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555-605 Level|   Algebra|   Word Problems|      
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honchos
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A store sells erasers for 0.23$ per piece and pencil for 0.1 26 Jul 2014, 10:36
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B
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Difficulty:

25% (medium)

Question Stats:

74% (01:34) correct 26% (01:11) wrong based on 722 sessions

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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
(2) She spent total of 1.70$
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B
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Bunuel
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A store sells erasers for 0.23$ per piece and pencil for 0.1 26 Jul 2014, 10:47
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Bunuel
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.

Similar questions to practice:
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html

Hope it helps.
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A store sells erasers for 0.23$ per piece and pencil for 0.1 21 Mar 2018, 23:11
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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.

Bunuel
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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.

Similar questions to practice:
https://gmatclub.com/forum/eunice-sold-s ... 09602.html
https://gmatclub.com/forum/martha-bought ... 00204.html
https://gmatclub.com/forum/a-rental-car- ... 05682.html
https://gmatclub.com/forum/joe-bought-on ... 06212.html
https://gmatclub.com/forum/a-certain-fru ... 01966.html
https://gmatclub.com/forum/joanna-bought ... 01743.html

Hope it helps.
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A store sells erasers for 0.23$ per piece and pencil for 0.1 26 Jul 2014, 10:47
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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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A store sells erasers for 0.23$ per piece and pencil for 0.1 01 Apr 2018, 22:44
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honchos
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
(2) She spent total of 1.70$

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
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A store sells erasers for 0.23$ per piece and pencil for 0.1 24 Jul 2019, 23:31
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Quote:
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
(2) She spent total of 1.70$


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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A store sells erasers for 0.23$ per piece and pencil for 0.1 26 Jul 2019, 11:46
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Bunuel
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.

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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A store sells erasers for 0.23$ per piece and pencil for 0.1 18 Oct 2019, 21:05
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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!
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A store sells erasers for 0.23$ per piece and pencil for 0.1 02 Aug 2020, 12:43
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Bunuel
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.

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?

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)
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A store sells erasers for 0.23$ per piece and pencil for 0.1 15 Jun 2025, 08:08
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