1. There are two trains, one is102m long, and it runs 20m per second. The length of the train is120m, and it runs at the speed of17m per second. Two cars are driving in the same direction. How many seconds does it take from the first train to catch up with the second train to the departure of the two cars?
2. Someone walks at a speed of 2 meters per second. A train came behind, which took 10 seconds longer than him. As we all know, this train is 90 meters long. Find the speed of the train.
At present, two trains run in the same direction at the same time. 12 seconds later, the express train overtook the local train. The express train runs18m per second, and the local train runs10m per second. If the tails of two trains are flush and driving in the same direction at the same time, the express train will overtake the local train after 9 seconds. Find the body length of two trains.
4. It takes 40 seconds for a train to cross a 440m bridge and 30 seconds for a 3 10/0m tunnel at the same speed. What is the speed and body length of this train?
Xiaoying and Xiao Min took two stopwatches to measure the speed and length of the passing train. Xiaoying used her watch to record that the time that the train passed in front of her was 15 seconds. Xiao Min used another watch to record that it took him 20 seconds to cross the first telephone pole in front and the second telephone pole in the back. It is known that the distance between two poles is 100 meters. Can you help Xiaoying and Xiao Min calculate the total length and speed of the train?
6. It takes 40 seconds for a train to cross a 530-meter bridge and 30 seconds to cross a 380-meter cave at the same speed. Find the speed and body length of this train.
7. The two started from two places along the path next to the railway line and walked at the same speed. A train came, 10 seconds. The whole train passed by A. Three minutes later, B met the train, and the whole train only took 9 seconds to pass by B. How long did the train leave B before they met?
8. Two trains, one with a length of120m and a speed of 20m per second; The other train is160m long and runs at a speed of15m per second. The two cars are driving in opposite directions. How many seconds does it take from the front meeting to the back leaving?
9. Someone walks at a speed of 2 meters per second. The train overtook him from behind 10 seconds. As we all know, the length of this train is 90 meters. Find the speed of the train.
10. Party A and Party B walk along the railway at the same speed. It took 8 seconds for a train to pass by Party A, and only 7 seconds to pass by Party B after leaving Party A for 5 minutes. How many minutes after Party B met the train?
Second, answer the question.
1 1. Fast train length182m, traveling 20m per second, slow train length1034m, traveling 8m per second. The two cars are parallel in the same direction. How long does it take for the express train to cross the local train when the rear of the express train meets the rear of the local train?
12. The length of the express train is182m, the length of the local train is1034m, and the speed of the local train is18m per second. The two cars are parallel in the same direction. How many seconds can an express train pass through the local train when the heads of the two cars are aligned?
13. A person is running along the railway at a speed of 120 meters per minute. A 288-meter-long train came from the opposite side. It took him 8 seconds to find the speed of the train.
14. A train is 600 meters long. It passes through a 200-meter-long tunnel at a speed of 10 meter per second. How long does it take to leave the tunnel from the front to the rear?
———————————— Answer the case ————————
Fill in the blanks
120m
102m
17x meter
20x meters
tail
tail
head
head
1. This topic is the catch-up problem of "two trains". Here "catching up" means that the head of the first train catches up with the tail of the second train, and "leaving" means that the tail of the first train leaves the head of the second train.
Suppose it takes x seconds from the first train catching up with the second train to the departure of the two trains, and the equation is:
102+ 120+ 17x = 20x
x =74。
2.
Assuming that the speed of the train is x meters per second, the equation is obtained.
10 x =90+2× 10
x = 1 1。
3.(
Then the express delivery length:18×12-10×12 = 96 (meters).
(2) the rear of the car is in a straight line and driving in the same direction at the same time, which is an express train.
Then the length of the local train is 18×9- 10×9=72 (meters).
4.( 1) The train speed is: (440-310) ÷ (40-30) =13 (m/s).
(2) The body length is: 13×30-3 10=80 (m).
5.( 1) The train speed is:100 ÷ (20-15) × 60× 60 = 72,000 (m/h).
(2) The body length is 20× 15=300 (m).
6. Set the train body to be x meters long and y meters long.
①②
solve
7. Let the train body be x meters long, Party A and Party B each walk y meters per second, and the train travels z meters per second. According to the meaning of the question, list the equations and get.
①②
①-②, so:
After the train left B, they met:
(seconds) (minutes)
8. Solution: The sum of the distances traveled by two cars is exactly the sum of the distances of two train conductors, so the time required to encounter the problem is: (120+60)? (15+20)=8 (seconds).
9. Think of it this way: when the train passes by people, their distance difference is the conductor. Divide the distance difference (90 meters) by the crossing time (10 second) to get the speed difference between the train and people. This speed difference plus the walking speed of people is the speed of the train.
90÷10+2 = 9+2 =11(m)
A: The speed of the train is 1 1 meter per second.
10. It is required that when A and B meet in a few minutes, the relationship between the distance between A and B and their speed must be found, which is related to the movement of the train. The distance between a and b can only be found by the movement of the train. The running time of the train is known, so it is necessary to find out its speed, at least the proportional relationship between it and the speeds of A and B. Because this question is more difficult.
① Find the relationship between the train speed and the speed of Party A and Party B, and let the train length be L, then:
(I) It takes 8 seconds for a train to pass through A, and this process is to catch up with the problem:
Therefore; ( 1)
(i i) It takes seven seconds for the train to pass through B. This process is a meeting problem:
Therefore. (2)
From (1) and (2),
So ...
(2) The distance between locomotive encounter A and train encounter B is:
.
③ Find the distance between A and B when the locomotive meets B. 。
After the locomotive meets A, it takes (8+5×60) seconds for the locomotive to meet B. Therefore, when the locomotive meets B, the distance between A and B is:
(4) How many minutes will A and B meet?
(seconds) (minutes)
A: In another minute, Party A and Party B will meet.
Second, answer the question.
11.1034÷ (20-18) = 91(seconds)
12.182 ÷ (20-18) = 91(seconds)
13.288 ÷ 8-120 ÷ 60 = 36-2 = 34 (m/s)
The speed of the train is 34 meters per second.
14. (600+200)10 = 80 (seconds)
Answer: It takes 80 seconds from the front of the car entering the tunnel to the rear leaving the tunnel.
Average problem
1. In the final exam, Cai Chen scored 89 points in political language, mathematics and English biology, 9 points in political mathematics, 84 points in Chinese English, 86 points in political English, and 65,438 points more in English than in Chinese.
2. Two cotton fields, A and B, with an average yield per mu 1.85 kg. A cotton field is 5 mu, with an average yield of 203 kg per mu. B Average yield per mu of seed cotton in cotton field170kg. B How many acres of cotton fields are there?
3. Given that the sum of eight consecutive odd numbers is 144, find these eight consecutive odd numbers.
4. 8.8 yuan per kilogram of sugar, 7.2 yuan per kilogram of sugar B, and how much sugar B is mixed into 5 kilograms of sugar A to make the price of sugar per kilogram reach 8.2 yuan?
I bought five sheep in the canteen, weighed two sheep at a time, and got ten different weights (kilograms): 47, 50, 5 1, 52, 53, 54, 55, 57, 58, 59. How much do these five sheep weigh?
arithmetic series
1, here is a series of numbers arranged according to the law. What number is 1995?
Answer: 2,5,8, 1 1, 14, ... According to the law, this is a arithmetic progression, the first term is 2, and the tolerance is 3, so the term 1995 = 2+3× (/kloc-0.
2. Of the natural numbers starting from 1, what is the100th number that cannot be divisible by 3?
Answer: We found that every three numbers in 1, 2, 3, 4, 5, 6, 7, ... are grouped from 1, and the first two of each group cannot be divisible by 3. If there are two groups, 100 will have 100 ÷ 2.
3. If 1988 is expressed as the sum of 28 consecutive even numbers, what is the largest even number?
Answer: 28 even numbers are grouped into 14 groups, and 2 symmetrical numbers are grouped into groups, that is, the minimum number and the maximum number. The sum of each group is:1988 ÷14 =142, and the difference between the minimum number and the maximum number is 28- 1=27 tolerance.
4. In the integer greater than 1000, find all the numbers with equal quotient and remainder after division by 34, so what is the sum of these numbers?
Answer: Because 34× 28+28 = 35× 28 = 980 < 1000, there are only the following figures:
34×29+29=35×29
34×30+30=35×30
34×3 1+3 1=35×3 1
34×32+32=35×32
34×33+33=35×33
The sum of the above numbers is 35× (29+30+31+32+33) = 5425.
5. There is a 1, 2, 3, ... which says 134 and 135 respectively. Take some cards out of the box at random, calculate the sum of the numbers on these cards and divide by 17, then write the remainder on another yellowcard and put it back in the box.
Answer: It is difficult to grasp several times at a time. It is better to consider the whole situation and take a step back to a simple situation analysis: suppose there are two numbers, 20 and 30, and divide their sum by 17 to get the number of yellow cards. If calculated separately, it is 3 and 13, and then divide the sum of 3 and 13 by 65436. That is to say, no matter how many numbers are added, the remainder of the sum divided by 17 remains unchanged, and we return to the topic1+2+3+...+134+135 =136×136. The remainder is 0, and19+97 =1116 ÷17 = 6 ... 65438+.
6. The following formulas are arranged regularly:
1+ 1, 2+3, 3+5, 4+7, 1+9, 2+ 1 1, 3+ 13, 4+ 15, 65438
Solution: First find out the law: each formula is added by two numbers, the first number is the cycle of 1, 2, 3, 4, and the second number is the continuous odd number starting from 1 Because 1992 is even, the second of the two addends must be odd, so the first one must be odd, so it is 1 or 3. If it is 1: then the second number is1992-1=191. 199 1 is (1 991+1) ÷ 2 = 996, and1is always odd and inconsistent, so this formula is 3+/kloc-0.
7. As shown in the figure, the upper and lower rows in the table are arithmetic progression, so what is the minimum difference (reduction of large numbers) between two numbers in the same column?
Answer: From left to right, their differences are: 999, 992, 985, ... 12, 5. From right to left, their differences are: 1332, 1325,13/kloc.
8. There are 19 formulas:
So what are the results on the left and right sides of equation 19?
Answer: Because the left and right sides are equal, we might as well just consider the situation on the left and solve two problems: How many are used in the first 18 formula? All numbers are 5, 7, 9, ..., 18, and 5+2× 17 = 39, 5+7+9+...+39 = 396, so the equation of 19 starts from 397; How many numbers are added to the formula 19? The numbers on the left are 3, 4, 5, ... and 19 should be 3+ 1 × 18 = 2 1, so the result of 19 formula is 397+398+399+...+.
9. The number of two known columns: 2, 5, 8, 1 1, …, 2+(200-1) × 3; 5、9、 13、 17、……、5+(200- 1)×4。 They are all 200 projects. How many pairs of the same number of items are there in these two columns?
A: It is easy to know that the first such number is 5. Note that in the first series, the tolerance is 3, and in the second series, the tolerance is 4. That is to say, the second logarithm minus 5 is a multiple of 3, which is a multiple of 4, so the number converted into arithmetic numbers is calculated with the tolerance of12,5,17,29, ... The maximum number of the second sequence is 5+(200-1) × 4 = 80. The maximum number of new series cannot exceed 599, because 5+ 12× 49 = 593, 5+ 12× 50 = 605, so there are 50 pairs of * *.
10, as shown in the figure, has a lower triangle with a side length of 1 m. Starting from the vertex of each side, take a point every 2 cm, and then take these points as the endpoint to divide the large regular triangle into many small regular triangles with a side length of 2 cm as parallel lines. Find (1) the number of small regular triangles with a side length of 2 cm, and (2) the total length of parallel lines.
Answer: (1) From top to bottom, * * has 100÷2=50 lines, the first line has 1, the second line has 3, the third line has 5, ... and the last line has 99, so * * * has (/kloc. There are three parallel lines, which are the same. There are 49 lines * * * in the horizontal direction, the first one is 2cm, the second one is 4cm, the third one is 6cm, ..., and the last one is 98 cm, so the length of * * * is (2+98)×49÷2×3=7350 cm.
1 1, a factory 1 1 is busy in October and does not rest on Sundays. And since the first day, the same number of workers have been sent from the general factory to the branch factory every day. By the end of the month, there were 240 workers left in the main factory. If at the end of the month, the workload of workers in the general factory is 8070 working days (one person works 1 working day) and no one is absent, how many workers will the general factory send to work in the branch factory this month?
Answer: 165438+ 10 has 30 days. According to the meaning of the question, the number of people in general factories is decreasing every day, and finally it is 240. The number of people every day constitutes arithmetic progression. According to the nature of arithmetic progression, the sum of the number of people on the first day and the last day is equivalent to 8070÷ 15=538, that is to say, there are 538-240=298 workers on the first day, and (298-240) II is scheduled every day.
12, Xiao Ming read an English book. When he first read it, he read 35 pages on the first day, and then read 5 more pages every day than the day before. As a result, he only read 35 pages on the last day. The second time, I read 45 pages on the first day, and then I read 5 more pages every day than the day before. As a result, I only need to read 40 pages on the last day. How many pages are there in this book?
Answer: The first scheme: 35, 40, 45, 50, 55, ... 35 The second scheme: 45, 50, 55, 60, 65, ... 40 The secondary scheme is adjusted as follows: the primary scheme: 40, 45, 50, 55, ... 35+35. The second plan: 40, 45, 50, 55, ... (The last day is on the first day) So the second plan must be 40, 45, 50, 55, 60, 65, 70, ***385 pages.
13 and 7 teams * * * planted trees 100, and the number of trees in each team is different. Among them, the team with the most trees planted 18 trees, and the team with the least trees planted at least how many trees?
Answer: We know that the other six teams planted 100- 18=82 trees in order to make Li Ru-Nan's "What's the value of neon" stamp? As many as possible, including:17+16+15+14+13 = 75 trees, so the minimum team should plant at least 82-75=7 trees.
14. Line up 14 different natural numbers from small to large. It is known that their total number is 170. If the maximum and minimum numbers are removed, the remaining total is 150. What is the second number in the original arrangement order?
A: The sum of the maximum and minimum numbers is 170- 150 = 20, so the maximum number is 20- 1 = 19. When the maximum number is 19, there are19+18+17+16+15+14+13+. Yes18+17+16+15+13+12+/kloc-0.
Periodic problem
Basic exercises
1, (/kloc-0 1)○□□□□□□△□□…… The 20th number is (□).
(2) The thirty-ninth chess piece is (sunspot).
2. Xiaoyu practices calligraphy. She wrote the sentence "I love the great motherland" over and over again, and the 60th word should be written as (big).
Class 3.2 (1) participated in the tug-of-war competition of the school. Their participating teams are arranged in a row according to "three men and two women", and the 26th student is (male).
4. There is a column number: 1, 3,5, 1, 3,5, 1, 3,5 ... The 20th number is (3), and the sum of these 20 numbers is (58).
5. There are three kinds of beads with the same size *** 100, which are discharged continuously according to the requirements of 3 red and 2 white 1 black.
……
(1) The 52nd one is a (white) bead.
(2) There are (17) white beads in the first 52 beads.
6.a Q B: Today is Friday, and it will be Sunday in 30 days.
B Q A: If 16 is Monday, then the 3 1 day of this month is Tuesday.
May 1 2006 is Monday, so the 28th of this month is Sunday.
A, B, C and D play poker. Party A puts "Wang" in the middle of 54 cards, and the 37th card is counted from top to bottom. Party C thought about it, confidently grabbed the card first, and finally caught the "king". Do you know how C is worked out? ※? (37 ÷ 4 = 9 ... 1 The first person to get the card must catch the "king". )
answer
1、( 1)□。
(2) sunspots.
2. It's very big.
3. Male students.
The 20th number is (3), and the sum of these 20 numbers is (58).
5、
(1) The 52nd one is a (white) bead.
(2) There are (17) white beads in the first 52 beads.
6. (days) (2). (days).
(37 ÷ 4 = 9 ... 1) The first person to get the card will definitely catch the "king". ※
Improve practice
1, (1) ○□□□□□□□□……… The 20th number is (□).
(2) ○○○○○○○○○○○○○○○○○○○○○○○○○○○○967
2. There is a row of colorful flags on the sports field, with 34 faces, arranged according to "three reds, one green and two yellows", and the last face is (green flag).
3. "Love mathematics since childhood, love mathematics since childhood ..." The 33rd word is (love).
4. The class (1) participated in the school tug-of-war. The teams in their competition are arranged in the order of "three men and two women", and the 26th student is (male).
5. There is a column number: 1, 3,5, 1, 3,5, 1, 3,5 ... The 20th number is (3), and the sum of these 20 numbers is (58).
6.a Q B: Today is Friday, and it will be Sunday in 30 days.
B Q A: If 16 is Monday, then the 3 1 day of this month is Tuesday.
May 1 2006 is Monday, so the 28th of this month is Sunday.
A, B, C and D play poker. Party A puts "Wang" in the middle of 54 cards, and the 37th card is counted from top to bottom. Party C thought about it, confidently grabbed the card first, and finally caught the "king". Do you know how C is worked out? ※?
37 ÷ 4 = 9 ... 1 (The first person to get the card will definitely catch the "king"). ※
answer
1、( 1)□。
(2)○。
2. Green flag.
3. Love.
4.( 1) male students.
5. The 20th number is (3), and the sum of these 20 numbers is (58).
6. (days) (2). (days).
37 ÷ 4 = 9 ... 1 (The first person to get the card will definitely catch the "king"). ※
Fast Skill Calculation of Decimals (2)
First, the vacuum problem.
1. Calculate 4.75-9.64+8.25-1.36 = _ _ _.
2. Calculate 3.17-2.74+4.7+5.29-0.26+6.3 = _ _ _.
3. Calculate (5.25+0. 125+5.75) 8 = _ _ _ _.
4. Calculate 34.5 8.23-34.5+2.77 34.5 = _ _ _.
5. Calculate 6.25 0.16+264 0.0625+5.26.25+0.625 20 = _ _ _.
6. Calculate 0.035935+0.035+30.035+0.0761.5 = _ _ _.
7. Calculate19.9837-199.81.9+19980.82 = _ _ _.
8. Calculate13.59.9+6.510.1= _ _ _.
9. Calculate 0.125 0.25 0.564 = _ _ _.
10. Calculation11.8 43-860 0.09 = _ _ _.
Second, answer the question.
1 1. Calculation 32.14+64.28 0.5378 0.25+0.5378 64.28 0.75-8 64.28 0.125 0.5378.
12. Calculate 0.888 125 73+999 3.
13. Calculation1998+199.8+19.98+1.998.
14. There are two decimal places below:
a=0.00…0 125 b=0.00…08
1996 0 2000 0
Try a+b, a-b, a b, a b.
———————————— Answer the case ————————
1.2
Original formula =(4.75+8.25)-(9.64+ 1.36)
= 13- 1 1
=2
2. 17
Original formula = (3.71+5.29)+(4.7+6.3)-(2.74+0.26)
=9+ 1 1-3
= 17
3.89
Original formula =(5.25+5.75+0. 125) 8.
=( 1 1+0. 125) 8
= 1 1 8+0. 125 8
=88+ 1
=89
4.345
Original formula =34.5 (8.23+2.77- 1)
=34.5 10
=345
5.62.5
Original formula = 6.25 0.16+2.64 6.25+5.26.25+6.25 2
=6.25 (0. 16+2.64+5.2+2)
=6.25 10
=62.5
6.35
7. 1998
8. 199.3
The original formula =13.5 (10-0.1)+6.5 (10+0.1).
= 13.5 10- 13.5 0. 1+6.5 10+6.5 0. 1
= 135- 1.35+65+0.65
=( 135+65)-( 1.35-0.65)
=200-0.7
= 199.3
9. 1
Original formula =0. 125 0.25 0.5 (8 4 2)
=(0. 125 8) (0.25 4) (0.5 2)
= 1 1 1
= 1
10.430
Original formula = 1 1.8 43-43 20 0.09
= 1 1.8 43-43 1.8
=43 ( 1 1.8- 1.8)
=43 10
=430
1 1.
Original formula = 32.14+64.28 0.5378 (0.25+0.75-8 0.125)
=32. 14+64.28 0.5378 0
=32. 14
12.
The original formula = 0.11(8125) 73+11(93).
= 1 1 1 73+ 1 1 1 27
= 1 1 1 (73+27)
= 1 1 1 100
= 1 1 100
13.
Original formula = (2000-2)+(200-0.2)+(20-0.02)+(2-0.002)
=2222-2.222
=2222-( 10-7.778)
=2222- 10+7.778
=22 19.778
14.a+b, a+B, A has 1998 digits after the decimal point, and B has 2000 digits after the decimal point. The decimal addition needs to be aligned and then calculated according to the integer addition rule, so
a+b = 0.00…0 12508 = 0.00…0 12508
2000 bits 1996 zero
, method with a+b, digit alignment, pay attention to abdication to fill zero, because
A=0.00…0 125, b=0.00…08, from 12500-8= 12492, so
1998 bit 2000 bit
a-b = 0.00… 12492 = 0.00…0 12492
2000 bits 1996 zero
A b, A b should have 1998+2000 digits after the decimal point, but 125 8= 1000, so
a b = 0.00…0 1000 = 0.00…0 1
1998+2000 bits, 3995 zeros
A b, expand A and B at the same time 100…0 times, and get
2000 zero
a b= 12500 8= 1562.5
Calculation of geometric knowledge area
1. The playground of Renmin Road Primary School is 90 meters long and 45 meters wide. After transformation, the length is increased by10m, and the width is increased by 5m. How many square meters has the playground area increased now?
Subtract the original area of the playground for thinking navigation to get the increased area. The current area of the playground is (90+ 10)×(45+5)=5000 (square meters), and the original area of the playground is: 90×45=4050 (square meters). So now it is 5000-4050=950 square meters more than before.
(90+10) × (45+5)-(90× 45) = 950 (square meter)
Exercise (1) has a rectangular board with a length of 22 decimeters and a width of 8 decimeters. If the length and width are reduced by 10 decimeter and 3 decimeter respectively, how many square decimeters will the area be reduced?
Exercise (2) A rectangular plot is 80 meters long and 45 meters wide. If the width is increased by 5 meters, how many meters should the length be reduced to keep the area unchanged?
2. If the width of a rectangle remains the same and its length increases by 6 meters, its area will increase by 54 square meters. If the length remains the same and the width is reduced by 3 meters, its area will be reduced by 36 square meters. What is the original area of this rectangle?
The train of thought navigation shows that its width is 54÷6=9 (meters), which means: "The width is unchanged, the length is increased by 6 meters, and its area is increased by 54 square meters"; From "the length is constant, the width is reduced by 3m, so its area is reduced by 36m", we can know that its length is 36÷3= 12 (m), so the area of this rectangle is 12x9 = 108 (m). (36÷3)×(54÷9)= 108 (square meter)
Exercise (1) a rectangle. If its width remains the same and its length is reduced by 3 meters, its area will be reduced by 24 square meters. If its length remains the same and its width increases by 4 meters, its area will increase by 60 square meters. What is the original area of this rectangle?
Exercise (2) If the width of a rectangle remains the same, its length will be increased by 5 meters, and its area will be increased by 30 square meters; If the length remains the same and the width increases by 3 meters, its area will increase by 48 square meters. What is the original area of this rectangle?
Exercise (3) A rectangle, if its length is reduced by 3 meters, or its width is reduced by 2 meters, then its area is reduced by 36 square meters. Find the original area of this rectangle.
3. The picture below shows a rectangular chicken farm surrounded by a fence16m long. How big is the floor space?
According to the meaning of the question, because a wall is used on one side, two lengths plus one width is equal to 16m, and the width is 4m, so the length is (16-4)÷2=6 (m). Therefore, it covers an area of 6×4=24 (square meters).
(16-4)÷2×4=24 (square meter)
Exercise (1) The picture below shows a professional poultry farmer enclosing a rectangular chicken farm with a fence13m long. How big is the chicken farm?
Exercise (2) Use a 56-meter-long wooden fence to form a rectangle with a length or width of 20 meters. How to maximize the enclosed area by using the fence on one side?
4, a square steel plate, first cut off the rectangle of 5 decimeters wide, and then cut off the rectangle of 8 decimeters wide (as shown in the figure below). The area is reduced by 18 1 square decimeter compared with the original square. What is the side length of the original square?
Cut out the shadow, put two small squares (as shown below) together, and add a small rectangle with a length of 8 decimeters and a width of 5 decimeters respectively. The area of this splicing rectangle is:181+8× 5 = 221(square decimeter), and the length is the original square. Therefore, the side length of the original square is 22 1÷ 13= 17 (decimeter).
(181+8× 5) ÷ (8+5) =17 (decimeter)
Exercise (1) One side of the square is reduced by 6 decimeters, and the other side is reduced by 10 decimeter to become a rectangle. The area of this rectangle is 260 square decimeters smaller than that of the square. Find the side length of the original square.
Exercise (2) If the length of a rectangular plate is reduced by 5 decimeters and the width is reduced by 2 decimeters, its area is reduced by 66 square decimeters, and the rest is just a square. Find the area of the original rectangle.
Exercise (3) After a square piece of glass is cut off by 8 cm, the remaining square is 448 cm less than the original one. What is the original area of this square glass?
God, I'm so tired!