Question 4 - Case Based Questions (MCQ) - Chapter 3 Class 12 Matrices
Last updated at August 8, 2026 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)
Made by
Davneet Singh
Davneet Singh is an IIT Kanpur graduate and has been teaching for 16+ years. At Teachoo, he breaks down Maths, Science and Computer Science into simple steps so students understand concepts deeply and score with confidence.
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