Question

Andre has $425 in 14 bills in his wallet, including ones, twenties, and hundreds. If the number of twenty dollar bills is two less than the combined number of ones and hundreds, how many of each denomination does Andre have in his wallet?

A. 3 ones, 5 twenties, 6 hundreds

B. 4 ones, 6 twenties, 4 hundreds

C. 5 ones, 3 twenties, 6 hundreds

D. 6 ones, 4 twenties, 4 hundreds

Answer 1

Andre has 3 ones, 4 twenties, and 6 hundreds in his wallet.

Step-by-step explanation:

To solve this problem, we can set up a system of equations. Let's denote the number of ones as x, the number of twenties as y, and the number of hundreds as z.

First, we know that Andre has 14 bills in total, so we can write the equation x + y + z = 14.

We also know that the number of twenties is two less than the combined number of ones and hundreds, so we can write the equation y = x + z - 2.

Now we can solve this system of equations for x, y, and z. By substituting the second equation into the first equation, we have x + (x + z - 2) + z = 14. Simplifying this equation, we get 2x + 2z = 16. Now we can solve this equation to find that x = 3 and z = 6. Substituting these values back into the second equation, we find that y = 4.

Therefore, Andre has 3 ones, 4 twenties, and 6 hundreds in his wallet, which corresponds to answer choice B.