题目信息
Merle's spare change jar has exactly 16 U.S. coins, each of which is a 1-cent coin, a 5-cent coin, a 10-cent coin, a 25-cent coin, or a 50-cent coin. If the total value of the coins in the jar is 288 U.S. cents, how many 1-cent coins are in the jar?
A:Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B:Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C:BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D:EACH statement ALONE is sufficient.
E:Statements (1) and (2) TOGETHER are NOT sufficient.
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正确答案：
D:EACH statement ALONE is sufficient.
Algebra Simultaneous equations
Let a, b, c, d, and e be the number, respectively, of 1-cent, 5-cent, 10-cent, 25-cent, and 50-cent coins. We are given the two equations shown below. Determine the value of a.
a + b + c + d + e = 16
a + 5b + 10c + 25d + 50e = 288
We are given that c = 6, d = 5, and e = 2. Substituting these values into the two equations displayed above and combining terms gives a + b = 3 and a + 5b = 3. Subtracting these last two equations gives 4b = 0, and therefore b = 0 and a = 3; SUFFICIENT. We are given that c = 2a. Substituting c = 2a into the two equations displayed above and combining terms gives the following two equations.
3a + b + d + e = 16
21a + 5b + 25d + 50e = 288
From the first equation above we have 3a = 16 − b − d − e. Therefore, 3a ≤ 16, and it follows that the value of a must be among 0, 1, 2, 3, 4, and 5. From the second equation above we have 5(b + 5d + 10e) = 288 − 21a, and thus the value of 288 − 21a must be divisible by 5.
The table above shows that a = 3 is the only nonnegative integer less than or equal to 5 such that 288 − 21a is divisible by 5; SUFFICIENT.
The correct answer is D;each statement alone is sufficient.
Let a, b, c, d, and e be the number, respectively, of 1-cent, 5-cent, 10-cent, 25-cent, and 50-cent coins. We are given the two equations shown below. Determine the value of a.
a + b + c + d + e = 16
a + 5b + 10c + 25d + 50e = 288
We are given that c = 6, d = 5, and e = 2. Substituting these values into the two equations displayed above and combining terms gives a + b = 3 and a + 5b = 3. Subtracting these last two equations gives 4b = 0, and therefore b = 0 and a = 3; SUFFICIENT. We are given that c = 2a. Substituting c = 2a into the two equations displayed above and combining terms gives the following two equations.
3a + b + d + e = 16
21a + 5b + 25d + 50e = 288
From the first equation above we have 3a = 16 − b − d − e. Therefore, 3a ≤ 16, and it follows that the value of a must be among 0, 1, 2, 3, 4, and 5. From the second equation above we have 5(b + 5d + 10e) = 288 − 21a, and thus the value of 288 − 21a must be divisible by 5.
| a | 288 − 21a |
| 0 | 288 |
| 1 | 267 |
| 2 | 246 |
| 3 | 225 |
| 4 | 204 |
| 5 | 183 |
The table above shows that a = 3 is the only nonnegative integer less than or equal to 5 such that 288 − 21a is divisible by 5; SUFFICIENT.
The correct answer is D;each statement alone is sufficient.
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