If don't want to use algebra, just logical thinking...
The sum of $5 notes needs to end in 0 (not in 5), because the sum of $2 notes can't be end in impair number. => we have even-numbered $5 notes => we have even-numbered $2 notes.
possible variations for a $5 notes (to make it less than 50):
A. 2 pieces: $10
B. 4 pieces: $20
C. 6 pieces: $30
D. 8 pieces: $40
E. 10 pieces: $50
Based on the 2nd point => the sum of $2 notes need to end also in 0. => the possible variations:
a. 5 pieces: $10
b. 10 pieces:$20
c. 15 pieces: $30
d. 20 pieces: $40
e. 25 pieces: $50
For a total amount of $50, we need:
A,d
B, c
C, b
D, a
E
e
Of the above 6 options, only in one case is the total number of banknotes 16 (C, b)
2*x + 5*(16-x) = 50
2x + 80 - 5x = 50
30 - 3x = 0
30 = 3x
x = 10
Ergo, 10 * $2 and (16-10 = 6) * $5
We don’t teach our pri school kids to use algebra to solve word problems, but of course, your method is elegant.
I gotta teach my son to use algebra next time!
10 notes x 2$ = 20$ 6 notes x 5$ = 30$
I bet you kinda rolled your eyes at the inane things my education system is testing kids haha
9 years old kids?
I guess you only have a 5$ ticket, it has 32$ in 2$ bills and the remaining (13) that is not mentioned will be 1$ tickets
Effort for trying!
6
Good job
Did you post anonymously on that forum?~~
No.
I used my real name to post my blog articles to drive traffic to my blog haha.
I took a screenshot of that question because I wanted to consolidate concepts about money on my blog haha
If don't want to use algebra, just logical thinking...
For a total amount of $50, we need:
The answer: Harry have 6 pieces of $5 notes.