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Bunuel
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How I would do it mentally:

Firstly, I planned to use smart numbers. Did not want to get into algebra.
My aim was come up with a 't' that would give an integer value for all options.
To get that, the LCM of denominators of all options - 2, 3, 4, 5 i.e. 60.

Now t = 60, in 60 mins candle A will burn up entirely. Additionally, since we are asked a T for which h(B) = 2*h(A), T has to be more than t/2. Therefore, A and C are eliminated.

Now I have to test values. Starting with option B, which gives a value of 40, candle A will be 1/3 of its original height. Similarly, candle B will be 2/3 of its original height. Therefore, OA is B.
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An algebraic approach to this one.

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Ratio of time, A:B= 1:2
Ratio of burning speed, A:B= 2:1

Means,
A burns at a rate of 2 unit/min
B burns at a rate of 1 unit/min

let the height of both candles be 2 unit.
let x = minutes to take for the height of B to be twice the height of A.


[2A = B]
2 * (2 - 2x) = 2 - 1*x
x =2/3 (of t)

Answer: B
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There maybe a faster graphical method to solve this or this kind of questions.

Candle A |-------------------------------------------------------|base

Candle B |-------------------------------------------------------|base

Both have same height (see in the horizontal manner)
Candle a is 2x faster.


So let's see how much time 't' has elapsed as given in the problem.

1. At this time Candle A has burnt twice (like at any pt in the given timeline) than B
2. And it remaining height is half of remaining height of B

Candle A |------------------------------------|-------------------|base. Faster

Candle B |------------------|-------------------------------------|base. Slower

Thus at this time
1. Burning of B has to be half of burning of A
2. Remaining of A is half of remaining of B (given)

Thus visualise the burning as follows

Candle A |---------t/3---------!---------t/3---------|----------t/3---------|base. Faster

Candle B |----------t/3--------|----------t/3--------!-----------t/3--------|base. Slower


Hence twice the t/3 time spent. Answer 2/3t.
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