At Supersonic Corporation, the time required for a machine to complete

I agree with manpreet's solution, but hopefully this adds a bit clarity:

First thing you need to know for this type of problem is the work problem rules (if you have the Official Guide, look at section 4.4.2). The formula for solving work problems where two machines work together is:

\(\frac{1}{r} + \frac{1}{s} = \frac{1}{h}\)

r = number of hours it takes machine A to complete a unit;
s = number of hours it takes machine B to complete a unit;
h = number of hours it takes both machines working together

This video explains the simplification that Manpreet did.
[youtube]https://www.youtube.com/watch?v=1oiy9OjwxLg[/youtube]

Now that we have our new, simplified formula, we apply it to our current problem:
t(A) = time for machine A to complete a unit = r
t(B) = time for machine B to complete a unit = s
t(h) = time for both machines working together


(\(\sqrt{8}-\sqrt{8-1}\)) + (\(\sqrt{7}-\sqrt{7-1}\))

= (\(\sqrt{8}-\sqrt{7}\)) + (\(\sqrt{7}-\sqrt{6}\))

=(\(\sqrt{8}-\sqrt{6}\)) = \(\frac{1}{h}\)

Now, multiply by (\(\sqrt{8}+\sqrt{6}\)) / (\(\sqrt{8}-\sqrt{6}\)) so that you can simplify the answer to match the answer choices.



You found your answer, choice B.

You may be thinking: what if you don't know exactly how to factor these equations during the exam. It may end up taking a lot of time. I don't recommend it as the primary way, but if you can get close to the answer, but it isn't formatted in a way reflected by the answer choices, try to solve it out (if it can be done quickly) or approximate a number and find the answer choice that matches.

Alternatively, if you're REALLY short on time, you can try to eliminate some answer choices early on by partially solving the problem and thereby improve your chances of guessing correctly.

Thank you for elaborating the above solution. i believe it will help other students as well.

Calculation of rate of work of machines A and B:-
\(t=\sqrt{w}+\sqrt{(w-1)}\)
Machine A=> \(t\) units in \(\sqrt{8}+\sqrt{7}\)
therefore \(1\) unit in \(1/\sqrt{8}+\sqrt{7}\).
With rationalisation, that is multiplying and dividing by \(\sqrt{8}-\sqrt{7}\) machine A does \(1\) unit of work in \(\sqrt{8}-\sqrt{7}\) hours.

Similarly, Machine B does \(1\) units of work in\(\sqrt{7}-\sqrt{6}\).
Adding the per unit rate of work of both the machines, we get \(\sqrt{8}-\sqrt{6}\).

\(\frac{1}{\sqrt{8}-\sqrt{6}}\).

With rationalisation we get option c as the answer.

could you please explain, does this property always hold true?

Absolutely!! Let me explain.......

Consider \(\frac{1}{\sqrt{a+1} + \sqrt{a}}\)

Multiply both numerator & denominator by \(\sqrt{a+1} - \sqrt{a}\)

\(\frac{1}{\sqrt{a+1} + \sqrt{a}} * \frac{\sqrt{a+1} - \sqrt{a}}{\sqrt{a+1} - \sqrt{a}}\)

\(= \frac{\sqrt{a+1} - \sqrt{a}}{(\sqrt{a+1} + \sqrt{a}) * (\sqrt{a+1} - \sqrt{a})}\)

\(= \frac{\sqrt{a+1} - \sqrt{a}}{(\sqrt{a+1})^2 - (\sqrt{a})^2}\)

\(= \frac{\sqrt{a+1} - \sqrt{a}}{a+1-a}\)

\(= \sqrt{a+1} - \sqrt{a}\)

still don't understand where the 1/2 is coming from...

Lets go to the end result

Combined time consumed \(= \frac{1}{\sqrt{8} - \sqrt{6}}\)

\(= \frac{1}{\sqrt{8} - \sqrt{6}} * \frac{\sqrt{8} + \sqrt{6}}{\sqrt{8} + \sqrt{6}}\)

\(= \frac{\sqrt{8} + \sqrt{6}}{(\sqrt{8})^2 - (\sqrt{6})^2}\)

\(= \frac{\sqrt{8} + \sqrt{6}}{8 - 6}\)

\(= \frac{\sqrt{8} + \sqrt{6}}{2}\)

Answer = B

what a horrible question...I tried to solve it algebraically..then gave up and used approximations.

A -> sqrt(8) + sqrt(7) = > sqrt(8) is slightly less than sqrt3 so smth 3-; sqrt(7) - again, slightly less than 9 and slightly less than sqrt(8), so together, A finishes the job in slightly less than 6 hours. suppose 6.
B -> sqrt(7) + sqrt(6) -> slightly less 3 + slightly less than 3. as a matter of fact, sqrt(6) can be rewritten as sqrt(3)*sqrt(2) =1.7*1.4 aprox 2.38. so all together, will be aprox 5 hours.

now rates:
1/6+1/5 = 11/30 or 30/11 hours to finish the job. this is less than 3 hours. 2hours and 9/11. aprox.

A: 6/(7-1.7) = 7/5.3 -> that is slightly more than 1. not good.
B: 1/2*(sqrt(8)+sqrt(6)) -> 1/2 * slightly less 3 + 2.38 => 1/2 * slightly more 5. that is 2 hours and smth. ok, looks good.
C: 1/3 * (6 - sqrt(3)) = 1/3* 6-1.7 = 1/3*4.3 -> that is less than 2 hours, out.
D: 3*(1.7+1.4) = 3*3.1 -> way more, so out.
E: slightly less 3+ 2*slightly less 3+ 2.38 -> way more.

as you can see, only B gives an answer that is close to my approximations.

At Supersonic Corporation, the time required for a machine to complete a job is determined by the formula:
t = √w + √(w – 1), where w = the weight of the machine in pounds and t = the hours required to complete
the job. If machine A weighs 8 pounds, and machine B weighs 7 pounds, how many hours will it take the
two machines to finish one job if they work together?

amit10th


please provide answer options

But for instance below is way to solve

time for A = √8+ √7
time for B = √7+ √6

thus their combined rate = 1/(√8 + √7) + 1/(√7 + √6)

solving we get (√8- √7) + (√7- √6) = (√8- √6)

thus time required = 1/ (√8- √6).........

your ans. choices will help me solve further

Merging topics. Please refer to the discussion above. Also, please follow our rules (https://gmatclub.com/forum/rules-for-po ... 33935.html) when posting a question. Thank you.

8 pound machine taskes \(\sqrt{8}+\sqrt{7}\)
work rate for it is \(W_8 = \frac{1}{(\sqrt{8}+\sqrt{7}})\) = \(\sqrt{8}-\sqrt{7}\)

7 pound machine taskes \(\sqrt{7}+\sqrt{6}\)
work rate for it is \(W_7 = \frac{1}{(\sqrt{7}+\sqrt{6}})\) = \(\sqrt{7}-\sqrt{6}\)

combined time t = \(\frac{1}{(W_8+W_7)}\)

\(t= \frac{1}{(\sqrt{8}-\sqrt{7}+\sqrt{7}-\sqrt{6}}\)
\(t = \frac{1}{\sqrt{8}-\sqrt{6}}\)
\(t = \frac{1}{2}*(\sqrt{8}+\sqrt{6})\)

B is correct

Hi PareshGmat I have a question (probably faulting at my concepts). To find the combined time to finish a job why do we need to find the combined rate first? why couldn't I just add the time it takes A with the time it takes B to finish the job?

Does this property hold only when the numbers are "a" and (a-1)? Thank you!!

Actually, you can apply the similar formula with numbers are "a" and a+n
Apply above solution, change 1 to n to find \(\frac{1}{\sqrt{a+n} + \sqrt{a}}\)
then we have
\(\frac{1}{\sqrt{a+n} + \sqrt{a}} = \frac{\sqrt{a+n} - \sqrt{a}}{n}\)

We have to find out the value of: 1/(sqrt(8)+sqrt(7)) this is nothing but sqrt(8)-sqrt(7) similarly 1/(sqrt(7)+sqrt(6)) will be sqrt(7)-sqrt(6)
further adding the rates will yield in sqrt(8)-sqrt(6) now time will be reciprocal of rate that is 1/(sqrt(8)-sqrt(6))
or we can write it as (sqrt(8)+sqrt(6))/2

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