[email protected] wrote:The value of an investment increases by x% during January and decreases by y% during February. If the value of the investment is the same at the end of February as at the beginning of January, what is y in terms of x ?

(A) 200x/(100+2x)
(B) x(2+x)/(1+x)^2
(C) 2x/(1 + 2x)
(D) x(200 + x)/10000
(E) 100 - 10000/(100+x)

I added some brackets to your answer choices to avoid ambiguity.


IMPORTANT CONCEPT: x% = x/100
So, to calculate an increase of x%, multiply by (1 + x/100)
Likewise, to calculate an decrease of x%, multiply by (1 - x/100)

Here's an algebraic approach:

Let I = initial investment (at the beginning of January)

value of an investment increases by x% during January
Value at end of January = I(1 + x/100)

value of decreases by y% during February
Value at end of February = I(1 + x/100)(1 - y/100)

The value of the investment is the same at the end of February as at the beginning of January
We get: I = I(1 + x/100)(1 - y/100) . . . Solve for y
Divide both sides by I: 1 = (1 + x/100)(1 - y/100)
Rewrite both brackets as entire fractions: 1 = [(100 + x)/100][(100 - y)/100]
Multiply both sides by 100/(100 + x) to get: 100/(100 + x) = (100 - y)/100
Multiply both sides by 100 to get: 10,000/(100 + x) = 100 - y
So, y = 100 - 10,000/(100 + x)

Answer = E

Cheers,
Brent

The value of an investment increases by x% during January and decreases by y% during February. If the value of the investment is the same at the end of February as at the beginning of January, what is y in terms of x ?

(A) 200x/(100+2x)
(B) x(2+x)/(1+x)^2
(C) 2x/(1 + 2x)
(D) x(200 + x)/10000
(E) 100 - 10000/(100+x)

Here's the plug in numbers approach.

Let the initial investment (at the beginning of January) = $100

value of an investment increases by x% during January
Let x = 25
So, at the end of January, the investment is worth $125

value of decreases by y% during February AND the value of the investment is the same at the end of February as at the beginning of January
So, the value must decrease from $125 back to $100
This represents a 20% decrease.
In other words, y = 20

So, all of the conditions are met when x = 25 and y = 20

Now we'll plug x = 25 into each answer choice, and see which one outputs a y-value of 20

You will find that only answer choice E works

(A) 200x/(100+2x) = 200(25)/100+ 2(25) = 33 1/3 (NOPE)
.
.
.
(E) 100 - 10000/(100+x) = 100 - 10000/(100 + 25) = 20 PERFECT!

Answer: E

Cheers,
Brent

[email protected] wrote: Below I am posting the way I did the question:
Let no. = 100
Jan - 100+x
Feb - 100-y

Sure, if the initial investment is $100, then at the end of January the value is 100 + x
But now what?

We have $(100+x) at the end of January and we are told that the value of decreases by y% during February. So, we need to take the value at the end of January, $(100+x) and decrease it by y%.

You have not done that.
In your solution, you are taking $100 and decreasing it by y% to get $(100-y)

I hope that helps

Cheers,
Brent

[email protected] wrote:Brent,

Thanks for your patience. I am unable to understand when the unitary method works for successive increase or decrease why cant we achieve the same with 100.

Also, I thought by multiplying (100-y) with (100+x), I am taking the decrease of 100+x. Why doesn't this concept work with 100 when it works with the unitary method?

When it comes to conclusions one might draw concerning algebraic expressions, we can always test those conclusions. So, here`s part of your solution:

Let no. = 100
Jan - 100+x
Feb - 100-y

So, let`s see what happens if we let x = 25 and let y = 20 (these are the values I chose for my original solution. In my solution, we had $125 at the end of Jan and we had $100 at the end of Feb.

Let`s see what happens with your expressions:
End of Jan: 100+x = 100 + 25 = $125 PERFECT so far
End of Feb: 100-y = 100 - 20 = $80 INCORRECT. We`re supposed to have $100 at the end of Feb.

Your expression 100-y does not work because, at the beginning of Feb, we don`t have $100. We have something else, because there was an increase of x% the previous month.

I hope that helps.

Cheers,
Brent