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)
Ayoosh said:
1 decade ago
For all having probs in the 60x + 40x - x^2!!!!
See road is a rect. and they are intercrossing. let the width be x, then for one road area will be 40x and for the other road will be 60x since 60 and 40 will be the lengths of road.
now, we area of roads is 291 m2, so
ar(road1) + ar(road2) - ar(intercrossing)
60x + 40x - x^2 = 291 and then on you can find the ans
See road is a rect. and they are intercrossing. let the width be x, then for one road area will be 40x and for the other road will be 60x since 60 and 40 will be the lengths of road.
now, we area of roads is 291 m2, so
ar(road1) + ar(road2) - ar(intercrossing)
60x + 40x - x^2 = 291 and then on you can find the ans
(9)
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)
Ganesh Kumar said:
2 years ago
Why - x^2? can anyone explain this?
(5)
Ammu said:
2 years ago
Thank you for explaining.
(3)
Bhargavi said:
8 years ago
When two roads intersecting like a +.
So road 1 area is l*b =60*x rectangle road 2 area is l*b =40*x rectangle intersecting place is in a square shape so side =x.; the area is x^2.
In road 1 and road2 we are calculating intersecting point area two times so we have to Subtract that square shape intersecting point.
So road 1 area is l*b =60*x rectangle road 2 area is l*b =40*x rectangle intersecting place is in a square shape so side =x.; the area is x^2.
In road 1 and road2 we are calculating intersecting point area two times so we have to Subtract that square shape intersecting point.
(3)
Jerusha said:
4 years ago
Thanks a lot @Ayoosh.
(2)
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)
Ayoosh said:
1 decade ago
If we dont subtract the X^2 THEN the area of intercrossing will be doubled since its accountd in area of both roads !! thats why we subtract x^2
(2)
Sk Rahad Mannan said:
7 years ago
Thanks @Ayoosh, @Tanya.
(2)
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