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[1] On 3 consecutive days, a vehicle used 24(1)/(2)l,20(1)/(4)l , and 18(2)/(3) I and of gasoline . How much gasoline did the vehicle use in the 3 days?

Pergunta

[1] On 3 consecutive days, a vehicle
used 24(1)/(2)l,20(1)/(4)l , and 18(2)/(3) I and
of gasoline . How much gasoline
did the vehicle use in the 3 days?

[1] On 3 consecutive days, a vehicle used 24(1)/(2)l,20(1)/(4)l , and 18(2)/(3) I and of gasoline . How much gasoline did the vehicle use in the 3 days?

Solução

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To find the total amount of gasoline used by the vehicle over the 3 days, we need to add up the amounts of gasoline used each day.<br /><br />Given:<br />- On the first day, the vehicle used $24\frac {1}{2}l$ of gasoline.<br />- On the second day, the vehicle used $20\frac {1}{4}l$ of gasoline.<br />- On the third day, the vehicle used $18\frac {2}{3}l$ of gasoline.<br /><br />To add these fractions, we first convert them to improper fractions:<br /><br />$24\frac {1}{2} = \frac{49}{2}$<br /><br />$20\frac {1}{4} = \frac{81}{4}$<br /><br />$18\frac {2}{3} = \frac{56}{3}$<br /><br />Now, we find a common denominator to add these fractions. The least common multiple of 2, 4, and 3 is 12.<br /><br />$\frac{49}{2} = \frac{49 \times 6}{12} = \frac{294}{12}$<br /><br />$\frac{81}{4} = \frac{81 \times 3}{12} = \frac{243}{12}$<br /><br />$\frac{56}{3} = \frac{56 \times 4}{12} = \frac{224}{12}$<br /><br />Now, we can add these fractions:<br /><br />$\frac{294}{12} + \frac{243}{12} + \frac{224}{12} = \frac{761}{12}$<br /><br />Finally, we convert the improper fraction back to a mixed number:<br /><br />$\frac{761}{12} = 63\frac{5}{12}$<br /><br />Therefore, the vehicle used a total of $63\frac{5}{12}l$ of gasoline over the 3 days.
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