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Here goes the first question.
Ques 1: If a/b = 1/3, b/c = 2, c/d = 1/2, d/e = 3 and e/f = 1/4, then what is the value of abc/def ?
(1) 3/8
(2) 27/8
(3) 3/4
(4) 27/4
(5) 1/4
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Neha
Member •
Dec 1, 2006
aashima
Here goes the first question.
Ques 1: If a/b = 1/3, b/c = 2, c/d = 1/2, d/e = 3 and e/f = 1/4, then what is the value of abc/def ?
(1) 3/8
(2) 27/8
(3) 3/4
(4) 27/4
(5) 1/4
My answer to this question would be option 1 i.e 3/8..right??
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Neha
My answer to this question would be option 1 i.e 3/8..right??
Lets others also try Neha. Between try to add your explanations as well with the answers.
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Neha
Member •
Dec 1, 2006
aashima
Lets others also try Neha. Between try to add your explanations as well with the answers.
Sure!!
I'll post the explanation when others have replied.
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Hi Neha,
My answer is also 3/8.can I give explanation or give others also a try?
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Do we have any more takers for this??? I expected a good response from CEans on this!
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May be many thought, OK the problem is simple. And it is in fact so. The actual quant stuff in CAT which is a head ache is the Speed-distance-time part and Work related problems.
Anyway answer is 3/8 as all of you would have got.
Explanation : Sorry, I am usurping someone's wish to post the explanation.
here goes -
a/d = a/b*b/c*c/d = 1/3*2*1/2 = 1/3
b/e = b/c*c/d*d/e = 2*1/2*3 = 3
c/f = c/d*d/e*e/f = 1/2*3*1/4 = 3/8
abc/def = a/d*b/e*c/f = 1/3*3*3/8 = 3/8
Simple right?
Anyway, let me take this opportunity to propose something.
May be people who gave their CAT exams this year, can post in questions to which they couldn't find answers. And we others here can exercise our brains to find the solution.
What do the others say?
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Nice Idea. Aashima has already taken an initiative to start the related threads. CAT takers may put the difficult questions in these threads.
Way to go, Guys! 😁
-The Big K-
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reachrkata
Explanation : Sorry, I am usurping someone's wish to post the explanation.
No problems with that Reachrkata, besides answering correctly, answering in minimum possible time is also a factor that counts. So everybody should come up with the explanation.
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Its hard to find out here, but can anyone get the answer without actually solving the problem?
... i mean to say, sometimes just by looking at some of these ratios, u can rationalize abc/def = c/f = 3/8
heres its reasoning,
a/b = 1/3 ... ab/b^2=1/3 and d/e = 3 ... de/e^2=3 ... e^2/de = 1/3 = ab/b^2 ... thus ab=de
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Kidakaka has a point. CAT is not about tough questions. Actually the questions are very basic.
The CAT gets tougher because of the time constraint. So, the idea is to crack the questions without actually solving step-by-step.
Short-cuts! Anyone? 😉
-The Big K-
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I would concur partially.
It totally depends on what your strength is.
If one is very good at manipulating variables to get the final answer, its good.
But for me, it looks straightforward that a/d is the multiplication of a/b, b/c and c/e. So if i am very fast in fraction arithmetic, this problem I think is totally solvable in 15 secs.
So there is never "a" shortcut method for the problem. You choose the method that works the fastest for you.
Just my $0.02 !!
By the way, any more problems?
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Acccording to me the answer is 3/8.
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Well the correct answer indeed is 3/8. And thats good to know that all of the CEans cracked it! Great going!
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Okay lets have next question now.
Ques 2:
If x = –0.5, then which of the following has the smallest value?
(1) 2^1/x
(2) 1/x
(3) 1/x^2
(4) 2^X
(5) 1/ −x^1/2
Flood in with your answers CEans, All the best!
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Neha
Member •
Jan 7, 2007
My answer to this question would be option 2.
(Fingers crossed)
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Hi,
Good start.
My answer is option 2 (1/x)
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Guys try to add your explanations for proper verification.
Between any more takers??? Cme on CEans..
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Well the correct option is (2).
Sol. Go by option, put x=-1/2
(1) 2^(-2) = 1/4
(2) 1/x => 1/ (-1/2) = -2
(3) 1/ (x^2) => 1/ (-1/2)^2 = 4
(4) 2^ (-1/2) = 1/ 2^1/2
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Good going Neha and Mahesh.
Ques:
A survey was conducted of 100 people to find out whether they had read recent issues of Golmal, a monthly magazine. The summarized information regarding readership in 3 months is given below:
Only September: 18; September but not August: 23; September and July: 28;September: 28; July: 48; July and August: 10;
None of the three months: 24.
What is the number of surveyed people who have read exactly two consecutive issues (out of the three)?
(1) 7
(2) 9
(3) 12
(4) 14
(5) 17
All the best!
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Hallo,
I do not get ans.s of my Q.Which I post 5 days ago.My posting headline is Apti Problems from MBA under CE Higher Education.Please guide me.I think U people will be interested to solve those.As I find them heard.
Bye
Suparna
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@supernaday
yar i can give u a detailed explanation about those probs and how to handle them.But i dont have time to post it here.Stay touch with me thorugh PM/or instant messaging.
my id is : nsit.vipul [@]gmail[dot]com
PS: I wil post the solution here after some time,when i wil get time to do so.
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Yes, please do post the solutions over here. Explaining problems through email or over IM will help one, but posting the solution on CE will help many.
-The Big K-
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nyny
Member •
Dec 14, 2007
aashima
Good going Neha and Mahesh.
Ques: A survey was conducted of 100 people to find out whether they had read recent issues of Golmal, a monthly magazine. The summarized information regarding readership in 3 months is given below:
Only September: 18; September but not August: 23; September and July: 28;September: 28; July: 48; July and August: 10;
None of the three months: 24.
What is the number of surveyed people who have read exactly two consecutive issues (out of the three)?
(1) 7
(2) 9
(3) 12
(4) 14
(5) 17
All the best!
Do you know the solution to this? I'm stuck...
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This problem can be solved using the venn diagram approach. Draw three intersecting circle with one representing each month. Then add the values. You can easily eliminate options 1 and 2 since july and august is already given as 10. So you now just need to find out how many people read august and sept but not july. This can be found out by substracting from 100 Only July readers, then readers of all the three issues, then only august readers, only september readers and july and sept readers. You may try it now. Do post if you have some more difficulties.
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