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705-805 (Hard)|   Geometry|      
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Bunuel
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In circle P, the two chords intersect at point X, with the lengths as 16 Feb 2016, 01:07
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95% (hard)

Question Stats:

42% (02:19) correct 58% (02:16) wrong based on 104 sessions

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In circle P, the two chords intersect at point X, with the lengths as indicated in the figure. Which could not be the sum of lengths a and b, if a and b are integers?

A. 49
B. 30
C. 26
D. 16
E. 14

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B
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chetan2u
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GMAT Focus 1: 735 Q90 V89 DI81
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In circle P, the two chords intersect at point X, with the lengths as 16 Feb 2016, 02:01
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Bunuel

In circle P, the two chords intersect at point X, with the lengths as indicated in the figure. Which could not be the sum of lengths a and b, if a and b are integers?

A. 49
B. 30
C. 26
D. 16
E. 14

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Hi,
To do this the CHORD INTERSECTING THEOREM is the simplest..
Quote:
Given a point X in the interior of a circle, pass two lines through X that intersect the circle in points A and B and, respectively, D and C. Then AX·BX = DX·CX
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How do you get it or if you do not know the theorem, how do you get to the answer..

Another property of CHORDS
Quote:
A chord will subtend the same angle from any point on the circumference on the same side of chord
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so see the att sketch..



ACX and DBX are similar triangles and we get AX*BX=DX*CX..
substitute values of AX and BX...
6*8= DX*CX..
so a*b= 48...
we have to find the sum of a and b, so lets test the values..

A. 49..ab=48 = 1*48.. so POSSIBLE if a and b are 1 and 48, a+b=49
B. 30..ab=30... no values of a and b possible as integer
C. 26..ab=26 = 2*24.. so POSSIBLE if a and b are 2 and 24, a+b=26
D. 16..ab=16 = 4*12.. so POSSIBLE if a and b are 4 and 12, a+b=16
E. 14..ab=14 = 6*8.. so POSSIBLE if a and b are 6 and 8, a+b=14
ans B..

also we can check values by factoring..
48=1*2*2*2*2*3...
largest sum possible if it is product of 1 and the integer itself, 48*1, so sum = 1+48=49..
next largest sum is when it is product of 2nd smallest factor,2=24*2.. and sum =2+24=26..

here we can see 30 is larger than 26, so not possible..
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In circle P, the two chords intersect at point X, with the lengths as 22 Feb 2016, 16:27
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E: 14 because the sum of the shorter cord is 14, so the longer chord cannot be that.
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In circle P, the two chords intersect at point X, with the lengths as 29 Feb 2016, 08:16
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Bunuel

In circle P, the two chords intersect at point X, with the lengths as indicated in the figure. Which could not be the sum of lengths a and b, if a and b are integers?

A. 49
B. 30
C. 26
D. 16
E. 14

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Attachment:
2016-02-14_1430.png
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Hi Bunuel,

Should D be 19?

We know that the triangle is having a defined ratio

6 -> b = 3 or 2 or 1 (divide by 2, 3 and 6 respectively since b must be int)
8 -> a = 16 or 24 or 48 ( times by 2, 3 and 6 following the above)

hence possible values 14, 19, 26 or 49... I don't see how we can get a+b=16 and the other possible answer is 30...
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In circle P, the two chords intersect at point X, with the lengths as 29 Feb 2016, 08:23
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Answered this by substitution in 20 seconds.

6x8=48
So now get the combinations for 48:

1 and 48; sum is 49
2 and 24; sum is 26
3 and 16; sum is 19
4 and 12; sum is 16
6 and 8 ;sum is 14

The only one missing is 30 > Answer B
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In circle P, the two chords intersect at point X, with the lengths as 31 Dec 2021, 06:42
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Hi,

use Power of Point Theorem:

a*b=48

-> (a,b) e {(48,1), (29,2), (16,3), (12,4), (8,6), (4,12), (3,16)...}

Taking the sums, we see that (B) is the outlier
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