During two 7-day promotional weeks, Week 1 and Week 2, a bike-share

Question

During two 7-day promotional weeks, Week 1 and Week 2, a bike-share service tracked the total number of bicycles rented. If in Week 1, 20 percent of the bicycles rented were rented on Sunday, while in Week 2, 15 percent of the bicycles rented were rented on Sunday, were more bicycles rented in Week 2 than in Week 1?

(1) On each day from Monday through Saturday in Week 2, the number of bicycles rented was at least 7 percent higher than the number rented on the corresponding day in Week 1.
(2) The number of bicycles rented on Sunday in Week 2 was greater than 76 percent of the number rented on Sunday in Week 1.

Bunuel · Accepted Answer

GMAT Club Official Solution:

During two 7-day promotional weeks, Week 1 and Week 2, a bike-share service tracked the total number of bicycles rented. If in Week 1, 20 percent of the bicycles rented were rented on Sunday, while in Week 2, 15 percent of the bicycles rented were rented on Sunday, were more bicycles rented in Week 2 than in Week 1?

Let W1 be the total number of bicycles rented in Week 1, and let W2 be the total number rented in Week 2.

Since 20 percent of Week 1 rentals were on Sunday, Sunday rentals in Week 1 were 0.20W1.
So Monday through Saturday rentals in Week 1 were 0.80W1.

Since 15 percent of Week 2 rentals were on Sunday, Sunday rentals in Week 2 were 0.15W2.
So Monday through Saturday rentals in Week 2 were 0.85W2.

We need to determine whether W2 > W1.

(1) On each day from Monday through Saturday in Week 2, the number of bicycles rented was at least 7 percent higher than the number rented on the corresponding day in Week 1.

So the total for Monday through Saturday in Week 2 was at least 1.07 times the total for Monday through Saturday in Week 1.

Thus,

0.85W2 >= 1.07(0.80W1)

0.85W2 >= 0.856W1

W2/W1 >= 0.856/0.85

So W2 > W1.

Sufficient.

(2) The number of bicycles rented on Sunday in Week 2 was greater than 76 percent of the number rented on Sunday in Week 1.

So,

0.15W2 > 0.76(0.20W1)

0.15W2 > 0.152W1

So W2 > W1.

Sufficient.

Answer: D.

BongBideshini · Answer

Let total rentals:
Week 1 =  T1 
 Week 2 =  T2 
Sunday rentals:
 Week 1:  0.2T1 
 Week 2:  0.15T2 
We need:  Is T2 > T1?

(1)
Mon–Sat (6 days) in Week 2 are each >=  7% higher  than Week 1.
So total Mon–Sat:
 Week 1:  0.8T1 
 Week 2: >=  1.07 × 0.8T1 = 0.856T1 
Add Sunday:
 T2 ≥ 0.856T1 + 0.15T2
Rearrange:
 T2 − 0.15T2 >=0.856T1
 0.85T2 >= 0.856T1
 T2 >=(0.856 / 0.85)T1 = 1.007T1

So  T2 > T1 always 
 Sufficient

(2)
Week 2 Sunday > 76% of Week 1 Sunday
0.15T2 > 0.76 × 0.2T1
 0.15T2 > 0.152T1
 T2 > (0.152 / 0.15)T1 =  1.013T1

So  T2 > T1 always 
 Sufficient

Answer:  D

thisisakr · Answer

Let W1 = total bicycles rented in Week 1, W2 = total bicycles rented in Week 2.
Sunday rentals: Week 1 = 0.20 × W1, Week 2 = 0.15 × W2.
Monday through Saturday rentals: Week 1 = 0.80 × W1, Week 2 = 0.85 × W2.
 W2 > W1?

Statement 1: Each Mon–Sat day in Week 2 was at least 7% higher than the corresponding day in Week 1
This means total Mon–Sat rentals in Week 2 ≥ 1.07 × (0.80 × W1) = 0.856 × W1.
So 0.85 × W2 ≥ 0.856 × W1, giving W2 ≥ (0.856/0.85) × W1 => 1.007 × W1.
This means W2 > W1 in all cases satisfying the condition.
——— SUFFICIENT .

Statement 2: Sunday rentals in Week 2 were greater than 76% of Sunday rentals in Week 1
0.15 × W2 > 0.76 × (0.20 × W1)
0.15 × W2 > 0.152 × W1
W2 > (0.152/0.15) × W1 ==> 1.013 × W1.
This also means W2 > W1 in all cases satisfying the condition.
——-  SUFFICIENT

D

Adit_ · Answer

IMO D.

We can find a direct linkage with the x/y ratio in both statements 1 and 2.
This way we know if x/y is less than 1 then y<x otherwise x>y.
Each statement sufficient on its own.

Answer:  Option D

During two 7-day promotional weeks, Week 1 and Week 2, a bike-share

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