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Martina earns one-sixth of her annual income during the mont

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Martina earns one-sixth of her annual income during the mont [#permalink] New post 21 Jul 2010, 21:22
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58% (03:51) correct 42% (02:28) wrong based on 142 sessions
Martina earns one-sixth of her annual income during the month of June and one-eighth in August. Pam earns one-third of her annual income in June and one-fourth in August. Martina's earnings for June and August equal Pam's earnings for the same period. What portion of their combined annual income do the two girls earn during the ten months NOT including June and August?

(A) 1/8
(B) 7/24
(C) 7/18
(D) 11/18
(E) 7/8
[Reveal] Spoiler: OA

Last edited by Bunuel on 23 Nov 2013, 05:29, edited 1 time in total.
RENAMED THE TOPIC.
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Re: Martina earns [#permalink] New post 21 Jul 2010, 22:52
IMO D

rohitgoel15 wrote:
Martina earns one-sixth of her annual income during the month of June and one-eighth in August. Pam earns one-third of her annual income in June and one-fourth in August. Martina's earnings for June and August equal Pam's earnings for the same period. What portion of their combined annual income do the two girls earn during the ten months NOT including June and August?
(A) 1/8
(B) 7/24
(C) 7/18
(D) 11/18
(E) 7/8


Solution:

Lets assume that Martina earns 6000 in the entire year and Pam earns 3000 during the annual year.
So, Martina:
June = 1000 (1/6)
August = 750 (1/8)

For Pam:
June = 1000 (1/3)
August = 750 (1/4)

Since both their earnings are same for these 2 months (If you carefully observe, the ratios for Martina in those 2 months have just been halved for respective months as compared to Pam's). so net earnings for the entire year would be double for the income in those 2 months to be same.

So total earnings for both in the annual year = 9900
Earnings from June & August = 1750 + 1750 = 3500
Earnings during rest of the year = 5500


= 5500/9000 = 11/18.

I hope everything is clear.. :)
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Re: Martina earns [#permalink] New post 21 Jul 2010, 23:08
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Martina's earnings for June and August equal Pam's earnings for the same period

=> x/8 + x/6 = y/3 + y/6 where x = martina' earning and y = pam's

=> 7x/24 = 7y/12 => x=2y
Their combined income = x+y = 3y

For ten months their combined income = x-\frac{7x}{24}+ y-\frac{7y}{12}= \frac{17x}{24}+ \frac{5y}{12}
= \frac{17y}{12}+ \frac{5y}{12}= \frac{22y}{12}

Required ratio = (22y/12 )/3y =\frac{11}{18}
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Re: Martina earns [#permalink] New post 23 Nov 2013, 05:27
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Re: Martina earns one-sixth of her annual income during the mont [#permalink] New post 31 Mar 2014, 23:45
\frac{1}{6} th & \frac{1}{8} th Income of Martina = \frac{1}{3} rd & \frac{1}{4} th Income of Pam

\frac{1}{6} th = Half of \frac{1}{3} rd &

\frac{1}{8} th = Half of \frac{1}{4} th

Above itself means that Income of Martina is twice that of Pam

Say Income of Pam = 120

Then income of Martina = 240

\frac{1}{3} rd & \frac{1}{4} th of 120 = 40 + 30 = 70;

Balance for Pam = 120 - 70 = 50 ........... (1)

Balance for Martina = 240 - 70 = 170......... (2)

Total 10 months combined = (1) + (2)

= 220

Fraction = \frac{220}{360} =\frac{11}{18}

Answer = D
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Re: Martina earns one-sixth of her annual income during the mont   [#permalink] 31 Mar 2014, 23:45
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