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In option 1
T = M +2. Does it not prove that time takes more time than Mike given M cannot be negative? Can we have the official solution?
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Bunuel
Working alone at respective constant rates , Tim and Mike made pots. Did Tim take more time than Mike to make 100 pots?

(1) Tim takes 4 hours more than Mike to produce twice the number of pots.
(2) Tim takes twice the time as Mike to produce thrice the number of pots.


­

Language of the statements seems a little confusing as it is actually comparing Tim's twice/thrice the work to Mike's once the work but the way it's put out, it seems that the comparison is happing for same number of pots ie. both Tim and Mike were producing twice or thrice the pots and just the time taken was different for same amount of output.

To produce 1 pot, let's assume Tim's work rate is \(\frac{1}{T}\) and Mike's work rate is \(\frac{1}{M}\), and
\(W=R*T\) (Work = Rate x Time)
which means, it takes Tim \(T\) hours to make 1 pot and it takes Mike \(M\) hours to make 1 pot.

We are asked - For 100 pots did Tim take more time than Mike? Algebraically => is \(100T > 100M\) => \(T>M\)?

Statement 1 - Tim takes 4 hours more than Mike to produce twice the number of pots.

Time taken by Tim to produce 2 pots = 2*T
Time taken by Mike to produce 1 pot = M

Time taken by Tim to produce 2 pots = Time taken by Mike to produce 1 pot + 4
\(2T = M + 4\)
\(T = \frac{M}{2} + 2\)

\(T>M\) if T=3 and M=2
\(T<M\) if T=5 and M=6

This statement is insufficient.

Statement 2 - Tim takes twice the time as Mike to produce thrice the number of pots.

Time taken by Tim to produce 3 pots = 3*T
Time taken by Mike to produce 1 pot = M

Time taken by Tim to produce 3 pots = 2*Time taken by Mike to produce 1 pot
\(3T = 2M\)
\(T = \frac{2M}{3}\)

\(T<M\) for all values of M and T.

Is \(100T > 100M\) => \(T > M\)? => No

This statement is sufficient.

Answer: B
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