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Math Expert V
Joined: 02 Sep 2009
Posts: 57020
How much tea worth $0.93 per pound must be mixed with tea worth$0.75  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 77% (01:52) correct 23% (02:29) wrong based on 119 sessions

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 _________________ Verbal Forum Moderator V Status: Greatness begins beyond your comfort zone Joined: 08 Dec 2013 Posts: 2388 Location: India Concentration: General Management, Strategy Schools: Kelley '20, ISB '19 GPA: 3.2 WE: Information Technology (Consulting) Re: How much tea worth$0.93 per pound must be mixed with tea worth $0.75 [#permalink] ### Show Tags 1 1 Let weight of tea worth .75$ = 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

scanning the choices quickly, we need the resulted to have tea of $0.93 a bit higher than 5 pounds. Answer: D Senior SC Moderator V Joined: 22 May 2016 Posts: 3243 How much tea worth$0.93 per pound must be mixed with tea worth $0.75 [#permalink] ### Show Tags 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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Re: How much tea worth $0.93 per pound must be mixed with tea worth$0.75  [#permalink]

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

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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.
Intern  Joined: 25 Sep 2016
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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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1
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 -------------------------------------------------------------------------------------------- +1 Kudos if you find this post helpful. Thanks! SVP  V Status: It's near - I can see. Joined: 13 Apr 2013 Posts: 1689 Location: India Concentration: International Business, Operations Schools: INSEAD Jan '19 GPA: 3.01 WE: Engineering (Real Estate) How much tea worth$0.93 per pound must be mixed with tea worth $0.75 [#permalink] ### Show Tags 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.
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3019
Re: How much tea worth $0.93 per pound must be mixed with tea worth$0.75  [#permalink]

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) Regards, Saquib e-GMAT Quant Expert _________________ Re: How much tea worth$0.93 per pound must be mixed with tea worth $0.75 [#permalink] 31 Dec 2017, 09:23 Display posts from previous: Sort by # How much tea worth$0.93 per pound must be mixed with tea worth \$0.75

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