Answer:
a. 9.5x + 6.5(x+c) < 8 Â when c>0
b. Must be one child more than the no. of adults.
Step-by-step explanation:
For Cinema 1:
for adult = $9.50
for child = $6.50
For Cinema 2:
Per person regardless of age = $8.00
First of all, we will find out the condition when per person rates in both cinema are equal.
Assume x = no. of adults
y = no. of children
Rate per person in Cinema I = Rate per person in Cinema II
(9.5x + 6.5y)/(x+y) Â = Â 8
9.5x + 6.5y = 8(x+y)
9.5x + 6.5y = 8x + 8y
9.5x-8x = 8y-6.5y
=> x = y
So rates are equal when no. of adults equals no. of children
For Cinema I to have better rates, no. of children should be atleast 1 more than the no. of adult. In this way the rate per person of Cinema I will be less than 8
Hence we form an inequality when y = x+c and c > 0
9.5x + 6.5(x+c) < 8 Â when c>0
Hence there must be 1 more children than the no. of adults attending Cinema I for it to be a better deal.
Answer:
9.50a+6.50c < 8.00(a+c), 6 children
Step-by-step explanation:
Let the number of adults be a
the number of children be c
Part 1 :
Cinema 1:
Cost for adults = $9.50 x a = 9.50a
Cost of children = $6.50 x c = 6.50c
Total cost = 9.50a+6.50c
Cinema 2 :
Total cost = 8.00(a+c)
The inequality which shows that the cinema I is cheaper
9.50a+6.50c < 8.00(a+c)
9.50a+6.5c<8a+8c
9.5a-8a<8c-6.5c
1.5a<1.5c
a<c
Case 2:
6 adults goes to cinema , let they are accompanied by c number of children
Cinema 1
Total cost = 9.5 x 6 + 6.5 x c
for cinema 2 the total cost will be
8 ( 6+c)
for cinema 1 to be a better deal
9.5 x 6 + 6.5 x c  < 8(6+c)
57+6.5c<48+8c
57-48<8c-6.5c
9<1.5c
c>6
Hence for Cinema 1 to be a better deal , there must be 6 children accompanying them
