GMAT TIGER wrote:
tejal777 wrote:
Which of the following is closest to \frac{(998*0.02 + 4.97)}{(1002*0.02 + 5.03)} ?
(C) 2008 GMAT Club - m10#22
* 0.50
* 0.89
* 0.98
* 1.02
* 1.05
The OE is not clear..i did it without rounding it off and not getting the answer:(
Let’s use some logic..
By observation, numerator is very close too denominator.
D/E are out as numerator cannot be > denominator.
A is also out because numerator is substantially > 1/2(denominator).
B is equivalent to 90% i.e. numerator < 90% (denominator) where as .
So left with C.
Alternatively:
=
\frac{(998*0.02 + 4.97)}{(1002*0.02 + 5.03)}=
\frac{(1000 - 2) *0.02 + 4.97)}{(1000+2)*0.02 + 5.03)}=
\frac{(20 - 0.04) + 4.97)}{(20 + 0.04) + 5.03)}=
\frac{(19.96 + 4.97)}{(20.04) + 5.03)}=
\frac{(24.93)}{(25.07)} = equivalent to 1 but < 1, which is 0.98....
In 30 second, it is known that OA is C....
Almost did it in similar fashion but with less computation i guess....
=
\frac{(998*0.02 + 4.97)}{(1002*0.02 + 5.03)}=
\frac{(1000 - 2) *0.02 + (5 - .03)}{(1000+2)*0.02 + (5 + .03)]}Now Let 1000 be replaced with 'X' and 5 with 'Y', then the resultant fraction is:
=
\frac{(X - 0.04) + (Y - .03)}{(20 + 0.04) + 5.03)}=
\frac{(X + Y - .07)}{(X + Y + .07)}= 1 -
\frac{(.21)}{(X + Y + .07)} = 1 - ~.0001 => 0.9998
So the closest value is 0.98...
Answer: (c)
_________________
-- Because beauty Lies in the Eyes.. So donate those eyes 
KUDOS if like the post ... 
close
