Official Quant thread for CAT 2013

MBA CAT 2024
25328
@techgeek2050 said:
sir, can u plz tell me where am i wrong if i solve like thistotal number of ways of selecting 5 = 14C5number of ways in which none is consecutive = 10C5number of ways in which 3, 4 or 5 are consecutive = 12+11+10 = 33number of ways where exactly 2 are consecutive = 14C5 -(10C5 + 33) ???
bhai number of ways in which 3 4 5 consecutive that part is wrong
yes 3 consecutive is 12 but u need to select 5 numbers in total how will u select next 2
similaRLY FOR 4
25329
@abhishek.2011 said:
bhai number of ways in which 3 4 5 consecutive that part is wrong yes 3 consecutive is 12 but u need to select 5 numbers in total how will u select next 2similaRLY FOR 4
Ya i got that bro.. deleted that post :)
25330
@bodhi_vriksha said:
All these no's are divisible by 3 . so hcf=3 in your case and we need to find a triplet with hcf=1 .. so here goes my approach :820 is a multiple of lcm .. so let our lcm be 420 which is the largest it can be ..and we want it to be the largest bcuz we want to maximise sum of three no's..so one no is fixed at 420 .. if we take second no as 210 ..then we cant take a multiple of 2 or 3 or 5 or 7 in third no ..so we take our second no as 140 .. now we can afford to take our third no as 3 but we cant take 7.. so lets take 3 as third no so three nos are 420,140 and 3 which sum up to 563 Team BV--Pratik Gauri
yes sir i edited that :)
25331
@techgeek2050 said:
sir, can u plz tell me where am i wrong if i solve like thistotal number of ways of selecting 5 = 14C5number of ways in which none is consecutive = 10C5number of ways in which 3, 4 or 5 are consecutive = 12+11+10 = 33number of ways where exactly 2 are consecutive = 14C5 -(10C5 + 33) ???
When 5 are consecutive, then 10 ways (this is correct)

When 4 are consecutive
a X b XXXX c (I'm representing those which are selected by X)
OR
a XXXX b X c

a + b + c = 9 and b is atleast one
=> C(10, 2) ways

So, 2*C(10, 2) ways

When 3 are consecutive
a XXX b X c X d
OR
a X b XXX c X d
OR
a X b X c XXX d

a + b + c + d = 9, and b and c are atleast 1
=> C(10, 3) ways

So, 3*C(10, 3) ways

Then there will be cases when 3 are together and 2 together & when two together and two other together. So, it will be very lengthy
25332
@chillfactor said:
When 5 are consecutive, then 10 ways (this is correct)When 4 are consecutivea X b XXXX c (I'm representing those which are selected by X)ORa XXXX b X ca + b + c = 9 and b is atleast one=> C(10, 2) waysSo, 2*C(10, 2) waysWhen 3 are consecutivea XXX b X c X dORa X b XXX c X dORa X b X c XXX da + b + c + d = 9, and b and c are atleast 1=> C(10, 3) waysSo, 3*C(10, 3) waysThen there will be cases when 3 are together and 2 together & when two together and two other together. So, it will be very lengthy
Roger that sir! 😃
25333
@heenalove said:
Its 890=dk..ans is 89 bt hw u got it hereabhishek.2011
SUBTRACT THE TWO EQUATIONS THAT I WROTE IN MY EXPLANATION
890 = dk
now as d and k are integers so they must be a factor of 890
factorise 890 it is 2*5*89
so d can be 2 k will be 89*5
d can be 5 , k will be 89*2
d can be 89, k will be 2*5
so no unique solution
25334
@ChirpiBird said:
563 h kya??A=420B=140C=3LCM = 420.
Sumit committed a mistake in finding the LCM of three distinct positive integers greater than 1 namely A, B and C, and found it to be 840, which is a common multiple of A, B and C all, but is not the lowest. The HCF of A, B and C is 1. Find the maximum possible value of A + B + C.


OA : 563
25335
@bodhi_vriksha said:
This one is today'sQuantitative Question of the Day (QQAD). Let's see how many of you are up for this challenge.Let A be a subset of {1, 2, 3, ... , 27} such that no two subsets of A have the same sum. What is the largest possible sum for A?(1) 101 (2) 121 (3)129 (4) 135 (5) none of theseThis problem will also be flashed on our FB page. Team BV -Vineet
Solution:
To find A with sum of the elements having max sum, we can start with 27 and 26 safely. 25 can also be taken without any harm. Now, 27 = 27, 26 = 27-1, 25 = 27-2. The next number will be 27-4 as if next were 27-3, then 27-3 + 27 = 27-1 + 27-2 [condition violated]. Now, we have 4 numbers with us.

Each of the next number below 27-4 i.e. 27-5 can't be in our set as 27-5 + 27 = 27-1 + 27-4, 27-6 can't also be there as 27-6 + 27 = 27-2 + 27-4. But we have no such problems with 27-7. The set till now is {27, 26, 25, 23, 20}.

Thus -> 0, 1, 2, 4, 7 became seed (for the namesake) elements which needed to be subtracted from 27.

The next seed element will be 13 as any number between [8, 12] will violate condition. e.g. 8+0 = 7+1 or 11+0+1 = 7+4+1 or 12+0+1=2+4+7-> note that we must have same number of summands on either side to do this kind of generation.

Any new element 27-x, where x > 13 can be combined with some of the elements in {27, 26, 25, 23, 20, 14} whose sum will be equal to some of the rest elements
.
The initial few seed elements can be found out to be 0, 1, 2, 4, 7, 13, 24, 46, 86.

=> Choice (4) is the right answer

Team BV - Vineet
25336
A person starts from the origin of the coordinate axis . He travels in this pattern . 1 unit to right , (1/2) units up , (1/4) units right, (1/8) units down , and continues the above pattern . At what point(coordinates) will he ultimately come to stop?
25337
@pankaj1988 : confuse between 2 answers:

4/3, 2/5

4/3,2/3
25338

@ziddiarmaan said:
@pankaj1988 : confuse between 2 answers:4/3, 2/54/3,2/3
25339
@pankaj1988 : it will be Infinite GP ..

in the right right direction : 1/1-1/4= 4/3

whereas for up down .. i think we have to make series : 1/2 3/8 7/16 13/32 ... still dont understand the pattern
25340
@ziddiarmaan said:
@pankaj1988 : it will be Infinite GP .. in the right right direction : 1/1-1/4= 4/3 whereas for up down .. i think we have to make series : 1/2 3/8 7/16 13/32 ... still dont understand the pattern
X axis: 1+1/4+1/16......=4/3
Y axis: 1/2-1/8+1/32-1/128..........=2/5
25341
@pankaj1988 : http://www.gmatgeniescore.com/http:/www.kmtutor.com/category/gmat-questions/

hey i got sumthng
25342
@ziddiarmaan said:
Ok. Here ia the question.
Well looks like page contains a nice set on concepts. Thanks for sharing.
25343
@pankaj1988 : i am still not satisfied wirth what u have posted for Y axis : :(...

anyways the answr must be

:)
25344
@ziddiarmaan said:
@pankaj1988 : i am still not satisfied wirth what u have posted for Y axis : ...anyways the answr must be
what's doubt? I am not getting your confusion.
25345
@ziddiarmaan said:
Looks like all the files are password protected..
25346
@albiesriram
If the number is (a b c d ) then , it means
a.1000 + b.100 + c.10 +d = 9(b.100 + c.10 + d)
=> a.1000 = 8(b.100 + c.10 + d)
that means if we divide the number (a.1000) by 8 we will get the respective three digit numbers. Now 'a' can take the vales of (1,2,3,4,5,6,7,8,9).
If a=1 , the three digit no. will be =1x1000/8 =125
If a=2 , the three digit no. will be =2x1000/8 =250
If a=3 , the three digit no. will be =1x1000/8 =375
If a=4 , the three digit no. will be =1x1000/8 =500
If a=5 , the three digit no. will be =1x1000/8 =625
If a=6 , the three digit no. will be =1x1000/8 =750
If a=7 , the three digit no. will be =1x1000/8 =875
Therefore according to me the answer must be 7.
Please correct if I am wrong.
25347

@albiesriram

If the number is (a b c d ) then , it means
a.1000 + b.100 + c.10 +d = 9(b.100 + c.10 + d)
=> a.1000 = 8(b.100 + c.10 + d)
that means if we divide the number (a.1000) by 8 we will get the respective three digit numbers. Now 'a' can take the vales of (1,2,3,4,5,6,7,8,9).
If a=1 , the three digit no. will be =1x1000/8 =125
If a=2 , the three digit no. will be =2x1000/8 =250
If a=3 , the three digit no. will be =1x1000/8 =375
If a=4 , the three digit no. will be =1x1000/8 =500
If a=5 , the three digit no. will be =1x1000/8 =625
If a=6 , the three digit no. will be =1x1000/8 =750
If a=7 , the three digit no. will be =1x1000/8 =875
Therefore according to me the answer must be 7.
Please correct if I am wrong.