Aptitude - Area - Discussion
Discussion Forum : Area - General Questions (Q.No. 5)
5.
A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?
Answer: Option
Explanation:
Area of the park = (60 x 40) m2 = 2400 m2.
Area of the lawn = 2109 m2.
Area of the crossroads = (2400 - 2109) m2 = 291 m2.
Let the width of the road be x metres. Then,
60x + 40x - x2 = 291
x2 - 100x + 291 = 0
(x - 97)(x - 3) = 0
x = 3.
Video Explanation: https://youtu.be/R3CtrAKGxkc
Discussion:
75 comments Page 2 of 8.
Priyanka said:
9 years ago
We can also subtract the width of the road from length and breadth of the road and multiply the values to get the area of the lawn i.e.
Let the width be x.
Length of lawn = 60 - x.
Breadth of lawn = 48 - x.
(60 - x) (40 - x) = 2109. As given in question.
X^2 - 100x - 291 = 0.
X = 3 and 97.
The value 3 is there in the given option. So, width of the road is 3m.
Let the width be x.
Length of lawn = 60 - x.
Breadth of lawn = 48 - x.
(60 - x) (40 - x) = 2109. As given in question.
X^2 - 100x - 291 = 0.
X = 3 and 97.
The value 3 is there in the given option. So, width of the road is 3m.
(1)
Diego said:
1 decade ago
So it is a rectangle of 60x40cm. It has 2 crossroads, therefore the area of the roads is the horizontal one: 60 times the with (x). -->60*x plus the vertical road 40*x. This results in 60x+40x but you are adding 2 times the area where the roads intersect (one time for the vertical road and one time for the horizontal road). The double counted area is x*x = x^2. So if you subtract this area from 60x+40x you will eliminare 60x+40x -x^2.
Hope this helps you better understand the problem.
60cm
_______________________________
I LAWN I RD I LAWN I
I______________I____I________________I
I ROAD I I ROAD I
I_____________ IX^2I________________I 40cm
I LAWN I I LAWN I
I______________I RD I________________I
Hope this helps you better understand the problem.
(1)
GAYATRI said:
9 years ago
@All.
Please draw the picture first.
Two roads intersecting each other like " + " sign not like "x" sign. One road has an area of 60x other one has 40x. There is a common area of x^2.
Please draw the picture first.
Two roads intersecting each other like " + " sign not like "x" sign. One road has an area of 60x other one has 40x. There is a common area of x^2.
(1)
Himanshu said:
7 years ago
Why only square of width is taken and not length?
(1)
Swastik said:
6 years ago
Use formula,
X+Y+(X*Y/100) for an increase in length and breadth of any quadrilateral.
Which is,
=20+20+(20*20/100)
=20+20+4
=44%.
X+Y+(X*Y/100) for an increase in length and breadth of any quadrilateral.
Which is,
=20+20+(20*20/100)
=20+20+4
=44%.
(1)
Sreekanth said:
7 years ago
@All.
Take:
(60-x) * (40-x).
We gets the remaining area,
Already in the questions, it gives that remaining area is 2109 m2,
Verify the equation with 3.
The equation will be satisfied.
Take:
(60-x) * (40-x).
We gets the remaining area,
Already in the questions, it gives that remaining area is 2109 m2,
Verify the equation with 3.
The equation will be satisfied.
(1)
Sly said:
7 years ago
Thanks all for explaining the answer in detail.
(1)
Puskar prasun said:
10 years ago
Guys! it should be 2x^2 instead of simply x^2? don't you all agree?
ARAVIND said:
9 years ago
Great explanation @Ayoosh.
Arjun said:
9 years ago
(60x + 40x -x^2 this is the area of cross roads alone. 60 and 4O are lengths of cross roads and x is the breadth.
X^2 is subtracted because the meeting of crossroads, the calculated area overlaps. So - X^2.
X^2 is subtracted because the meeting of crossroads, the calculated area overlaps. So - X^2.
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