Question 1206098: o, who works at a fast food restaurant, received $9.05 in tips one afternoon, all in quarters, dimes, and nickels. There were ten less dimes than quarters and five less nickels than dimes. How many of each kind of coin was there?
Big thank you for anyone who answers this.
Found 4 solutions by josgarithmetic, math_tutor2020, ikleyn, mananth:
Answer by josgarithmetic(39620)   (Show Source):
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
d = number of dimes
d-5 = number of nickels
d+10 = number of quarters
10d = value of just the dimes in cents
5(d-5) = value of just the nickels in cents
25(d+10) = value of just the quarters in cents
$9.05 = 905 cents
dimes+nickels+quarters = 905 cents
10d+5(d-5)+25(d+10) = 905
10d+(5d-25)+(25d+250) = 905
40d+225 = 905
40d = 905-225
40d = 680
d = 680/40
d = 17
Then,
d-5 = 17-5 = 12
d+10 = 17+10 = 27
Conclusion:
27 quarters
17 dimes
12 nickels
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Check:
A = 27 quarters = 27*25 = 675 cents = $6.75
B = 17 dimes = 17*10 = 170 cents = $1.70
C = 12 nickels = 12*5 = 60 cents = $0.60
D = A+B+C = $6.75 + $1.70 + $0.60 = $9.05
The answers are confirmed.
Answer by ikleyn(52810)  (Show Source):
You can put this solution on YOUR website! .
o, who works at a fast food restaurant, received $9.05 in tips one afternoon,
all in quarters, dimes, and nickels. There were ten less dimes than quarters
and five less nickels than dimes. How many of each kind of coin was there?
Big thank you for anyone who answers this.
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This problem can be solved MENTALLY.
Below I show you how to do it.
Mentally, take off 10 quarters from the collection and add 5 nickels.
You will get then an updated collection, which is worth (905-10*25+5*5) = 680 cents
and consists of equal number of dimes, quarters and nickels.
So, in the updated collection, you can group coins mentally, placing one quarter,
one dime and one nickel in each group. Each such a group is worth 25+10+5 = 40 cents,
so the number of groups is 680/40 = 17.
From here, the number of dimes in the real collection is 17; the number of quarters is 17+10 = 27
and the number of nickels in the real collection is 17-5 = 12.
ANSWER. 12 nickels, 17 dimes and 27 quarters.
Solved mentally, without using equations.
Answer by mananth(16946)  (Show Source):
You can put this solution on YOUR website! o, who works at a fast food restaurant, received $9.05 in tips one afternoon, all in quarters, dimes, and nickels. There were ten less dimes than quarters and five less nickels than dimes. How many of each kind of coin was there?
There were ten less dimes than quarters and five less nickels than dimes.
quarters q numbers
dimes q-10
nickels q-10-5=q-15
25q+10(q-10)+5(q-15)=905
25q+10q-100+5q-75 =905
40q = 905+100+75
40q = 1080
q= 27 quarters
dimes q-10 =27-10=17
nickels q-15=27-15=12
27*25+17*10+12*5=905 cents
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