A beaker contains 100 milligrams of a solution of salt and water that

Question

A beaker contains 100 milligrams of a solution of salt and water that is x% salt by weight such that  x < 90 . If the water evaporates at a rate of  y  milligrams per hour, how many hours will it take for the concentration of salt to reach  (x + 10) %, in terms of  x  and  y ?

(A)  x^x y^y=\frac{24^3}{2}

(B)  \frac{1000 + 99x}{xy+10y}

(C)  \frac{100y+100xy}{x+10}

(D)  \frac{1000}{xy+10y}

(E)  \frac{\sqrt{x+y}}{2}

ScottTargetTestPrep · Accepted Answer

Solution:

Initially, the beaker contains x milligrams of salt. Since salt does not evaporate, the amount of sat in the solution is x at all times.

Let the salt concentration reach (x + 10)% after n hours. Then, ny milligrams of water have evaporated and the solution is now 100 - ny milligrams in total. The concentration of salt in the solution is x/(100 - ny) and this must be equal to (x + 10)%. Thus,

x/(100 - ny) = (x + 10)/100

100x = (x + 10)(100 - ny)

100x = 100x -nxy + 1000 - 10ny

nxy - 10ny = 1000

n(xy - 10y) = 1000

n = 1000/(xy - 10y)

So, the solution will reach a concentration of (x + 10)% in 1000/(xy - 10y) hours.

Answer: D

nick1816 · Answer

Amount of solution after 'k' hours = 100-ky

\frac{x}{100-ky} = \frac{x+10}{100}

100x = 100x+1000-kyx -10ky

k(xy+10y) = 1000

k = \frac{1000}{(xy+10y)}

Fdambro294 · Answer

(1st) Let the Amount of Water that we need to have Evaporate (in mg) = W

Question asks us to Solve for the Number of Hours it takes for the needed water to evaporate ----> Let this time = H

At a Rate of Y (mg per 1 hour) ------> the Water is Evaporating from the Solution

If we need W mg to evaporate in H hours at a Rate of Y (mg per 1 hour) ----> then:

H = W / Y

or

W = H * Y

(2nd)Originally we have (X)% Salt in 100 mg Solution --------> Water is going to Evaporate such that the New Solution will have (100 - W) mg and (X + 10)% Salt

NOTE: only Water is evaporating, so the Quantity of Salt that is in the Solution BEFORE = Quantity of Salt in the Solution AFTER the Water evaporates

Set up the Proportion:

(Amount of Salt) / (Total Amount of Solution - W mg of water that will evaporate) = (X + 10)%
*****************************************

Amount of Salt = (X)% * 100 mg

Total Amount of Solution = 100 mg

After W mg of Water Evaporates, the NEW Total Amount of Solution = (100 - W) mg
*****************************************

/   = (X + 10)%

Finally, from above at (1st) ------> at Y mg per 1 hour for a Total of H hours, the Quantity of Water evaporating is:

W = Y * H ------> Substitute into Above equation

/   = (X + 10)% ——(this is the Equation we need to Manipulate and solve for H-hours)

(3rd) We need to Isolate H to answer the Question

/   = (X + 10)%

/   = (X + 10) / (100)

---Cross-Multiply-----

100X = (100 - YH) * (X + 10)

100X = 100X + 1,000 - XYH - 10YH

--Cancel 100X on Each Side and Group Like Terms and Factor Out H-----

XYH + 10YH = 1,000

H * (XY + 10Y) = 1,000

H = 1,000 / (XY + 10Y)

-D- is the Correct Answer

RandomStoryteller · Answer

Total Solution (T1) = 100 mg
Salt(s1) = x % of Total Solution(100 mg) = x
Water(w1) = 100 - x

After evaporation, the salt content remains the same. But, now the salt constitutes (x +10) per cent of the solution.
x = (x+10)% of New total(T2) (This new total will be formed once the water is evaporated from solution)
New total (T2) = 100x /(x +10)

Composition of Water and Salt in New total:
Total Solution(T2) = 100x/(x+10)
Salt (S2) = x (Salt is not evaporated, the content will remain the same).
Water(W2) = New Total (T2) - Salt (S2) = 100x/(x+10) - x.

Now, we need to find the time that was needed for evaporation= Total Water Evaporated/ Rate of Evaporation.
Total Water Evaporated = Original Water (W1) - New Water(W2) = (100 - x) -   = 1000/(x+10)
Rate of evaporation = y mg/hour.
Total Time = 1000/(x+10) divided by y = 1000/xy +10y.
Option D.

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=82432 ___ GMAT-Club-Forum-ofp72j5y.png

axeofflane · Answer

Smart numbers worked for me, really easy here

Let Salt % = 10 = x

Let y = 5

So salt to be 20% (Which is x+10%), we need the water level to be 40 ml from the original 90ml, as the 100mg was split into 90 mg water, 10 mg salt

Therefore for 90 water to reach 40, it would take 50/5 = 10 hrs

Now plug in the x and y into the answers and see which gives 10

(D) Comes out pretty easy

Advait01 · Answer

The question seems pretty hard after initial reading but is actually pretty easy if u form an algebric equation

pappal · Answer

given 100 mg of salt water solution of which salt is x%=x mg and water=100-x mg
water evaporates at y mg/hr 
 let in t hrs the required (x+10)% of salt is reached 
so amount of water evaporated=ty mg
 amount of solution left =100-ty mg
while the amount of salt remains same i.e. x mg
so concentration of salt after evaporation=x/(100-ty) which is given as (x+10)/100
100x=100x+1000-tyx-10ty
ty(x+10)=1000
or t=1000/(xy+10y) D

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