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An alloy of copper and aluminum has 40% copper. An alloy of Copper and Updated on: 09 Nov 2018, 20:50
Originally posted by doomedcat on 07 Nov 2018, 08:47.
Last edited by chetan2u on 09 Nov 2018, 20:50, edited 1 time in total.
Corrected the OA
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37% (02:58) correct 63% (03:18) wrong based on 262 sessions

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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%
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C
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An alloy of copper and aluminum has 40% copper. An alloy of Copper and 12 Feb 2020, 01:46
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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%
Show more

Alloy1
Cu:Al = 40:60 = 2 : 3 = 14 : 21

Alloy2
Cu:Zn = 2 : 7 = 6 : 21

If we mix 35 parts of alloy1 with 27 parts of alloy2, we will have 21 parts each of Al and Zn and (14 + 6 =) 20 parts of Cu.
So Cu will constitute 20/(35+27) = 20/62 = 32.25%

If we want more Al, we need alloy1 more which means the percentage of copper will increase because it has higher concentration of copper.

Answer must be (C)

Also note that since we want more Al, the entire mixture can be only alloy1 which will give 40% Cu and that is how you get the maximum value.
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An alloy of copper and aluminum has 40% copper. An alloy of Copper and 09 Nov 2018, 11:15
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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%
Show more

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
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An alloy of copper and aluminum has 40% copper. An alloy of Copper and 09 Nov 2018, 11:43
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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%
Show more

Other way would be..
(2a+2b)/(5a+9b)=X/100

MAx Value of X ... Reduce denominator, do put b=0
Hence 2a/5b=x/100....X=40%

Min value
When a=7b/3, so (2*7b/3+2b)/(5*7b/3+9b)=20b/62b=20/62=10/31=X/100
x=1000/31 ~32.25

So range is 32.25% ≤ x ≤ 40%

C

OA is wrong, it has to be C
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An alloy of copper and aluminum has 40% copper. An alloy of Copper and 09 Nov 2018, 12:12
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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
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An alloy of copper and aluminum has 40% copper. An alloy of Copper and 15 Nov 2018, 05:18
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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
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An alloy of copper and aluminum has 40% copper. An alloy of Copper and Updated on: 16 Nov 2018, 12:15
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.
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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%
Show more

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
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An alloy of copper and aluminum has 40% copper. An alloy of Copper and 15 Nov 2018, 23:49
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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%


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
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An alloy of copper and aluminum has 40% copper. An alloy of Copper and 11 Feb 2020, 04:13
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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
Show more
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 :)
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An alloy of copper and aluminum has 40% copper. An alloy of Copper and 12 Feb 2020, 06:26
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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 :)
Show more

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!!!
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An alloy of copper and aluminum has 40% copper. An alloy of Copper and 01 Jul 2025, 02:10
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Could you please explain why you have taken 35 parts of alloy1 and 27 parts of alloy2? I could not understand the reasoning behind that. Thanks
KarishmaB
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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%

Alloy1
Cu:Al = 40:60 = 2 : 3 = 14 : 21

Alloy2
Cu:Zn = 2 : 7 = 6 : 21

If we mix 35 parts of alloy1 with 27 parts of alloy2, we will have 21 parts each of Al and Zn and (14 + 6 =) 20 parts of Cu.
So Cu will constitute 20/(35+27) = 20/62 = 32.25%

If we want more Al, we need alloy1 more which means the percentage of copper will increase because it has higher concentration of copper.

Answer must be (C)

Also note that since we want more Al, the entire mixture can be only alloy1 which will give 40% Cu and that is how you get the maximum value.
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An alloy of copper and aluminum has 40% copper. An alloy of Copper and 01 Jul 2025, 20:52
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I want to make the ratio of Al and Zn equal to give me the extreme value.

I get Alloy1
Cu:Al = 40:60 = 2 : 3 = 14 : 21
This alloy has 14 parts copper for 21 parts Al i.e. the alloy has total 14+21 = 35 parts.

Alloy2
Cu:Zn = 2 : 7 = 6 : 21
This alloy has 6 parts copper for 21 parts zinc i.e. the alloy has total 6 + 21 = 27 parts

If we mix 35 parts of alloy 1 with 27 parts of alloy 2, then quantity of Al is equal to that of Zinc.






Ilanchezhiyan
Could you please explain why you have taken 35 parts of alloy1 and 27 parts of alloy2? I could not understand the reasoning behind that. Thanks
KarishmaB
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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%

Alloy1
Cu:Al = 40:60 = 2 : 3 = 14 : 21

Alloy2
Cu:Zn = 2 : 7 = 6 : 21

If we mix 35 parts of alloy1 with 27 parts of alloy2, we will have 21 parts each of Al and Zn and (14 + 6 =) 20 parts of Cu.
So Cu will constitute 20/(35+27) = 20/62 = 32.25%

If we want more Al, we need alloy1 more which means the percentage of copper will increase because it has higher concentration of copper.

Answer must be (C)

Also note that since we want more Al, the entire mixture can be only alloy1 which will give 40% Cu and that is how you get the maximum value.
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An alloy of copper and aluminum has 40% copper. An alloy of Copper and 12 Aug 2026, 09:04
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The condition states Aluminum is strictly greater than Zinc (not greater than or equal to); shouldn't the lower boundary be strictly greater than 32.25%?
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An alloy of copper and aluminum has 40% copper. An alloy of Copper and 12 Aug 2026, 10:20
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The condition states Aluminum is strictly greater than Zinc (not greater than or equal to); shouldn't the lower boundary be strictly greater than 32.25%?
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Yes. Since Aluminum must be strictly greater than Zinc, equality at the boundary is not allowed. The exact boundary is 10/31 = 32.258...%, so x must be strictly greater than 10/31.

Thus, C is the intended answer, but its lower bound is not stated precisely.
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