If |x| + 3|y| = 30, then find the largest distance

Question

If |x| + 3|y| = 30, then find the largest distance (in units) between any two points on the graph of the given equation.

A. 60 
B. 90
C. 30
D. 45
E. 75

from Unacademy

gmatophobia · Answer

The minimum value of any modulus operation is zero. Now in this equation, the sum of the distances of x from 0 and y from 0 is fixed (i.e. equals 30)

|x| + 3|y| = 30

We need to maximize the difference (distance) between x and y.

In order to do so, we need to allocate the minimum value possible to y. Because for any value we put in y, 3|y| results in three times the value. So if we can make 3|y| minimum, the value of |x| can be maximum.

The minimum value of 3|y| = 0 when y = 0. Maximum value of |x| = 30.

x =  \pm 30

Hence maximum distance = 30 - (-30) = 60

Option A

P.S. - The options in GMAT are always in either ascending or descending order and not randomly placed. Hence, we can use the options as well to logically arrive at the solution.

Oppenheimer1945 · Answer

Pls check the OA. It should be 30+30=60 .

KarishmaB  ,  GMATNinja  Pls confirm

Bunuel · Answer

Please provide a link to this question or a screenshot from the source. Thank you!

KarishmaB · Answer

Yes, that's correct. The required distance is the distance between (30, 0) and (-30, 0) which is 60.

pintukr · Answer

error on my part..

the correct answer is A..

amended the OA.. Thanks

Posted from my mobile device

Bunuel · Answer

Once again, please provide a link to this question or a screenshot from the source. Thank you!

pintukr · Answer

Have added the screenshot (as required).

The question was obtained from newly launched Unacademy portal.

Thanks

Bunuel · Answer

___________________
Thank you very much!

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If |x| + 3|y| = 30, then find the largest distance

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If |x| + 3|y| = 30, then find the largest distance

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