Each of the cucumbers in 100 pounds of cucumbers is composed

Responding to a pm:

There are two ways of getting to the answer here.

Method 1:
Let's go forward to back. Say we have 98% mixture (2% pure cucumber extract and 98% water). Say, we also have plain water (100% water). If we mix these two, we get 100 pounds of 99% mixture (1% pure cucumber extract and 99% water). We want to find the amount of pure water that was mixed. (Usually, we are given two components and we have to mix them. This question does the opposite. It separates out the components of the mixture but overall it doesn't matter to us at all.)

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (100 - 99)/(99 - 98) = 1/1

The 98% mixture and the pure water must have been mixed in 1:1 ratio. Since total mixture is 100 pounds, the amount of pure water must be 50 pounds and the 98% mixture must have been 50 pounds.
Answer (B)

Method 2:
Another way to think about this problem: The amount of pure cucumber extract stays the same. Only the amount of water reduces. The amount of pure cucumber extract was 1 pound in the 99% mixture and it stayed 1 pound in the 98% mixture.
In 100 pounds of 99% cucumber mix, this pure cucumber extract was 1% and in 98% cucumber mix, this same pure cucumber extract was 2% of the cucumber mix.

Say, the total weight of 98% cucumber mix is x pounds.
1 pound = (2/100)*x
x = 50 pounds

The answer is a little un-intuitive since you don't expect so much water to disappear to just make it go down from 99% to 98% but instead, think that you need to make the amount of pure cucumber extract to go up from 1% to 2% i.e. you need it to double in proportion. For the same amount (1 pound) to be double in proportion, you need to reduce the total weight by half.

Nice explanation Bunuel

Hi Bunnel,
Would it be possible explain your solution? For some reason I am not getting it that how cucumber will be 50 pound after reduction of 1% water.
Thanks

Before some of the water evaporates: 100 pounds of cucumbers is composed of 99% water (= 99 pounds of water) AND 1% non-water (= 1 pound of non-water);

After some of the water evaporates: 98% water by weight AND the same 1 pound of non-water, which did not evaporate. This 1 pound of non-water now composes 100%-98%=2% of the total weight of cucumbers. So, 1 pound = 2% of the total weight of cucumbers --> total weight of cucumbers = 1/0.02 = 50 pounds.

Hope it's clear.

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I don't think any formula is needed in such questions.

It is pure logic which governs these questions.

Approach:

When one thing is getting replaced or removed just CONCENTRATE on the CONCENTRATION of that thing.

Concentrate here on 'Water'

99 Pounds = Water
1 pound = non water

Let X be evaporated so that concentration changes to 98%

99 - X = Water in new thing = 98% of total

99-X = 98/100 (100 - X)

X=50

Actually, it is easier to concentrate on the thing that does not change. See approach 2 above. But whatever works for you!

Hi Karishma,

how did you get to the right side of "1/1"?
Why did you state that both of the weights are the same?

The cucumbers left after some water has evaporated is 98% water. The evaporated water is 100% water.
These two were mixed together in the original 100 pounds of cucumber. The original 100 pounds of cucumbers had 99% water.

The formula of the scale method gives us:
So w1/w2 = (A2 - Aavg)/(Aavg - A1)
A1 - Water concentration of cucumber leftover = 98%
A2 - Water concentration of pure water = 100%
w1 - weight of cucumber leftover
w2 - weight of water

w1/w2 = (100 - 99)/(99 - 98) = 1/1

So we can say that the weight of cucumber leftover and weight of water should be the same i.e. they both must be 50 pounds each.

1 portion is non water ---- 99 portions are water
1% 99%
now after vaporization
The 1 portion became 2% ------> we have to calculate 1 is 2% of what?
x * 0.02= 1 ----> x= 100/2 ----> x=50 "the new whole.
hope this one is simple enough!

My 2 cents on mixtures and allegations..

Always deal with the quantity that does not change..

Mass content here does not change..hence play with this info

Say Water is added to milk...Content of milk does not change..play with milk info..

weight - 99-x (water), 1(other)

ratio - 98(water), 2(other)

(99-x)/1 = 98/2
x=50 i.e, 50 pounds of water evaporates
Remaining is 50 pounds.

Sol:



100-x/x =1/1

2x= 100

x=50

Logically:

We should concentrate on non-water component as not changing in number
1% of non-water ---> 2% of non-water
doubling the concentration that can be only if we halve the total weight, so 100/2=50

Algebraic:

(99-x/100-x)*100=98
100=2x
x=50, so 100-50=50

B

100 pounds of cucumbers have 99 pounds water and 1 pound solid.

Now 1 pound solid is 2% of the total cucumber weight.

So 98% of the cucumber which is water is 49 pounds. [ 2% -> 1 what is 98% ? ......98% is 49]

Therefore the total cucumber weight is 50 pounds.

100 pound cucumber Haas 99% water I.e. 1% is the pulp in it

On water evaporation, pulp remains same I.e. 1 pound pulp now should serve the purpose of 2% of total weight of leftover cucumber now because in new weight water is 98%.

1 pound = (2/100)*New weight
I.e. New Weight = 50 Pound

Answer : Option B

weight * concentration = constant

w1*c1 = w2*c2

100 * 1/100 = w2 * 2/100

w2 = 50 pounds.