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

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Math Expert
Joined: 02 Sep 2009
Posts: 46237
A collection of quality cigars went on sale, 3/7 of which were sold at [#permalink]

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10 May 2017, 02:19
2
4
00:00

Difficulty:

55% (hard)

Question Stats:

68% (01:37) correct 32% (02:48) wrong based on 111 sessions

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: Cheers, Brent _________________ Brent Hanneson – Founder of gmatprepnow.com CEO Joined: 12 Sep 2015 Posts: 2564 Location: Canada Re: A collection of quality cigars went on sale, 3/7 of which were sold at [#permalink] ### Show Tags 10 May 2017, 06:39 Top Contributor 1 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: Cheers, Brent _________________ Brent Hanneson – Founder of gmatprepnow.com Director Joined: 14 Nov 2014 Posts: 667 Location: India Schools: Rotman '20 (S) GMAT 1: 700 Q50 V34 GPA: 3.76 Re: A collection of quality cigars went on sale, 3/7 of which were sold at [#permalink] ### Show Tags 10 May 2017, 10:40 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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Joined: 26 Feb 2016
Posts: 2831
Location: India
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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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10 May 2017, 11:10
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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12 May 2017, 13:35
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 _________________ Jeffery Miller Head of GMAT Instruction GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Target Test Prep Representative Status: Head GMAT Instructor Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2570 Re: A collection of quality cigars went on sale, 3/7 of which were sold at [#permalink] ### Show Tags 15 May 2017, 17:21 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.

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Re: A collection of quality cigars went on sale, 3/7 of which were sold at   [#permalink] 15 May 2017, 17:21
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