1. Brianna's coins:
Let:
n = number of nickels
d = number of dimes
q = number of quarters
We know:
n + d + q = 19 (total number of coins)
d = n + q - 11
5n + 10d + 25q = 295 (total value in cents)
Substitute the second equation into the third equation:
5n + 10(n + q - 11) + 25q = 295
15n + 10q - 110 + 25q = 295
40n + 35q = 405
8n + 7q = 81
We have two independent equations:
n + d + q = 19
8n + 7q = 81
Solve for n, d, and q using elimination or substitution.
Solution: n = 6, d = 4, q = 9
2. Mason's stamps:
Let:
x = number of 49¢ stamps
y = number of 20¢ stamps
z = number of 3¢ stamps
We know:
x + y + z = 56 (total number of 49¢ and 20¢ stamps)
z = y + 9
49x + 20y + 3z = 2355 (total value in cents)
Substitute the second equation into the third equation:
49x + 20y + 3(y + 9) = 2355
49x + 20y + 3y + 27 = 2355
49x + 23y + 27 = 2355
49x + 23y = 2328
We have two independent equations:
x + y + z = 56
49x + 23y = 2328
Solve for x, y, and z using elimination or substitution.
Solution: x = 40, y = 16, z = 25
3. Theater ticket prices:
Let:
g = cost of general admission ticket
s = cost of senior citizen ticket
c = cost of child ticket
We know:
9g + 2s + 3c = 170 (Mark's purchase)
5g + 4s + 7c = 173 (Sarah's purchase)
4g + s + 6c = 116.50 (Kyle's purchase)
We have three independent equations. Solve for g, s, and c using elimination or substitution.
Solution: g = $13.50, s = $11.50, c = $8.50
4. Investor's accounts:
Let:
x = amount deposited in the 9% account
y = amount deposited in the 6% account
z = amount deposited in the 4% account
We know:
x + y + z = 15,000 (total amount deposited)
y = z + 2,000
0.09x + 0.06y + 0.04z = 980 (total annual interest)
Substitute the second equation into the first equation:
x + (z + 2,000) + z = 15,000
x + 2z + 2,000 = 15,000
x + 2z = 13,000
Substitute the second equation into the third equation:
0.09x + 0.06(z + 2,000) + 0.04z = 980
0.09x + 0.06z + 0.12 + 0.04z = 980
0.09x + 0.1z + 0.12 = 980
0.09x + 0.1z = 968
We have two independent equations:
x + 2z = 1
Problem 5: Relief Organization Purchases
Define variables:
b = cost of one blanket
c = cost of one cot
l = cost of one lantern
Set up equations based on the given information:
Week 1: 15b + 5c + 10l = 1250
Week 2: 20b + 10c + 15l = 2000
Week 3: 10b + 15c + 5l = 1625
Solve for b, c, and l:
Eliminate l: Multiply equation 1 by 3 and equation 3 by 2:
45b + 15c + 30l = 3750
20b + 30c + 10l = 3250
Subtract the second equation from the first: 25b - 15c = 500
Solve for b and c:
Express c in terms of b: c = (500 + 25b) / 15
Substitute this expression for c in equation 2: 20b + 10((500 + 25b) / 15) + 15l = 2000
Simplify and solve for b: 20b + 166.67 + 15l = 2000
20b + 15l = 1833.33
Use equation 1 to solve for b: 15b + 5c + 10l = 1250
Substitute the expression for c and solve for b: 15b + 5((500 + 25b) / 15) + 10l = 1250
Solve for b: b = 25
Substitute b back into the equation for c:
c = (500 + 25b) / 15
c = (500 + 25 * 25) / 15
c = 75
Substitute b and c back into the equation for l:
15b + 5c + 10l = 1250
15 * 25 + 5 * 75 + 10l = 1250
l = 50
Solution:
Cost of one blanket: b = $25
Cost of one cot: c = $75
Cost of one lantern: l = $50