Introduction to Logic in Reasoning

Introduction to Logic in Reasoning

How do we know that tomorrow sun rises in the east? How do we know that if we touch the fire it burns us? Eventhough there are scientifiic proves, mostly we argue based on our past experiences.

Let us take one example:

Proposition:  Mr.X is driving a rolls royce car
Conclusion:  so Mr.X is a rich person

You might argue that this argument is not valid because Mr.X can be driver or the car or may be it is a rented car. But If you are living in a developed country where Rolls Royce cars can be seen every where, you are most likely to agree that Mr.X is a rich person as he is driving the Rolls royce.

This reasoning is called Inductive reasoning or probability.  This reasoning is proposed by David Hume.  He suggested that the people who won't agree with this type of reasoning must be starved to death as this is one of the most important way of convincing people and to draw valid conclusions.

Let us take another Argument:

Proposition: Sachin is a great batman
Conclusion: So India will win the match

Some of you again may not agree with the varacity of the conclusion but If I may add another proposition "Great batsman help teams to win matches" then this argument looks like below

Proposition: Sachin is a great batman
Proposition: Great batsman help teams to win matches
Conclusion: So India will win the match

Now this is more convincing.  This type of reasoning is called Deductive reasoning.  This is proposed by Aristotle.  He studied 216 different structures of Deductive reasoning and found that only 16 structures give valid conclusions.

Let us look at a couple of the false arguments

1. Propositions are valid but conclusion is false
Proposition: If Bills gates has kohinoor diamond then He is rich
Proposition: Bill Gates does not have Kohinoor
Conclusion: So he is not rich

We know this is a false conclusion, even though both the propositions are true

2. Propositions are false but conclusion is true
Proposition: All Rats are CATs
Proposition: All CAT are Dogs
Conclusion: All Rats are Dogs

Clearly both the propositions are false but given the propositions true, the conclusion is true.

We will see lot of arguments in our daily life.  These arguments range from convincing our friend to make him watch your favorite hero movie to whether Bihar state has to be given or not.  Critical reasoning is the use of logic to evaluate arguments.

 Logic is defined as the study of methods and principles used to distinguish good (correct) reasoning from bad (incorrect) reasoning.  Let us have a look at some technical terms

Argument:

An argument is a group of statements (propositions) where the statements follow one another and ultimately give a final statement known as the conclusion or inference.  The group of all these statements including the conclusion is known as an argument.

Most questions in Logical reasoning are based on whether the student is capable of testing the validity of an Argument, the first thing one has to clearly understand is the concept of the Argument.  For the purposes of understanding the concept of the Argument fully, it would help to get acquainted with a few key terms.

Elements of an Argument:

Proposition: A proposition is the basic units of an argument.  A typical proposition has a relationship spelled out between a subject and an object in the form a sentence.
Illustration: of Proposition

Eg:  All Andhrites are Indians
Here “Andhrites is the subject and “Indians” is predicate

Premise:  The term premise is applied to the proposition that gives rise to the conclusion or the inference.  Unless the premise is valid, the conclusion will not be valid.

Conclusion or inference:  The conclusion or inference of an argument is the final proposition that is affirmed on the basis of other propositions of the same argument.

Argument = Proposition 1 (Premise) + Proposition 2 (Premise) + Proposition 3 (Conclusion)
Eg:  All Students are good
       Rama is a student
       Rama is good

Types of arguments:

Deductive argument:  A deductive argument is one whose conclusion is claimed to follow from its premises with absolute necessity or certainty, this certainty not being a matter of degree and not depending in any way on anything else.  Therefore a deductive argument has to be either valid or invalid.  There is no grey area in between.

In competitive exams questions on this area comes under the header "syllogisms or Statements and conclusions"

Inductive Argument:  An inductive argument is one whose conclusion is claimed to follow from its premises only with probability,  Inductive arguments, therefore, cannot be absolutely valid or invalid, the way deductive arguments are.  But most of the arguments we make in our life are based on inductive reasoning.  We may not convince others purely based on deduction, but by giving some past examples.

In  most of the competitive exams questions on this area comes under the header "Critical reasoning".

Non-Verbal Reasoning (Analogy) Interview questions

Non-Verbal Reasoning (Analogy) Interview questions 

Analogy means relationship.  Let us have a look at an example: 

Teacher : Pen  : : Soldier : _________   

What should come in the blank? If teacher's main tool is pen, Soldier's main tool is a gun.  

Similarly, we have to identify the relationship between in figures A and B so that to identify the option which got relationship with figure C. 

Just look at few examples and you can easily understand these problems:

1. 

The square in PF(A) rotated 90$^0$ clockwise along with dot.  So option 5 is correct. 

2. 

 Pentagon in PF(A) became small and circumscribed with Square in PF(B).  So If a square has to become small and to be circumscribed with triangle.  So option 1 is correct.  option 5 is rules out as the square rotated 45$^0$ instead of 90$^0$.

3. 

Here the hexagon becae pentagon and the dots came out of the diagram, and a new darkened dot appeared in the middle.  So PF(C) should become triangle and two dots must come out and a darkened dot must appear in the iddle.  So correct Option 5

4. 

  Here Bottom square became big, and the figure above it, came into it and pentagon appeared in the triangle.  So in PF(C) pentagon must become big, and square must be inside it and a hexagon should appear in it.  so correct option 2

5. 

  Simple one.  Two circles became a single square, and the square became two squares. So two triangles must become single triangle and circle must become two circles. So answer option 4.

6. 

 Another simple one.  the directions of the arrows changed their positions.  So answer option 4.

7. 

 In PF(1), top half darkened rectangle turned 90$^0$ clockwise, middle half darkened rectangle turned anti-clockwise 90$^0$ and bottom half darkened rectangle turned clockwise by 90$^0$.  So turn the rectangles in PF(C), clockwise, anti-clockwise, and clockwise. So correct Option 1

8. 

 In PF(A), the square has three dots each at the middle of its sides.  In PF(B), square became pentagon, and number of dots got increased by one and one dot occupied the vertex.  So PF(C) must become hexagon and there must be 5 dots and one dot should occupy the vertex. Correct option 4

9. 

 The entire diagnol rotated by about 180$^0$ clock wise and the open circle became darkened.  So the square must be darkened and the entire diagnol should rotate by 180$^0$.  So correct option 3

10. 

 Simple observation.  Pentagon at the top became bigger and square came inside of it.  So hexagon in PF(C) should become big and circle should enter into it.  So correct option 3

Questions for Practice

11. 

 Answer: Option 2

12. 

 Answer: Option 3

13. 

 Answer: Option 5

14. 

 Answer: Option 4

15. 

 Answer: Option 4

16. 

 Answer: Option 4

17. 

 Answer: Option 4

18. 

 Answer: Option 3

19. 

 Answer: Option 1

20. 

 Answer: Option 5

21. 

 Answer: Option 2

22. 

 Answer: Option 5

23.

 Answer: Option 1

24.

Answer: Option 3

Amcat-previous-papers

Amcat-previous-papers

• Amcat previous questions set - 5
• Amcat previous questions set - 4
• Amcat previous questions set - 3
• Amcat previous questions set - 2
• Amcat previous questions set - 1

More sets are coming up...

Non - verbal reasoning (Series)Interview questions

Non - verbal reasoning (Series)Interview questions 

Non-Verbal reasoning appears in Bank exams, Infosys, MAT exams constantly. There are 5 Problem Figures (PF) will be given with 5 Answer Figures (AF).  We need to determine the next figure in the series.  There are certain rules which make solving these problems easy.  So study the rules and solved examples.

How to answer these questions: 

Step 1: 

For all the series problems the following rules apply.  If problem figures A and E are equal our answer is problem figure B. Similarly, the other rules as follows. 

1. PF(A) = PF(E) then answer is PF(B)

2. PF(D) = PF(E) then answer is PF(C)

3. PF(A) = PF(C) = PF(E) then answer is PF(B)

4. PF(A) = PF(D), PF(B) = PF(E) then answer is PF(C)

5. PF(D) = inverse of PF(A) and PF(E) = inverse of PF(2) then answer is inverse of PF(C)

Step 2: 

In general, the items in the box takes different positions in the subsequent figures.  They may rotate certain degrees either clock wise or anti-clockwise.  Look at the following diagram. In some problems new items add to the existing figures and some existing figures vanish.  

Solved Examples

1. 

In this problem If PF(A) = PF(E) then answer is PF(B).  In the answer options AF(4) is same as PF(B) so option 4

2. 

 Here PF(C) and PF(E) are equal.  So Answer figure should be PF(B).  So correct option is c.

3. 

The arrow is changing its positions clock wise 90^o, 45^o, 135^o, 45^o, ....next should be 180^o. So option 3.

4. 

 Simple one.  A new arrow and a new line are adding alternatively.  In PF(E) a new line has added.  So in the next figure a new arrow must be added.  And total lines should be 6.  Option 5

5. 

 Small hand is moving anticlock wise 90^o, 45^o, 90^o, 45^o,... and Big hand is moving clock wise 135^o constantly.  So in the next figure, small hand must move 90^o anti clockwise, and big hand must move 135^o.  So option 4

 6.

Here the symbol is changing positions anti clockwise by  45^o and every time a new symbol is adding. The "C"s in the middle are rotating clock wise by 90^o. So the next figure must be option 4

7. 

 This is a simple analogy.  There is a relationship between 1 and 2, 3 and 4.  the small figures in the first diagram are getting bigger and vice-versa.  So Option 3

8. 

 All the three symbols in the dice are rotating clockwise.  So option 3

Alternative method:

We know that if  PF(A) = PF(D), PF(B) = PF(E) then answer is PF(C).  So option 3

9. 

 A new symbols is appearing in the middle of the previous figure and the previous figure is getting bigger. So option 4 is the right option.  3 and 5 options are ruled out as the figures in the middle are appeared already.

10. 

 A dot and line are adding constantly to the figures in left and right sides alternatively.  So option 3

11.

There appears to be no pattern on immediate look, but his problem can be solved by simple observation.  Have a look at the diagram below..

The positions of two symbols are not changing in 2 consecutive figures.  So option 5

12. 

 the arrow and small line inside the small square are rotating constantly anti clockwise and clockwise respectively by 90^o, 45^o, 90^o, 45^o,... and 45^o, 90^o, 45^o, 90^o .  So next figure would be option 3.

13.

 The line is rotating anti clock wise by 90^o, 180^o, 270^o, 360^o  so next figure should be 90^o from figure E and a new symbol must appear.  So option 1 is the correct.

14. 

 Symbols X is rotating clockwise by 45^o, 90^o, 45^o, 90^o.  So our options will be either 1 or 3 as in the next figure symbol X must move 45^o.  A new symbols is being added to X each time one at front and next time at back.  So option 3 is right one.

15.

 Simple.  Observe PF(A) and PF(E) are equal.  So next figure will be PF(B).  So option 5

16. 

 the symbols are changing constantly in clockwise direction and a new symbol is being added.

The red rounded circle is a place whenever a symbol appear in that position must not appear in the next.  And remaining positions are moving clockwise by 90^o.  A new symbol must come at the place shown by green arrow.  So our option will be 1.  Option 2 is ruled out as + symbol appeared earlier.

17.

 Circle is moving diagonally and triangle is moving clockwise by 90^o. So option 1 is correct one.


18. 

Here you can easily observe that the lines are rotating 90^o clockwise. also in PF(B) and PF(D), half line has added at the right most side and in figures PF(C) and PF(E) a new line has added. So in our answer half line has to be added and lines should rotate 90^o.  So answer option 2.

19. 

 Simple one.  Figures A and B changed their symbols opposite them.  C and D also did So.  So option 1

20. 

Symbols in  A, B are same except Symbols at bottom.  A new symbol is coming there.  Similarly in C, D.  So option 3.  Option 2 is ruled out as C appeared earlier. 

Puzzles Interview questions and answer

Puzzles Interview questions and answer

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 letter’s position in the alphabet, raised to the power of the previous constant. That means that B =21$2^1$, C = 321$3^{2^1}$ ... 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⎞⎠⎟$\left(X={24}^{{23}^{{22}^{{21}^{{.}^{{.}^{{.}^{3^{2^1}}}}}}}}\right)$
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. That’s 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 doesn’t 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 it’s 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 remaining 98 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 they’re 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 (let’s say Sharukh - it doesn’t 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. (That’s 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 A’s 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 day’s 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 hasn’t 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 hand over the four - unit piece and get the two smaller ones as change. Use them to pay the worker on the fifth, sixth, and seventh days.

8. Five men crash-land their airplane on a deserted island in the South Pacific. On their first day they gather as many coconuts as they can find into one big pile. They decide that, since it is getting dark, they will wait until the next day to divide the coconuts.
The night each man took a turn watching for rescue searchers while the others slept. The first watcher got bored so he decided to divide the coconuts into five equal piles. When he did this, he found he had one remaining coconut. He gave this coconut to a monkey, took one of the piles, and hid it for himself. Then he jumbled up the four other piles into one big pile again.
To cut a long story short, each of the five men ended up doing exactly the same thing. They each divided the coconuts into five equal piles and had one extra coconut left over, which they gave to the monkey. They each took one of the five piles and hid those coconuts. They each came back and jumbled up the remaining four piles into one big pile.
What is the smallest number of coconuts there could have been in the original pile?

Solution:
Assume total number of Coconuts are N.
When these coconuts are divided into 5 equal parts one is remaining so N = 5A+1 (Here A is number of cococonuts in the smaller pile)
After given one coconut to monkey and has taken first man his share, then 4A coconuts are remaining.
When the second person did the same thing to the remaining coconuts we can write
4A = 5B + 1
4B = 5C +1
4C = 5D +1
4D = 5E + 1

Now Add 4 to the both sides 
N+4 = 5A + 1 +4 ⇒$\Rightarrow$ N+4 = 5 (A+1)
4A + 4 = 5B + 1 +4 ⇒$\Rightarrow$ 4(A+1) = 5 (B+1) ⇒(A+1)=54×(B+1)$\Rightarrow \left(A+1\right)={\displaystyle \frac{5}{4}\times \left(B+1\right)}$


4B + 4 = 5C + 1 +4 ⇒$\Rightarrow$ 4(B+1) = 5 (C+1) ⇒(B+1)=54×(C+1)$\Rightarrow \left(B+1\right)={\displaystyle \frac{5}{4}\times \left(C+1\right)}$


4C + 4 = 5D + 1 +4 ⇒$\Rightarrow$ 4(C+1) = 5 (D+1) ⇒(C+1)=54×(D+1)$\Rightarrow \left(C+1\right)={\displaystyle \frac{5}{4}\times \left(D+1\right)}$

4D + 4 = 5E + 1 +4 ⇒$\Rightarrow$ 4(D+1) = 5 (E+1) ⇒(D+1)=54×(E+1)$\Rightarrow \left(D+1\right)={\displaystyle \frac{5}{4}\times \left(E+1\right)}$

Now N + 4 = 5 (A +1) ⇒5×54×(B+1)$\Rightarrow 5\times {\displaystyle \frac{5}{4}\times \left(B+1\right)}$
Similarly we can substitute remaining values in this equation so

N+4=5×54×54×54×54(E+1)$N+4=5\times {\displaystyle \frac{5}{4}\times {\displaystyle \frac{5}{4}\times \frac{5}{4}\times \frac{5}{4}\left(E+1\right)}}$

We know that N is an intezer so E +1 must be a multiple of 44$4^4$. The N+4 = 55$5^5$
So N = 55$5^5$ - 4 = 3121


9. A woman took a certain number of eggs to the market and sold some of them. A next day, through the industry of her hens, the number left over had been doubled, and she sold the same number as the previous day.
On the third day the new remainder was tripled, and she sold the same number as before.
On the fourth day the remainder was quadrupled, and her sales were the same as before.
On the fifth day what had been left over were quintupled, yet she sold exactly the same as on all the previous occasions and so disposed of her entire stock.
(i) What is the smallest number of eggs she could have taken to market the first day?
(ii) How many did she sell daily? (The answer is not zero).

Solution:
Let us make small table to understand this question clearly.

Right hand columns tells us the number of eggs sold.
Number of eggs remaining = (24x−41y)×5−Y=0⇒120x−206Y=0⇒xy=10360$(24x-41y)\times 5-Y=0\Rightarrow 120x-206Y=0\Rightarrow {\displaystyle \frac{x}{y}={\displaystyle \frac{103}{60}}}$


10.  Doo-Bee-Doo had born on 1468 B.C. He had lived one-fourth of his life as a boy, one-third of his life as a youth, one-fifth of his life as a man and the remaining 52 years as an old man. Which year did Doo-Bee-Doo die?
Solution:
x4+x3+x5+52=x${\displaystyle \frac{x}{4}+\frac{x}{3}+\frac{x}{5}+52=x}$

⇒13x60=52$\Rightarrow {\displaystyle \frac{13x}{60}=52}$ ⇒$\Rightarrow$ x = 210

Year of Death = 1468BC - 240 = 1228 BC ( In BC the next year is 1 year less than the previous year)