Sebastian has been saving part of his allowance in a bag every month. The bag containing $2 and $10 notes had a total value of $2280. 4 pieces of $10 notes were exchanged for $2 notes of the same value. In the end, the number of $10 notes was the same as the $2 notes. How many $2 notes were there at first?

(i don understand algebra so pls explain it in another way but its ok if u don no other ways u can just use algebra)

Guest Mar 3, 2020

edited by
Guest
Mar 3, 2020

#1**+1 **

Let the number of $2 bills =B

The number of $10 bills =T

2B + 10T = 2280.................(1)

T - 4 =B + 20.......................(2), solve for B, T

B = 70 - number of $2 bills originally

T =194 - number of $10 bills originally.

Note: When he exchanged 4 $10 bills for $2 bills, it means he lost 4 x $10 =$40 in 10-dollar bills. This $40 was exchanged into $40/$2 =20 2-dollar bills. In other words:4 x $10 =20 x $2. And that is what equation (2) means.

Guest Mar 3, 2020

#4**+1 **

Call the original number of $10 notes, M

Call the original number of $2 notes , N

So

10M + 2N = 2280 (1)

If 4 pieces of $10 notes were exchanged for $2 notes of the same value....then the number of $2 notes must increase by 20 since after the exchange and the number of $10 notes must decreases by 4....and since we have the same number after the exchanges.....then

So

M - 4 = N + 20

M = N + 24 (2)

Sub (2) into (1) and we have that

10 (N + 24) + 2N = 2280 simplify

10N + 240 + 2N = 2280 subtract 240 from both sides

12N = 2040 divide both sides by 12

N = 170 = number of $2 notes originally

Proof

There were 170 $2 notes originally and

10M + 2 (170) = 2280

10M + 340 = 2280

10m = 2280 - 340 =

10M = 1940

M = 194 $10 notes originally

After the exchange there are 170 + 20 = 190 $2 notes and 194 - 4 = 190 $10 notes

CPhill Mar 3, 2020