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) _________________

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