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

Last updated at April 16, 2024 by Teachoo

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.

Question 1

(i) x + y + z = _______.

(a) 800

(b) 900

(c) 1000

(d) 1200

Question 2

(ii) 4x + y + 3z = _______.

(a) 1600

(b) 2300

(c) 900

(d) 1200

Question 3

The value of y is _______.

(a) 200

(b) 250

(c) 300

(d) 350

Question 4

The value of 2x + 3y is _______.

(a) 1000

(b) 1100

(c) 1200

(d) 1300

Question 5

y – x = _______.

(a) 100

(b) 200

(c) 300

(d) 400

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)

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.

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