1. Good post? |

## value of expression

From Set10 #20 (before solving scroll down...)

If x > 0, then 1/[v(2x)+vx] =

A. 1/v(3x)
B. 1/[2v(2x)]
C. 1/(xv2)
D. (v2-1)/vx
E. (1+v2)/vx

This is how it is mentioned in set 10.pdf problem # 20

I am still figuring out what the expression means!

SPOILER: Not A

...

OKAY figured out whats wrong...

v is actually NOT (letter v) but SQRT - someone tried to use v in place of writing SQRT or ROOT or it's PDF convertors fault that couldn't load proper character for the symbol (the one we draw as )...

So question must look like this...

If x > 0, then 1/[SQRT(2x)+SQRTx] =

A. 1/SQRT(3x)
B. 1/[2SQRT(2x)]
C. 1/(xSQRT2)
D. (SQRT2-1)/SQRTx
E. (1+SQRT2)/SQRTx

I am sure everyone will get to Official Answer now...

SPOILER: D

This applied to some other questions as well in the GMAT Sets e.g. Set 10 # 23 ... Watch out while doing SETS.

2. Good post? |

1/v[2x+x] = 1/v[3x]

3. Good post? |
1/[SQRT(2x)+SQRTx]
= [SQRT(2x)-SQRTx] /{[SQRT(2x)+SQRTx] [SQRT(2x)-SQRTx]}
= [SQRT(2x)-SQRTx]/[2x-x]
= [SQRT(2x)-SQRTx]/x
= [SQRT 2-1]/SQRTx

4. Good post? |
Ans must be "D"

solu: multiply denominator and numerator by ( sqrt2 + 1) and you see the ans

5. Good post? |
D
mtd exactly as animesh

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•

SEO by vBSEO ©2010, Crawlability, Inc.