Word Problem

[Moy12]

The 12-hour digital clock above shows one example of a time at which the sum of the digits representing the time is equal to 20 (time shown is 8:57). During a 12-hour period, starting at noon, for how many minutes would the sum of the digits displayed be greater than or equal to 20?

I made a chart running from 12-noon to 12 am.

12 sum of digits can't be equal or greater than 20
1 "dido"
2 "dido"
3 "dido"
4 "dido"
5 "dido"
6 from 6:59-7:00 we get 1 minute
7 from 7:58-8:00 we get 2 minutes
8 from 8:57-9:00 we get 3 mins
9 from 9:56-10:00 we get 4 mins
10 sum of digits can't be equal or greater than 20
11 dido
12 dido

Sum is then 1+2+3+4=10

but according to my book, the answer is 20. Where am I going wrong?

 

[Moy12]

Got it

I found the explanation for it in my book: I guess I didn't think hard enough.

7 7:49 is another minute that counts
8 8:48-8:49 and 8:39 also count
9 9:47-9:49 and 9:38-9:39 and 9:29 also count


The total minutes are actually 20.

 

[soroban]

Hello, Moy12!

The 12-hour digital clock above shows one example of a time at which
the sum of the digits representing the time is equal to 20 (time shown is 8:57).
During a 12-hour period, starting at noon, for how many minutes
would the sum of the digits displayed be greater than or equal to 20?

Note: The number \(\displaystyle 5\!:\!37\) represents one minute,
. . . . .the time from \(\displaystyle 5\!:\!37\!:\!00\) through \(\displaystyle 5\!:\!37\!:\!59\).
\(\displaystyle \text{Then we have: }\;\;\begin{Bmatrix}6\!:\!59\end{Bmatrix} \; 1 \qquad \begin{Bmatrix}7\!:\!49 \\ 7\!:\!58 \\ 7\!:\!59\end{Bmatrix} \; 3 \qquad \begin{Bmatrix}8\!:\!39 \\ 8\!:\!48 \\ 8\!:\!49 \\ 8\!:\!57 \\ 8\!:\!58\\8\!:\!59 \end{Bmatrix} \; 6 \qquad \begin{Bmatrix}9\!:\!29 \\ 9\!:\!38 \\ 9\!:\!39 \\ 9\!:\!47 \\ 9\!:\!48 \\ 9\!:\!49 \\ 9\!:\!56 \\ 9\!:\!57 \\ 9\!:\!58 \\ 9\!:\!59 \end{Bmatrix}\; 10 \)
\(\displaystyle \text{Therefore: }\:1+3+6+10 \:=\:20\text{ minutes.}\)