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Snowcat [4.5K]
3 years ago
7

Brine is a solution of salt and water. If a tub contains 50 pounds of 5% solution of brine, how much water (to the nearest tenth

lb.) must evaporate to change it to an 8% solution?

Mathematics
1 answer:
antiseptic1488 [7]3 years ago
5 0
The only important thing here is the ratio between the salt and the entire solution. The salt makes up 5% of the entire solution, which is 50lbs. 5% of 50lbs is 2.5lbs. So, we want to know how much brine must be left to create a 8% solution with 2.5lbs of salt. To do this, make the amount of brine solution a variable, x. So, you know that 2.5 should be 8% of x. This can be modeled by the equation 2.5/x = 0.08. Solve for x. Now that you've got the amount of brine, 31.25lbs, you've got to subtract that from the original to see how much water has evaporated. 50 - 31.25 = 18.75. Round up to 18.8lbs and there's your answer!

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What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?
olasank [31]

First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.

We have

130 = 2 • 5 • 13

231 = 3 • 7 • 11

so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.

To verify the claim, we try to solve the system of congruences

\begin{cases} 17x \equiv 1 \pmod{130} \\ 17y \equiv 1 \pmod{231} \end{cases}

Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:

130 = 7 • 17 + 11

17 = 1 • 11 + 6

11 = 1 • 6 + 5

6 = 1 • 5 + 1

⇒   1 = 23 • 17 - 3 • 130

Then

23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)

so that x = 23.

Repeat for 231 and 17:

231 = 13 • 17 + 10

17 = 1 • 10 + 7

10 = 1 • 7 + 3

7 = 2 • 3 + 1

⇒   1 = 68 • 17 - 5 • 231

Then

68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)

so that y = 68.

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Combine like terms.

<h3>4t^2 + 3t - 3 is the simplified form of the given expression.</h3>
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