GMAT Club Olympics 2024 (Day 10): An alloy is made by mixing certain

Question

An alloy is made by mixing certain quantities of iron, copper, and lead. If the average (arithmetic mean) cost of the alloy is $20 per kilogram, what is the ratio of the weight of iron to the weight of copper to the weight of lead in the alloy?

(1) The cost per kilogram of iron, copper, and lead is $10, $20, and $30, respectively.

(2) The cost of the copper used in the alloy equals the cost of the lead used in the alloy.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D. EACH statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Bunuel · Accepted Answer

GMAT Club Official Explanation:

An alloy is made by mixing certain quantities of iron, copper, and lead. If the average (arithmetic mean) cost of the alloy is $20 per kilogram, what is the ratio of the weight of iron to the weight of copper to the weight of lead in the alloy?

Assuming x kilograms of iron, y kilograms of copper, and z kilograms of lead were used to make the alloy, the question asks us to find the ratio x : y : z.

(1) The cost per kilogram of iron, copper, and lead is $10, $20, and $30, respectively.

Since the average cost of the alloy is $20 per kilogram, from the above statement we can deduce that (10x + 20y + 30z)/(x + y + z) = 20. Simplifying this equation gives x = z. Thus, we have the relationship x : z = 1 : 1. However, we still don't know the relationship between y and x, or y and z to get the final ratio of x : y : z. Not sufficient.

(2) The cost of the copper used in the alloy equals the cost of the lead used in the alloy.

This statement alone is clearly not sufficient to determine the desired ratio.

(1) + (2) Since from (1) the cost per kilogram of copper and lead is $20 and $30, respectively, then from (2) we can deduce 20y = 30z, which gives y : z = 3 : 2. Therefore, combining with x = z from statement (1), we get x : y : z = 2 : 3 : 2. Sufficient.

Answer: C.

Kinshook · Answer

Given : An alloy is made by mixing certain quantities of iron, copper, and lead.
Asked: If the average (arithmetic mean) cost of the alloy is $20 per kilogram, what is the ratio of the weight of iron to the weight of copper to the weight of lead in the alloy?

Let the ratio of the weight of iron to the weight of copper to the weight of lead in the alloy = i : c : l

(1) The cost per kilogram of iron, copper, and lead is $10, $20, and $30, respectively.
The average (arithmetic mean) cost of the alloy = (10i + 20c + 30l)/(i + c + l) = $20
10i + 20c + 30l = 20i + 20c + 20l
10i = 10l
i = l
The ratio of the weight of iron to the weight of copper to the weight of lead in the alloy = l : c : l
Since ratio of c & l can not be determined.
NOT SUFFICIENT

(2) The cost of the copper used in the alloy equals the cost of the lead used in the alloy.
Since the cost of the copper & cost of the lead $/kg are not provided.
NOT SUFFICIENT

(1) + (2) 
(1) The cost per kilogram of iron, copper, and lead is $10, $20, and $30, respectively.
The average (arithmetic mean) cost of the alloy = (10i + 20c + 30l)/(i + c + l) = $20
10i + 20c + 30l = 20i + 20c + 20l
10i = 10l
i = l
(2) The cost of the copper used in the alloy equals the cost of the lead used in the alloy.
20c = 30l 
c = 1.5l
The ratio of the weight of iron to the weight of copper to the weight of lead in the alloy = l : 1.5l : l = 1: 1.5: 1 = 2:3:2
SUFFICIENT

IMO C

meet459 · Answer

An alloy is made by mixing certain quantities of iron, copper, and lead. If the average (arithmetic mean) cost of the alloy is $20 per kilogram, what is the ratio of the weight of iron to the weight of copper to the weight of lead in the alloy?

(1) The cost per kilogram of iron, copper, and lead is $10, $20, and $30, respectively.

(2) The cost of the copper used in the alloy equals the cost of the lead used in the alloy.

Let us assume the weight of iron be i
Let us assume the weight of copper be c
Let us assume the weight of lead be l

(1) The cost per kilogram of iron, copper, and lead is $10, $20, and $30, respectively. => (i*10 + c*20 + l*30) / (i + c + l) = 20
=> i + 2c + 3l = 2i + 2c + 2l => i = l (But we don't know anything about c) Hence INSUFF

(2) The cost of the copper used in the alloy equals the cost of the lead used in the alloy. => 2c = 3l, clearly INSUFF (we don't have info about i)

(1) + (2)

i = l and 2c = 3l SUFF (3 : 2 : 3)

Hence answer is C

AthulSasi · Answer

Let qty of iron, copper and lead be x,y and z

statement 1 
(1) The cost per kilogram of iron, copper, and lead is $10, $20, and $30, respectively.

This impllies multiple possibilities. x and z has to be same and y can be anything. 
Mathematically,
avg cost = \frac{10x+20y+30z}{x+y+z) = 20
10x+20y+30z = 20x+20y+20z
x = z
We know x: z = 1:1  but we dont the relationship of x&z to y. y can be anything,  Hence not sufficient.

statement 2 
(2) The cost of the copper used in the alloy equals the cost of the lead used in the alloy.
This only gives the cost of copper and lead in the alloy to be equal. since we dont know the cost per kilogram, we cannot find the weight ratio.  Hence not sufficient.

statement 1&2 together 
From 2, 20y = 30z
2y = 3z
y:z = 3:2

from 1, we already established x = z

hence x:y:z = 2:3:2
 Hence sufficient.

Answer C

divi22 · Answer

From Statement 1,
Let the amount of lead used per kg of alloy = x
Let the amount of copper used per kg of alloy = y
=> the amount of iron used per kg of alloy = 1-y-x
 => 10(1-y-x)+20y+30x = 20
=> 2x+y = 1
=> The ratio will be 1:y/x:1
Since the information is not provided, the statement is insufficient.

Statement 2 is insufficient as the cost of material used is given, but not the general cost of material to find the amount used in making the alloy.

Both Statement together:
30x = 20y
=>y/x = 3/2

Hence, the ratio will be 2:3:2

So, the answer is C.

prashantthawrani · Answer

An alloy is made by mixing certain quantities of iron, copper, and lead. If the average (arithmetic mean) cost of the alloy is $20 per kilogram, what is the ratio of the weight of iron to the weight of copper to the weight of lead in the alloy?

Let x, y and z be the weight proportion of iron, copper, and lead respectively.
 x+y+z=1..........(i)

Let A, B and C be the cost per kg of iron, copper, and lead respectively.
 (Ax+By+Cz)=20..........(ii)

(1)  The cost per kilogram of iron, copper, and lead is $10, $20, and $30, respectively.

From (ii) 
 10x + 20y + 30z=20 .........(ii-a) 
(i) and (ii-a) could not be solved as there are two equations and three variables(x,y,z).

Not sufficient.

(2)  The cost of the copper used in the alloy equals the cost of the lead used in the alloy.

By=Cz..........(iii)

Subtituting (iii) in (ii)
(Ax+2By)=20............. (ii-b) 
(i) and (ii-b) could not be solved as there are two equations and four variables(A,B,x,y).
Not sufficient.

(1)+(2):  
 A=10, B=20, C=30;

By=Cz  -> 20y=30z    ->  y=(1.5)z

From (ii)
(Ax+By+Cz)=20
10x + 30z + 30z=20
 10x+60z=20.............(ii-c)

From (i) and y=(1.5)z
x+y+z=1
x+(1.5)z +z =1
 x+(2.5)z =1.............(i-c)

(i-c) & (ii-c) can be solved 
 Ans. C. Sufficient together

kavyavishnoi02 · Answer

Let, A, B and C be the cost per kg of Iron, Copper and Lead, respectively.
And, x, y and z be the weights of Iron, Copper and Lead, respectively used in the alloy.
Thus, (Ax + By + Cz)/ (x+y+z) = $20/ kg

We need to  find x:y:z

Let's analyse each statement now:

I . (10x + 20y + 30z)/ (x+y+z) = 20
10x + 20y + 30z = 20 (x+y+z)
10x + 20y + 30z = 20x + 20y + 20z
10x + 30 z = 20x + 20z
10z = 10x
 z = x

Yet, this doesn't give us the relation of y with any of x and z. Hence,  insufficient .

II . By=Cz
(Ax + 2Cz)/ (x+y+z) = 20

Apart from this, it doesn't give us any additional relation or info.  Insufficient .

Combining I and II :
20y = 30z
 y= 3z/2

Now,  x : y : z 
= z : 3z/2 : z 
= 2z : 3z : 2z
=  2:3:2

Thus, both statements together are sufficient
Therefore, the answer stands option (C)

rsrobin864 · Answer

Let 
                  Price               Weight
For Iron         a                       b
For Copper     c                       d
For Lead        e                       f

​​​​​​​avg cost=20= a*b+c*d+e*f/ b+d+f

To find b:d:f

1--a=10, c=20, e=30

20= 10*b+ 20*d+ 30*f/ b+d+f

20b+20d+20f= 10b+20d+30f

10b=10f
b=f

Statement 1 Not suff

2-- cd=ef

Statement 2 Not suff

1+2-- cd=ef

20d= 30f
 2d=3f
d=3f/2

b:d:f= f: 3f/2: f
b:d:f= 1:3/2:1
b:d:f= 2:3:2

So Statement 1 and 2 together suff

Ans is C Statement 1 and 2 together suff

captain0612 · Answer

Choice C

Given:
1. Average cost of an alloy per kilogram = 20$
2. Alloy contains : Iron, Copper and Lead

Question: ratio of the weights used in the alloy

Let the cost and weights of individual metals in the alloy be : Ci, Wi, Cc, Wc, Cl and Wl respectively for metals Iron, Copper and Lead.

Average cost = (Ci * Wi + Cc * Wc + Cl * Wl)/ ( Wi + Wc + Wl) = 20 ------ (1)
 We need to find : Wi : Wc : Wl

Statement 1:

Individual cost per kilogram for Iron, Copper and Lead is 10$, 20$ and 30$ respectively.
Substitute these costs in equation (1) to get

(10 * Wi + 20 * Wc + 30 * Wl)/ ( Wi + Wc + Wl) = 20  Multiply both sides with the total weight of individual metals used in the alloy

10 * Wi + 20 * Wc + 30 * Wl = 20 * ( Wi + Wc + Wl )
10 * Wi + 20 * Wc + 30 * Wl = 20 * Wi + 20 * Wc + 20 *Wl
10 * Wi + 30 * Wl = 20 * Wi + 20 * Wl

=> Wi = Wl  We get that the weight of Iron and Weight of Lead in the alloy are equal, But we know nothing about the weight of Copper used.  Insufficient

Statement 2:

Cost of copper used = Cost of lead used

=> Cc * Wc = Cl * Wl

Again we arrive at the similar equation we got in Statement 1. Hence,  Insufficient

Statement 1 and Statement 2:

We know that Wi = Wl from Statement 1
and Substituting the cost of of individual metals in statement 2 we arrive at:

Cc * Wc = Cl * Wl => 20 * Wc = 30 * Wl

hence, Wc/Wl = 3/2

We already know that Wi = Wl = 3/2Wc

hence the ratio of weights used is : 3 : 2 : 3 respectively for Iron, Copper and Lead Sufficient

bumpbot · Answer

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