Latest Capgemini Aptitude Question SOLUTION: There is a rectangular Garden whose length and width are 60m X 20m.There is a walkway of uniform width around garden. Area of walkway is 516m^2. Find width of walkway i.1 ii.2 iii.

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There is a rectangular Garden whose length and width are 60m X 20m.There is a walkway of uniform width around garden. Area of walkway is 516m^2. Find width of walkway
i.1
ii.2
iii.3
iv.4
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Asked In Capgemini arjun (12 years ago)

Unsolved Read Solution (14)

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CHANDAN KR SINGH  (12 years ago)

ANSWER : III (3m)

SOLUTION :

Area of Garden = 60*20 = 1200 sq m.
Total Area of garden and walkway = 1200+516= 1716 sq m.
Let assume width of walk way be x
So (60+2x)(20+2x) = 1716
Or x = 3, -43
3 is correct answer
Since 3 is positive.

Is this solution Helpfull? Yes (65) | No (6)  

priya  (9 years ago)

why 60+2x and 20+2x ?

Is this solution Helpfull? Yes (12) | No (1)  

Dr.Dipin Singh  (12 years ago)

Area of Garden = 60*20 = 1200 sq m.
Total Area of garden and walkway = 1200+516= 1716 sq m.

It means dimensions of final rectangular area are 66*26.

so width of walkway = 3 m ... option (iii)

Is this solution Helpfull? Yes (9) | No (11)  

Ashwath Hegde  (12 years ago)

Answer:3m
Area of Garden = 60*20 = 1200 sq m.
Total Area of garden and walkway = 1200+516= 1716sq m.
Let us assume that width of the walk way be x
therefore(60+2x)(20+2x) = 1716
x = 3, -43
since width can't be negative
3 is correct answer

Is this solution Helpfull? Yes (8) | No (4)  

pritam kumar soni  (7 years ago)

let the width of rectangle be x
so the length & breath is increased by 2x(draw on a paper
then only u'll understand)
so new total area along with walkway is (60+2x)*(20+2x)
so (60+2x)*(20+2x)-60*20=516
=>(60+2x)*(20+2x)=1716
=>(30+x)*(10+x)=429=33*13
=>x=3
which is da width of gateway

Is this solution Helpfull? Yes (7) | No   

Dr.Dipin Singh  (12 years ago)

when options are given and final area is known, there is no need to go for complete solution.
Area is 1716 sq m.
Only possible if digit at unit place of width and breadth is 6 .
only possible when walkway is having uniform width of 3 mtrs.

Is this solution Helpfull? Yes (2) | No (13)  

arjun  (12 years ago)

PLEASE CAN U SIMPLY YOUR ANSWER MORE..............

Is this solution Helpfull? Yes (1) | No (5)  

Phaneendra Sandeep  (12 years ago)

area of walkway=516=sum of areas of new rectangles formed by walkway.
=>(60+2x)*x*2+20*x*2=516
=> x^2+40x-129=0
=> x=3

Is this solution Helpfull? Yes (1) | No (4)  

sravani  (8 years ago)

there is a formula for these type ques.if the fence is around the rectangle then area of fence width (w) is 2w(l+b+2w)

Is this solution Helpfull? Yes (1) | No   

teja priya  (7 years ago)

let, x be the pathway length and width
area of garden =60*20=1200m^2
area of pathway = 516m^2 (given)
now, the total area i.e, area of garden + pathway = 1200+516= 1716
this total area has length of 60+2x and width of 20+2x(if u draw a clear diagram u can easily understand)
therefore, (60+2x)(20+2x)=1716;
by solving we get x^2 +40x-129=0
x^2+43x-3x-129=0 (129=43*3)
x(x+43)-3(x+43)=0
(x-3)(x+43)=0
x=3 or x=-43
length cannot be negative so answer is 3
so answer is 3

Is this solution Helpfull? Yes (1) | No   

arjun  (12 years ago)

thanks now i understand the concept...........

Is this solution Helpfull? Yes | No (4)  

sowmya  (7 years ago)

(2*60w) + (2*20w) + 4w^2 = 516
120w + 40w + 4w^2 = 516
160w + 4w^2 = 516
(w-3)(w+43) = 0
w = 3
so option iii) is correct

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Quamrul Siddiqui  (6 years ago)

let whole width of garden=60+2x
let whole length of garden=20+2x
where x=common width of gateway
thus,
(60+2x)(20+2x)-1200=516
=>1200+120x+40x+4x^2-1200=516
4x^2+160x-516=0
=>x=3

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Monojit Sadhukhan  (4 years ago)

Total Area is=(60*20)+516=1716 m^2.
(60+2x)(20+2x)=1716
=> (30+x) (10+x)=429
if we put 3 then it satisfy the above equation.

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