AMC8 1991

AMC8 1991 · Q4

AMC8 1991 · Q4. It mainly tests Arithmetic misc.

If $991 + 993 + 995 + 997 + 999 = 5000 - N$, then $N =$

如果 $991 + 993 + 995 + 997 + 999 = 5000 - N$，则 $N = $

(A) 5 5

(B) 10 10

(D) 20 20

(E) 25 25

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Each of the five numbers on the left side of the equation is approximately equal to 1,000. Thus $N$ can be found by computing the difference between 1,000 and each number, so $N = 9 + 7 + 5 + 3 + 1 = 25$.

答案（E）：等式左边的五个数每个都约等于 1,000。因此，可以通过计算 1,000 与每个数的差来求得 $N$，所以 $N = 9 + 7 + 5 + 3 + 1 = 25$。

Topics

AR.015 Arithmetic misc