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04 May 2019, 02:46
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E
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AB, CD and EF are three parallel lines, in that order. Let d1 end d2 be the distances from CD to AB and EF respectively. d1 and d2 are integers, where d1:d2 = 2:1, P is a point on AB, Q and S are points on CD and R is a point on EF, If the area of the quadrilateral PQRS is 30 square units, what is the value of QR when value of SR is the least?
A. slightly less than 10 units
B. 10 units
C. slightly greater than 10 units
D. slightly less than 20 units
E. slightly greater than 20 units
A. slightly less than 10 units
B. 10 units
C. slightly greater than 10 units
D. slightly less than 20 units
E. slightly greater than 20 units
05 May 2019, 08:22
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Manas1212
Look at the attached sketch
Least value of SR means it is the altitude between CD and EF, so d2..
Now d2 is an integer and obviously a positive one, so least value =1. Thus d1=2...
Area of parallelogram = A(PQS)+A(QSR)=\(\frac{1}{2}*QS*2+\frac{1}{2}*QS*1=\frac{3}{2}*QS=30....QS=20\)
Now QR is the hypotenuse of triangle PQS where QS =20 and SR =1..
so QR=\(\sqrt{20^2+1^2}=\sqrt{402}\)..Slightly more than \(\sqrt{400}\), that is slightly more than 20.
E
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1201.png [ 40.16 KiB | Viewed 3775 times ]
udaypratapsingh99
Joined: 12 Jan 2019
Last visit: 17 May 2026
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Concentration: Strategy, Leadership
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GMAT 1: 660 Q47 V34
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06 May 2019, 04:37
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Manas1212
d1:d2
QR =?
SR is least
If SR is least, it means SR =1. Let’s take QS =x. Also PQ =2.
Hence
[(1/2) * x * 2] + [(1/2) * x * 1] = 30
x = 20
So, QR = squareroot(20^2 + 1^2) = squareroot(401)
This value is slightly greater than 20 units.
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5b9b71df02f9890c839d62be1b1bb3a3_view.png [ 28.09 KiB | Viewed 3593 times ]
