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30 Jan 2020, 06:15
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C
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A book new released include paperback and hardback version. The sales price for a hardback book is twice that for a paperback book. On the first week, if a total of 1200 books were sold out, how many of them were hardback books?
(1) The sales price for one hardback book and one paper back book sums to $60.
(2) The total revenue on the sales of these books was $30,000.
(1) The sales price for one hardback book and one paper back book sums to $60.
(2) The total revenue on the sales of these books was $30,000.
30 Jan 2020, 20:03
Kudos
Bookmarks
Hi,
I have taken the following approach to solve this question. Please correct me if I am wrong.
Let H = No. of Hardback version books
P = No. of Paperback version books
h = SP(Sales Price) of one hardback version book
p = SP(Sales Price) of one paperback version book
Total no. of books sold out => H + P = 1200 (Given)
h = 2p (Given)
Statement 1:-
h + p =$60
2p + p = $60
p = $20
therefore h = 2p = $40
But we don't have any information about the total cost for purchasing 1200 books because we don't know the total count of each type of book.
So. this statement is insufficient. We can eliminate options A and D.
Statement 2:-
Total Revenue from the sales of the books is $30,00 i.e
hH + pP = 30,000
Here, we don't know the values of h and p, so we cannot solve this equation.
So this statement is insufficient. We can eliminate option B.
Considering both statements together,
h=$40, p=$20
So substituting these values in the above equation,
40H + 20P=30,000
H + P =1200 (Given in the question)
We have 2 variables and 2 equations, so we can solve for the 2 variables(H & P).
So both statements together are sufficient.
Option C is the answer.
I have taken the following approach to solve this question. Please correct me if I am wrong.
Let H = No. of Hardback version books
P = No. of Paperback version books
h = SP(Sales Price) of one hardback version book
p = SP(Sales Price) of one paperback version book
Total no. of books sold out => H + P = 1200 (Given)
h = 2p (Given)
Statement 1:-
h + p =$60
2p + p = $60
p = $20
therefore h = 2p = $40
But we don't have any information about the total cost for purchasing 1200 books because we don't know the total count of each type of book.
So. this statement is insufficient. We can eliminate options A and D.
Statement 2:-
Total Revenue from the sales of the books is $30,00 i.e
hH + pP = 30,000
Here, we don't know the values of h and p, so we cannot solve this equation.
So this statement is insufficient. We can eliminate option B.
Considering both statements together,
h=$40, p=$20
So substituting these values in the above equation,
40H + 20P=30,000
H + P =1200 (Given in the question)
We have 2 variables and 2 equations, so we can solve for the 2 variables(H & P).
So both statements together are sufficient.
Option C is the answer.
31 Jan 2020, 03:56
Kudos
Bookmarks
Since hardbacks cost twice as much as paperbacks, Statement 1 tells us they cost $40 and $20 respectively.
Statement 2 tells us the average price per book was $30,000/1200 = $25.
Taking both Statements together, we now have a standard weighted average situation: the average cost of paperbacks is $20, the average cost of hardbacks is $40, and the average of the two combined is $25. Since the overall average is "three times as close" to the paperback average, we must have three times as many paperbacks (using alligation principles). So 3/4 of the books sold were paperbacks, and 1/4 of the 1200 books sold, or 300 books, were hardbacks. So the answer is C.
Statement 2 tells us the average price per book was $30,000/1200 = $25.
Taking both Statements together, we now have a standard weighted average situation: the average cost of paperbacks is $20, the average cost of hardbacks is $40, and the average of the two combined is $25. Since the overall average is "three times as close" to the paperback average, we must have three times as many paperbacks (using alligation principles). So 3/4 of the books sold were paperbacks, and 1/4 of the 1200 books sold, or 300 books, were hardbacks. So the answer is C.
