Aptitude - Area - Discussion
Discussion Forum : Area - General Questions (Q.No. 12)
12.
The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?
Answer: Option
Explanation:
Let original length = x and original breadth = y.
Original area = xy.
| New length = | x | . |
| 2 |
New breadth = 3y.
| New area = |  |
x | x 3y |  |
= | 3 | xy. |
| 2 | 2 |
 Increase % = |
|
1 | xy x | 1 | x 100 | % |
= 50%. |
| 2 | xy |
Discussion:
25 comments Page 2 of 3.
Naga said:
1 decade ago
Assume L = 100, B = 80, L*B = 8000, Given data half length so, -50 breadth-3s = 3*80 = 240.
50*240 = 12000, change percentage = 12000-8000/80 = 50.
50*240 = 12000, change percentage = 12000-8000/80 = 50.
Medha said:
1 decade ago
If we take l = 10 nd b = 10.
Area = l*b = 100.
Now the new length l/2 = 5.
And new breadth 3b = 30.
New area = 5*30 = 150.
Old area was 100 nd new is 150 so it increases by 50 which is double(new-Old = 150-100 = 50 it is 50% with respect to 100).
Means increases 50%.
Area = l*b = 100.
Now the new length l/2 = 5.
And new breadth 3b = 30.
New area = 5*30 = 150.
Old area was 100 nd new is 150 so it increases by 50 which is double(new-Old = 150-100 = 50 it is 50% with respect to 100).
Means increases 50%.
Devender said:
1 decade ago
In the formula x+y+xy/100. Y value should be 300. How it is taken 200. I am not getting answer.
Kanmani said:
1 decade ago
Increase% = (new area-original)/original.
Himanshuvijayvargia said:
1 decade ago
x + y + x*y/100 = % increase or decrease in the area of an image when there is any increase or decrease in its sides.
Here x =(-50%); y = (200%).
-50 + 200 + (-50*200/100) = +50% answer.
Here x =(-50%); y = (200%).
-50 + 200 + (-50*200/100) = +50% answer.
Kavish said:
1 decade ago
@Abc.
The area is obviously increasing in this case :
Take for example as given in this question.
At normal condition. When, L = x and B = y dn the area = xy,
While as per the conditions in question, L = x/2 and B = 3y dn the area = 3/2 (xy).
Dat means the area increases by 3/2 times of original area.
Further as you can see, change in area or increment in area = 3/2 (xy) -xy = 0.5xy.
Increase in percent = 0.5xy/xy *100 = 50%.
That's it.
The area is obviously increasing in this case :
Take for example as given in this question.
At normal condition. When, L = x and B = y dn the area = xy,
While as per the conditions in question, L = x/2 and B = 3y dn the area = 3/2 (xy).
Dat means the area increases by 3/2 times of original area.
Further as you can see, change in area or increment in area = 3/2 (xy) -xy = 0.5xy.
Increase in percent = 0.5xy/xy *100 = 50%.
That's it.
Abc said:
1 decade ago
How to know the area increasing or decreasing ?
Ketan Mehta said:
1 decade ago
@Anshu
Nice Dude!
Nice Dude!
Anshu said:
1 decade ago
ORIGINAL
LENGTH = X
BREADTH = Y
AREA = XY
----------------------------
LENGHT = 0.5X
BREADTH = 3Y
AREA = 1.5XY
THERE WAS INCREASE OF O.5 XY
0.5XY / XY * 1OO = 50%
SHORT AND CRISP METHOD
LENGTH = X
BREADTH = Y
AREA = XY
----------------------------
LENGHT = 0.5X
BREADTH = 3Y
AREA = 1.5XY
THERE WAS INCREASE OF O.5 XY
0.5XY / XY * 1OO = 50%
SHORT AND CRISP METHOD
Sundar said:
1 decade ago
We can solve this question by simply taking an example. Let me explain.
Assume a rectangle, lxb = 4x2 and area = 8 (4*2).
New Length (after halved 4/2) = 2
New Breadth (after tripled 2*3) = 6
New area of rectangle = 2x6 = 12.
Change in area = 12 - 8 = 4.
Now % of change is 4/8*100 = 50%.
Hope this will help you. Have a nice day!
Assume a rectangle, lxb = 4x2 and area = 8 (4*2).
New Length (after halved 4/2) = 2
New Breadth (after tripled 2*3) = 6
New area of rectangle = 2x6 = 12.
Change in area = 12 - 8 = 4.
Now % of change is 4/8*100 = 50%.
Hope this will help you. Have a nice day!
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