| Case Study : Q. 16 to 20 |
| The residents of a group housing society at Jaipur decided to build a rectangular garden to beautify the garden. |
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| One of the members of the society made some calculations and informed that if the length of the rectangular garden is increased by 2 m and the breadth reduced by 2 m, the area gets reduced by 12 sq. m. If, however, the length is decreased by 1 m and breadth increased by 3 m, the area of the rectangle is increased by. |
| Based on the above information, give the answer of the following questions: |
A) Length = 6 m, breadth = 4 m
B) length = 10 m, breadth = 6 m
C) length =10 m, breadth =4 m
D) length = 6 m, breadth = 2m
Correct Answer: B
Solution :
| Let the length and breadth of the rectangular garden be denoted by x m and y m respectively. The area of the rectangular garden\[=\text{xy sq}.\text{ m}\]. |
| According to the question, |
| \[(x+2)\,\,\,(y-2)\,=xy-12\] |
| and \[(x-1)\,\,\,(y+3)\,=xy+12\] |
| Simplifying the above equations, we get |
| \[xy-2x+2y-4=xy-12\] |
| or \[-2x+2y=-8\] |
| or \[x-y=4\] ...(1) |
| Also, \[xy+3x-y-3=xy+21\] |
| or \[3x-y=24\] ...(2) |
| Let us now solve the eqs. (1) and (2) by the method of substitution. |
| From eq. (1). \[x=y+4\] ...(3) |
| Substituting in eq. (2), |
| \[3(y+4)-y=24\] |
| or \[3y+12-y=24\] |
| or \[2y=12\,\,\,\,\,\Rightarrow \,\,\,\,\,\,y=6\] |
| Substituting in (3), \[x=10\] |
| Therefore, length \[=10\,m\] and breadth \[=6\,m\]. |
| So, option [b] is correct. |
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