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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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.

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

Hence B
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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!