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Maths Case study (4 Marks)

Pair of Linear Equations in Two variables · CBSE STD 10 (CBSE - English Medium). 5 questions.

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Case study (4 Marks)

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5 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
Raman usually go to a dry fruit shop with his mother. He observes the following two situations.
On 1st day: The cost of 2 kg of almonds and 1 kg of cashew was ₹1600.
On 2nd day: The cost of 4 kg of almonds and 2 kg of cashew was ₹3000.
Denoting the cost of 1 kg almonds by x and cost of 1 kg cashew by y, answer the following questions
  1. Linear equations represented by day-I and day -II situations, are
  2. Represent algebraically the situation of day-II.
  3. The linear equation represented by day-1, intersect the x axis at
    Or
    The linear equation represented by day-II, intersect the y-axis at
Answer
1. parallel

2. 2x+y=1500

3. (800, 0)
Or
(0, 1500)

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Question 24 Marks
Mr Manoj Jindal arranged a lunch party for some of his friends. The expense of the lunch are partly constant and partly proportional to the number of guests. The expenses amount to 650 for 7 guests and 970 for 11 guests.
Denote the constant expense by ₹ x and proportional expense per person by ₹ y and answer the following questions.
  1. What is the system of linear equations representing both the situations?
  2. Represent both the situations algebraically.
  3. What is the Proportional expense for each person?
    Or
    The fixed (or constant) expense for the party is?
Answer
1. unique solution

2. x+7y=650, x + 11y=970

3. ₹80
Or
₹90

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Question 34 Marks
Points A and B representing Chandigarh and Kurukshetra respectively are almost 90 km apart from each other on the highway. A car starts from Chandigarh and another from Kurukshetra at the same time. If these cars go in the same direction, they meet in 9 hours and if these cars go in opposite direction they meet in 9/7 hours. Let X and Y be two cars starting from points A and B respectively and their speed be x km/hr and y km/hr respectively.
Then, answer the following questions.
  1. When both cars move in the same direction, then what situation can be represented as algebraically?
  2. When both cars move in opposite direction, then what situation can be represented as algebraically?
  3. what is the speed of car X?
    Or
    what is the speed of car Y?
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Question 44 Marks
From Bengaluru bus stand, if Riddhima buys 2 tickets to Malleswaram and 3 tickets to Yeswanthpur, then total cost is ₹46; but if she buys 3 tickets to Malleswaram and 5 tickets to Yeswanthpur, then total cost is ₹74.
Consider the fares from Bengaluru to Malleswaram and that to Yeswanthpur as x and y respectively and answer the following questions.


  1. what is the system of linear equations represented by both situations?
  2. 1st situation can be represented algebraically as?
  3. 2nd situation can be represented algebraically as?
    Or
    What is the Fare from Bengaluru to Malleswaram?
Answer
1. unique solution

2. $1^{ ts }$ situation can be representedalgebraically as
$
2 x+3 y=46
$

3.$2^{\text {nd }}$ situation can be represented algebraically as
$3 x+5 y=74$
Or
Putting the value of $x$ in equation (i), we get $3 y=46-2 \times 8=30 \Rightarrow y=10$
$\therefore \quad$ Fare from Bengaluru to Yeswanthpur is $₹ 10$.

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Question 54 Marks
A part of monthly hostel charges in a college is fixed and the remaining depends on the number of days one has taken food in the mess. When a student Anu takes food for 25 days, she has to pay 4500 as hostel charges, whereas another student Bindu who takes food for 30 days, has to pay 5200 as hostel charges.
Considering the fixed charges per month by x and the cost of food per day by y, then answer the following questions.
  1. Represent algebraically the situation faced by both Anu and Bindu.
  2. Which system of linear equations is represented in above situations.
  3. What is the cost of food per day?
    Or
    What is the fixed charges per month for the hostel?
Answer
1. For student Anu:
Fixed charge + cost of food for 25 days $=₹ 4500$
i.e., $x+25 y=4500$
For student Bindu:
Fixed charges + cost of food for 30 days $=₹ 5200$
i.e., $x+30 y=5200$2.From above, we have $a_1=1, b_1=25$,
$c_1=-4500$ and $a_2=1, b_2=30, c_2=-5200$
$\therefore \frac{a_1}{a_2}=1, \frac{b_1}{b_2}=\frac{25}{30}=\frac{5}{6}, \frac{c_1}{c_2}=\frac{-4500}{-5200}=\frac{45}{52}$
$\Rightarrow \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
Thus, system of linear equations has unique solution.
3. We have, $x+25 y=4500$ ...(i)
and $x+30 y=5200$ ...(ii)
Subtracting (i) from (ii), we get
$5 y=700 \Rightarrow y=140$
$\therefore \quad$ Cost of food per day is $₹ 140$
Or
We have, $x+25 y=4500$
$\Rightarrow x=4500-25 \times 140$
$\Rightarrow x=4500-3500=1000$
$\therefore \quad$ Fixed charges per month for the hostel is $₹ 1000$
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Case study (4 Marks) - Maths STD 10 Questions - Vidyadip