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A collection of quality cigars went on sale, 3/7 of which were sold at

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A collection of quality cigars went on sale, 3/7 of which were sold at [#permalink]

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A collection of quality cigars went on sale, 3/7 of which were sold at $20.50 each while 1/4 were sold at $35 each. If the total revenue was $491, how many cigars were left unsold?

A. 9
B. 11
C. 14
D. 19
E. 28
[Reveal] Spoiler: OA

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A collection of quality cigars went on sale, 3/7 of which were sold at [#permalink]

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Bunuel wrote:
A collection of quality cigars went on sale, 3/7 of which were sold at $20.50 each while 1/4 were sold at $35 each. If the total revenue was $491, how many cigars were left unsold?

A. 9
B. 11
C. 14
D. 19
E. 28


Nice question!

We can actually solve it WITHOUT using the information about total revenue being $491

Let x = TOTAL number of cigars.

3/7 sold at $20.50 each and 1/4 were sold at $35 each
So, (3/7)x sold for $20.50 each, and (1/4)x sold for $35 each
Total # of cigars SOLD = (3/7)x + (1/4)x
= 3x/7 + x/4
= 12x/28 + 7x/28
= 19x/28
= (19/28)x
So, we can see that 19/28 of the x cigars were sold

This means that 9/28 of the x cigars were NOT sold
In other words, the # of cigars left UNSOLD = 9x/28
Since we know that the # of cigars must be an INTEGER, we can conclude that x is divisible by 28
Also, since we can write 9x/28 as 9(x/28), we can see that the # of cigars left UNSOLD must be a multiple of 9
Check the answer choices...only 1 is a multiple of 9

Answer:
[Reveal] Spoiler:
A


Cheers,
Brent
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Re: A collection of quality cigars went on sale, 3/7 of which were sold at [#permalink]

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Bunuel wrote:
A collection of quality cigars went on sale, 3/7 of which were sold at $20.50 each while 1/4 were sold at $35 each. If the total revenue was $491, how many cigars were left unsold?

A. 9
B. 11
C. 14
D. 19
E. 28


Now let's solve the question by using the information about total revenue being $491

Let x = TOTAL number of cigars.

3/7 sold at $20.50 each...
So, (3/7)x = # of cigars sold for $20.50 each
So, (3/7)(x)($20.50) = total sales from $20.50-cigars

...and 1/4 were sold at $35 each
So, (1/4)x = # of cigars sold for $35 each
So, (1/4)(x)($35) = total sales from $35-cigars

The total revenue was $491
So, (3/7)(x)($20.50) + (1/4)(x)($35) = $491
To eliminate the fractions, multiply both sides by 28 to get: (12)(x)(20.50) + (7)(x)(35) = (491)(28)
Simplify: 246x + 245x = (491)(28)
Simplify: 491x = (491)(28)
Divide both sides by 491 to get: x = 28
So, we STARTED with 28 cigars.

3/7 of 28 = 12, so 12 cigars were sold for $20.50 each
1/4 of 28 = 7, so 7 cigars were sold for $35 each
So, total cigars SOLD = 12 + 7 = 19
So, the # of cigars left UNSOLD = 28 - 19 = 9

Answer:
[Reveal] Spoiler:
A


Cheers,
Brent
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Re: A collection of quality cigars went on sale, 3/7 of which were sold at [#permalink]

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New post 10 May 2017, 09:40
Bunuel wrote:
A collection of quality cigars went on sale, 3/7 of which were sold at $20.50 each while 1/4 were sold at $35 each. If the total revenue was $491, how many cigars were left unsold?

A. 9
B. 11
C. 14
D. 19
E. 28


As number of Cigar will be a integer ...
The lCM of 7 and 4 will be 28,..so number of cigar can be 28,56 etc
let take 28 .
3/7 * 28 * 20.50 + 1/4 * 28 * 35 will give 491...
So total number is 28 ..
We will get left out = 9 .
A
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Re: A collection of quality cigars went on sale, 3/7 of which were sold at [#permalink]

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New post 10 May 2017, 10:10
If the total collection of quality cigars which went on is x,
the unsold quality cigars can be got by the following equation
x - \(\frac{3x}{7}\) - \(\frac{x}{4}\)

Hence, the unsold cigars are \(\frac{9x}{28}\)

Value of x can be a multiple of 28.
If x=1, unsold cigars are 9
If x=2, unsold cigars are 18
If x=3, unsold cigars are 27.

Hence, the unsold cigars were 9(Option A)
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Re: A collection of quality cigars went on sale, 3/7 of which were sold at [#permalink]

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New post 12 May 2017, 12:35
Bunuel wrote:
A collection of quality cigars went on sale, 3/7 of which were sold at $20.50 each while 1/4 were sold at $35 each. If the total revenue was $491, how many cigars were left unsold?

A. 9
B. 11
C. 14
D. 19
E. 28


We can let n = the number of cigars. We are given that 3/7 or 12/28 were sold for 20.5 per cigar and 1/4 or 7/28 were sold for 35 per cigar. Thus:

(12/28)n(20.5) + (7/28)n(35) = 491

Let us multiply each side of the equation by 28 (we will not actually multiply the right hand side unless we really have to):

12n(20.5) + 7n(35) = 491(28)

246n + 245n = 491(28)

491n = 491(28)

Cancelling 491 from each side of the equation, we get that n = 28. So, there were 28 cigars to begin with. We know that 3/7 of the cigars, or 28(3/7) = 12 cigars were sold for 20.50 and 1/4 of the cigars, or 28(1/4) = 7 cigars were sold for 35; therefore 12 + 7 = 19 cigars were sold. This leaves us 28 - 19 = 9 cigars unsold.

Answer: A
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Re: A collection of quality cigars went on sale, 3/7 of which were sold at [#permalink]

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New post 15 May 2017, 16:21
Bunuel wrote:
A collection of quality cigars went on sale, 3/7 of which were sold at $20.50 each while 1/4 were sold at $35 each. If the total revenue was $491, how many cigars were left unsold?

A. 9
B. 11
C. 14
D. 19
E. 28


We can let n = the number of cigars. We are given that 3/7 were sold for $20.50 per cigar and 1/4 were sold for $35 per cigar, for a total revenue of $491. Thus:

(3/7)n(20.5) + (1/4)n(35) = 491

Let’s multiply each side of the equation by 28 (we will not actually multiply the right-hand side unless we have to):

12n(20.5) + 7n(35) = 491(28)

246n + 245n = 491(28)

491n = 491(28)

Dividing each side of the equation by 491, we obtain n = 28. Thus, there were 28 cigars initially. We know that 3/7 of the cigars, or 28(3/7) = 12 cigars, were sold for $20.50 and 1/4 of the cigars, or 28(1/4) = 7 cigars, were sold for $35; therefore 12 + 7 = 19 cigars were sold. This leaves 28 - 19 = 9 cigars unsold.

Answer: A
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Re: A collection of quality cigars went on sale, 3/7 of which were sold at   [#permalink] 15 May 2017, 16:21
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