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Re: The Official CAT 2007 Quant Thread -
12-11-2007, 10:47 PM
Blacof bhai..i too do it the way Guarishankar has done it..
yes (n-1) should be a multiple of 17 coz here we are dealing with integers and hence the average should be an integer...but since the reduced avg is a fraction...so n-1 should be such that the sum of n-1 terms should be an integer...avg*no. of terms.... hence n-1 should be a multiple of 17....
hope u get it...
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DayZero-Saint Gobain ;)
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Re: The Official CAT 2007 Quant Thread -
12-11-2007, 10:56 PM
Quote:
Originally Posted by blacof
assumes initially there were k +1 numbers then original sum is
(K+1)(k+2) /2 - x = K ( 602/17)
k^2 + 3k + 2 /2 - K ( 602/17) = x therefore
we get 34 ( x-1 ) = K ( 17k-1153)
this solves for k = 68 => x =7
@ Gauri I dint get the fact that n-1 should be a multiple of 17
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blacof bhai hw did u sole the final equation u got in your solution..can u please elucidate...
There's no escaping reason, no denying purpose - because as we both know, without purpose, we would not exist.
It is purpose that created us, Purpose that guides us, That drives us, It is purpose that defines us, Purpose that binds us.
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Re: The Official CAT 2007 Quant Thread -
13-11-2007, 05:30 PM
Quote:
Originally Posted by gaurishankar
0.wxyz.........=wxyz/9999=wxyz/3*3*11*101
now DED can be either 101, 303 or 909
but since the no. is less than 1 hence it should be 909
also numerator should be divisible by 11 to make denominator 9999
check for 11^2=121 , 242,363 ,484
we get numerator 484
hence 484/909
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Hi gaurishankar....
Can you please explain the bold parts still more clearly....
i dint get the concept behind it....
Thanks in advance...
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The Following User Says NO Thank You to Manoj M R For This Un-useful Post:
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Re: The Official CAT 2007 Quant Thread -
13-11-2007, 10:04 PM
Can anyone help me out with this Ratio & Proportion problem -
A diamond falls down and breaks into three pieces whose weights are in the ratio 2:3:5. The value of the diamond is directly proportional to the cube of its weight. If the original diamond is worth Rs 343,000, what is the loss in the value due to the breakage?
1) Rs 212,660
2) Rs 309,043
3) Rs 275,429
4) Rs 288,120
5) Rs 224,000
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Re: The Official CAT 2007 Quant Thread -
13-11-2007, 10:10 PM
Quote:
Originally Posted by milind20
Can anyone help me out with this Ratio & Proportion problem -
A diamond falls down and breaks into three pieces whose weights are in the ratio 2:3:5. The value of the diamond is directly proportional to the cube of its weight. If the original diamond is worth Rs 343,000, what is the loss in the value due to the breakage?
1) Rs 212,660
2) Rs 309,043
3) Rs 275,429
4) Rs 288,120
5) Rs 224,000
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ans is option 4?
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Re: The Official CAT 2007 Quant Thread -
13-11-2007, 10:15 PM
Quote:
Originally Posted by milind20
Can anyone help me out with this Ratio & Proportion problem -
A diamond falls down and breaks into three pieces whose weights are in the ratio 2:3:5. The value of the diamond is directly proportional to the cube of its weight. If the original diamond is worth Rs 343,000, what is the loss in the value due to the breakage?
1) Rs 212,660
2) Rs 309,043
3) Rs 275,429
4) Rs 288,120
5) Rs 224,000
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Value=k*w^3
let wt be 10 gms
k=343000/1000=343
New value =343(2^3+3^3+5^3)=343*160
loss in value=343000-343*160=288120
IF IT IS TO BE, IT IS UP TO ME !
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Re: The Official CAT 2007 Quant Thread -
13-11-2007, 10:41 PM
Quote:
Originally Posted by sharang
Value=k*w^3
let wt be 10 gms
k=343000/1000=343
New value =343(2^3+3^3+5^3)=343*160
loss in value=343000-343*160=288120
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yaa.. the correct answer is 4... Thanks...
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Re: The Official CAT 2007 Quant Thread -
14-11-2007, 01:41 AM
Hey Puys...!!
Can sum1 crack this one....
The remainder when 10^10+10^100+10^100.......+10^10000000000 is divided by 7....
Best...!!!
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Re: The Official CAT 2007 Quant Thread -
14-11-2007, 02:01 AM
Please help on this one as well....
Remainder when 32^32^32 divided by 7
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Re: The Official CAT 2007 Quant Thread -
14-11-2007, 08:56 AM
Quote:
Originally Posted by parikhdhruv
Please help on this one as well....
Remainder when 32^32^32 divided by 7
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Remainder theorem concept
(32^32^32) / 7
(4^32^32) / 7
((4^3)^10*4^2)^32 / 7
16^32/7
2^32/7
(2^6)^5*2^2/7
64*2^2/7
1*4/7
4....remainder
Last edited by Manoj M R; 14-11-2007 at 10:12 AM.
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