Chapter 4 Class 12 Determinants
Serial order wise

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.

 

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Question 1

(i) x + y + z =  _______.

(a) 800  

(b) 900 

(c) 1000  

(d) 1200

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Question 2

(ii) 4x + y + 3z =  _______.

(a) 1600  

(b) 2300

(c) 900  

(d) 1200

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Question 3

The value of y is  _______.

(a) 200  

(b) 250

(c) 300  

(d) 350

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Question 4

The value of 2x + 3y is  _______.

(a) 1000  

(b) 1100

(c) 1200  

(d) 1300

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Question 5

y – x =  _______.

(a) 100  

(b) 200

(c) 300  

(d) 400

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Transcript

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)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.