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Minimum mathematical problem
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