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Jim’s weight is 140 percent of Marcia’s weight. Bob’s weight is 90 per

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Jim’s weight is 140 percent of Marcia’s weight. Bob’s weight is 90 per [#permalink]

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New post 05 Nov 2017, 01:40
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Jim’s weight is 140 percent of Marcia’s weight. Bob’s weight is 90 percent of Lee’s weight. Lee weighs twice as much as Marcia. What percentage of Jim’s weight is Bob’s weight?

(A) 64 2/7
(B) 77
(C) 90 7/9
(D) 128 4/7
(E) 155 5/9
[Reveal] Spoiler: OA

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Jim’s weight is 140 percent of Marcia’s weight. Bob’s weight is 90 per [#permalink]

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New post 05 Nov 2017, 01:59
Let Marcia's weight be 100. Therefore, Jim weighs 140(140% of Marcia's weight)
Bob's weight(180) is 90% of Lee's weight which is 200, twice as much as Marcia's weight.

Hence we need to find what percentage is Jim's weight to Bob's weight

Bob's weight is 180 and is x% of 140(Jim's weight) OR 180 is x% of 140

\(180 = \frac{140x}{100}\)

\(x = \frac{18000}{140} = 128 \frac{4}{7}\) (Option D)
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Jim’s weight is 140 percent of Marcia’s weight. Bob’s weight is 90 per [#permalink]

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New post 05 Nov 2017, 02:15
Bunuel wrote:
Jim’s weight is 140 percent of Marcia’s weight. Bob’s weight is 90 percent of Lee’s weight. Lee weighs twice as much as Marcia. What percentage of Jim’s weight is Bob’s weight?

(A) 64 2/7
(B) 77
(C) 90 7/9
(D) 128 4/7
(E) 155 5/9



just incase you are having trouble in some calculations...
\(J=\frac{140}{100}*M\)...
\(B=\frac{90}{100}*L=\frac{90}{100}*2M=\frac{180}{100}M\)...

now B is > J so \(\frac{B}{J} >100\)%
only D and E left
\(\frac{B}{J}*100=\frac{180}{140}*100\)...

easier calculation is 150% of 140 and D is < 150 and E>150....
\(140*\frac{150}{100} = 140*(\frac{100}{100}+\frac{50}{100})= 140+\frac{140}{2}=140+70=210\)..
but we are looking for 180 so ans has to be LESS than 150%
ONLY D left.
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Kudos [?]: 5816 [0], given: 117

Jim’s weight is 140 percent of Marcia’s weight. Bob’s weight is 90 per   [#permalink] 05 Nov 2017, 02:15
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