Just cant work the correct answer out for this type of question. Any advice would be a great help. Picture attached.

Question:

You are overhead ‘A’ at A015 on track to ‘B’ where ATC instructs you to climb to A055.
If the planned groundspeed during climb is 110Kt & you are to remain within class CHARLIE airspace, the minimum rate of climb required is closest to.
A. 430 fpm
B. 340 fpm
C. 370 fpm
D. 250 fpm

My calculation:
110kts in 10Nm = 5.5mins
1500 feet (500 above LL2500) in 5.5 mins =
= B. 273 FPM

What am I doing wrong here?

Thanks

The wording of your example is ambiguous. But I am assuming that the question implies that the instruction to climb was given when you were over A.

I don't see anything wrong with your reasoning. You have 10nm to the first CTA step and you should be 500ft above that when you get there. 10nm @ 110kt = 5.45mins (to be exact). You have to gain 1500ft in that time so you will need a rate of climb of 275ft/min. The other CTA steps beyond 10nm are irrelevant, if you clear the first one, you will certainly clear the others.

Thanks for the reply

Though a little confused in why you would need to gain 3000ft in 10nm?
Why wouldnt it be 1500ft in 10nm?
(2500 + 500) - 1500.

Thinking about this one some more... If the distances are from point ‘A’, wouldnt the 15nm step be the most critical since being the steepest/ shortest step.
I have attached my working so you can see what I am thinking.
Its the only way I can work out to get to the correct answer of B- 340fpm.
2500ft /8.2mins = 305fpm

Thanks

You are correct. On first reading of your diagram, I didn't notice that the 15nm is actually from A. So that does make it only 5nm to the next CTA step.

Thanks for the verification Bob, appreciate it!