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

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For each order, a certain company charges a delivery fee d [#permalink]

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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.


Need help to tackle such questions under 1.5 mins.
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[Reveal] Spoiler: OA

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Last edited by Bunuel on 10 Dec 2012, 01:15, edited 1 time in total.
Renamed the topic and edited the question.
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Re: For each order [#permalink]

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New post 10 Dec 2012, 01:01
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
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Re: For each order [#permalink]

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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.
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Re: For each order [#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 [#permalink]

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New post 18 Jan 2013, 03:51
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 [#permalink]

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New post 26 Mar 2013, 06:36
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 [#permalink]

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New post 14 May 2013, 11:17
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 [#permalink]

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New post 14 May 2013, 15:55
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 [#permalink]

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New post 30 Jul 2013, 05:25
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 [#permalink]

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New post 12 Jan 2014, 06:12
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 [#permalink]

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Re: For each order, a certain company charges a delivery fee d   [#permalink] 01 Dec 2015, 13:11
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