Problems on Quadratic Equations — Case-Study Based Questions

Case Study I

Some students planned a picnic. The total budget for hiring a bus was ₹ 1440. Later on, eight of them refused to go and instead paid their total share of money towards the fee of one economically weaker student of their class and thus, the cost for each member who went for picnic is increased by ₹ 30.

1. If x students planned for the picnic, then the share for hiring the bus per student who went for the picnic, was :

1. ₹ 30x

2. ₹ 1440x

3. ₹ $\dfrac{1440}{x}$

4. ₹ $\dfrac{1440}{x - 8}$

2. The algebraic representation of the given information in the form of a quadratic equation is:

1. x² − 8x − 384 = 0

2. x² + 8x − 384 = 0

3. x² − 8x − 184 = 0

4. x² + 8x − 184 = 0

3. How many students went for the picnic?

1. 24

2. 16

3. 32

4. 2

4. How much money was paid towards the fee?

1. ₹ 280

2. ₹ 340

3. ₹ 420

4. ₹ 480

5. What would be the share of each student if all the students had attended the picnic?

1. ₹ 90

2. ₹ 30

3. ₹ 60

4. none of these

Answer

1. Let x be the number of students planned a picnic.

Given,

Budget of hiring a bus = ₹ 1440

After 8 students refused, number of students those who went = x - 8

Share for hiring the bus per student who went for the picnic = $\dfrac{1440}{x - 8}$

Hence, option (4) is the correct option.

2. In first case:

Initially share per student for hiring the bus = $\dfrac{1440}{x}$

Share per student after 8 students refused to go to the picnic = $\dfrac{1440}{x - 8}$

According to question,

Cost per student increases by ₹ 30.

$\Rightarrow \dfrac{1440}{x - 8} - \dfrac{1440}{x} = 30 \\[1em] \Rightarrow \dfrac{1440x - 1440(x - 8)}{x(x - 8)} = 30 \\[1em] \Rightarrow \dfrac{1440x - 1440x + 1440 \times 8}{x^2 - 8x} = 30 \\[1em] \Rightarrow 1440 \times 8 = 30(x^2 - 8x) \\[1em] \Rightarrow \dfrac{1440 \times 8}{30} = x^2 - 8x \\[1em] \Rightarrow 384 = x^2 - 8x \\[1em] \Rightarrow 0 = x^2 - 8x - 384 \\[1em] \Rightarrow x^2 - 8x - 384 = 0$

Hence, option (1) is the correct option.

3. Solving,

⇒ x² - 8x - 384 = 0

⇒ x² - 24x + 16x - 384 = 0

⇒ x(x - 24) + 16(x - 24) = 0

⇒ (x + 16)(x - 24) = 0

⇒ (x + 16) = 0 or (x - 24) = 0     [Using zero-product rule]

⇒ x = -16 or x = 24

⇒ x = 24 [Number of students cannot be negative]

Number of students who went for picnic = x - 8 = 24 - 8 = 16.

Hence, option (2) is the correct option.

4. Share for hiring the bus per student who planned picnic = $\dfrac{1440}{x} = \dfrac{1440}{24}$ = ₹ 60.

Number of students who did not go to picnic = 8

Share of eight persons = 60 × 8 = ₹ 480.

Hence, option (4) is the correct option.

5. Share of each student (initially) = ₹ 60.

Hence, option (3) is the correct option.

Case Study II

A bus travels at a certain average speed for a distance of 75 km and then it travels a distance of 90 km at an average speed of 10 km/hr more than the original speed. If it takes 3 hours to complete the total journey, then based on this information, answer the following questions:

1. If the original speed of the bus be x km/hr, then time taken by the bus to travel the next given distance is:

1. $\dfrac{75}{x}$ hours

2. $\dfrac{90}{x}$ hours

3. $\dfrac{90}{x + 10}$ hours

4. $\dfrac{90}{x - 10}$ hours

2. The quadratic equation for the given information, if the original speed of the bus be x km/hr, is:

1. x² + 45x − 250 = 0
2. x² − 45x − 250 = 0
3. x² − 75x − 450 = 0
4. x² − 45x + 250 = 0

3. The original speed of the bus is:

1. 50 km/hr
2. 40 km/hr
3. 75 km/hr
4. 60 km/hr

4. The speed of the bus during which it travels the distance of 90 km is:

1. 70 km/hr
2. 50 km/hr
3. 60 km/hr
4. 85 km/hr

5. The time taken by the bus to travel a distance of 510 km with the new speed is:

1. 8 hours

2. 8 $\dfrac{1}{2}$ hours

3. 10 $\dfrac{1}{5}$ hours

4. 12 $\dfrac{3}{4}$ hours

Answer

1. Given,

Original speed of the bus = x km/hr

Next given distance = 90

Speed for next distance = (x + 10) km/hr

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

Time taken for next distance = $\dfrac{90}{(x + 10)}$.

Hence, option (3) is the correct option.

2. Time taken by bus to travel 75 km = $\dfrac{75}{x}$

Time taken by bus to travel 90 km = $\dfrac{90}{x + 10}$

Given,

Total time taken to complete the journey = 3 hours

$\Rightarrow \dfrac{75}{x} + \dfrac{90}{x + 10} = 3 \\[1em] \Rightarrow \dfrac{75(x + 10) + 90x}{x(x + 10)} = 3 \\[1em] \Rightarrow \dfrac{75x + 750 + 90x}{x^2 + 10x} = 3 \\[1em] \Rightarrow 750 + 165x = 3(x^2 + 10x) \\[1em] \Rightarrow 0 = 3x^2 + 30x - 165x - 750 \\[1em] \Rightarrow 3x^2 - 135x - 750 = 0 \\[1em] \Rightarrow 3(x^2 - 45x - 250) = 0 \\[1em] \Rightarrow x^2 - 45x - 250 = 0.$

Hence, option (2) is the correct option.

3. Solving,

⇒ x² - 45x - 250 = 0

⇒ x² + 5x - 50x - 250 = 0

⇒ x(x + 5) - 50(x + 5) = 0

⇒ (x + 5)(x - 50) = 0

⇒ (x + 5) = 0 or (x - 50) = 0     [Using zero-product rule]

⇒ x = -5 or x = 50

⇒ x = 50 [As speed cannot be negative]

Speed = 50 km/hr

Hence, option (1) is the correct option.

4. The new speed of bus = x + 10 = 50 + 10 = 60 km/hr.

Hence, option (3) is the correct option.

5. Given,

Distance = 510 km

New speed = 60 km/hr.

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

= $\dfrac{510}{60} = 8\dfrac{1}{2}$ hrs.

Hence, option (2) is the correct option.

Case Study III

Two water taps together fill a tank in 1 $\dfrac{7}{8}$ hours. The tap with larger diameter takes 2 hours less than the tap with smaller one to fill the tank completely. Based on the above information, answer the following questions:

1. If time taken by the tap with smaller diameter to fill the tank alone be x hours, then part of the tank filled by the tap with larger diameter alone in 2 hours is:

1. 2(x + 2)

2. 2(x - 2)

3. $\dfrac{2}{x + 2}$

4. $\dfrac{2}{x - 2}$

2. The quadratic equation representing the given information is:

1. 4x² − 23x + 15 = 0
2. 2x² − 23x + 15 = 0
3. 4x² + 23x − 15 = 0
4. 2x² + 23x − 15 = 0

3. Time taken by the larger tap to fill the tank alone is:

1. 5 hours
2. 3 hours
3. 7 hours
4. none of these

4. The part of the tank which can be filled by the smaller tap in 3 hours is:

1. $\dfrac{1}{3}$

2. $\dfrac{1}{5}$

3. $\dfrac{3}{5}$

4. $\dfrac{3}{7}$

5. The part of the tank which can be filled by the larger tap in $1\dfrac{1}{4}$ hours is:

1. $\dfrac{5}{12}$

2. $\dfrac{3}{5}$

3. $\dfrac{5}{8}$

4. $\dfrac{3}{8}$

Answer

1. Let time taken by smaller tap to fill tank = x hours

Part of tank filled by smaller tap in one hour = $\dfrac{1}{x}$

Time taken by larger tap to fill tank = (x - 2) hours

Part of tank filled by larger tap in one hour $\dfrac{1}{x - 2}$

Part of tank filled by larger tap in two hour $\dfrac{2}{x - 2}$

Hence, option (4) is the correct option.

2. Given,

Together smaller and larger tap take $1\dfrac{7}{8} = \dfrac{15}{8}$ hours to fill tank.

Rate at which the taps fill tank in one hour = $\dfrac{1}{\dfrac{15}{8}} = \dfrac{8}{15}$ tank/hr

From question 1 we have,

Part of tank filled by smaller tap in one hour = $\dfrac{1}{x}$

Part of tank filled by larger tap in one hour = $\dfrac{1}{x - 2}$

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{x - 2} = \dfrac{8}{15} \\[1em] \Rightarrow \dfrac{x - 2 + x}{x^2 - 2x}= \dfrac{8}{15} \\[1em] \Rightarrow 15(2x - 2) = 8(x^2 - 2x) \\[1em] \Rightarrow 30x - 30 = 8x^2 - 16x \\[1em] \Rightarrow 0 = 8x^2 - 16x - 30x + 30 \\[1em] \Rightarrow 8x^2 - 46x + 30 = 0 \\[1em] \Rightarrow 2(4x^2 - 23x + 15) = 0 \\[1em] \Rightarrow 4x^2 - 23x + 15 = 0$

Hence, option (1) is the correct option.

3. Solving equation from question 2,

⇒ 4x² - 23x + 15 = 0

⇒ 4x² - 20x - 3x + 15 = 0

⇒ 4x(x - 5) - 3(x - 5) = 0

⇒ (4x - 3)(x - 5) = 0

⇒ (4x - 3) = 0 or (x - 5) = 0     [Using zero-product rule]

⇒ x = $\dfrac{3}{4}$ or x = 5

Case 1 : x = $\dfrac{3}{4}$

Larger tap take (x - 2) hours = $\dfrac{3}{4} - 2 = \dfrac{3 - 8}{4} = \dfrac{-5}{4}$

= -1.25 hours, which is not possible.

Case 2 : x = 5

The time taken by the smaller tap is x = 5 hours.

The time taken by the larger tap is x − 2 = 5 - 2 = 3 hours.

Hence, option (2) is the correct option.

4. The smaller tap fills the tank in x = 5 hours.

Part of tank filled by smaller tap in one hour = $\dfrac{1}{5}$

Part of tank filled by smaller tap in 3 hours = $\dfrac{3}{5}$

Hence, option (3) is the correct option.

5. The larger tap fills the tank in 3 hours.

Part of tank filled by larger tap in one hour $\dfrac{1}{3}$

Part of tank filled by larger tap in $1\dfrac{1}{4} = \dfrac{5}{4}$ hours

= $\dfrac{\dfrac{5}{4}}{3}$

= $\dfrac{5}{12}$

Hence, option (1) is the correct option.

Case Study IV

A motorboat whose speed in still water is 24km/hr, takes 1 hour more to go 32 km upstream than to return downstream to the same spot . Based on this information answer the following questions

1. What is the speed of the motorboat in going upstream, if speed of the stream is x km/hr?

1. (x − 24) km/hr
2. (24 − x) km/hr
3. (x + 24) km/hr
4. none of these

2. The quadratic equation which represents the given information is:

1. x² − 64x − 576 = 0
2. x² + 64x + 576 = 0
3. x² + 64x − 576 = 0
4. x² − 64x + 576 = 0

3. Speed of the motorboat in going downstream is:

1. 16 km/hr
2. 8 km/hr
3. 28 km/hr
4. 32 km/hr

4. Time taken by the motorboat to go 272 km downstream is:

1. $8\dfrac{1}{2}$ hours

2. 17 hours

3. $12\dfrac{1}{2}$ hours

4. $6\dfrac{1}{2}$ hours

5. Time taken by the motorboat to go 80 km upstream and then to return back to the same spot is:

1. $5\dfrac{1}{2}$ hours

2. $6\dfrac{1}{2}$ hours

3. $7\dfrac{1}{2}$ hours

4. $8\dfrac{1}{2}$ hours

Answer

1. Given,

The speed of the motorboat in still water is 24 km/hr

The speed of the stream be x km/hr

The speed of the motorboat in going upstream = boat speed in still water − stream speed = (24 - x) km/hr

Hence, option (2) is the correct option.

2. The speed of the motorboat in going downstream = boat speed in still water + stream speed = (24 + x) km/hr

Distance to be covered by motorboat = 32km

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

In first case:

Time taken by motorboat in going upstream = $\dfrac{32}{24 - x}$

In second case:

Time taken by motorboat in going downstream = $\dfrac{32}{24 + x}$

The difference in time taken between first case and second case = 1 hour more

$\Rightarrow \dfrac{32}{24 - x} - \dfrac{32}{24 + x} = 1 \\[1em] \Rightarrow \dfrac{32(24 + x) - 32(24 - x)}{(24 - x)(24 + x)}= 1 \\[1em] \Rightarrow \dfrac{768 + 32x - 768 + 32x}{24^2 - x^2}= 1 \\[1em] \Rightarrow \dfrac{64x}{576 - x^2}= 1 \\[1em] \Rightarrow 64x = 576 - x^2 \\[1em] \Rightarrow x^2 + 64x - 576 = 0.$

Hence, option (3) is the correct option.

3. Solving equation from question 2,

⇒ x² + 64x - 576 = 0

⇒ x² - 8x + 72x - 576 = 0

⇒ x(x - 8) + 72(x - 8) = 0

⇒ (x + 72)(x - 8) = 0

⇒ (x + 72) = 0 or (x - 8) = 0     [Using zero-product rule]

⇒ x = -72 or x = 8

⇒ x = 8km/hr [speed of the stream cannot be negative]

Speed of the motorboat downstream = 24 + x = 24 + 8 = 32 km/hr.

Hence, option (4) is the correct option.

4. The downstream speed of motorboat is 32 km/hr (from Question 3).

The time taken by the motorboat to go 272 km downstream = $\dfrac{272}{32} = \dfrac{17}{2}$ = 8.5 hours

Hence, option (1) is the correct option.

5. Speed of the motorboat upstream = 16 km/hr

The time taken by the motorboat to go 80 km upstream = $\dfrac{80}{16}$ = 5 hours

The time taken by the motorboat to return 80 km downstreamstream = $\dfrac{80}{32}$ = 2.5 hours

Total time = 5 + 2.5 = 7.5 hours

Hence, option (3) is the correct option.