he has 4 pennies
2 nickels=10 cents
3 dime=30 cents
20 cents
40 +4 pennies =44 cents
Answer:
Keith has 4 pennies, 3 dimes and 2 nickels.
Step-by-step explanation:
Let the pennies be = p
Let the nickels be = n
Let the dimes be = d
Total coins Keith has = 9
So, 1st equation form :
[tex]p+n+d=9[/tex] ....(1)
Keith has one more dime than nickels, means [tex]d=n+1[/tex] .....(2)
Total value of these coins is $0.44
So, third equation forms:
[tex]0.01p+0.05n+0.10d=0.44[/tex] .....(3)
Substituting (2) in (1)
[tex]p+n+n+1=9[/tex]
[tex]p+2n=8[/tex] .....(4)
Similarly substitution (2) in (3)
[tex]0.01p+0.05n+0.10(n+1)=0.44[/tex]
[tex]0.01p+0.05n+0.10n+0.10=0.44[/tex]
[tex]0.01p+0.15n=0.34[/tex] ....(5)
Now multiplying (4) by 0.01 and subtracting by (5)
[tex]0.01p+0.02n=0.08[/tex] subtracting this from (5) we get
[tex]0.13n=0.26[/tex]
So, we get n = 2
[tex]p+2n=8[/tex]
[tex]p+2(2)=8[/tex]
[tex]p+4=8[/tex]
p = 4
[tex]p+n+d=9[/tex]
[tex]4+2+d=9[/tex]
[tex]6+d=9[/tex]
d = 3
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Therefore, Keith has 4 pennies, 3 dimes and 2 nickels.
