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76% (02:11) correct
24%
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An alloy of copper and nickel contains 65 % copper. A second alloy contains copper and nickel in the ratio 17 : 3. In what ratio should the two alloys be mixed so that the new mixture contains 4 times as much copper as nickel?
A. 4 : 5
B. 5 : 4
C. 1 : 3
D. 2 : 3
E. None of these
A. 4 : 5
B. 5 : 4
C. 1 : 3
D. 2 : 3
E. None of these
IanStewart
Expert
GMAT Tutor
Joined: 24 Jun 2008
Last visit: 04 Sep 2026
Posts: 4,146
Location: United States
GMAT 1: 780 Q51 V47
Expert reply
GMAT 1: 780 Q51 V47
Posts: 4,146
We are mixing a 65% copper alloy with an 85% copper alloy (17/20 = 85%) and want to arrive at an 80% alloy (4/5 = 80%). Putting these three 'averages' on a number line:
--65---------------80-----85----
the ratio of the two components must equal the ratio of the two distances to the middle average of 80, so they must be mixed in a 15 to 5, or 3 to 1 ratio. Since we need to use more of the 85% alloy (since 80 is closer to 85 than it is to 65), we need to mix the 65% alloy and the 85% alloy in a 1 to 3 ratio.
There's a problem with the wording of the question -- it isn't clear whether the answer should be 1 to 3 or 3 to 1, since the question doesn't tell us which of the two alloys is supposed to go first in this ratio.
_________________
--65---------------80-----85----
the ratio of the two components must equal the ratio of the two distances to the middle average of 80, so they must be mixed in a 15 to 5, or 3 to 1 ratio. Since we need to use more of the 85% alloy (since 80 is closer to 85 than it is to 65), we need to mix the 65% alloy and the 85% alloy in a 1 to 3 ratio.
There's a problem with the wording of the question -- it isn't clear whether the answer should be 1 to 3 or 3 to 1, since the question doesn't tell us which of the two alloys is supposed to go first in this ratio.
_________________
now teaching gmat focus
ianstewartgmat.com
ianstewartgmat.com
Let C1 be the copper in first alloy and C2 be the copper in second alloy. It is given that C1=65%
And from the ratio of C2:N2=17:3, we can say that C2=85%.
The copper in mixture of two alloys should be 4 times that of nickel, that is C = 4N, and from this ratio we can say that copper in resultant mixture should be 80%.
We have % of copper in two allows and the desired %, therefore applying allegation method
C2---85--------------15
--------------80--------
C1---65---------------5
Therefore, C1:C2=5:15=1:3
OA - C
-------------------
Kudos if helpful!
And from the ratio of C2:N2=17:3, we can say that C2=85%.
The copper in mixture of two alloys should be 4 times that of nickel, that is C = 4N, and from this ratio we can say that copper in resultant mixture should be 80%.
We have % of copper in two allows and the desired %, therefore applying allegation method
C2---85--------------15
--------------80--------
C1---65---------------5
Therefore, C1:C2=5:15=1:3
OA - C
-------------------
Kudos if helpful!
Schools: LBS '21 Anderson '21
Posts: 68
dharam44
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dharam44
let x and y=respective amounts of alloys 1 and 2 to be mixed
.65x+.85y=.8(x+y)→
x/y=1/3
x:y=1:3
C
I reasoned this way:
65% of Copper means that C1 is composed of 65x of Copper and 35x of Nikel and so (simplifying) of:
- \(13x\) of Copper
- \(7x\) of Nikel
While C2 is composed of:
- \(17x\) of Copper
- \(3x\) of Nikel
Now I can build the equation:
Taking \(a\) as the amount of C1 and \(b\) as the amount of C2
\(\frac{((13x)a + (17x)b)}{((7x)a+(3x)b)} = 4\) --> \(15xa = 5xb \)
So \(a = \frac{5}{15}b\)
65% of Copper means that C1 is composed of 65x of Copper and 35x of Nikel and so (simplifying) of:
- \(13x\) of Copper
- \(7x\) of Nikel
While C2 is composed of:
- \(17x\) of Copper
- \(3x\) of Nikel
Now I can build the equation:
Taking \(a\) as the amount of C1 and \(b\) as the amount of C2
\(\frac{((13x)a + (17x)b)}{((7x)a+(3x)b)} = 4\) --> \(15xa = 5xb \)
So \(a = \frac{5}{15}b\)
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