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(#M40014215) MBA Numerical Ability question number systems, progressions Keep an EYE

Find the lowest of the three numbers as described: If the cube of the first number exceeds their product by 2, the cube of the second number is smaller than their product by 3 , and the cube of the third number exceeds their product by 3.

1) 3^1/3
2) 9^1/3
3) 2
4) Any of these
5) None of these

Asked In MBA (11 years ago)

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sai teja  (8 years ago)

Let a, b, c be the numbers.

a3 = (abc + 2), b3 = (abc - 3), c3 = (abc + 3). Multiply all three equations and keep abc = x.

x3 = (x + 2)(x - 3)(x + 3) --> 2x2 - 9x - 18 = 0 --> (2x + 3)(x - 6) = 0 --> x = -3/2 or x = 6. Taking abc = 6 and noting that b is the smallest number b3 = 6 - 3 = 3 --> b = 31/3

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fqgqwrggr  (7 years ago)

x=2
y=3^(1/3)
z=9^(1/3)

Reasoning: We know that:
x^3 = xyz + 2
y^3 = xyz - 3
z^3 = xyz + 3

Multiplying all equations gives:
(xyz)^3 = (xyz)^3 - 9(xyz) + 2(xyz)^2 - 18

Let a = xyz. Then the above equation means:

2a^2 -9a - 18 = 0

Solutions are a = 6, a = -1.5. Let's try a=xyz = 6:

x^3 = xyz + 2 = 6 + 2 = 8 ===> x = 2
y^3 = xyz - 3 = 6-3 = 3 ===> y = 3^(1/3)
z^3 = xyz + 3 = 6 + 3 = 9 ===> z = 9^(1/3)

Then verify that this indeed works.

***
In fact, the only other possibility uses the root a=-1.5:
x = (0.5)^(1/3)
y = (-4.5)^(1/3)
z = (1.5)^(1/3)

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