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A child bought candy that cost 85 cents. If the child only had quarter
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29 Mar 2017, 23:44
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29% (02:13) correct 71% (02:32) wrong based on 158 sessions
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A child bought candy that cost 85 cents. If the child only had quarters, dimes, and nickels, did he pay using at least two quarters? (quarter = $0.25, dime = $0.1 and nickel = $0.05) (1) The child used 8 coins. (2) The child used 5 of one type of coin.
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Re: A child bought candy that cost 85 cents. If the child only had quarter
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Updated on: 30 Mar 2017, 07:28
lets start with easiest statement..
5 one type of coin implies...quarter cant be used 5 times,, case 1 : dime 5*10 + 35 = 85
35 can be made with one quarter and 2 nickels or 7 nickels
case 2 : nickels :
5*5 + 65 = 85
65 can be made using one quarters as well as 2 quarters..
not suff
stat 1 : several combinations possbile,,
IMO ans E
Originally posted by mohshu on 30 Mar 2017, 04:58.
Last edited by mohshu on 30 Mar 2017, 07:28, edited 2 times in total.



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Re: A child bought candy that cost 85 cents. If the child only had quarter
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Updated on: 30 Mar 2017, 10:03
Answer : [E] total: 85 cents Case 1: 3 quarters + 1 dime; total :4 coins Case 2: 2 quarters + 3 dimes + 1 nickel; total :6 coins Case 3: 2 quarters + 2 dimes + 3 nickel; total :7 coins Case 4: 2 quarters + 1 dime + 5 nickels; total :8 coins Case 5: 1 quarter + 6 dimes; total :7 coins Case 4: 1 quarter + 5 dimes+ 2 nickels; total :8 coins etc. Stmnt 1 : The child used 8 coins. 2 cases already there. Not Sufficient. Stmnt 2: The child used 5 of one type of coin. 2 cases already there. Not Sufficient. Combining both Stmnt : Not Sufficient
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Originally posted by Spartan85 on 30 Mar 2017, 06:03.
Last edited by Spartan85 on 30 Mar 2017, 10:03, edited 1 time in total.



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Re: A child bought candy that cost 85 cents. If the child only had quarter
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30 Mar 2017, 09:43
Bunuel wrote: A child bought candy that cost 85 cents. If the child only had quarters, dimes, and nickels, did he pay using at least two quarters?
(1) The child used 8 coins. (2) The child used 5 of one type of coin. Q25,D10,N5 At least two quarters: either 2*25 or 3*25 is possible 1) 8 coins 1*25+5*10+2*5 2*25+1*10+5*5 Both the cases are possible hence Not suff 2) 5 of one type of coin even with this it can either be 5 Dimes or 5 Nickels. Not suff 1+2 also doesnt give us a solution Ans: E
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Re: A child bought candy that cost 85 cents. If the child only had quarter
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23 Jun 2018, 02:06
Bunuel wrote: A child bought candy that cost 85 cents. If the child only had quarters, dimes, and nickels, did he pay using at least two quarters?
(1) The child used 8 coins. (2) The child used 5 of one type of coin. ]the currencies are.. Nickel 5 cents, dimes 10 and quarter25 I) child used 8 coins. Surely there is a quarter as all dimes are also less than 85 cents. So 25x+10y+5z=85... 5x+2y+z=17... So X=2, y=1 and z=5...will fit in X=1, y=5 and z=2... Will also fit in Inspired II)child used 5 of one type of coin.. 5 of quarter not possible.. So if 5 of dimes are there, rest can be nickels, so NO quarter.. 2 quarters means 50.. add 5 nickels so 25 more and 1 dime.. Again insufficient Combined.. Ways in statement I still stands.. Quarte2, dimes1 and nickel5 OR Quarter1, dimes5 and nickel 1 E
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Re: A child bought candy that cost 85 cents. If the child only had quarter
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15 Jul 2018, 14:50
Given: 25q + 10d + 5n = 85 Find: Is q greater than or equal to 2?
(1) Child uses 8 coins. Before trying any combos, notice that if all 8 coins are dimes, that will fall short of the 85 cent price. And we can't have 8 coins of dimes and nickels only because the price will be even less. This means that we'll have at least 1 quarter. Let's draw a table for some combos.
 q  d  n  total price   1  5  2  85   2  1  5  85 
Note that once we find a combo that works using 1 quarter, we can move on to looking for a combo that works for 2 quarters. This is because there will be just 1 combo that works for 1 quarter. Here's why: if we start with the combo that works (1q, 5d, 2n), if we remove dimes and add nickels, then that's a net negative, so price will be less than 85. And if we add dimes and remove nickels, that will be a net positive, so price will be greater than 85 cents.
So, table above shows 2 combos that work. One uses 1 quarter, one uses 2 quarters. Not sufficient
(2) Because of the table in (1), we can see that this is insufficient in 5 seconds.
(1) + (2) Again because of the table in (1), we can quickly see that both statements combined are insufficient.
Answer: E



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Re: A child bought candy that cost 85 cents. If the child only had quarter
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15 Jul 2018, 15:15
Bunuel wrote: A child bought candy that cost 85 cents. If the child only had quarters, dimes, and nickels, did he pay using at least two quarters?
(1) The child used 8 coins. (2) The child used 5 of one type of coin. Could you add the value of these currencies? I think people from outside the US might not be 100% certain of the values of nickel/dime
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Re: A child bought candy that cost 85 cents. If the child only had quarter
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15 Jul 2018, 20:43
Arro44 wrote: Bunuel wrote: A child bought candy that cost 85 cents. If the child only had quarters, dimes, and nickels, did he pay using at least two quarters?
(1) The child used 8 coins. (2) The child used 5 of one type of coin. Could you add the value of these currencies? I think people from outside the US might not be 100% certain of the values of nickel/dime ___________________ Done. Thank you.
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Re: A child bought candy that cost 85 cents. If the child only had quarter
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19 Aug 2019, 01:15
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Re: A child bought candy that cost 85 cents. If the child only had quarter
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