abhasjha wrote:Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate, produces y widgets every 5 minutes. If machines M and N working simultaneously at their respective constant rates for 20 minutes, does machine M produce more widgets than machine N in that time ?

1) x > 0.8y
2) y = x + 1

Machine M produces x widgets in 4 minutes
Machine M produces x/4 widgets in 1 minutes

Machine N produces y widgets in 5 minutes
Machine N produces y/5 widgets in 1 minutes

Both Machines Together in 1 minute Produce = (x/4)+(y/5) = (5x+4y)/20
Both Machines Together in 1 minute Produce = (x/4)+(y/5) = (5x+4y)/20
Both Machines Together in 20 minute Produce = (x/4)+(y/5) = (5x+4y)

Question : Does M produce more widgets than Machine N?

Question Rephrased : Is 5x>4y ? i.e. IS x > (4/5)y? i.e. Is x > 0.8y ?

Statement 1) x > 0.8y
SUFFICIENT

Statement 1) y = x + 1
i.e. 0.8y = 0.8x + 0.8 but this can't be compared with x therefore
INSUFFICIENT

Answer: Option A

abhasjha wrote:Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate, produces y widgets every 5 minutes. If machines M and N working simultaneously at their respective constant rates for 20 minutes, does machine M produce more widgets than machine N in that time ?

1) x > 0.8y
2) y = x + 1

Both machines work for the same amount of time.
To produce more widgets than N, M must work faster than N.

Question rephrased:
Is M faster than N?

Statement 1: x > 0.8y
Let y=10, implying that x>8.
Since y=10, N produces 10 widgets every 5 minutes, implying that N's rate = 10/5 = 2 widgets per minute.
Since x>8, M produces MORE than 8 widgets every 4 minutes, implying that M's rate = (more than 8)/4 = MORE than 2 widgets per minute.
Thus, M is faster than N.
SUFFICIENT.

Statement 2: y = x+1
Case 1: x=1, y=2
Since y=2, N produces 2 widgets every 5 minutes, implying that N's rate = 2/5 widget per minute.
Since x=1, M produces 1 widget every 4 minutes, implying that M's rate = 1/4 widget per minute.
In this case, N is faster than M.

Case 2: x=100, y=101
Since y=101, N produces 101 widgets every 5 minutes, implying that N's rate = 101/5 ≈ 20 widgets per minute.
Since x=100, M produces 100 widgets every 4 minutes, implying that M's rate = 100/4 = 25 widgets per minute.
In this case, M is faster than N.

Since N is faster than M in Case 1 but slower than M in Case 2, INSUFFICIENT.

The correct answer is A.

Alternate approach:
Since M produces x widgets every 4 minutes, M's rate = x/4 widgets per minute.
Since N produces y widgets every 5 minutes, N's rate = y/5 widgets per minute.
Since M and N work for the same amount of time, M will produce more widgets than N if M's rate is greater than N's rate.
For M's rate to be greater than N's rate, the following must be true:
x/4 > y/5
x > (4/5)y.

Question stem, rephrased: Is x > (4/5)y?

Statement 1: x > (0.8)y
Thus, x > (4/5)y.
SUFFICIENT.

Statement 2: y = x+1
If x=1 and y=2, then (4/5)y = 8/5.
In this case, x < (4/5)y.
If x=100 and y=101, then (4/5)y ≃ 80.
In this case, x > (4/5)y.
INSUFFICIENT.

The correct answer is A.

abhasjha wrote:Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate, produces y widgets every 5 minutes. If machines M and N working simultaneously at their respective constant rates for 20 minutes, does machine M produce more widgets than machine N in that time ?

1) x > 0.8y
2) y = x + 1

\[M\,\,\, - \,\,\,4\,\,{\text{min}}\,\,\, - \,\,\,x\,\,{\text{widgets}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,M\,\,\, - \,\,\,20\,{\text{min}}\,\,\, - \,\,\,5x\,\,{\text{widgets}}\]
\[N\,\,\, - \,\,\,5\,\,{\text{min}}\,\,\, - \,\,\,y\,\,{\text{widgets}}\,\,\,\,\, \Rightarrow \,\,\,\,\,N\,\,\, - \,\,\,20\,{\text{min}}\,\,\, - \,\,\,4y\,\,{\text{widgets}}\]
\[5x\mathop > \limits^? \,4y\]

\[\left( 1 \right)\,\,\,x > \frac{4}{5}y\,\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,5} \,\,\,\,\,5x > 4y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \]

\[\left( 2 \right)\,\,\,\,y = x + 1\,\,\,\,\left\{ \begin{gathered}
\,\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {1,2} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\, \hfill \\
\,\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {5,6} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\, \hfill \\
\end{gathered} \right.\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.