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Viktor [21]
2 years ago
10

A librarian is expanding some sections of the city library. He buys book at a special price from a dealer who charges one price

for any hardback book and another price for any paperback books. For the children’s section, Mr. Kent purchased 12 new hardcover books and 73 new paperback books, which cost a total of $594. He also purchased 11 new hardcover books and 87 new paperback books for the adult fiction section, spending a total of $665. What is the special price for each type of book?
Mathematics
1 answer:
stiks02 [169]2 years ago
7 0

Answer:

The special price of 1 hardcover book = $13

The special price of 1 paperback book = $6

Step-by-step explanation:

Here, let us assume the special price of 1 hardcover book = $ m

and the the special price of 1 paperback  book = $ n

Now, the  price of 12 hardcover books  = 12 x ( Price of 1 hardcover)  = 12 m

and  the  price of  73 paperback books  = 73 x ( Price of 1 paperback)  = 73 n

Also, The price paid for  12 hardcover and  73 paperback books is $594

⇒ 12 m +  73  n = 594  .......  (1)

Similarly:

The  price of 11 hardcover books  = 11 x ( Price of 1 hardcover)  = 11 m

The  price of  87 paperback books  = 87 x ( Price of 1 paperback)  = 87 n

Also, The price paid for  11 hardcover and  87 paperback books is $665.

⇒ 11 m +  87  n = 665  .......  (2)

Now, solving both equation for the value of m and n , we get:

12 m +  73  n = 594       x 11          

11 m +  87  n = 665    x   (-12)

we get:  132 m +  803 n =  6534

             - 132 m  - 1044 n = - 7980

adding both equation, we get:

132 m +  803 n   - 132 m  - 1044 n = - 7980 +   6534

or,  - 241  n =  - 1446

or, n = 1446/241 =  6, or n = 6

Now,  Substituting n = 6 in 12 m +  73  n = 594 ,

we get: 12 m +  73 ( 6)  = 594, m =   13

Hence, The special price of 1 hardcover book = $13

and The special price of 1 paperback book = $6

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As Per Provided Information

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On substituting the value in above formula we obtain

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\boxed{\bf \: Perimeter_{(Square)} = 4 \times side}

Substituting the value we obtain

\qquad\longrightarrow\sf  \:Perimeter_{(Square)} \:  = 4 \times 4 \\  \\  \\ \qquad\longrightarrow\sf  \:Perimeter_{(Square)} = 16 \: cm

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\qquad\longrightarrow\sf  \:Area_{(Square)} \:  =  {4}^{2}  \\  \\  \\ \qquad\longrightarrow\sf  \:Area_{(Square)} = 16 \:  {cm}^{2}

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