Latest Elitmus Aptitude Question SOLUTION: given a rectangle ABCD , there's a point x on side AB and two points P and Q are on side BC and DE respectively, what is the ratio of area of figure PCDQ and and area of triangle

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(#M40038845) ELITMUS Logical Reasoning question Data sufficiency Keep an EYE

given a rectangle ABCD , there's a point x on side AB and two points P and Q are on side BC and DE respectively, what is the ratio of area of figure PCDQ and and area of triangle XCD
i> BP=QD
ii>AQ=PC

Asked In Elitmus rahul kumar (9 years ago)

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Gaurav Dhimankar  (9 years ago)

I am Extremely Sorry, Previously I did It Wrongly So You Can Ignore It.
A____X____B
| |
Q| |D
|_________|
C D

From S1: BP = QD & From S2: AQ = PC .
From S1 & S2 --> PC/BP = AQ/QD = x/y(ratio)
So Area of Triangle(XCD) / Area of Rectangle(PCDQ) = 0.5(AD*DC)/(QD*DC)
=> (0.5 AD)/QD => 0.5[(AQ+QD)/QD] => 0.5[(AQ/QD)+ 1] => 0.5[(x/y) +1 ] by. Putting the value of AQ/QD [Note:Inverse the above to get the exact Answer]
So Answer is C) The data in both statements I and II together are necessary to answer the question.
Note: Try To Ignore my Previous Post , This one is 100% Right Solution.

Is this solution Helpfull? Yes (22) | No (1)  

MUNISH YADAV  (9 years ago)

Since BP=QD & AQ=PC; it is clear that if point P divides side BC in the ratio x:y then Q divides side DA in y:x. Thus, area of PCDQ=PC*CD=y*CD and that of triangle XCD=(CD*AD)/2=(CD*(x+y))/2.
Their areas are in the ratio 2y/(x+y)

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veer kumar  (9 years ago)

DRAWing the figure first and let length=l and breadth=b
area of triangle(CDX)=1/2*b*l(bcoz length will be the height and breadth will be base of the triangel )
now take 1st condition BP=QD(suppose point P and Q are mid point)
the area of fig PCDQ= breadth(b)*length(l/2)
now if we move the point P towards B, then we should move Q towards D such that BP=QD so they are moving proportionally
let P coinsides with B and Q coinsides with D forming a triangle having base b and height l,so area =1/2*b*l
thus the area of the fig PCDQ will be (l*b)/2 just by the condition BP=QD independent of the point P and Q
so (area of PCDQ)/(area of XCD) will be 1/1

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Gaurav Dhimankar  (9 years ago)

The Figure will look like a Triangle(XCD) inscribed in a Rectangle (ABCD)
Thus to find out the PCDQ:XCD, we need to find out AQ:QD & BP:PC ,then only we can find out the areas of XCD and PCDQ,the easy way to do it if we know that Q and P are the mid points of AD & BC.Lets check that first
(Minor Corrections P and Q lies on BC & AD)

From S1: BP = QD it can't gives us the ratio
From S2: AQ = PC it can't gives us the ratio

Putting S1 and S2 together:
AQ + QD = BP + PC [Its irrelevant actually]
Suppose if BP = 1 & PC = 1 So , AQ = 1 & QD = 1 then ratio will be AQ:QD = 1:1 & BP:PC = 1:1
but wait what if BP = 2 & PC = 1 [It can be possible], So AQ = 1 & QD = 2 then the ratio will be AQ:QD = 1:2 & BP:PC = 2:1.
Again lets take BP = 7 & PC = 3 ,So AQ = 3 & QD = 7 then the ratio will be AQ:QD = 3:7 & BP:PC = 7:3.

Damn it ,getting variable Ratios Thus Everything Goes Wrong ,So data is insufficient.
Thus Answer will be d)Both statements I and II together are not sufficient to answer the question.

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shrishti singh  (8 years ago)

well i think answer is either statement is sufficient to answer the question as beacause if statement 1 is true then 2 is automatically true and vice versa

if i take the first one BP=DQ
BC-CP=AD-AQ ( AD=BC)
So it implies CP = AQ (AD & BC will cancel out)
so P and Q are mid points of BC and AD
area of PCDQ=DC*PC=1/2(DC*BC)=(1/2)area of ABCD
AREA OF DXC=(1/2)*DC*BC=(1/2)area of ABCD

so both area are equal and ratio is 1:1

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chandan  (9 years ago)

Area of PCDQ= (QD+PC)*DC/2
since PC= AQ SO QD+PC=AD
Hence Ar(PCDQ)= AD*DC/2
Ar(XCD)=(1/2)*(DC)*AD
Ar(PCDQ)/Ar(XCD)=1:1

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vikas kumar rai  (9 years ago)

bhai thanks yehi sab question main bhi post karne wala tha

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rahul kumar  (9 years ago)

if you remember other questions , do post , bhai

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richa pathak  (9 years ago)

ii>AQ=PC since both are on opposite side

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vikas kumar rai  (9 years ago)

i remember the logarithm question in which log (base 3)N =K AND N

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pallavi kumari  (9 years ago)

1>BP=QD by assuming P & Q are center point

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Tarun Sharma  (8 years ago)

GAURAV DHIMANKAR your solution is again worng
! area of triangle between a rectangle is half the of the area of rectangle.
!! and area of figure ABPQ= Area of figure QPCD
so Area of QPCD = half of rectangle
so Ratio of triangle and quadrilaateral =1:1
so ANY of 1and 2 equation is enough to answer the question

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