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by ronnie1985 » Thu Mar 29, 2012 7:21 am
(5/4)j = (1/4)k; (3/2)k = (1/2)i; (7/4)i = (3/4)m.
Solving for m, m = 35j => (m/5) = (7/2)*(2j) => 350% of 2j.
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by Alok Jha » Fri May 04, 2012 1:18 am
5j= k, 3k=l,7l= 3m
=>5*3*7j = 3m
=>35j =m
=> (20m/200j) = (35j*20/200j)= 350 ..!!
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by anujan007 » Sat Jul 14, 2012 8:14 pm
Solved this by the longer way of substituting each value by base referencing with j=100. took significantly longer than 2 mins.

Should have plugged in j=4 as mentioned previously. That would have been quicker.
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by Lifetron » Thu Sep 06, 2012 7:27 am
Didn't plug values.
Did the hard way.
At last, got 350 !
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by rajeshsinghgmat » Sun Apr 14, 2013 2:09 am
(D) 350

(1.25j)(1.5k)(1.75l)=(0.25k)(0.5l)(0.75m)

(5/4)(3/2)(7/4)(jkl)=(1/4)(1/2)(3/4)(klm)

i.e. 35j=m
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by rintoo22 » Sun Apr 14, 2013 11:33 am
If 125% of j is equal to 25% of k, 150% of k is equal to 50% of l, and 175% of l is equal to 75% of m, then 20% of m is equal to what percent of 200% of j ?
Thanks for all the good approaches. I will like to post one more.

125% of j is equal to 25% of k => 5j=k ....(1)
150% of k is equal to 50% of l => 3k=l .... Substituing values from (1) 15j=l .... (2)
175% of l is equal to 75% of m => 7l=3m .... Substituing values from (2) 35j=m .... (3)
20% of m is equal to what percent of 200% of j ?
If 35j=m. Then 20% of m is equal to 7j.
So 'What' percent of 200 % of j should be '7j'
Equation should be 7j = x/100 (200/100 * j)
x = 350.

350 .... Answer D
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by jaspreetsra » Sun Dec 28, 2014 4:13 am
If 125% of j is equal to 25% of k, 150% of k is equal to 50% of l, and 175% of l is equal to 75% of m, then 20% of m is equal to what percent of 200% of j?

(A) 0.35
(B) 3.5
(C) 35
(D) 350
(E) 3500

My approach:
125*j = 25*k
or 5j = k
let j be 100
then k=500
l=1500
m=3500
Now 20m = 200j =>1m =10j
So,
m/j = (3500*10)/100 (j is assumed as 100)
350%

IMO:D
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by nikhilgmat31 » Thu Aug 27, 2015 3:15 am
Answer is D

In last m = 35 J

so 20 % of m = 1/5 * 35 J = 7J

7J of 200% of J = 7J/2J * 100 = 350

Answer D
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