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15 Dec 2019, 00:35
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Three students Alex, Brian and Clara have got a combined total of 88 marks in a test and the marks each get is a different integer. If Alex has got more than two times the marks of Brian and Brian has got more than two times the marks of Clara, then what is the least possible marks that Alex could have got?
(A) 49
(B) 50
(C) 51
(D) 52
(E) 84
(A) 49
(B) 50
(C) 51
(D) 52
(E) 84
Chetan's question
15 Dec 2019, 01:29
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B>2C
B=2C+x
A>2B
or A= 2B+y= 4C+2x+y, where x and y are positive integers.
A+B+C=88
7C+3x+y=88
To minimize A, we gotta maximize C or minimize x and y.
Minimum value that x and y can take is 1.
7C+4=88
C=12
A= 12*4+3=51
B=2C+x
A>2B
or A= 2B+y= 4C+2x+y, where x and y are positive integers.
A+B+C=88
7C+3x+y=88
To minimize A, we gotta maximize C or minimize x and y.
Minimum value that x and y can take is 1.
7C+4=88
C=12
A= 12*4+3=51
chetan2u
Three students Alex, Brian and Clara have got a combined total of 88 marks in a test. If Alex has got more than two times the marks of Brian and Brian has got more than two times the marks of Clara, then what is the least possible marks that Alex could have got?
(A) 49
(B) 50
(C) 51
(D) 52
(E) 84
(A) 49
(B) 50
(C) 51
(D) 52
(E) 84
Chetan's question
PoojanB
Joined: 09 May 2017
Last visit: 08 Dec 2020
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Location: India
Schools: S.P.Jain Institute (II)
GMAT 1: 610 Q42 V33
GMAT 2: 620 Q45 V31
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WE:Engineering (Energy)
15 Dec 2019, 03:09
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Hi nick1816
How can we say that minimum value x & y can take is 1?can't it be 0.5? or when not mentioned we should consider only integer values in such case?
How can we say that minimum value x & y can take is 1?can't it be 0.5? or when not mentioned we should consider only integer values in such case?
nick1816
B>2C
B=2C+x
A>2B
or A= 2B+y= 4C+2x+y, where x and y are positive integers.
A+B+C=88
7C+3x+y=88
To minimize A, we gotta maximize C or minimize x and y.
Minimum value that x and y can take is 1.
7C+4=88
C=12
A= 12*4+3=51
B=2C+x
A>2B
or A= 2B+y= 4C+2x+y, where x and y are positive integers.
A+B+C=88
7C+3x+y=88
To minimize A, we gotta maximize C or minimize x and y.
Minimum value that x and y can take is 1.
7C+4=88
C=12
A= 12*4+3=51
chetan2u
Three students Alex, Brian and Clara have got a combined total of 88 marks in a test. If Alex has got more than two times the marks of Brian and Brian has got more than two times the marks of Clara, then what is the least possible marks that Alex could have got?
(A) 49
(B) 50
(C) 51
(D) 52
(E) 84
(A) 49
(B) 50
(C) 51
(D) 52
(E) 84
Chetan's question
