Lets see what you think

x = tail
x + 4 = heads

1) x + x + 4 =24
2x = 20
x=10

10 tails
14 heads

SUFF

2)
x + 3(x+4) = 52
x + 3x + 12 = 52
4x + 12 = 52
4x = 40
x = 10

10 tails
14 heads

SUFF

I choose D

Firstly, according to the OG, you are correct.

BUT the question asks How many times DID heads result? Not, how many times was heads LIKELY to result. I feel like the assumption that heads will arise half of the time is just that, an assumption. And hence we cannot make any concrete conclusions about what the outcome will be. What do you think?

We know concretely that the coin was flipped 24 times and that with those 24 flips heads DID result 4 more times than tails.

This concretely tells us that heads must result 14 times and tails only 10.

From the stem:
heads = tails + 4; h=t+4

1) 24 = h + t
24 = t+4+t
24 = 2t + 4
20 = 2t
t = 10
h = 10 + 4
h = 14
1 is sufficient

2) 52 = 3h + t
52 = 3(t+4) +t
52 = 3t + 12 +t
52 = 4t+12
40= 4t
10 = t
h = t + 4
h = 14
2 is sufficient

Answer = D

-Carrie

saitjordan wrote:I feel like the assumption that heads will arise half of the time is just that, an assumption.

None of the solutions given assume that heads will occur half the time. If heads occured half the time, then, when there are 24 coin tosses, 12 of them would be heads.

The stem said that "4 more heads than tails resulted."

From that, we know that total = tails+4 +tails

If the coin had been flipped 4 times, there would have been 4 heads and 0 tails - hardly heads arising half the time.

The only thing in the question that seems necessary is, "Heads or tails resulted each time the player tossed the coin." A coin is a 2-sided object. Anytime it is flipped, it will land on either one side or the other.

-Carrie

For your clarity-
forget abt probability.
Let the no. of times head comes be A.
the no. of times tail comes be B

Given A-B = 4

1) A+B = 24;hence A is 14
2) 3A +B =52:hence A is 14

So,both 1) and 2) are sufficient.
D