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* Selected Questions on Reasoning from various past PO papers appearing very common.

Question : How many such pairs of letters are there in the word CHRONICLE each of which has as many letters in between them as in the english alphabets ?
a) None
b) One
c) Two
d) Three
e) More than three

First we must understand what the question intend to convey.
It says that in this word CHRONICLE, you take any two alphabets and then see how many letters are in between them. And if you see in your english alphabets a,b,c,d,e,f...........z and these very two same alphabets, then count how many letters occur in between them. Are they two equal. If yes then this becomes a pair. So how many such pairs can you get ?

There is no fixed formulae to check this it is just that your ability to see that two alphabets are so close or far and compare it in the word there.

Like suppose If I say take R & N, it surely cannot make a pair because even without counting we can see between R & N more than one letters will come. So it is not a pair. So actually it will check how fast you can scroll your eyes on all the possibilities.

Lets see between CLE - ( as it comes CDE in english alphabets)
HRONIC - ( CDEFGH ) four letters come in between C & H in both cases.
& ON - ( NO in elnglish alphabets ) they occur together so no letter ( zero letter) between them.

Hence only three pairs can be formed and not more.

Question : If '+' means '/' , '*' means '-' , '-' means '+' , '/' means'*'
then what is the value of
20 – 16 + 4 * 3 / 2 = ?
a) 16
b) 30
c) 18
d) 24

Solution : 18

Question : Ramesh said, “The son of my maternal grandfather's only child is engineer”. About whom is he talking ?
a) His Brother
b) Himself
c) Cousin Brother
d) Data Inadiquate

Solution : Here Data is inediquate. Remember it is saying maternal Grandfather's only child's son. So we dont know if there is only one son or more sons & daughters.

Question :

Statements :
All monkeys are parrots.
No Parrot is crow.
Some crows are horses.
All horses are tigers.

I. Some tigers are parrots.
II. Some Crows are monkeys.
III. No Tiger is parrot.
IV. Some horses are parrots.

Which of the conclusions are correct.
a) None follows.
b) Only II follows
c) Only III follows
d) Only I follows
e) Either I or III follows.

Solution : In such type of questions, it says all monkeys are parrots but there are some extra parrots also which are not monkeys.

No parrot is crow

Some crows are there which are equal to horses but not all crows. And at the same time horses are there which are not crow. Means some extra horses are there which are not equal to crow.

All horses are tigers.

NOW ........
when it says some tigers are parrots. Can it be drawn as conclusion from above ?
No Not necessarily. No link is there between tigers and parrots in the statements. So it may or may not be true.

For option A. It will be last to check if all other possibilities are ruled out.

Some Crows are Monkeys. No it is not possible. Because the link is broken when it say ALL MONKEY ARE PARROT AND NO PARROT IS CROW. So no monkey is crow. II does not follow. Option B is not possible.

Next, No tiger is parrot. Cant say that . Because might or might not be possible since we dont know. Because it is not mentioned clearly any relation. So option C is not correct because of lack of certainity.

Next, Some tigers are parrot. Hmmm ......... again we are not sure may be possible or may not be because no mention about this thing is there. So we will not go with this option also. Hence D is also rejected.
But we see that some tigers are parrot if we reject this then it means that no tiger is parrot. And if we reject the statement that no tiger is parrot. Then it will mean that yes there might be some tigers which are parrot.
If you are getting confused then read the above slowly and with more concentration.
Now you know that either of above must be correct because they are just opposite things to each other. So option is E .

Question : A, B,C,D,E,F,G and H are sitting in a circle facing towards center. D is fourth to the right of H & second to the left of B. F is fourth to the right of B. C is fourth to the right of E who is not an immediate neighbour of B or D. A is not an immediate neighbour of D.

Now answer following questions...

Well here I dont ask any question because they are so simple like who is sitting next to this , alternate of so and so, between so and so.... etc etc etc...
Main thing is that we must atleast find the order of sitting then we can answer all the questions.

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