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Tanya [424]
3 years ago
6

A dump truck brought 1⁄3 of a ton of rock on the first trip, 1⁄2 of a ton on the second trip, but on the third trip, had to take

back 4⁄5 of a ton. What was the total weight of the rock left at the construction site?
Mathematics
2 answers:
Allisa [31]3 years ago
7 0

Answer:

1/30 of a ton

Step-by-step explanation:

Write this as an expression:

1/3 of a ton and 1/2 of a ton were taken to the site.

4/5 of a ton were remove from the site.

Amount of a ton at the site  =   \frac{1}{3}+ \frac{1}{2}- \frac{4}{5}

Solve the equation by finding <u>common denominators</u> (when the bottom numbers are the same).

(\frac{1}{3}+ \frac{1}{2})- \frac{4}{5}       Focus on the first part. The least common multiple (LCM) of 3 and 2 is "6", which will become the denominator.

For <u>1/3 to become ?/6</u>, multiply top and bottom by 2.

For <u>1/2 to become ?/6</u>, multiply top and bottom by 3.

(\frac{2}{6}+ \frac{3}{6})- \frac{4}{5}

Now we can add the numerators together over the same denominator.

\frac{5}{6}- \frac{4}{5}

Do the same thing as before and change the denominators. The LCM of 6 and 5 is "30".

For <u>5/6 to become ?/30</u>, multiply top and bottom by 5.

For <u>4/5 to become ?/30</u>, multiply top and bottom by 6.

\frac{25}{30}- \frac{24}{30}

Subtract the numerators:

\frac{1}{30}                  Answer in tons

Therefore there was 1/30 of a ton of rock left at the construction site.

bija089 [108]3 years ago
5 0
Larry received 1/3 pound of candy from his grandmother and 2/5 pound of candy from his best friend. Larry's sister ate 1/2 pound of Larry's candy. How many pounds of candy does Larry have left?
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The percentage shows that they were 28% more likely to come from Riverdale High school.

<h3>How to calculate the percentage?</h3>

From the information given, the percentage of those from Rockport will be:

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Answer:

33 copies were paperback and 12 were hardcover.

Step-by-step explanation:

Let h represent the number of hardcover copies and p represent the number of paperback copies.

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h+p = 45

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\left \{ {{5(h+p=45)} \atop {7h+5p=249}} \right. \\\\\left \{ {{5h+5p=225} \atop {7h+5p=249}} \right.

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