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netwizio
11-10-2004, 10:43 AM
Quantitative Comparison
x=3 y=10

Column A

y ^ x

Column B

x ^ y

________________________________________

in this question ^=power

Now this question is easy to solve (Ans=Column B) as the values are not large
but what if the values are large

let us say x=19 and y=20

What is the trick or logic behind these questions

Dingus
11-10-2004, 03:16 PM
I think this tests the binomial theorem. (I know this because I'm trying to brush up long forgotten high-school math now, after my GRE is over. :o) There MUST be some other way of doing this but here is mine.
Compare 19^20 and 20^19

--> can be written as: 19^19 * 19 ? (19 +1)^19
--> 19^19 * 19 ? [19 (1+1/19)]^19
--> 19^19 * 19 ? 19^19 * (1+1/19)^19 { as (ab)^x = a^x * b^x }
--> cancel 19^19 from both sides.
--> we are left with: 19 ? (1+1/19)^19
--> Using the Binomial theorem notation ie.
(1 + x)^n = 1 + nx + n(n-1)/2! * x^2 + n(n-1)(n-2)/3! * x^3 + ........up to x^n

Thus the right hand side of the equation --> (1 + 1/19)^19 can be expressed as:
1 + 19 * 1/19 + 19 * 18 / 2! *19^2 + ......
--> 1 + 1 + value < 1 + value << 1 + value <<<1 ....
Hence right hand side is 2.something
Left hand side = 19
19 > 2.something

Hence => 19^20 > 20^19

danielman
11-10-2004, 05:25 PM
I like Dingus aproach.

Another way to look at this problem is using logarithms:

x^y y^x

ln(x^y) ln(y^x) (logarithms preserve order relationships for x>0)

yln(x) xln(y)

ln(x)/x ln(y)/y

the function f(x) = ln(x)/x is always decrecient for x > e (e = 2.71828...)
Just an idea. I find Dingus aproach simpler and more elegant.

panther_vip
11-10-2004, 05:54 PM
Quantitative Comparison
x=3 y=10

Column A

y ^ x

Column B

x ^ y

________________________________________

in this question ^=power

Now this question is easy to solve (Ans=Column B) as the values are not large
but what if the values are large

let us say x=19 and y=20

What is the trick or logic behind these questions

I will use different aproach to solve it:

Suppose the column is like this:

Col A Col B

19 ^ 20 20 ^ 19

First I will divide col B by col A , which comes up with a result of {(20/19)^19}/19

As 20/19 gives 1.05.....then (1.05...) ^ 20 will be definitely less than 19

So col A will be greater than Col B

I think this will help.....

-Panther

sabna
11-11-2004, 05:01 AM
Hi panther
can u make ur way of doing a bit more clear..i cudnt get how u got the result like{(20/19)^19}/19..after dividing 20^19 by 19^20..can u explain it..?

prayers
sabna

panther_vip
11-11-2004, 05:18 AM
Hi panther
can u make ur way of doing a bit more clear..i cudnt get how u got the result like{(20/19)^19}/19..after dividing 20^19 by 19^20..can u explain it..?

prayers
sabna

Sure Sabna,
Here it is:

19 ^ 20 means,19 * (19^19) as x^(a+b) means (x^a) * (x^b).
If we want to devide a ^ n by b ^ (n+1) then definitely we can write b ^ (n+1) as (b ^ n) * b and then we can write {(a/b)^n}/b.

I think now it will be clear to you.

-Panther

sabna
11-11-2004, 06:06 AM
hi..
one more Q..Find the numbers of ways in which 4 boys and 4 girls can be seated in a
row of 8 seats if they sit alternately !!!!..just try..and please explain how u got it..!!!!

sabna
11-11-2004, 06:10 AM
..a lot of thanks Panther!!!...it was nice and i got it now..!!!just give an answer to my above Q..also..!!
hope to get it soon!!
regards n prayers

sabna

sabna
11-11-2004, 07:02 AM
still , panther i have another doubt, hope that it wont be a sign of my ignorance.
i am from little maths background, so can u please tell me, how could u write 19^20
( 19 * 19^19) as [19 (1+ 1/19)]^19. i tried a lot ..but still no bulb is shining in my brain!!!
prayers
sabna

netwizio
11-11-2004, 09:13 AM
thanks dingus
ur technique is lucid enough for me

panther_vip
11-11-2004, 03:58 PM
still , panther i have another doubt, hope that it wont be a sign of my ignorance.
i am from little maths background, so can u please tell me, how could u write 19^20
( 19 * 19^19) as [19 (1+ 1/19)]^19. i tried a lot ..but still no bulb is shining in my brain!!!
prayers
sabna
Hi Sabna,

Just think of it...
( 19 * 19^19)= (19^1)*(19^19)=19^(1+19)=19^20

as (x^a) * (x^b)= x^(a+b)

-Panther

panther_vip
11-11-2004, 04:08 PM
hi..
one more Q..Find the numbers of ways in which 4 boys and 4 girls can be seated in a
row of 8 seats if they sit alternately !!!!..just try..and please explain how u got it..!!!!

Hi sabna,
What's the ans of it?

I think it's 576(not sure)
Is it correct?
Plz let me know.I will then try to explain.

-Panther

netwizio
11-11-2004, 04:38 PM
hi..
one more Q..Find the numbers of ways in which 4 boys and 4 girls can be seated in a
row of 8 seats if they sit alternately !!!!..just try..and please explain how u got it..!!!!

i think that answer is 576*2

in my logic there are to possible ways of sitting
BGBGBGBG

and GBGBGBGB

For each way of sitting mentioned above the no of arrangements can be found by multiplicative rule
as
4*4*3*3*2*2*1*1=576

Ans=576+576=1152

panther_vip
11-11-2004, 05:14 PM
i think that answer is 576*2

in my logic there are to possible ways of sitting
BGBGBGBG

and GBGBGBGB

For each way of sitting mentioned above the no of arrangements can be found by multiplicative rule
as
4*4*3*3*2*2*1*1=576

Ans=576+576=1152

I think you are right. I did not take the possibility of two combinations into consideration .I just calculated it by taking one combination.

Thanks
-Panther

manasi4gre
11-12-2004, 02:00 PM
hello dingus!
I have jst started preparation for the GRE.could you give me any tips/important ques.with answers for the quant.section.I plan to take the GRE in 5-6 months time.Any suggestions for gre(verb) & toefl are also welcome.
your methods of solving math problems is quite appreciated by me.
thank you!
have you got admission into a good college?keep posting.

Dingus
11-12-2004, 03:49 PM
Hi manasi!
I'm still in the process of applying. I suggest you check out the "Just gave my GRE" forum. Each and every thread in there is a gem, as they have not only real questions from the exams, they have test experiences, tips and tricks for tackling all the three sections as well. I'd suggest you take the time and go through the entire Math forum, each and every thread in it. Every TMagician has a unique way of solving the questions and more often than not, one ends up learning new and more easy methods to attack them everyday! Same goes for TOEFL prep.

Cheers and all the best
Dingus

This is my "Just gave my GRE" thread:

sabna
11-18-2004, 11:35 AM
To
Panther and Netwizio,

prayers
sabna

swethav12
11-18-2004, 07:24 PM
hi dingus
I should say u r a genius man
anyways thanks for that easy solution .
All The Best for ur admissions

lorcar
11-22-2004, 01:17 AM
I will use different aproach to solve it:

Suppose the column is like this:

Col A Col B

19 ^ 20 20 ^ 19

First I will divide col B by col A , which comes up with a result of {(20/19)^19}/19

As 20/19 gives 1.05.....then (1.05...) ^ 20 will be definitely less than 19

So col A will be greater than Col B

I think this will help.....

-Panther
but i do not understand

you should not divide BOTH the columns?
so you have 1 in column A and (1.05)^20 in B, that has not to be compared with 19...

surans
11-23-2004, 04:45 PM
here is my sol:

suppose we have 19^20 in col A and 20^19 in col B

div both col by 19^20
we will have 1 in col A and 20^19/19^20 in col B
write 19^20 in col B as 19^(19+1) . That will become (19^19)*(19^1)
now we have (20/19)^19*(1/19) in col B
20/19 is 1.05 . so we have 1.05^19 *(1/19)
now mul both col by 19
now col A has 19 and col B has (1.05)^19
col B > col A

hope i'm clear..

lorcar
11-23-2004, 09:38 PM
here is my sol:
now col A has 19 and col B has (1.05)^19
col B > col A

hope i'm clear..
you are clear,
but are you SURE???????
1.05^19 = 2.52 < 19

surans
11-23-2004, 10:03 PM
sorry...yes, in final step i was wrong..

now 19>(1.05)^19 the value of B come somewhere around 1.102...that means col A comes out to be greater than col A..

thx for correcting me..

ofcourse the ans is A, right?

surans
11-23-2004, 10:04 PM
oops again!!! col A> col B there was a typo as col A