MathDB

13

Part of 1989 AMC 8

Problems(1)

1989 AJHSME Problem 13

Source:

6/25/2011
97×53=\frac{9}{7\times 53} =
(A) .9.7×53(B) .9.7×.53(C) .9.7×5.3(D) .97×.53(E) .09.07×.53\text{(A)}\ \frac{.9}{.7\times 53} \qquad \text{(B)}\ \frac{.9}{.7\times .53} \qquad \text{(C)}\ \frac{.9}{.7\times 5.3} \qquad \text{(D)}\ \frac{.9}{7\times .53} \qquad \text{(E)}\ \frac{.09}{.07\times .53}