It takes 3 tons of copper ore and 5 tons of iron ore to make o

Question

It takes 3 tons of copper ore and 5 tons of iron ore to make one ton of alloy A. How many tons of alloy A can be made from 60 tons of copper ore and 90 tons of iron ore?

(A) 18
(B) 18 3/4
(C) 20
(D) 30
(E) 50

KarishmaB · Accepted Answer

Yes, you need copper ore:iron ore in the ratio 3:5. Total 8 tons of the mix in this ratio will give 1 ton of alloy A. 
If you have 60 tons of copper ore, it is enough for 60/3 = 20 tons of alloy A.
If you have 90 tons of iron ore, it is enough for 90/5 = 18 tons of alloy A.

Since iron ore is available for only 18 tons of alloy A, you can make only 18 tons of alloy A. 
The leftover copper ore alone cannot make any alloy A and hence will be leftover only.

Answer must be 18.

(A)

Note: This reminds me of our maximum power of 6 in 40! kind of questions. We have many 2s but fewer 3s. We can make only as many 6s as the number of 3s we have. The leftover 2s alone cannot make a 6. Makes sense?

TimeTraveller · Answer

3c + 5i = a
60c + 90i = ?

20(3c + 5i) = 60c + 100i = 20a, so the answer has to be less than 20 since we are dealing with 90i.

But, 18(3c + 5i) = 54c + 90i = 18a, here we get 54c < 60c. So answer must be 18 3/4 (B).

applebus · Answer

why not A
if you produce 18 3/4 the copper is not enough.

mejia401 · Answer

IMO, this is a max/min problem, where the amount of alloy A is limited to how much material we have. Since we need more iron ore to make alloy A than we do from copper ore, iron ore is limiting factor, and therefore, we have enough iron ore only to make 18 tons of alloy A. Answer A.

TimeTraveller · Answer

Yes, none of the given options works perfectly. But the answer must be between 18 and 20, so I chose 18 3/4, although it doesn't fit all too well.

Mascarfi · Answer

It's an alloy, so you need both metals to make it. If you had 150 tons of copper ore you would not be able to make any amount of the alloy, you would just have copper ore. Therefore, the answer is A.

stne · Answer

Struggled with this for almost 15 minutes , I also tried to take the average of 18 tons of alloy and 20 tons of alloy . When no answer matched I thought I was doing something wrong .
Then I finally scrolled down and behold , I saw Karishma' s post , then a bulb lit up and every thing became clear . True to her name Karishma is indeed a magician !

onlyPlanA · Answer

I tried the allegation approach but it didn't work, and why would copper and iron be taken into 3:5 ratio?

3 5
 1
4 2

This means they are in the ratio - 2:1
Hence, 2/3 * 60 = 40 and 1/3 * 90 = 30
Total = 70 tons

And 70 tons is not given in the options. But, what is wrong with this method? It is not given in the question that all of the copper and iron must be used. In fact, the method that brings us to 18 also says the same thing.

egmat · Answer

Hi onlyPlanA,

You're right that copper and iron sit in a  3:5  ratio, and you're right that not every ton of ore has to be used. But the trouble starts one step earlier:  alligation is the wrong tool for this question , so the numbers it produces don't mean anything here.

Why alligation doesn't apply

Alligation answers one specific kind of question:  "In what proportion should I blend two things of different strengths to hit a target average?"  It needs an  averaging  situation - two concentrations pulling toward a mean.

This problem has no averaging at all. It hands you a  fixed recipe : every single ton of alloy A eats exactly  3  tons of copper and  5  tons of iron. The  3:5  ratio isn't something you get to choose or balance - it's locked. So there's no "target" to alligate toward, and your  3 5 / 1 / 4 2 → 2:1  step is mixing quantities that were never meant to be averaged. That's why  70  falls outside the choices - the method was never valid here.

The right frame: how many full batches?

Think of it as building batches, each needing  3  copper  and5  iron together:

- Copper on hand:  60 ÷ 3 = 20  batches
- Iron on hand:  90 ÷ 5 = 18  batches

Every batch needs  both , so whichever runs out first caps you. Iron runs out at  18 . You then use  54  tons of copper and have  6  tons of copper left over - which matches your instinct that not all the ore is used. That leftover just can't become alloy on its own.

A quick parallel to lock it in

Say each sandwich needs  2  bread slices and  1  cheese slice, and you have  10  bread and  3  cheese. How many sandwiches?

You wouldn't alligate bread against cheese - the recipe is fixed. Bread allows  5 , cheese allows  3 , so cheese limits you to  3  sandwiches, with  4  bread left over. Same logic, every time:  fixed recipe → the scarcest ingredient sets the limit.

Answer: A

onlyPlanA · Answer

Ah, now I get it. I didn't realize that the alloy  should  have 3 parts copper and 5 parts iron. Thank you for clarifying!

It takes 3 tons of copper ore and 5 tons of iron ore to make o

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It takes 3 tons of copper ore and 5 tons of iron ore to make o

Prep Toolkit

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