David bought greater than 10 paperback books that cost $8 each and gre
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21 Sep 2021, 07:32
Take paperback books \(= x\) & hardcover books \(= y\)
The cost is between \(240 - 300\)
\(240 < 8x + 20y < 300\)
\(60 < 2x + 5y < 75\)
Now, \(x > 10\) & \(y > 8\)
Now from these, we can solve it.
Let's say
if \(y = 9\), then \(x\) can be \(8, 9, ... , 14\)
BUT \(x > 10\) so x can be \(11/ 12/ 13/ 14\) so the sum will be \(x + y = 20 / 21 / 22 / 23\)
if \(x = 11\), then y can be \(8, 9 , 10\)
BUT \(y > 8\) so y can be \(9 / 10\) so the sum will be \(x + y = 20 / 21 \)
20 / 21 / 22 / 23
Answer D E F G