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805+ (Hard)|   Algebra|   Word Problems|      
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Bunuel
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A child bought candy that cost 85 cents. If the child only had quarter 29 Mar 2017, 23:44
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E
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95% (hard)

Question Stats:

36% (02:40) correct 64% (02:35) wrong based on 390 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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E
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mohshu
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A child bought candy that cost 85 cents. If the child only had quarter Updated on: 30 Mar 2017, 07:28
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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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
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A child bought candy that cost 85 cents. If the child only had quarter Updated on: 30 Mar 2017, 10:03
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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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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A child bought candy that cost 85 cents. If the child only had quarter 30 Mar 2017, 09:43
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Bunuel
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.
Show more

Q-25,D-10,N-5

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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A child bought candy that cost 85 cents. If the child only had quarter 23 Jun 2018, 02:06
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Bunuel
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.
Show more


]the currencies are..
Nickel 5 cents, dimes -10 and quarter-25


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..
Quarte-2, dimes-1 and nickel-5 OR
Quarter-1, dimes-5 and nickel -1

E
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A child bought candy that cost 85 cents. If the child only had quarter 15 Jul 2018, 14:50
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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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A child bought candy that cost 85 cents. If the child only had quarter 15 Jul 2018, 15:15
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Bunuel
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.
Show more

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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A child bought candy that cost 85 cents. If the child only had quarter 15 Jul 2018, 20:43
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Bunuel
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
Show more
___________________
Done. Thank you.
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A child bought candy that cost 85 cents. If the child only had quarter 24 Jan 2026, 13:39
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Picking values would be the fastest approach efficiency

Approach I used for testing values ---> make a small vertical table, easier to plug values this way

Value Coins Total value
25 1 25
10 5 50
5 2 10
-------
85 No

Value Coins Total value
25 2 50
10 1 10
5 5 25
---------
85 Yes

Since these values satisfy both statments ---> Answer is E
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