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Three students Alex, Brian and Clara have got a combined total of 88 m

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chetan2u
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Three students Alex, Brian and Clara have got a combined total of 88 m [#permalink] New post   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


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nick1816
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Re: Three students Alex, Brian and Clara have got a combined total of 88 m [#permalink] New post   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



chetan2u wrote:
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


Chetan's question
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PoojanB
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Re: Three students Alex, Brian and Clara have got a combined total of 88 m [#permalink] New post   15 Dec 2019, 03:09
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?


nick1816 wrote:
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



chetan2u wrote:
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


Chetan's question
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nick1816
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Re: Three students Alex, Brian and Clara have got a combined total of 88 m [#permalink] New post   15 Dec 2019, 03:38
PoojanB
You're correct brother but that doesn't affect the answer tho, as A>\(88*\frac{4}{7}\) or 50.28.

As 50.29 is not among the options, that's why i considered only integral values.



PoojanB wrote:
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?


nick1816 wrote:
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



chetan2u wrote:
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


Chetan's question
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PoojanB
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Re: Three students Alex, Brian and Clara have got a combined total of 88 m [#permalink] New post   15 Dec 2019, 04:22
nick1816

So we have to check this math every-time in case of such question?

[quote="nick1816"]PoojanB
You're correct brother but that doesn't affect the answer tho, as A>\(88*\frac{4}{7}\) or 50.28.

As 50.29 is not among the options, that's why i considered only integral values.
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Re: Three students Alex, Brian and Clara have got a combined total of 88 m [#permalink] New post   15 Dec 2019, 04:46
First, I assume Alex has got EXACTLY two times the marks of Brian and Brian has got EXACTLY two times the marks of Clara.

Assume C gets C points, we have 4C + 2C + C = 88 => C = 12.57....

Secondly, using C = 12.57... as guide, check what if C is indeed 12
12 + 25 = 37
88 - 37 = 51
This matches the actual condition of the question. So, the answer is C
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Re: Three students Alex, Brian and Clara have got a combined total of 88 m [#permalink] New post   15 Dec 2019, 05:14
chetan2u wrote:
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


Chetan's question


Imo C

Let the marks by Alex, Brain and Clara be denoted by A,x, and y

A+x+y=88

To minimize Alex marks we have to maximize marks by Brain and Clara.

Per this question

Brain marks > 2 * Clara marks = X > 2*Y or 2y + 1

Alex marks >2 * Brain marks = A>2*(2y+1) or 2*(2y+1) + 1 = 4y +3

Now

Alex + Brain + Clara = 88

4y +3 + 2y + 1 +y = 88

7y = 84 or y = 12

A = 4y +3 = 4*12 +3 = 51
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KarishmaB
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Re: Three students Alex, Brian and Clara have got a combined total of 88 m [#permalink] New post   16 Dec 2019, 00:33
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chetan2u wrote:
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


Chetan's question


A + B + C = 88
A > 2B
B > 2C

We need to minimise A. So A = 2B + 1.
B should be as small as possible for A to be minimum so B = 2C + 1.
A = 2(2C + 1) + 1 = 4C + 3

Then, 4C + 3 + 2C + 1 + C = 88
7C = 84
C = 12
A = 4C + 3 = 51
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Re: Three students Alex, Brian and Clara have got a combined total of 88 m [#permalink] New post   16 Dec 2019, 03:00
VeritasKarishma wrote:
chetan2u wrote:
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


Chetan's question


A + B + C = 88
A > 2B
B > 2C

We need to minimise A. So A = 2B + 1.
B should be as small as possible for A to be minimum so B = 2C + 1.
A = 2(2C + 1) + 1 = 4C + 3

Then, 4C + 3 + 2C + 1 + C = 88
7C = 84
C = 12
A = 4C + 3 = 51


....................................................Thanks.............................
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Re: Three students Alex, Brian and Clara have got a combined total of 88 m [#permalink]
16 Dec 2019, 03:00
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