represents true statement and F represents false statement. We have to check possibilities and contradictions by assuming one person speaking truth and others will say truth or lie alternatively. Assuming Bruce to speak truth

Above mentioned possibility satisfies the conditions as others give contradictions. So, Albert passed the goods. Bruce created diversion. Clive took goods out of shop.

Solved Example 6: Directions: On an Island there live three types of tribes Sachcha, Jhutha and Lota. Sachchas always tell the truth, Jhuthas always lie and Lotas tell the truth and lie alternating (they can tell truth first or lie first). Three persons (of different tribes) from this Island give these statements. GOOD: UGLY is of Sachcha tribe: I am of Lota tribe BAD : GOOD is of Jhutha tribe; I am of Sachcha Tribe UGLY: BAD is of Jhutha tribe; I am of Lota tribe. GOOD belongs which tribe? (a) Sachcha (b) Jhutha (c) Lota (d) either (a) or (c) (e) Cannot say If we assume Good is of Sachcha tribe person, His both statements should be true. But one of his statement Ugly is of sachcha tribe should be wrong as there is only one shachcha tribe person. Now assume BAD is of sacha tribe person. Now his second statement is obviously true and His first statement indicates that Good is of Jutha type which implies that Ugly is of Lota type. Now checking of the truthfullness of the statements of Good and Ugly, we get Good's both the statements are wrong and Ugly's one statements is correct and one is wrong. So Good Belong to Jutha tribe. Solved Example 7: Directions: Chatia, Matia and Toni participated in a race and on of them won the race. They belong to three different communities - Sororian, Nororian and Cororian. Sororians always speak the truth, Nororians always lie and Cororians always tell the truth and lie alternatively. (Each of Chatia, Matia and Toni belongs to one community.) After the race they gave these statements. Chatia: 1. I would have won the race if Toni had not obstructed me at the last moment. 2. Toni always speaks the truth. 3. Toni is the winner. Matia: 1. Chatia won the race. 2. Toni is not a Nororian. Toni: 1. I hadnt obstructed Chatia at the last moment. 2. Matia won the race. Toni belongs to which community? (a) Sororian (b) Nororian (c) Cororian (d) Either b or c (e) Cannot say Solved Example 8: Who won the race? (a) Matia (b) Toni (c) Sororian (d) Chatia (e) Cannot say Sol: Assume Matia is truth teller So he is a Sororian. Then chatia is the winner and Toni is Cororian (Alternator) Which implies Chatia is a false sayer (Nororian) If we check the truthfullness of the Chatia, We get his all statements are wrong and Toni's one statement is wrong. So Toni belongs to Cororian and Chatia won the race

Calendars Today is 15 August 1995. And you are asked to find the day of the week on 15 August 2001. If you dont know the method, it will prove a tough job for you. This type of question is sometimes asked in competitive exams. The process of finding it lies in obtaining the number of odd days. So, we should be familiar with odd days. The number of days more than the complete number of weeks in a given period, are called odd days. For example : (1) In an ordinary year (of 365 days) there are 52 weeks and one odd day. (2) In a leap year (of 366 days) there are 52 weeks and two odd days. What is the Leap and Ordinary year ? Every year which is exactly divisible by 4 such as 1988, 1992, 1996 et. is called a leap year. Also every 4th century is a leap year. The other centuries, although divisible by 4, are not leap years. Thus, for a century to be a leap year, it should be exactly divisible by 400. For example : (1) 400, 800, 1200, etc are leap years since they are exactly divisible by 400. (2) 700, 600, 500 etc are not leap years since they are not exactly divisible by 400. How to find number of odd days :An ordinary year has 365 days. If we divide 365 by 7, we get, 52 as quotient and 1 as remainder. Thus, we may say that an ordinary year of 365 days has 52 weeks and 1 day. Since, the remainder day is left odd-out we call it odd day. Therefore, an ordinary year has 1 odd day. A leap year has 366 days, i.e. 52 weeks and 2 days. Therefore, a leap year has 2 odd days. A century, ie, 100 years has : 76 ordinary years and 24 leap years. = [(76 X 1 day] + [(24 X 2 days] = 124 days if we devide 124 with 7 there are 17 full weeks and 5 odd days remains. Therefore, 100 years contain 5 odd days. Now, (i) 200 years contain 10 odd days, ie, 3 odd days. (ii) 300 years contain 8 odd days, ie, 1 odd day. (iii) 400 years contain 6 + 1 = 21, ie, no odd day. (Nore: 400th year is a leap year therefore, one additional day is added, So number of odd days in year 301 to 400 = 6 instead of 5) Similarly, 800, 1200 etc contain no odd day.

Practice Problems 1. January 1, 1992 was a Wednesday. What day of the week will it be on January 1, 1993 Ans: 1992 being a leap year, it has 2 odd days. So, the first day of the year 1993 will be two days beyond Wednesday. ie it will be Friday 2. On January 12, 1980, it was Saturday. The day of the week on January 12, 1979 was : Ans: The year 1979 being an ordinary year, it has 1 odd day. So, the day on 12th January 1980 is one day beyond the day on 12th January, 1979. But, January 12, 1980 being Saturday. January 12, 1979 was Friday 3. On July 2, 1985, it was Wednesday. The day of the week on July 2, 1984 was : Ans: Let us calculate the number of odd days between these two dates. July month of 1984 has

29 days left. So odd days are 1. Now august 1984 to June 1985 we have 3 + 2 + 3 + 2 + 3 + 2 + 0 + 3 + 2 + 3 + 2. Now July, 1985 till 2nd July contains 2 odd days. Total 29 odd days. or 1 odd day. So July 2, 1984 is 1 day before Wednesday. It is Tuesday. (2nd July 1985 is in fact Tuesday, and 2nd July 1984 is Monday) 4. Monday falls on 4th April, 1988. What was the day on 3rd November, 1987 ? Ans: Counting the number of days after 3rd November, 1987 we have : Nov Dec Jan Feb March April days 27 + 31 + 31 + 29 + 3 + 4 = 153 days containing 6 odd days i.e., (7-6) = 1 day beyond the day on 4th April, 1988. So, the day was Tuesday. 5. Today is 1st August. The day of the week is Monday. This is a leap year. The day of the week on this day after 3 years will be : Ans: This being a leap year none of the next 3 years is a leap year. So, the day of the week will be 3 days beyond Monday ie, it will be Thursday. 6. January 16, 1997 was a Thursday. What day of the week will it be on January 4, 2000 ? Ans: First we look for the leap years during this period. 1997, 1998, 1999 are not leap years. 1998 and 1999 together have net 2 odd days. No, of days remaining in 1997 = 365 - 16 = 349 days = 49 weeks 6 odd days. Total no. of odd days = 2 + 6 + 4 = 12 days = 7 days (1 week) + 5 odd days Hence, January 4, 2000 will be 5 days beyond Thursday ie it will be on Tuesday. 7. February 20, 1999 was Saturday. What day of the week was on December 30, 1997 ? Ans: The year during this interval was 1998 and it was not a leap year. Now, we calculate the no. of odd days in 1999 up to February 19 : January 1999 gives 3 odd days 19 February 1999 gives 5 odd days 1998, being ordinary year, gives 1 odd day In 1997, December 30 and 31 give 2 odd days Total no. of odd days = 3 + 5 + 1 + 2 = 11 days = 4 odd days Therefore, December 30, 1997 will fall 4 days before Saturday ie on Tuesday. 8. March 5, 1999 was on Friday, what day of the week will be on March 5, 2000 ? Ans: Year 2000 is a leap year No. of remaining days in 1999 = 365 - [31 days in January + 28 days in February + 5 days in March] = 301 days = 43 weeks ie 0 odd day. No. of days passed in 2000 = January (31 days) gives 3 odd days. February (29 days, being a leap year) gives 1 odd day March (5 days) gives 5 odd days Total no. of odd days = 0 + 3 + 1 + 5 = 9 days ie 2 odd days. Therefore, March 5, 2000 will be two days beyond Friday, ie on Sunday. 9. On which week day August 15, 1947 falls? Ans: We know that odd days upto 1600 years are zero. For the years 1601 to 1700 there exist 5 odd days, 1701 to 1800 there exist 5 odd days, 1801 to 1900 there are another 5 odd days. So upto 1900 there are 15 odd days or 1 odd day Now from 1901 to 1946 there are 11 leap and 36 non leap years. So number of odd days for these 46 years will be 11 X 2 + 35 X 1 = 57 . After deviding this with 7 we get 1 odd day. Now we entered into year 1947. January contains 3 odd days, february 2, march 3, april 2, may 3, june 2, july 3, august 15 = 3 +0+3+2+3+2+3+15 = 31 = 3 odd days So total odd days = 1 + 1+3 = 5 If odd days are 0 then it is sunday, 1 monday,........... so It is friday

Short Cut: Remember this shortcut technique: Month Code: 033 614 625 035 Century Code: 6420 15 - 08 - 1947

Century code explanation: We must consider every 400 years as a set, and of these if the given years fall in between first 100 years then CC = 6, 101 to 200 then CC = 4... So on. For 1947 we should take 1601 to 2000 as a 400 year set. Of these 1947 fall in the last century Total odd days = 1 + 2+ 0+ 11 + 5 = 19 = 5 5 odd days means friday. 10. When do we use same calender of the year 1968? Ans: To solve this problem we need to calculate the odd days for consecutive years upto the odd days become 0. This is a tedius job. Shortcut: If the given year is a leap year then add 28 If the given year is an year next to a leap year then add 6 For other years add 11. Here 1968 is a leap year so add 28 to it. So we can use the same calender for year 1996 Finding the weekday of a date when a reference date is given To find on which weekday the given day falls is a bit tricky to calculate by using odd day method. We have to find the number of odd days between the given two days and add to the given reference weekday. To find the odd days we need to divide the number of days between the given days by 7. But by using a simple technique we can solve this problem easily.

Clocks (Updated) The dial of a clock is a circle whose circumference is divided into 12 parts, called hour spaces. Each hour space is further divided into 5 parts, called minute spaces. This way, the whole circumference is divided into 12 X 5 = 60 minute spaces. The time taken by the hour hand (smaller hand) to cover a distance of an hour space is equal to the time taken by the minute hand (longer hand) to cover a distance of the whole circumference. Thus, we may conclude that in 60 minutes, the minute hand gains 55 minutes on the hour hand. Note : The above statement (underlined) is very much useful in solving the problems in this chapter, so it should be remembered. The above statement wants to say that : In an hour, the hour-hand moves a distance of 5 minute spaces whereas the minute-hand a distance of 60 minute spaces. Thus the minute-hand remains 60 - 5 = 55 minute spaces ahead

of the hour-hand. Some other facts : 1. In every hour, both the hands coincide once. 2. When the two hands are at right angle, they are 15 minute spaces apart. This happens twice in every hour. 3. When the hands are in opposite directions, they are 30 minute spaces apart. This happens once in every hour. 4. The hands are in the same straight line when they are coincident or opposite to each other. 5. The hour hand moves around the whole circumference of clock once in 12 hours. So the minute hand is twelve times faster than hour hand. 6. The clock is divided into 60 equal minute divisions.

7. 1 minute division = 360060=60 apart 8. The clock has 12 hours numbered from 1 to 12 serially arranged.

9. Each hour number evenly and equally separated by five minute divisions 560=300 apart.

10. In one minute, the minute hand moves one minute division or 60.

11. In one minute, the hour hand moves 120

12. In one minute the minute hand gains 5120 more than hour hand.

13. When the hands are together, they are 00 apart. Hence, Too Fast And Too Slow : If a watch indicates 9.20, when the correct time is 9.10, it is said to be 10 minutes too fast. And if is said to be 10 minutes too fast. And if it indicates 9.00, when the correct time is 9.10, it is said to be 10 minutes too slow. 2 important shortcut techniques: The angle between the hour hand and minute hand at a given time H:MM is given

by =30H112MM

The time after H hours, hour hand and minute hand are at degrees = MM=211(30H) Remember any angle less than 180 degrees comes 2 times in 24 hours.

Practice Problems 1. At what time, in minutes, between 3o clock and 4oclock, both the needles will coincide each other At 3oclock, the minute hand is 15 min. spaces apart from the hour hand. To be coincident, it must gain 15 min. spaces. 55 min. are gained in 60 min.

15 min. are gained in (155560) =16411 min

The hands are coincident at 16411 min. past 3. Alternate method: We can also solve this problem using degrees. At 3'O clock, Hour hand and minute hand are seperated by 90 degrees. Now to meet the Hour hand minute hand has to gain 90

degrees. We know that for every minute, minute hand gains 5120. To gain 90 degree it

takes 9005120=90112=90211=1801116411 minutes Alternate method:

Use formula degrees = MM=211(30H)

MM=211(3030)=18011=16411 min 2. At what time between 7 and 8oclock will the hands of a clock be in the same straight line but, not together ? When the hands of the clock are in the same straight line but not together, they are 30 minute spaces apart. At 7oclock, they are 25 min. spaces apart. Minute hand will have to gain only 5 min. spaces 55 min. spaces are gained in 60 min.

5 min. spaces are gained in (55560)5511 min. past 7 Alternate method: At 7'O clock minute hand and hour hand are 150 degrees apart. To be in the same line minute

hand has to gain another 30 degrees. But we know that miute hand gains 5120 degrees per

minute. So 3005120=30112=30211=60115511 Alternate method:

Use formula degrees = MM=211(30H)

MM=211(307180)=21018011=39011or3011

We regect 39011 as hour hand and minute hand are at 0 degrees. 3. The minute hand of a clock overtakes the hour hand at intervals of 65 minutes of correct time. How much does the clock gain or lose per day?

If a clock is running on time, Its hour hand and minute hand meets exactly for every 65511 min. But in this clock both hand are meeting at intervals of 65 min. so this clock is gaining time.

or 65 minutes in the correct clock = 65511 minutes in this clock. Or for every 65 minutes this clock is gaining 5/11 minutes.

for every minute this clock is gaining 511165

In 24 hours or 1440 min it gains = 5111652460 = 1010143 Alternatively: The minute hand of a clock overtakes the hour hand at intervals of M minutes of correct time.

The clock gains or loses in a day by (72011M)(6024M) minutes. 4. A watch which gains time uniformly is 5 minutes slow at 8'O clock in the morning on Sunday and is 5 minutes 48 seconds fast at 8 PM the following Sunday. When was it correct? This sunday morning at 8:00 AM, the watch is 5 min. Slow, and the next sunday at 8:00PM it

becomes 5 min 48 sec fast. The watch gains 5 + 54860 = 545 min in a time of (724)+12 = 180 hours. To show the correct tine, it has to gain 5 min.

545 min -------180 hours 5 min -------?

5545180 8313hrs=72hrs+1113hrs=3days+11hrs+20min So the correct time will be shown on wednesdy at 7:20 PM 5. A clock is set right at 8 AM. The clock gains 10 minutes in 24 hours. What will be the true time when the clock indicates 1 PM the following day? Between 8 AM and 1 PM total 29 hours have passed.

This clock shows 24 hr 10 min or 1456 for 24 hours in correct clock. In 29 hours in this clock

= 29145624 hours in actual clock = 1445 = 2845 = 28 hrs 48 min. So actual time is 12: 48 PM

6. How much does a watch lose per day, if its hands coincide every 64 minutes? Ans: If a clock is running on time, Its hour hand and minute hand meets exactly for

every 65511 min. But in this clock both hand are meeting at intervals of 64 min. So this clock is gaining time.

or 64 minutes in the correct clock = 65511 minutes in this clock.

Or for every 64 minutes this clock is gaining 1511 minutes. or 1611 minutes

For every minute this clock is gaining 1611164

In 24 hours or 1440 min it gains = 16111641440 = 32811 minutes Alternatively: The minute hand of a clock overtakes the hour hand at intervals of M minutes of correct time.

The clock gains or loses in a day by (72011M)(6024M) minutes. So (72011M)(6024M)=(7201164)(602464) = 1611(6038)=211(603) = =36011=32811 Level - 2 7. A person who left home between 4 p.m. and 5 p.m. returned between 5 p.m. and 6 p.m. and found that the hands of his watch had exactly changed places. When did he go out? We know that the dial of the clock has 60 equal divisions (Minute divisions). In one hour the minute hand makes one complete revolution, i.e., it moves through 60 divisions, and the hour hand moves trough 5 divisions,. Suppose that when the man went out the hour-hand was x divisions ahead after 4'O clock. Also suppose that when the man came back, the hour hand was y divisions ahead of 5'O clock.

Since the minute-hand and hour-hand exactly interchanged places during the interval that the man remained out, it is clear that when the man went out, the minute-hand was at y and hour-hand was at x, and when the man came back the minute-hand was at x and the hour-hand was at y. We know that the speed of the hour hand and minute hand are in the ratio 1 : 12. From the above diagram, In the time hour hand moves x divisions, hour hand moves 25 + y divisions. (calculate from 4'O clock)

112=x25+y ----------- (1) Also in the time hour hand moves y divisions, minute hand moves 10 + x divisions (calculate from 5'O clock)

112=y20+x ----------- (2)

From equation (1) we get 25 + y = 12x and from equation (2) we get 20 + x = 12y or x = 12 y - 20 Substituting x value in equation (1)

25 + y = 12 (12y - 20) 25 + y = 144y - 240 143 y = 265 y = 1122143

So the person went out y divisions after 5. So 25 + 1122143

So the time he went out = 4 hours 26122143 Alternate method:

From the above diagram it is clear that Hour hand and Minute hand together covered 60 minute spaces. We know that the speeds of hour hand and minute hand are in the ratio 1 : 12. So our of these

60 minute spaces hour hand would have covered 60113=4813 minute spaces.

i.e., Minute hand is 4813 minute spaces ahead of hour hand when the man went out. At 4'O clock Minute hand is 20 minute spaces behind hour hand.. When the man went out it

was 4813 spaces ahead of hour hand. So it has gained 20 + 4813 = 20 + 6013 = 32013 But we know that minute hand gains 55 minute spaces over the hour hand in 60 minutes.

It gains 1 minute space in 6055 minutes.

To gain 32013 minute spaces it takes 320136055=26122143 minutes.

So the man went out at 4 hours 26122143 minutes. 8. A person who left home between 2 p.m. and 3 p.m. returned between 4 p.m. and 5 p.m. and found that the hands of his watch had exactly changed places. How much time did he out?

In this questions, Hour hand and minute hand together covered 120 minute spaces together. (2:mm to 4:mm) Of these minute hand would have covered 12/13 part.

So total time he went out given by 1201213=1101013 minutes. 9. Between 5 and 6 a lady looked at her watch and mistaking hour hand for the minute hand, she thought that she was 57 minutes earlier than the correct time. When was the correct time?

Let hour hand is x minute spaces ahead of 5. As we know hour hand speed is 12 times of hour hand, Minute hand moved 12x minute spaces. So correct time = 5: 12x or (300 + 12 x) minutes But she mistook this time and assumed 4 : (25 + x) or (265 + x) minutes

(300 + 12x) - (265 + x) = 57 minutes x = 12 min So correct time = 5: 24 minutes. 10. When asked about the time, Amit replied; "If you add one quarter of the time from midnight till now to half the time from now till the next midnight, you get the time". what is the time now?

Let the time be "t" hours. From mid night till this time "t" hours passed. From now to next midnight there are (24-t) hours. Now

t4+24t2=tt+482t4=t 48t=4t 5t=48t=485hours or 935 hours 11. The inhabitants of planet Rahu measure time in hours and minutes which are different from the hours and minutes of our earth. Their day consists of 36 hours with each hour having 120 minutes. The dials of their clocks show 36 hours. What is the angle between the hour hand and the minutes hand of a Rahuian clock when it shows a time of 9:48? [Rahuians measure angles in degrees the way we do on earth. But for them, the angle around a point is 720 degrees instead of 360 degrees. ] In rahuian degrees the minutes hand travels a full circle in 1 hour. i.e., 120 minutes i.e., 720 degrees in 120 min or 6 degrees per min and the hours hand travels 720/36= 20 degrees per hour and 20/120 = 1/6 degrees/minute. At 9:48 the hours: Angle covered by hour hand = 9 x 20 + 48 x 1/6 = 188 deg Angle covered by min hand = 48 x 6 = 288 deg Angle between hour hand and minute hands = 288 - 188 = 100 degrees

Directions To solve directions related questions effectively, we first understand the follow diagram

TIP: When ever you are solving a direction problem, you must assume that you are at the intersection point. A man starts walking from his house towards north and covers 15.5 km and then turns left and walks 7 km. He then turns left again and after walking 31 km again turns left and stops after 7 km. How far is he from his house?

From the above diagram we know that CD is equal to BE. and AE = BE - AB = 31 - 15.5 = 15.5 Vasu facing west moved 50 m, then he took a left-turn and moved another 100 m. He then took a left-turn and moved another 70 m. After that he took another right-turn and moved 120 m. How far is he from the starting point?

From the above diagram We need to calculat OS which is the short distance. Now OY = 100 + 120 = 220. and XR = QR - OP = 70- 50 = 20m

Now OS=OY2+YS2 =(220)2+(20)2 =48400+400 =48800 = 220.9 meters If a bear walks ten miles south, turns left and walks ten miles to the east and then turns left again and walks ten mile north and arrives at its original position, what is the color of the bear. Tricky one. A bear which took 2 turns after starting and came to the same position may happen only at Earth poles. Now the bear moves towards south means it started at north pole. In the north pole bears are in white color. Directions for 1 to 5: 1. Agra is north of Erode and west of Calcutta. 2. Bombay is north of Agra and west of Federicktown. 3. Delhi is south and east of Agra

4. Erode is north of Faridabad and east of Delhi.5. Faridabad is north of Delhi and west of Agra. 6. Calcutta is south of Faridabad and west of Delhi. 1. Which of the towns mentioned is furthest to the northwest ? a. Agra b. Bombay c. Calcutta d. Erode e. Faridabad 2. Which of the following must be both north and east of Faridabad ? I. Agra II. Calcutta III. Erode a. I only b. II only c. III only d. I and II e. I and III 3. Which of the following towns must be situated both south and west of at least one other town ? a. Agra only b. Agra and Faridabad c. Delhi and Faridabad d. Delhi, Calcutta, and Faridabad e. Calcutta, Delhi, and Erode 4. Which of the following statements, if true, would make the information in the numbered statements more specific ? a. Calcutta is north of Delhi. b. Erode is north of Delhi c. Agra is east of Bombay. d. Calcutta is east of Faridabad. e. Bombay is north of Faridabad. 5. Which of the numbered statements gives information that can be deduced from one or more of the other statements ? a. 1 b. 2 c. 3 d. 4 e. 6 Your first instinct may be to draw a map and try to place the towns on it directly. Youll go hopelessly wrong if you try. (This is also true for other puzzles that contain two sets of ranked variables - John runs faster and jumps higher than Tom, and so on.) First place the towns on a north-south scale and on a separate east-west scale (Diagram 1). Then, if you wish, combine these into a two-dimensional map. This isnt necessary, but it may make the questions a little-easier. Weve included it (Diagram 2).

1.. B Bombay is both farthest north and farthest west. 2. E Agra and Erode are north of Faridabad, while Calcutta is to the south. All three towns are east of Faridabad. 3. D This one may be easier to read from the two-dimensional map, but you can also read it from the two separate scales. Calcutta. Delhi, and Faridabad are all south and west of Erode. Faridabad is also south and west of Agra. Bombay is not such of any town. Agra is south of Bombay, but east of it. 4. A The only ambigous information in the statements concerns the north-south position of Delhi. Statements (5) and (6) tell us that Delhi and Calcutta are both south of Faridabad, but not their position in relation to each other. A would clear this up. Choices B-E can all be deduced from the statements as given. 5. C Delhis north-south position with respect to Agra can be deduced from statement (1), (4), and (5), without statement (3). Delhis east-west position with respect to Agra can be deduced from statements (1) and (6). each of the other choices is necessary to place the town it mentions either on the north-south scale on the east-west scale, or on both.

Logical Consistency Suppose your father promised you a new bike, if you get good marks in your engineering. What if, you don't get good marks? Our analysis explore possibilities of your father buying a new bike for you, even if you don't get good marks! This type of reasoning is classified under a head called "Logical Consistency" If ..., Then ....:

Let us take an example: If it rains, It will be cloudy

Let us explore the above statement in various cases Case 1: It rained then we say, It should be cloudy. So If x happened then y should happen. x

y Case 2: There are no clouds, So there is no rain. y x Case 3: It is not raining. Uncertain. As there may be clouds or may not be. Case 4: It is cloudy. Uncertain. As it may rain may not rain. So of the above 4 cases Case 1 and Case 2 holds good. Other Structures: Only If: Let us take an example, Only If you work hard, you will be successful. Write the above statement like below. whenever there is 'onlyif' make sure it is in the middle of the two given statements.

Now to become successful, there is only one condition. To work hard. So we say, If one is succeeded means he must have work hard. So x y Also you did not work hard means, you are not succeeded. y x When / Whenever: When / Whenever is same as If. When x then y. So possible conclusions are 1. x y 2. y x Unless : Unless means "If not" Unless you work hard, you fail = If you don't work hard, then you fail. Again x y and y x are true Either / or : Take the Proposition: Either I will drink Pepsi or I will eat a sandwich. Let 'I will drink Pepsi' be 'X' and 'I will eat a sandwich' be 'Y'. I drank Pepsi, then one cannot say whether I ate sandwich or not. But If did not drink Pepsi, then one can say that I must have eaten sandwich. So Possible conclusions: 1. x y 2. y x

Solved Examples

1. Sam is either black or white. A. Sam is not white B. Sam is white C. Sam is black. D. Sam is not black. a. CB b. BA c. DB d. DC Solution: We know that If not black then White or If not white then black. So AC or DB correct. Correct option C. 2. Rohit is in the class when Puneet is in the lab. A. Puneet is in the lab. B. Rohit is in the park.

C. Puneet is not in the lab. D. Rohit is in the class. a. CA b. AD c. BC d. BD Solution: When X then Y. So When puneet is in the lab, then Rohit is in the class or Rohit is not in the class then Puneet is not in the lab. So AD is correct.

3. You will add more value to the brand if strategic planning is done. A. Stratigic planning was done. B. More value was not added to the brand. C. More value was added to the brand. D. Stratigic planning was not done. a. BD b. DB c. BC d. DC Solution: If strategic planning was done then you added more value to the brand or you did not add more value then Strategic planning was not done. So AC or BD correct. So choice A.

4. She sleeps only when her boss is away from the office. A. The boss is away B. She did not sleep. C. She slept. D. The boss ins in the office. a. DB b. AB c. DC d. BC Solution: Only when X then Y means Y happen then X happens or its contra positive X did not happen then Y did not happen. So We say She slept means boss is away, or Boss is not away then She did not sleep. Option A.

5. If Berty and Oly are selected in that order, Phil and Santhi cannot be selected. A. Phil and Santhi are selected in that order. B. Oly and Berty are selected in that order. C. Berty and Oly are selected in that order. D. Phil and Santhi are not selected. a. BC b. CD c. BD d. DB Solution: this is called compound hypothetical. If A and B then not C and D, Then C and D then not A or not B. Option B

Level 2 6. My house has got a number. If it is a multiple of 3, then it is in between 50 and 59. If it is not a multiple of 4, then it is in between 60 and 69 If it is not a multiple of 6, then it is in between 70 to 79 What is my house number? Solution: If the house number has to be in 50 to 59, then "If "conditions 2nd and 3rd statements should not happen. i.e., It is a multiple of 4 and 6. Now we know that if a number is a multiple of both 4 and 6, then it is a multiple of 12. But no 12 multiple exists between 50 to 59. So house number should not be in between 50 to 59 If the house number has to be in 60 to 69, then "if" conditions of 1st and 3rd statements should not happen. i.e., the number should not be a multiple of 3 but multiple of 6. All multiple of 6 should be multiples of 3. So no number exists in between 60 to 69 So the house number should exists between 70 to 79. Then It should not be a multiple of 3 but multiple of 4. Between 70 to 79, 72 and 76 are multiples of 4 but only 76 is not a multiple of 3. So my house number is 76

Forcefulness of Arguments We know that an argument consists of a few premises, unstated premises (Assumptions) and a conclusion. But not all the the arguments are strong enough to convince others to agree with the author. So students will be tested on his capability of judging whether a given argument is strong enough or not. Model question: Should women be provided more job opportunities. Argument 1: No. They will be given household jobs to manage Argument 2: Yes. They should also go into the outside world. Solving Argument questions: Step 1: Remove the options based on the preliminary screening as some arguments are just too simple, or ambiguous and may not contain any substance to convince others clearly why this has to be followed. Example: One should enjoy one's life to the fullest extent as tomorrow one has to die. Argument 1: No because, one should strive to achieve a goal Argument 2: No. This philosophy hardly enables us to do anything. Analysis: Argument 1 seems to be good but it is not suggesting how achieving a goal is a priority than enjoying one's life and one can achieve a goal in his life and simultaneously enjoy his life also. So argument 1 is weak. Argument 2 is just try to oppose the statement and not suggesting any course of action. This is a simple opinion rather than an argument. Example: Love marriages should be encouraged compared to arranged marriages Argument 1: No. Both are having their good points as well as bad points Argument 2: Yes. Arranged marriages are of no use in these days. Analysis: Argument 1 is just an elusive answer rather than taking a stance. It is not addressing the core issue. Argument 2 is also a mere statement and not saying why arranged marriages are of no use in these days. So both are weak arguments. Step 2: Check whether the result follows or not if the said argument holds good. A result follows in the following cases 1. Established fact / Prevailing notions of truth 2. Experiences predict that the result will follow 3. Logically, the result will follow Similarly we can reject an argument based on the following 1. Established fact suggests that the result may not follow 2. Experience predict that the result may not follow 3. Logically it is impossible

4. If it is an individual perception 5. If based on analogy or an example Example: Government must give more funds to midday meal system to reduce the dropouts from schools Argument 1: Yes. It act as an incentive to poor families to send their children to schools Argument 2: No. It increases additional burden on the government Analysis: Experience shows that midday meal system improves the attendance rate if properly executed. We are not arguing here that whether it is really successful or not, but we know that it is surely act as an incentive to many poor families to send their children to school. But second one is not so strong. Though it increases burden on the government exchequer, this action is desirable as education is a fundamental right and government must take necessary actions to make people enjoy this right. Step 3: Check whether the result is desirable to follow? Some arguments which pass the first two steps appear to be good arguments but they may not bring the desired benefits. Even though it gives the desired benefit, the course of doing it may bring more trouble or expenditure. The best way to check the validity of argument is to ask yourself "if it is true, why many people or institutions or government is not following it?" . Example: Military training must be made compulsory in schools Argument : Yes. It brings discipline to students Analysis: Military discipline may improve discipline as it is an established fact that as in the areas of army, navy etc. it is highly desirable and results are proven. But the suggested argument is like killing a mosquito with an ax. There are other proven methods are in place to bring discipline to students.

Solved Examples

Statement : Should there be no place for interview in selections? Arguments 1: Yes. It is very subjective in assessment. Arguments 2: No, It is only instrument to judge the candidates motives and personality. I is strong as a subjective mode of selection is not desirable. II is of course, right. Statement : Should higher education be completely stopped for some time? Arguments 1: No. It will hamper the countrys progress. Arguments 2: Yes. It will reduce educated unemployment. None is strong. Temporary stopping of higher education will not hamper the nations progress. It will reduce educated unemployment but so what? It will then increase uneducated unemployment. Statement : Should all news be controlled by the government in a democracy? Arguments 1: Yes. Variety of new only confuses people. Arguments 2: No. Controlled news loses credibility. Second is strong. First argument is debatable while the second is an established fact. Statement : Should there be students union in college / university? Arguments 1: No. This will create a political atmosphere in the campus. Arguments 2: Yes. It is very necessary. Students are the future political leaders. I is true as it is based on experiences. And political atmosphere in the campus is not really desirable as the campus is a place of learning not politics. II is also true because tomorrows leaders will come from todays students and it is good that they get some political training early.

Statement : Should there be only one university throughout India? Arguments 1: Yes. This is the only way to bring about uniformity in educational standards. Arguments 2: No. This is administratively impossible. Second is strong. First is weak because it is not correct. (Is it the only way?) Second is perhaps correct, on logical thinking. Statement : Should all the remote parts of a country be connected by road? Arguments 1: No. It will disturb peaceful simple life of the villages. Arguments 2:. Yes. It must be done immediately. None is strong. I is rejected because it may not happen (will not follow). II is rejected because it is too simple and does not have argumentative substance. Statement : Should government jobs in rural areas have more incentives? Arguments 1: Yes. Incentives are essential for attracting government servants there. Arguments 2: No, Rural areas are already cheaper, healthier and less complex than big towns. So, why offer extra incentives! Both are strong. Incentives do lure people. Second is also an established fact. Statement : Should religion be taught in our schools? Arguments 1: No. Ours is a secular state. Arguments 2: Yes. Teaching religion helps inculcate moral values among children. Second is strong. First is not very clear. If the state is secular, it means it is against religious bias but not against religion as such. Second is in consonance with the prevailing notions of truth. Statement : Should mercy death be legalised? Arguments 1: Yes. Patients undergoing terrible suffering and having absolutely no chance of recovery should be liberated from suffering through mercy death. Arguments 2: No. Even mercy death is a sort of killing and killing can never be legalized. Both are strong. They mention a positive and a negative feature of merry death; both these features are desirable/harmful respectively and both are related with important aspects of the topic of mercy death. Statement : Should there be a world government? Arguments 1: Yes. It will help in eliminating tensions among the nations. Arguments 2: No. Then only the developed countries will dominate in the government. Both are strong. On logical thinking both look probable, both are desirable (harmful in the case of second) and both touch significant aspects of the issue. Statement : Should the institution of marriages be abolished? Arguments 1: Yes. It is already showing cracks. Arguments 2: No. It is necessary for the survival of society. Second is strong. First is weak as you cannot abolish a system simply because it is showing cracks. Second is an accepted truth. Statement : Should telecasting feature films be stopped? Arguments 1: Yes. Young children are misguided by feature films. Arguments 2: No. This is the only way to educate the masses. First is strong. First is an acceptable piece of truth backed by evidence. Second is weak as it is not true. Statement : Should agriculture in rural India be mechanised? Arguments 1: Yes. It would lead to higher production. Arguments 2: No. It would lead to rural unemployment. Both are strong. Both are true (both will follow). The first is really desirable while the second it really harmful.

Statement : Should the illiterate be debarred from voting? Arguments 1: Yes. They are easily misguided. Arguments 2: No. It is their Constitutional right. Second is strong. First talks of a negative feature which goes exist but is not sufficient enough to go for such a drastic action as disallowing for vote. Second is an established fact. Statement : Can pollution be controlled? Arguments 1: Yes. If every one realizes the hazard it may create and cooperates to be rid of it, pollution may be controlled. Arguments 2: No. The crowded highways, factories and industries and an ever-growing population eager to acquire more land for constructing houses are beyond control. Both are strong. Although both arguments contradict each other, yet both are based upon sound assumptions or facts and hence, independently, both are strong.

Assumptions Every argument contains a few propositions which act as a base. But there are some propositions, an author deliberately leaves as they are obvious to the reader. For example, If A suggests B to go to ENT specialist as B is suffering from Throat infection, A assumes that a specialist doctor may diagnose better than normal physician. So an Assumption is an unstated premise.

In finding assumption to any question, we need to search for the reason which gives strength to the argument and without which the entire argument may not hold good. Note: Always remember, Assumption is an unstated premised from the author point of view which may not be true for the reader. For example, A suggests B, Let us go to XYZ movie, as Mahesh acts in that. Here A assumes if Mahesh acts in a movie, they are worth watching. But from B point of view, It may not be true.

Solved Examples Directions : In each question below is given a statement followed by two assumptions numbered I and II. An assumption is something supposed or taken for granted. You have to consider the statement and the following assumptions and decide which of the assumptions is implicit in the statement. Give answer (a) if only assumption I is implicit; (b) if only assumption II is implicit; (c) if either I or

II is implicit; (d) if neither I nor II is implicit; and (e) if both I and II are implicit. Statement : A taxi is required on rent. - an advertisement. Assumptions : I. All types of vehicles are available on rent. II. People will respond to the advertisement. b. The statement says that a taxi is required on rent. It does not say that any vehicle is required. So the statement cannot be said to have assumed that any vehicle is available for rent. But II is obviously implicit; whenever an advertisement is made it is assumed that people will respond to it. Statement : Buy pure ghee of company ABC. - an advertisement in a news paper Assumptions : I. No other company supplies pure ghee. II. People read advertisements. b. I is definitely not mentioned in the advertisement. II is implicit, otherwise Company ABC wouldnt have given the advertisements. Statement : Of all the TV sets manufactured in India, XYZ brand has the largest sale. Assumptions : I. The sale of all the TV sets manufactured in India is known. II. The manufacturing of no other TV set in India is an large as XYZ brand TV. a. Unless the sale of all TV sets manufactured in India was known, the statement could not have been made. Hence I is implicit. II is not implicit because we do not know about manufacturing; we know only about sales. XYZ brand has the largest sale but it may not be the largest manufacturer of TV sets. May be Y company manufactured more sets than XYZ does but it exports all its sets. In that case Y is a bigger manufacturer but its sale in India would be lesser than that of XYZ. Statement : Rams advice to Gopal - Go to Tirupathi via Gudur - the shortest route. Assumptions : I. Gopal wishes to go to Tirupathi. II. Ram gives advice to everybody. a. Unless Gopal would be going to Tirupathi, Ram would not have advised him this. Hence I is implicit. But it is not certain that Ram gives advice to everybody. Maybe Ram is giving advice to Gopal because Gopal is Rams friend. Statement : Cricket matches have become indispensable for the entertainment of people. Assumptions : I. Cricket matches are the only medium of entertainment. II. People enjoy Cricket matches. b. Cricket matches have become indispensable but it does not imply that they are the only medium of entertainment. But it is certain that people enjoy Cricket matches. Hence II is implicit. Statement : Do not lean out of the door of the bus. - a warning in a school bus. Assumptions : I. Leaning out of a running bus is dangerous. II. Children do not pay any heed to such warnings. a. Leaning out of a running bus must be dangerous, otherwise the warning would not have been

there. Hence I is implicit. But II is not implicit. If the authorities would have assumed that children do not pay any need to such warning, they would not have put it up there. Statement : If you are a software engineer, we want you as our programmer. - an advertisement by company XYZ. Assumptions : I. Software engineers are expected to be better performers by company XYZ. II. The company XYZ needs programmers. b. I is not implicit. The company wants software engineers. One reason could be that the company expects software engineers to be good performers, as I suggests. But there could be another reason: for example, the companys programmer job could be such that only a software engineer could perform it. But one thing is certain. The advertisement was for programmers; this means programmers are needed. Hence II is implicit. Statement : Be humble ever after being victorious. Assumptions : I. Many people the after being victorious. II. Generally, people are not humble. d. The statement asks a man to be humble ever after being victorious. This implies that people are usually not humble after victory. I is just the opposite of it. II is not implicit because it generalises the statement. Generally, people may be humble; the point is if they are humble or not after victory. Statement : A sentence in the letter to the candidates called for written examinations ---- You have to bear your expenses on travel etc. Assumptions : I. If not clarified, all the candidates may claim reimbursement of expenses. II. Many organisations reimburse expense on travel to candidates called for written examinations. e. If the letter mentions expenses to be borne by candidates, those who sent the letter must have assumed that the candidates may demand for reim-bursement if the point is not clarified to them. Also, the candidates would not demand reimbursement if it was not a prevalent practice. So I and II both are implicit. Statement : One of the opposition leaders said that the time had come for like-minded opposition parties to unite and dislodge the corrupt government. Assumptions : I. Like minded opposition parties should unite only when they have to dislodge a corrupt government. II. Opposition parties are not corrupt. d. To dislodge a corrupt government has been mentioned as the present purpose for the call of unity. But this does not mean that this is the only purpose. So I is not implicit. Further, the leader asks like-minded parties to unite against the government and not the entire opposition. So we cannot generalise that (all) opposition parties are non-corrupt. Hence II is not implicit. Statement : Bus charges have been increased to meet the deficit. Assumptions : I. The present charges are very low. II. If the charges are not increased, the deficit cannot be met.

b. Bus charges have been increased. The cause: to meet the deficit. This never means that the present are low. If the price of goods increase, it is not necessary that the earlier price was low. But the tone of the statement clearly implies that Bus charges have been increased out of compulsion: so II is implicit. Statement : If degrees are de-linked from jobs, students will think twice before joining college. Assumptions : I. Students join college education to get jobs. II. A degree is of no use in getting a job. a. The statement says that if degrees are delinked from jobs, students will not join colleges. This implies that jobs are a major reason for them to join college. So, I is implicit. Now, if I is implicit, II is not because II is just the opposite of I. Statement : Present-day education is in a shambles and the country is going to the dogs. Assumptions : I. A good educations system is essential for the well-being of a nation. II. A good education alone is sufficient for the well-being of a nation. a. The statement uses a tone that implies that if education is in a shambles, then the country deteriorates. This means that a good education is needed for the well-being of a nation. But this does not mean that a good education alone is sufficient for it. So, I is implicit but II is not. Statement : The next Annual general meeting of the Company will be held after one year. Assumptions : I. The Company will remain in function after one year. II. The Governing Board will be dissolved after one year. a. Obviously, the author assumes that the Company will be functioning after one year, otherwise he would not have fixed the date of the meeting one year later. But there is no hint that the Board will be dissolved after that. So I is implicit, II is not. Statement : Computer education should start at schools itself. Assumptions : I. Learning computers is easy. II. Computer educations fetches jobs easily. a. If one says that computers should be taught at schools he must have assumed that it is an easy subject, because schools are a place of elementary education; tougher things are taught at colleges. But the statement does not say anything about jobs. So I is implied, II is not. Statement : The new education policy envisages major modifications in the education system. Assumptions : I. Present education system is inconsistent with national needs. II. Present education system needs change. e. If major modifications are being envisaged, it must have been assumed that the present educational system is inconsistent with what the nation needs. So I is implicit. Again, if I is implicit, II also is, because II says the same thing as I. Statement : Srirams advice to Krishna - If you want to study Management, join Institute Y. Assumptions : I. Institute Y provides good Management education. II. Krishna listens to Srirams advice. e. If Sriram advises Krishna to join a particular institute, Sriram must have assumed that the particular institute was a good institute. So I is implicit. While advising Krishna, Sriram must also

have thought that Krishna would listen to Srirams advice; so II is implicit. Statement : Drop this letter in the letter in the letter-box on your way to school. - A mother to her daughter. Assumptions : I. The child knows the address of the person to whom the letter is being sent. II. The child will comply with the orders of his mother. b. The mother only asks her son to drop the letter. I would have been implicit if she had asked her son to drop the letter and write the address also. But II is implicit. Had the mother not assumed that her son would comply, she wouldnt have asked him to do the job. Statement : Everybody loves reading romantic stories. Assumptions : I. Romantic stories are the only reading material. II. Nobody loves reading any other material. d. People love reading romantic stories but this does not means that they have nothing else to read. Nor does it mean that they do not like to read anything else So, both I and II are not implicit. Statement : Read this notice before entering the office. Assumptions : I. People coming to the office are literate. II. No blind person comes to the office. e. When some says read this notice he must have assumed that the other person can read. This would be possible only if the person is literate and not blind. So both I and II are implicit.

Number Series Generally, two kinds of series are asked in the examination. One is based on numbers and the other based on alphabets. In questions based on series, some numbers or alphabets are arranged in a particular sequence. You have to decipher that particular sequence of numbers or alphabets and on the basis of that deciphered sequence, find out the next number or alphabet of the series. Although there is no limit of logics which can be used to build a series, here are some important examples given which highlight the type of series asked in the examination. How to solve number series problems: Step 1: Observer are there any familier numbers in the given series. Familier numbers are primes numbers, perfect squares, cubes ... which are easy to identify. Step 2: Calculate the differences between the numbers. Observe the pattern in the differences. If the differences are growing rapidly it might be a square series, cube series, or multiplicative series. If the numbers are growing slowly it is an addition or substration series. If the differences are not having any pattern then 1. It might be a double or triple series. Here every alternate number or every 3rd number form a series 2. It might be a sum or average series. Here sum of two consecutive numbers gives 3rd number. or average of first two numbers give next number Step 3: Sometimes number will be multiplied and will be added another number So we need to check those patterns

TYPES : I. Prime number Series : Example (1) : 2,3,5,7,11,13, ........... Answer : The given series is prime number series . The next prime number is 17. Example (2) :2,5,11,17,23,...........41. Answer: The prime numbers are written alternately. II. Difference Series : Example (1): 2,5,8,11,14,17,...........,23. Answer: The difference between the numbers is 3. (17+3 = 20) Example (2): 45,38,31,24,17,...........,3. Answer: The difference between the numbers is 7. (17-7=10). III. Multiplication Series: Example (1) : 2,6,18,54,162,.........,1458. Answer: The numbers are multiplied by 3 to get next number. (162x3 = 486). Example: (2) : 3,12,48,192,............,3072. Answer : The numbers are multiplied by 4 to get the next number. (192x4 =768). IV. Division Series: Example (1): 720, 120, 24, .........,2,1 Answer: 720/6=120, 120/5=24, 24/4=6, 6/3=2, 2/2=1. Example (2) : 32, 48, 72, 108, .........., 243. Answer: 2. Number x 3/2= next number. 32x3/2=48, 48x3/2=72, 72x3/2=108, 108x3/2=162.

V. n2 Series: Example(1) : 1, 4, 9, 16, 25, ......., 49 Answer: The series is 12, 22, 32, 42, 52, .... The next number is 62=36; Example (2) : 0, 4, 16, 36, 64, ........ 144. Answer :The series is 02, 22, 42, 62, etc. The next number is 102=100.

VI. n21 Series : Example : 0, 3, 8, 15, 24,35, 48, .........., Answer : The series is 12-1, 22-1, 32-1 etc. The next number is 82-1=63. Another logic : Difference between numbers is 3, 5, 7, 9, 11, 13 etc. The next number is (48+15=63).

VII.n2+1 Series : Example : 2, 5, 10, 17, 26, 37, .........., 65. Answer : The series is 12+1, 22+1, 32+1 etc. The next number is 72+1=50.

VIII. n2+n Series (or)n2n Series : Example : 2, 6, 12, 20, ............, 42. Answer : The series is 12+1, 22+2, 32+3, 42+4 etc. The next number = 52+5=30. Another Logic : The series is 1x2, 2x3, 3x4, 4x5, The next number is 5x6=30. Another Logic : The series is 22-2, 32-3, 42-4, 52-5, The next number is 62-6=30.

IX. n3 Series : Example : 1, 8, 27, 64, 125, 216, ......... . Answer : The series is 13, 23, 33, etc. The missing number is 73=343.

X. n3+n Series :

Example : 2, 9, 28, 65, 126, 217, 344, ........... Answer : The series is 13+1, 23+1, 33+1, etc. The missing number is 83+1=513.

XI. n31 Series : Example : 0, 7, 26, 63, 124, ............, 342. Answer:The series is 13-1, 23-1, 33-1 etc The missing number is 63-1=215.

XII. n3+n Series : Example : 2, 10, 30, 68, 130, .............., 350. Answer : The series is 13+1, 23+2, 33+3 etc The missing number is 63+6=222.

XIII. n3n Series : Example :0, 6, 24, 60, 120, 210, .............., Answer : The series is 13-1, 23-2, 33-3, etc. The missing number is 73-7=336. Another Logic : The series is 0x1x2, 1x2x3, 2x3x4, etc. The missing number is 6x7x8=336.

XIV. n3+n2 Series : Example : 2, 12, 36, 80, 150, ............, Answer: The series is 13+12,23+22,33+32etc. The missing number is 63+62=252

XV. n3n2 Series: Example: 0,4,18,48,100,................., Answer : The series is 13-12,23-22,33-32 etc. The missing number is 63-62=180 XVI. xy, x+y Series: Example: 48,12,76,13,54,9,32,..............., Answer :2. 4+8=12, 7+6=13, 5+4=9 .: 3+2=5. XVII. Image Series or (Interchange Series): Example : 34, 81, 72, 47, 74, 27, 18, ................ Answer : (47,74,), (72,27), (81,18), are images. :. Image of 34 is 43.

Puzzles 1. You have three baskets filled with fruits. One has mangoes, one has bananas, and

the third has a mixture of mangoes and bananas. You cannot see the fruit inside the baskets. Each basket is clearly labelled. Each label is wrong. You are permitted to close your eyes and pick one fruit from one basket, then examine it. How can you

determine what is in each basket?

Solution: Observe the table:

Pick up a fruit from the basket labelled Mangoes and Bananas. If the fruit is banana

then we must be certain that that box contains only Bananas as it must not contains Mangoes and Bananas mixture. Now from the table possbile combinations of remaining two baskets can be seen. Basket 1 should not contain Mangoes but

contain Bananas / Mango and banana mixture. But we already confirmed that basket 3 contains all bananas. So basket 1 must have mangoes and bananas mixture. Basket 2 must contain Mangoes.

Similary logic we can apply for the scinario if we pick up Mango. So The fruit must be picked up from the basket labelled Mangoes and Bananas

mixture 2. You have 26 consonants, labelled A through Z. Let A equal 1. The other constants

have values equal to the letters position in the alphabet, raised to the power of the previous constant. That means that B = 21, C = 321 ... and so on. Find the exact numerical value for the following expression: (x - A) x (x - B) x (x - C) x ... (X - Y) x (X - Z)

Solution: X is the twenty - fourth letter of the alphabet. The constant X must equal 24 raised

to the power of the previous constant, W. Since W is the twenty - third letter, W equals 23 to the power of U, which is 22 to the power of T, which is 21 to the power

of ...X=24232221...321 What all this means is that X is going to be 24 raised to the power of 23 to the power of 22 to the power of 21 ... and so on, all the way upto 3 to the power of 2 to the power of 1. Thats 23 nested exponentiations. X is a very big number. However the correct answer is zero. Among the 26 terms which are being multiplied, there will be (X - X). Which is 0. Hence answer is 0. It doesnt matter what all the other terms are. Multiply anything by 0 and you get 0.

3. You have a bag full of feathers of three colours - red, green, and blue. With your eyes closed, you have to reach the bag and take out two feathers of the same colour. How many feathers do you have to take to be certain of getting two of the

same colour ? Solution:

Four. Pick just three feathers, and its possible you have one of each colour and therefore no match. With four feathers, at least two have to be of same colour.

4. Five robbers have one hundred gold coins to split among themselves. They divide the coins as follows : The senior robber proposes a division, and everyone votes on

it. Provided that at least half the robbers vote including himself for the proposal i.e, he has to get 50% of the votes, they split the coins that way. If not, they kill the senior robber and start over. The most senior (surviving) robber proposes his own

division plan, and they vote by the same rules and either divide the coins or kill the senior robber, as the case may be. The process continues until one plan is accepted. Suppose you are the senior robber. What division do you propose ? (The robbers are

all extremely logical and greedy, and all want to live.) Solution:

Initially you may think that, the coins should be divided equally or apportion more coins for other robberers so that the senior may save his skin. But we follow a smart approach to solve this puzzle

Assume there is only one robber. Then he votes himself so he gets 100% of the votes and takes all the 100 coins. If there are two robbers, D and E. Say D is senior. In this case D votes for himself

and gets 50% of votes and takes all coins. Observe If D is the senior E gets nothing.

If there are 3 robbers, C, D and E. Say C is senior. Now C has to get atleast one more vote to survive. As his votes consists only 33.33% of total votes. He thinks like this. If D is the senior, E gets nothing. So let us offer "One gold coin" to E. Now E

obviously votes for C Because, if D is the senior, he gets nothing. If there are 4 robbers, B, C, D, and E. B has to get the support of atleast one robber to achieve his 50% vote target. He thinks like this, If C is the senior, B gets nothing.

So He offers one coin to D and gets his support If there are 5 robbers, A, B, C, D and E. A has to get atleast two more votes to survive. If D is the senior, C and E both gets nothing. So he offers one coin each to

get their votes and keep remaining98 coins with him. Very interesting. is it not?!

5. There is an old bridge over river Ganga. Four people wants to cross the bridge at night. Many plants are missing, and the bridge can hold only two people at a time

(any more than two, and the bridge collapses). The travellers must use a torch to guide their steps; otherwise theyre sure to step through a missing space and fall to their death. There is only one torch. The four people each travel at different speeds.

Sharukh can cross the bridge in one minute; Aamir in two minutes; Salman takes five minutes; and the slowest person, Saif, takes ten minutes. The bridge is going to collapse in exactly seventeen minutes. How can all four people cross the bridge?

Solution: Round - trip one: The fastest pair, Sharukh and Aamir cross, taking two minutes.

One of them (lets say Sharukh - it doesnt matter) immediately returns with the torch (one minute). Elapsed time: there minutes. Round - trip two : the slow pair, Salman and Saif, cross taking ten minutes. As soon

as they reach the farside of the bridge, they hand the torch to the faster person who is already there. (Thats Aamir, assuming that Sharukh returned in the first round - trip). Aamir returns the torch to the nearside (two minutes). Elapsed time : fifteen

minutes. Final, one - way trip : the fast pair is now reunited on the nearside. They cross for the second and last time (two minutes). Elapsed time : seventeen minutes.

6. You have two candles. Each will burn for exactly one hour. But the candles are not identical and do not burn at a constant rate. There are fast-burning sections and slow-burning sections. How do you measure forty-five minutes using only the

candles and a lighter? Solution:

At time zero, light both ends of candle A and one end of candle B. The candles must not touch each other. It takes thirty minutes for candle As two flames to meet. When they do, there is exactly thirty minutes left on candle B. Instantly light the

other end of (still-burning) candle B. The two flames will now meet in fifteen minutes, for an elapsed time of forty-five minutes.

7. One of your female employees insists on being paid daily in silver. You have a silver bar whose value is that of seven days salary for this employee. The bar is already segmented into seven equal pieces. If you are allowed to make just two cuts

in the bar, and must settle with the employee at the end of each day, how do you do it ?

Solution: You need a one-unit piece to pay the employee for the first days work. You lop one unit off the end and hand it to the employee. This leaves you with a six - unit bar and one more permitted cut. Cut off a two - unit piece. At the end of the second day, you hand over the two -

unit piece to the employee and get the one - unit piece back as change. (You have to assume that the employee hasnt already spent it.) This leaves you with a four - unit bar, the one - unit piece you got in change, and no

more cuts. On the third day, you return the one - unit piece. On the fourth day, you