Alternative method for this sum?

[sushantsahaji]

0.99999999/1.0001-0.99999991/1.0003=
A. 10^-8
B. 3 (10^-8)
C. 3(10^-4)
D. 2(10^-4)
E. 10^-4

Following is my approach, but have done some approximation.

Question
= (99999999/10001- 99999991/10003)*10^-4

(Let # = 99999999 or 99999991= 10^8)
=(#/10001-#/10003)*10^-4
=#( 1/10001-1/10003)*10^-4
=#{(10003-10001)/10^8}*10^-4 ( 10003*10001=10^€)
= 10^8(2/10^8)*10^-4
=[spoiler]2*10^-4[/spoiler]


1.whether the approximation done is allowed in general in GMAT?
2. Suggest different approach to this type of sum.

[Brent@GMATPrepNow]

0.99999999/1.0001 - 0.99999991/1.0003=

A. 10^-8
B. 3(10^-8)
C. 3(10^-4)
D. 2(10^-4)
E. 10^-4

One approach is to combine the fractions and then use some approximation.

First combine the fractions by finding a common denominator.
(9999.9999)/(10001) - (9999.9991)/(10003)
= (9999.9999)(10003)/(10001)(10003) - (9999.9991)(10001) /(10003)(10001)
= [(10003)(9999.9999) - (10001)(9999.9991)] / (10001)(10003)
= [(10003)(10^4) - (10001)(10^4)] / (10^4)(10^4) ... (approximately)
= [(10003) - (10001)] / (10^4) ... (divided top and bottom by 10^4)
= 2/(10^4)
= 2*10^(-4)
= D

Cheers,
Brent

[Brent@GMATPrepNow]

0.99999999/1.0001 - 0.99999991/1.0003 =

A. 10^-8
B. 3(10^-8)
C. 3(10^-4)
D. 2(10^-4)
E. 10^-4

Another approach is to recognize that both 9999.9999 and 9999.9991 can be rewritten as differences of squares.

First, 0.99999999 = 1 - 0.00000001
= (1 - 0.0001)(1 + 0.0001)

Similarly, 9999.9991 = 1 - 0.00000009
= (1 - 0.0003)(1 + 0.0003)

Original question: 0.99999999/1.0001 - 0.99999991/1.0003
= (1 - 0.0001)(1 + 0.0001)/(1.0001) - (1 - 0.0003)(1 + 0.0003)/(1.0003)
= (1 - 0.0001)(1.0001)/(1.0001) - (1 - 0.0003)(1.0003)/(1.0003)
= (1 - 0.0001) - (1 - 0.0003)
= 1 - 0.0001 - 1 + 0.0003
= 0.0002
= 2 x 10^(-4) = D

Cheers,
Brent

[OptimusPrep]

0.99999999/1.0001-0.99999991/1.0003=
A. 10^-8
B. 3 (10^-8)
C. 3(10^-4)
D. 2(10^-4)
E. 10^-4

1.whether the approximation done is allowed in general in GMAT?
2. Suggest different approach to this type of sum.

1, Yes the approximation is definitely allowed in the GMAT, infact in questions involving big numbers, you SHOULD use approximations.
However, there is a caveat. Use approximations only in the cases when the answers choices are not too close to each other.

2. Coming back to the question
Whenever you see something in the form of 0.9999... or 1.0001...., see if they can be broken in the from of (a+b)(a-b)
In this case,

0.99999999 = 1 - 0.00000001 = 1 - 10^(-8)
= [1 - 10^(-4)]*[1 + 10^(-4)]

9999.9991 = [1 - 9*10^-(8)]
= [1 - 3*10^(-4)]*[1 + 3*10^(-4)]

As soon as we get to this, our lives become a lot simpler.

Substituting the values in the original equations, we have
= (1 - 0.0001)(1 + 0.0001)/(1.0001) - (1 - 0.0003)(1 + 0.0003)/(1.0003)
= (1 - 0.0001) - (1 - 0.0003)
= -10^(-4) + 3*10^(-4)
= 2*10^(-4)

Correct Option: D

[sushantsahaji]

0.99999999/1.0001 - 0.99999991/1.0003 =

A. 10^-8
B. 3(10^-8)
C. 3(10^-4)
D. 2(10^-4)
E. 10^-4

Another approach is to recognize that both 9999.9999 and 9999.9991 can be rewritten as differences of squares.

First, 0.99999999 = 1 - 0.00000001
= (1 - 0.0001)(1 + 0.0001)

Similarly, 9999.9991 = 1 - 0.00000009
= (1 - 0.0003)(1 + 0.0003)

Original question: 0.99999999/1.0001 - 0.99999991/1.0003
= (1 - 0.0001)(1 + 0.0001)/(1.0001) - (1 - 0.0003)(1 + 0.0003)/(1.0003)
= (1 - 0.0001)(1.0001)/(1.0001) - (1 - 0.0003)(1.0003)/(1.0003)
= (1 - 0.0001) - (1 - 0.0003)
= 1 - 0.0001 - 1 + 0.0003
= 0.0002
= 2 x 10^(-4) = D

Cheers,
Brent

This is what I would have loved to do on my first attempt! Thank you.

[chetan.sharma]

0.99999999/1.0001-0.99999991/1.0003=
A. 10^-8
B. 3 (10^-8)
C. 3(10^-4)
D. 2(10^-4)
E. 10^-4

Following is my approach, but have done some approximation.

Question
= (99999999/10001- 99999991/10003)*10^-4

(Let # = 99999999 or 99999991= 10^8)
=(#/10001-#/10003)*10^-4
=#( 1/10001-1/10003)*10^-4
=#{(10003-10001)/10^8}*10^-4 ( 10003*10001=10^€)
= 10^8(2/10^8)*10^-4
=[spoiler]2*10^-4[/spoiler]


1.whether the approximation done is allowed in general in GMAT?
2. Suggest different approach to this type of sum.

another approach would be to make use of the choices available..
since the Q does not mention approx value, the equation 0.99999999/1.0001-0.99999991/1.0003 must be an integer or terminating decimal..
lets use this property..
BOTH 0.99999999/1.0001 and 0.99999991/1.0003 are ODD/ODD so would result in ODD terms..
so ODD - ODD should be EVEN...
Only D gives you the last integer as EVEN..
so D

Many times we can just make use of the choices and save a lot of time..