Question

A jar contains 19 coins consisting of quarters and dimes. The total value of the coins is $3.70. Determine the number of quarters and the number of dimes in the jar.

# of quarters-______________ # of dimes-______________

A large box of apricots cost $7 and a small box of apricots cost $4. Chad purchases a total of 26 large and small boxes. The total cost of the boxes is $146. How many large and small boxes did Chad purchase?
# of large boxes-_______________ # of small boxes-_______________

Answer 1

# of quarters: 12 # of dimes: 7

# of large boxes: 14 # of small boxes: 12

Explanation:

Problem 1: Dimes and Quarters

Let q represent the number of quarters
Let d represent the number of dimes

We have as a first equation
q + d = 19 (1)

Each quarter is worth $0.25 so q quarters worth = 0.25q

Each dime is worth $0.10 so d dimes worth = 0.10d = 0.1d

Total value of q quarters and d dimes
0.25q + 0.1d = 3.70 (2)

We have two equations in 2 variables which can be solved as follows

• Multiply equation (2) by 4 so that the coefficients of q in both equations are the same, namely 1
  4(0.25q + 0.1d) = 4 x 3.70
  q + 0.4d = 14.8 (3)
• Subtract equation 3 from equation 1 to eliminate q term
  q + d - (q + 0.4d) = 19 - 14.8
  q + d - q - 0.4d = 4.2
  0.6d = 4.2
  d = 4.2/0.6 = 7
• Substitute this value of d in equation 1
  q + 7 = 19
  q = 19 - 7 = 12
  
• Therefore there are 12 quarters and 7 dimes

Verify:
12 x 0.25 + 7 x 0.10 = 3.70

Problem 2: Apricots

Solution strategy is the same as Problem 1 so I am skipping lengthy explanations

Let L be the number of large boxes, S be the number of small boxes

We have
L + S = 26 (1)

Total cost of boxes

7L + 4S = 146 (2)

Multiply equation (1) by 4 :
4L + 4S = 4 x 26 = 104 (3)

Subtract (2) from (3):
7L + 4S - (4L + 4S) = 146 - 104
3L = 42
L = 42/3 = 14

Substitute in (1)
14 + S = 26
S = 26-14
S = 12

So there are 14 large and 12 small boxes

Check:
14 x 7 + 12 x 4 = 146