.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:
1 2 3 4
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Answer / chinmaya panda
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 Answer Correct ? | 174 Yes | 10 No |
Answer / anuja malviya
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
ANS....................
| Is This Answer Correct ? | 44 Yes | 2 No |
Answer / karthikeyan
the area of walk way with width X is 2(60*x)+2*(20*x) = 516
120X+40X = 516
160 X = 516
X= 3.225 = 3 (approx)
| Is This Answer Correct ? | 28 Yes | 15 No |
Answer / sachin chauhan
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
4(x^2+40x-129)=0
x^2+40x-129=0
x^2+43x-3x-129=0
x(x+43)-3(x+43)=0
(x+43)(x-3)=0
=>x=3,-43
but width can never be -ve
so answer is 3 meter
| Is This Answer Correct ? | 11 Yes | 1 No |
Answer / hdhd
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 Answer Correct ? | 5 Yes | 1 No |
Answer / ravindra sharma
total area = 1200
area of walk way = 516 (given)
area of remaning part(R)(with out walkway path area)
=1200-516
=684
let,
length= l
width= b
area or R=>l*b =684
but,
l=60-2x (x is width of the walk way)
b=20-2x
so,
(60-2x)*(20-2x)=684
=> (x^2)-40x+129=0
=> x=36.46 (not possibal)
and
x=3.5379(so ans is 3.5)
| Is This Answer Correct ? | 13 Yes | 12 No |
Answer / prasad
formulae for area of path=
a=2w(l+b+2w)
where a is path area,
l=lenth
b=brieth
w =width
516=2w(60+20+2w)
finally we get 4w^2+160w-516=0
from options 3 is satisfied.
| Is This Answer Correct ? | 3 Yes | 3 No |
Answer / arunkumar
ground is rectangle shaped so if I consider the width the new length will be original length+2* width
therefore new L=60+2(3)==66
b=20+2(3)==26
(66*26)-(60*20)=516
condition satisfied ...Ans will be 3
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / shreyas
length : breadth is 6:2
6+2=8
512/8=64
64*6=384
64*2=132
384/60=6.5
132/20=6.5
The width of the pathways are 3.25
| Is This Answer Correct ? | 1 Yes | 4 No |
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