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For each order, a certain company charges a delivery fee d [#permalink]
09 Dec 2012, 23:33

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35% (medium)

Question Stats:

75% (02:28) correct
25% (01:51) wrong based on 120 sessions

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. source: Gmat Prep

Re: For each order [#permalink]
10 Dec 2012, 00:01

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]
10 Dec 2012, 00:11

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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.

Re: For each order [#permalink]
10 Dec 2012, 00:22

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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. _________________

Re: For each order [#permalink]
18 Jan 2013, 02:51

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 _________________

Re: For each order [#permalink]
26 Mar 2013, 05:36

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

Re: For each order [#permalink]
14 May 2013, 10:17

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 _________________

Re: For each order [#permalink]
30 Jul 2013, 04:25

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

Re: For each order, a certain company charges a delivery fee d [#permalink]
12 Jan 2014, 05:12

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

gmatclubot

Re: For each order, a certain company charges a delivery fee d
[#permalink]
12 Jan 2014, 05:12

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...