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\(x_1\): 8
\(x_4\): 2
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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For each positive integer k, the quantity \(x_k\) is defined such that \(x_k = x_{k+1} * ( x_{k+2})^3\)
In addition, \(x_3 = 1\). In the table, select values for \(x_1\) and \(x_4\) that are jointly compatible with these conditions. Select only two values, one in each column.
In addition, \(x_3 = 1\). In the table, select values for \(x_1\) and \(x_4\) that are jointly compatible with these conditions. Select only two values, one in each column.
| \(x_1\) | \(x_4\) | Values |
| 0 | ||
| 2 | ||
| 5 | ||
| 8 | ||
| 16 | ||
| 64 |
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Official Answer
\(x_1\): 8
\(x_4\): 2
arpanpatnaik
GMAT 1: 640 Q42 V37
Posts: 101
07 May 2013, 11:07
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Bunuel
From the above it can be deduced, that x1 = x2 = (x4)^3. Hence, for values of x1, x4 only 2,8 satisfy the same. If the above values apply, I believe the expression x-k cannot be said always an integer. since x-5 = (1/2)^(1/3). But since the question mentions it a 'quantity', it seems that fits the question as well.
Please correct me if I am wrong
Regards,
Arpan
16 May 2013, 20:18
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X3 = 1
therefore x1=x2*1 or x1=x2 and
x2 = 1*(x4)^3
X1 = (x4)^3
only combinations for x1 and x4 where x1 can be cube of x4 is 8 for x1 and 2 for x4
And X1 = 8 / X4 = 2
therefore x1=x2*1 or x1=x2 and
x2 = 1*(x4)^3
X1 = (x4)^3
only combinations for x1 and x4 where x1 can be cube of x4 is 8 for x1 and 2 for x4
And X1 = 8 / X4 = 2
