Problem Solving

(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.

5j= k, 3k=l,7l= 3m
=>5*3*7j = 3m
=>35j =m
=> (20m/200j) = (35j*20/200j)= 350 ..!!

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.

Didn't plug values.
Did the hard way.
At last, got 350 !

(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

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

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

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