There are 2 candles with the same height. It takes t mins for A to bur

Question

There are 2 candles with the same height. It takes t mins for A to burn, and it takes 2t mins for B to burn. When the candles are burnt at the same time, how many minutes did it take for the height of B to be twice the height of A?

A. \(\frac{t}{3}\)

B. \(\frac{2t}{3}\)

C. \(\frac{t}{2}\)

D. \(\frac{3t}{4}\)

E. \(\frac{4t}{5}\)

Show Answer

Official Answer

B

A.  \frac{t}{3}

B.  \frac{2t}{3}

C.  \frac{t}{2}

D.  \frac{3t}{4}

E.  \frac{4t}{5}

Diablobuster · Accepted Answer

woah woah woah let's not complicate this. Think logically.

A burns 2x faster than B. Logically, find a fraction of the height LEFT of A that can double and equal B. In other words, find a fraction of B that has burned and is 1/2 of what has burned off of candle A.

1/4 burned off of B, 1/2 burned off of A. Nope. 1/5 off B, 2/5 off A? Nope. We're going backward.

1/3 burned off of B, 2/3 burned off of A. This means 2/3 left of B, 1/3 left of A. Bingooooo.

It takes t time to burn all of A. We burned 2/3 of it. This means 2t/3 time has passed.

Aks111 · Answer

Lets assume it takes 'x' mins for height of B to be twice the height of A.
Let height of candle be H.

A takes t mins to burn full H height
So in 'x' mins A will burn =  \frac{Hx }{ t}  
Height of A remaining =  H - \frac{Hx }{ t} = H(1 - \frac{x}{t})

B takes 2t mins to burn.
So in 'x' mins B will burn =  \frac{ Hx }{ 2t }  
Height of B remaining =  H - \frac{Hx }{ 2t} = H(1 - \frac{x}{2t})

Height of B remaining = 2*Height of A remaining
 H (1-\frac{x}{2t}) = 2H (1-\frac{x}{t})

\frac{(2t-x) }{ 2t }= 2\frac{(t-x)}{t}

2t-x = 4t-4x

x= 2t/3

Answer: B

LamboWalker · Answer

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.

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=82671 An algebraic approach to this one. ___ Harsha GMAT-Club-Forum-8emnnc7u.png

shishirtu · Answer

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.

2 * ( 2  -  2 x) =  2  -  1 *x
x =2/3 (of t)

Answer: B

SteelBranch · Answer

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.

GMATGuruNY · Answer

See the figure at bottom.

Let x = the amount of candle that B loses.

Since A burns in 1/2 the time that B burns, A burns twice as fast as B.
Thus:
If B loses x inches, then A loses 2x inches, as shown in the figure below.

Let y = the amount of candle that A has left.

B's remaining candle must be twice as high as A's.
Thus:
If y inches of A remains, then 2y inches of B must remain, as shown in the figure below.

Since the height of each candle is the same in each case, we get:
2x + y = x + 2y
x = y

Thus, the total height of A = 2x + y = 2x + x = 3x.
Since the fraction that A loses =  \frac{2x}{3x }= \frac{2}{3},  candle A requires 2/3 of its total time to burn down to the required level.

B

There are 2 candles with the same height. It takes t mins for A to bur

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There are 2 candles with the same height. It takes t mins for A to bur

Prep Toolkit

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