Asked by vincent
Mae open her coin purse and found pennies,nickels and dimes with a total value of $2.85.If there are twice as many pennies as there are nickels and dimes combined.How many pennies,dimes and nickels are there if she has a collection of 90 coins
given:
penny= 1 cent =$0.01
nickel= 5 cents =$0.05
dine= 10 cents =$0.10
quarter= 25 cents =$0.25
half= 50 cents =$0.50
dollar= 100 cents =$1
given:
penny= 1 cent =$0.01
nickel= 5 cents =$0.05
dine= 10 cents =$0.10
quarter= 25 cents =$0.25
half= 50 cents =$0.50
dollar= 100 cents =$1
Answers
Answered by
Reiny
number of nickels --- n
number of dimes ---- d
number of pennies --- n+d
n + d + n+d = 90
2n+2d = 90
n+d = 45 or d = 45-n
value equation:
5n + 10d + 1(n+d) = 285
6n + 11d = 285
using substitution:
6n + 11(45-n) = 285
6n + 495 - 11n = 285
-5n = -210
n = 42
d = 45-42 = 3
so we have 42 nickels, 3 dimes and 45 pennies
check:
do they add up to 90 ? YEAH
what is their value?
42(5) + 3(10) + 45(1) = 285. YEAHHH!!!
number of dimes ---- d
number of pennies --- n+d
n + d + n+d = 90
2n+2d = 90
n+d = 45 or d = 45-n
value equation:
5n + 10d + 1(n+d) = 285
6n + 11d = 285
using substitution:
6n + 11(45-n) = 285
6n + 495 - 11n = 285
-5n = -210
n = 42
d = 45-42 = 3
so we have 42 nickels, 3 dimes and 45 pennies
check:
do they add up to 90 ? YEAH
what is their value?
42(5) + 3(10) + 45(1) = 285. YEAHHH!!!
Answered by
the better Reiny
by Reiny's formation of equations,
eqn.1. should look like n + d + 2(n + d)=90
3n + 3d=90
3(n + d)=90
n + d=30
n=30 - d
eqn.2. 5n + 10d + 2(n + d)=285
7n + 12d=285
7(30 - d) + 12 d=285
5d= 70
[d=15]
then lets find out n:
n=30-15
[n=15]
lets check if it all adds to 90 coins:
n + d + 2(n + d)
15 + 15 + 2(15 + 15)
30 + 60
90 coins checks out
eqn.1. should look like n + d + 2(n + d)=90
3n + 3d=90
3(n + d)=90
n + d=30
n=30 - d
eqn.2. 5n + 10d + 2(n + d)=285
7n + 12d=285
7(30 - d) + 12 d=285
5d= 70
[d=15]
then lets find out n:
n=30-15
[n=15]
lets check if it all adds to 90 coins:
n + d + 2(n + d)
15 + 15 + 2(15 + 15)
30 + 60
90 coins checks out
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