Question 4 - Case Based Questions (MCQ) - Chapter 3 Class 12 Matrices (Term 1)

Last updated at July 28, 2021 by Teachoo

On her birth day, Seema decided to donate some money to children of an orphanage home. If there were 8 children less, everyone would have got Rs. 10 more. However, if there were 16 children more, everyone would have got Rs. 10 less. Let the number of children be x and the amount distributed by Seema for one child be y (in Rs.).

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

Question On her birth day, Seema decided to donate some money to children of an orphanage home. If there were 8 children less, everyone would have got Rs. 10 more. However, if there were 16 children more, everyone would have got Rs. 10 less. Let the number of children be x and the amount distributed by Seema for one child be y (in Rs.). Based on the information given above, answer the following questions:Question 1 The equations in terms x and y are (a) 5x – 4y = 40 5x – 8y = –80 (b) 5x – 4y = 40 5x – 8y = 80 (c) 5x – 4y = 40 5x + 8 y = –80 (d) 5x + 4y = 40 5x – 8y = –80(a) 5x – 4y = 40 5x – 8y = –80
Let the Number of children = x
Amount distributed by Seema for one child = Rs y
Now,
Total money = xy
Given that
If there were 8 children less, everyone would have got Rs. 10 more.
Total money now = Total money before
(x − 8) × (y + 10) = xy
x(y + 10) − 8 (y + 10) = xy
xy + 10x − 8y − 80 = xy
10x − 8y − 80 = 0
10x − 8y = 80
Dividing both sides by 2
5x − 4y = 40
Also,
if there were 16 children more, everyone would have got Rs. 10 less
Total money now = Total money before
(x + 16) × (y − 10) = xy
x(y − 10) + 16(y − 10) = xy
xy − 10x + 16y − 160 = xy
−10x + 16y − 160 = 0
10x − 16y + 160 = 0
10x − 16y = −160
Dividing both sides by 2
5x − 8y = −80
Thus, the equations are
5x – 4y = 40
5x – 8y = –80
So, the correct answer is (a)
Question 2 Which of the following matrix equations represent the information given above? (a) [■8(5&4@5&8)] [■8(𝑥@𝑦)] = [■8(40@−80)] (b) [■8(5&−4@5&−8)] [■8(𝑥@𝑦)] = [■8(40@80)] (c) [■8(5&−4@5&−8)] [■8(𝑥@𝑦)] = [■8(40@−80)] (d) [■8(5&4@5&−8)] [■8(𝑥@𝑦)] = [■8(40@−80)]
Since the equations are
5x – 4y = 40
5x – 8y = –80
We write it as
[■8(5&−4@5&−8)] [■8(𝑥@𝑦)] = [■8(40@−80)]
So, the correct answer is (c)
Question 3 The number of children who were given some money by Seema, is (a) 30 (b) 40 (c) 23 (d) 32
We need to find x
The equation is
5x – 4y = 40 …(1)
5x – 8y = –80 …(2)
Subtracting (1) and (2)
(5x – 4y) − (5x – 8y) = 40 − (−80)
−4y + 8y = 120
4y = 120
So, the correct answer is (a)
y = 120/4
y = 30
Putting y = 30 in (1)
5x – 4y = 40
5x − 4(30) = 40
5x − 120 = 40
5x = 40 + 120
5x = 160
x = 160/5
x = 32
Thus,
Number of children = x = 32
So, the correct answer is (d)
Question 4 How much amount is given to each child by Seema? (a) Rs. 32 (b) Rs. 30 (c) Rs. 62 (d) Rs. 26
Amount given to each child = Rs y
= Rs. 30
So, the correct answer is (b)
Question 5 How much amount Seema spends in distributing the money to all the students of the Orphanage? (a) Rs. 609 (b) Rs. 960 (c) Rs. 906 (d) Rs. 690
Total Amount = Number of students × Money spent per student
= xy
= 32 × 30
= Rs 960
So, the correct answer is (b)

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.