Candle A and Candle B have equal heights but different volumes. While

Question

Candle A and Candle B have equal heights but different volumes. While the candles are burning, the height of each candle decreases at its own individual rate, with Candle A taking a total of t minutes to completely burn down and Candle B taking a total of 2t minutes to completely burn down. If both candles begin burning at the same time, in terms of t, how many minutes will it take for Candle B's height to be twice Candle A's height?

A) 1/4 t
B) 1/3 t
C) 2/3 t
D) 3/4 t
E) t

A.  \frac{1}{4} t

B.  \frac{1}{3} t

C.  \frac{2}{3} t

D.  \frac{3}{4} t

E.  t

ManishaPrepMinds · Accepted Answer

candle A   
Total Burning Time t mins
Total Height( Assume ) 1
Rate of Burning  \frac{1}{t }units/min  
  candle B   
Total Burning Time 2t mins
Total Height( Assume ) 1
Rate of Burning  \frac{1}{2t}  units/min

Let's take x as the number of mins when height of B is twice of A's height
that means we need to figure out what amount of height burnt in x mins
For    candle A   
  Amount of height burnt in x mins   = x* \frac{1}{t } =  \frac{x}{t} 
For    candle B   
  Amount of height burnt in x mins   = x*  \frac{1}{2t }= \frac{x}{2t}  
Height Remaining A=1- \frac{x}{t} 
Height Remaining B = 1-  \frac{x}{2t} 
  As mentioned after x mins height of B is twice of A's height, hence  
 2(1-\frac{x}{t}) = 1- \frac{x}{2t} 
2-1=  \frac{x}{t}   
1 =  \frac{3x}{2t} 
x= \frac{ 2t}{3}

Answer C

richdoggo · Answer

I did it by process of elimination. A will always melt by the fraction presented, and B will melt by half of the fraction presented.

A: 1/4 T
 
 A: Has 3/4 remaining because 1/4 is gone 
 B: Has 3.5/4 remaining because 0.5/4 is gone (half of what A melted)

B: 1/3 T

A: Has 2/3 remaining because 1/3 is gone 
 B: Has 2.5/3 remaining because 0.5/3 is gone

C: 2/3 T -  (correct answer) 
 
 A: Has 1/3 remaining because 2/3 is gone 
 B: Has 2/3 remaining because 1/3 is gone  
2 is double 1.

D: 3/4 T
 
 A: Has 1/4 remaining because 3/4 is gone 
 B: Has 2.5/4 remaining because 1.5/4 is gone

E: 1 T
 
 A is completely melted, so cannot be correct

Bunuel · Answer

Let's assume that time interval is x minutes.

In x minutes, the height of Candle A reduces to (t - x)/t of its original height, and Candle B's height becomes (2t - x)/(2t). For instance, if it takes t = 10 minutes for Candle A to burn down completely, then after x = 1 minute, its height is reduced to (10 - 1)/10 = 9/10 of the original height.

To find when Candle B's height is twice that of Candle A's, we set up the equation:

2*\frac{t - x}{t} = \frac{2t - x}{2t}

4*(t - x) = 2t - x

4t - 4x = 2t - x

2t = 3x

x = \frac{2t}{3} .

Answer: C.

unraveled · Answer

Candle A and Candle B have equal heights but different volumes. While the candles are burning, the height of each candle decreases at its own individual rate, with Candle A taking a total of t minutes to completely burn down and Candle B taking a total of 2t minutes to completely burn down. If both candles begin burning at the same time, in terms of t, how many minutes will it take for Candle B's height to be twice Candle A's height?

A) 1/4 t
B) 1/3 t
C) 2/3 t
D) 3/4 t
E) t

A) 1/4 t
B) 1/3 t 
C) 2/3 t 
D) 3/4 t 
E) t
A little brainstorming helps here.

Let's take choice at hand first if algebra doesn't clicks.
The time t can't be possibly our answer as in that case A is already burnt out. E's out
Now, let's say A get half of its initial height and that happens in  \frac{t}{2}  time. Out of simple mental calculation we can see that this is not the answer but it gives us an anchor point.

Here B's burning speed is half of A's.
So, B would have decreased by  \frac{t}{2}/2t  i.e.  \frac{1}{4}^{th}  of original height 
OR
B's height would be a multiple of factor  \frac{3t}{4}  since  \frac{t}{2}  is  \frac{3}{4}^{th}  of total time 2t of B.

Hence any time less than  \frac{t}{2}  time are eliminated. 
OR 
The time must be between  \frac{t}{2}  and t.

Thus, we are left with C and D. 
Checking for  \frac{2t}{3}  time we have A's height of  1 - \frac{2}{3} = \frac{1}{3}^{th}  of original height and B would have decreased by  \frac{2t}{3}/2t = \frac{1}{3}^{th}  of original height OR after  \frac{2t}{3}  time it's height is a factor of  1 - \frac{1}{3} = \frac{2}{3}^{th}  of original height.

Therefore B being twice the height of A.

Answer D.

ostrick5465 · Answer

H  is the original height of two candles.
 x  is the number of minutes we need to find.
Burning rate of candle A is  \frac{H}{t} 
Burning rate of candle B is  \frac{H}{2t} 
The height of candle A at a particular time in terms of H and t :  H-\frac{H}{t}*x 
The height of candle A at a particular time in terms of H and t :  H-\frac{H}{2t}*x 
=> When Candle B's height is twice Candle A's height:  (H-\frac{H}{t}*x) : (H-\frac{H}{2t}*x) = 1 : 2 
=>  2-\frac{2x}{t} = 1-\frac{x}{2t} 
=>  1 = \frac{3x}{2t} 
=>  x = \frac{2t}{3} 
=> Choice C

KarishmaB · Answer

Here is the video explanation to this problem: https://youtube.com/shorts/M2pPLTp6mpA?feature=shared Assume when candle A's height is x, candle B's height is 2x. Since time taken to burn fully is in the ratio t:2t, their rates of burning are 2:1. Hence 2x of candle A must have burnt and x of candle B must have burnt (so that they started with the same height) For candle A, 3x height was burnt in t mins. So 2x height would be burnt in t*2/3 mins Answer (C) Ratios and work rate are discussed here: Ratios: https://youtu.be/5ODENGG5dvc Work Rate: https://youtu.be/88NFTttkJmA

av722 · Answer

KarishmaB  how do you know to divide the candle's height into three pieces? couldn't it just as easily be four or five pieces?

Kinshook · Answer

@Khk_BeastmodeGiven: Candle A and Candle B have equal heights but different volumes. While the candles are burning, the height of each candle decreases at its own individual rate, with Candle A taking a total of t minutes to completely burn down and Candle B taking a total of 2t minutes to completely burn down.

Asked: If both candles begin burning at the same time, in terms of t, how many minutes will it take for Candle B's height to be twice Candle A's height?
Let the height of each candle be h cm.

Candle A: - 
h = kt: where k is burning rate of candle A in cm/minute
After time m minutes
Height of candle A = h - km = kt - km = k(t-m)

Candle B: - 
h = kt = k'2t: k' = k/2; where k' is burning rate of candle B in cm/minute
After time m minutes
Height of candle B = h - k'm = kt - km/2 = k(t-m/2)

Let us assume that after m minutes, Candle B's height will become twice of Candle A's height.
k(t-m/2) = 2k(t-m)
kt - mk/2 = 2kt - 2km
t = 2m - m/2 = 3m/2
m = 2t/3 minutes

IMO C

KarishmaB · Answer

I don't. There are 2 parts - a part that is burnt and the part remaining. 
If x and 2x are remaining, and rates of burning are 2:1 (height burnt per unit of time), it means 2a and a have burnt in the same time. Because both had the same height, x = a.

rak08 · Answer

does volume have no relevance then?

Bunuel · Answer

Volume has no relevance here because the question is only about  height .

Burning rate is given in terms of time to burn the full height, so the volume or thickness of the candles doesn’t affect the answer.

Only the decrease in  height  over time matters for this question.

MartyMurray · Answer

Candle A and Candle B have equal heights but different volumes. While the candles are burning, the height of each candle decreases at its own individual rate, with Candle A taking a total of t minutes to completely burn down and Candle B taking a total of 2t minutes to completely burn down. If both candles begin burning at the same time, in terms of t, how many minutes will it take for Candle B's height to be twice Candle A's height?

A) 1/4 t
B) 1/3 t
C) 2/3 t
D) 3/4 t
E) t

Let the candles' height be  1 .

The rate at which Candle A burns is  \frac{1}{t} .

The rate at which Candle B burns is  \frac{1}{2t} .

Let  T  be the amount of time it will take for Candle B's height to be twice Candle A's height.

1 - (T × \frac{1}{2t}) = 2(1 - (T × \frac{1}{t}))

1 - \frac{T}{2t} = 2 - \frac{2T}{t}

2 - \frac{T}{t} = 4 - \frac{4T}{t}

\frac{3T}{t} = 2

T = \frac{2}{3}t

A) 1/4 t
B) 1/3 t 
C) 2/3 t 
D) 3/4 t 
E) t

Answer:   C

Dream009 · Answer

How I solved it was to assume this as a speed and time problem.
Candle A = Assume speed of burn is 20 and time taken is t. Distance to cover is 20T. We have to find an x by which distance left if multiplied by 2 is twice of distance left of candle be- let that time be x. 20t-20x (since 20 is the speed of travel) 
Candle B = Assume speed of burn is 10, time taken is 2t. Distance to cover is 20t. At same x value candle B would have remaining distance to cover of 20t-10x

Now we want to establish candle A distance left to travel * 2 = Candle B distance left to travel
=> 2 (20t-20x) = 20t - 10x
=> 40t - 40x = 20t - 10x
=> 20t = 30x
=> 2/3t = x = which is your answer.

Adit_ · Answer

Here the work done is the reduction of height not the increase of height.

So the equation firstly is:

For A and B respectively"

t  = H.
2t  = H.

Final equation we need is:

Height of B = 2 
=  H - Hx/2t = 2  
What it says is that the height = Total initial height - amount of work done to reduce that height by candle.

Solving the equation we get x =  2t/3 .

Answer:  Option C

Alasoy · Answer

first of all, to finisih this problem, what you need to note is that

1. Height for both candle is the same
2. Velocity of burning is different
3. time used is stated in the question
4. (most important) the end result after burning (where candle b = 2 candle a heights) is coming from Original Heights - Candle Burns (which is formula -> Final Height = Original Height - Burn Height

then we have the variables here

Original Height (candle a) = Va x T
Original Height (candle b) = Vb x 2T

Question, how many minutes (further will be addressed as x) will it takes for candle B Final Height is equals to candle A Final Heights

How to Finish:

FH B = 2 FH A

turns every variable into our 4th principle: Final Heights = Original Heights - Burns (velocity * times needed) 
OH - Vb * x = 2 (OH - Va * x)

change every velocity variable into OH/Times needed to fully burn 
OH - (OH/2T * x) = 2 (OH - OH/T * x)

Eliminates Original Heights 
1 - x/2T = 2 - 2x/T

substitute

2x/T - x/2T = 1

x = 2T/3

AndyRugger · Answer

This might be my all-time favorite GMAT quant question, mostly because A) I totally flubbed it on my practice test and B) I learn something new every time I come back to it.

If you miss the clever bit of algebra that the remaining height of each candle can be modeled as   -  * , which comes out to H-H(x/t) and H-H(x/2t) for Candles A and B, respectively, then this problem can be really confusing to solve.

You can get a good intuitive sense of the relationship by plotting the height of the candles as a function of time on an X-Y graph, but it still doesn't jump out at you when the height of Candle B must be twice the height of Candle A.

However, you can actually solve this question in about two minutes by picking a "smart denominator" for the variable "t," and then numerically checking the answer choices. I have quite slow arithmetic skills, but even I solved it in about 2:10 minutes.

- Let T=10 minutes (to make fractions easier). In that case, Candle A burns down all the way in 10 minutes; Candle B burns down all the way in 20 minutes.
- In 5 minutes, Candle A has 1/2 height remaining, Candle B has 3/4 remaining. Candle B is less than twice Candle A. Not the answer.
- In 8 minutes, Candle A has 1/5 height remaining; Candle B has 3/5 remaining. Candle B is more than twice Candle A. Not the answer, but you're honing in!

Obviously, 5/10 = 1/2 and 8/10 = 4/5, so you know the final answer must be between those ratios. It's either Answer C or D. Let's try T=7 minutes, which will split the two.

- In 7 minutes, Candle A has 3/10 height remaining; Candle B has 13/20 remaining. Candle B is barely more than twice Candle A. So the answer must be barely less than 7/10 and greater than 5/10, and the only answer in that interval is .... ANSWER C: 2/3

(Granted, the algebraic way is definitely better, but this a more intuitive brute-force approach if you don't "see" the algebraic relationship between height of the candles and work done.)

Candle A and Candle B have equal heights but different volumes. While

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