She just doesn't know the answer and is looking to change the subject.

Since I have no life:

y = Younger brother's position
o = Older brother's position
t = time

y = 30t

When t<6:
o = 0

When t=6 or t>6:
o = 60(t-6)

To solve the problem, we need to find t for when y=o.
y = o
30t = 60(t-6) = 60t-360
30t = 360
t = 12

Solution: The brothers will meet up after 12 minutes.

mrtonton said:

You kinda overcomplicated that problem pretty hard. It's just basic algebra.

60x = (30 * 6) + (30 * x) "simplify"
60x = 30 * (6 + x) "divide 30"
2x = 6 + x "subtract x"
x = 6

But, it's not real clear whether it's asking for total time elapsed since the start of the problem or time taken after the big brother takes off. Even then, it would be (x + 6) for total time.

Hoobajoob said:

Take a closer look at the last four lines I wrote. It's almost literally the same exact equation that you wrote except it's just rearranged, lol. The only difference is that you're using 30*6 (the distance the younger brother travels in 6 minutes) whereas I use the number 360, which came from 60(t-6) (as in, the older brother's distance only goes above 0 after 6 minutes have already passed), which gives us two different starting points ("time starts when the younger brother leaves" vs "time starts when the older brother leaves"). Everything before that is just an explanation of where the numbers are coming from.

Either way, the easiest way to solve the problem is probably to just forget about math altogether and just use logic. Since the older brother is twice as fast as the younger brother, it would only make sense that he would catch up with the younger brother after double the time that he waited before leaving. Since he waited 6 minutes, the answer must be the double of that. Therefore, he meets up with the younger brother 12 minutes after the younger brother left.

im not good with formula so i often use this method instead of try to recall the formulas

younger bro : 30 m/minutes
older bro : 60m/minutes

younger bro leave 6 minutes earlier than older bro
30 m/minutes x 6 = younger bro is 180m ahead

the moment older bro leave :
younger bro is already 180m ahead
meanwhile older bro just leave the house so 0m [?]
then i add the number that i got from the question till i meet the same number for both sides

180m + 30m + 30m + 30m + 30m + 30m + 30m = 360m away from house

0m + 60m + 60m + 60m + 60m + 60m + 60m = 360m away from house

so the answer is , they should have meet in 6 minutes after younger bro left

What if the school is just across the street?

FJH said:

What if the school is just across the street?

A black hole swallows the universe.