Envelopes

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Ankitaverma: Senior | Next Rank: 100 Posts; Posts: 72; Joined: Tue Nov 05, 2013 4:35 pmFollowed by: 4 members

by Ankitaverma » Wed Nov 20, 2013 1:30 pm

Envelopes can be purchased for $1.50 per pack of 100, $1.00 per pack of 50, or $0.03 each. What is the greatest number of envelopes that can be purchased for $7.30?

(A) 426
(B) 430
(C) 443
(D) 460
(E) 486

Q/A-C can someone explain

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Source: Beat The GMAT — Problem Solving | Wed Nov 20, 2013 1:30 pm

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Wed Nov 20, 2013 3:01 pm

Ankitaverma wrote:Envelopes can be purchased for $1.50 per pack of 100, $1.00 per pack of 50, or $0.03 each. What is the greatest number of envelopes that can be purchased for $7.30?

(A) 426
(B) 430
(C) 443
(D) 460
(E) 486

Q/A-C can someone explain

The correct answer is actually D

The 3 options:
$1.50 per pack of 100 = 1.5 cents per envelope
$1.00 per pack of 50 = 2 cents per envelope
$0.03 per individual envelope = 3 cents per envelope

Since the packs of 100 are the best value, let's buy as many of these as we can.

4 packs of 100 costs $6.00 (we still have $1.30 left to spend)
1 packs of 50 costs $1.00 (we still have $0.30 left to spend)
10 individual envelopes costs $0.30 (we've spent all of the money)

Number of envelopes purchased = 400 + 50 + 10 = [spoiler]460 = D[/spoiler]

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Thu Nov 21, 2013 12:33 am

Are you sure of the OA?

We have:
$1.50 per pack of 100 ==> Price per envelope 0.015
$1.00 per pack of 50 ==> Price per envelope 0.02
$0.03 each

To find maximum envelopes we must be INT Quotient for cheapest price per envelope
= 7.3/1.5 = 4
So, 4x100 = 400

Now, 7.3 - 1.5*4 = 1.3
Divide by next highest = 1.3/1 = 1
So, 1x50 = 50

At last, 1.3 - 1 = 0.3 => 0.3/0.03 = 10
So, 10x1 = 10

Total = 400 + 50 + 10 = 460

Answer [spoiler]{D}[/spoiler]

R A H U L

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Mathsbuddy: Master | Next Rank: 500 Posts; Posts: 447; Joined: Fri Nov 08, 2013 7:25 amFollowed by: 1 member; Thanked: 25 times

by Mathsbuddy » Thu Nov 21, 2013 9:54 am

I 100% agree that the answer is (D) 460.

However, for those of you who like to think outside the box:

First buy 5 * 100 envelopes at $1.50 per pack = $7.50
Then return with 7 envelopes for a refund at $0.03 each = -$0.21
Resultant spend = $7.29

Now you will have 493 envelopes plus 1 cent for $7.30! (only 1 digit different to answer B)

Alternatively:
First buy 6 * 100 envelopes at $1.50 per pack = $9.00
Then return with 49 and then another 8 envelopes (=57) for a refund at $0.03 each = -$1.71
Resultant spend = $7.29 again

Now you will have 543 envelopes plus 1 cent for $7.30! (also only 1 digit different to answer B)

Of course you could keep going on and on this way, for example:
First buy 30 * 100 envelopes at $1.50 per pack = $45.00
Then return with 1257 envelopes (in batches less than 50) for a refund at $0.03 each = -$37.71
Resultant spend = $7.29 again

Now you will have 1743 envelopes plus 1 cent for $7.30

If you could buy partial packs pro-rata at the rate of $1.50 per 100 envelopes, then the maximum number would be 836 for $7.29.

As $7.30 is not a multiple of 0.03 and $1.50 is such a multiple, it is impossible to spend exactly $7.30 by overspending and being refunded at the single envelope rate.

Therefore 460 envelopes seems to be the only solution to $7.30 exactly.

However if we allow refunding with a limit of 49 envelopes refunded at a time (i.e. at a rate of 1 refund per purchase), we are capped at 493 envelopes for $7.29 for a single transaction pair (sale + refund). Therefore, if you went in with 2 friends and all three of you did this, between you you would have 3 cents remaining for one of you to buy just one more envelope. Hence, 2 people would have 493 envelopes each and 1 person would have 494 envelopes. This seems to be the maximum for the minimum number of people and optimum refund policy, albeit cheating!

In short: 460, 493 (and 494?) could all be reasonable answers depending on the rules of the game.
Some small shop keepers would find it quite fair and reasonable to refund just 7 envelopes, so in real-life I believe 493 envelopes is acceptable.

OK, it's not GMat, but a possibly interesting distraction nonetheless!

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[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members; Location: Palo Alto, CA; Thanked: 2867 times; GMAT Score:800

by [email protected] » Thu Nov 21, 2013 12:53 pm

Hi Mathsbuddy,

Your approach is amusing, but as you admit, NOT GMAT. Be very careful about how you interpret questions in both the Quant and Verbal sections. You're given all the information that you need to have to answer the question within the "confines" of what's given. The type of thinking that you've shown in your post will get you killed in CR.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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Mathsbuddy: Master | Next Rank: 500 Posts; Posts: 447; Joined: Fri Nov 08, 2013 7:25 amFollowed by: 1 member; Thanked: 25 times

by Mathsbuddy » Thu Nov 21, 2013 11:39 pm

[email protected] wrote:Hi Mathsbuddy,

Your approach is amusing, but as you admit, NOT GMAT. Be very careful about how you interpret questions in both the Quant and Verbal sections. You're given all the information that you need to have to answer the question within the "confines" of what's given. The type of thinking that you've shown in your post will get you killed in CR.

GMAT assassins aren't born, they're made,
Rich

I'm glad you saw the humour, it was not meant be a serious. 460 is the obvious answer

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