ascent4cat : Messages : 3732-3761 of 4167

Given 3^x=5^y=(75) ^z.

then find xy/2x+y..?

options-

1. z

2. z/2

3. 2z

4. 1

solution:- 3^x=5^y=(75) ^z=k
x=log(k) / log 3   ; y=log(k) / log 5
; z=log(k) / log75
xy/2x+y = {log(k)/log3} {log(k)/log3}
                  """""""""""""""""""""""""""""""""'
                 2[log(k)/log3] + [log(k)/log5]
log(k) is canceled from neu and den and take lcm and solve it to
=   log(k) / 2log5 + log3 = log(k)/log [(5^2)*3]
= log(k)/log 75 = z

answer is 1. z

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#3733

From: ambarish kulkarni <ambarish_22@...>
Date: Sat Mar 6, 2010 11:49 am
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)

ambarish_22

@ smitha is right accordin to me.

It is the % change that we hav to find.

So the answer for that is 2400 %,accordin to me.

--- On Thu, 4/3/10, Anand Pal (Alcatel-Lucent) <varshneyinfo@...> wrote:

From: Anand Pal (Alcatel-Lucent) <varshneyinfo@...>
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)
To: ascent4cat@yahoogroups.com
Date: Thursday, 4 March, 2010, 4:46 PM

This is wrong solution, & Solution provided by chayank is correct , he is asking for % change in results.

On Wed, Mar 3, 2010 at 9:46 PM, Smita Dutta Chowdhury <smitadutta.chowdhur y@...> wrote:

hey i think it wil go as follows:
let the no be x

therefore x/5 instead of 5x

(5x- x/5)/x*100
= (25x-x)/x*100
= 2400

thereis another way of doing it too.

let the no. be 100.
as the no. shud be multiplied by 5 so 500
but divide dby mistake so 20

therefore
((500-20)/20) *100
2400

hope am right if not please send the correct solution

From: the boss <keerthi_indya@ yahoo.com>
To: ascent4cat@yahoogro ups.com
Sent: Tue, 2 March, 2010 2:49:56 PM

Subject: [2IIM CAT Prep] Help me out Guys!! (percentages)

Hi all,

1.A number is mistakenly divided by 5 instead of multiplied by 5.Find the percentage
change in result due to this mistake?
how to calculate the percentage change from the mistaken result or the expected result?..
gimme some ideas on this kind of sums.thank u in advance

Regards,
keerthivasan

The INTERNET now has a personality. YOURS! See your Yahoo! Homepage.

Your Mail works best with the New Yahoo Optimized IE8. Get it NOW!.

--
Thanks & Regards
Anand Pal
Alcatel-Lucent, Gurgoan
+919990255835
"Laugh so hard that even sorrow smiles at you, Live life so well that even death loves to see you live, Fight so hard that even fate accepts its defeat."

The INTERNET now has a personality. YOURS! See your Yahoo! Homepage.

#3734

From: ambarish kulkarni <ambarish_22@...>
Date: Sat Mar 6, 2010 12:00 pm
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)

ambarish_22

sorry i wrote wrong ans 1st, although i had gt it 1st

Acordin ot me it is 480 %.

I feel the solution is this way

Let th no b x.

=> % change after multiplyin by 5 = (5x-x)*100/x

=> % change aftr dividin by 5 = (x-(x/5))*100/x

=> total change = (5x-x+x-(x/5))*100/x

= 480

i hav tried d soln fr nos 20 n 30. n both satisy d ans.

So i feel the ans is this.

regards

Ambarish

--- On Wed, 3/3/10, chayank agrawal <chayankag@...> wrote:

From: chayank agrawal <chayankag@...>
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)
To: ascent4cat@yahoogroups.com
Date: Wednesday, 3 March, 2010, 10:47 PM

Hi

Let the number be 25.

Original resultant should be 25x5= 125

But it is divided by 5

So new no. would be 25/5 = 5

Error = 125-5= 120

% error = (120/125)*100= 96%

From: the boss <keerthi_indya@ yahoo.com>
To: ascent4cat@yahoogro ups.com
Sent: Tue, March 2, 2010 2:49:56 PM
Subject: [2IIM CAT Prep] Help me out Guys!! (percentages)

Hi all,

1.A number is mistakenly divided by 5 instead of multiplied by 5.Find the percentage
change in result due to this mistake?
how to calculate the percentage change from the mistaken result or the expected result?..
gimme some ideas on this kind of sums.thank u in advance

Regards,
keerthivasan

The INTERNET now has a personality. YOURS! See your Yahoo! Homepage.

Your Mail works best with the New Yahoo Optimized IE8. Get it NOW!.

#3735

From: Sampath Kumar <sampy.friend@...>
Date: Sat Mar 6, 2010 6:29 pm
Subject: Re: [2IIM CAT Prep] hiiii....

sampy_143

hey, i dont think you can imply that...�

if 1 day work of X1 men is 1/D, then

1 day work of X2 men should be X2/(X1*D)...�

X1 and X2 are variables like x and y...�

just think about it...�

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#3736

From: Vaishali Dube <dubevaishali@...>
Date: Sat Mar 6, 2010 4:23 pm
Subject: problem solution

dubevaishali

hie ........
accordin to me the solution is as follows:
As per data:3^x=5^y=(75) ^z.
i.e.....3^x=5^y=(3^z+5^2z)

equating both sides:
3^x=3^z
hence x=z

5^y=5^2z
hence y=2z

now xy/(2x+y)
put values of x & y in terms of z..........answer comes out as z/2

i hope this solution is satisfactory.....

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Reply | Messages in this Topic (3)

#3737

From: santanu pradhan <santanupradhan2010@...>
Date: Sun Mar 7, 2010 11:44 am
Subject: problem on surds..

santanupradh...

3^x=5^y=75^z
xy/(2x+y)?
Answer of this Q as fallows
log(3^x)=log(75^z)
x=zlog75/log3
log(5^y)=log(75^z)
y=zlog75/log5
((zlog75/log3)*(zlog75/log5))/((2*zlog75/log3)+zlog75/log5)=z

so answer=z;

santanu

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#3738

From: JiM <iiitm_jitesh@...>
Date: Wed Mar 10, 2010 8:00 am
Subject: Re: [2IIM CAT Prep] problem solution

iiitm_jitesh

Hey ^ sign is for raise to power not simple multiplication...
BUt I think u have manipulated the laws of Maths ...

On Sat, Mar 6, 2010 at 9:53 PM, Vaishali Dube <dubevaishali@...> wrote:

�

hie ........
accordin to me the solution is as follows:
As per data:3^x=5^y=(75) ^z.
i.e.....3^x=5^y=(3^z+5^2z)

equating both sides:
3^x=3^z
hence x=z

5^y=5^2z
hence y=2z

now xy/(2x+y)
put values of x & y in terms of z..........answer comes out as z/2

i hope this solution is satisfactory.....

The INTERNET now has a personality. YOURS! See your Yahoo! Homepage.

--
Thanks & Regards

Jitesh Maharwal
+91 9239678001
+91 9804275135

Reply | Messages in this Topic (3)

#3739

From: Rahul Bhat <rahulbhat2000@...>
Date: Wed Mar 10, 2010 7:37 am
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)

rahulbhat2000

hey ans shouwld be 96%..

�

ifi take 100...teh it will be 20 when divided and 500 when multiplied...

�

so ( (500-20)/500 )* 100 = 96%

On Sat, Mar 6, 2010 at 5:30 PM, ambarish kulkarni <ambarish_22@...> wrote:

�

sorry i wrote wrong ans 1st, although i had gt it 1st

Acordin ot me it is 480 %.

�

I feel the solution is this way

�

Let th no b x.

�

=> % change after multiplyin by 5 = (5x-x)*100/x

=> % change aftr dividin by 5 = (x-(x/5))*100/x

�

=> total change = (5x-x+x-(x/5))*100/x

��= 480

�

i hav tried d soln fr nos 20 n 30. n both satisy d ans.

�

So i feel the ans is this.

�

regards

Ambarish

--- On Wed, 3/3/10, chayank agrawal <chayankag@...> wrote:

From: chayank agrawal <chayankag@...>
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)
To: ascent4cat@yahoogroups.com
Date: Wednesday, 3 March, 2010, 10:47 PM

�

Hi

Let the number be 25.

Original resultant should be 25x5= 125

But it is divided by 5

So new no. would be 25/5 = 5

�

Error�= 125-5= 120

% error = (120/125)*100= 96%

�

From: the boss <keerthi_indya@ yahoo.com>
To: ascent4cat@yahoogro ups.com
Sent: Tue, March 2, 2010 2:49:56 PM
Subject: [2IIM CAT Prep] Help me out Guys!! (percentages)

�

Hi all,

�� 1.A number is mistakenly divided by 5 instead of multiplied by 5.Find the percentage
change in result due to this mistake?
how to calculate the percentage change from the mistaken result or the expected result?..
gimme some ideas on this kind of sums.thank u in advance

Regards,
keerthivasan

�

The INTERNET now has a personality. YOURS! See your Yahoo! Homepage.

Your Mail works best with the New Yahoo Optimized IE8. Get it NOW!.

#3740

From: vikas veshan <vikasveshan@...>
Date: Fri Mar 12, 2010 5:35 am
Subject: Re: [2IIM CAT Prep] problem solution

vikasveshan...

Aree Bhai ,

�

�How is This possible :-

�

�If a=b= c+d

�

Then how come

� a=c , and b = d

�

If it had been written like this

� 3^x + 5^y = 75 ^ z

�

Then u can solve with the said method...

�

i hope u getting my point .

�

Regards

Vicky Veshan

�

On 3/10/10, JiM <iiitm_jitesh@...> wrote:

�

Hey ^ sign is for raise to power not simple multiplication...
BUt I think u have manipulated the laws of Maths ...

On Sat, Mar 6, 2010 at 9:53 PM, Vaishali Dube <dubevaishali@...> wrote:

�

hie ........
accordin to me the solution is as follows:
As per data:3^x=5^y=(75) ^z.
i.e.....3^x=5^y=(3^z+5^2z)

equating both sides:
3^x=3^z
hence x=z

5^y=5^2z
hence y=2z

now xy/(2x+y)
put values of x & y in terms of z..........answer comes out as z/2

i hope this solution is satisfactory.....

The INTERNET now has a personality. YOURS! See your Yahoo! Homepage.

--
Thanks & Regards

Jitesh Maharwal
+91 9239678001
+91 9804275135

Reply | Messages in this Topic (3)

#3741

From: ambarish kulkarni <ambarish_22@...>
Date: Fri Mar 12, 2010 7:15 am
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)

ambarish_22

To keerthi

can u post the correct answer or atleast the options.

->I feel the % wil b greater than 100 %.

if a number is doubled, then there is change of 100%.

so here no is being multiplied 5 times ...so cange in % has to b greater than 100 for sure.

ambarish

--- On Wed, 10/3/10, Rahul Bhat <rahulbhat2000@...> wrote:

From: Rahul Bhat <rahulbhat2000@...>
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)
To: ascent4cat@yahoogroups.com
Date: Wednesday, 10 March, 2010, 1:07 PM

hey ans shouwld be 96%..

ifi take 100...teh it will be 20 when divided and 500 when multiplied.. .

so ( (500-20)/500 )* 100 = 96%

On Sat, Mar 6, 2010 at 5:30 PM, ambarish kulkarni <ambarish_22@ yahoo.co. in> wrote:

sorry i wrote wrong ans 1st, although i had gt it 1st

Acordin ot me it is 480 %.

I feel the solution is this way

Let th no b x.

=> % change after multiplyin by 5 = (5x-x)*100/x

=> % change aftr dividin by 5 = (x-(x/5))*100/ x

=> total change = (5x-x+x-(x/5) )*100/x

= 480

i hav tried d soln fr nos 20 n 30. n both satisy d ans.

So i feel the ans is this.

regards

Ambarish

--- On Wed, 3/3/10, chayank agrawal <chayankag@yahoo. com> wrote:

From: chayank agrawal <chayankag@yahoo. com>
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)
To: ascent4cat@yahoogro ups.com
Date: Wednesday, 3 March, 2010, 10:47 PM

Hi

Let the number be 25.

Original resultant should be 25x5= 125

But it is divided by 5

So new no. would be 25/5 = 5

Error = 125-5= 120

% error = (120/125)*100= 96%

From: the boss <keerthi_indya@ yahoo.com>
To: ascent4cat@yahoogro ups.com
Sent: Tue, March 2, 2010 2:49:56 PM
Subject: [2IIM CAT Prep] Help me out Guys!! (percentages)

Hi all,

1.A number is mistakenly divided by 5 instead of multiplied by 5.Find the percentage
change in result due to this mistake?
how to calculate the percentage change from the mistaken result or the expected result?..
gimme some ideas on this kind of sums.thank u in advance

Regards,
keerthivasan

The INTERNET now has a personality. YOURS! See your Yahoo! Homepage.

Your Mail works best with the New Yahoo Optimized IE8. Get it NOW!.

Your Mail works best with the New Yahoo Optimized IE8. Get it NOW!.

#3742

From: deepan adhi <deepanadhi@...>
Date: Tue Mar 16, 2010 11:38 am
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)

deepanadhi

Hi Guys,

The Ans would be 96%.

For Example take the number as 5.

It should have been 5*5 = 25.

But mistakenly it was divided by 5, i.e 5/5 = 1.

SO 25 have become 1 here.

Change in percentage. (original value – changed value/original value)*100.

= (25-1)/25 * 100

= (24/25)*100

= 96 % (change in negative sense) - is percentage change (for 25 to become 1).

Cheers,
Deepan Mahendran.

From: ambarish kulkarni <ambarish_22@...>
To: ascent4cat@yahoogroups.com
Sent: Fri, 12 March, 2010 2:15:04 AM
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)

To keerthi

can u post the correct answer or atleast the options.

->I feel the % wil b greater than 100 %.

if a number is doubled, then there is change of 100%.

so here no is being multiplied 5 times ...so cange in % has to b greater than 100 for sure.

ambarish

--- On Wed, 10/3/10, Rahul Bhat <rahulbhat2000@ gmail.com> wrote:

From: Rahul Bhat <rahulbhat2000@ gmail.com>
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)
To: ascent4cat@yahoogro ups.com
Date: Wednesday, 10 March, 2010, 1:07 PM

hey ans shouwld be 96%..

ifi take 100...teh it will be 20 when divided and 500 when multiplied.. .

so ( (500-20)/500 )* 100 = 96%

On Sat, Mar 6, 2010 at 5:30 PM, ambarish kulkarni <ambarish_22@ yahoo.co. in> wrote:

sorry i wrote wrong ans 1st, although i had gt it 1st

Acordin ot me it is 480 %.

I feel the solution is this way

Let th no b x.

=> % change after multiplyin by 5 = (5x-x)*100/x

=> % change aftr dividin by 5 = (x-(x/5))*100/ x

=> total change = (5x-x+x-(x/5) )*100/x

= 480

i hav tried d soln fr nos 20 n 30. n both satisy d ans.

So i feel the ans is this.

regards

Ambarish

--- On Wed, 3/3/10, chayank agrawal <chayankag@yahoo. com> wrote:

From: chayank agrawal <chayankag@yahoo. com>
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)
To: ascent4cat@yahoogro ups.com
Date: Wednesday, 3 March, 2010, 10:47 PM

Hi

Let the number be 25.

Original resultant should be 25x5= 125

But it is divided by 5

So new no. would be 25/5 = 5

Error = 125-5= 120

% error = (120/125)*100= 96%

From: the boss <keerthi_indya@ yahoo.com>
To: ascent4cat@yahoogro ups.com
Sent: Tue, March 2, 2010 2:49:56 PM
Subject: [2IIM CAT Prep] Help me out Guys!! (percentages)

Hi all,

1.A number is mistakenly divided by 5 instead of multiplied by 5.Find the percentage
change in result due to this mistake?
how to calculate the percentage change from the mistaken result or the expected result?..
gimme some ideas on this kind of sums.thank u in advance

Regards,
keerthivasan

The INTERNET now has a personality. YOURS! See your Yahoo! Homepage.

Your Mail works best with the New Yahoo Optimized IE8. Get it NOW!.

Your Mail works best with the New Yahoo Optimized IE8. Get it NOW!.

#3743

From: Rahul Bhat <rahulbhat2000@...>
Date: Thu Mar 18, 2010 8:25 am
Subject: Coefficient of a^48

rahulbhat2000

Hey guyz..

�

how to find out the Coefficient of a^48 in (a-1)(a-2)(a-3)......(a-50)??

�

Plz explain the answer with the method....

�

Thanks

Rahul.

#3744

From: Sachin K <sachin.sk@...>
Date: Sun Mar 21, 2010 7:00 am
Subject: Re: [2IIM CAT Prep] Coefficient of a^48

sachin2082000

Hey Rahul,

�

I�will detail out the procedure here for you..

Any expression like this is a polynomial in 'a'.. and the numbers�within the bracket (e.g. 1, 2, 3,�...50) are the roots of this polynomial in 'a', and you can also write this polynomial in its expsnaded form as:

�

(a-1)(a-2)(a-3)......(a-50) = (a^50) - (SUM_OF_ROOTS)*(a^49) + (SUM_OF_ALL_PRODUCT_OF_TWO_ROOTS)*(a^48) - ....... + (PRODUCT_OF_ROOTS).

�

So, You can see yourself that the coefficient of a^48 is nothing but the "SUM_OF_ALL_PRODUCT_OF_TWO_ROOTS".

�

Answer =� (1.2 + 1.3 + 1.4 + ... + 1.50) + (2.3 + 2.4 + .... + 2.50) +�(3.4 + ... +3.50) + ...+ (48.49 + 48.50) + (49.50).

�

GOT IT !!

�

--Sachin Kumar

==================================================================

On Thu, Mar 18, 2010 at 1:55 PM, Rahul Bhat <rahulbhat2000@...> wrote:

�

Hey guyz..

�

how to find out the Coefficient of a^48 in (a-1)(a-2)(a-3)......(a-50)??

�

Plz explain the answer with the method....

�

Thanks

Rahul.

#3745

From: Rahul Bhat <rahulbhat2000@...>
Date: Sun Mar 21, 2010 2:10 pm
Subject: Re: [2IIM CAT Prep] Coefficient of a^48

rahulbhat2000

Thnx Sachin.....

Got it....

So the coefiicient will be the same for all ?? a^47,a^46 and so on??

Rahul

On Sun, Mar 21, 2010 at 12:30 PM, Sachin K <sachin.sk@gmail.com> wrote:

Hey Rahul,

�

I�will detail out the procedure here for you..

Any expression like this is a polynomial in 'a'.. and the numbers�within the bracket (e.g. 1, 2, 3,�...50) are the roots of this polynomial in 'a', and you can also write this polynomial in its expsnaded form as:

�

(a-1)(a-2)(a-3)......(a-50) = (a^50) - (SUM_OF_ROOTS)*(a^49) + (SUM_OF_ALL_PRODUCT_OF_TWO_ROOTS)*(a^48) - ....... + (PRODUCT_OF_ROOTS).

�

So, You can see yourself that the coefficient of a^48 is nothing but the "SUM_OF_ALL_PRODUCT_OF_TWO_ROOTS".

�

Answer =� (1.2 + 1.3 + 1.4 + ... + 1.50) + (2.3 + 2.4 + .... + 2.50) +�(3.4 + ... +3.50) + ...+ (48.49 + 48.50) + (49.50).

�

GOT IT !!

�

�

--Sachin Kumar

==================================================================

On Thu, Mar 18, 2010 at 1:55 PM, Rahul Bhat <rahulbhat2000@...> wrote:

�

Hey guyz..

�

how to find out the Coefficient of a^48 in (a-1)(a-2)(a-3)......(a-50)??

�

Plz explain the answer with the method....

�

Thanks

Rahul.

#3746

From: Revathi <revathivanga@...>
Date: Tue Mar 23, 2010 4:33 am
Subject: Re: [2IIM CAT Prep] Coefficient of a^48

revathivanga

"So the coefiicient will be the same for all ?? a^47,a^46 and so on??""
@ rahul ...

a^49 -- sum of all roots

a^48 -- sum of (product of two roots at a time).

a^47 -- sum of product of 3 roots at a time .

.... (if sign should consider then alternative + and _ signs from a^50).

for Eg : (x-k1)(x-k2) = 0 is equation in variable x and k1, k2 are roots ( as like a is var & 1,2 are roots in the above equation) then equation is x^2 - (k1+k2)x+k1k2 =0

i.e, x^2 coeffient is 1

x^1 coeffient is - sum of roots

x^0 coeffient is --- product of roots (or sum of product of roots two at a time ).... (as equation power is 2 it ended thr.).

--- On Sun, 21/3/10, Rahul Bhat <rahulbhat2000@...> wrote:

From: Rahul Bhat <rahulbhat2000@...>
Subject: Re: [2IIM CAT Prep] Coefficient of a^48
To: "Sachin K" <sachin.sk@...>
Cc: ascent4cat@yahoogroups.com
Date: Sunday, 21 March, 2010, 7:40 PM

Thnx Sachin.....

Got it....

So the coefiicient will be the same for all ?? a^47,a^46 and so on??

Rahul

On Sun, Mar 21, 2010 at 12:30 PM, Sachin K <sachin.sk@gmail.com> wrote:

Hey Rahul,

I will detail out the procedure here for you..

Any expression like this is a polynomial in 'a'.. and the numbers within the bracket (e.g. 1, 2, 3, ...50) are the roots of this polynomial in 'a', and you can also write this polynomial in its expsnaded form as:

(a-1)(a-2)(a- 3)......( a-50) = (a^50) - (SUM_OF_ROOTS) *(a^49) + (SUM_OF_ALL_ PRODUCT_OF_ TWO_ROOTS) *(a^48) - ....... + (PRODUCT_OF_ ROOTS).

So, You can see yourself that the coefficient of a^48 is nothing but the "SUM_OF_ALL_PRODUCT_ OF_TWO_ROOTS".

Answer = (1.2 + 1.3 + 1.4 + ... + 1.50) + (2.3 + 2.4 + .... + 2.50) + (3.4 + ... +3.50) + ...+ (48.49 + 48.50) + (49.50).

GOT IT !!

--Sachin Kumar

============ ========= ========= ========= ========= ========= =========

On Thu, Mar 18, 2010 at 1:55 PM, Rahul Bhat <rahulbhat2000@ gmail.com> wrote:

Hey guyz..

how to find out the Coefficient of a^48 in (a-1)(a-2)(a- 3)......( a-50)??

Plz explain the answer with the method....

Thanks

Rahul.

Your Mail works best with the New Yahoo Optimized IE8. Get it NOW!.

#3747

From: hardev Parmar <hardev_meet@...>
Date: Wed Mar 24, 2010 6:04 pm
Subject: Re: [2IIM CAT Prep] Coefficient of a^48

hardev_meet@...

Rahul,

Let it reiterate what thing you were narrating in your earlier email

a-1)(a-2)(a- 3)......( a-50) = (a^50) - (SUM_OF_ROOTS) *(a^49) + (SUM_OF_ALL_ PRODUCT_OF_ TWO_ROOTS) *(a^48) - ....... + (PRODUCT_OF_ ROOTS).

Muddle in Aforementioned line. pls make it terse

Hoping for a positive response

Thanks & Regards,
Hardev

From: Rahul Bhat <rahulbhat2000@...>
To: Sachin K <sachin.sk@...>
Cc: ascent4cat@yahoogroups.com
Sent: Sun, March 21, 2010 6:10:56 AM
Subject: Re: [2IIM CAT Prep] Coefficient of a^48

Thnx Sachin.....

Got it....

So the coefiicient will be the same for all ?? a^47,a^46 and so on??

Rahul

On Sun, Mar 21, 2010 at 12:30 PM, Sachin K <sachin.sk@gmail.com> wrote:

Hey Rahul,

I will detail out the procedure here for you..

Any expression like this is a polynomial in 'a'.. and the numbers within the bracket (e.g. 1, 2, 3, ...50) are the roots of this polynomial in 'a', and you can also write this polynomial in its expsnaded form as:

(a-1)(a-2)(a- 3)......( a-50) = (a^50) - (SUM_OF_ROOTS) *(a^49) + (SUM_OF_ALL_ PRODUCT_OF_ TWO_ROOTS) *(a^48) - ....... + (PRODUCT_OF_ ROOTS).

So, You can see yourself that the coefficient of a^48 is nothing but the "SUM_OF_ALL_PRODUCT_ OF_TWO_ROOTS".

Answer = (1.2 + 1.3 + 1.4 + ... + 1.50) + (2.3 + 2.4 + .... + 2.50) + (3.4 + ... +3.50) + ...+ (48.49 + 48.50) + (49.50).

GOT IT !!

--Sachin Kumar

============ ========= ========= ========= ========= ========= =========

On Thu, Mar 18, 2010 at 1:55 PM, Rahul Bhat <rahulbhat2000@ gmail.com> wrote:

Hey guyz..

how to find out the Coefficient of a^48 in (a-1)(a-2)(a- 3)......( a-50)??

Plz explain the answer with the method....

Thanks

Rahul.

#3748

From: "anirudhmshr" <anirudhmshr@...>
Date: Sun Mar 28, 2010 1:20 pm
Subject: NUMBER SYSTEM

anirudhmshr

WHAT IS THE REMAINDER WHEN 54 RAISE TO THE POWER 124 IS DIVIDED BY IS DIVIDED BY
17?

#3749

From: Rahul Bhat <rahulbhat2000@...>
Date: Thu Apr 1, 2010 8:11 am
Subject: Re: [2IIM CAT Prep] Coefficient of a^48

rahulbhat2000

@Sachin & @Revathi...

�

Thanks a ton both of you...Now i have got it properly!!......

�

Thanks again.

�

Rahul

On Tue, Mar 23, 2010 at 10:03 AM, Revathi <revathivanga@...> wrote:

�

"So the coefiicient will be the same for all ?? a^47,a^46 and so on??""
@ rahul ...

�

a^49 -- sum of all roots

�a^48 -- sum of (product of two roots at a time).

a^47 -- sum of product of 3 roots at a time .

.... (if sign should consider then alternative + and _ signs from a^50).

�

for Eg : (x-k1)(x-k2) = 0 is equation in variable x and k1, k2 are roots ( as like a is var & 1,2 are roots in the above equation) then equation is x^2 - (k1+k2)x+k1k2 =0

�

i.e, x^2 coeffient is 1

�� x^1 coeffient is - sum of roots

�� x^0 coeffient is --- product of roots (or sum of product of roots two at a time ).... (as equation power is 2 it ended thr.).��

��

--- On Sun, 21/3/10, Rahul Bhat <rahulbhat2000@...> wrote:

From: Rahul Bhat <rahulbhat2000@...>
Subject: Re: [2IIM CAT Prep] Coefficient of a^48
To: "Sachin K" <sachin.sk@gmail.com>
Cc: ascent4cat@yahoogroups.com
Date: Sunday, 21 March, 2010, 7:40 PM

�

Thnx Sachin.....

Got it....

So the coefiicient will be the same for all ?? a^47,a^46 and so on??

Rahul

On Sun, Mar 21, 2010 at 12:30 PM, Sachin K <sachin.sk@gmail.com> wrote:

Hey Rahul,

�

I�will detail out the procedure here for you..

Any expression like this is a polynomial in 'a'.. and the numbers�within the bracket (e.g. 1, 2, 3,�...50) are the roots of this polynomial in 'a', and you can also write this polynomial in its expsnaded form as:

�

(a-1)(a-2)(a- 3)......( a-50) = (a^50) - (SUM_OF_ROOTS) *(a^49) + (SUM_OF_ALL_ PRODUCT_OF_ TWO_ROOTS) *(a^48) - ....... + (PRODUCT_OF_ ROOTS).

�

So, You can see yourself that the coefficient of a^48 is nothing but the "SUM_OF_ALL_PRODUCT_ OF_TWO_ROOTS".

�

Answer =� (1.2 + 1.3 + 1.4 + ... + 1.50) + (2.3 + 2.4 + .... + 2.50) +�(3.4 + ... +3.50) + ...+ (48.49 + 48.50) + (49.50).

�

GOT IT !!

�

�

--Sachin Kumar

============ ========= ========= ========= ========= ========= =========

On Thu, Mar 18, 2010 at 1:55 PM, Rahul Bhat <rahulbhat2000@ gmail.com> wrote:

�

Hey guyz..

�

how to find out the Coefficient of a^48 in (a-1)(a-2)(a- 3)......( a-50)??

�

Plz explain the answer with the method....

�

Thanks

Rahul.

Your Mail works best with the New Yahoo Optimized IE8. Get it NOW!.

#3750

From: Sujith Menon <sujithmenonp@...>
Date: Thu Apr 1, 2010 9:17 am
Subject: Re: [2IIM CAT Prep] NUMBER SYSTEM

sujith0337

54^124 % 17= 3^124 % 17 = 81^31 %17 = (85-4)^31%17 =-4^31 %17 =-4* (17-1)^15 %17 = 4

On Sun, Mar 28, 2010 at 6:50 PM, anirudhmshr <anirudhmshr@...> wrote:

�

WHAT IS THE REMAINDER WHEN 54 RAISE TO THE POWER 124 IS DIVIDED BY IS DIVIDED BY 17?

#3751

From: Rahul Bhat <rahulbhat2000@...>
Date: Fri Apr 2, 2010 6:19 am
Subject: Re: [2IIM CAT Prep] NUMBER SYSTEM

rahulbhat2000

Hey All,

Is the answer 5 by any chance?

I did it as follows:-

(54^124)� = ( 2 * 27 ) ^124
�� = 2^124 * 27*124
�� = ( 2^10 )^12 * ( 2^4) * (27^4)^31
�� = 24^12�� *� 16�� * (41)^31
�� = 76�� * 16�� * 41
�� = 16 * 41
�� = 56

Now 56/17 gives remainder 5.....

Notes- 1-2^10 will always have last 2 digits as 24
�� 2-For no's ending in 3,7,9 express them as the last digit to be 1..hence 27^4
�� 3-24^Even power will always have 76 as the last 2 digits and 24^odd power will always have 24 as the last two digits.
�� 4- 76 multiplied by any power of 2 will have the last two digits of the power of 2.
��

On Fri, Apr 2, 2010 at 11:26 AM, Rahul Bhat <rahulbhat2000@...> wrote:

What r the options??

On Sun, Mar 28, 2010 at 6:50 PM, anirudhmshr <anirudhmshr@...> wrote:

�

WHAT IS THE REMAINDER WHEN 54 RAISE TO THE POWER 124 IS DIVIDED BY IS DIVIDED BY 17?

#3752

From: abhishek pandey <abat001_1983@...>
Date: Mon Apr 12, 2010 11:24 am
Subject: Re: [2IIM CAT Prep] NUMBER SYSTEM

abat001_1983

hi rahul,

plz help

tell me how 41 raise to the power 31 again comes 41 in 5th step.

& how 16*41will result to 5 raise to the power 6?

--- शुक्र, 2/4/10 को, Rahul Bhat <rahulbhat2000@...> ने लिखा:

द्वारा: Rahul Bhat <rahulbhat2000@...>
विषय: Re: [2IIM CAT Prep] NUMBER SYSTEM
To: ascent4cat@yahoogroups.com
दिनांक: शुक्रवार, 2 अप्रैल, 2010, 11:49 AM

Hey All,

Is the answer 5 by any chance?

I did it as follows:-

(54^124) = ( 2 * 27 ) ^124
              = 2^124 * 27*124
              = ( 2^10 )^12 * ( 2^4) * (27^4)^31
              = 24^12        * 16    * (41)^31
             = 76              * 16    * 41
             = 16 * 41
            = 56

Now 56/17 gives remainder 5.....

Notes- 1-2^10 will always have last 2 digits as 24
          2-For no's ending in 3,7,9 express them as the last digit to be 1..hence 27^4
         3-24^Even power will always have 76 as the last 2 digits and 24^odd power will always have 24 as the last two digits.
        4- 76 multiplied by any power of 2 will have the last two digits of the power of 2.


On Fri, Apr 2, 2010 at 11:26 AM, Rahul Bhat <rahulbhat2000@ gmail.com> wrote:

What r the options??

On Sun, Mar 28, 2010 at 6:50 PM, anirudhmshr <anirudhmshr@ yahoo.co. in> wrote:

WHAT IS THE REMAINDER WHEN 54 RAISE TO THE POWER 124 IS DIVIDED BY IS DIVIDED BY 17?

Send free SMS to your Friends on Mobile from your Yahoo! Messenger. Download Now! http://messenger.yahoo.com/download.php

#3753

From: abhishek pandey <abat001_1983@...>
Date: Mon Apr 12, 2010 12:44 pm
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)

abat001_1983

the correct answer is 96%.

Suppose the no. is x

then actual result (which should come) = 5x

and deviated result = x/5

now the formulla of percentage error is : (Actual value)-(Deviated value) * 100%

(Actual value)

hence answer is = (5x)-(x/5) *100 = (24x/5) *100 = 24x *100 = 96%.

(5x) (5x) 25x

--- मंगल, 16/3/10 को, deepan adhi <deepanadhi@...> ने लिखा:

द्वारा: deepan adhi <deepanadhi@...>
विषय: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)
To: ascent4cat@yahoogroups.com
दिनांक: मंगलवार, 16 मार्च, 2010, 5:08 PM

Hi Guys,

The Ans would be 96%.

For Example take the number as 5.

It should have been 5*5 = 25.

But mistakenly it was divided by 5, i.e 5/5 = 1.

SO 25 have become 1 here.

Change in percentage. (original value – changed value/original value)*100.

= (25-1)/25 * 100
= (24/25)*100
= 96 % (change in negative sense) - is percentage change (for 25 to become 1).

Cheers,
Deepan Mahendran.

From: ambarish kulkarni <ambarish_22@ yahoo.co. in>
To: ascent4cat@yahoogro ups.com
Sent: Fri, 12 March, 2010 2:15:04 AM
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)

To keerthi

can u post the correct answer or atleast the options.

->I feel the % wil b greater than 100 %.

  if a number is doubled, then there is change of 100%.

so here no is being multiplied 5 times ...so cange in % has to b greater than 100 for sure.

ambarish

--- On Wed, 10/3/10, Rahul Bhat <rahulbhat2000@ gmail.com> wrote:

From: Rahul Bhat <rahulbhat2000@ gmail.com>
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)
To: ascent4cat@yahoogro ups.com
Date: Wednesday, 10 March, 2010, 1:07 PM

hey ans shouwld be 96%..

ifi take 100...teh it will be 20 when divided and 500 when multiplied.. .

so ( (500-20)/500 )* 100 = 96%

On Sat, Mar 6, 2010 at 5:30 PM, ambarish kulkarni <ambarish_22@ yahoo.co. in> wrote:

sorry i wrote wrong ans 1st, although i had gt it 1st

Acordin ot me it is 480 %.

I feel the solution is this way

Let th no b x.

=> % change after multiplyin by 5 = (5x-x)*100/x

=> % change aftr dividin by 5 = (x-(x/5))*100/ x

=> total change = (5x-x+x-(x/5) )*100/x

                           = 480

i hav tried d soln fr nos 20 n 30. n both satisy d ans.

So i feel the ans is this.

regards

Ambarish

--- On Wed, 3/3/10, chayank agrawal <chayankag@yahoo. com> wrote:

From: chayank agrawal <chayankag@yahoo. com>
Subject: Re: [2IIM CAT Prep] Help me out Guys!! (percentages)
To: ascent4cat@yahoogro ups.com
Date: Wednesday, 3 March, 2010, 10:47 PM

Hi

Let the number be 25.

Original resultant should be 25x5= 125

But it is divided by 5

So new no. would be 25/5 = 5

Error = 125-5= 120

% error = (120/125)*100= 96%

From: the boss <keerthi_indya@ yahoo.com>
To: ascent4cat@yahoogro ups.com
Sent: Tue, March 2, 2010 2:49:56 PM
Subject: [2IIM CAT Prep] Help me out Guys!! (percentages)

Hi all,

       1.A number is mistakenly divided by 5 instead of multiplied by 5.Find the percentage
change in result due to this mistake?
how to calculate the percentage change from the mistaken result or the expected result?..
gimme some ideas on this kind of sums.thank u in advance

Regards,
keerthivasan

The INTERNET now has a personality. YOURS! See your Yahoo! Homepage.

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#3754

From: vishal babbar <babbar_vishal2005@...>
Date: Thu Apr 15, 2010 7:02 am
Subject: Re:

babbar_visha...

I think ans is 4..........

By euler theoram 54^112 divided by 17 gives rem as 1

we left with 54^12 div by 17

then (51+3)^12 div by 17

now 3 ^ 12 div by 17

we can easily calculate from here now

Regards

Vishal Babbar

09814770284

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Reply | Messages in this Topic (1)

#3755

From: neelima agrawal <neeli2101@...>
Date: Sun Apr 18, 2010 3:47 am
Subject: Re:

neeli2101

can u plz elaborate ur answer using euler's theorem....

Reply | Messages in this Topic (1)

#3756

From: "Neeli" <neeli2101@...>
Date: Mon Apr 19, 2010 12:13 pm
Subject: Work....

neeli2101

HI,
My concepts in Work are weak.Need Help.
Please provide elaborated answer for the following problem so that I can refer
the same approach in other questions as well....


A,B and C can do a job in 15 days.After working with B and C for 5 days,A
leaves.Then B and C finish it in 20 days more.In how many days can A do the job
alone?

Thanks in advance.
Neelima

#3757

From: "Rahul R. Gupta" <rrahulanker@...>
Date: Thu Apr 22, 2010 7:47 am
Subject: Re: [2IIM CAT Prep] Work....

rrahulanker

To neeli

1/a+1/b+1/c = 1/15 eqn 1
for 5 days work done = 5/15 = 1/3
work remaining = 1-1/3 = 2/3
2/3 of work has done by b + c in 20 day then
ful work will be done by them in 20*3/2 =30 day
1/b+1/c = 1/30 eqn 2

from eqn 1 & 2
1/a+1/30 = 1/15
1/a = 1/15

there fore "a" alone can complete same work in 15 days

--- On Mon, 19/4/10, Neeli <neeli2101@...> wrote:

From: Neeli <neeli2101@...>
Subject: [2IIM CAT Prep] Work....
To: ascent4cat@yahoogroups.com
Date: Monday, 19 April, 2010, 5:43 PM

HI,
My concepts in Work are weak.Need Help.
Please provide elaborated answer for the following problem so that I can refer the same approach in other questions as well....

A,B and C can do a job in 15 days.After working with B and C for 5 days,A leaves.Then B and C finish it in 20 days more.In how many days can A do the job alone?

Thanks in advance.
Neelima

#3758

From: "deepak1982_narayanan" <deepakraam@...>
Date: Thu Apr 22, 2010 9:25 am
Subject: Re: Work....

deepak1982_n...

Hi Neeli,
             Below is my approach for the problem

A+B+C = 15 days

So in 5 days they will complete 1/3rd of the job.
Remaining part of the job = 2/3rd.

B+C can complete 2/3 of job in  20 days.So they can complete full job in   30
days.

Now,A+B+C = 15 days of which B+C = 30 days.So A can complete full job in 30
days.

A = 30 days.

Please post the OA

-Deepak.

--- In ascent4cat@yahoogroups.com, "Neeli" <neeli2101@...> wrote:
>
> HI,
> My concepts in Work are weak.Need Help.
> Please provide elaborated answer for the following problem so that I can refer
the same approach in other questions as well....
>
>
> A,B and C can do a job in 15 days.After working with B and C for 5 days,A
leaves.Then B and C finish it in 20 days more.In how many days can A do the job
alone?
>
> Thanks in advance.
> Neelima
>

#3759

From: Ashish Anand <excusemeashish@...>
Date: Thu Apr 22, 2010 9:03 am
Subject: Re: [2IIM CAT Prep] Work....

excusemeashish

For all practical purposes, you can�treat�B and C to be the same guys.

So the question becomes

A and B can do the job in 15 days.�After working with B for 5 days, A leaves.Then B finishes the job in 20 days more. In how many days can A do the job alone?

Lets say A produces a units of work in a day and B produces b units of work in a day.

Then total work = 15a + 15b

After 5 days left over work = 10a + 10b

Therefore b will finish this in (10a+10b)/b days which is given to be 20.

(10a+10b)/b=20

=>10a +10b = 20b

=>10a=10b

=>a=b

=> A and B are equally efficient.

=> If A and B can do the job in 15 days, A can do the job in 7.5 days

Ans=7.5

On Mon, Apr 19, 2010 at 5:43 PM, Neeli <neeli2101@...> wrote:

�

HI,
My concepts in Work are weak.Need Help.
Please provide elaborated answer for the following problem so that I can refer the same approach in other questions as well....

A,B and C can do a job in 15 days.After working with B and C for 5 days,A leaves.Then B and C finish it in 20 days more.In how many days can A do the job alone?

Thanks in advance.
Neelima

#3760

From: "Raja Raman" <ksrajaraman27@...>
Date: Thu Apr 22, 2010 9:39 am
Subject: Re: [2IIM CAT Prep] Work....

ksrajaraman27

Hi Neelima,

Let us first understand the question point by point.

1) A, B C can finish a job in 15 days.
2) A,B C work together for the first 5 days. This means that 1/3 rd of the job is done, as this is arrived from the first point above.
3) This also means that 2/3rd of the job is left and that is completed by B C together in 20 days.
4) From the 3rd point, if B+C can do 2/3rd of a job in 20 days, then they can do the full job in 30 days.
5) As time and work problems are called as unitary problems, we will break them into 1 days job

A+B+C one days work is 1/15
B+C one days work is 1/30
So, A's one days work is 1/15 - 1/30 = 1/30
Therefore, A can complete the work alone in 30 days.

Hope this has helped you understand the question and answer. For any queries feel free to contact me thru mail.

For all of you out there preparing for competitive exams i have started a website called www.vedicaptitude.com. Kindly help me in my endeavour by pouring in your suggestions for me to improve the website. The site is ready in its framework, but yet to fill it with content. Kindly visit the site everyday for your learning.

On Thu, 22 Apr 2010 11:31:58 +0530 wrote

HI,

My concepts in Work are weak.Need Help.

Please provide elaborated answer for the following problem so that I can refer the same approach in other questions as well....

A,B and C can do a job in 15 days.After working with B and C for 5 days,A leaves.Then B and C finish it in 20 days more.In how many days can A do the job alone?

Thanks in advance.

Neelima

#3761

From: JiM <iiitm_jitesh@...>
Date: Thu Apr 22, 2010 12:51 pm
Subject: Re: [2IIM CAT Prep] Work....

iiitm_jitesh

hey how to find average quickly. Like for this : the no of student opting for finance in IIM a,b,c are 90,50,45. Find the Average Salary? data Given Finance Student in : IIM A �: 620, IIM B: 540, IIM C: 725. Options are 1. 615 2. 647 3. 624 4. 635

--
Thanks & Regards

Jitesh Maharwal
+91 9239678001
+91 9804275135