Machines A and B, working simultaneously at their respective constant

Hello all

My attempt:

Let the rate of \(Machine A\) be \(a\) units per hour and that of \(Machine B\) be \(b\) units per hour.

From first statement we have
\(a*1 + b*1 = m\)\(........equation I\)

From the second statement we have
\(2m/a - 2m/b = 3\)\(..........equation II\)

We need to find out the value of \(\frac{m}{a}\). Solving \(I\)and \(II\) and substituting \(x\) for \(\frac{m}{a}\) we get
\(2x^2 - 7x + 3 = 0\)
we get \(x = 0.5\) or \(3\)
and none of these exist in the options
so perhaps I am on the wrong path.
Any help appreciated.

My answers are \(0.5 hours\) or \(3 hours\)

The question is: how long does it take machine A to produce 5m units, not m units.

Thanks.

Then the answer is \(E\) \(15 hours\).

\(x=0.5 or 3\)
But we need \(5x\) therefore \(2.5 hours\) or \(15 hours\). We reject the solution \(x=0.5 hours\) because it invalidates the statement 1.

hi,
i think the answer should be 3 for A and 1.5 for B...
Since A and B together produce m units together in 1 hr so individually it has to be more than 1..
secondly, the difference in producing 2m is 3 , so in producing m the difference will be 1.5...
this means ans for A> 1+1.5 ie 2.5... only D and E are there which dont fit in...
let A take x hr , B will take x-1.5 hr.....so 1/x + 1/(x-1.5)=1... 3 fits in... answer should be 3...


since answer is for 5m ans is 5*3=15

together A and B in 1 Hour can produce M units

A's 1 Hr + B's 1 Hr = M --- Equation 1

Let A takes x Hours to Produce 2M => A in one hour can produce 2M/X
As per the statement
B takes X-3 hours to produce 2M => B in one hour can produce 2M/(X-3)

Putting A's 1 hour and B's 1 Hour in equation 1

2M/X + 2M/(X-3) = M

on simplifying we will get
X = 1 or X = 6

X cannot be equal to 1, since as per the condition B takes (X-3) hours i.e -2 Hours to produce 2 M (Not possible)

Hence X = 6

A takes 6 Hours to Produce 2M
and hence 15 Hours to produce 5M

Option E

BTW the question has 1-minute shortcut solution. Without any equations.

Would love to know some tricks! I am poor at these work type problems. I usually take the clean approach because I know this is not a strength area for me and thus my mind should not get muddled up after spending few seconds on the question.

Thanks!

Consider this: if together A and B produce 5m units in 5 hours (m in 1 hour). Can only A produce 5m in LESS than 5 hours? This eliminates A, B and C right away.

Now, can you tell me how can we eliminate the remaining incorrect answer?

This is how I approached the question.

combined rate is m units in 1 hr and B takes 3 hrs less then A to produce 2m units.

Combined rate will take 2 hr to produce 2m units ( just to make same number of units) . Say A takes x hrs to produce 2m units, so B will take x-3 hrs to produce 2m units.

so we have equation

(combined rate) 1/2 = 1/x-3 + 1/x ( individual rates of A and B)

solving this we have => x^2 -7x +6 = 0, so we have two values of x = 1 & 6 ( 1 is not possible as this will make B's rate -ve)

so we have x=6, this is the rate of A.

Now A takes 6 hrs to produce 2m units and for 5m units he will take 15 hrs.

Soln E

6 hours is also not possible, if A takes 6 hrs to produce 5m => to produce 2m, A needs 12/5 hrs = 2.4 hrs.. Now B needs 3 hours lesser than A => 2.4 - 3 = negative value, which is not possible

To check E, if A takes 15 hrs to produce 5m => to produce 2m, A needs 6 hrs, and B needs 3 hours. (to cross check, A's rate = 1/3 and B's rate = 2/3 => m/hour rate)

Hence the answer is E

hi,

Since A and B together produce m units together in 1 hr so individually it has to be more than 1..
secondly, the difference in producing 2m is 3 , so in producing 5m the difference will be 7.5...
this means ans for A> 1+7.5 ie 8.5... only E LEFT...
ans 15 E

Given: Machines A and B, working simultaneously at their respective constant rates produce m units in 1 hour.
Asked: If working independently it takes machine B 3 hours less than machine A to produce 2m units, how long does it take machine A to produce 5m units?

Let us assume machine A produces x units in 1 hour
Machine B produces (m-x) units in 1 hour

Time taken by machine A to produce 2m units = 2m/x hours
Time taken by machine B to produce 2m units = 2m/(m-x) hours

2m/x - 2m/(m-x) = 3
(2m(m-x) - 2mx)/x(m-x) =3
2m(m-2x) = 3x(m-x)
2m^2 - 4mx = 3mx - 3x^2
2m^2 - 7mx + 3x^2 = 0
2t^2 - 7t + 3 = 0 ; where t= m/x
2t^2 - 6t - t + 3 = 0
(2t-1)(t-3) = 0
t = 1/2 or t = 3

Definitely m>x ; t= m/x >1; t=3
t= m/x = 3
Take take by machine A to produce 5m units = 5m/x = 5t = 15 hours

IMO E

A and B working together takes 1 hour to produce M units --> This tells that working alone A would take more than 1 hour to complete M units.

So to produce 5M units, the time taken would definitely greater than 5 hours. We can eliminate options A,B and C.

Now in the remaining options, if we conider D, time taken to produce 5m units = 6 hours, then M units = 1.2 hours

So to produce 2m units, A would take 2.4 hours, but considering the information that B takes 3 hours less than A to produce 2m units, then B would take -0.4 hours to produce 2m units which is not possible.

So only possible option is E - 15 hours

Ans: E

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