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banksy
A bookman purchased 30 copies of a new book released recently, 10 of which are hardback and sold for $20 each, and rest are paperback and sold for $15 each. If 12 copies were sold and the total value of the remaining books was 280, how many paperback copies were sold?

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12

Since 10 of the 30 books are hardback, the remaining 20 books are paperback.
Since 12 of the 30 books are sold, the total number of unsold books = 30-12 = 18.
We can PLUG IN THE ANSWERS, which represent the number of paperback books that are sold.
When the correct answer is plugged in, the total value of the unsold books = 280.

B: 6
Since 6 of the 20 paperback books are sold, the number of unsold paperback books = 20-6 = 14.
Since the total number of unsold books = 18, the number of unsold hardback books = 18-14 = 4.
Since the 14 unsold paperback books are priced at $20 each, while the 4 unsold hardback books are priced at $15 each, the total value of unsold books = (14*20) + (4*15) = 280 + 60 = 340.
The total value of the unsold books is TOO HIGH.
To reduce the total value of the unsold books, the less expensive type of book -- paperback -- must constitute a greater proportion of the unsold books.
Implication:
Fewer paperback books must be sold, so that a greater number of paperback books remain UNSOLD.



Nice Approach.However I had a query.

When I read the question stem for the first time.

A bookman purchased 30 copies of a new book released recently, 10 of which are hardback and sold for $20 each, and rest are paperback and sold for $15 each. If 12 copies were sold and the total value of the remaining books was 280, how many paperback copies were sold?

I was confused if already 10 are hard back and sold $20 and remaining are paperpack and sold
for $15..Why does it again reiterate If 12 copies were sold...

Can you please help me understand the way it i constructed?
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banksy
A bookman purchased 30 copies of a new book released recently, 10 of which are hardback and sold for $20 each, and rest are paperback and sold for $15 each. If 12 copies were sold and the total value of the remaining books was 280, how many paperback copies were sold?

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12


Now let us look at different ways to do this question

(I) Weighted average
If you know this, I would surely recommend this
total books unsold = \(30-12=18\)
cost of these 18 books is 280..
Had all these been H 20 per book, cost would be 18*20=360
And Had all these been P @ 15 per book, cost would be 18*15=270
so Number of books of P = \(18*\frac{360-280}{360-270}=18*\frac{8}{9}=16\)
so 16 out of 20 paperback are unsold and remaining 4 are sold

(II) Substitution
If you are stuck, I would recommend this
let us take choice A - 4..
so 4 out of 20 paperback are sold, and remaining 16 are unsold, costing \(16*15=240\)\(\)
Total unsold is 18, so 18-16 or 2 are hardback, costing 2*20=40
total cost of unsold, thus, is \(240+40=280\) and given is also 280
so choice A is correct

(III) Algebraic
there can be two equations...
\(H+P=12.......H=12-P\)...(i)
\(20*(10-H)+15*(20-P)=280..........200-20H+300-15P=280..........20H+15P=220\)
substitute value of H in it
\(20(12-P)+15P=220.........240-20P+15P=220.........5P=20....P=4\)

A
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banksy
A bookman purchased 30 copies of a new book released recently, 10 of which are hardback and sold for $20 each, and rest are paperback and sold for $15 each. If 12 copies were sold and the total value of the remaining books was 280, how many paperback copies were sold?

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12


Now let us look at different ways to do this question

(I) Weighted average
If you know this, I would surely recommend this
total books unsold = \(30-12=18\)
cost of these 18 books is 280..
Had all these been H 20 per book, cost would be 18*20=360
And Had all these been P 20 per book, cost would be 18*15=270
so Number of books of P = \(18*\frac{360-280}{360-270}=18*\frac{8}{9}=16\)
so 16 out of 20 paperback are unsold and remaining 4 are sold

(II) Substitution
If you are stuck, I would recommend this
let us take choice A - 4..
so 4 out of 20 paperback are sold, and remaining 16 are unsold, costing \(16*15=240\)\(\)
Total unsold is 18, so 18-16 or 2 are hardback, costing 2*20=40
total cost of unsold, thus, is \(240+40=280\) and given is also 280
so choice A is correct

(III) Algebraic
there can be two equations...
\(H+P=12.......H=12-P\)...(i)
\(20*(10-H)+15*(20-P)=280..........200-20H+300-15P=280..........20H+15P=220\)
substitute value of H in it
\(20(12-P)+15P=220.........240-20P+15P=220.........5P=20....P=4\)

A


Chetan sir .Can you please help me with this


When I read the question stem for the first time.

A bookman purchased 30 copies of a new book released recently, 10 of which are hardback and sold for $20 each, and rest are paperback and sold for $15 each. If 12 copies were sold and the total value of the remaining books was 280, how many paperback copies were sold?

I was confused if already 10 are hard back and sold $20 and remaining are paperpack and sold
for $15..Why does it again reiterate If 12 copies were sold...

Can you please help me understand the way it i constructed?
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chetan2u
banksy
A bookman purchased 30 copies of a new book released recently, 10 of which are hardback and sold for $20 each, and rest are paperback and sold for $15 each. If 12 copies were sold and the total value of the remaining books was 280, how many paperback copies were sold?

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12


Now let us look at different ways to do this question

(I) Weighted average
If you know this, I would surely recommend this
total books unsold = \(30-12=18\)
cost of these 18 books is 280..
Had all these been H 20 per book, cost would be 18*20=360
And Had all these been P 20 per book, cost would be 18*15=270
so Number of books of P = \(18*\frac{360-280}{360-270}=18*\frac{8}{9}=16\)
so 16 out of 20 paperback are unsold and remaining 4 are sold

(II) Substitution
If you are stuck, I would recommend this
let us take choice A - 4..
so 4 out of 20 paperback are sold, and remaining 16 are unsold, costing \(16*15=240\)\(\)
Total unsold is 18, so 18-16 or 2 are hardback, costing 2*20=40
total cost of unsold, thus, is \(240+40=280\) and given is also 280
so choice A is correct

(III) Algebraic
there can be two equations...
\(H+P=12.......H=12-P\)...(i)
\(20*(10-H)+15*(20-P)=280..........200-20H+300-15P=280..........20H+15P=220\)
substitute value of H in it
\(20(12-P)+15P=220.........240-20P+15P=220.........5P=20....P=4\)

A


Chetan sir .Can you please help me with this


When I read the question stem for the first time.

A bookman purchased 30 copies of a new book released recently, 10 of which are hardback and sold for $20 each, and rest are paperback and sold for $15 each. If 12 copies were sold and the total value of the remaining books was 280, how many paperback copies were sold?

I was confused if already 10 are hard back and sold $20 and remaining are paperpack and sold
for $15..Why does it again reiterate If 12 copies were sold...

Can you please help me understand the way it i constructed?

Hi..
The give away is the verb ARE..
These are sold for 50$ would mean these are priced at 50$..
Had it been these were sold for 50$, yes it would mean they have already been sold...

.
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Hi..
The give away is the verb ARE..
These are sold for 50$ would mean these are priced at 50$..
Had it been these were sold for 50$, yes it would mean they have already been sold...

.[/quote]



So its more like Jeans are sold for $20.(MEANS its priced at $20)

Thank you so much.Just wanted to make sure I don't have the same GAP for new problems of this type.
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banksy
A bookman purchased 30 copies of a new book released recently, 10 of which are hardback and sold for $20 each, and rest are paperback and sold for $15 each. If 12 copies were sold and the total value of the remaining books was 280, how many paperback copies were sold?

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12
30 Books (10 Hardbacks and 20 Paperbacks)

Sold Price: $20 Hard and $15 Paper
Total: 10x$20 + 20x$15 = $500
500 - 280 = 220 (220 is total sold price)

Two Equations

1. H + P = 12 (sold) or H = 12-P

2. 20H + 15P = 220

Combine: 20(12-P) + 15P = 220
240 - 20P + 15P = 220
20 = 5P or P=4
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