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21 Oct 2019, 02:44
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Each of the numbers w, x, y, and z is equal to either 0 or 1. What is the value of w + x + y + z ?
(1) \(\frac{w}{2} + \frac{x}{4} + \frac{y}{8} + \frac{z}{16} = \frac{11}{16}\)
(2) \(\frac{w}{3} + \frac{x}{9} + \frac{y}{27} + \frac{z}{81} = \frac{31}{81}\)
(1) \(\frac{w}{2} + \frac{x}{4} + \frac{y}{8} + \frac{z}{16} = \frac{11}{16}\)
(2) \(\frac{w}{3} + \frac{x}{9} + \frac{y}{27} + \frac{z}{81} = \frac{31}{81}\)
21 Oct 2019, 05:05
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Bunuel
(1) \(\frac{w}{2} + \frac{x}{4} + \frac{y}{8} + \frac{z}{16} = \frac{11}{16}\)
\(\frac{8w+4x+2y+z}{16}=\frac{11}{16}......8w+4x+2y+z=11\)
As all terms 8w, 4x, 2y are even z has to be 1 to make the sum odd..
Thus 8w+4x+2y+1=11...8w+4x+2y=10....ONLY possibility is that w=1 and y=1, so w+x+y+z=1+0+1+1=3
(2) \(\frac{w}{3} + \frac{x}{9} + \frac{y}{27} + \frac{z}{81} = \frac{31}{81}\)
\(\frac{27w+9x+3y+z}{81}=\frac{31}{81}.....27w+9x+3y+z=31\)
Only way w=1, x=0, y=1, z=1...w+x+y+z=1+0+1+1=3
D
21 Oct 2019, 05:06
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Each of the numbers w, x, y, and z is equal to either 0 or 1. What is the value of w + x + y + z ?
STATEMENT (1)- \(\frac{w}{2}\)+ \(\frac{x}{4}\) + \(\frac{y}{8}\)+ \(\frac{z}{16}\) = \(\frac{11}{16}\)
8w+4x+2y+z = 11
given - w, x, y, and z is equal to either 0 or 1
now, w cant be = 0
if w = 0 then 4x+2y+z cant be = 11 (cause maximum value of 4x+2y+z = 7)
so, w = 1
x cant be = 1 cause 8w+4x will be greater than 11 which is not possible
so, x = 0
now for 8w+4x+2y+z to be = 11 only possible value of y and z = 1
w = 1, x = 0, y = 1, z = 1
value of w + x + y + z = 3
SUFFICIENT
STATEMENT (2) - \(\frac{w}{3}\) + \(\frac{x}{9}\) + \(\frac{y}{27}\) + \(\frac{z}{81}\) = \(\frac{31}{81}\)
27w + 9x + 3y + z = 31
given - w, x, y, and z is equal to either 0 or 1
now, w cant be = 0
if w = 0 then 9x+3y+z cant be = 31 (cause maximum value of 9x+3y+z = 13 if x = 1 y =1 z = 1 )
so, w = 1
x cant be = 1 cause 27w+9x will be greater than 31 which is not possible
so, x = 0
now for 27w+9x+3y+z to be = 31 only possible value of y and z = 1
w = 1, x = 0, y = 1, z = 1
value of w + x + y + z = 3
SUFFICIENT
D is the answer
STATEMENT (1)- \(\frac{w}{2}\)+ \(\frac{x}{4}\) + \(\frac{y}{8}\)+ \(\frac{z}{16}\) = \(\frac{11}{16}\)
8w+4x+2y+z = 11
given - w, x, y, and z is equal to either 0 or 1
now, w cant be = 0
if w = 0 then 4x+2y+z cant be = 11 (cause maximum value of 4x+2y+z = 7)
so, w = 1
x cant be = 1 cause 8w+4x will be greater than 11 which is not possible
so, x = 0
now for 8w+4x+2y+z to be = 11 only possible value of y and z = 1
w = 1, x = 0, y = 1, z = 1
value of w + x + y + z = 3
SUFFICIENT
STATEMENT (2) - \(\frac{w}{3}\) + \(\frac{x}{9}\) + \(\frac{y}{27}\) + \(\frac{z}{81}\) = \(\frac{31}{81}\)
27w + 9x + 3y + z = 31
given - w, x, y, and z is equal to either 0 or 1
now, w cant be = 0
if w = 0 then 9x+3y+z cant be = 31 (cause maximum value of 9x+3y+z = 13 if x = 1 y =1 z = 1 )
so, w = 1
x cant be = 1 cause 27w+9x will be greater than 31 which is not possible
so, x = 0
now for 27w+9x+3y+z to be = 31 only possible value of y and z = 1
w = 1, x = 0, y = 1, z = 1
value of w + x + y + z = 3
SUFFICIENT
D is the answer
