Imagine as many people as in Unit 1.
It turns out that this boy is 10/9.
It turns out that girls are 20/ 19.
370 is equivalent to 10/9+20/ 19.
370 ÷ (10/9+20/19) × 2 = 342 people.
or
Suppose the existing number is 2X, and now there are x boys and x girls, then there should be 10x/9 boys and10x/95 girls,10x/9+10x/95 = 370, so X = 65438.
All the observations are (n+ 1) 3-n, you can try it once!
1*2*3+2*3*4+3*4*5+ +25*26*27+26*27*28
= (2? - 2) + (3? - 3) + …… + (27? - 27)
= 1? + 2? + 3? + …… + 27? - ( 1+2+3+……+27)
Apply the summation formula of continuous cube and the summation formula of arithmetic sequence.
= ( 1+2+3+……+27)^2 - ( 1+27) * 27 / 2
= [( 1+27)*27/2]^2-378
=378^2-378
=378*377
= 142506
1x2+2x3+3x4+4x5+...+2002x2003
= 1/3* 1*2*3+ 1/3[2*3*4- 1*2*3]+ 1/3[3*4*5-2*3*4]+....+ 1/3[2002*2003*2004-200 1*2002*2003]
= 1/3*2002*2003*2004
=2678684008
Party A and Party B set out from AB 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*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
When Party A, Party B and Party C go shopping together, 1/2 of Party A's consumption is equal to 1/3 of Party B's consumption, and 3/4 of Party B's consumption is equal to 3/5 of Party C's consumption. As a result, Party C spent more money in 98 yuan than Party A, and asked them how much * * *?
98÷(3/4÷3/5- 1/3÷ 1/2)×( 1+ 1/3÷ 1/2+3/4÷3/5)
=98÷(5/4-2/3)×( 1+2/3+5/4)
=98÷7/ 12×35/ 12
= 168×35/ 12
=490 yuan
A and B ran the 100 meter race (assuming their speed remained the same). When A ran 75 meters, B ran 60 meters. So, how many meters did B run when A reached the finish line?
100×60/75
= 100×4/5
=80 meters
1+ 12 1+24 1+48 1+96 1+ 192 1.
= 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
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.
If the speed of the car is increased by 20%, it can arrive 1 hour earlier than the original time. If you drive at the original speed120km 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.
The distance between Party A and Party B is 45× 6 = 270km.
Class 4 (1) has an average score of 92 in the final math exam, and the number of boys taking the exam is 18, with an average score of 89 and the average score of girls is 94. Find the number of girls (by the method of the fourth grade of primary school).
(92-89)× 18÷(94-92)=27 people
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. Ask him how many kilometers of mountain roads he walked during this period.
(38 * 15-450)/(38-23)* 23 = 8 * 23 = 184km