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Bunuel
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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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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Bunuel
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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Bunuel
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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Bunuel
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

[solution 1]
cigars left unsold: \(1-3x/7+x/4=1-(12x+7x)/28=9x/28\);
so, the cigars left unsold must be a multiple of 9, and only (A) is a multiple of such.

Answer (A).

[solution 2]
\(3x/7•20.50+x/4•35=491… 12x•20.50+7x•35=491•28… 246x+245x=491•28…491x=491•28…x=28\);
\(x=total.cigars=28…sold.cigars=19x/28=19…unsold.cigars=9\)

Answer (A).
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