Check sibling questions
Chapter 4 Class 12 Determinants
Serial order wise

Manjit wants to donate a rectangular plot of land for a school in his village. When he was asked to give dimensions of the plot, he told that if its length is decreased by 50 m and breadth is increased by 50m, then its area will remain same, but if length is decreased by 10m & breadth is decreased by 20m, then area will decrease by 5300 m 2
Manjit Wants to Donate a Rectungler - Teachoo.jpg


Based on the information given above, answer the following questions:

Slide2.JPG

Question 1

The equations in terms of X and Y are

(a) x – y = 50, 2x – y = 550

(b) x – y = 50, 2x + y = 550

(c) x + y = 50, 2x + y = 550

(d) x + y = 50, 2x + y = 550

Slide3.JPG Slide4.JPG Slide5.JPG

 

Question 2

Which of the following matrix equation is represented by the given information

(a) [1 -1] [2 1] [x y] = [50 550] 

(b) [1 1] [2 1] [x y] = [50 550] 

(c) [1 1] [2 -1] [x y] = [50 550]

(d) [1 -2] [2 1] [x y] = [-50 -550] 

Slide6.JPG

 

Question 3

The value of x (length of rectangular field) is

(a) 150 m

(b) 400 m

(c) 200 m

(d) 320 m

 

Slide7.JPG

 

Question 4

The value of y (breadth of rectangular field) is

(a) 150m

(b) 200m

(c) 430m

(d) 350m

 

Slide8.JPG

Question 5

How much is the area of rectangular field?

(a) 60000 sq. m.

(b) 30000 sq. m.

(c) 30000 m

(d) 3000 m

Slide9.JPG

Learn Intergation from Davneet Sir - Live lectures starting soon!


Transcript

Question Manjit wants to donate a rectangular plot of land for a school in his village. When he was asked to give dimensions of the plot, he told that if its length is decreased by 50 m and breadth is increased by 50m, then its area will remain same, but if length is decreased by 10m & breadth is decreased by 20m, then area will decrease by 5300 m2 Based on the information given above, answer the following questions: Question 1 The equations in terms of X and Y are (a) x – y = 50, 2x – y = 550 (b) x – y = 50, 2x + y = 550 (c) x + y = 50, 2x + y = 550 (d) x + y = 50, 2x + y = 550 Let Length of plot = x m Breadth of plot = y m Now, Area of plot = xy m2 Given that if its length is decreased by 50 m and breadth is increased by 50m, then its area will remain same (Length − 50) × (Breadth + 50) = Area (x − 50) × (y + 50) = xy x (y + 50) − 50 (y + 50) = xy xy + 50x − 50y − 2500 = xy 50x − 50y − 2500 = 0 50x − 50y = 2500 Dividing both sides by 50 x − y = 50 Also, if length is decreased by 10m & breadth is decreased by 20m, then area will decrease by 5300 m2 (Length − 10) × (Breadth − 20) = Area − 5300 (x − 10) × (y − 20) = xy − 5300 x (y − 20) − 10 (y − 20) = xy − 5300 xy − 20x − 10y + 200 = xy − 5300 −20x − 10y + 200 = −5300 −20x − 10y = −5300 − 200 −20x − 10y = −5500 20x + 10y = 5500 Dividing both sides by 10 2x + y = 550 Thus, our equations are x – y = 50, 2x + y = 550 So, the correct answer is (b) Question 2 Which of the following matrix equation is represented by the given information (a) [■8(1&−[email protected]&1)] [■8(𝑥@𝑦)] = [■8([email protected])] (b) [■8(1&[email protected]&1)] [■8(𝑥@𝑦)] = [■8([email protected])] (c) [■8(1&[email protected]&−1)] [■8(𝑥@𝑦)] = [■8([email protected])] (d) [■8(1&[email protected]&1)] [■8(𝑥@𝑦)] = [■8(−[email protected]−550)] Since our equations are x – y = 50 2x + y = 550 Matrix equation will look like [■8(𝟏&−𝟏@𝟐&𝟏)] [■8(𝒙@𝒚)] = [■8(𝟓𝟎@𝟓𝟓𝟎)] So, the correct answer is (a) Question 3 The value of x (length of rectangular field) is (a) 150 m (b) 400 m (c) 200 m (d) 320 m Since our equations are x – y = 50 …(1) 2x + y = 550 …(1) Adding (1) and (2) (x – y) + (2x + y) = 50 + 550 3x = 600 x = 600/2 x = 200 m So, the correct answer is (c) Question 4 The value of y (breadth of rectangular field) is (a) 150m (b) 200m (c) 430m (d) 350m Putting x = 200 in (1) x − y = 50 200 − y = 50 200 − 50 = y 150 = y y = 150 m So, the correct answer is (a) Question 5 How much is the area of rectangular field? (a) 60000 sq. m. (b) 30000 sq. m. (c) 30000 m (d) 3000 m Area of rectangular field = Length × Breadth = 200 × 150 = 30,000 m2 = 30,000 sq. m So, the correct answer is (b)

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.