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This Lesson (Simple arithmetic problems to solve by a backward method) was created by by ikleyn(52879)
  About ikleyn: Simple arithmetic word problems to solve by a backward methodProblem 1I am thinking an integer number. If you add 1 to my number and then multiply the sum by 5,you will get 40. What is the numbers that I am thinking of? Solution Usually, even young students of the 4-th grade level can solve such problems using a backward method and making inverse operations in the reverse order.
Solving this way, you make two steps:
(1) You divide 40 by 5, and you conclude that the mysterious number increased by 1 is 8.
(2) Then the next and the last step is to subtract 1 from the 8, obtained in step (1).
Doing it, you conclude that your original numbers is 7.
ANSWER. The mysterious number is 7.
Problem 2When I divide 48 by the sum of 5 and a certain number, the result is 3. What is the number ?Solution It is a simple arithmetic problem. To find the answer, move backward: - first divide 48 by 3 to get 16; - then subtract 5 from 16 to get 11. The number of 11 is your ANSWER. Problem 341 is added to a mystery number. The result is multiplied by 9.After that 47 is subtracted. The final value is 979. What was the mystery number at the beginning? Solution Usually, even young students of the 4-th grade level can solve such problems using a backward method and making inverse operations in the reverse order.
Solving this way, you make three steps:
(1) You add 47 to 979, and you obtain 47 + 979 = 1026.
(2) Next you divide 1026 by 9, and you get 1026/9 = 114.
(3) Now your final step is to subtract 41 from 114.
By doing it, you obtain your miracle number 114 - 41 = 73.
ANSWER. The miracle number is 73.
Problem 4Eli has a collection of marbles. If he gives away 3/5 of his marbles to his friendand then buys 20 more marbles, he will have 100 marbles in total. How many marbles did Eli have originally? Solution Good problem to solve it MENTALLY by a BACKWARD method.
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| We start from the end and move backward to the beginning |
| step by step, using reasoning and making simple arithmetic. |
| No equations are needed, so this method is good even for |
| 4-th grade young students. |
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At the end, Eli has 100 marbles.
Hence, immediately before he bought 20 marbles, he had 100-20 = 80 marbles.
Hence, these 80 marbles were
Problem 5Every month, Andie spent 1/5 of his salary on food and 2/3 of the remaining salary on transport.After spending his salary on food and transport, he gave 50% of the rest of the salary to his parents and saved remaining $350. How much did he spend on food and transport every month? Solution
Let x be the Andie's total salary.
Make this table
(line 1) 1/5 of x is spent on food. The remainder is R1 = 4/5*x.
(line 2) 2/3 of R1 is spent on transport. The remainder is R2 = (1/3)*R1.
(line 3) 0.5 of R2 is given to parents. The remaining $350 go to saving.
Let' start analyze from the last line 3.
From the table, from its line 3, the remainder of $350 for saving is the same amount as given to parents.
So, $350 is given to parents.
Thus R2 is twice $350, or $350 + $350 = $700.
Next, from the table, from its line 2, R2 is 1/3 of R1.
R2 is $700, as we found out in line 3. Hence, R1 is 3 * $700 = $2100.
2/3 of R1 is 2/3 * 2100 = $1400. It is the amount for transport.
Now let's analyze in line 1.
R1 is $2100, as we found out in line 2.
R1 is 4/5 of x. Hence, x is 5/4*2100 = 2625 dollars and the spending on food is 1/5 * 2625 = 525 dollars.
Thus spending on food and transport together is $1400 + $525 = $1925. ANSWER
CHECK. Total salary is $2625.
1/5 is spent on food, 1/5 * 2625 = 525 dollars. The remainder is R1 = $2625 - $525 = $2100.
2/3 of R1 is spent on transport. 2/3 of $2100 is $1400, The remainder R2 = $2100 - $1400 = $700.
The remainder R2 = $700 is split evenly between parents and saving. ! correct !
Solved. --------------- I solved this problem using a backward method. It allows restore unknowns step by step MENTALLY, moving backward from the end to the beginning, without using equations. I consciously did not use the method of making and solving equations here, since my goal was to make you familiar with this BACKWARD method. ||||||||||||||||||||||||||||||||||| It is useful to know and to understand the following fact. If your approach to solving the problem is making and solving equations, then the technical difficulties of constructing these equations will quickly increase with increasing the number of subdivisions (of the length of the chain) in this problem. But if your approach is a backward method, then it is almost the same, if the length of the chain is 5, or 10, or 20. The formal algorithm of the backward method remains the same, with no changes. The backward method remains robust even for long chains of subdivisions (!) My other lessons on additional arithmetic word problems in this site are Simple problems on filling or emptying a pool or a reservoir Arithmetic problems on submerging solid bodies into water in containers Arithmetic word problems on mixtures Arithmetic mixture problems on dilution mixtures or making mixtures stronger Arithmetic word problems solved using fractions Clock hands arithmetic word problems Counting elements in finite sets and subsets Smart "trial and error" method solving arithmetic word problems Logic problems Entertainment arithmetic word problems solved in a right way Entertainment minimax problems that seem to be complicated, but in reality are simple OVERVIEW of the second group of lessons on arithmetic word problems To see the whole list of lessons on arithmetic problems, use this link Arithmetic problems - YOUR ONLINE TEXTBOOK It is your way to the entry page of the online textbook on Arithmetic problems. This lesson has been accessed 1574 times. |
