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.
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.
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)
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)
Rajan said:
1 decade ago
The road is in the shape of plus mark.
So area of 1st road is 60x.
So area of 2nd road is 40x.
But in the middle the square has been calculated twice above. So subtract that area x^2.
So area of 1st road is 60x.
So area of 2nd road is 40x.
But in the middle the square has been calculated twice above. So subtract that area x^2.
Prajwal said:
1 decade ago
I got it. Its because. 60 and 40 get multiplies to width and. Since there are two cross roads. There will be a common space for two roads. (width of two cross roads is width^2).
Nakul gowda said:
1 decade ago
60x + 40x includes the intersection part twice..so to include the intersection part only once, subtract area of intersection part once, ie, x^2....got it yet..
Rupam said:
1 decade ago
There is no intercrossing left. When we considered 60x +40x, aren't we considering the intercrossing too? Otherwise the lengths couldn't have been 60 and 40.
Lavanya said:
9 years ago
@Ayoosh,
60 and 40 are not the lengths of the roads. 60 is the length and 40 is the width. Still, have a confusion. Could anyone please explain it clearly?
60 and 40 are not the lengths of the roads. 60 is the length and 40 is the width. Still, have a confusion. Could anyone please explain it clearly?
Ashish Malviya said:
1 decade ago
Cosidering Road parallel to 60m side Area :- 60x.
Cosidering Road parallel to 40m side Area :- 40x.
Common area :- x*x.
Hence :- 291=60x+40x-x*x.
Cosidering Road parallel to 40m side Area :- 40x.
Common area :- x*x.
Hence :- 291=60x+40x-x*x.
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)
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