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A store had a total of a pens which it sold. The store first sold b of

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A store had a total of a pens which it sold. The store first sold b of  [#permalink]

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New post Updated on: 21 Jun 2018, 21:10
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A store had a total of a pens which it sold. The store first sold b of these pens for c dollars each and it sold the remaining pens for 2c/3 dollars each. Which of the following is equal to the total amount of money the store received for the a pens, in dollars?


A) (2a + b)c/3

B) (a + 2b)c/3

C) (2a + 3b)c/3

D) (2a + c)b/3

E) (3a + 2b)c/3

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Originally posted by Harshgmat on 21 Jun 2018, 19:44.
Last edited by Bunuel on 21 Jun 2018, 21:10, edited 2 times in total.
Renamed the topic and edited the question.
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A store had a total of a pens which it sold. The store first sold b of  [#permalink]

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New post 22 Jun 2018, 00:00
Harshgmat wrote:
A store had a total of a pens which it sold. The store first sold b of these pens for c dollars each and it sold the remaining pens for 2c/3 dollars each. Which of the following is equal to the total amount of money the store received for the a pens, in dollars?


A) (2a + b)c/3

B) (a + 2b)c/3

C) (2a + 3b)c/3

D) (2a + c)b/3

E) (3a + 2b)c/3


The total pens the store had was a pens. It sold the first b pens for c dollars.
Now, the remaining a - b pens that were left were sold for \(\frac{2c}{3}\) dollars each.

The total money recovered from the sale of pens is calculated as follows:

\(bc + (a-b)\frac{2c}{3} = \frac{3bc + 2ac - 2bc}{3} = \frac{bc + 2ac}{3} = \frac{c}{3}(2a + 3b)\)( after taking out \(\frac{c}{3}\) as common)

Therefore, the total amount of money the store received for the pens, in dollars is \((2a + 3b)\frac{c}{3}\) (Option C)
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Re: A store had a total of a pens which it sold. The store first sold b of  [#permalink]

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New post 22 Jun 2018, 07:35
Total number of pen is a.
Revenue from selling b pen is bc.
Remaining pen is a-b.
Revenue from selling those pen is (a-b)*2c/3.

Total revenue is bc + (a-b)*2c/3
<=> 3bc/3 + (2ac - 2bc)/3
<=> (3bc + 2ac - 2bc)/3
<=> (2ac + bc)/3

In this case, we are only taking c out, not c/3. So it should be (2ac + bc)/3 = (2a + b)c/3, i.e. A.

It's sometimes easy to make minor mistakes in simplifying fractions, so always important to double-check.
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Re: A store had a total of a pens which it sold. The store first sold b of  [#permalink]

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New post 22 Jun 2018, 09:14
A store had a total of a pens which it sold. The store first sold b of these pens for c dollars each and it sold the remaining pens for 2c/3 dollars each. Which of the following is equal to the total amount of money the store received for the a pens, in dollars?

Take SMART and easy numbers and substitute in choices..
c=3 (since the cost is 2c/3)
a=3, b=1..
So cost = 1*3+2*2=7..
Check choices


A) (2a + b)c/3.....(2*3+1)*3/3=6+1=7....yes

B) (a + 2b)c/3.....(3+2*1)*3/3=5...no

C) (2a + 3b)c/3....((2*3+3*2)*3/3=12...no

D) (2a + c)b/3.....(2*3+3)*1/3=3....no

E) (3a + 2b)c/3.....(3*3+2*1)*3/3=11...no

A
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: A store had a total of a pens which it sold. The store first sold b of  [#permalink]

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New post 25 Jun 2018, 11:53
Harshgmat wrote:
A store had a total of a pens which it sold. The store first sold b of these pens for c dollars each and it sold the remaining pens for 2c/3 dollars each. Which of the following is equal to the total amount of money the store received for the a pens, in dollars?


A) (2a + b)c/3

B) (a + 2b)c/3

C) (2a + 3b)c/3

D) (2a + c)b/3

E) (3a + 2b)c/3


The total made was:

bc + 2c/3(a - b)

bc + 2ca/3 - 2bc/3

bc/3 + 2ca/3

(b + 2a)c/3

Answer: A
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Re: A store had a total of a pens which it sold. The store first sold b of &nbs [#permalink] 25 Jun 2018, 11:53
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