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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
A. 60
B. 90
C. 30
D. 45
E. 75
from Unacademy
24 Jul 2023, 10:54
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pintukr
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
Joined: 16 Jul 2019
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GMAT Focus 1: 645 Q90 V76 DI80 
GPA: 7.81
Schools: INSEAD '26 (D) ISB '27 (D) IIMB '26 (A) IIMA '26 (A) HEC '26 (D) LBS '26 IESE '27 (D) Fuqua '27 (WL) Tepper '27 (WL)
GMAT Focus 1: 645 Q90 V76 DI80
Posts: 784
24 Jul 2023, 10:59
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