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If x is 20 percent more than y and y is 50 percent less than z, then x

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If x is 20 percent more than y and y is 50 percent less than z, then x [#permalink]

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New post 02 May 2016, 05:16
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If x is 20 percent more than y and y is 50 percent less than z, then x is what percent of z?

A. 500%
B. 250%
C. 500/3%
D. 125%
E. 60%


* A solution will be posted in two days.
[Reveal] Spoiler: OA

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Re: If x is 20 percent more than y and y is 50 percent less than z, then x [#permalink]

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New post 02 May 2016, 05:42
MathRevolution wrote:
If x is 20 percent more than y and y is 50 percent less than z, then x is what percent of z?

A. 500%
B. 250%
C. 500/3%
D. 125%
E. 60%


* A solution will be posted in two days.


Lets take values of x,y,z which can satisfy given equations.

Let Z= 200, so Y becomes 100 ( since Y is 50% less than Z),
X becomes 120 ( since 20% more than Y)

now req is (X/Z) * 100 = (120/200) *100 = 60 %
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Re: If x is 20 percent more than y and y is 50 percent less than z, then x [#permalink]

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New post 02 May 2016, 11:35
MathRevolution wrote:
If x is 20 percent more than y and y is 50 percent less than z, then x is what percent of z?

A. 500%
B. 250%
C. 500/3%
D. 125%
E. 60%


* A solution will be posted in two days.


Z = 100 ; Y = 50 so X = 60

X as % of Z = 60/100 * 100 => 60%

Answer will be (E)
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Re: If x is 20 percent more than y and y is 50 percent less than z, then x [#permalink]

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New post 04 May 2016, 07:27
From the question, we get x=1.2y and y=0.5z. By substitution, we can get x=1.2y=1.2(0.5)z=0.6z. Hence, the answer is 60% and E is the correct answer choice.

- Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
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Re: If x is 20 percent more than y and y is 50 percent less than z, then x [#permalink]

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New post 29 Aug 2016, 05:46
consider z = 100
so y is 50 % of z = 50
x is 20 % more than y = 1.2 * 50 = 60
x is 60 % of z
correct answer - E

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Re: If x is 20 percent more than y and y is 50 percent less than z, then x [#permalink]

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New post 30 Aug 2016, 14:15
As mentioned above this problem is fairly easy by using by z=100
For Pro Algebra users such as myself , here is how i did it
x=y*[1+20/100]
50/100*z=y

Equation 1 divided by equation 2 results in=>
2x/y=120/100=> x/y=60/100
now we are asked => x/z*100 = (60/100)*100=60
SMASH that E

P.S => Its way too easy by number plugging :)
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Re: If x is 20 percent more than y and y is 50 percent less than z, then x [#permalink]

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New post 30 Aug 2016, 14:54
x/y=12/10
y/z=10/20
x/z=12/20=60%

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Re: If x is 20 percent more than y and y is 50 percent less than z, then x [#permalink]

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Re: If x is 20 percent more than y and y is 50 percent less than z, then x   [#permalink] 14 Oct 2017, 08:08
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