math

Bess

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Jan 8, 2021
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Sandra invested a certain amount of money.
In the first year the investment reached 110% of its initial value.
She spent 20% of the profit to buy a tennis racket and 1/8 of what was left she spent on a tennis outfit.
140 euros remain from the profit.
How much had Sandra invested?
The answer is 2000 euros, but why?
What is the way to solve the problem
 
Sandra invested a certain amount of money.
In the first year the investment reached 110% of its initial value.
She spent 20% of the profit to buy a tennis racket and 1/8 of what was left she spent on a tennis outfit.
140 euros remain from the profit.
How much had Sandra invested?
The answer is 2000 euros, but why?
What is the way to solve the problem
What % of the profit was left after the purchase of the racket?

Hence,

how much was the profit?

How much was the initial investment?
 
Can you try to solve your problem? No one here will solve it for you but helpers here will help you until you get the answer.
 
Sandra invested a certain amount of money.
In the first year the investment reached 110% of its initial value.
She spent 20% of the profit to buy a tennis racket and 1/8 of what was left she spent on a tennis outfit.
140 euros remain from the profit.
How much had Sandra invested?
The answer is 2000 euros, but why?
What is the way to solve the problem
One way to better understand a problem (one meaning of "why") is to check the answer.

She invests 2000 euros.

After a year it is worth 110% of that, which is 1.10*2000 = 2200.

The profit is 2200-2000 = 200. This is a likely misunderstanding.

The tennis racket costs 20% of 200 = 40. That leaves 200 - 40 = 160 euros of the profit.

The tennis outfit costs 1/8 of that 160, which is another 20 euros. She is left with 160 - 20 = 140 euros.

So the answer is correct.

There are several ways you might solve this, such as algebra or just working backward. We can't tell you how to solve it without some clue as to what methods you are capable of using, or are naturally inclined toward. But one algebraic approach is to do just what I did here, but starting with x in place of 2000, in order to write an equation at the end.
 
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