Problem Solving

If two pipes A and B together can fill a cistern in x minutes and if A alone can fill it in a minutes more than x minutes and B alone can fill it in b minutes more than x minutes, then what is x equal to?
(A) √ (a^2 + b^2)
(B) √ (a^2 - b^2)
(C) a b
(D) √ (a b)
(E) a b - a^2 - b^2

1/(a+x)+1/(b+x)=1/x

(a+x)(b+x)/(a+b+2x)=x

ax+bx+2x^2=ab+ax+bx+x^2

x^2=ab

x=√ab

sanju09 wrote:If two pipes A and B together can fill a cistern in x minutes and if A alone can fill it in a minutes more than x minutes and B alone can fill it in b minutes more than x minutes, then what is x equal to?
(A) √ (a^2 + b^2)
(B) √ (a^2 - b^2)
(C) a b
(D) √ (a b)
(E) a b - a^2 - b^2

since this relation is a generic one, it should be satisfied by all the values.

Let's take equal and simple values, and then go about it.

let A and B each fill the cistern in 4 mins. therefore, together they will take half the time , i.e. 2 mins to fill the cistern.

according to the question X= 2, a=2, b=2

put the values, check for correctness.

harshavardhanc wrote:

sanju09 wrote:If two pipes A and B together can fill a cistern in x minutes and if A alone can fill it in a minutes more than x minutes and B alone can fill it in b minutes more than x minutes, then what is x equal to?
(A) √ (a^2 + b^2)
(B) √ (a^2 - b^2)
(C) a b
(D) √ (a b)
(E) a b - a^2 - b^2

since this relation is a generic one, it should be satisfied by all the values.

Let's take equal and simple values, and then go about it.

let A and B each fill the cistern in 4 mins. therefore, together they will take half the time , i.e. 2 mins to fill the cistern.

according to the question X= 2, a=2, b=2

put the values, check for correctness.

heheheh!! U logic wont work bro....Try for other values of X where a is not equal to b...

Only the straight method wrks as posted by @truplayer256.

gmatmachoman wrote:

harshavardhanc wrote:

sanju09 wrote:If two pipes A and B together can fill a cistern in x minutes and if A alone can fill it in a minutes more than x minutes and B alone can fill it in b minutes more than x minutes, then what is x equal to?
(A) √ (a^2 + b^2)
(B) √ (a^2 - b^2)
(C) a b
(D) √ (a b)
(E) a b - a^2 - b^2

since this relation is a generic one, it should be satisfied by all the values.

Let's take equal and simple values, and then go about it.

let A and B each fill the cistern in 4 mins. therefore, together they will take half the time , i.e. 2 mins to fill the cistern.

according to the question X= 2, a=2, b=2

put the values, check for correctness.

heheheh!! U logic wont work bro....Try for other values of X where a is not equal to b...

Only the straight method wrks as posted by @truplayer256.

Gmatmachoman..buddy, I know you are very experienced on this forum and I think you have a strong prep too.

But, this CR has made me so adamant that I cannot take a plain no. I need solid reasoning for any argument.

now coming to my previous response, there is NO logic involved in my first post. None at all.

it's like saying : if you have an equation of line as X+Y = 2, sum of every set of value on this line will be 2 . Simple.

Now let me show you how this is true. I'll take some difficult numbers, calculation verification for which is left to you .

Let A take 4 mins to fill the cistern and B take 5 mins. So working together, they well take 20/9 mins to fill the cistern

Hence, per the question 'a' becomes 4-(20/9) = (16/9) and 'b' becomes 5-(20/9) = 25/9

therefore X = 20/9 = sqrt( 16/9 * 25/9)

no option other than D will give you the correct answer.

plugging in the values ALWAYS works. Sometimes it gives two probable answers. You then need to take another set which eliminates the wrong choice.

agreed Bro!!

Seems some Hitch in my statement!!

Apologies from my end bro!!

gmatmachoman wrote:agreed Bro!!

Seems some Hitch in my statement!!

Apologies from my end bro!!

not a prob ...to err is human

harshavardhanc wrote:

gmatmachoman wrote:

harshavardhanc wrote:

sanju09 wrote:If two pipes A and B together can fill a cistern in x minutes and if A alone can fill it in a minutes more than x minutes and B alone can fill it in b minutes more than x minutes, then what is x equal to?
(A) √ (a^2 + b^2)
(B) √ (a^2 - b^2)
(C) a b
(D) √ (a b)
(E) a b - a^2 - b^2

since this relation is a generic one, it should be satisfied by all the values.

Let's take equal and simple values, and then go about it.

let A and B each fill the cistern in 4 mins. therefore, together they will take half the time , i.e. 2 mins to fill the cistern.

according to the question X= 2, a=2, b=2

put the values, check for correctness.

heheheh!! U logic wont work bro....Try for other values of X where a is not equal to b...

Only the straight method wrks as posted by @truplayer256.

Gmatmachoman..buddy, I know you are very experienced on this forum and I think you have a strong prep too.

But, this CR has made me so adamant that I cannot take a plain no. I need solid reasoning for any argument.

now coming to my previous response, there is NO logic involved in my first post. None at all.

it's like saying : if you have an equation of line as X+Y = 2, sum of every set of value on this line will be 2 . Simple.

Now let me show you how this is true. I'll take some difficult numbers, calculation verification for which is left to you .

Let A take 4 mins to fill the cistern and B take 5 mins. So working together, they well take 20/9 mins to fill the cistern

Hence, per the question 'a' becomes 4-(20/9) = (16/9) and 'b' becomes 5-(20/9) = 25/9

therefore X = 20/9 = sqrt( 16/9 * 25/9)

no option other than D will give you the correct answer.

plugging in the values ALWAYS works. Sometimes it gives two probable answers. You then need to take another set which eliminates the wrong choice.

great advocacy of plugging in, this helps you foresee without pen sometimes, but a fair practice that is enough to decide where to apply plugging in policy, is always must.

sanju09 wrote: great advocacy of plugging in, this helps you foresee without pen sometimes, but a fair practice that is enough to decide

where to apply plugging in policy, is always must.

yes, I missed the *conditions apply part in my post. Thanks for supplementing my response with it !

Anyway, as the normal equation method had already been posted, I chose to reply with the plugging-in method. In that way, the readers are aware of another method which, in some cases, might be faster.

But as Sanju said, it's totally depends on the users discretion whether to apply that method in the actual exam.

harshavardhanc wrote:

gmatmachoman wrote:

harshavardhanc wrote:

sanju09 wrote:If two pipes A and B together can fill a cistern in x minutes and if A alone can fill it in a minutes more than x minutes and B alone can fill it in b minutes more than x minutes, then what is x equal to?
(A) √ (a^2 + b^2)
(B) √ (a^2 - b^2)
(C) a b
(D) √ (a b)
(E) a b - a^2 - b^2

since this relation is a generic one, it should be satisfied by all the values.

Let's take equal and simple values, and then go about it.

let A and B each fill the cistern in 4 mins. therefore, together they will take half the time , i.e. 2 mins to fill the cistern.

according to the question X= 2, a=2, b=2

put the values, check for correctness.

heheheh!! U logic wont work bro....Try for other values of X where a is not equal to b...

Only the straight method wrks as posted by @truplayer256.

Gmatmachoman..buddy, I know you are very experienced on this forum and I think you have a strong prep too.

But, this CR has made me so adamant that I cannot take a plain no. I need solid reasoning for any argument.

now coming to my previous response, there is NO logic involved in my first post. None at all.

it's like saying : if you have an equation of line as X+Y = 2, sum of every set of value on this line will be 2 . Simple.

Now let me show you how this is true. I'll take some difficult numbers, calculation verification for which is left to you .

Let A take 4 mins to fill the cistern and B take 5 mins. So working together, they well take 20/9 mins to fill the cistern

Hence, per the question 'a' becomes 4-(20/9) = (16/9) and 'b' becomes 5-(20/9) = 25/9

therefore X = 20/9 = sqrt( 16/9 * 25/9)

no option other than D will give you the correct answer.

plugging in the values ALWAYS works. Sometimes it gives two probable answers. You then need to take another set which eliminates the wrong choice.

I too used this plugging approach; but it is very time consuming and the numbers are messy. Is there an easier and faster method we can use?

Thanks

tgou008 wrote:

harshavardhanc wrote:

gmatmachoman wrote:

harshavardhanc wrote:

sanju09 wrote:If two pipes A and B together can fill a cistern in x minutes and if A alone can fill it in a minutes more than x minutes and B alone can fill it in b minutes more than x minutes, then what is x equal to?
(A) √ (a^2 + b^2)
(B) √ (a^2 - b^2)
(C) a b
(D) √ (a b)
(E) a b - a^2 - b^2

since this relation is a generic one, it should be satisfied by all the values.

Let's take equal and simple values, and then go about it.

let A and B each fill the cistern in 4 mins. therefore, together they will take half the time , i.e. 2 mins to fill the cistern.

according to the question X= 2, a=2, b=2

put the values, check for correctness.

heheheh!! U logic wont work bro....Try for other values of X where a is not equal to b...

Only the straight method wrks as posted by @truplayer256.

Gmatmachoman..buddy, I know you are very experienced on this forum and I think you have a strong prep too.

But, this CR has made me so adamant that I cannot take a plain no. I need solid reasoning for any argument.

now coming to my previous response, there is NO logic involved in my first post. None at all.

it's like saying : if you have an equation of line as X+Y = 2, sum of every set of value on this line will be 2 . Simple.

Now let me show you how this is true. I'll take some difficult numbers, calculation verification for which is left to you .

Let A take 4 mins to fill the cistern and B take 5 mins. So working together, they well take 20/9 mins to fill the cistern

Hence, per the question 'a' becomes 4-(20/9) = (16/9) and 'b' becomes 5-(20/9) = 25/9

therefore X = 20/9 = sqrt( 16/9 * 25/9)

no option other than D will give you the correct answer.

plugging in the values ALWAYS works. Sometimes it gives two probable answers. You then need to take another set which eliminates the wrong choice.

I too used this plugging approach; but it is very time consuming and the numbers are messy. Is there an easier and faster method we can use?

Thanks

In 1 minute, A alone can fill 1/ (x + a) of the cistern, B alone can fill 1/ (x + b) of the same cistern, whereas A and B together can fill 1/x of the same cistern.

That way


1/ (x + a) + 1/ (x + b) = 1/x

» (2 x + a + b)/ (x^2 + a x + b x + a b) = 1/x

» 2 x^2 + a x + b x = x^2 + a x + b x + a b

» x^2 = a b

» [spoiler]x = √ (a b)


D[/spoiler]

1/x = 1/(x+a) + 1/x+b
=> x(x+a+x+b) = (x+a)(x+b)
> x^2 = ab > x = sqrt(ab).

The easiest way is to plug in.

Formula: Ra+Rb=Rx
<=> 1/(a+x) + 1/(b+x) = 1/x
=> x = sqrt(ab)

The answer to this question is option D picking numbers would be the bets possible approach if anyone know which values would approximately nearby to the squareroot xy