= 1/6×( 1+ 1/2+ 1/4+ 1/8+ 1/ 16+ 1/32)
= 1/6×( 1- 1/32)
= 1/6- 1/ 192
=3 1/ 192
2. 1+ 1/( 1+2)+ 1/( 1+2+3)+……+ 1/( 1+2+3+……+n)
= 2 * 1/2+ 1/2 *( 1+2)+ 1/2 *( 1+2+3)+……+ 1/2 *( 1+2+3+……+n)
= 2 * 1- 1/2+ 1/2- 1/3+ 1/3- 1/4+……+ 1/n- 1/(n+ 1)
=2* 1- 1/(n+ 1)
=2n/(n+ 1)
3. 1/ 1*3+ 1/2*4+ 1/3*5+ 1/4*6+ 1/5*7...... 1/98* 100+ 1/99* 10 1
=( 1- 1/3+ 1/2- 1/4+ 1/3- 1/5+ 1/4- 1/6+ 1/5- 1/7+……+ 1/98- 1/ / kloc-0/00+ 1/99- 1/ 10 1)÷2
=( 1+ 1/2- 1/ 100- 1/ 10 1)÷2
= 15049/ 10 100÷2
= 15049/20200
4. If the speed of a car from A to B is increased by 20%, it can arrive 1 hour earlier than the original time. If you drive at the original speed of120km and then increase the speed by 25%, you can arrive 40 minutes earlier. How many kilometers is it between A and B?
40 minutes =2/3 hours
Scheduled time1÷1-1(1+20%) = 6 hours.
The original speed is120-120/(1+25%) ÷ 6-2/3-6/(1+25%) = 24 ÷ 8//kloc-0.
5. Chen Ming travels by bike, walking 38 kilometers every day on the flat road and 23 kilometers every day on the mountain road. He/kloc-walked 450 kilometers in 0/5 day. How many kilometers of mountain roads did he walk during this period?
(38* 15-450)/(38-23)*23
=8*23
=184km
6. A batch of fruits with water content of 90% was delivered from the warehouse, 100kg, and it was found that the water content became 80% after one week. What is the total weight of this batch of fruit now?
100*(80%/90%)
=800/9
=88 8/9 kg
7. Party A and Party B cooperate to complete a job. Therefore, the work efficiency of Party A is improved by110, and the work efficiency of Party B is improved by 1/5. If Party A works alone for 6 hours and needs 1 1 hour, how many hours does Party B need to work alone?
11/6-11* (1+10%) =15 hours.
1115/(1+1/5) =18 hours.
B It takes one person 18 days.
8./kloc-Party A and Party B can cooperate in 0/0 day, and Party B and Party C can cooperate in 8 days. Now, after four days of cooperation among Party A, Party B and Party C, the remaining works will be completed by Party B alone in 1 1/2 days.
( 1 1/2-4)÷ 1-( 1/ 10+ 1/8)×4
= 1.5÷ 1/ 10=
15 days
9. There is a fish tail weighing 5 kilograms. The mass of the fish head is equal to half the mass of the fish tail plus the mass of the fish body, and the mass of the fish body is equal to the mass of the fish head plus the mass of the fish tail. How much does this fish weigh?
Fish (5+5)/(1-1/2) = 20kg.
Fish head 5+20/2 = 15kg
Fish weight 5+20+ 15 = 40kg.
10. Party A and Party B set out from AB respectively and walked towards each other at the same time. When they set off, their speed ratio was 3: 2. After the meeting, Party A's speed increased by 1/5, and Party B's speed increased by 2/5. When A arrives at B, B is still 26 kilometers away from A ... What is the distance between these two places?
Let AB be x kilometers apart.
[2/(3+2)x]/[3×( 1+ 1/5)]=[3/(3+2)x-26]/[2×( 1+2/5)]
x/9 = 3x/ 14- 130/ 14
13x/ 126 = 130/ 14
x = 90°
1 1. If Party A keeps driving for 4 hours and Party B keeps driving 1 hour after the first encounter between the two cars, how many hours did they drive when the two cars met for the 15th time (regardless of the meeting times between AB and AB)?
Let the first meeting be x hours.
4/(x+4)+ 1/(x+ 1)= 1
4(x+ 1)+x+4 =(x+4)(x+ 1)
4x+4+x+4=x^2+5x+4
x^2=4
x=2
The fifteenth encounter between the two cars went through 29 complete journeys.
So driving 2×29=58 hours.
12. Two cars, A and B, leave each other at the same time. The speed ratio of the two cars is 5:4. After the two cars meet, the speed of car A remains unchanged, and car B travels more than before18km. The result is that two cars reach each other's starting point at the same time. How many kilometers does a car travel per hour?
Armored vehicles travel x kilometers per hour.
[4/(5+4)]/x =[5/(5+4)]/(4/5x+ 18)
5/9x=4/9(4/5x+ 18)
5x=4(4/5x+ 18)
5x=3.2x+72
1.8x=72
x=40
13 How many minutes? 9, the hour hand and the minute hand are at the same distance from 9, on both sides of 9 respectively?
30×9÷(6+0.5)
=270÷6.5
=540/ 13
=4 1 and 7/ 13 points
After 9 o'clock, at 4 1 and 713, the distance between the hour hand and the minute hand from 9 is equal, on both sides of 9 respectively.
14.1* 2 * 3 * 4 * 5 * 6 * ... * How many zeros were there at the end of 2008? Important steps
The number of factors 5 determines the number of zeros at the end.
2008÷5=40 1 (rounded)
2008÷25=80 (rounded)
2008÷ 125= 16 (rounded)
2008÷625=3 (rounded)
40 1+80+ 16+3=500
1* 2 * 3 * 4 * 5 * 6 * ... * There were 500 zeros at the end of 2008.
15. There are 85 workers in a workshop. On average, each person can process 8 large gears or 10 small gears per day. It is known that 1 large gears and 3 small gears can be matched into a set. So how to arrange the labor force to make the products produced complete? (List the equations and explain the steps in detail)
Set up x people to process large gears.
16x: 10(85-x)=2:3
48x=20(85-x)
48x= 1700-20x
68x= 1700
x=25
85-x=85-25=60
There are 25 people processing large gears and 60 people processing small gears.