byRituparna Nath Content Writer at Study Abroad Exams
Question - The variable x is inversely proportional to the square of the variable y. If y is divided by 3a, then x is multiplied by which of the following?
- 1/9a
- 1/9a^2
- 1/3a
- 9a
- 9a^2
‘The variable x is inversely proportional to the square of the variable y’ - is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Official Guide Quantitative Review". To solve GMAT Problem Solving questions a student must have knowledge about a good number of qualitative skills.
In GMAT Problem Solving section, examiners measure how well the candidates make analytical and logical approaches to solve numerical problems. In this section, candidates have to evaluate and interpret data from a given graphical representation. In this section, mostly one finds out mathematical questions. Five answer choices are given for each GMAT Problem solving question.
Solution and Explanation:
Approach Solution 1:
It is Given that x is inversely proportional to the square of the variable y.
So,
x\(y^2\)= k, here k is constant
Putting in original values
x = 1
y = 6
with the result that k= 1∗ \(6^2\)= 36
Updated values:
Divide y by 3a if a=2,
then we will get:
y = \(\frac{6}{[3*2]}\)
y = 1
Now, Substituting y=1 and k=36 into x\(y^2\) = k
x * \(1^2\)=36
New x = 36
The value of x is multiplied by 36 because old x = 1 and new x = 36.
Thus, when a=2, 36 must be the result of the right solution.
Putting in every option we get E as the right one
9\(a^2\)= 9*\(2^2\)
=9\(a^2\)= 9*4
=9\(a^2\)= 36
The answer is E, which is 9\(a^2\)
Correct Answer: E
Approach Solution 2:
There is another approach to solve this question which is pretty easy
And uses algebra method
Let x = k*\(\frac{1}{y^2}\)
Here k is constant
x = k*\(\frac{1}{[3a]^2}\)
x = k*\(\frac{1}{9a^2}\)
k = [9\(a^2\)]x
The answer is E, which is 9\(a^2\)
Correct Answer: E
Approach Solution 3:
Let , x=k∗1/y^2(here K is constant)
x=k∗1/(3a)^2= k∗1/9(a)^2
=> k=(9a2)x
Ans E.
The answer is E, which is 9\(a^2\)
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