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Re: For each order, a certain company charges a delivery fee d that depend [#permalink]
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MacFauz wrote:
1) Obviously insufficient. The other order could have any value.

2) We can easily create a situation for which the value is more than $499. So lets try to create a situation for which the value is lesser than $499 to prove insufficiency.

We can see that atleast one package has to be more than $100. To minimize the value of this , the first package has to be worth $100. So, for insufficiency, the second package can be at most only $399 and so "d" can only be a maximum of nearly $6. But in such a case, the toatl will not add up to 10. Hence the second package HAS to be more than $399.
Sufficient


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.
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Re: For each order, a certain company charges a delivery fee d that depend [#permalink]
Bunuel wrote:
For each order, a certain company charges a delivery fee d that depends on the total price x of the merchandise in the order, where d and x are in dollars and

d = 3, if 0<x<=100
d = 3 + (x-100)/100, if 100<x<=500
d = 7, if x>500

If George placed two separate orders with the company, was the total price of the merchandise in the two orders greater than $499?


Notice that the highest charge for the delivery with the second formula is for x=500, thus it equals to d=3+(500-100)/100=$7.

(1) The delivery fee for one of the orders was $3. The price of the merchandise for that order is less than or equal to 100, but we know nothing about the second order. Not sufficient.

(2) The sum of the delivery fees for the two orders was $10. If the delivery charges for the two orders are $3 and $7, then the price of the second merchandise must be more than or equal to $500, which means that the total price must be greater than $499. Sufficient.

Now, if both delivery charges were calculated with the second formula, then we'd have \((3+\frac{x_1-100}{100})+(3+\frac{x_2-100}{100})=10\) --> \(x_1+x_2=600>499\).

So, we have that in both possible cases the total price of the merchandise in the two orders is greater than $499. Sufficient.

Answer: B.

Hope it's clear.


Thanks this was really helpful, I missed out on the second case. where there can be 2 values within the range of d=3
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Re: For each order, a certain company charges a delivery fee d that depend [#permalink]
Bunuel wrote:
For each order, a certain company charges a delivery fee d that depends on the total price x of the merchandise in the order, where d and x are in dollars and

d = 3, if 0<x<=100
d = 3 + (x-100)/100, if 100<x<=500
d = 7, if x>500

If George placed two separate orders with the company, was the total price of the merchandise in the two orders greater than $499?


Notice that the highest charge for the delivery with the second formula is for x=500, thus it equals to d=3+(500-100)/100=$7.

(1) The delivery fee for one of the orders was $3. The price of the merchandise for that order is less than or equal to 100, but we know nothing about the second order. Not sufficient.

(2) The sum of the delivery fees for the two orders was $10. If the delivery charges for the two orders are $3 and $7, then the price of the second merchandise must be more than or equal to $500, which means that the total price must be greater than $499. Sufficient.

Now, if both delivery charges were calculated with the second formula, then we'd have \((3+\frac{x_1-100}{100})+(3+\frac{x_2-100}{100})=10\) --> \(x_1+x_2=600>499\).

So, we have that in both possible cases the total price of the merchandise in the two orders is greater than $499. Sufficient.

Answer: B.

Hope it's clear.

Thanks
Its really so helpful.I was confused of the statement 2.Now its clear
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Re: For each order, a certain company charges a delivery fee d that depend [#permalink]
Bunuel wrote:
[b]Now, if both delivery charges were calculated with the second formula, then we'd have \((3+\frac{x_1-100}{100})+(3+\frac{x_2-100}{100})=10\) --> \(x_1+x_2=600>499\)


can you please clarify this part more? thanks in advance
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Re: For each order, a certain company charges a delivery fee d that depend [#permalink]
WOW! I didnt even know where to start i was off and in the end i took a guess and went for B
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Re: For each order, a certain company charges a delivery fee d that depend [#permalink]
TheNona wrote:
Bunuel wrote:
[b]Now, if both delivery charges were calculated with the second formula, then we'd have \((3+\frac{x_1-100}{100})+(3+\frac{x_2-100}{100})=10\) --> \(x_1+x_2=600>499\)


can you please clarify this part more? thanks in advance




Cross multiply you will get

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

After we solve its 600...
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Re: For each order, a certain company charges a delivery fee d that depend [#permalink]
Marcab wrote:
For each order, a certain company charges a delivery fee d that depends on the total price x of the merchandise in the order, where d and x are in dollars and

d = 3, if 0<x<=100
d = 3 + (x-100)/100, if 100<x<=500
d = 7, if x>500

If George placed two separate orders with the company, was the total price of the merchandise in the two orders greater than $499?

(1) The delivery fee for one of the two orders was $3.
(2) The sum of the delivery fees for the two orders was $10.



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
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Re: For each order, a certain company charges a delivery fee d that depend [#permalink]
Marcab wrote:
For each order, a certain company charges a delivery fee d that depends on the total price x of the merchandise in the order, where d and x are in dollars and
d = 3, if 0<x<=100
d = 3 + (x-100)/100, if 100<x<=500
d = 7, if x>500
If George placed two separate orders with the company, was the total price of the merchandise in the two orders greater than $499?

Is x > 499?

Quote:
(1) The delivery fee for one of the two orders was $3.
(2) The sum of the delivery fees for the two orders was $10.


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.
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Re: For each order, a certain company charges a delivery fee d that depend [#permalink]
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Marcab wrote:
For each order, a certain company charges a delivery fee d that depends on the total price x of the merchandise in the order, where d and x are in dollars and

d = 3, if 0<x<=100
d = 3 + (x-100)/100, if 100<x<=500
d = 7, if x>500

If George placed two separate orders with the company, was the total price of the merchandise in the two orders greater than $499?

(1) The delivery fee for one of the two orders was $3.
(2) The sum of the delivery fees for the two orders was $10.



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
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Re: For each order, a certain company charges a delivery fee d that depend [#permalink]
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!
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Re: For each order, a certain company charges a delivery fee d that depend [#permalink]
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TheNightKing wrote:
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.
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Re: For each order, a certain company charges a delivery fee d that depend [#permalink]
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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For each order, a certain company charges a delivery fee d that depend [#permalink]
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Don't get into the weeds, (2) is 2 orders that add up to 10 dollars:

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