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30 Oct 2020, 00:15
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
47% (02:07) correct
53%
(01:57)
wrong
based on 79
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A jogger runs along the track PABCQRCDASP. How many times will he need to run around this track to cover 5 miles? Note that ABCD is a square.
(1) ∠PSD = ∠QRD = 90° and PA = AB = CQ
(2) PS = QR = BD = 0.35 miles
Attachment:
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30 Oct 2020, 04:28
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Given ABCD is Square
AB=BC=CD=DA
(1) ∠PSD = ∠QRD = 90° and PA = AB = CQ
AB=BC=CD=DA=CQ=PA
and ∠ASP = ∠APS = ∠CQR = ∠CRQ =45 (similar triangle property)
so track PABCQRCDASP is nothing but 2 squares with 2 diagonals
but we don't know the length of these squares side.
(Insufficient)
(2) PS = QR = BD = 0.35 miles
so value of side of square ABCD = diagonal/√2
=> 0.35/1.4 => 0.25
but what about the length of PA, PS, CQ, CR?
(Insufficient)
Combining both
track length = 2*square perimeter + 2*diagonal
=> 2*4*0.25+2*0.35
=> 2+0.7
=> 2.7
so the runner need to run almost 2 times to cover 5 miles distance
Answer is C
AB=BC=CD=DA
(1) ∠PSD = ∠QRD = 90° and PA = AB = CQ
AB=BC=CD=DA=CQ=PA
and ∠ASP = ∠APS = ∠CQR = ∠CRQ =45 (similar triangle property)
so track PABCQRCDASP is nothing but 2 squares with 2 diagonals
but we don't know the length of these squares side.
(Insufficient)
(2) PS = QR = BD = 0.35 miles
so value of side of square ABCD = diagonal/√2
=> 0.35/1.4 => 0.25
but what about the length of PA, PS, CQ, CR?
(Insufficient)
Combining both
track length = 2*square perimeter + 2*diagonal
=> 2*4*0.25+2*0.35
=> 2+0.7
=> 2.7
so the runner need to run almost 2 times to cover 5 miles distance
Answer is C
30 Oct 2020, 06:32
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if angle pab is 90 then angle sad will also be 90 verticle opposite angle
line sb is 180 one angle is 90 so angle pas will be 90 so now we know that all angle are 90 that will only be in rhombus and we know that all side of rhombas is equal
so option B is sufficient
line sb is 180 one angle is 90 so angle pas will be 90 so now we know that all angle are 90 that will only be in rhombus and we know that all side of rhombas is equal
so option B is sufficient
