how to solve these quickly?

[infinityzero]

How do you solve these 2 problems quickly?

 

1)

Column A = (4^6 - 4^4) / (4^5 - 3^3)

Column B = 1

 

2)

S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + 1/9 - 1/10

 

Column A = S

Column B = 1

[kronique]

How do you solve these 2 problems quickly?

 

1)

Column A = (4^6 - 4^4) / (4^5 - 3^3)

Column B = 1

 

 

remove 4^4 from num. & 4^3 from the den.

 

hecne we have [4*(4^2 - 1)]/[4^2 - (3/4)^3]

 

Now since 3/4 is 4^2 -1 hence the fraction

[4^2 -1]/[4^2 - (3/4)^3] 1[/b]

 

hence Col. A > Col. B

[mansoor316]

How do you solve these 2 problems quickly?

 

2)

S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + 1/9 - 1/10

 

Column A = S

Column B = 1

 

S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + 1/9 - 1/10

=> S = 1 - (1/2 - 1/3) - (1/4 - 1/5) - (1/6 - 1/7) - (1/8 - 1/9) - (1/10)

 

All terms in the brackets are positive.

=> S is 1 subtracted by five positive numbers

Hence, S is less than 1

 

Answer: B

[manwiththemission2005]

Thats right and this q has been recurrin in many threads..

[hace]

How do you solve these 2 problems quickly?

 

1)

Column A = (4^6 - 4^4) / (4^5 - 3^3)

Column B = 1

 

2)

S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + 1/9 - 1/10

 

Column A = S

Column B = 1

 

S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + 1/9 - 1/10

 

this is a part of a convergent sequence converges to 1 so since it is truncated it is less than 1

[mansoor316]

S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + 1/9 - 1/10

 

this is a part of a convergent sequence converges to 1 so since it is truncated it is less than 1

 

Hace, it is true that S is a part of a famous convergent series. However, it converges to ln 2 = 0.693 and not 1.

[infinityzero]

S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + 1/9 - 1/10

=> S = 1 - (1/2 - 1/3) - (1/4 - 1/5) - (1/6 - 1/7) - (1/8 - 1/9) - (1/10)

 

All terms in the brackets are positive.

=> S is 1 subtracted by five positive numbers

Hence, S is less than 1

 

Answer: B

 

Hi,

Thanks. Is there a rule that says that if 1 is subtracted by a certain amount of numbers that S will be less than 1? Or is it just safe to assume that since 1 is being subtracted by all of these fractions the end result will be a number less than 1?

 

remove 4^4 from num. & 4^3 from the den.

 

hecne we have [4*(4^2 - 1)]/[4^2 - (3/4)^3]

 

Now since 3/4 is 4^2 -1 hence the fraction

[4^2 -1]/[4^2 - (3/4)^3] 1[/b]

 

hence Col. A > Col. B

 

Hi,

I don't understand where the 3/4 came from? I did what you said, but how can you remove 4^4 from the numer. & 4^3 from the denom.? This is what I have and was hoping it could be shortened somehow. I removed 4^2 from the numerator and trying to figure out how to reduce the denom down:

4^2 (4^3 - 4^2) / 4^5 - 4^3

 

There is probably a better technique, like yours, but can you explain again?

[kronique]

Hi,

Thanks. Is there a rule that says that if 1 is subtracted by a certain amount of numbers that S will be less than 1? Or is it just safe to assume that since 1 is being subtracted by all of these fractions the end result will be a number less than 1?

Infinity, maybe this will help you understand it.

We have:

S = 1 - (1/2 - 1/3) - (1/4 - 1/5) - (1/6 - 1/7) - (1/8 - 1/9) - (1/10)

 

All terms in the brackets are positive.

Hence say S = 1 - K where K is a positive quantity(not equal to 0 here.) = all the bracket terms...

hence 1-S = K.

Now K > 0

hence 1 - S > 0

or S

 

 

I don't understand where the 3/4 came from? I did what you said, but how can you remove 4^4 from the numer. & 4^3 from the denom.?

Well consider the numertor & denominator to be two seperate parts i.e

consider (4^6 - 4^4)....from this in the usual way we can remove 4^ 4. hence we have 4^4(4^2 -1)

 

Now consider denominator seperately:

(4^5 - 3^3).

 

Now if we remove 4^ 3 from here... the first term will be reduced to 4^ 2.( since 4^3*4^2= 4^5.) and the second term...be careful here....since there is no 4 term here...we divide it by the common term removed i.e 4^3..so we get (3/4)^3.

 

Now the den. becomes: 4^3( 4^2 - (3/4)^3.)...to cross check if you have done this correctly...open the bracket now...i.e 4^3*4^2 - 4^3* (3^3/4^3.).....On solving this gives us 4^5 - 3^3.( since 4^3 gets cancelled.for the second term.)..This is our original denominator. Hence the interchanging done by us is correct.

 

 

Hope this is clear.:tup:

Now try to solve the way mentioned before, am sure u'll find it easy.

If u still face a problem.....post it....Hope we can clear it for u.:)

[hace]

Hace, it is true that S is a part of a famous convergent series. However, it converges to ln 2 = 0.693 and not 1.

 

sorry probabily i remembered wrong thnx for correction

[infinityzero]

Kronique,

Thank you. I understand now. I realized there was a typo made. The denom in the real problem is actually 4^5 - 4^3 (not 3^3), which is why I wondered where the

(3/4)^3 came from. Either way, I understand now.

[kronique]

. The denom in the real problem is actually 4^5 - 4^3 (not 3^3), which is why I wondered where the

(3/4)^3 came from. Either way, I understand now.

 

Ahhh..then its even easier.:) ....

[mansoor316]

Hi,

Thanks. Is there a rule that says that if 1 is subtracted by a certain amount of numbers that S will be less than 1? Or is it just safe to assume that since 1 is being subtracted by all of these fractions the end result will be a number less than 1?

 

Subtracting positive numbers from any number amounts to 'removing' or 'reducing'. The result is bound to be less than the number in question.

 

picture such scenarios in terms of apples being added/removed...

[infinityzero]

To make sure I did this right (don't have orig. answer), this is the original question for the first one:

Column A:

(4^6 - 4^4) / (4^5-4^3) =

4^4(4^2-1)/ 4^3 (4^2 - 1) = 4^4/4^3 = 1

 

Column B = 1

 

 

Then the answer is C...?

[kronique]

To make sure I did this right (don't have orig. answer), this is the original question for the first one:

Column A:

(4^6 - 4^4) / (4^5-4^3) =

4^4(4^2-1)/ 4^3 (4^2 - 1) = 4^4/4^3 = 1 ( 4^4/4^3 = 4.)

 

Column B = 1

 

 

Then the answer is C...?

 

Hence ans. is Col. A is greater.:)

[infinityzero]

Oh whoops. Yes. :blush:

 

 

(I think I'm thinking too much on this)

[kronique]

Oh whoops. Yes. :blush:

 

 

(I think I'm thinking too much on this)

 

lolz...its good that u've got the concept clear buddy !!:tup:

[comp]

i am having a tough time with graph problems in GRE

 

please suggest me ways to crack them

 

At present i am able to do only 1 correct problem from Graph problem