Arnold had $1.70 in dimes and quarters. He had 3 more dimes than quarters. How many of each coin did he have?

Arnold had $1.70 in dimes and quarters, with 3 more dimes than quarters. To solve this problem, let's define quarters as 'q' and dimes as 'q + 3' (since Arnold has 3 more dimes than quarters). Remember that a quarter is worth 25 cents and a dime is worth 10 cents. The total value equation in cents is thus: 25q + 10(q + 3) = 170.
Solving for 'q' we get: 25q + 10q + 30 = 170, which simplifies to 35q + 30 = 170, and then 35q = 140 when we subtract 30 from both sides. Dividing both sides of the equation by 35 gives us q = 4. This means Arnold has 4 quarters. To find the number of dimes, we substitute q back into 'q + 3' to get 4 + 3, meaning he has 7 dimes.

Answered by JorjaFox | 2024-06-24

x − n u mb er o f q u a r t ers x + 3 − n u mb er o f d im es 0.25 x + 0.1 ( x + 3 ) = 1.7 0.25 x + 0.1 x + 0.3 = 1.7 0.25 x + 0.1 x = 1.7 − 0.3 0.35 x = 1.4 ∣ : 0.35 x = 4 x + 3 = 4 + 3 = 7 He ha d 4 q u a r t ers an d 7 d im es .

Answered by luana | 2024-06-25

Arnold has 4 quarters and 7 dimes. We found the number of coins by setting up an equation based on the total value of the coins. We solved the equation step-by-step to determine both quantities.
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Answered by luana | 2025-01-24