Question 3 - Case Based Questions (MCQ) - Chapter 4 Class 12 Determinants

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Question Two schools Oxford and Navdeep want to award their selected students on the values of sincerity, truthfulness and helpfulness. Oxford wants to award ₹ x each, ₹ y each and ₹ z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹ 1600. Navdeep wants to spend ₹ 2300 to award its 4, 1 and 3 students on the respective values (by giving the same amount to the three values as before). The total amount of the award for one prize on each is ₹ 900. Given that Value of award for sincerity = Rs x Value of award for truthfulness = Rs y Value of award for helpfulness = Rs z Given that Total amount of the award for one prize on each is ₹ 900 x + y + z = 900 Also, Oxford wants to award ₹ x each, ₹ y each and ₹ z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹ 1600 3x + 2y + z = 1600 And, Navdeep wants to spend ₹ 2300 to award its 4, 1 and 3 students on the respective values. 4x + y + 3z = 2300 Question 1 (i) x + y + z = _______. (a) 800 (b) 900 (c) 1000 (d) 1200 From (1) x + y + z = 900 So, the correct answer is (B) Question 2 (ii) 4x + y + 3z = _______. (a) 1600 (b) 2300 (c) 900 (d) 1200 From (3) 4x + y + 3z = 2300 So, the correct answer is (B) Question 3 The value of y is _______. (a) 200 (b) 250 (c) 300 (d) 350 Now, our equations are x + y + z = 900 3x + 2y + z = 1600 4x + y + 3z = 2300 Writing equation as AX = B [■8(1&1&1@3&2&1@4&1&3)] [■8(𝑥@𝑦@𝑧)] = [■8(900@1600@2300)] Hence A = [■8(1&1&1@3&2&1@4&1&3)], X = [■8(𝑥@𝑦@𝑧)] & B = [■8(900@1600@2300)] Calculating |A| |A| = |■8(1&1&1@3&2&1@4&1&3)| = 1 |■8(2&1@1&3)| – 1 |■8(3&1@4&3)| + 1 |■8(3&2@4&1)| = 1(6 − 1) − 1 (9 − 4) + 1 (3 − 8) = 1 (5) − 1 (5) + 1 (–5) = 5 − 5 − 5 = −5 Thus, |A| ≠ 0 ∴ The system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Calculating A-1 A-1 = 1/(|A|) adj (A) adj (A) = [■8(A_11&A_12&A_13@A_21&A_22&A_23@A_31&A_32&A_33 )]^′ = [■8(A_11&A_21&A_31@A_12&A_22&A_23@A_13&A_32&A_33 )] And, A = [■8(1&1&1@3&2&1@4&1&3)] M11 = [■8(2&1@1&3)] = 6 − 1 = 5 M12 = [■8(3&1@4&3)] = 9 − 4 = 5 M13 = [■8(3&2@4&1)] = 3 − 8 = −5 M21 = [■8(1&1@1&3)] = 3 − 1 = 2 M22 = [■8(1&1@4&3)] = 3 − 4 = −1 M23 = [■8(1&1@4&1)] = 1 − 4 = −3 M31 = [■8(1&1@2&1)] = 1 − 2 = −1 M32 = [■8(1&1@3&1)] = 1 − 3 = −2 M33 = [■8(1&1@3&2)] = 2 − 3 = −1 Now, A11 = (–1)1+1 . M11 = (–1)2 . (5) = 5 A12 = (–1)1+2 . M12 = (–1)3 . (5) = −5 A13 = (–1)1+3 . M13 = (–1)4 . (−5) = −5 A21 = (–1)2+1 . M21 = (–1)3 . (2) = −2 A22 = (–1)2+2 . M22 = (–1)4 . (−1) = −1 A23 = (–1)2+3 . M23 = (–1)5 . (−3) = 3 A31 = (–1)3+1 . M31 = (–1)4 . (−1) = −1 A32 = (–1)3+2 . M32 = (–1)5 . (−2) = 2 A33 = (–1)3+3 . M33 = (–1)6 . (−1) = −1 Thus, adj (A) =[■8(5&−2&−1@−5&−1&2@−5&3&−1)] Now, A-1 = 1/(|A|) adj A Putting values = 1/(−5) [■8(5&−2&−1@−5&−1&2@−5&3&−1)] = 1/5 [■8(−5&2&1@5&1&−2@5&−3&1)] Also, X = A-1 B Putting values [█(■8(𝑥@𝑦)@𝑧)] = 1/5 [■8(−5&2&1@5&1&−2@5&−3&1)][■8(900@1600@2300)] [█(■8(𝑥@𝑦)@𝑧)] = 1/5 [■8(−5(900)+2(1600)+1(2300)@5(900)+1(1600)+(−2)(2300)@5(900)+(−3)(1600)+1(2300) )] [█(■8(𝑥@𝑦)@𝑧)] = 1/5 [■8(−4500+3200+2300@4500+1600−4600@4500−4800+2300)] = 1/5 [█(■8(1000@1500)@2000)] [█(■8(𝑥@𝑦)@𝑧)] = [█(■8(200@300)@400)] Hence, x = 200, y = 300 & z = 400 Since y = 300 So, the correct answer is (C) Question 4 The value of 2x + 3y is _______. (a) 1000 (b) 1100 (c) 1200 (d) 1300 Since, x = 200, y = 300 & z = 400 Thus, 2x + 3y = 2(200) + 3(300) = 400 + 900 = 1300 So, the correct answer is (d) Question 5 y – x = _______. (a) 100 (b) 200 (c) 300 (d) 400 Since, x = 200, y = 300 & z = 400 Thus, y − x = 300 − 200 = 100 So, the correct answer is (A)