An unknown number of coins have been scattered on a table.

Question

An unknown number of coins have been scattered on a table. What percent of them are tails?
I) By turning only four of the coins the number of heads and tails will be equal.
II) By removing 4 coins (2 heads and 2 tails) from the table, 30 percent of the coins will be heads.
A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D) EACH statement ALONE is sufficient to answer the question asked.
E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

MahmoudFawzy · Answer

From statement (1) 
after turning 4 coins, T = H.
lets say 8 coins. if after turning we have (HHHH,TTTT),
so before turning they may be (HHHT,HHHT), or (HHTT,HHTT)
so T% could be 25% (2/8) or 50% (4/8) -->  insufficient .
actually, (1) give no info at all except that the coins are  ≥ 4

From statement (2)
 \frac{H-2}{{T-2 + H-2}} = \frac{3}{10}  ---------->  7H = 3T + 8  ---------->  T = \frac{{7H-8}}{3} 
(H,T) pair can be (5,9) or (8,16) or ...
so T% could be 64% (9/14) or 66% (16/24) -->  insufficient

combining (1) and (2) has no use so  E

Shrey9 · Answer

Statement 2 looks sufficient to me..
They say by removing 4 coins we have a 30-70 split

Then we, without solving further, can find the percentage.. 
I'm just not sure how Statement 1 is sufficient, as the OA is given to be D

Posted from my mobile device

IanStewart · Answer

This is not as straightforward a problem as the solutions posted already suggest. Using both statements, the information is almost sufficient, since we can only get two different answers to the problem.

We know removing two H and two T from the table, 30% of our coins are H. So after that removal, the ratio of H to T is 3 to 7. That's reduced, so the total number of coins (after removing four of them) must be a multiple of 10. Note this is not sufficient alone, because we might have exactly 3 H and 7 T, and thus after adding back the four coins, 5/14 ~ 36% could be Heads, or we might have 3 trillion Heads and 7 trillion Tails, and adding back just four coins will make almost no difference, and we'll have negligibly more than 30% Heads.

So the total number of coins is 4 greater than some multiple of 10. From Statement 1, we also know that by flipping just 4 coins, we can make the number of Heads and Tails equal. We can do that with 5 Heads and 9 Tails -- we can flip one H and one T, keeping the numbers at 5 and 9, and then flip two of the T's to get 7 of each. We can also do that with 8 Heads and 16 Tails -- flipping four of the Tails, so they become Heads, gives us 12 of each. But that's the biggest change we can make: if the difference in the number of Tails and the number of Heads exceeds 8, we'd never be able to flip only four coins and arrive at equal numbers of Heads and Tails. And if we have 34 or more coins, the difference T-H will exceed 8. So using both statements, we can have only 14 coins or 24 coins in total, and the fraction that are tails can only be 9/14 or 16/24. But those are different, so the answer is E.

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An unknown number of coins have been scattered on a table.

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An unknown number of coins have been scattered on a table.

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