Company C has a machine that, working alone at its constant rate, proc

Efficiency of Machine A = 20 units/hour
Efficiency of New Machine A = e units/hour

Now, we have \(( 20 + e )*2 = 100\)

So, \(20 + e = 50\)

Or, \(e = 30\), Answer must be (C)
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The rate of the current machine, \(M_1\),
plus the rate of the new machine, M\(_2\),
must = \(\frac{100u}{2hrs}\).
Hence express M\(_1\)'s rate in "per 2 hours"

Work rate of M\(_1\)?
\(\frac{100}{5hrs}=\frac{20}{1hr}=\frac{40}{2hrs}\)

\(M_1 + M_2 =\frac{100u}{2hrs}\)

\(\frac{40}{2} + M_2 = \frac{100}{2}\)

\(40 + 2*(M_2) = 100\)
\(2*(M_2) = 60\)
\(M_2 = 30\)

Answer C

Alternatively, two machines will produce
\(\frac{100u}{2hrs}=\frac{50u}{1hr}=50\) units per hour
Find M\(_1\)'s rate in units per 1 hour

\(M_1=\frac{100u}{5hrs}=\frac{20u}{1hr}=20\) per hour
Total Units needed per hour = 50
\(M_2\) must make up the difference
\(M_2\) must make (50 - 20) = 30 units per hour

Answer C
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Company C has a machine that, working alone at its constant rate, processes 100 units of a certain product in 5 hours.

Rate = 1/5

Another machine = 1/x

Togather 1/5+1/x=1/2

x=3.333

constant rate, in units per hour, of the new machine = 100/3.33 = 30

Hence C

HI pushpitkc

Am'i Right?
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Yes NandishSS - you are absolutely right!

Posted from my mobile device

Machine A(old machine) can produce 100 units in 5 hour so in 1hour it can produce is 20 units.
Total requirement is of 100 units in 2 hours.. So 100- Out put of Machine A in 2 hrs=60 units.
To manufacture 60 units in 2 hrs new machine should produce 30 units ...
Ans:C 30units (new machine output/hr)

Given to us here the efficiency of Machine A = 20 units/hour

The efficiency of New Machine A = e units/hour

Given, the two machines, working together at their respective constant rates, are to process 100 units of this product in 2 hours

=>(20+e) * 2 =100

=>e=30 (option c)

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