Johnny is selling tickets to a school play. On the first day of ticket sales he sold 14 senior (S) citizen tickets and 4 child (C) tickets for a total of $200. On the second day of ticket sales he sold 7 senior (S) citizen tickets and 1 child (C) ticket for a total of $92. What is the price of one child ticket?
14S + 4C = 200
14S = 200 - 4C
S = (200 - 4C)/14
7S + 1C = 92
7S = 92 - C
S = (92 - C)/7
(200 - 4C)/14 = (92 - C)/7
7 x (200 - 4C) = 14 x (92 - C)
1400 - 28C = 1288 - 14C
1400 - 1288 = 28C - 14C
112 = 14C
C = 112/14 = 8
the price of one child ticket = $8
Answer:
Figure (i) and (iv)
Step-by-step explanation:
Given:
Optional figure is given in attached file.
We need to find two figures that are similar to the 5 by 10 figure.
All the given figure are
form.
Where m represent the number of rows and n represent the number of columns.
Solution:
Observe that in the given figure 5 by 10, the number of rows is 5 and number of columns is 10, that is, the number of columns is double of that the number of rows.
So we need to find two such figures whose number of columns is double of the number of rows.
From the given figures, figure (i) the number of rows is 2 and number of columns is 4, which is double of number of rows. so it is similar to 5 by 10 figure.
Similarly in figure (iv), the number of rows is 4 and number of columns is 8. so the number of columns is double the number of rows, so it is similar to the figure 5 by 10.
Therefore, the two figures that are similar to 5 by 10 figure are given in attached file such as (i) and (iv).
You could subtract the denominator fro the numerator which would give you a total sum of 1 thousand an fifty five