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705-805 Level|   Algebra|   Word Problems|      
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A beaker contains 100 milligrams of a solution of salt and water that 29 Oct 2020, 10:12
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D
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58% (03:04) correct 42% (03:02) wrong based on 468 sessions

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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) \(\frac{\sqrt{x+y}}{2}\)
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D
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A beaker contains 100 milligrams of a solution of salt and water that 05 Nov 2020, 12:34
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Sajjad1994
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) \(\frac{\sqrt{x+y}}{2}\)
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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
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A beaker contains 100 milligrams of a solution of salt and water that 29 Oct 2020, 13:41
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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

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


Sajjad1994
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) \(\frac{\sqrt{x+y}}{2}\)
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A beaker contains 100 milligrams of a solution of salt and water that Updated on: 31 Oct 2020, 07:44
Originally posted by Fdambro294 on 29 Oct 2020, 23:08.
Last edited by Fdambro294 on 31 Oct 2020, 07:44, edited 2 times in total.
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(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)% * 100 ] / [100 - W] = (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%) * 100 ] / [100 - YH] = (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%) * 100] / [100 - YH] = (X + 10)%

[ (X/100) * 100] / [100 - YH] = (X + 10) / (100)

[X] / [100 - YH] = (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
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A beaker contains 100 milligrams of a solution of salt and water that 30 Oct 2020, 00:04
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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) - [100x/(x+10) - x] = 1000/(x+10)
Rate of evaporation = y mg/hour.
Total Time = 1000/(x+10) divided by y = 1000/xy +10y.
Option D.
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A beaker contains 100 milligrams of a solution of salt and water that Updated on: 14 Jul 2025, 07:05
Originally posted by HarshavardhanR on 08 Jul 2024, 03:21.
Last edited by HarshavardhanR on 14 Jul 2025, 07:05, edited 1 time in total.
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A beaker contains 100 milligrams of a solution of salt and water that 13 Jul 2025, 07:13
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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
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