Time remaining:
##### need help to solve this

label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5

chris has \$2.77 in her coin jar- all in pennies, nickels, and dimes. if she has 53 coins in all and four more nickels than dimes, find how many of each tupe of coins she has

Oct 19th, 2017

Let number of dimes = d

Then number of nickels = 4 + d    ( 4 more than number of dimes)

Let number of pennies = p

total number of coins =d + 4+ d +p = 2d + 4 + p

But  2d + 4 +p = 53

2d+ p = 49

So p = 49 - 2d    (*)

value of the dimes in cents  =10 x d = 10d

value of the nickels in cents  =5 x (4 + d) = 20 + 5d

Total value of nickels and dimes = 20 + 5d + 10 d = 20 + 15d

Total value of all coins = 20 + 15d + p = 277

Now replace p by 49 - 2d  from (*) gives

20 + 15d + 49 -2d = 277

13d + 69 = 277

13d = 277-69 =208

d = 208/13 = 16

So he has 16 dimes, 20 nickels (4 more than dimes) and 53- 36 = 17 pennies

(check value: 160 + 100 + 17 = 277 = \$ 2.77)

Oct 19th, 2014

...
Oct 19th, 2017
...
Oct 19th, 2017
Oct 20th, 2017
check_circle