# IPT question help (1 Viewer)

#### LegoSlaughter

##### New Member
It involves counting in binary, which I'm pretty sure isn't in this syllabus, but basically:
000 = 0
001 = 1
010 = 2
011 = 3
100 = 4

and so on until it reaches the maximum value for 3 bits which is 111 = 7.
For this question it asks you to take a look at 2, 3 and 4 seconds and it samples it at each dotted point, which correlates to a number for height on the y axis. So you merely convert each value to binary and that's your answer.
At 2 seconds height is 2 so the first value is 010
At 3 seconds height is 4 so the second value is 100
And at 4 seconds height is one so the last value is 001

However I really wouldn't worry about it, I HIGHLY doubt that will be in the exam - nothing to do with sampling and converting to binary anyway. Could be questions on it in the MM topic if you do that.

• cossine and queenb_3

#### queenb_3

##### Active Member
It involves counting in binary, which I'm pretty sure isn't in this syllabus, but basically:
000 = 0
001 = 1
010 = 2
011 = 3
100 = 4

and so on until it reaches the maximum value for 3 bits which is 111 = 7.
For this question it asks you to take a look at 2, 3 and 4 seconds and it samples it at each dotted point, which correlates to a number for height on the y axis. So you merely convert each value to binary and that's your answer.
At 2 seconds height is 2 so the first value is 010
At 3 seconds height is 4 so the second value is 100
And at 4 seconds height is one so the last value is 001

However I really wouldn't worry about it, I HIGHLY doubt that will be in the exam - nothing to do with sampling and converting to binary anyway. Could be questions on it in the MM topic if you do that.
Thank you so much! You explained it really well!

• LegoSlaughter

#### seremify007

##### Junior Member
I would've thought counting in base 2 (binary) was something more likely to pop up in a 2u maths paper than IPT!