few more - how to solve?

[infinityzero]

http://i41.photobucket.com/albums/e289/infinity10/b03cc1fc.jpg

30. x and y are positive integers and x > y.

Col A: (x^2)/(y^3)

Col B: (y^3) / (x^2)

OA is C, but I chose D. Can someone explain?

 

For #29 especially, what is a fast way to solve these types of questions?Thanks.

[krishnakanthc]

soln 15 > the hypot of the right angle triangles are sqrt2 and sqrt2 .

 

{sqrt( 1 + 1) }

 

 

i.e 2(sqrt2) + 2

 

 

soln 20 >

 

answer is A . |x| gives positve values always

 

i.e | -x| = x and |x| = x

 

by elimination you get A

 

soln29 > this prob was posted few days ago

 

Anyways, sum of first N numbers is N*(N+1) / 2

 

for N =50 Sum = 1275

 

for N=100 Sum=5050

 

for 100 - 50 numbers 5050 - 1275

 

3775 i.e D

 

soln 30> the answer is D .

 

consider numbers x=4 y=3

 

we get 16/ 27

 

27/16

 

i.e B

 

consider numbers x=2 y=1

 

4/1 and 1/4 i.e A

 

therefore it is D (not C)

[bscout]

I agree with Krishna.

[ltoeyl]

soln 15 > the hypot of the right angle triangles are sqrt2 and sqrt2 .

 

{sqrt( 1 + 1) }

 

 

i.e 2(sqrt2) + 2

 

 

soln 20 >

 

answer is A . |x| gives positve values always

 

i.e | -x| = x and |x| = x

 

by elimination you get A

 

soln29 > this prob was posted few days ago

 

Anyways, sum of first N numbers is N*(N+1) / 2

 

for N =50 Sum = 1275

 

for N=100 Sum=5050

 

for 100 - 50 numbers 5050 - 1275

 

3775 i.e D

 

soln 30> the answer is D .

 

consider numbers x=4 y=3

 

we get 16/ 27

 

27/16

 

i.e B

 

consider numbers x=2 y=1

 

4/1 and 1/4 i.e A

therefore it is D (not C)

I 'm not really understand about no#29.

Could you explain me again plzz 'O'.

[rpgre2005]

ans:d,a,d,d

[krishnakanthc]

{c,a,d,d }

[bscout]

I 'm not really understand about no#29.

Could you explain me again please 'O'.

 

You need the sum of the integers from 51 to 100 and you know that the first 50 sum up 1,275.

 

The formula for the sum of the first N number is, as Krishna wrote, N*(n+1)/2

 

For the first 50 you have= 50*(51)/25= 1,275

For the first 100 you have= 100*(101)/2= 5,050

 

But you need only the sum from 51 to 100, that is to said, you need to take away from the sum the first 50 integers.

 

Therefore, 5,050-1,275=3,775.

 

I hope this explanation helps you.

[rpgre2005]

hi infinityzero

can you please disclose answers to the above problems?

 

 

 

thnks

[sgsits]

{d,a,d,d}

The ans of first is 'd' coz nowhere it is given dat the highect vertex of triangle is in the line of mid point of the base (and note that if if assume it as midpoint then ans is 'c'.......but as a rule,in GRE u r not allowed to make assumptions until stated for figures!)

[htownstreetracer]

29. Play with the numbers with something easier.

 

Tell me if my method is wrong, but establishing this relationship with numbers such that you can mirror the relationship on a smaller scale.

 

Change the problem to sum of first 5 integers

 

1+2+3+4+5=15

 

so the sum of the integers 6-10?

6+7+8+9+10=40

 

the figure almost 3x....

 

 

 

try with smaller numbers too...

 

sum of the first four integers

1+2+3+4=10

 

integers 5-8?

5+6+7+8=26

 

almost 3x....:)

 

 

try 1275 x 3=3825.....BALLPARK with small numbers? >_

 

answer is 3775

[htownstreetracer]

so is the first one is D?

 

if it said the triangle is inscribed then it is C?

[pavelbuet]

Q.29

why don't we solve it without the formula n(n+1)/2...?

 

51+52+.....+100

=>(50+1) + (50+2)+.....+(50+50)

=>50*50 + (1+2+....+50)

=>2500 + 1275

=>3775

 

if u can guess it quickly then the steps come down to...

50*50 + 1275

=> 2500 + 1275

=> 3775

 

one easy multiplication and addition.....but i should admit that using a well known formula like n(n+1)/2 makes one feel confident

[dipen01]

Yeah i guess....well we could have tried another method..but what if Options are real close....

 

and this method is way to easy ...I didnt need Pen/paper..when one can do it orally..why take unnecessary risk...!!!

[bscout]

Pavel, in this question your method is very useful but what if you don't have the information about how many 1+2+...+50 is? Is for this kind of situation that I think this formula is very useful and it's a good opportunity to become familiar with it.