So the possible numbers are multiples of 6 smaller than 24:
6
12
18
24 is not ok, since he won both kinds of trophies, so if he won 24 soccer trophies, he'd have no baseball trophies!
if he had 12 of one kind of trophies, he'd have 24-12=12 also 12 of the other trophies, and we know that he has more soccer trophies, so we reject this
so he has 6 and 18 trophies of one kind. Since he has more soccer trophies, this means that he has 18 soccer trophies and 6 baseball trophies!
Answer:
Number of each ticket is;
$10 tickets = 1115
$20 tickets = 1251
$30 tickets = 934
Step-by-step explanation:
Let x,y and z represent the number of $10,$20 and $30 tickets sold.
Given;
Total number of tickets n = 3300
x+y+z = 3300 .....1
Total sales = $64,190
10x + 20y + 30z = 64,190 .....2
It has sold 136 more $20 tickets than $10 tickets
y = x +136 ........3
Substituting equation 3 into equation 1 and 2;
For 1;
x+y+z = 3300
x+(x+136)+z = 3300
2x + z = 330-136
2x + z = 3164 ........4
For 2;
10x + 20y + 30z = 64,190
10x + 20(x+136) + 30z = 64,190
10x + 20x + 2720 + 30z = 64190
30x + 30z = 64190-2720
30x+30z = 61470
divide through by 30
x+z = 2049 ......5
Subtract equation 5 from 4
2x-x +z-z = 3164-2049
x = 1115
From equation 3
y = x + 136 = 1115+136
y = 1251
From equation 1;
z = 3300 - (x+y)
z = 3300- (1115 + 1251)
z = 934
Number of each ticket is;
$10 tickets = 1115
$20 tickets = 1251
$30 tickets = 934
Yes because you can multiply each sides number by three and it equals the larger figures side numbers.
Answer:
the answer is c .113r10 try the app math