GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 17 Aug 2019, 04:39

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

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

Show Tags

New post 24 Jan 2016, 06:29
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

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

HideShow timer Statistics

Verbal Forum Moderator
User avatar
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)
GMAT ToolKit User Reviews Badge CAT Tests
Re: How much tea worth $0.93 per pound must be mixed with tea worth $0.75  [#permalink]

Show Tags

New post 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
SVP
User avatar
V
Joined: 26 Mar 2013
Posts: 2302
Reviews Badge CAT Tests
How much tea worth $0.93 per pound must be mixed with tea worth $0.75  [#permalink]

Show Tags

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
Senior SC Moderator
avatar
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

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.

_________________
SC Butler has resumed!
Get two SC questions to practice, whose links you can find by date, here.

What we do now echoes in eternity.—Marcus Aurelius
Manager
Manager
User avatar
B
Joined: 24 Jul 2014
Posts: 89
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

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
_________________
The Mind ~ Muscle Connection
My GMAT Journey is Complete.Going to start the MBA in Information Management from 2016
Good Luck everyone.
Intern
Intern
avatar
B
Joined: 16 Feb 2016
Posts: 49
Concentration: Other, Other
GMAT ToolKit User Reviews Badge
Re: How much tea worth $0.93 per pound must be mixed with tea worth $0.75  [#permalink]

Show Tags

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.
Intern
Intern
avatar
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

New post 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!
SVP
SVP
User avatar
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)
Reviews Badge
How much tea worth $0.93 per pound must be mixed with tea worth $0.75  [#permalink]

Show Tags

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 = \(\frac{50}{9}\)= \(5\) \(\frac{5}{9}\)
_________________
"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
User avatar
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]

Show Tags

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)


Regards,
Saquib
e-GMAT
Quant Expert
_________________
GMAT Club Bot
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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne