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stepladder [879]
3 years ago
5

A chef is going to use a mixture of two brands of Italian dressing. The first brand contains 6% vinegar, and the second brand co

ntains 11% vinegar. The chef wants to make 380 milliliters of a dressing that is 7% vinegar. How much of each brand should she use?
Mathematics
1 answer:
rewona [7]3 years ago
6 0

Answer: To make a 310mL mixture that contains 8% vinegar, use 186mL of the first brand and 124mL of the second brand.

Step-by-step explanation:We want the total mixture to be 310mL so: F + S = 310

 

Also, you want the total mixture to contain 8% vinegar so: .06F + .11S = .08(310)

 

Now, we can solve the system of equations using substitution.

 

F + S = 310

.06F + .11S = .08(310)

 

F = 310 - S      [solve 1st equation for F]

.06(310 - S) + .11S = .08(310)     [substitute into 2nd equation]

18.6 -.06S +.11S = 24.8

.05S = 6.2

S = 124

 

F + 124 = 310     [substitute answer for S back into original equation]

F = 186

 

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It looks like you have to find the value of the sum,

C = \displaystyle \frac1{21\times22\times23} + \frac1{22\times23\times24} + \cdots + \frac1{200\times201\times202}

so that the <em>n</em>-th term in the sum is

\dfrac1{(21+(n-1))\times(21+n)\times(21+(n+1))} = \dfrac1{(n+20)(n+21)(n+22)}

for 1 ≤ <em>n</em> ≤ 180.

We can then write the sum as

\displaystyle C = \sum_{n=1}^{180} \frac1{(n+20)(n+21)(n+22)}

Break up the summand into partial fractions:

\dfrac1{(n+20)(n+21)(n+22)} = \dfrac a{n+20} + \dfrac b{n+21} + \dfrac c{n+22}

Combine the fractions into one with a common denominator and set the numerators equal to one another:

1 = a(n+21)(n+22) + b(n+20)(n+22) + c(n+20)(n+21)

Expand the right side and collect terms with the same power of <em>n</em> :

1 = a(n^2+43n+462)+b(n^2+42n+440) + c(n^2+41n + 420) \\\\ 1 = (a+b+c)n^2 + (43a+42b+41c)n + 462a+440b+420c

Then

<em>a</em> + <em>b</em> + <em>c</em> = 0

43<em>a</em> + 42<em>b</em> + 41<em>c</em> = 0

462<em>a</em> + 440<em>b</em> + 420<em>c</em> = 1

==>   <em>a</em> = 1/2, <em>b</em> = -1, <em>c</em> = 1/2

Now our sum is

\displaystyle C = \sum_{n=1}^{180} \left(\frac1{2(n+20)}-\frac1{n+21}+\frac1{2(n+22)}\right)

which is a telescoping sum. If we write out the first and last few terms, we have

<em>C</em> = 1/(2×21) - 1/22 <u>+ 1/(2×23)</u>

… … + 1/(2×22) - 1/23 <u>+ 1/(2×24)</u>

… … <u>+ 1/(2×23)</u> - 1/24 <u>+ 1/(2×25)</u>

… … <u>+ 1/(2×24)</u> - 1/25 <u>+ 1/(2×26)</u>

… … + … - … + …

… … <u>+ 1/(2×198)</u> - 1/199 <u>+ 1/(2×200)</u>

… … <u>+ 1/(2×199)</u> - 1/200 + 1/(2×201)

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which we simplify further to

<em>C</em> = 1/42 - 1/44 - 1/402 + 1/404

<em>C</em> = 1,115/1,042,118 ≈ 0.00106994

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