SECTION - E (Case Study Based - 4 Marks Each)
Each case study has three sub-parts (i), (ii), (iii) of marks 1, 1, 2 respectively.
Q.36
Linear Equations
[4 Marks]
A student entrepreneur started a company that manufactures sanitizers in two sizes - small and large. The cost of a small bottle of sanitizer is ₹10 and that of a large bottle is ₹15. In June, the company sold 1000 bottles and recorded a total sale of ₹12,750. Seeing the increased demand, the company decided to increase the price of both sanitizer bottles by ₹2 each. In the next month, the company sold 2500 bottles and recorded a total sale of ₹34,250.
Based on the above information, answer the following questions:
(i) Form a linear equation in two variables representing the sale for June. [1]
(ii) Form a linear equation in two variables representing the sale for July. [1]
(iii)(A) How many sanitizer bottles of each type were actually sold in June? [2]
(i) Form a linear equation in two variables representing the sale for June. [1]
(ii) Form a linear equation in two variables representing the sale for July. [1]
(iii)(A) How many sanitizer bottles of each type were actually sold in June? [2]
OR
(iii)(B) How many sanitizer bottles of each type were sold in July? [2]
💡 Hint: Let $x$ be the number of small bottles and $y$ be the number of large bottles. You will need TWO equations for each month: one for the total quantity ($x + y$) and one for the total sales amount ($P_1 x + P_2 y = \text{Total}$). (Ref: www.maheshkumarsir.in)
Solution 36 (i, ii & iii-A)
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Let the number of small bottles sold be $x$ and large bottles sold be $y$.
Part (i): June Sales Equation
Prices in June: Small = ₹10, Large = ₹15.
Total bottles: $x + y = 1000$ ...(1)
Total sales: $10x + 15y = 12750$
Dividing by 5: $2x + 3y = 2550$ ...(2) [1 Mark]
Part (ii): July Sales Equation
New prices (increased by ₹2): Small = ₹12, Large = ₹17.
Total bottles: $x + y = 2500$
Total sales: $12x + 17y = 34250$ [1 Mark]
Part (iii)(A): Bottles sold in June
From eq (1), $y = 1000 - x$.
Substitute in eq (2):
$2x + 3(1000 - x) = 2550$
$2x + 3000 - 3x = 2550$
$-x = 2550 - 3000 = -450$
$x = 450$ [1 Mark]
Substitute $x=450$ into $y = 1000 - x$:
$y = 1000 - 450 = 550$.
$\therefore \text{Small} = 450, \text{Large} = 550$ [1 Mark]
Solution 36 (iii-B)
www.maheshkumarsir.in
For July, let small bottles be $a$ and large bottles be $b$.
Equations from Part (ii):
$a + b = 2500 \Rightarrow b = 2500 - a$
$12a + 17b = 34250$
Substitute the value of $b$:
$12a + 17(2500 - a) = 34250$
$12a + 42500 - 17a = 34250$ [1 Mark]
$-5a = 34250 - 42500$
$-5a = -8250$
$a = \frac{8250}{5} = 1650$
Substitute $a = 1650$ to find $b$:
$b = 2500 - 1650 = 850$.
$\therefore \text{Small} = 1650, \text{Large} = 850$ [1 Mark]
Q.37
Circles
[4 Marks]
In a technology park in Hyderabad, a company installed a circular meditation garden for its employees as shown in the figure. Arjun, a designer, stood at a point P outside the garden and drew two tangents PA and PB to the circular boundary. The radius of the garden is $9\text{ m}$, and the distance of point P from the centre O is $18\text{ m}$. For preparing safety guidelines, Arjun needed the lengths of the tangents. During a demonstration to interns, he asked them to compute the angle formed between the two tangents using the properties of circle.
Based on the above information, answer the following questions:
(i) What is the measure of $\angle OAP$? [1]
(ii) Calculate the length of the tangent $PA$. [1]
(iii)(A) Calculate the measure of $\angle AOB$. [2]
(i) What is the measure of $\angle OAP$? [1]
(ii) Calculate the length of the tangent $PA$. [1]
(iii)(A) Calculate the measure of $\angle AOB$. [2]
OR
(iii)(B) In the above design, if PA and PB were inclined at $60^\circ$, then what would be their length? [2]
Solution 37 (i, ii & iii-A)
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Part (i): Measure of $\angle OAP$
Since the radius is always perpendicular to the tangent at the point of contact:
$\angle OAP = 90^\circ$ [1 Mark]
Part (ii): Length of tangent PA
In right-angled $\Delta OAP$ (right angled at $A$):
$OP^2 = OA^2 + PA^2$ (Pythagoras Theorem)
$18^2 = 9^2 + PA^2$
$324 = 81 + PA^2 \Rightarrow PA^2 = 243$
$PA = \sqrt{243} = \sqrt{81 \times 3}$
$PA = 9\sqrt{3}\text{ m}$ [1 Mark]
Part (iii)(A): Measure of $\angle AOB$
In right $\Delta OAP$:
$\sin(\angle APO) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{OA}{OP} = \frac{9}{18} = \frac{1}{2}$
Since $\sin(30^\circ) = \frac{1}{2}$, $\angle APO = 30^\circ$. [1 Mark]
The line joining the external point to the centre bisects the angle between the tangents.
$\therefore \angle APB = 2 \times 30^\circ = 60^\circ$.
In quadrilateral $OAPB$, the sum of opposite angles $\angle AOB + \angle APB = 180^\circ$.
$\angle AOB = 180^\circ - 60^\circ$.
$\therefore \angle AOB = 120^\circ$ [1 Mark]
Solution 37 (iii-B)
www.maheshkumarsir.in
Given angle between tangents, $\angle APB = 60^\circ$.
The line $OP$ bisects $\angle APB$.
$\Rightarrow \angle APO = \frac{60^\circ}{2} = 30^\circ$. [1 Mark]
In right-angled $\Delta OAP$:
We need the adjacent side $PA$, and we know the opposite side $OA = 9\text{ m}$.
$\tan(\angle APO) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{OA}{PA}$
$\tan(30^\circ) = \frac{9}{PA}$
$\frac{1}{\sqrt{3}} = \frac{9}{PA}$
$PA = 9\sqrt{3}\text{ m}$ [1 Mark]
Note: Since lengths of tangents from an external point are equal, $PB = 9\sqrt{3}\text{ m}$ as well.
Q.38
Statistics
[4 Marks]
A school conducted a weekly test for Class X students, before the commencement of the pre-board examination and recorded their scores (out of 50). To analyse performance patterns, the academic coordinator grouped the marks into intervals. The grouped frequency distribution is as below:
The academic coordinator computed the central tendencies to judge overall learning level, the most common performance range and to understand consistency across the batch for planning the remedial sessions.
| Marks Obtained | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 |
|---|---|---|---|---|---|
| Number of students | 3 | 6 | 12 | 15 | 14 |
The academic coordinator computed the central tendencies to judge overall learning level, the most common performance range and to understand consistency across the batch for planning the remedial sessions.
Based on the above information, answer the following questions:
(i) Identify the class with the most common performance range. [1]
(ii) Find the class interval containing the median. [1]
(iii)(A) What is the average performance of the students? [2]
(i) Identify the class with the most common performance range. [1]
(ii) Find the class interval containing the median. [1]
(iii)(A) What is the average performance of the students? [2]
OR
(iii)(B) Find the mode of the data. [2]
💡 Hint: "Most common" means the Modal Class (highest frequency). For the Median class, find the cumulative frequencies (CF) and look for the class where CF is just greater than $N/2$. For "average performance", calculate the Mean using $\sum f_i x_i / \sum f_i$.
Solution 38 (i, ii & iii-A)
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Part (i): Most common performance range
This refers to the Modal Class (the class with the highest frequency).
Highest frequency is $15$.
$\therefore \text{Modal Class is } 30 - 40$ [1 Mark]
Part (ii): Class interval containing median
Cumulative Frequencies (CF): $3, 9, 21, 36, 50$.
Total students $N = 50 \Rightarrow \frac{N}{2} = 25$.
The CF just greater than $25$ is $36$, which corresponds to the class $30 - 40$.
$\therefore \text{Median Class is } 30 - 40$ [1 Mark]
Part (iii)(A): Average (Mean) performance
Class Marks ($x_i$): $5, 15, 25, 35, 45$.
$f_i x_i = (3 \times 5) + (6 \times 15) + (12 \times 25) + (15 \times 35) + (14 \times 45)$
$f_i x_i = 15 + 90 + 300 + 525 + 630 = 1560$ [1 Mark]
$\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1560}{50} = 31.2$
$\therefore \text{Average performance is } 31.2\text{ marks}$ [1 Mark]
Solution 38 (iii-B)
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Modal Class is $30 - 40$ (Highest frequency $= 15$).
Lower limit ($l$) $= 30$
Class size ($h$) $= 10$
Frequency of modal class ($f_1$) $= 15$
Frequency of preceding class ($f_0$) $= 12$
Frequency of succeeding class ($f_2$) $= 14$ [1 Mark]
Mode Formula:
$\text{Mode} = l + \left[ \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right] \times h$
$= 30 + \left[ \frac{15 - 12}{2(15) - 12 - 14} \right] \times 10$
$= 30 + \left[ \frac{3}{30 - 26} \right] \times 10$
$= 30 + \left[ \frac{3}{4} \right] \times 10$
$= 30 + 0.75 \times 10$
$= 30 + 7.5 = 37.5$
$\therefore \text{Mode is } 37.5\text{ marks}$ [1 Mark]
