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ms_cs
Member •
Mar 9, 2009
Nice, ...I am having R.S.Aggarwal Book...I can post the questions from that book...
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hello ms_cs
if you are preparing for cat exam RS aggarwal is no longer useful.
please get a book by arun sharma or another stuff, basically RS aggarwal is a easy version not much useful for CAT takers.
😎
feel free to post your questions.
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ms_cs
Member •
Mar 12, 2009
Whether Agarwal book is helpful for quantitative aptitude test conducted during interview..?.I thought that it will useful for both.
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Hello People,
Here i have an idea , instead of you people posting questions i will post the difficult questions.
initially i will start posting stuff about the chapter say numbers some topics and you people can read them and if i find any interesting questions i will post it here.
We will discuss the chapter for some N no of days depending upon its weight age in CAT exam.
Wat say guys?
😎😎
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Trishna's Quantitative ability is a very good book.has many difficult questions.Is there weightage for specific chapters in cat quant?
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Hey,
I am ready. Start asking questions.
Lets crack CAT 2009😁
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This thread went dead.
Ok lets see...
Tell me what are the kinds of numbers?
i mean the classification of Numbers?
😎
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Numbers are mainly classified as Complex, Real, Rational, Integer, and Natural Numbers.
CAT is not a subjective paper I guess.. 😁
-Crazy
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ok.
let me put a question here.
Q: Log[(tan 2)(tan 4)(tan 6)....(tan 86)(tan 88)] = ?
a) 0
b) 2
c) 1
d) infinity
I think it is easy .
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Raviteja.g
ok.
let me put a question here.
Q: Log[(tan 2)(tan 4)(tan 6)....(tan 86)(tan 88)] = ?
a) 0
b) 2
c) 1
d) infinity
I think it is easy .
tan a = (sin a/cos a)
Sin 2 = Cos 88
similarly cos 2 = sin 88
So [(tan 2)(tan 4)(tan 6)....(tan 86)(tan 88)] = 1
and Log 1 = 0, so answer is
a)
Correct me if I am wrong?
-Crazy
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crazyboy
Numbers are mainly classified as Complex, Real, Rational, Integer, and Natural Numbers.
CAT is not a subjective paper I guess.. 😁
-Crazy
Thats Correct.
Its not subjective , but if we can classify numbers according to it types we will be getting the behavior of numbers .
say for example we can have many questions on Prime numbers.
number missing questions will all come under Whole or Natural numbers.
If we go for complex numbers we will have complex multiplication and division which is a bit tough one .
i expected that we can discuss some easier methods on this!!!
😎😎
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crazyboy
tan a = (sin a/cos a)
Sin 2 = Cos 88
similarly cos 2 = sin 88
So [(tan 2)(tan 4)(tan 6)....(tan 86)(tan 88)] = 1
and Log 1 = 0, so answer is a)
Correct me if I am wrong?
-Crazy
this is also right.
but i did it in another way as below:
we all know that
tanAtanB = 1-(tanA+tan B)/tan(A+B)
Putting A = 2 and B= 88
tan2tan88 = 1-(tan2+tan88)/tan 90 = 1-0 =1(tan 90 = infinity)
we get 1 for all the pairs and we are left with tan 45 with no pair.
Now tan 45 = 1
so it comes down to log(1.1.1...) = log 1 = 0.
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again one more question to solve.
Q) a, b, c are positive integers forming an increasing geometric sequence, b-a is a square, and (loga + logb + logc)/log6 = 6. Then a + b + c is
(a) 54
(b) 78
(c) 111
(d) 36
e) None of these
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Raviteja.g
again one more question to solve.
Q) a, b, c are positive integers forming an increasing geometric sequence, b-a is a square, and (loga + logb + logc)/log6 = 6. Then a + b + c is
(a) 54
(b) 78
(c) 111
(d) 36
e) None of these
(Log (abc) / Log abc) = 6
so we got take 36 as one of the terms .. then.. we can take the other terms. 6^4.. so we know 27*48 .. =6^4.. answer is c.
so answer is 27, 36, 48 = 111
-Crazy
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(Log (abc) / Log abc) = 6
i can't get you crazyboy
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Raviteja.g
i can't get you crazyboy
Ok Let me explain in detail...
= Log a + log b + log c
= Log abc
so on the same principle
6=(loga + logb + logc)/log 6
6=log (abc)/log 6
abc are in GP so b^2 = ac
log (b^3) = 6 (log 6)
log (b^3) = log (6^6)
b^3 = 6^6
b = 36
I hope i am suounding much better now...
-Crazy
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ya. I got it
and its correct and the way i did is
log a + log b + log c = 6 log 6--------(1)
a,b,c are in GP. so b^2 = ac => 2 log b = log a + log c------(2)
from 1 and 2, we get
3 log b = 6 log 6 => b = 36
As (b - a) is a square, a can be 32, 27, 20, 11. Among these only for a = 27, we can have an integral value for c which is 48.
hene a+b+c+ = 27+36+48 = 111. so option c....
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Raviteja.g
ya. I got it
and its correct and the way i did is
log a + log b + log c = 6 log 6--------(1)
a,b,c are in GP. so b^2 = ac => 2 log b = log a + log c------(2)
from 1 and 2, we get
3 log b = 6 log 6 => b = 36
As (b - a) is a square, a can be 32, 27, 20, 11. Among these only for a = 27, we can have an integral value for c which is 48.
hene a+b+c+ = 27+36+48 = 111. so option c....
Sorry to say but if you think the way I have solved the question is not a correct way 😁 then give a serious thoughts about your math skills.
CAT is all about how quickly you can solve the question. Its nothing like by proper method you can solve it or not.
No offense means!!
-Crazy
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CAT is all about how quickly you can solve the question. Its nothing like by proper method you can solve it or not.
yes you are correct.
but here it didn't seem that the paths you solved the problems are time taking.
and I think the above problems are not up to the mark for testing of fastness. Right?
so ,don't worry.
what makes you to think in that way?
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Raviteja.g
yes you are correct.
but here it didn't seem that the paths you solved the problems are time taking.
and I think the above problems are not up to the mark for testing of fastness. Right?
so ,don't worry.
what makes you to think in that way?
Absolutely correct, even I didnt say/think that somebody is testing speed of somebody.
I was just saying that you should not comment that which is the right method. You solved it by your method and I solved it by my method.
-Crazy
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All are correct 😁.
So please guys post questions and answers its interesting.
Dont argue even friendly argument also a big NO .
i argued earlier and i am sorry about it.
so we will continue with questions.
😎
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Here i poses a simple question.
Find the sum till n terms of the series where T(n)= 2^(n-1) × (2^n- 1) ?
its very easy.
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Hey this thread again went dead
Crazyboy i think you can solve this problem
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Few tips to solve the problems based on series.
The questions will be mostly on these types::
For Finding missed or next number in the series The series may in the form of,
Even Numbers,Odd Numbers,Composite Numbers,Prime Numbers,Multiplyers ,Divisors,
In the form of Arithmatic Progession, or In the form of Geometric Progression,
Succssively divided by 8,6,4,2 like that,
Addition of the consecutive numbers,
Subtraction of the consecutive numbers,
Multiplication of the consecutive numbers,
Division of the consecutive numbers,
Consecutively Adding or Subtracting or Multiplying orDividing by Even numbers or Odd numbers or prime numbers,
Square series ,cube series
Square+1 or -1
Square+2 or -2 or *2 or/2
Square+3 or-3 or *3 or /3Like that or
Cube+1 or -1
Cube+2 or -2 or *2 or /2
Cube+3 or -3 or *3 or /3 Like that
Middle number is addition or subtraction or multiplication or division of the other two numbers
Sometimes starts with two series you can guess and
Add any number to first number subtract any number from second number
Multiply any number to first number divide the second number with any number
Some time here you can do in the reverse format also
Every number multiplied by (.5 or1.5)
Every number divided by(.5or1.5)
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Here is an interesting question.
All of us know 4! = 24
What is the value of 4!! and 4!!!
anyone??
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I don't get it buddy.
4!! will be 24! and 4!!! will be 24!! I guess..
I don't know how to go about it..
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Any takers ??
i have read in some site , i too wondered by seeing this answer,
i dont know the explanation but the answer is ..
4!! = 4 * 2 = 8
The difference is 2 here, so if you want 7!! = 7 * 5 * 3 * 1
4!!! = 4* 1 = 4
For 7!!! here = 7 * 4 * 1
I guess it should be useful for everyone here..
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oh, is this the meaning of the notation?? I missed that.
I guess 24! would be (4!)! and not 4!!. Right??
Anyhow, nice one ES.
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Its d infinity
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Yes 24! is (4!) ! not 4!!
I am not sure why the answer is like this ? need to refer some books or some sites , anyone with clear explanation is expected here.
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English-Scared
Any takers ??
i have read in some site , i too wondered by seeing this answer,
i dont know the explanation but the answer is ..
4!! = 4 * 2 = 8
The difference is 2 here, so if you want 7!! = 7 * 5 * 3 * 1
4!!! = 4* 1 = 4
For 7!!! here = 7 * 4 * 1
I guess it should be useful for everyone here..
I am just guessing one explanation from my side. 😕
If number of factorials sign is more than or equal to one then subtract the count from the previous number and keep on repeating and multiplying until you get it zero.
Ex
4! = 4 * 3 [(4-1=3) as 1 factorial sign is there ] * 2 [3-1=2 ] * 1 [2-1]
4!! = 4 * 2 [(4-2=2)as 2 factorial signs are there]
4!!! = 4 * 1 [(4-3=1)as 3 factorial signs are there]
7!! = 7 * 5 [(7- 2=5 ) as 2 factorial signs are there ] * 3 [ 5-2=3 ] * [3-2=1]
7 !!! = 7 * 4 [(7- 3=4 ) as 3 factorial signs are there ] * 1 [ 4-3=1 ]
Unfortunately my stupid trick is working for your provided cases somehow. Any corrections to my logic, Most Welcome 😀
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I too thought of the same procedure.
I guess we need to investigate more on what is Factorial ?
Will post here if i get any solution , though i too agree with shalini's solution as it gives proper answer to all the factorials.
IF any one has a proper solution please post here.
Thanks Everyone!!
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