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There is $500 in each of three envelopes. The first envelope contains only $10 bills, the second envelope contains only $20 bills, and the third envelope contains only $50 bills. One bill is taken out of one of the envelopes, two bills are taken out of another envelope, and three bills are taken out of the remaining envelope (in any order). What is the least possible amount, in dollars, that can be taken out?

Respuesta :

Answer:

order:

-3 bills from first envelope (-30 dollars)

-2 bills from second envelope (-40 dollars)

-1 bill from third envelope (-50 dollars)

total least possible amount that can be taken out: 120

Step-by-step explanation:

Any other ways results in more than 120 being taken away. Need to have the least taken away from the 50 bill envelope because 50 is the largest bill. Take away the most from the 10 bill envelope because 10 is the smallest bill.

wergompol23869

Answer:

$120

Step-by-step explanation:

So this question isn't hard either

We can cross out excess information and just find the main points.

a. 3 Envolopes containing 10 bills, 20 bills, and 50 bills

b. 1-3 bills are taken out of each envolope

c. It wants you to find the least possible amt. that could have been token out.

So since we want the least, we must take the smallest out of the largest

In order, 1 from 50 (largest)

2 from 20 (middle)

and 3 from 10 (smallest)

this leaves us the equation (1x50) + (2x20) + (3x10) => 50+40+30 => 120

Thus, we have our answer 120

Hope this helps!

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