|
This Lesson (Solving problems by the backward method) was created by by ikleyn(52776)
  About ikleyn: Solving problems by the backward methodProblem 1A breeder sold half of his horses, then sold another 2.He then sold half of his remaining horses, then another 2. He then had 1 horse left. How many did he have at first ? Solution This problem can be solved by different methods, and I will show you several basic methods. First method is to solve the problem MENTALLY, without using equations The method of solving such problems mentally is to go back from the end to the beginning.
In order to better represent the selling process, I will reformulate the problem.
Step 1. Sold the half of the horses.
Step 2. Sold another 2 horses.
Step 3. Sold the half of remaining horses.
Step 4. Solved another 2 horses.
Step 5. Had 1 horse left.
Now I will move from the end to the beginning.
(a) Before step 4, he had 1+2 = 3 horses.
(b) This 3 horses are half of what he had before step 3.
Hence, before step 3, he had 2*3 = 6 horses.
(c) Before step 2, he had 6+2 = 8 horses
(d) These 8 horses are the half of what he had before step 1.
Hence, initially, the breeder had 2*8 = 16 horses.
ANSWER. Initially, the breeder had 16 horses.
Second method is to solve the problem using equations This method is to go back from the end to the beginning, again.
Consider the selling process in steps as described in the problem.
Step 1. A breeder sold half of his horses, then sold another 2.
Step 2. He then sold half of his remaining horses, then another 2.
Step 3. He then had 1 horse left. How many did he have at first.
Let "n" be number of horses he had before step 2.
Then from the description, we have this equation
Third method is to solve the problem using equations This method is to construct an equation which wraps the whole problem. Let x be the original number of horses. After 1st selling, Thus you learned thee methods solving this problem. Next problem is similar to the first one. Problem 2Donna bought a notebook for $3.50 and then spent half of her remaining money for a textbook.Next she bought sheet music for $4 and then spent half of the remaining money on a calculator. She has $8 left. How much money did she start with? It is a nice problem to solve it in the backward manner.
Make the sheet of her purchases, step by step:
1. bought a notebook for $3.50
2. spent half of her remaining money for a textbook
3. bought the sheet music for $4.
4. spent half of the remaining money on a calculator.
5. she has $8 left.
Now we will move from the bottom to the top.
- before step 4, she had twice $8, i.e 16 dollars.
- before step 3, sha had 16 + 4 = 20 dollars.
- before step 2, she had twice $20, i.e. 40 dollars.
- before step 1, she had 40 + 3.50 = 43.50 dollars.
It is what she started with.
You may check that my solution and my answer are correct at each step, moving from step 1 to the end.
Nice method, isn't it ? Problem 3Natalie, Rachel and Winnie were given a box of beads.Natalie took 5/9 of the beads and then put 14 beads back. Rachel took 1/5 of the remaining beads and then put 8 beads back. Winnie then took half of the remaining beads. The rest of the beads in the box were packed equally into 12 bags. There were 26 beads in each bag. How many beads were given to the girls? Solution
To make the solution totally clear, I will present the problem's description in steps.
(1) Natalie took 5/9 of the beads.
(2) then Natalie returned 14 beads back.
(3) then Rachel took 1/5 of the remaining beads
(4) then Rachel returned 8 beads back
(5) then Winnie took half of the remaining beads.
(6) then the rest beads were packed.
OK, very good. It will help me and you very much, as you will see from what follows.
Now I will solve the problem moving backward.
The number of the packed beads is 12*26 = 312.
Hence, the number of beads immediately before step 5 was 312*2 = 624.
the number of beads immediately before step 4 was 624 - 8 = 616.
the number of beads immediately before step 3 was
Problem 4Basket G and basket H have a total of 284 lollipops.After 43 lollipops were transferred from basket G to basket H, there were 3 times as many lollipops in basket H as in basket G. How many lollipops were there in basket G and basket H respectively at first? Solution Since the transfer was between baskets G and H and nothing was transferred outside, as well as nothing was transferred from outside, the total sum of 284 lollipops was unchangeable. So after transfer, there were total 284 lollipops in basket G and basket H, and were there 3 times as many lollipops in basket H as in basket G. It implies that after transfer, were there 284/4 = 71 lollipops in basket G and 71*3 = 213 lollipops in basket H. Hence, BEFORE the transfer, were there 71 + 43 = 114 lollipops in basket G and 213 - 43 = 170 lollipops in basket H. ANSWER. At first, there were (a) 114 lollipops in basket G and (b) 170 lollipops in basket H. Problem 5At a farm 2 more than half of a number of eggs were taken out of a basket.Later 2 fewer than half of the remaining number of eggs were taken out again. If 20 eggs were left, how many were there at first ? Solution
Let x be the number of remaining eggs after the first taking out.
About this amount of x, we know that after taking out "2 fewer than half of x"
20 eggs were left.
It gives us an equation to find x
x -
This solution is a kind of the backward methodology, although in the hidden/(latent) form. Problem 6An inheritance is split among 5 brothers.The first receives half of the inheritance plus $1. The second receives half of the remainder plus $2. The third receives half of the remainder plus $3. The fourth receives half of the remainder plus $4. The last brother receives the remaining $500. What is the total amount of the inheritance? Solution
Let the brothers be B1, B2, B3, B4, and B5.
Let b[0] be the entire inheritance.
Brother B1 got half of inheritance + $1, or b[0]/2 + 1 dollars. The remainder is r[1] = b[0]/2 - 1. (1)
Brother B2 got half of the remainder r[1] plus $2, or r[1]/2 + 2. The remainder is r[2] = r[1]/2 - 2. (2)
Brother B3 got half of the remainder r[2] plus $3, or r[2]/2 + 3. The remainder is r[3] = r[2]/2 - 3. (3)
Brother B4 got half of the remainder r[3] plus $4, or r[3]/2 + 4. The remainder is r[4] = r[3]/2 - 4. (4)
r[4] is what brother B5 got: r[4] = 500 = r[3]/2 - 4.
From it, we have r[3]/2 = 500 + 4 = 504, which gives r[3] = 2*504 = 1008.
+----------------------------------------------------------------------+
| Now I will solve equations (3), (2), (1) in this backward order. |
+----------------------------------------------------------------------+
From (3), we have r[2]/2 - 3 = r[3] = 1008, or r[2]/2 = 1008+3 = 1011, which gives r[2] = 2*1011 = 2022.
From (2), we have r[1]/2 - 2 = r[2] = 2022, or r[1]/2 = 2022+2 = 2024, which gives r[1] = 2*2024 = 4048.
From (1), we have b[0]/2 - 1 = r[1] = 4048, or b[0]/2 = 4048+1 = 4049, which gives b[0] = 2*4049 = 8098.
ANSWER. The whole inheritance was 8098 dollars.
Solved. ----------------- The method which I used is called a " backward solution method ". On the way forward I establish the equations. On the way back I solve these equations, one after another. Nice method, isn't it ? On solving single linear equations and relevant word problems see the lessons - HOW TO solve a linear equation - Simple word problems to solve using a single linear equation - More complicated word problems to solve using a single linear equation - Typical word problems to solve using a single linear equation - Typical problems on buying and selling items - Typical investment problems - Advanced word problems to solve using a single linear equation - HOW TO algebraize and solve these problems using one equation in one unknown - Challenging word problems to solve using a single linear equation - Selected word problems to solve by reducing to single linear equation - Solving some business-related problems - HOW TO solve these simple word problems MENTALLY without using equations - Using time equation to solve some Travel and Distance problems - Using price equation to solve some business related problems - Solving more complicated problems by the backward method - Solving entertainment problems on shortage of money - OVERVIEW of lessons on solving linear equations and word problems in one unknown in this site. This lesson has been accessed 1182 times. |
