a. A chemist needs 200ml of a 1.5% acidic solution. Two other acidic solutions - 0.8% and 3.2% - are available for mixing. How much of each solution should be mixed to produce the desired solution?
b. A hotel has 260 rooms. Some are singles, and some are doubles. The singles cost $35 and the doubles cost $60. Because of a math teachers' convention, all of the hotel rooms are occupied. The sales for this night are $14,000. How many of each type of room does the hotel have?
(a) 0.8x + 3.2(200-x) = 1.5(200)
(b)
s+d = 260
35s+60d = 14000
now just solve the equations.
volume of the 0.8% solution -- x ml
volume of the 3.2% solution = 200-x ml
.08x + .032(200-x) = .015(200)
times 1000
80x + 32(200-x) = 15(200)
solve for x
b) number of single rooms --- x
number of double rooms = 260-x
35x + 60(260-x) = 14000
solve for x
$35 for a motel room??
Time to update those textbooks
a. To solve this problem, we can use the concept of mixtures. Let's assume that the chemist needs to mix x ml of the 0.8% solution and y ml of the 3.2% solution to obtain the desired 1.5% acidic solution.
The first step is to set up a system of equations based on the given information:
Equation 1: x + y = 200 (since the chemist needs a total of 200 ml)
Equation 2: 0.008x + 0.032y = 0.015 * 200 (since the final concentration is 1.5%)
To solve this system of equations, we can use either substitution or elimination method. Let's use the elimination method:
Multiply Equation 1 by 0.008 to make the coefficients of x in both equations equal:
0.008x + 0.008y = 0.008 * 200
Now, subtract Equation 2 from the new equation:
0.008x + 0.008y - 0.008x - 0.032y = 0.016
-0.024y = 0.016
y = 0.016 / -0.024 = -0.666...
Since we cannot have negative quantities, this solution is not valid. This implies that there is no combination of the two given solutions that can be mixed to obtain the desired 1.5% acidic solution.
b. Let's denote the number of single rooms as x and the number of double rooms as y.
From the given information, we can set up a system of equations:
Equation 1: x + y = 260 (since there are a total of 260 rooms)
Equation 2: 35x + 60y = 14,000 (since the sales from the room bookings are $14,000)
To solve this system of equations, we can again use either substitution or elimination method. Let's use the substitution method:
Rearrange Equation 1 to solve for x:
x = 260 - y
Substitute this value of x into Equation 2:
35(260 - y) + 60y = 14,000
9,100 - 35y + 60y = 14,000
25y = 14,000 - 9,100
25y = 4,900
y = 4,900 / 25
y = 196
Now substitute this value of y back into Equation 1 to solve for x:
x + 196 = 260
x = 260 - 196
x = 64
Therefore, the hotel has 64 single rooms and 196 double rooms.