Application of solving fractional equation A 20km wide shelterbelt has been built in a county, so that the average height of sand passing through the shelterbelt is reduced by 50%, and the time of sand passing through the same distance is increased by 20 minutes. What is the height of sand before it reaches the shelterbelt? Let's solve it

Application of solving fractional equation A 20km wide shelterbelt has been built in a county, so that the average height of sand passing through the shelterbelt is reduced by 50%, and the time of sand passing through the same distance is increased by 20 minutes. What is the height of sand before it reaches the shelterbelt? Let's solve it


"Average height" should be "average speed"
Suppose the speed of sand before it reaches the shelter forest is x km / h
20/[(1-50%)X]-20/X=20/60
X=60
So the speed of sand before it reaches the shelter forest is 60 km / h



The translation efficiency of an electronic receiver is 75 times that of manual translation. It takes 2 hours and 28 minutes less to translate 3 000 words than manual translation
A. 78 000,1 200B. 12 000,78 000C. 97 500,13 000D. 90 000,1 200


Suppose that the number of words per minute translated by human is x, then the number of words per minute translated by electronic receiver is 75X. According to the meaning of the title, 3000x-300075x = 2 × 60 + 28, and the solution is x = 20. After testing, x = 20 is the root of the original equation, 75X = 1500. Therefore, the number of words per hour translated by this receiver is 1500 × 60 = 90000, and the number of words per hour translated by human is 60 × 20 = 1200 So D



An applied problem of fractional equation in elementary school
How much water is needed to dilute 50 kg of 90% alcohol to 75% alcohol?
en


Let x kg of water be needed, 90% * 50 = 75% * x, and then calculate X