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(#M40040641) TCS Numerical Ability question easy? Keep an EYE

a ladder is 100 m long and distance between bottom of the ladder and wall is 60m.what is the maximum size of the cube that can be placed between the wall and the ladder.

Asked In TCS ABHISHEK KUMAR (10 years ago)

Solved Argha Banerjee Read Solution (15)

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Argha Banerjee  (10 years ago) Best Solution

side of the cube is 40m

AB-WALL
BC- DISTANCE BETWEEN WALL AND LADDER FOOT=60m
AC-LADDER=100m
SO AB=SQ.RT (AC^2-BC^2)=80m

Area of triangle ABC=1/2*b*h=2400 sq. m

lasrgest possible square will be of area 1600sq. m
so side =40m

Is this solution Helpfull? Yes (25) | No (16)  

neha kumari  (10 years ago)

let the length of cube be x
By using similarity of triangle
(80-x)/80=x/60
x=34.28

Is this solution Helpfull? Yes (9) | No (1)  

MANISH  (10 years ago)

(80-x)/80=x/60 will give the correct answer ie 34.28

Is this solution Helpfull? Yes (4) | No (2)  

Nidhi Rai  (10 years ago)

ladder =100m , wall =80m distance between bottom of the ladder and wall is = 60m
so ,here see the least distance that is 60 half of 60 is 30 ,if we take any value greater than 30 then it will coincide with any of the other 2 distance hence, triangel will not be formed,so a cube of side 30 m can b placed

Is this solution Helpfull? Yes (3) | No (3)  

RANJAN KUMAR  (10 years ago)

#ARGHA# plz explain..?

bcoz , according to me,

Ans is 34.28m(using similarity of triangle).
and, if we take 40m thn
area of sqr. = 1600
taking side 40m and calculating area of two triangles formed after placing the cube are as:
area of upper triangle = (1/2)*40*40=800.
area of lower(right side to the sqr.)triangle = (1/2)*40*20=400
so,total area of both triangles are = 800+400=1200

here total area captured by two triangles & 1 sqr is = 1600(sqr.)+1200(triangles)=2800.
this is greater than total area of triangle formed by ladder &wall
i.e = (1/2)*80*60=2400

hence 40m is wrong answer.

Is this solution Helpfull? Yes (2) | No (1)  

Sarabeshwar  (10 years ago)

The answer will be the distance how far the altitude of ladder(triangle) will be from the wall.

Formula to find altitude h=2(a/b)

Here, a= 60 m
b=100m

that gives you 1.2 m.

so you can place the cube of size = 1.2m

Is this solution Helpfull? Yes (1) | No (6)  

Abhijeet Srivastava  (10 years ago)

The triangles formed above and adjacent to the square must be S.A.S. congruent. When equating the ratios of their sides and performing substitution, we get an equation yielding side of square equal to 34.286.

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SUMAN GHOSH  (10 years ago)

height of the wall=(100^2-60^2)^0.5=80m
area of the triangle form=1/2*80*60=2400sqm
area of largest cube=6*(a^2)
area of one face =a^2
a^2

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Saurabh Mishra  (10 years ago)

34.2 is the ans

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latha.m  (10 years ago)

the other side of the triangle is 80 m . if the cube is to be placed then it should have all sides equal. from this we will come to know that the cube should be placed at midpoints of triangle to get all sides equal of cube.Let x be the sides of cube. so 80/80-x=60/x (from similar triangle property) on solving for x we get x=240

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Madhu Kalyan Reddy  (10 years ago)

Ans: 30
consider that the ladder is placed in such a way that the height of the wall is 80m. the max size of cube is possible only if one corner of cube is touching the hypotenuse at the middle point. from similar triangles we can find out the side of cube

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ramesh chandi  (10 years ago)

base= 100^2 - 60^2
= 80m
half of the length of the wall is 30
so, we can place maximum of 30m side of the cube

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govind dwivedi  (10 years ago)

distance from top of ladder to bottom of wall=80mt.
so,we find mid point of triangle b/w wall & ladder
that gives apprx 30 mt max. size of cube.
ans is 30 mt.

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latha.m  (10 years ago)

sorry . after solving equation 80/80-x=60/x . then 80x=(80-x)60 then 80x=4800-60x then 140x=4800 then x=34.28

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kumari megha jagannath  (9 years ago)

which is the correct ans??

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