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Complex decimal problem - any shortcuts?

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Complex decimal problem - any shortcuts?

by rachitakapoor » Tue Sep 11, 2012 6:58 am
Q. 199 on OG 13

(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

The right answer is D. Is there any shortcut for this question? The solution in the OG is tedious.
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by rijul007 » Tue Sep 11, 2012 8:11 am
rachitakapoor wrote:Q. 199 on OG 13

(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

The right answer is D. Is there any shortcut for this question? The solution in the OG is tedious.
(0.99999999 ÷ 1.0001) - (0.99999991 ÷ 1.0003)
(9999.9999/10001) - (9999.9991/10003)
(9999 * 1.0001/10001) - (9997 * 1.0003/10003)
(9999 * 10^-4) - (9997 * 10^-4)
0.9999 - 0.9997
0.0002

Option D
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by GMATGuruNY » Tue Sep 11, 2012 8:41 am
rachitakapoor wrote:Q. 199 on OG 13

(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

The right answer is D. Is there any shortcut for this question? The solution in the OG is tedious.
Approach 1:

(9999.9999)/(10001) - (9999.99991)(10003)

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

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

Only C, D and E include 10^-4.
Thus, the correct answer must be C, D or E, implying three possibilities:
99999999/10001 - 99999991/10003 = 3.
99999999/10001 - 99999991/10003 = 2.
99999999/10001 - 99999991/10003 = 1.

Since their difference is an integer, the quotients here are almost certainly integer values themselves.
Focus on the UNITS DIGITS:
The units digit of 99999999/10001 must be 9, since 10001 must be multiplied by a units digit of 9 to yield 99999999.
The units digit of 99999991/10003 must be 7, since 10003 must be multiplied by a units digit of 7 to yield 99999991.
Since 9-7=2:
99999999/10001 - 99999991/10003 = 2.

The correct answer is D.

Approach 2:

(9999.9999)/10001 - (9999.9991)/10003

= [10003(9999.9999) - 10001(9999.9991)] / (10001)(10003)

≈ [10003(10^4) - 10001(10^4)] / (10^4)(10^4)

≈ (10003-10001)/(10^4)

≈ 2(10^-4).

The correct answer is D.

Approach 3:
Try a simpler version of the problem.

.99/1.1 - .91/1.3

= 99/110 - 91/130

= ( 99/11 - 91/13) (1/10)

= (9-7)(10^-1)

= 2(10^-1).

This result suggests that the correct answer almost certainly is D.
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by rachitakapoor » Tue Sep 11, 2012 11:46 pm
rijul007 wrote:
rachitakapoor wrote:Q. 199 on OG 13

(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

The right answer is D. Is there any shortcut for this question? The solution in the OG is tedious.
(0.99999999 ÷ 1.0001) - (0.99999991 ÷ 1.0003)
(9999.9999/10001) - (9999.9991/10003)
(9999 * 1.0001/10001) - (9997 * 1.0003/10003)
(9999 * 10^-4) - (9997 * 10^-4)
0.9999 - 0.9997
0.0002

Option D
Many thanks, rijul007! :)
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by rachitakapoor » Tue Sep 11, 2012 11:51 pm
GMATGuruNY wrote:
rachitakapoor wrote:Q. 199 on OG 13

(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

The right answer is D. Is there any shortcut for this question? The solution in the OG is tedious.
Approach 1:

(9999.9999)/(10001) - (9999.99991)(10003)

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

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

Only C, D and E include 10^-4.
Thus, the correct answer must be C, D or E, implying three possibilities:
99999999/10001 - 99999991/10003 = 3.
99999999/10001 - 99999991/10003 = 2.
99999999/10001 - 99999991/10003 = 1.

Since their difference is an integer, the quotients here are almost certainly integer values themselves.
Focus on the UNITS DIGITS:
The units digit of 99999999/10001 must be 9, since 10001 must be multiplied by a units digit of 9 to yield 99999999.
The units digit of 99999991/10003 must be 7, since 10003 must be multiplied by a units digit of 7 to yield 99999991.
Since 9-7=2:
99999999/10001 - 99999991/10003 = 2.

The correct answer is D.

Approach 2:

(9999.9999)/10001 - (9999.9991)/10003

= [10003(9999.9999) - 10001(9999.9991)] / (10001)(10003)

≈ [10003(10^4) - 10001(10^4)] / (10^4)(10^4)

≈ (10003-10001)/(10^4)

≈ 2(10^-4).

The correct answer is D.

Approach 3:
Try a simpler version of the problem.

.99/1.1 - .91/1.3

= 99/110 - 91/130

= ( 99/11 - 91/13) (1/10)

= (9-7)(10^-1)

= 2(10^-1).

This result suggests that the correct answer almost certainly is D.
Thanks a tonne for all the approaches! Looking at various approaches does help a lot and I'm gonna try finding atleast 2 of them for the tough problems.
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by Brent@GMATPrepNow » Wed Sep 12, 2012 6:47 am
rachitakapoor wrote: Looking at various approaches does help a lot and I'm gonna try finding at least 2 of them for the tough problems.
Yes!! This is an awesome strategy that all students should practice while preparing for the GMAT (for all math questions).

Cheers,
Brent
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by KateG1302 » Sat Oct 20, 2012 5:11 am
rijul007 wrote:
rachitakapoor wrote:Q. 199 on OG 13

(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

The right answer is D. Is there any shortcut for this question? The solution in the OG is tedious.
(0.99999999 ÷ 1.0001) - (0.99999991 ÷ 1.0003)
(9999.9999/10001) - (9999.9991/10003)
(9999 * 1.0001/10001) - (9997 * 1.0003/10003)
(9999 * 10^-4) - (9997 * 10^-4)
0.9999 - 0.9997
0.0002

Option D
Hi,

sorry but I just don't understand how to get from:

(9999.9999/10001) - (9999.9991/10003)

to

(9999 * 1.0001/10001) - (9997 * 1.0003/10003)


I see how the third approach works (i.e. using "simplified" numbers) but I am worried that I will not recognise this pattern during the real test, so I'd like to understand the calculations properly.

Thanks in advance !

Kate
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by KateG1302 » Tue Oct 23, 2012 4:59 am
Hi all,


can nobody help with the above?

Thanks a lot!
Kate
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by Brent@GMATPrepNow » Tue Oct 23, 2012 6:04 am
KateG1302 wrote: Hi,

sorry but I just don't understand how to get from:

(9999.9999/10001) - (9999.9991/10003)

to

(9999 * 1.0001/10001) - (9997 * 1.0003/10003)


I see how the third approach works (i.e. using "simplified" numbers) but I am worried that I will not recognise this pattern during the real test, so I'd like to understand the calculations properly.

Thanks in advance !

Kate
Hi Kate,

The step from (9999.9991/10003)
to
(9997 * 1.0003/10003)
is a gigantic step. Don't worry, the vast majority of students won't see it either.

I believe that the test maker wants us 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)
= 0.0002
= 2 x 10^(-4) = D

Cheers,
Brent
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