An alloy of copper and aluminum has 40% copper. An alloy of Copper and : Problem Solving (PS)

L

chetan2u

RC & DI Moderator

Joined: 02 Aug 2009

Status:Math and DI Expert

Posts: 11238

Own Kudos [?]: 32433 [3]

Given Kudos: 301

Send PM

Re: An alloy of copper and aluminum has 40% copper. An alloy of Copper and [#permalink]

09 Nov 2018, 11:15

2

Kudos

1

Bookmarks

Expert Reply

doomedcat wrote:

An alloy of copper and aluminum has 40% copper. An alloy of Copper and Zinc has Copper and Zinc in the ratio 2: 7. These two alloys are mixed in such a way that in the overall alloy, there is more aluminum than Zinc, and copper constitutes x% of this alloy. What is the range of values x can take?

A)30% ≤ x ≤ 40%
B)33.33% ≤ x ≤ 40%
C)32.25% ≤ x ≤ 40%
D)32.5% ≤ x ≤ 42%
E)35.5% ≤ x ≤ 42%

C:A = 40:60=2:3=2a:3a
C:Z=2:7=2b:7b....
Now 3a>7b.....a>7b/3
Also %of copper or C= (2a+2b)/(5a+9b)=x/100..

Now let us make use of choices...
If 42 fits in , A,B and C will get eliminated else D and E will get eliminated.
(2a+2b)/(5a+9b)=42/100......200a+200b=210a+378b......-10a=178b....
Negative value hence not possible, eliminate D and E as they contain 42%
Now let's try 33.33% or 100/3%
So (2a+2b)/(5a+9b)=100/100*3=1/3........6a+6b=5a+9b...a=3b >7b/3... possible
Or let X=30
(2a+2b)/(5a+9b)=30/100=3/10.....10a+10b=15a+27b.....-5a=27b... Not possible..
Eliminate A
Try a=33%
(2a+2b)/(5a+9b)=33/100.....200a+200b=165a+297b....35a=97b...a=97b/35.. possible
Eliminate B

Ans C

L

GMATinsight

GMAT Club Legend

Joined: 08 Jul 2010

Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator

Posts: 5975

Own Kudos [?]: 13465 [3]

Given Kudos: 124

Location: India

GMAT: QUANT+DI EXPERT

Schools: IIM (A) ISB '24

GMAT 1: 750 Q51 V41

WE:Education (Education)

Send PM

Re: An alloy of copper and aluminum has 40% copper. An alloy of Copper and [#permalink]

15 Nov 2018, 23:49

3

Kudos

Expert Reply

Archit3110 wrote:

doomedcat wrote:

An alloy of copper and aluminum has 40% copper. An alloy of Copper and Zinc has Copper and Zinc in the ratio 2: 7. These two alloys are mixed in such a way that in the overall alloy, there is more aluminum than Zinc, and copper constitutes x% of this alloy. What is the range of values x can take?

A)30% ≤ x ≤ 40%
B)33.33% ≤ x ≤ 40%
C)32.25% ≤ x ≤ 40%
D)32.5% ≤ x ≤ 42%
E)35.5% ≤ x ≤ 42%

GMATinsight : sir could you help in giving an alternate solution to this mixture problem..

An alloy of copper and aluminum has 40% copper

i.e. Copper in first alloy = 40%
i.e. every 10 kg of alloy has 4 kg copper and 6 kg of Aluminium

An alloy of Copper and Zinc has Copper and Zinc in the ratio 2: 7
i.e. Copper in Second alloy = (2/9)*100 = 22.22%
i.e. every 10 kg of alloy has 2.22 kg copper and 7.78 kg of Zinc

As question mentions "there is more aluminum than Zinc"
for first alloy to have 7.78 kg of aluminium the total weight of first alloy = 7.78(10/6) = 12.96 kg
i.e. first alloy must be more than 12.96 kg if weight of he second alloy is 10 kg
and weight of copper in 12.96 kg of first alloy = 5.19

i.e. If total weight of two alloys = 10+12.96 = 22.96 kg
then weight of copper = 5.19+2.22 = 7.4 kg

minimum percentage of copper in final alloy = (7.4/22.96)*100 = 32.25%

Also, Maximum percentage of copper can't exceed 40% even if the second alloy becomes 0%

i.e. 32.25 < x < 40

Although I don' think that Option C should include equal to sign alongwith < sign, but that seems the best answer available hence

Answer: Option C

Archit3110

L

ShankSouljaBoi

Director

Joined: 21 Jun 2017

Posts: 637

Own Kudos [?]: 532 [2]

Given Kudos: 4092

Location: India

Concentration: Finance, Economics

Schools: Rotman '23 (D) IIMA PGPX'22 (D) Desautels '23 (D) Schulich Sept '23 (D) WBS (D) IIML IPMX'25 (A)

GMAT 1: 660 Q49 V31

GMAT 2: 620 Q47 V30

GMAT 3: 650 Q48 V31

GPA: 3.1

WE:Corporate Finance (Non-Profit and Government)

Send PM

Re: An alloy of copper and aluminum has 40% copper. An alloy of Copper and [#permalink]

09 Nov 2018, 12:12

1

Kudos

1

Bookmarks

Let the quantity of 1st alloy be y -----> Cu:Al= (2/5)y : (3/5)y
Let the proportionality constant for second alloy be p ----> Cu:Zn = 2p:7p

The two alloys are mixed with condition : Al> Zn -----> 3/5y > 7p or 3/35y>p therefore max p can be 3/35y

Now, in the mixture the quantity of Cu is x%
-----> x = {Cu quantity / total quantity}*100 = {(2/5y + 2p) / ((2/5y + 2p)+3/5y + 7p)}*100= 20(2y+10p)/(y+9p).
or x= 20 + 20 (y+p)/ (y+9p).

Lets analyze (y+p)/ (y+9p). Max when p = 0 -----> x = 40 Min when p is max or p = 3/35y ------> x=32.5

Ans C

G

eabhgoy

Manager

Joined: 12 Apr 2011

Posts: 115

Own Kudos [?]: 198 [2]

Given Kudos: 86

Location: United Arab Emirates

Concentration: Strategy, Marketing

Schools: CBS '21 Yale '21 INSEAD Jan'21 Intake IMD '21 (A)

GMAT 1: 670 Q50 V31

GMAT 2: 720 Q50 V37 Verified score

GPA: 3.2

WE:Marketing (Telecommunications)

Send PM

Re: An alloy of copper and aluminum has 40% copper. An alloy of Copper and [#permalink]

15 Nov 2018, 05:18

2

Kudos

Two mixtures can be written in units as:

1) 40% Copper + 60% Aluminum
i.e. 40 units Copper & 60 units Aluminum
2) 2:7 ratio
i.e. 22.22% Copper + 77.78% Zinc
i.e. 22.22 units Copper and 77.78 units Zinc

Now we need to calculate the ranges i.e. we need to calculate the two ends of the spectrum.

Since 1st mixture has less concentration of Aluminum when compared to the concentration of Zinc in the 2nd mixture. Hence we will need to mix more of mixture 1 to take up the % of aluminum in the final mixture.

One of the ranges is quite straightforward i.e. if we keep on adding more of the 1st mixture then the maximum concentration for copper can be 40% i.e. only A, B, C remain. (D & E are eliminated).

Next, for the lower range, let’s take the units of Aluminum just above the units of Zinc i.e. multiply the first mixture by 1.3 i.e:

1st Mixture = 40*1.3 & 60*1.4 i.e. 52 units of copper and 78 units of aluminum

By doing the above we take the units of aluminum (78) just above that of Zinc (77.78)

Now calculate the total concentration of copper in the final mixture = \(\frac{(52+22.22)}{(130+100)}\) = \(\frac{74.22}{230}\) ~ 32.25%

Hence 32.25% =< X =< 40%
i.e. the correct answer is C

L

GMATinsight

GMAT Club Legend

Joined: 08 Jul 2010

Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator

Posts: 5975

Own Kudos [?]: 13465 [2]

Given Kudos: 124

Location: India

GMAT: QUANT+DI EXPERT

Schools: IIM (A) ISB '24

GMAT 1: 750 Q51 V41

WE:Education (Education)

Send PM

Re: An alloy of copper and aluminum has 40% copper. An alloy of Copper and [#permalink]

12 Feb 2020, 06:26

2

Kudos

Expert Reply

dave13 wrote:

GMATinsight wrote:

doomedcat wrote:

An alloy of copper and aluminum has 40% copper. An alloy of Copper and Zinc has Copper and Zinc in the ratio 2: 7. These two alloys are mixed in such a way that in the overall alloy, there is more aluminum than Zinc, and copper constitutes x% of this alloy. What is the range of values x can take?

A)30% ≤ x ≤ 40%
B)33.33% ≤ x ≤ 40%
C)32.25% ≤ x ≤ 40%
D)32.5% ≤ x ≤ 42%
E)35.5% ≤ x ≤ 42%

GMATinsight : sir could you help in giving an alternate solution to this mixture problem..

An alloy of copper and aluminum has 40% copper

i.e. Copper in first alloy = 40%
i.e. every 10 kg of alloy has 4 kg copper and 6 kg of Aluminium

An alloy of Copper and Zinc has Copper and Zinc in the ratio 2: 7
i.e. Copper in Second alloy = (2/9)*100 = 22.22%
i.e. every 10 kg of alloy has 2.22 kg copper and 7.78 kg of Zinc

As question mentions "there is more aluminum than Zinc"
for first alloy to have 7.78 kg of aluminium the total weight of first alloy = 7.78(10/6) = 12.96 kg
i.e. first alloy must be more than 12.96 kg if weight of he second alloy is 10 kg
and weight of copper in 12.96 kg of first alloy = 5.19

i.e. If total weight of two alloys = 10+12.96 = 22.96 kg
then weight of copper = 5.19+2.22 = 7.4 kg

minimum percentage of copper in final alloy = (7.4/22.96)*100 = 32.25%

Also, Maximum percentage of copper can't exceed 40% even if the second alloy becomes 0%

i.e. 32.25 < x < 40

Although I don' think that Option C should include equal to sign alongwith < sign, but that seems the best answer available hence

Answer: Option C

Archit3110

Hi GMATinsight

can you pleaseprovide logical insight into this --- > why are we multiplying 7.78 by (10/6)

why are multiplying by 10/6 and not by 6/10

shouldnt concentration of aluminium be on top of fraction

thanks

dave13

7.78 = Al
Al = (6/10) total weight
i.e. 7.78 = (6/10) total weight

i.e. Total Wt. = (10/6)*7.78

I hope this helps!!!

L

Archit3110

GMAT Club Legend

Joined: 18 Aug 2017

Status:You learn more from failure than from success.

Posts: 8025

Own Kudos [?]: 4119 [0]

Given Kudos: 242

Location: India

Concentration: Sustainability, Marketing

GMAT Focus 1:
545 Q79 V79 DI73 Verified score

GPA: 4

WE:Marketing (Energy and Utilities)

Send PM

Re: An alloy of copper and aluminum has 40% copper. An alloy of Copper and [#permalink]

15 Nov 2018, 03:52

doomedcat wrote:

An alloy of copper and aluminum has 40% copper. An alloy of Copper and Zinc has Copper and Zinc in the ratio 2: 7. These two alloys are mixed in such a way that in the overall alloy, there is more aluminum than Zinc, and copper constitutes x% of this alloy. What is the range of values x can take?

A)30% ≤ x ≤ 40%
B)33.33% ≤ x ≤ 40%
C)32.25% ≤ x ≤ 40%
D)32.5% ≤ x ≤ 42%
E)35.5% ≤ x ≤ 42%

GMATinsight : sir could you help in giving an alternate solution to this mixture problem..

P

gracie

VP

Joined: 07 Dec 2014

Posts: 1071

Own Kudos [?]: 1572 [0]

Given Kudos: 27

Send PM

An alloy of copper and aluminum has 40% copper. An alloy of Copper and [#permalink]

Updated on: 16 Nov 2018, 12:15

doomedcat wrote:

An alloy of copper and aluminum has 40% copper. An alloy of Copper and Zinc has Copper and Zinc in the ratio 2: 7. These two alloys are mixed in such a way that in the overall alloy, there is more aluminum than Zinc, and copper constitutes x% of this alloy. What is the range of values x can take?

A)30% ≤ x ≤ 40%
B)33.33% ≤ x ≤ 40%
C)32.25% ≤ x ≤ 40%
D)32.5% ≤ x ≤ 42%
E)35.5% ≤ x ≤ 42%

let a and b=respective weights of alloys
because aluminum must exceed zinc in mixture,
3/5*a>7/9*b
a/b>35/27
35/27=1.296
to ensure that aluminum barely exceeds zinc,
let a/b=1.3/1
let x=% of copper in alloy mixture
.4*1.3+(2/9)*1=x*(1.3+1)
x=32.26%
C

Originally posted by gracie on 15 Nov 2018, 16:11.
Last edited by gracie on 16 Nov 2018, 12:15, edited 1 time in total.

L

Archit3110

GMAT Club Legend

Joined: 18 Aug 2017

Status:You learn more from failure than from success.

Posts: 8025

Own Kudos [?]: 4119 [0]

Given Kudos: 242

Location: India

Concentration: Sustainability, Marketing

GMAT Focus 1:
545 Q79 V79 DI73 Verified score

GPA: 4

WE:Marketing (Energy and Utilities)

Send PM

Re: An alloy of copper and aluminum has 40% copper. An alloy of Copper and [#permalink]

16 Nov 2018, 10:19

GMATinsight wrote:

Archit3110 wrote:

doomedcat wrote:

An alloy of copper and aluminum has 40% copper. An alloy of Copper and Zinc has Copper and Zinc in the ratio 2: 7. These two alloys are mixed in such a way that in the overall alloy, there is more aluminum than Zinc, and copper constitutes x% of this alloy. What is the range of values x can take?

A)30% ≤ x ≤ 40%
B)33.33% ≤ x ≤ 40%
C)32.25% ≤ x ≤ 40%
D)32.5% ≤ x ≤ 42%
E)35.5% ≤ x ≤ 42%

GMATinsight : sir could you help in giving an alternate solution to this mixture problem..

An alloy of copper and aluminum has 40% copper

i.e. Copper in first alloy = 40%
i.e. every 10 kg of alloy has 4 kg copper and 6 kg of Aluminium

An alloy of Copper and Zinc has Copper and Zinc in the ratio 2: 7
i.e. Copper in Second alloy = (2/9)*100 = 22.22%
i.e. every 10 kg of alloy has 2.22 kg copper and 7.78 kg of Zinc

As question mentions "there is more aluminum than Zinc"
for first alloy to have 7.78 kg of aluminium the total weight of first alloy = 7.78(10/6) = 12.96 kg
i.e. first alloy must be more than 12.96 kg if weight of he second alloy is 10 kg
and weight of copper in 12.96 kg of first alloy = 5.19

i.e. If total weight of two alloys = 10+12.96 = 22.96 kg
then weight of copper = 5.19+2.22 = 7.4 kg

minimum percentage of copper in final alloy = (7.4/22.96)*100 = 32.25%

Also, Maximum percentage of copper can't exceed 40% even if the second alloy becomes 0%

i.e. 32.25 < x < 40

Although I don' think that Option C should include equal to sign alongwith < sign, but that seems the best answer available hence

Answer: Option C

Archit3110

GMATinsight , sir thanks understood the working, but doing such a question under 2 mins is definitely a challenge ..

L

dave13

VP

Joined: 09 Mar 2016

Posts: 1157

Own Kudos [?]: 1018 [0]

Given Kudos: 3851

Send PM

Re: An alloy of copper and aluminum has 40% copper. An alloy of Copper and [#permalink]

11 Feb 2020, 04:13

GMATinsight wrote:

Archit3110 wrote:

doomedcat wrote:

An alloy of copper and aluminum has 40% copper. An alloy of Copper and Zinc has Copper and Zinc in the ratio 2: 7. These two alloys are mixed in such a way that in the overall alloy, there is more aluminum than Zinc, and copper constitutes x% of this alloy. What is the range of values x can take?

A)30% ≤ x ≤ 40%
B)33.33% ≤ x ≤ 40%
C)32.25% ≤ x ≤ 40%
D)32.5% ≤ x ≤ 42%
E)35.5% ≤ x ≤ 42%

GMATinsight : sir could you help in giving an alternate solution to this mixture problem..

An alloy of copper and aluminum has 40% copper

i.e. Copper in first alloy = 40%
i.e. every 10 kg of alloy has 4 kg copper and 6 kg of Aluminium

An alloy of Copper and Zinc has Copper and Zinc in the ratio 2: 7
i.e. Copper in Second alloy = (2/9)*100 = 22.22%
i.e. every 10 kg of alloy has 2.22 kg copper and 7.78 kg of Zinc

As question mentions "there is more aluminum than Zinc"
for first alloy to have 7.78 kg of aluminium the total weight of first alloy = 7.78(10/6) = 12.96 kg
i.e. first alloy must be more than 12.96 kg if weight of he second alloy is 10 kg
and weight of copper in 12.96 kg of first alloy = 5.19

i.e. If total weight of two alloys = 10+12.96 = 22.96 kg
then weight of copper = 5.19+2.22 = 7.4 kg

minimum percentage of copper in final alloy = (7.4/22.96)*100 = 32.25%

Also, Maximum percentage of copper can't exceed 40% even if the second alloy becomes 0%

i.e. 32.25 < x < 40

Although I don' think that Option C should include equal to sign alongwith < sign, but that seems the best answer available hence

Answer: Option C

Archit3110

Hi GMATinsight

can you pleaseprovide logical insight into this --- > why are we multiplying 7.78 by (10/6)

why are multiplying by 10/6 and not by 6/10

shouldnt concentration of aluminium be on top of fraction

thanks

bumpbot

Non-Human User

Joined: 09 Sep 2013

Posts: 32933

Own Kudos [?]: 828 [0]

Given Kudos: 0

Send PM

Re: An alloy of copper and aluminum has 40% copper. An alloy of Copper and [#permalink]

26 Feb 2024, 10:48

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

gmatclubot

Re: An alloy of copper and aluminum has 40% copper. An alloy of Copper and [#permalink]

26 Feb 2024, 10:48

GMAT Problem Solving (PS) Questions

An alloy of copper and aluminum has 40% copper. An alloy of Copper and