clothing design problem

Expert replies

jessicamuniz: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Thu Jul 17, 2008 1:02 pm

by jessicamuniz » Tue Sep 09, 2008 8:22 pm

Here is a problem...
I hope somebody can help me...

A clothing designer buys cloth in boltd of 10 yards each.
For the first 100 yards purchased, the cost is $6.00 per yard. For the next 100 yards purchased, the cost decreases to $5.00 per yard. For all cloth purchased after 200 yards, the cost decreases to $4.00 per yard.
How many bolts of cloth does the clothing designer buy in order X?

1) In order X, the clothing designer spends a total of $1,420
2) In order X, the clothing designer spends $320 on the cloth that costs $4.00 per yard.

Thanks in advance

Source: Beat The GMAT — Data Sufficiency | Tue Sep 09, 2008 8:22 pm

LSB: Master | Next Rank: 500 Posts; Posts: 229; Joined: Thu Nov 08, 2007 3:47 pm; Thanked: 14 times; GMAT Score:750

by LSB » Tue Sep 09, 2008 8:51 pm

Statement 1:

Assume Z is number of yards of cloth

1420 = 100*6 + 100*5 + (Z-200)*4

This is an equation with one unknown. You can solve for Z and divide by 10 to arrive at the bolts of cloth. So, this statement is sufficient.

For clarity, here is the actual solution:

1420 = 100*6 + 100*5 + (Z-200)*4
1420 = 600 + 500 + 4Z – 800
1420 = 300 + 4Z
1120 = 4Z
Z = 280

# of Bolts = Z / 10 = 28

Statement 2:

Assume Y is the number of yards of cloth that costs 4$

320 = 4Y
Y = 80

We know from the question stem that in order to be eligible to purchase the 4$ cloth, the designer had to purchase 100 yards of the 6$ cloth and 100 yards of the 5$ cloth. This means that a total of 280 yards of cloth were purchased.

This puts us right to the end of our calculations from statement 1.

Z = 280
# of Bolts = Z / 10 = 28

This statement is also sufficient.

Since both statements are sufficient, the answer is D

Note that for DS you don’t need to calculate the entire solution. As soon as you’re comfortable that the information you have is sufficient (or insufficient) you can stop calculating.

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