A store sells erasers for 0.23$ per piece and pencil for 0.11$ per piece. How many eraser and pencils did Jessica buy?

(1) She bough 5 erasers. Clearly insufficient.

(2) She spent total of 1.70$ --> \(0.23e+0.11p=1.70\) --> \(23e+11p=170\) --> by trial and error we can find that the only non-negative integer solution for this equation is e=5 and p=5. Sufficient.

Answer: B.
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Hope it helps.
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Trial and error is too long and is not exhaustive enough.

There is a simpler solution.

Now, we know that 23x + 11y = 170

The units digit of 23 and 11 should add up to a 0.

So taking the units digit only ->

23x --> 3, 6, 9, 2, 5, 8, 1, 4, 7, 0
11y --> 1, 2, 3, 4, 5, 6, 7, 8, 9, 0

There are a lot of combinations, but trying the easiest first helps ->
(0,0)
23*10 + 11*10 = 340 !! (Voila) This is exactly double of what we need. Halve it.

So, x and y are 5 each.

(Cross checking with Option 1, we see that x is 5 is given, so we can be sure that this is the only answer)

Takes 45 seconds.

But, to be doubly sure, take some other combination in random, Say, (6, 4)

23*2 + 11*4 = 46 + 44 = 90 (Too low)

A bit higher is (5,5), almost confirmed. Have a leap of faith and mark B.

Given : P(Eraser) = 0.23 , P(Pencil) = 0,11
To find No. of Erasers and No. of Pencils

Statement 1 :
No of erasers = 5 , nut we cannot get the No of pencils --> Insufficient

Statement 2 :
1.70 = 0.23*E + 0.11*P , we don't know values of E and P --> Sufficient
Plus , E and P > 0
E = 5 P = 5

Option B

A. NO info about pencils. Insufficient
B. only 5 erasers and 5 pencil satisfy the equation : 0.23e+0.11p=1.70

Hence B

Hello Bunuel Sir,
Will it be right to say that since 23 and 11 are co-prime hence there will be only one unique solution and hence sufficient?
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Hi,

I see that in questions of similar types, all the explanations involve a trial and error method. I was wondering if there is a way to be sure that the result we stumble upon is actually the only possible combination. Even if I arrive at integer values that fit the equation, I still need to check that other numbers don't. And this process always takes more than the allotted time for a question(2 mins). I would be glad if anyone can help me with a way to ensure that no other solution exist once I arrive at a solution, that would save some precious seconds.

Thank you!

I was really excited to see this comment! I thought it'd be a nice trick, but after testing it out I'm not sure it works, even in best-case scenario. I'm curious to hear other thoughts, because I may be mis-understanding.

let's say erasers are 0.03, and pencils are 0.05, and the total bill was 1.50
e.g.
rearrange to get 3E + 5P = 150
3 and 5 are co-prime, as in the OP question
several solutions exist (E=5 & P=27, or E=10 & P=24, etc.)

When if 3*5 is NOT a multiple of the total price, then is still doesn't work
e.g.
3E + 5P = 160
multiple solutions still exist (E=5 & P=29, or E=10 and P=26)

Even if the individual prices are not factors in the total bill, then I think the trick about them being "co-prime is sufficient" still does not work
e.g.
3E + 5P = 112 (which = 2^4*7)
multiple solutions still exist (E=34 & P=2, or E=9 and P=17)