Adam has 3 coins in his pocket and they are all different from each other. Ben has 3 coins in his pocket and they are all the same as each other. Adam has half as much money as Ben. What are the coins they each have?

The possible coins are:

P = penny = $0.01

N = nickel = $0.05

D = dime = $0.10

Q = quarter = $0.25

H = half dollar = $0.50

We know that Adam has 3 coins in his pocket, and all of them are different, then the money that Adam has can be represented with (A + B + C)

Behn has 3 coins, and all of them are the same, then if his coin has a value X, he has a total of 3*X

We know that Adam has half as much money as Ben, this can be written as:

(A + B + C) = 3*X/2

The easier way to solve this, is to play with different values of X, and see if we can find the values of A, B and C.

For example, if X = $0.10

then:

(A + B + C) = $0.10*(3/2) = $0.15

We can find 3 different value of coins for this equation? No, we cant, so X = $0.10 is also discarded.

if x = $0.50 then:

(A + B + C) = (3/2)*$0.50 = $0.75

Here we could have:

A = $0.50

B = $0.25

then: A + B = $0.75

But we still have the coin C.

If we take X = $0.25 then:

(A + B + C) = (3/2)*$0.25 = $0.375

We could round the right part to $0.40

then the coins in the left part would be:

A = $0.25

B = $0.10

C = $0.05

A + B + C =  $0.25 + $0.10 +  $0.05 = $0.40

This is a strech, but is the only thing we can make with the given problem.

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0.16 or 16%

step-by-step explanation:

lets take a coin for example.

if you toss it twice, it has a 1/4 chance of landing on heads both times because 1/2 * 1/2 is 1/4.

with the cards, the chance of picking dog is 0.4. having two people pick one is like picking a card twice in a row. so 0.4 * 0.4 is 0.16, or 16%