CAT 2008: Quantitative Questions a Day 51-100 -The discussions-Brought to you by SMOT

dewan_iitr

Quote:

Originally Posted by mba.yodha

@dewan_iitr
Dude can u plz explain y r we assuming trngl ABC as isosceles.... will the result still hol if it is not...

yes it will
no need to assume it isosceles
actually i din used it anywhere in my sol.

ravishah17

Quote:

Originally Posted by implex

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Quantitative Question # 063
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Let ABC be a triangle and D and E be internal points on BC and AC respectively. BD/DC = EA/CE = 1/2. If the area of triangle ABC is 2 sq. unit, then area of quadrilateral ODCE (in sq. unit) is

(1) 4/5 (2) 14/15 (3) 16/15 (4) 6/5 (5) none of these

From the figure,
Here tria CED ~ tria CAB
where ratio is 2/3

Therefore
ar( tria CED)=4/9 ar(tria CAB)= 4/9 *2
ar(tria CED)= 8/9----(I)

Also
tria AOB ~ tria DOE
where ratio is 2/3 (from DE=2/3 BA)

ar(tria DOE)= 4/9 ar(tria AOB)

Now, altitude of AOB+altitude of DOE= 1/3 altitude of ABC
altitude of AOB+2/3 altitude of AOB= 1/3 altitude of ABC
5/3 (altitude of AOB)= 1/3 altitude of ABC

Therefore altitude of AOB=1/5 altitude of ABC

ar(tria AOB)=1/5 ar(tria ABC)=2/5

So, ar(tria DOE)= 4/9 *2/5= 8/45 --- (II)

ar OECD= 8/9+8/45= 8(6/45)= 8(2/15)=16/15

Therefore answer is (3) 16/15

hi.shivani

Quote:

Originally Posted by mba.yodha

@dewan_iitr
Dude can u plz explain y r we assuming trngl ABC as isosceles.... will the result still hol if it is not...

he didnt assumed d triangle to b isoceles..it just came coz of the similarity in the triangle..(condition given..wl lead to the conclusion dat triangles r similar)

sanyo

Quantitative Question # 063

Answer option 3) 16/15

soln:
let h be he altitude of the triangle from C to AB.

now area(CED) = 1/2 *2/3 AB* 2/3 h,
also 1/2 * AB* h = 2 (given)
=> area(CED) = 2*4/9 = 8/9
and area of ABDE = 2-8/9 = 10/9.

consider quad ABDE, here triangle ABO is similar to traingle DEO,
let area of ABO= A,
=> area of DEO = 4/9 A (similar triangles)
=> area BOD = area AOE = 1/2 *AB* h/3 - A
= 2/3 -A
therefore A + 4/9 A + 2(2/3 -A) = 10/9.
=> A = 2/5

=> area of DEO = 4/9 *2/5 = 8/45
hence area DCEO = 8/45+ 8/9 = 16/15

implex

Quote:

Originally Posted by Aarav

The point O is the intersection point of BE and AD. Solve now.

oops at aarav .. I went to sleep, dreams me bhi yeh sawal aa raha tha !!

anyways will solve now

implex

Quote:

Originally Posted by implex

oops at aarav .. I went to sleep, dreams me bhi yeh sawal aa raha tha !!

anyways will solve now

AREA ODCE=ARE CDE+ AREA ODE
area CDE is easy to prove 8/9

the next part is to find the area of ODE
here we just need the altitude of oDE
which is easy to prove again

AREA ODE =2/5.4/9=8/45
adding
8/9+8/45=48/45=16/15

neynetr · 02:35 PM

wow.... the question has already been torn apart...
my solution.. though not much different but will post it:

tr(ABC) ~ tr(CDE)
and ratio of sides is 2:3
so area(CDE) = 2 * (4/9) = 8/16 sq. units

now DE || AB so AEDB trapezium and ratio of opposite sides (AB and DE) is 2:3
area(AEBD) = 2 - 8/9 = 10/9
=> area(BDE) = 2/5 * (10/9) = 4/9
=> area(OED) = 2/5 * (4/9)= 8/45

so area(OEDC) = 8/9 + 8/45 = 16/15

option (3)

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one more question in mind after solving this though....
is it possible to find the solution if BD/DC != AE/EC...?? i.e. if the triangles r not similar...???

thevoid

HI PGUYS...

first of all ..thanks to aarav for clearing the doubt....

got lost in my office work so posting it late...

but happy to see so many approaches especially the nbangalore,implex and neynetr....

hats off to u guys...

my approach was similar with that of ravishah so not posting it again...
hence answer 3...
Q.E.D.>

srikar2097

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Quantitative Question # 064
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Katrina and Deepika have some marbles with each of them, such that the number of marbles with Deepika is thrice that with Katrina. If Katrina distributes her marbles equally among certain number of bags, then she is left with 31 extra marbles. If Katrina and Deepika were to pool the marbles and then distribute the total marbles equally among the same number of bags as Katrina did, they will be left with 16 marbles. The number of marbles with Deepika is the largest possible three digit number. How many bags are needed to equally divide all the marbles with Deepika, if the number of those bags is the smallest possible two digit number?

(1) 11 (2) 13 (3) 29 (4) 31 (5) none of these

22baba

Answer is E.