Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Pamela drove her car 99 kilometers and used 9 liters of fuel. She wants to know how many kilometers (k) she can drive with 12 liters of fuel. She assumes the relationship between kilometers and fuel is proportional. How many kilometers can Pamela drive with 12 liters of fuel? km

Pamela drove her car 9999 kilometers and used 99 liters of fuel. She wants to know how many kilometers (k) (k) she can drive with 1212 liters of fuel. She assumes the relationship between kilometers and fuel is proportional.\newlineHow many kilometers can Pamela drive with 1212 liters of fuel?\newlinekm \mathrm{km}

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Grade 8right chevron icon
Solve proportions: word problemsright chevron icon

Full solution

Q. Pamela drove her car 9999 kilometers and used 99 liters of fuel. She wants to know how many kilometers (k) (k) she can drive with 1212 liters of fuel. She assumes the relationship between kilometers and fuel is proportional.\newlineHow many kilometers can Pamela drive with 1212 liters of fuel?\newlinekm \mathrm{km}
  1. Establish values and relationship: Establish the known values and the relationship between kilometers and fuel.\newlinePamela drove 9999 kilometers using 99 liters of fuel. We assume a proportional relationship between kilometers driven and liters of fuel used.
  2. Set up proportion: Set up the proportion to find out how many kilometers kk Pamela can drive with 1212 liters of fuel.\newlineThe proportion is based on the known values: 9999 km for 99 liters is equal to kk km for 1212 liters.\newlineSo, we write the proportion as 999=k12\frac{99}{9} = \frac{k}{12}.
  3. Cross-multiply for kk: Cross-multiply to find the value of kk.99×12=9×k99 \times 12 = 9 \times k1188=9×k1188 = 9 \times k
  4. Solve for kk: Solve for kk by dividing both sides of the equation by 99.11889=(9⋅k)9\frac{1188}{9} = \frac{(9 \cdot k)}{9}132=k132 = k

More problems from Solve proportions: word problems

Question
Elio makes candles that are 14 cm 14 \mathrm{~cm} tall. Each candle burns 88 hours before going out. He is wondering how many hours a 21 cm 21 \mathrm{~cm} tall candle can burn for.\newlineHe assumes that the relationship between the height of a candle and number of hours it burns (h) (h) is proportional.\newlineHow long can a 21 cm 21 \mathrm{~cm} tall candle burn for?\newlinehours
Get tutor helpright-arrow

Posted 3 months ago

Question
37=h14−27\frac{3}{7} = \frac{h}{14} - \frac{2}{7} solve for hh
Get tutor helpright-arrow

Posted 2 months ago

Question
Yesterday, between noon and midnight, the temperature went down by 28.7∘F 28.7^{\circ} \mathrm{F} . If the temperature was −13.5∘F -13.5^{\circ} \mathrm{F} at midnight, what was it at noon?\newlineAnswer: □\square ∘F { }^{\circ} \mathrm{F}
Get tutor helpright-arrow

Posted 2 months ago

Question
Yesterday the temperature at noon was 20.4∘F 20.4^{\circ} \mathrm{F} . By midnight, it had gone down by 28.9∘F 28.9^{\circ} \mathrm{F} . What was the temperature at midnight?\newlineAnswer: □\square ∘F { }^{\circ} \mathrm{F}
Get tutor helpright-arrow

Posted 2 months ago

Question
From midnight to 99:0000 am, the temperature rose 35∘C \frac{3}{5}^{\circ} \mathrm{C} each hour. If the temperature at midnight was −9∘C -9^{\circ} \mathrm{C} , what was the temperature at 99:0000 am?\newlineAnswer: □\square ∘C { }^{\circ} \mathrm{C}
Get tutor helpright-arrow

Posted 2 months ago

Question
A 134 1 \frac{3}{4} -inch candle burns down in 77 hours. Assuming the candles are the same thickness and make (that is, directly proportional), how long would it take a 11-inch candle to burn down?\newlineAnswer: hours\newlineSubmit Answer
Get tutor helpright-arrow

Posted 2 months ago

Question
A 312 3 \frac{1}{2} -inch candle burns down in 77 hours. Assuming the candles are the same thickness and make (that is, directly proportional), how long would it take a 412 4 \frac{1}{2} -inch candle to burn down?\newlineAnswer: hours
Get tutor helpright-arrow

Posted 2 months ago

Question
A 312 3 \frac{1}{2} -inch candle burns down in 77 hours. Assuming the candles are the same thickness and make (that is, directly proportional), how long would it take a 55 -inch candle to burn down?\newlineAnswer: hours
Get tutor helpright-arrow

Posted 2 months ago

Question
A 134 1 \frac{3}{4} -inch candle burns down in 77 hours. Assuming the candles are the same thickness and make (that is, directly proportional), how long would it take a 11-inch candle to burn down?\newlineAnswer: hours
Get tutor helpright-arrow

Posted 2 months ago

Question
(88) How many liters of pure water should be added to 150150 liters of a 35%35\% salt water saline solution to make it a 25%25\% salt water saline solution? Let =\quad = \quad and Let 11=\quad 11 =\quad
Get tutor helpright-arrow

Posted 4 hours ago

View all (31)