RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11 (Updated for 2024)

RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11: The main focus of this exercise is on various applications of linear equations. The RD Sharma Class 10 Solutions created by Kopykitab experts is a powerful tool for students to clear their concerns and analyze their weak areas. Students can also refer to the RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.11 PDF, which is available below.

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RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11

Access answers to RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.11- Important Question with Answers

1. If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the area increases by 33 square units. Find the area of the rectangle.

Solution:

Let’s assume the length and breadth of the rectangle be x units and y units, respectively.

Hence, the area of rectangle = xy sq. units

From the question we have the following cases,

Case 1:

Length is increased by 2 units ⇒ now, the new length is x+2 units

Breadth is reduced by 2 units ⇒ now, the new breadth is y-2 units

And it’s given that the area is reduced by 28 square units i.e., = xy – 28

So, the equation becomes

(x+2)(y−2) = xy − 28

⇒ xy − 2x + 2y – 4 = xy − 28

⇒ −2x + 2y – 4 + 28 = 0

⇒ −2x + 2y + 24 = 0

⇒ 2x − 2y – 24 = 0 ……… (i)

Case 2:

Length is reduced by 1 unit ⇒ now, the new length is x-1 units

Breadth is increased by 2 units ⇒ now, the new breadth is y+2 units

And, it’s given that the area is increased by 33 square units i.e. = i.e. = xy + 33

So, the equation becomes

(x−1)(y+2) = xy + 33

⇒ xy + 2x – y – 2 = x + 33

⇒ 2x – y − 2 − 33 = 0

⇒ 2x – y −35 = 0 ……….. (ii)

Solving (i) and (ii),

By using cross multiplication, we get

x = 46/2

x = 23

And,

y = 22/2

y = 11

Hence,

The length of the rectangle is 23 units.

The breadth of the rectangle is 11 units.

So, the area of the actual rectangle = length x breadth,

= x×y

= 23 x 11

= 253 sq. units

Therefore, the area of rectangle is 253 sq. units.

2. The area of a rectangle remains the same if the length is increased by 7 metres and the breadth is decreased by 3 metres. The area remains unaffected if the length is decreased by 7 metres and the breadth is increased by 5 metres. Find the dimensions of the rectangle.

Solution:

Let’s assume the length and breadth of the rectangle be x units and y units, respectively.

Hence, the area of rectangle = xy sq.units

From the question we have the following cases,

Case 1

Length is increased by 7 metres ⇒ now, the new length is x+7

Breadth is decreased by 3 metres ⇒ now, the new breadth is y-3

And it’s given, the area of the rectangle remains the same i.e. = xy.

So, the equation becomes

xy = (x+7)(y−3)

xy = xy + 7y − 3x − 21

3x – 7y + 21 = 0 ………. (i)

Case 2:

Length is decreased by 7 metres ⇒ now, the new length is x-7

Breadth is increased by 5 metres ⇒ now, the new breadth is y+5

And it’s given that, the area of the rectangle still remains the same i.e. = xy.

So, the equation becomes

xy = (x−7)(y+5)

xy = xy − 7y + 5x − 35

5x – 7y – 35 = 0 ………. (ii)

Solving (i) and (ii),

By using cross-multiplication, we get,

x = 392/14

x = 28

And,

y = 210/14

y = 15

Therefore, the length of the rectangle is 28 m. and the breadth of the actual rectangle is 15 m.

We have provided complete details of RD Sharma Class 10 Solutions Chapter 3 Exercise 3.11. If you have any queries related to CBSE Class 10, feel free to ask us in the comment section below.