A fruit shop transported 25 baskets of pears and 60 baskets of apples from other places, with a total weight of 2625 kg. It is known that each basket of pears weighs 20 kg more than each basket of apples. What are the weights of each basket of pears and apples?

A fruit shop transported 25 baskets of pears and 60 baskets of apples from other places, with a total weight of 2625 kg. It is known that each basket of pears weighs 20 kg more than each basket of apples. What are the weights of each basket of pears and apples?


Let x kg apples in each basket, then x + 20 kg pears in each basket. According to the meaning of the question, we can get the equation: 60x + 25 (x + 20) = 2625, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 85x + 500 = 2625, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 85



If they start at the same time and meet for the first time when B runs 100 meters, and for the second time when a runs 60 meters away, what is the length of the track______ Rice


(100 × 3-60) × 2, = 240 × 2, = 480 (m). A: then the length of the runway is 480 M



If you go to Jiadi by bike, it's 10 minutes earlier if it's 15 kilometers per hour. If it's 12 kilometers per hour, it's 5 minutes earlier. How many kilometers is this journey?


The solution of the equation is as follows: let the normal time be x.15 (X-10) = 12 (X-5) 15x-150 = 12x-6015x-90 = 12x3x = 90x = 3015 * (30-10) = 15 * 20 = 300 (km). The constant solution is as follows: 15 * {[(15 * 10-12 * 5) / (15-12)] - 10} = 15 * {[(150-60) / 3] - 10} = 15



Please help solve the physics calculation problem. Thank you
For a voltmeter with two ranges, the range is 10V when a and B are used, and 100V when a and C are used. The internal resistance of the voltmeter is known to be RG, 500 ohm, and the full bias current Ig is 1mA. Calculate the value of the partial resistance R1 and R2?
There is a picture of this road, but it seems that it can not be passed on, who can help me to answer, to the process, thank you


Let R1 be the partial resistance between AB, the voltage between AB is UAB, and the voltage between AC is UAC,
Obviously there are:
Ig=1mA=0.001A
R1 = UAB / Ig RG = 9500 ohm
R2 = UAC / Ig rg-r1 = 90000 ohm