jc114 wrote:If the sum of 2 positive integers is 24 and the difference of their squares is 48, what is the product of the 2 integers?
A. 109
B. 118
C. 128
D. 135
E. 143

a+b = 24

a^2 - b^2 = 48
(a+b)(a-b) = 48
a-b = 2

a+b = 24

Solving for a and b we get 13 and 11.

So product = 143

jc114 wrote: On a Saturday night, each of the rooms at a certain motel was rented for either $40 or $60. If 10 of the rooms that were rented for $60 had instead been rented for $40, then the total rent the motel charged for that night would have been reduced by 25%. What was the total rent the motel actually charged for that night.
A. 600
B.800
C. 1000
D.1600
E. 2400

10*60 - 10*40 = 200.

200 is 25% of total rent which implies total rent = 800

If the sum of 2 positive integers is 24 and the difference of their squares is 48, what is the product of the 2 integers?
A. 109
B. 118
C. 128
D. 135
E. 143

x + y = 24
x^2 - y^2 = 48
Now x^2 - y^2 = (x+y) (x-y)
So x-y = 48/24 = 2

x+y = 24
x-y = 2
So 2x = 26 and x = 13
y = 11
x*y = 11*13 = 143

On a Saturday night, each of the rooms at a certain motel was rented for either $40 or $60. If 10 of the rooms that were rented for $60 had instead been rented for $40, then the total rent the motel charged for that night would have been reduced by 25%. What was the total rent the motel actually charged for that night.
A. 600
B.800
C. 1000
D.1600
E. 2400

Let total revenue be x
Since 10 rooms which have been switched from $60 to $40 they represent a loss of $200 @$20 per room
$200 represents a loss of 25% on the old revenue
25x/100 = 200
x = 800