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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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24 Jan 2016, 06:29
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77% (01:52) correct 23% (02:29) wrong based on 120 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 Status: Greatness begins beyond your comfort zone Joined: 08 Dec 2013 Posts: 2448 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 $0.75 [#permalink] Show Tags 24 Jan 2016, 09:18 1 1 Let weight of tea worth .75$ = 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
has to be more than 50% of 10 pound , i.e 5 pounds . If w1=w2 , then resultant mix would have been of .84 $. We can also eliminate option E because then the resultant mix will be very close to .93 , only marginally less . _________________ When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long SVP Joined: 26 Mar 2013 Posts: 2344 Concentration: Operations, Strategy Schools: Erasmus '21 (M$)
How much tea worth $0.93 per pound must be mixed with tea worth$0.75  [#permalink]

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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.

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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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30 Dec 2017, 23:05
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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. _________________ SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here. Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it's the only thing that ever has -- Margaret Mead Manager Joined: 24 Jul 2014 Posts: 88 Location: India WE: Information Technology (Computer Software) Re: How much tea worth$0.93 per pound must be mixed with tea worth $0.75 [#permalink] Show Tags 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 _________________ The Mind ~ Muscle Connection My GMAT Journey is Complete.Going to start the MBA in Information Management from 2016 Good Luck everyone. Intern Joined: 16 Feb 2016 Posts: 49 Concentration: Other, Other Re: How much tea worth$0.93 per pound must be mixed with tea worth $0.75 [#permalink] Show Tags 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. Intern Joined: 25 Sep 2016 Posts: 5 Schools: MBS '18, LBS MiM"18 WE: Analyst (Consulting) Re: How much tea worth$0.93 per pound must be mixed with tea worth $0.75 [#permalink] Show Tags 06 Oct 2016, 04:33 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!
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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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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 = $$\frac{50}{9}$$= $$5$$ $$\frac{5}{9}$$ _________________ "Do not watch clock; Do what it does. KEEP GOING." 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 Joined: 04 Jan 2015 Posts: 3230 Re: How much tea worth$0.93 per pound must be mixed with tea worth $0.75 [#permalink] Show Tags 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)

Regards,
Saquib
e-GMAT
Quant Expert
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Re: How much tea worth $0.93 per pound must be mixed with tea worth$0.75   [#permalink] 31 Dec 2017, 09:23
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