Introduction:
Distance word problems, repeatedly entitled as uniform rate problems, engage something travelling at set and stable speed or as well moving at an average speed. The formulas that used in distance and rate problems are two basic formulas that relate distance, rate and time.
Distance =speed * time
Speed = `(Distance) / (time )`
I like to share this algebra 2 formulas list with you all through my article.
Distance word problem - Example 1:
Ram completes a journey in 10 hours. He travels first half of the journey at the rate of 21 miles/hour and second half at the rate of 24 miles/hour. Find the total journey in miles.
Solution:
Let as assume total journey = x
From the given data we form the equation,
`((1/2) x) / (21)` + `((1/2) x) / (24)` = 10
=` (x/21)` + `(x/24)` = 20
= Taking LCM and solve the equation we get,
15 x = 168 * 20
Therefore x = `(168 * 20) / (15)`
= 224 miles
Understanding What is Hemisphere is always challenging for me but thanks to all math help websites to help me out.
Distance word problem - Example 2:
Jack travels 20 miles extra if he walks 14 miles per hour instead of 10 miles per hour. Find the actual distance travelled by him.
Solution:
Let the actual distance travelled by Jack be Y.
We know the formula that relates distance speed and time is
Time = distance / speed.
Jack travels Y miles at the rate of 10 miles per hour.
Jack travels 20 more miles if he travels at the rate of 14 miles per hour.
Time1 = `Y/ 10`
Time 2 = Y + `20 / 14.`
Time in both cases are same so time1 = time2.
So
`Y/ 10` = Y+ `20 / 14.`
By solving this we can find the actual distance travelled by Jack.
`Y/ 10` = Y + `20 / 14.`
Multiply by 14 on both sides of the above equation.
14 * `Y/ 10` = `(Y + 20) / 14` * 14
Crossing out 14 and 14 on one side we get the simplified as
14 * `Y/ 10` = (Y + 20)
Now multiply by 10 on both side of the above equation.
10 * `(14 * Y) / 10` = 10 * (Y + 20).
Crossing out 10 and 10 on one side we get the simplified as
14 *Y = 10 * (Y + 20).
14Y = 10Y + 10*20.
14 Y = 10Y + 200.
Now subtract 10 on both sides of the above equation.
14Y – 10Y = 10Y + 200 – 10Y.
4Y = 200
Now divide it by 4 on both sides.
`(4Y) / 4` = `200/ 4.`
By simplifying it we get the answer as
Y = 50
Therefore the answer is 50
Distance word problems, repeatedly entitled as uniform rate problems, engage something travelling at set and stable speed or as well moving at an average speed. The formulas that used in distance and rate problems are two basic formulas that relate distance, rate and time.
Distance =speed * time
Speed = `(Distance) / (time )`
I like to share this algebra 2 formulas list with you all through my article.
Distance word problem - Example 1:
Ram completes a journey in 10 hours. He travels first half of the journey at the rate of 21 miles/hour and second half at the rate of 24 miles/hour. Find the total journey in miles.
Solution:
Let as assume total journey = x
From the given data we form the equation,
`((1/2) x) / (21)` + `((1/2) x) / (24)` = 10
=` (x/21)` + `(x/24)` = 20
= Taking LCM and solve the equation we get,
15 x = 168 * 20
Therefore x = `(168 * 20) / (15)`
= 224 miles
Understanding What is Hemisphere is always challenging for me but thanks to all math help websites to help me out.
Distance word problem - Example 2:
Jack travels 20 miles extra if he walks 14 miles per hour instead of 10 miles per hour. Find the actual distance travelled by him.
Solution:
Let the actual distance travelled by Jack be Y.
We know the formula that relates distance speed and time is
Time = distance / speed.
Jack travels Y miles at the rate of 10 miles per hour.
Jack travels 20 more miles if he travels at the rate of 14 miles per hour.
Time1 = `Y/ 10`
Time 2 = Y + `20 / 14.`
Time in both cases are same so time1 = time2.
So
`Y/ 10` = Y+ `20 / 14.`
By solving this we can find the actual distance travelled by Jack.
`Y/ 10` = Y + `20 / 14.`
Multiply by 14 on both sides of the above equation.
14 * `Y/ 10` = `(Y + 20) / 14` * 14
Crossing out 14 and 14 on one side we get the simplified as
14 * `Y/ 10` = (Y + 20)
Now multiply by 10 on both side of the above equation.
10 * `(14 * Y) / 10` = 10 * (Y + 20).
Crossing out 10 and 10 on one side we get the simplified as
14 *Y = 10 * (Y + 20).
14Y = 10Y + 10*20.
14 Y = 10Y + 200.
Now subtract 10 on both sides of the above equation.
14Y – 10Y = 10Y + 200 – 10Y.
4Y = 200
Now divide it by 4 on both sides.
`(4Y) / 4` = `200/ 4.`
By simplifying it we get the answer as
Y = 50
Therefore the answer is 50
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