Problem Solving

Q20:
If x > 0, then 1/[√(2x)+√x] =

A. 1/√(3x)
B. 1/[2√(2x)]
C. 1/(x√2)
D. (√2-1)/√x
E. (1+√2)/√x
------------------------------------------------------------------------------------------------------------
Q2:
A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the 3 cars each time that passenger rides the roller coaster. If a certain passenger is to ride the roller coaster 3 times, what is the probability that the passenger will ride in each of the 3 cars?

A. 0
B. 1/9
C. 2/9
D. 1/3
E. 1
Answer:
------------------------------------------------------------------------------------------------------------
Q3:
If x and y are positive, is 4x > 3y?
(1) x > y - x
(2) x/y < 1

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
Answer:
------------------------------------------------------------------------------------------------------------

If x > 0, then 1/[&#8730;(2x)+&#8730;x] = (&#8730;2-1)/&#8730;x

Just multiply the equation with [&#8730;(2x)- &#8730;x] / [&#8730;(2x) - &#8730;x]

------------------------------------------------------------------------------------------------------------
Q2:
A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the 3 cars each time that passenger rides the roller coaster. If a certain passenger is to ride the roller coaster 3 times, what is the probability that the passenger will ride in each of the 3 cars?

so let say cars are labeled as A B and C

You can list them in 3x2x1 = 6 ways

well each time the chances are 1/3x1/3x1/3 = 1/27

so 6/27 = 2/9
------------------------------------------------------------------------------------------------------------
Q3:
If x and y are positive, is 4x > 3y?
(1) x > y - x
(2) x/y < 1

in other words: is x/y > 3/4 ?

1 ) x / y > 2 /4

we are not sure whether it is bigger than 3/4 INSUF

2) x / y < 4/4

we are not sure whether it is bigger than 3/4 INSUF

E
------------------------------------------------------------------------------------------------------------[/quote]

logitech wrote:If x > 0, then 1/[&#8730;(2x)+&#8730;x] = (&#8730;2-1)/&#8730;x

Just multiply the equation with [&#8730;(2x)- &#8730;x] / [&#8730;(2x) - &#8730;x]

Please, post full solution
I used the same approach but came to an answer that is not indicated here

agree with 4meonly

logitech can you explain in detail how you got (rt2-1) in the numerator.

I used the same method you did and got the answer as (sqrt(2x)-sqrtx)/(x)
and this option is not included.

oops sorry u had to simplify the fraction further

[sqrt(2x)-sqrt(x)]/x-------multiply the numerator and denominator with sqrt(x).

[sqrt(2x)-sqrt(x)]*sqrtx/(x*sqrtx).


[x*sqrt2-x]/x*sqrtx.

Taking out x from the numerator and canceling it with the denominator x gives:

[sqrt2-1]/sqrtx.

mals24 wrote:oops sorry u had to simplify the fraction further

[sqrt(2x)-sqrt(x)]/x-------multiply the numerator and denominator with sqrt(x).

[sqrt(2x)-sqrt(x)]*sqrtx/(x*sqrtx).


[x*sqrt2-x]/x*sqrtx.

Taking out x from the numerator and canceling it with the denominator x gives:

[sqrt2-1]/sqrtx.

You got it!

I don't understand how you simplified question 1 from the beginning. Work with me here:

Initial question: 1 / [sqrt(2x) + sqrt(x)]

Multiplying the top/bottom by sqrt(2x) - sqrt(x): 1 / [sqrt(2x) + sqrt(x)] * {[sqrt(2x) - sqrt(x)]/[sqrt(2x) - sqrt(x)]}

I get [sqrt(2x) - sqrt(x)] / (2x + x)

I get 2x + x because sqrt(2x)sqrt(2x) = 2x and sqrt(x)sqrt(x) = x


Where are you getting the x from in the denominator.

Sorry - I figured this out. Careless mistakes.

I'm curious, but with these problems, how do you know how to be solving them because how do you know which type of answer your trying to make this look like? YOu can't just take trial/error into account and just start telling yourself "I'm just going to multiple the numerator and denominator by sqrt(x) and make an attempt to see what happens and how far it matches the answer choice".

Isn't there a full proof way to finding an answer and then looking at the answer choices? It seems like there is a lot of guessing and time waste.

piyushdabomb wrote:Sorry - I figured this out. Careless mistakes.

I'm curious, but with these problems, how do you know how to be solving them because how do you know which type of answer your trying to make this look like? YOu can't just take trial/error into account and just start telling yourself "I'm just going to multiple the numerator and denominator by sqrt(x) and make an attempt to see what happens and how far it matches the answer choice".

Isn't there a full proof way to finding an answer and then looking at the answer choices? It seems like there is a lot of guessing and time waste.

How ? 4 years in Engineering school help

But the more question you solve in quant, the more you realize that GMAC actually has a scope for its questions.

This one is a classical root problem where you you have 1/ (A+B) or 1 / (A-B)

when you see a problem like this always keep in mind that:

(a+b) x (a-b) = a^2 - b^2

I got a PM to clarify my solution:

Let's go:

If x > 0, then 1/[&#8730;(2x)+&#8730;x] =

the trick with these questions is to use (a+b)(a-b)=a^2-b^2

this way lets say:

(&#8730;a+&#8730;b)(&#8730;a-&#8730;b) = a-b

This is a great method to take the numbers out of their roots

lets apply this to our problem

Lets multiply the equation with

&#8730;2x-&#8730;x/&#8730;2x-&#8730;x

So actually we multiply the both parts with same number!

we will have

&#8730;2x-&#8730;x
-----------
2x-x


&#8730;2x-&#8730;x
-------------
x

now lets take the top part and:

&#8730;x ( &#8730;2-1)
-------------
&#8730;x &#8730;x

( &#8730;2-1)
---------
&#8730;x

Hope all is clear now. If not, please stop at financing department and see Cramya and tell him that I sent you

its clear now

Usually answers in GMAT questions do not contain roots in denumerator...

Q20:
If x > 0, then 1/[&#8730;(2x)+&#8730;x] =

A. 1/&#8730;(3x)
B. 1/[2&#8730;(2x)]
C. 1/(x&#8730;2)
D. (&#8730;2-1)/&#8730;x
E. (1+&#8730;2)/&#8730;x

In terms of Q20 above, is there anyone who tried to solve this problem by plugging in numbers? Many of the major courses such as PR or Kap suggest this technique as an alternative to traditional algebra, in order to avoid possible mistakes in calculations. Unfortunately, it did not work well for me here. Can anyone provide their results if they used plug-ins? I am well aware that sqt(2) = approx 1.4 and sqt(3) = 1.7. Would it work for a problem like this? Thanks in advance!

jnellaz wrote:Q20:
If x > 0, then 1/[&#8730;(2x)+&#8730;x] =

A. 1/&#8730;(3x)
B. 1/[2&#8730;(2x)]
C. 1/(x&#8730;2)
D. (&#8730;2-1)/&#8730;x
E. (1+&#8730;2)/&#8730;x

In terms of Q20 above, is there anyone who tried to solve this problem by plugging in numbers? Many of the major courses such as PR or Kap suggest this technique as an alternative to traditional algebra, in order to avoid possible mistakes in calculations. Unfortunately, it did not work well for me here. Can anyone provide their results if they used plug-ins? I am well aware that sqt(2) = approx 1.4 and sqt(3) = 1.7. Would it work for a problem like this? Thanks in advance!

X=1

1/1&#8730;2+1 = 1/2.4 = 0.42 (approx)

a) too big
b) 1/2.8 AHA
c)1/1.4 nope
d) 0.4/1.4 = pretty close
e) too big

Now as you see it is down to B and D



You can test them with another number until you find which one closes the deal.

Do you really want to this ? Try X=3 for B and D

Thanks Logitech. Sorry to split hairs, but you wrote:

1/1&#8730;2+1 = 1/2.8 = 0.36 (approx)

Unless I am missing something, how did you get to this? I thought it would be 1/1&#8730;2+1 = 1/2.4 - (since &#8730;2 = 1.4 )