For each order, a certain company charges a delivery fee d that depend

Notice that we can have a case when charges for both deliveries are calculated with the second formula, for example x1=$250 and x2=$350, in this case the total price still would be $10.

Thanks this was really helpful, I missed out on the second case. where there can be 2 values within the range of d=3

Thanks
Its really so helpful.I was confused of the statement 2.Now its clear

can you please clarify this part more? thanks in advance

WOW! I didnt even know where to start i was off and in the end i took a guess and went for B

Cross multiply you will get

300 + x1 - 100 + 300 + x2 - 100 = 1000

After we solve its 600...

Right off the bat, you need to know if: 2*x > 499

1) This one tells you that one of the orders is >= 100, it's insufficient because the other order could be any value

2) This tells you that the total delivery was > 500 + (another value between 1 and 100, inclusive), because there are no other combinations of 3 and 7 other than 3 + 7 that yield 10. So this is clearly sufficient.

Answer is B

Is x > 499?



1) d = 3 for 1 of the orders.
The other could be 3, or could be 7.
x will vary.
Insufficient.

2) d = 10
=> x > 500 for d = 7

OR
10 = 3 + (x-100)/100 + 3 + (y-100)/100
=> 10 = 6 + (x + y - 200)/100
=> 1000 = 400 + x + y
--> x+y=600>499
Sufficient.

B is the answer.

We need to determine whether the total price of the merchandise in two separate orders is greater than $499 or not.

Statement One Alone:

The delivery fee for one of the two orders was $3.

Thus we know price of one of the orders is at least $1 and at most $100. However, since we don’t know anything about the other order, we can’t answer the question. Statement one alone is not sufficient.

Statement Two Alone:

The sum of the delivery fees for the two orders was $10.

If the sum of the delivery fees was $10, it is possible that one order is at most $100 (thus has a delivery fee of $3) and the other order is at least $501 (thus has a delivery fee of $7). In this case the total price of the two orders will be greater than $499 since one of them is already greater than $499.

However, it is also possible that each of the orders has a price that is more than $100 but no more than $500.

So let’s let a = price of the one order and b = price of the other order where both a and b are more than $100 but no more than $500. We can create the following equation:

3 + (a - 100)/100 + 3 + (b - 100)/100 = 10

(a - 100)/100 + (b - 100)/100 = 4

(a - 100) + (b - 100) = 400

a + b = 600

We can see that if each order is more than $100 but no more than $500, then the total price of the two orders is always $600, which is greater than $499.

Answer: B

Can we addd the tag "GMATPrep" if it is actually a GMAT Prep question? Just for people who intend to avoid prep questions. Thank you!

Added the tag. Thank you.

Given
d =
3 ,if 0<x<=100 - assume (case-1)
3+(x-100)/100 or 3.01<d<=7 if 100<x<=500 -assume (case-2)
7 if x>500 - assume (case-3)

-->Many of them above proved that option 1 is not sufficient. Hope you understood.
--> I am proving 2nd stmt is sufficient.

Assume my goal to know minimum value of x1+x2 , considering the possible case of d1+d2 =10

possibility cases - case 1 , case 3 (d1=3 , d2=7). ( reverse possibilities not mentioning - result will be same)
case 1, case 2 (d1=3, d2=7)
case 2, case 2. ( d1=3.01 , d2=7) - not possible to get 10


possibility -1 : case 1 , case 3
minimum value of x1+x2 = 1( case1)+501(case3) = 502 >499
possibility -2: case 1 , case 2
minimum value of x1+x2 = 1( case1) +500(case2) =501 >499



So stmt2 is sufficient. Option B is correct.

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