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The average value of all the pennies, nickels, dimes,
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11 Mar 2019, 00:36
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50% (01:03) correct 50% (02:55) wrong based on 14 sessions
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The average value of all the pennies, nickels, dimes, and quarters in Paula’s purse is 20 cents. If she had one more quarter, the average value would be 21 cents. How many dimes does she have in her purse? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4
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The average value of all the pennies, nickels, dimes,
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Updated on: 16 Mar 2019, 11:28
hi @ ur way is too complex and time consuming . if u want algebraic way do this way Let the total value, in cents, of the coins Paula has originally be v and the number of coins she has be n so, v/n = 20 v= 20n and , v+25/n+1=21 Substituting yields: 20n+25=21(n+1), so n=4 thats mean total number of coin = 4 so v= 20x4 v= 80 cents Then, we see that the only way Paula can satisfy this rule is if she had 3 quarters and 1 nickel in her purse.(3X25+5) so no dimes
Originally posted by Noshad on 16 Mar 2019, 11:13.
Last edited by Noshad on 16 Mar 2019, 11:28, edited 1 time in total.



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The average value of all the pennies, nickels, dimes,
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16 Mar 2019, 10:16
hi stneIf the new coin was worth 20 cents, adding it would not change the mean. The additional 5 cents raise the mean by 1, thus the new number of coins must be 5. Therefore there were 4 coins worth a total of 4X20=80 cents. only way to get 80 cents using 4 coins is 25+25+25+5. Thus, having three quarters, one nickel, and no dimes answer is A



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Re: The average value of all the pennies, nickels, dimes,
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16 Mar 2019, 10:48
Noshad wrote: hi stneIf the new coin was worth 20 cents, adding it would not change the mean. The additional 5 cents raise the mean by 1, thus the new number of coins must be 5. Therefore there were 4 coins worth a total of 4X20=80 cents. only way to get 80 cents using 4 coins is 25+25+25+5. Thus, having three quarters, one nickel, and no dimes answer is A Thank you, but really not able to understand the solution. Can you help me with the algebraic solution , I tried this way: \(\frac{1P +5N+10D+25Q}{P+N+D+Q}=20\) (I) \(\frac{1P +5N+10D+25(Q+1)}{P+N+D+Q}=21\) (II) solving I we get 5Q=19P+15N+10D solving II we get 4Q+25=20P+16N+11D III ( Subtracting I from II) Q+25=P+N+D or P+N+D+Q =25 After this, how do we deduce the solution?
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Re: The average value of all the pennies, nickels, dimes,
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16 Mar 2019, 11:22
Noshad wrote: hi @ ur way is too complex and time consuming . if u want algebraic way do this way Let the total value, in cents, of the coins Paula has originally be v and the number of coins she has be n so, v/n = 20 v= 20n and , v+25/n+1=21 Substituting yields: 20n+25=21(n+1), so n=4 thats mean total number of coin = 4 so v= 20x4 v= 80 cents Then, we see that the only way Paula can satisfy this rule is if she had 3 quarters and 1 nickel in her purse.(3X20+10) so no dimes Thank you +1, great solution. But there is small typo at the end , you may wish to edit (3X20+10) to (3*25 +5). Thanks.
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Re: The average value of all the pennies, nickels, dimes,
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16 Mar 2019, 11:22






