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7th Grade Math : Operations

Study concepts, example questions & explanations for 7th Grade Math

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Example Questions

Example Question #4 : 7th Grade Math

Andrew spends every Saturday at the gym working out. He can complete $\frac{1}{12}$ $\displaystyle \frac{1}{12}$ of his workout in $\frac{1}{6}$ $\displaystyle \frac{1}{6}$ of an hour. If he continues at this rate, how much of his workout does Andrew complete per hour?

Possible Answers:

$\frac{1}{2}\textup{ of his workout}$ $\displaystyle \frac{1}{2}\textup{ of his workout}$

$\frac{1}{6}\textup{ of his workout}$ $\displaystyle \frac{1}{6}\textup{ of his workout}$

$\frac{1}{3}\textup{ of his workout}$ $\displaystyle \frac{1}{3}\textup{ of his workout}$

$\frac{1}{4}\textup{ of his workout}$ $\displaystyle \frac{1}{4}\textup{ of his workout}$

Correct answer:

$\frac{1}{2}\textup{ of his workout}$ $\displaystyle \frac{1}{2}\textup{ of his workout}$

Explanation:

The phrase "per hour" gives us a clue that we are going to divide. In this problem, we can replace the word "per" with a division sign; therefore, we will have the portion of his workout, $\frac{1}{12}$ $\displaystyle \frac{1}{12}$, divided by hours, $\frac{1}{6}$ $\displaystyle \frac{1}{6}$:

$\frac{\ \frac{1}{12}\ \textup{of a workout}}{\ \frac{1}{6}\ \textup{hours}}$ $\displaystyle \frac{\ \frac{1}{12}\ \textup{of a workout}}{\ \frac{1}{6}\ \textup{hours}}$

Remember that when we divide fractions, we can simply multiply by the reciprocal of the denominator to solve.

$\frac{1}{6}\rightarrow \frac{6}{1}$ $\displaystyle \frac{1}{6}\rightarrow \frac{6}{1}$

Therefore:

$\frac{1}{12}\times\frac{6}{1}=\frac{6}{12}=\frac{1}{2}$ $\displaystyle \frac{1}{12}\times\frac{6}{1}=\frac{6}{12}=\frac{1}{2}$

Andrew can complete $\frac{1}{2}$ $\displaystyle \frac{1}{2}$ of his workout per hour.

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Example Question #1 : Operations

Solve:

-12+4 $\displaystyle -12+4$

Possible Answers:

$\displaystyle -4$

$\displaystyle 4$

$\displaystyle -8$

$\displaystyle 8$

Correct answer:

$\displaystyle -8$

Explanation:

In order to solve this problem, we need to start at $\displaystyle -12$ on the number line.

Next, we have $\displaystyle +4$ which means we need to move $\displaystyle 4$ places to the right on the number line. When we have an addition sign (+) $\displaystyle (+)$ we move to the right because that is towards the positive side of the number line. When we have a subtraction sign (-) $\displaystyle (-)$ we move to the left because that is towards the negative side of the number line.

The orange arrow moved $\displaystyle 4$ places to the right, and ended at $\displaystyle -8$; thus,

$\displaystyle -12+4=-8$

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Example Question #2 : Operations

Solve:

$-35\div-5$ $\displaystyle -35\div-5$

Possible Answers:

$\displaystyle -40$

$\displaystyle 40$

$\displaystyle 7$

$\displaystyle -7$

Correct answer:

$\displaystyle 7$

Explanation:

We know the following information:

$35\div5=7$ $\displaystyle 35\div5=7$

In this particular case, do the negative numbers change our answer? . There are a couple of rules that we need to remember when multiplying with negative numbers:

A negative number divided by a positive number will always equal a negative number, and a positive number divided by a negative number will always equal a negative number.
A negative number divided by a negative number will always equal a positive number

Thus,

$-35\div-5=7$ $\displaystyle -35\div-5=7$

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7th Grade Math : Operations

Study concepts, example questions & explanations for 7th Grade Math

All 7th Grade Math Resources

Example Questions

Example Question #4 : 7th Grade Math

Example Question #1 : Operations

Example Question #2 : Operations

All 7th Grade Math Resources

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