SOLUTION: Cycle shop has a reduction and lost 1/11 of its used bikes. If 220 of its used bikes were left after the reduction, how many used bikes were there originally?

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Question 474167: Cycle shop has a reduction and lost 1/11 of its used bikes. If 220 of its used bikes were left after the reduction, how many used bikes were there originally?
Answer by karaoz(32) About Me  (Show Source):
You can put this solution on YOUR website!

A good way to deal with these types of problems is to ask yourself:
"What do I need to find out?"
The answer should be: "the number of bikes that were in the shop before reduction".
The answer to this question can be usually used directly to define the variable, whose value you will eventually want to find out based on the rest of the information given.
So, x = "the number of bikes that were in the shop before reduction".

Now, translate the problem statement using x instead of "the number of bikes that were in the shop before reduction". Translation:
Cycle shop has a reduction and lost 1/11 of x. If 220 of its used bikes were left after the reduction, what is x?
Now, keep only the parts that matter for finding x and note that "bikes left" can be expressed as "bikes before reduction" less "bikes lost", or x - "bikes lost":
x/11 is "bikes lost". if x - "bikes lost" = 220, then what is x?
Clearly, the first sentence tells us that "bikes lost" = x/11, which we can use to substitute in the second sentence. Now, using only algebra and disregarding not needed information:
x - x/11 = 220. x = ?
This is linear equation with one unknown that can be solved for x:
  x    - x/11 = 220
11x/11 - x/11 = 220
       10x/11 = 220
         x    = 220*11/10
         x    = 242
So, the answer is:
There were 242 bikes originally.

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You will note that the main difficulty was in figuring out that "bikes left" can be expressed as "number of bikes before reduction" - "bikes lost", which was useful thing to do since we had a name for "bikes before reduction" (x) and we knew that "bikes lost" is equal to x/11. If you are uncomfortable about setting these things up, you can always use more than one variable. Here is perhaps an easier way to solve the same problem:

Let x be "the number of bikes lost" and let y be "number of bikes before reduction" and even let z be "the number of bikes left" (why not?).

The logic tells us that z = y - x.
The first sentence of the problem statement tells us that x = y/11.
The second sentence of the problem statement tells us that z = 220.
Also, the second sentence of the problem statement reminds us that we are really concerned only about the value of y.

This will be now a system of three equation with three unknowns, which certainly sounds a bit more difficult to solve than one equation with one unknown. However, the system will be rather simple to solve (which is almost always the case when we use more variables than what we need) while we did not suffer too much to find three equations needed to solve the system with three variables. (2 were obtained directly from the problem formulation and 1 from the logical relationship between the variables: namely, what is left is always equal to what was before less what was lost). Now, solve the system to find the value of y:
x = y/11
z = 220
z = y - x
Use the first two equations to substitute into the third and solve for y.
220 = y - y/11
220 = 10y/11
  y = 220*11/10
  y = 242.
So, the answer is again the same:
There were 242 bikes originally.