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by jain2016 » Tue Mar 15, 2016 8:38 am
If a stationery store owner buys 50% more identically-priced calendars than she usually purchases, she will be given a 20% discount off the standard price. Her total cost would then be 120 times the dollar value of the standard price of one calendar. How many calendars does she usually purchase?

A) 40

B) 80

C) 100

D) 120

E) 140

OAC

Hi Experts ,

Please check and let me know.

Let c = the number of calendars she usually buys

Let p = the dollar value of the standard price of one calendar

Then (1.5c)(0.8p) is total cost after discount right?

Then what should I do after this?

Please explain .

Many thanks in advance.

SJ
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by DavidG@VeritasPrep » Tue Mar 15, 2016 8:46 am
jain2016 wrote:If a stationery store owner buys 50% more identically-priced calendars than she usually purchases, she will be given a 20% discount off the standard price. Her total cost would then be 120 times the dollar value of the standard price of one calendar. How many calendars does she usually purchase?

A) 40

B) 80

C) 100

D) 120

E) 140

OAC

Hi Experts ,

Please check and let me know.

Let c = the number of calendars she usually buys

Let p = the dollar value of the standard price of one calendar

Then (1.5c)(0.8p) is total cost after discount right?

Then what should I do after this?

Please explain .

Many thanks in advance.

SJ

We're told that this amount (1.5c * .8p) is equal to 120 times the original price, or 120p. So now we have 1.5c * .8p = 120p; 1.2pc = 120p; p's cancel out, so 1.2c = 120; c = 100. Answer is C
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by [email protected] » Tue Mar 15, 2016 8:47 am
Hi jain2016,

You've translated part of the prompt, but you still have to create an equation (otherwise there's nothing to 'manipulate.' If you're going to approach this prompt algebraically, then you have to create algebraic equations.

We're told that by buying 50% more calendars, there will be a 20% discount in price AND that new total will be 120 times the regular price of a calendar. The equation would then be...

(1.5c)(0.8p) = 120p

From here, you can combine like terms and simplify:

1.20(c)(p) = 120p
(c)(p) = 100p
c = 100

Final Answer: C

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by GMATGuruNY » Tue Mar 15, 2016 8:53 am
If a stationery store owner buys 50% more identically-priced calendars than she usually purchases, she will be given a 20% discount off the standard price. Her total cost would then be 120 times the dollar value of the standard price of one calendar. How many calendars does she usually purchase?

A)40
B)80
C)100
D)120
E)140
Let the standard price = $10.

She will be given a 20% discount off the standard price.
Discounted price = 10 - 20% of 10 = $8.

Her total cost would then be 120 times the dollar value of the standard price of one calendar.
Total cost = 120*10 = $1200.

For a total cost of $1200, the number of calendars that can be purchased at the discounted price of $8 = 1200/8 = 150.

A stationery store owner buys 50% more identically-priced calendars than she usually purchases.
Since the 150 calendars purchased at the discounted price represent 50% more than the number usually purchased, the number usually purchased = 100.

The correct answer is C.
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by [email protected] » Tue Mar 15, 2016 8:53 am
Hi jain2016,

This prompt can also be solved with a combination of TESTing VALUES.

Here's how we can TEST VALUES using 'round numbers':

Regular price = $10 each
Discount price = $8 each

To get the discounted price, we have to buy 50% more calendars... and that total spent would be 120 times $10...

(1.5X)($8) = (120)($10)
$12X = $1200
X = 100 calendars

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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