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How much tea worth $0.93 per pound must be mixed with tea worth $0.75

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How much tea worth $0.93 per pound must be mixed with tea worth $0.75  [#permalink]

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New post 24 Jan 2016, 06:29
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How much tea worth $0.93 per pound must be mixed with tea worth $0.75 per pound to produce 10 pounds worth $0.85 per pound?

A. 2 2⁄9
B. 3 1⁄2
C. 4 4⁄9
D. 5 5⁄9
E. 9 1⁄2

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Re: How much tea worth $0.93 per pound must be mixed with tea worth $0.75  [#permalink]

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New post 24 Jan 2016, 09:18
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1
Let weight of tea worth .75 $ = w1
and weight of tea worth .93 $ = w2
w1 + w2 = 10 --- equation 1

.85 = ( w1*.75 + w2* .93 )/(w1+ w2 )
=> .85 w1 + .85 w2 = .75 w1 + .93 w2
=> .10 w1 -.08 w2 = 0
=> 10 w1 - 8 w2 = 0 --- equation 2
From 1 ,
8 w1 + 8 w2 = 80 --- equation 3
From equations 2 and 3 , we get
18 w1 = 80
=> w1 = 40/9
and w2 = 50/9
Answer D , which is in mixed fraction.

Alternatively , we can use scale method
.75 -----.85 ---- .93
.85 is at a distance of .1 from .75 and .08 from .93
The distances of .85 from .75 and 93 are in ratio of 10 : 8 , that is 5:4
Therefore , the weights will be in inverse proporation 4:5
Therefore , w2 = 5/(4+5) * 10 = 50/9
Alternatively, We can also use estimation here and eliminate options A, B and C as the weight of .93 $ variant
has to be more than 50% of 10 pound , i.e 5 pounds . If w1=w2 , then resultant mix would have been of .84 $.
We can also eliminate option E because then the resultant mix will be very close to .93 , only marginally less .
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Re: How much tea worth $0.93 per pound must be mixed with tea worth $0.75  [#permalink]

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New post 25 Jan 2016, 09:59
1
Let X and Y pounds to mixed
so X+ Y =10
Again from price we have
0.93X+0.75Y=0.85X+0.85Y
=>0.08X=0.1Y
=>4X=5Y
=>Y=4X/5

Since X+Y=10 =>X+4X/5 =10
=>X=5 5/9
Answer = D
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How much tea worth $0.93 per pound must be mixed with tea worth $0.75  [#permalink]

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New post 25 Jan 2016, 12:01
1
if we use equal amount (i.e 5 pounds from each tea type), then the resulted mix will equal to the mean of two types which is $0.84

scanning the choices quickly, we need the resulted to have tea of $0.93 a bit higher than 5 pounds.

Answer: D
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Re: How much tea worth $0.93 per pound must be mixed with tea worth $0.75  [#permalink]

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New post 08 Mar 2016, 00:22
0.93x+(10-x)*0.75=10*0.85
Multiply everything by 100

93x+750-75x=850

x= (850-750)/(93-75)
x=100/18 ~ 100/20
answer is slightly over 5.
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Re: How much tea worth $0.93 per pound must be mixed with tea worth $0.75  [#permalink]

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New post 06 Oct 2016, 04:33
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How much tea worth $0.93 per pound must be mixed with tea worth $0.75 per pound to produce 10 pounds worth $0.85 per pound?

A. 2 2⁄9
B. 3 1⁄2
C. 4 4⁄9
D. 5 5⁄9
E. 9 1⁄2
--------------------------------------------------------------------------------------------
Cost of Tea A = $0.93 per pound
Cost of Tea B = $0.75 per pound
Cost of Tea A and Tea B Mixture = $0.85 per pound

Using Alligation Method

Tea A------------------Tea B

$0.93------------------$0.75
-----------$0.85------------
$0.10------------------$0.08

Ratio of Tea A to Tea B is 10:8 or 5:4

In order to find the weight of Tea A in the mixture
= Ratio of Tea A / Ratio of Tea A + Ratio of Tea B * Total weight of Tea A and Tea B mixture
= 5 / 5+4 * 10 pounds
= 5 / 9 * 10
= 50 / 9
= 5 5⁄9
--------------------------------------------------------------------------------------------
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How much tea worth $0.93 per pound must be mixed with tea worth $0.75  [#permalink]

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New post 30 Dec 2017, 23:05
1
Bunuel wrote:
How much tea worth $0.93 per pound must be mixed with tea worth $0.75 per pound to produce 10 pounds worth $0.85 per pound?

A. 2 2⁄9
B. 3 1⁄2
C. 4 4⁄9
D. 5 5⁄9
E. 9 1⁄2


A weighted average approach, where one kind of tea is worth an average of $0.93 per pound and another kind of tea is worth an average of .$75 per pound.

One variable in the weighted average formula

Let \(A\) = # of pounds of tea worth $0.93/lb
Let \(B\) = # of pounds of tea worth $0.75/lb
\(A + B = 10\) pounds
\(B = (10 - A)\) pounds

\(.93(A) + .75(B) = .85(A+B=10)\)

\(.93A +.75(10 - A) = .85(10)\)

\(.93 A + 7.5 - .75A = 8.5\)

\(.18A = 1\)


\(A=\frac{1}{.18}=\frac{100}{18}=\frac{50}{9}= 5\frac{5}{9}\)

ANSWER D

Two variables in the weighted average formula

You could also find the ratio of A to B and then use total weight to find number of pounds of A:

\(.93(A) + .75(B) = .85(A + B)\)

\(.93A + .75B = .85A + .85B\)

\(.08A = .10B\)

\(8A = 10B\)

\(B = \frac{8}{10} = \frac{4}{5}A\)


We know that together the two teas weigh 10 pounds.
\(A + B = 10\)

\(A +\frac{4}{5}A = 10\)
\(5A + 4A = 50\)

\(9A = 50\)

\(A = \frac{50}{9} = 5\frac{5}{9}\)


ANSWER D


*where\(A = w_1, 0.93 = A_1, B = w_2, 0.75 = A_2\)

From this weighted average formula

\(\frac{w_1}{w_2} = \frac{(A_2 – A_{avg})}{(A_{avg} – A_1)}\),

we can derive this weighted average formula

\(A_{avg}= \frac{(A_1*w_1) + (A_2*w_2)}{(w_1 + w_2)}\)

And we can multiply both sides of that formula by the RHS denominator to derive yet another weighted average formula:

\((A_1)(w_1) + (A_2)(w_2) = A_{avg}(w_1 + w_2)\).

In these formulas, an average can be a straightforward average, a percentage, a concentration (e.g. of vinegar in a solution), a rate (e.g. an interest rate), and more. Preferences vary. But they are all weighted average formulas.

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How much tea worth $0.93 per pound must be mixed with tea worth $0.75  [#permalink]

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New post Updated on: 31 Dec 2017, 23:28
Bunuel wrote:
How much tea worth $0.93 per pound must be mixed with tea worth $0.75 per pound to produce 10 pounds worth $0.85 per pound?

A. 2 2⁄9
B. 3 1⁄2
C. 4 4⁄9
D. 5 5⁄9
E. 9 1⁄2



Method - 1 (Making equations)

Say, Tea A = $ 0.93/pound

Tea B = $ 0.75/pound

We know A + B = 10 -------> (I); as we want the quantity of A to be mixed with B, better to solve for A

Therefore B = 10 - A

Equation becomes, 0.93A + 0.75B = 0.85 (A + B)

93A + 75B = 85 (A+B) ------>(II)

From (I) & (II ) we get 93A + 75(10 - A) = 85 * 10

93A + 750 - 75A = 850
18A = 100

A = 100/18 = \(5\) \(\frac{5}{9}\)

(D)

Method -2 (Weighted average)

\(\frac{Wt A}{Wt B}\) = \(\frac{0.93 - 0.85}{0.85 - 0.75}\)

\(\frac{Wt A}{Wt B}\)= \(\frac{8}{10}\) = \(\frac{4}{5}\)

\(\frac{Wt A}{Wt B}\) (weights will be in inverse proportion) =\(\frac{5}{4}\)

(We have total 5 + 4 = 9 units. average 0.85 is more closer to 0.93 than to 0.75. This means we have more A than B)

Wt A = \(\frac{5}{9}\)*10 = \(\frac{50}{9}\)= \(5\) \(\frac{5}{9}\)
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Originally posted by AkshdeepS on 31 Dec 2017, 00:58.
Last edited by AkshdeepS on 31 Dec 2017, 23:28, edited 2 times in total.
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Re: How much tea worth $0.93 per pound must be mixed with tea worth $0.75  [#permalink]

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New post 31 Dec 2017, 09:23
Bunuel wrote:
How much tea worth $0.93 per pound must be mixed with tea worth $0.75 per pound to produce 10 pounds worth $0.85 per pound?

A. 2 2⁄9
B. 3 1⁄2
C. 4 4⁄9
D. 5 5⁄9
E. 9 1⁄2


Method ( Using Alligation)



Let X be the first variety of Tea : \($0.93\) per pound and

Let Y be the second variety of Tea : \($ 0.75\) per pound.

The final mixture Z costs \($ 0. 85\) per pound

The ratio in which X and Y are mixed \(= (0.85 - 0.75) : (0.93 - 0.85) = 10 : 8 = 5 : 4\)

Thus, the amount of X in the 10 pound of variety Y is \(\frac{5}{{(5+4)}} * 10 = 5 \frac{5}{9}\) (Option D)


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Re: How much tea worth $0.93 per pound must be mixed with tea worth $0.75 &nbs [#permalink] 31 Dec 2017, 09:23
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