CAT 2007: Quantitative Questions a Day 101 to 133 - The Discussions

krsh.vik

Quote:

Originally Posted by vineet.nitd

Many people say If it were a DS(Data Sufficiency) problem we could have safely told that the data is insufficient.
Here we had cannot be determined which played the same role.
So always be very cautious with cannot be determined.There is no scope for assumptions.

Should we also assume that the project should be completed in 30 days? It isn't mentioned anywhere in the question ..

Let me say again if it is cat I will not assume this and would have straight away ticked cannot be determined

swarnali

answers to PART (A)
is option(A)

PART (B) is undoubtedly option(D) cannot be determined

richa_nishu

Quote:

Originally Posted by vineet.nitd

We have plans to give you 2 more tests and they will be very close to the actual difficulty level of cat.

As a result we expect people to attempt more number of questions with increased accuracy.Our intention is to make you comfortable wrt guessing cat cut off of 96%ile which would help you in these last few days gauge the right level of application you need to do.

Next one month will also see a dip in the toughness level of qqad questions,so that with the DDay nearing you all could increase your speed up and improve on the number of attempts and get used to it,but ofcourse not at the cost of accuracy.

Im new to PGUY..could u tell me how to take the tests uu r talking abt...

vineet.nitd

Quote:

Originally Posted by richa_nishu

Im new to PGUY..could u tell me how to take the tests uu r talking abt...

4 tests have already been given and 3 more are yet to be given.Subscribe to the newsletter and you shall get the next 3 as and when we would send them.For the previous 4 I shall pm you the links by tomorrow evening

Aarav

Quote:

Originally Posted by vineet.nitd

4 tests have already been given and 3 more are yet to be given.Subscribe to the newsletter and you shall get the next 3 as and when we would send them.For the previous 4 I shall pm you the links by tomorrow evening

Vineet, just try if we can have archives for the QQAD problems and the tests like last year. This will save lot of effort at our end also.

Waise, the zip is still very much there for the tests (post #499)
http://www.pagalguy.com/forum/quanti...e-test-50.html (CAT 2007: QQAD, Practice Test Discussions)

Aarav

Don't get intimidated by big numbers, the questions are fairly simple and involves no complex calculations.

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Quantitative Question # 117 (The answer to both Part A and Part B is in the same set of options)
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Question(s):(Part A)

GOI directive proposed formation of more new IIMs. Fearing lack of good faculty and geographical distances as disadvantageous, IIMs decide to have one for all campus. The School of Planning and Architecture proposed 2 new projects for the new Housing campus. In each project, the campus is designed to have several identical dorimiatory buildings, with the same number of 1 bedroom apartment in each building. In the first project there are 12096 apartments in total. There are 8 more buildings in the second project than the the first and each building has more apartments which raises the total of apartments in the project to 23625. Let N be the number of buildings the second project require. Then N is divisible by

(Part B)

The number P = 10101...0101 is written using (n+1) ones and n zeros. Let N be the number of digits in least possible P such that P is divisible by 9999. Then N is divisible by

(a) 5 (b) 7 (c) 9 (d) at least two of the foregoing (e) none of the foregoing

pratyush_sinha · 09:47 AM

Quote:

Originally Posted by Aarav

Don't get intimidated by big numbers, the questions are fairly simple and involves no complex calculations.

------------------------------------------------------
Quantitative Question # 117 (The answer to both Part A and Part B is in the same set of options)
------------------------------------------------------

Question(s):(Part A)
GOI directive proposed formation of more new IIMs. Fearing lack of good faculty and geographical distances as disadvantageous, IIMs decide to have one for all campus. The School of Planning and Architecture proposed 2 new projects for the new Housing campus. In each project, the campus is designed to have several identical dorimiatory buildings, with the same number of 1 bedroom apartment in each building. In the first project there are 12096 apartments in total. There are 8 more buildings in the second project than the the first and each building has more apartments which raises the total of apartments in the project to 23625. Let N be the number of buildings the second project require. Then N is divisible by

(Part B)The number P = 10101...0101 is written using (n+1) ones and n zeros. Let N be the number of digits in least possible P such that P is divisible by 9999. Then N is divisible by

(a) 5 (b) 7 (c) 9 (d) at least two of the foregoing (e) none of the foregoing

part (b)

P= 10^0 + 10^2....10^2n (total terms = n+1) = (100^(n+1) -1)/99

Thsi has to be divisible by 9999.

100^(n+1) = 9999k *99 +1 = 989901*k +1 (this has to be a power of 100)

81 * 121* 101 k +1

k can either be 9,19,29....

part(a)

12096 => 2^6 * 3^3 * 7
23625= 5^3 * 3^3 *7

then a = 27 and a+8 =35 makes

(d) as the answer

targetABC · 10:08 AM

PART A.

12096 = 3^3*7*2^6
23625 = 3^3*7*5^6

Let A = 12096 = (N -

* (no. of apartments in P1)
B = 23625= N * (no. of apartments in P2)

N-8 should be a factor of A and N a factor of B.
N-8 cant have any of the powers of 2 as in that case N is even and B has no even factors.
N-8 can be 7,21,63,189 or 27.
If N-8 = 21, N= 29(not a factor of B).
Same is the case with 63 and 189.
N-8 can only be 7 or 27 => N = 15 or 35. Divisible by 5.

Answer option. a

targetABC · 10:08 AM

PART B.

P = 10101010....101 to be divisible by 9999, it should be individual divisible by 9,11,101 ( co-prime factors of 9999).
For divisibility by 9 and 11 the number of 1s should a multiple of 99, say 99k => no. of digits in P = 198k - 1.
For divisibility by 101 it shoulde be in the form of 101 0101 0101 ...i.e. 101 followed by some numbers of 0101. Total number of digits in P = 4l + 3.
4l + 3 = 198k - 1.
l = ( 198k - 4 ) / 4. l to be integer minimum k = 2. No. of digits in P = 198k - 1 = 295.
P is divisible by 5.

Answer option . a

swarnali

PART B

P=10101....1

since P is divisible by 9999 = 3^2 * 11 * 101

so,sum of digits of P should be divisible by 9 --(1)
if we take 9 1's and 8 0's,then

P = 10101010101010101 which is div by 9
but to make it divisible by 11,we need 11n 1's ---(2) (where n =1,2,3...)
to make it div by 101,we need even no. of 1's ---(3)

99 doesn't satisfy the eqn (3)

so,we need atleast 198 1's and 197 0's
so,N = no. of digits in P = 198+197 = 395

so,N is div by 5 only

so answer(a)