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26 Jul 2014, 10:36
4
13
00:00

Difficulty:

25% (medium)

Question Stats:

76% (01:29) correct 24% (01:06) wrong based on 333 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$_________________ Like my post Send me a Kudos It is a Good manner. My Debrief: http://gmatclub.com/forum/how-to-score-750-and-750-i-moved-from-710-to-189016.html Math Expert Joined: 02 Sep 2009 Posts: 58428 Re: A store sells erasers for 0.23$ per piece and pencil for 0.1  [#permalink]

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26 Jul 2014, 10:47
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. _________________ Math Expert Joined: 02 Sep 2009 Posts: 58428 Re: A store sells erasers for 0.23$ per piece and pencil for 0.1  [#permalink]

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26 Jul 2014, 10:47
Bunuel wrote:
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. _________________ Intern Joined: 17 Jan 2018 Posts: 44 Schools: ISB '20 (A) Re: A store sells erasers for 0.23$ per piece and pencil for 0.1  [#permalink]

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21 Mar 2018, 23:11
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 wrote:
Bunuel wrote:
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: http://gmatclub.com/forum/eunice-sold-s ... 09602.html http://gmatclub.com/forum/martha-bought ... 00204.html http://gmatclub.com/forum/a-rental-car- ... 05682.html http://gmatclub.com/forum/joe-bought-on ... 06212.html http://gmatclub.com/forum/a-certain-fru ... 01966.html http://gmatclub.com/forum/joanna-bought ... 01743.html Hope it helps. Intern Joined: 21 Jan 2018 Posts: 49 Location: India GPA: 2.85 WE: Consulting (Consulting) A store sells erasers for 0.23$ per piece and pencil for 0.1  [#permalink]

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01 Apr 2018, 22:44
1
honchos wrote:
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$Given : P(Eraser) = 0.23 , P(Pencil) = 0,11 To find No. of Erasers and No. of Pencils Statement 1 : No of erasers = 5 , nut we cannot get the No of pencils --> Insufficient Statement 2 : 1.70 = 0.23*E + 0.11*P , we don't know values of E and P --> Sufficient Plus , E and P > 0 E = 5 P = 5 Option B _________________ Regards , Dhruv _________________ Kudos will encourage many others, like me. Please Give +1 Kudos if you liked the post!! Thanks Manager Joined: 26 Jan 2016 Posts: 179 Re: A store sells erasers for 0.23$ per piece and pencil for 0.1  [#permalink]

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24 Jul 2019, 23:31
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$A. NO info about pencils. Insufficient B. only 5 erasers and 5 pencil satisfy the equation : 0.23e+0.11p=1.70 Hence B _________________ Your Kudos can boost my morale..!! I am on a journey. Gradually I'll there..!! Director Joined: 18 Dec 2017 Posts: 525 Location: United States (KS) Re: A store sells erasers for 0.23$ per piece and pencil for 0.1  [#permalink]

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26 Jul 2019, 11:46
Bunuel wrote:
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? _________________ ----The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long ---- Software Tester currently in USA ( ) Intern Joined: 07 Sep 2019 Posts: 1 Re: A store sells erasers for 0.23$ per piece and pencil for 0.1  [#permalink]

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18 Oct 2019, 21:05
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!