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P=price, Q=quantity
Price*quantity=75000
Price percentage increase=30%
Quantity percentage decrease=30%
Net percentage increase/decrease on PQ= 30%-30%-(30%*30%)/100..........(effective percentage formula)
Net percentage effect= -9%(or 9% decrease).Therefore PQdecrease by 9%
Now PQ on 100%=75000
Then PQ on 91%=x
Cross multiply to get x=75000*91/100=68250.
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Approximation technique maybe mightily helpful in this question.

We know that a cycle of successive %age increase-decrease of a quantity results in the overall decrease. Thus, options A and B are gone.

Now, %age increase in price of soda bottle = 30% --> 1.3x
Similarly, quantity of soda bottles sold = 30% less than last year --> 0.7y

Multiplying the above two equations, we get (1.3)(0.7)xy = 0.91xy.

We know that the sales for last year was $75,000. Using this in the above equation, we get 0.91(75,000). We know that 90% of 750 is 675, so the closest value to it is our answer.

Thus, D.
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Bunuel
A rural country store's annual sales of bottled soda totaled $75,000 last year. This year the price of a bottle of soda was 30 percent higher than last year but the number of bottles sold was 30 percent lower. What is value of store's total sales of bottled soda for this year?

A. $82,500
B. $75,000
C. $72,750
D. $68,250
E. $38,750


Hi Bunuel,

here is my solution. can you please advice if i made a mistake ? :? i got correct answer but i think i mixed up variables or not :? :) Thanks!

Let total number of bottles sold last year be n
Total value received 75 000
Then cost of each bottle \(\frac{75000}{n}\) = x (first equation)
where x is cost of each bottle

Let total value of bottles sold this year be R
Total number of bottles is n
Total number of bottles sold this year decreased by 30 %
--- > n-0.30n
The cost of each bottle increased by 30% ---- >
X(1.3)

So here we have second equation X(1.3) * (n-0.30n) = R

Now plug in first equation into the second one:

\(\frac{75 000(1.3)}{n}\) * \((n-0.30n) = R\)

\(\frac{97 500}{n}\)* \((n-0.30n) = R\)

\(\frac{(97 500n – 29 250 n)}{n}\) =\(R\)

\(\frac{68 250n}{n}\) = \(R\)

P.S niks18, perhaps you can explain :-) Bunuel is kinda busy:)
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dave13
Bunuel
A rural country store's annual sales of bottled soda totaled $75,000 last year. This year the price of a bottle of soda was 30 percent higher than last year but the number of bottles sold was 30 percent lower. What is value of store's total sales of bottled soda for this year?

A. $82,500
B. $75,000
C. $72,750
D. $68,250
E. $38,750


Hi Bunuel,

here is my solution. can you please advice if i made a mistake ? :? i got correct answer but i think i mixed up variables or not :? :) Thanks!

Let total number of bottles sold last year be n
Total value received 75 000
Then cost of each bottle \(\frac{75000}{n}\) = x (first equation)
where x is cost of each bottle

Let total value of bottles sold this year be R
Total number of bottles is n
Total number of bottles sold this year decreased by 30 %
--- > n-0.30n
The cost of each bottle increased by 30% ---- >
X(1.3)

So here we have second equation X(1.3) * (n-0.30n) = R

Now plug in first equation into the second one:

\(\frac{75 000(1.3)}{n}\) * \((n-0.30n) = R\)

\(\frac{97 500}{n}\)* \((n-0.30n) = R\)

\(\frac{(97 500n – 29 250 n)}{n}\) =\(R\)

\(\frac{68 250n}{n}\) = \(R\)

P.S niks18, perhaps you can explain :-) Bunuel is kinda busy:)

Hi dave13

Yes your method is correct. :thumbup: however you can use a simple approach here -

if \(n\) is the initial number of bottles and \(r\) is the initial rate per bottle, then we have

\(n*r=75000\)

Now after \(30\)% increase in rate and \(30\)% reduction in number of bottles we will have new revenue \(R_1\), as -

\(R_1=0.7n*1.3r=0.91nr=0.91*75000=68250\)
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