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GRE Math Challenge #92-Solve 3 - 2[ 5 - 7*(3+2) ]=
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09 May 2015, 11:37
09 May 2015, 11:37
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Question Stats:
66% (00:27) correct
33% (00:32) wrong based on 18 sessions
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Re: GRE Math Challenge #92-Solve 3 - 2[ 5 - 7*(3+2) ]=
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18 Dec 2017, 06:33
18 Dec 2017, 06:33
1
sandy wrote:
\(3 - 2[ 5 - 7*(3+2) ]=\)
(A) –30
(B) –10
(C) 23
(D) 63
(E) 77
(A) –30
(B) –10
(C) 23
(D) 63
(E) 77
Here we need to follow the BODMAS RULE
B = BRACKET (...)
O = ORDERS(numbers involving powers or square roots)
M = MULTIPLICATION
A = ADDITION
S= SUBTRACTION
Hence by solving the above equation comes = 63. i.e option D
Re: GRE Math Challenge #92-Solve 3 - 2[ 5 - 7*(3+2) ]=
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08 Mar 2018, 00:53
08 Mar 2018, 00:53
2
I think the ans should be A.
I assume \(*\) here is multiplication. then,
\(3+2 = 5*7 = 35\)
\(5-35 = -30\)
Outside bracket we have \(3-2 =1 so 1*(-30) = -30\)
Just to make sure I put the given syntax as it is in the desktop calculator and it gives \(-30\)
I assume \(*\) here is multiplication. then,
\(3+2 = 5*7 = 35\)
\(5-35 = -30\)
Outside bracket we have \(3-2 =1 so 1*(-30) = -30\)
Just to make sure I put the given syntax as it is in the desktop calculator and it gives \(-30\)
