#1: 214 tickets split 9 ways does not work out to an integer number. Nobody wants 2/3 of a ticket, right?
So the trick here is to determine the first multiple of 9 greater than 214.
Note that 214/9 = 23.8. No good. Not an integer.
Multiply 23 by 9; we get 207. Multiply 24 by 9; we get 216.
So, given that we have 214 tickets already, we need 2 more, to make 216.
Check: Is 216 divisible by 9 without a remainder? Yes. Then each of the 9 students will get 24 tickets.
The rest of the problems are somewhat similar to this one.
See whether you can do the next one, #2.
If you'd share your results, I'd be happy to comment on what you've done.
Answer:
c = 100, a = 50, s = 20
Step-by-step explanation:
following information is missing from the question
6c + 7a + 2s = 990
3c + 3a + 4s = 430
2c + 4a + 4s = 480
If you solve the equation by putting in the values given in the option you will deduce that only option d matches LHS to the RHS
Lets see how,
6c + 7a + 2s = 990
where, c = 100, a = 50, s = 20
6(100) + 7(50) + 2(20) = 990
600 + 750 + 40 = 990
990 = 990 hence true
2c + 4a + 4s = 480 where, c = 100, a = 50, s = 20
2 (100) + 4 ( 50) + 4 (20) = 480
200 + 250 + 80 = 480
See for the other option too with the same steps done above to have a more clear understanding :)
Answer:
I dont know what letter but its 12i^2+i+6
Step-by-step explanation:
cross multiply
