Acollection of dimes and quarters is worth $9.55. if the quarters were dimes and the dimes were quarters, the total value would be 7.60. find the number of each coin.

Number of dimes are 18 and number of quarters are 31

Solution:

Let "d" be the number of dimes

Let "q" be the number of quarters

value of 1 dime = $ 0.10

value of 1 quarter = $ 0.25

A collection of dimes and quarters is worth $9.55

value of 1 dime x number of dimes + value of 1 dime x number of quarters = 9.55

0.10d + 0.25q = 9.55 eqn 1

If the quarters were dimes and the dimes were quarters, the total value would be 7.60

quarters were dimes means , q = d

dimes were quarters means d = q

0.25d + 0.10q = 7.60 eqn 2

Let us solve eqn 1 and eqn 2 to find "d" and "q"

Multiply eqn 1 by 2.5

0.25d + 0.625q = 23.875 eqn 3

Subtract eqn 2 from eqn 3

0.25d + 0.625q = 23.875

0.25d + 0.10q = 7.60

( - )

0.525q = 16.275

q = 31

Substitute q = 31 in eqn 1

0.10d + 0.25q = 9.55

0.10d + 0.25(31) = 9.55

0.10d + 7.75 = 9.55

0.10d = 1.8

d = 18

Thus dimes are 18 and number of quarters are 31

i needed the answer to this too!! have a good life

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18