A bank contains $2.80 in nickels and quarters . there r 20 coins in all . how many of each kind are there ?
plz explane to me how to solve it . thank you very much .
2007-11-06 14:14:35 · 5 answers · asked by tt88754 1 in Education & Reference ➔ Homework Help
you just simply start by trying out amounts of coins and seeing which ones add up to 2.80
there could be several answers
2007-11-06 14:23:54 · answer #1 · answered by Laur 1 · 0⤊ 0⤋
how much is a nickel? i think it's five cents yeah? so. it can't be 20 nickels, cause that's only $1
is it 19 nickels and 1 quarter? no, that's only $1.20.
18 nickels and 2 quarters? $1.40
17 nickels and 3 quarters? $1.60
16 nickels and 4 quarters? $1.80
15 and 5? $2
14 and 6? $2.20
13 + 7? $2.40
12 + 8? $2.60
11 + 9? $2.80!!!!!!!!!!!!!
There fore, there are 11 nickels and 9 quarters.
How did i solve it? I added the amount of money up, and i reasoned it out one step at a time.
2007-11-06 14:26:36 · answer #2 · answered by A derka der 7 · 0⤊ 0⤋
There are 9 quarters and 11 nickels... 9 quarters = $2.25 + 11 nickels = $.fifty 5 it rather is rather approximately logic. the least perplexing thank you to start is with the cents ($.80). Having the cents be all nickels may be too many nickels and not adequate quarters. So I extra 1 / 4 to the cents which left $.fifty 5 cents in nickels and from there i substitute into able verify what proportion funds there have been of all varieties.
2016-09-28 12:06:27 · answer #3 · answered by ? 4 · 0⤊ 0⤋
.05N + .25(20-N) = 2.80
.05N + 5 -.25 N =2.8
2.2 = .2N
11 = N
So, there are 11 nickels. And 9 quarters.
2007-11-06 14:22:50 · answer #4 · answered by hottotrot1_usa 7 · 1⤊ 0⤋
Use two equations and substitute for one variable:
.05n+.25q=2.80 , since the total is $2.80
n+q=20, since there a total of 20 coins
q=20-n, solve for one variable
.05n+.25(20-n)=2.80, substitute and solve
.05n+4-.05n=2.80
n=11
q=9
2007-11-06 14:30:16 · answer #5 · answered by 999theone999 1 · 0⤊ 0⤋