The weight of a glass of jar is 20% of the weight of the jar

Initial weight of a jar with coffee beans: C
Weight of an empty jar: J = 0.2C
Initial weight of the coffee beans: = 1C - 0.2C = 0.8C

Weight of a jar with coffee beans after the removal: 0.6C
Weight of the cofee beans after the removal: 0.6C - 0.2C = 0.4C

So we know now that initially we had 0.8C and now only 0.4C: \(\frac{0.4}{0.8} = \frac{1}{2}\)

Answer: D

fraction part of the beans remain in the jar=(\(\frac{60}{100}\)-\(\frac{20}{100}\))/\(\frac{80}{100}\)=\(\frac{40}{100}\)*\(\frac{100}{80}\)=\(\frac{1}{2}\)

Correct Answer D

The trick to this problem is realizing that the glass jar is representative of the glass jar + coffee beans. So once you figure out the weight of the jar, you can subtract it from the weight of jar + beans to figure out weight of the beans.

Say weight of jar + coffee = 100lb --> weight of jar = 20lb --> coffee = 80lb

weight of jar + coffee changes to 60lb --> weight of jar = 20 --> coffee = 40lb

40lb/80lb = 1/2

Given:
1. The weight of a glass of jar is 20% of the weight of the jar filled with coffee beans.
2. After some of the beans have been removed, the weight of the jar and the remaining beans is 60% of the original total weight.

Asked: What fraction part of the beans remain in the jar?


The weight of a glass of jar is 20% of the weight of the jar filled with coffee beans.
Let the weight of coffee beans be x and weight of jar be y
y = 20% (x+y)
80%y = 20%x
y = .25x

After some of the beans have been removed, the weight of the jar and the remaining beans is 60% of the original total weight.
Let the fraction of beans remaining in the jar be k
y + kx = 60% (y+x)
.25x + kx = .6x + .6*.25x = .75x
kx = .5x
k = .5 = 1/2

IMO D

shahkevin99 please see the solution

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