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 1 of 8.
Chandinee Moyya said:
2 years ago
Consider the width of road as "x".
Now, the area of the lawn = (60 - x)(40 - x) = 2109.
=> 2400 - 60x - 40x + x^2 = 2109,
=> -60x - 40x + x^2 = -2400 + 2109.
=> -(60x + 40x) + x^2 = -291.
=> x^2 - 100x + 291 = 0.
=> (x - 97)(x - 3) = 0
=>x = 3.
Now, the area of the lawn = (60 - x)(40 - x) = 2109.
=> 2400 - 60x - 40x + x^2 = 2109,
=> -60x - 40x + x^2 = -2400 + 2109.
=> -(60x + 40x) + x^2 = -291.
=> x^2 - 100x + 291 = 0.
=> (x - 97)(x - 3) = 0
=>x = 3.
(10)
Ganesh Kumar said:
2 years ago
Why - x^2? can anyone explain this?
(5)
Ammu said:
2 years ago
Thank you for explaining.
(3)
Jerusha said:
4 years ago
Thanks a lot @Ayoosh.
(2)
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)
Sekhar said:
7 years ago
60x+40x-291 = x*x.
Then what about 291?
Can you explain?
Then what about 291?
Can you explain?
(2)
Sk Rahad Mannan said:
7 years ago
Thanks @Ayoosh, @Tanya.
(2)
Sly said:
7 years ago
Thanks all for explaining the answer in detail.
(1)
Harika said:
7 years ago
The intersection part area should be subtracted it's area = x^2.
Area of one road along length = 40x.
Area of one road along breadth = 60x.
But the area of roads is 2400-2109 = 291.
So 40x+60x-x^2 = 291.
Solve, we get x=3.
Area of one road along length = 40x.
Area of one road along breadth = 60x.
But the area of roads is 2400-2109 = 291.
So 40x+60x-x^2 = 291.
Solve, we get x=3.
(9)
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)
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