Respuesta :
c = 10 , a = 12
4c + 6a < = 120...4(10) + 6(12) < = 120...40 + 72 < = 120....112 < = 120 (true)
4c + 4a < = 100...4(10) + 4(12) < = 100....40 + 48 < = 100...88 < = 100 (true)
yes....they can build 10 child bikes and 12 adult bikes
4c + 6a < = 120...4(10) + 6(12) < = 120...40 + 72 < = 120....112 < = 120 (true)
4c + 4a < = 100...4(10) + 4(12) < = 100....40 + 48 < = 100...88 < = 100 (true)
yes....they can build 10 child bikes and 12 adult bikes
Answer: The company can build 10 child bikes and 12 adult bikes in the week .
Step-by-step explanation:
Let 'c' be the number of child bike and 'a' be the number of adult bike.
According to the problem
The restriction of building time for a week is 4c+6a≤120 hours.........(1)
and the restriction of testing time for a week is 4c+4a≤100 hours...............(2)
Lets check whether company can build c=10 and a=12 bikes in a week by putting this value in (1) and (2)
(1)......4(10)+6(12)=40+72=112≤120 ⇒Restriction of building time is satisfied.
(2)......4(10)+4(12)=40+48=88≤100⇒Restriction of testing time is satisfied.
Hence, the company can build 10 child bikes and 12 adult bikes in the week,as order is meeting with the restrictions.