2019 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 1 Easy

Alicia had two containers. The first was \frac{5}{6} full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was \frac 34 full of water. What is the ratio of the volume of the first container to the volume of the second container? (2019 AMC 10B Problem, Question#1)

  • A.

    \frac{5}{8}

  • B.

    \frac{4}{5}

  • C.

    \frac{7}{8}

  • D.

    \frac{9}{10}

  • E.

    \frac{11}{12}

Answer:D

Let the first jar's volume be A and the second's be B. It is given that \dfrac{3}{4}A= \dfrac{5}{6}B. We find that\dfrac{B}{A}=\dfrac{\left( \dfrac{3}{4} \right)}{\left( \dfrac{5}{6} \right)}= \left( \text {D}\right)\dfrac{9}{10}.

We already know that this is the ratio of the smaller to the larger volume because it is less than 1.

We can set up a ratio to solve this problem. If x is the volume of the first container, and y is the volume of the second container, then:\frac{5}{6}x= \frac{3}{4}y,

Cross-multiplying allows us to get\frac{x}{y}= \frac{3}{4} \cdot \frac{6}{5}= \frac{18}{20}= \frac{9}{10}. Thus the ratio of the volume of the first container to the second container is \left( \text {D}\right)\dfrac{9}{10}.

An alternate solution is to plug in some maximum volume for the first container - let's say 72, so there was avolume of 60 in the first container, and then the second container also has a volume of 60, so you get 60\cdot \frac 43=80 Thus to answer is \frac{72}{80}=\left( \text {D}\right)\dfrac{9}{10}.

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