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All questions of Quadratic Equation for Banking Exams Exam

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x² – 31x + 234 = 0
y² – 34y + 285 = 0
  • a)
    X > Y
  • b)
    X < Y
  • c)
    X ≥ Y
  • d)
    X ≤ Y
  • e)
    X = Y or relation cannot be established
Correct answer is option 'E'. Can you explain this answer?

Kishore Kumar answered
X^2-31X+234=0 ;

By factor method,

The roots of the equation are 18,13.i.e.,X = 18 or 13 ;

Y^2-34Y+285=0 ;

By factor method,

The roots of the equation are 19,15.i.e.,Y = 19 or 15 ;

Now compare both the roots,

X = 18 < 19="" ;="" x="13" />< 19="" />
X = 18 > 15 ; X = 13 < 15="" />

While comparing X & Y , X or Y should be either greater or lesser than X & Y .

But here in this question , Both X > Y & X < y="" is="" present="" ,="" so="" x="y" or="" relationship="" cannot="" be="" established.="" y="" is="" present="" ,="" so="" x="Y" or="" relationship="" cannot="" be="" />

Ravi, Hari and Sanjay are three typists, who working simultaneously, can type 228 pages in four hours. In one hour, Sanjay can type as many pages more than Hari as Hari can type more than Ravi. During a period of five hours, Sanjay can type as many passages as Ravi can, during seven hours.
Quantity I: Number of pages typed by Ravi
Quantity II: Number of pages typed by Hari
  • a)
    Quantity I > Quantity II
  • b)
    Quantity I < Quantity II
  • c)
    Quantity I ≥ Quantity II
  • d)
    Quantity I ≤ Quantity II
  • e)
    Quantity I = Quantity II or relation cannot be established
Correct answer is option 'B'. Can you explain this answer?

Sagar Sharma answered
Let's assume that Ravi can type x pages per hour.

Since Hari can type more pages than Ravi, Hari can type (x + y) pages per hour, where y is a positive number.

And since Sanjay can type more pages than Hari, Sanjay can type (x + y + z) pages per hour, where z is a positive number.

According to the given information, Ravi, Hari, and Sanjay working simultaneously can type 228 pages in four hours. So, in one hour, they can type 228/4 = 57 pages.

We can set up the following equation based on the information given:

x + (x + y) + (x + y + z) = 57

Simplifying the equation, we get:

3x + 2y + z = 57 ------(1)

During a period of five hours, Sanjay can type as many pages as Ravi can. So, in five hours, they can type 5(x + y + z) pages.

During a period of seven hours, Sanjay can type as many pages as Ravi can. So, in seven hours, they can type 7(x + y) pages.

We can set up the following equation based on the above information:

5(x + y + z) = 7(x + y)

Simplifying the equation, we get:

5x + 5y + 5z = 7x + 7y

2x + 2y + 5z = 0 ------(2)

Now, we have two equations:

3x + 2y + z = 57 ------(1)
2x + 2y + 5z = 0 ------(2)

To solve these equations, we can eliminate y by subtracting equation (1) from equation (2):

(2x + 2y + 5z) - (3x + 2y + z) = 0 - 57

-x + 4z = -57

x - 4z = 57 ------(3)

Now, we have two equations:

x - 4z = 57 ------(3)
2x + 2y + 5z = 0 ------(2)

We can solve these equations to find the values of x and z.

Multiplying equation (3) by 2, we get:

2x - 8z = 114 ------(4)

Subtracting equation (4) from equation (2), we get:

(2x + 2y + 5z) - (2x - 8z) = 0 - 114

10z = -114

z = -11.4

Since z is a positive number, the assumption that Sanjay can type more pages than Hari is incorrect.

Therefore, we cannot determine the values of pages typed by Ravi and Hari using the given information.

Hence, the answer is (E) Quantity I cannot be determined from the information given.

x² – 37x + 322 = 0
y² – 25y + 156 = 0
  • a)
    X > Y
  • b)
    X < Y
  • c)
    X ≥ Y
  • d)
    X ≤ Y
  • e)
    X = Y or relation cannot be established
Correct answer is option 'A'. Can you explain this answer?

Sagar Sharma answered
Given Quadratic Equations:
- x² - 37x + 322 = 0
- y² - 25y + 156 = 0

Comparing the Equations:
- x² - 37x + 322 = (x - 19)(x - 17)
- y² - 25y + 156 = (y - 12)(y - 13)

Finding the Solutions:
- x = 19 or x = 17
- y = 12 or y = 13

Comparing x and y:
Since x can be either 19 or 17, and y can be either 12 or 13,
we can see that x is always greater than y in both cases.
Therefore, the correct answer is:

X > Y

Smallest side of a right-angled triangle is 13 cm less than the side of a square of perimeter 72 cm. The second largest side of the right angled triangle is 2 cm less than the length of the rectangle of area 112 cm² and breadth 8 cm.
Quantity I: Six more than the side of the right-angled triangle(not the smallest)
Quantity II: The Sum of Hypotenuse of the right-angled triangle and the smallest side
  • a)
    Quantity I > Quantity II
  • b)
    Quantity I < Quantity II
  • c)
    Quantity I ≥ Quantity II
  • d)
    Quantity I ≤ Quantity II
  • e)
    Quantity I = Quantity II or relation cannot be established
Correct answer is option 'E'. Can you explain this answer?

Rajeev Kumar answered
Side of square = 72/4 = 18 cm
Smallest side of the right angled triangle = 18 – 13 = 5 cm
Length of rectangle = 112/8 = 14 cm
Second side of the right angled triangle = 14 – 2 = 12 cm
Hypotenuse of the right angled triangle = √(25 + 144) = 13cm

x² = 121
y² – 46y + 529 = 0
  • a)
    X > Y
  • b)
    X < Y
  • c)
    X ≥ Y
  • d)
    X ≤ Y
  • e)
    X = Y or relation cannot be established
Correct answer is option 'B'. Can you explain this answer?

Sagar Sharma answered
Analysis:
The given equation is a quadratic equation in the form of ax^2 + bx + c = 0, where a = 1, b = -46, and c = 529.

Discriminant:
The discriminant of a quadratic equation is given by the formula Δ = b^2 - 4ac. If Δ > 0, the equation has two distinct real roots. If Δ = 0, the equation has one real root. If Δ < 0,="" the="" equation="" has="" no="" real="" />

Calculating Discriminant:
Δ = (-46)^2 - 4*1*529
Δ = 2116 - 2116
Δ = 0

Conclusion:
Since the discriminant is equal to zero, the equation has one real root. This means that X = Y.
Therefore, the correct answer is option 'B' which states that X < y.="" />

Two pipes A and B can fill a tank in 12 hours and 18 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom of the tank it took 48 minutes excess time to fill the cistern.
Quantity I: Due to leakage, time taken to fill the tank
Quantity II: Time taken to empty the full cistern
  • a)
    Quantity I > Quantity II
  • b)
    Quantity I < Quantity II
  • c)
    Quantity I ≥ Quantity II
  • d)
    Quantity I ≤ Quantity II
  • e)
    Quantity I = Quantity II or relation cannot be established
Correct answer is option 'B'. Can you explain this answer?

Kavya Saxena answered
Work done by the two pipes in 1 hour = (1/12)+(1/18) = (15/108).
Time taken by these pipes to fill the tank = (108/15)hrs = 7 hours 12 min.
Due to leakage, time taken to fill the tank = 7 hours 12 min + 48 min = 8 hours
Work done by two pipes and leak in 1 hour = 1/8.
Work done by the leak in 1 hour =(15/108)-(1/8)=(1/72).
Leak will empty the full cistern in 72 hours.

Ajith can do a piece of work in 10 days, Bala in 15 days. They work together for 5 days, the rest of the work is finished by Chand in two more days. They get Rs. 6000 as wages for the whole work.
Quantity I: What is the sum of Rs.100 and the daily wage of Bala?
Quantity II: What is the daily wage of Chand?
  • a)
    Quantity I > Quantity II
  • b)
    Quantity I < Quantity II
  • c)
    Quantity I ≥ Quantity II
  • d)
    Quantity I ≤ Quantity II
  • e)
    Quantity I = Quantity II or relation cannot be established
Correct answer is option 'E'. Can you explain this answer?

Rhea Reddy answered
Ajith’s 5 days work = 50%
Bala’s 5 days work = 33.33%
Chand’s 2 days work = 16.66%[100- (50+33.33)] Ratio of contribution of work of Ajith, Bala and Chand = 3 : 2 : 1
Ajith’s total share = Rs. 3000
Bala’s total share = Rs. 2000
Chand’s total share = Rs. 1000
Ajith’s one day’s earning = Rs.600
Bala’s one day’s earning = Rs.400
Chand’s one day’s earning = Rs.500

x² – 26x + 168 = 0
y² – 32y + 252 = 0
  • a)
    X > Y
  • b)
    X < Y
  • c)
    X ≥ Y
  • d)
    X ≤ Y
  • e)
    X = Y or relation cannot be established
Correct answer is option 'D'. Can you explain this answer?

Sagar Sharma answered
Quadratic Equations Analysis:
- The given quadratic equations are x² - 26x + 168 = 0 and y² - 32y + 252 = 0.

Finding the Roots:
- To determine the relationship between x and y, we need to find the roots of the given equations.
- The roots of x² - 26x + 168 = 0 are x = 14 and x = 12.
- The roots of y² - 32y + 252 = 0 are y = 18 and y = 14.

Comparing the Roots:
- We observe that the roots of y are greater than the roots of x.
- Therefore, the relationship between x and y can be established as x ≤ y.

Conclusion:
- The correct answer is option 'D' which states that x ≤ y.
- This conclusion is based on the comparison of the roots of the given quadratic equations.

A Cistern has an inlet pipe and outlet pipe. The inlet pipe fills the cistern completely in 1 hour 20 minutes when the outlet pipe is plugged. The outlet pipe empties the tank completely in 6 hours when the inlet pipe is plugged.
Quantity I: X = Inlet Pipe Efficiency
Quantity II: Y = Outlet Pipe Efficiency
  • a)
    Quantity I > Quantity II
  • b)
    Quantity I < Quantity II
  • c)
    Quantity I ≥ Quantity II
  • d)
    Quantity I ≤ Quantity II
  • e)
    Quantity I = Quantity II or relation cannot be established
Correct answer is option 'A'. Can you explain this answer?

Sagar Sharma answered
Given information:
- Inlet pipe fills the cistern in 1 hour 20 minutes when the outlet pipe is plugged.
- Outlet pipe empties the tank in 6 hours when the inlet pipe is plugged.

Calculating Efficiency:
- Let the capacity of the cistern be C.
- Inlet pipe fills the cistern in 1.33 hours (1 hour 20 minutes) when the outlet pipe is plugged, so the inlet pipe's efficiency = C/1.33.
- Outlet pipe empties the cistern in 6 hours when the inlet pipe is plugged, so the outlet pipe's efficiency = C/6.

Comparing efficiencies:
- Inlet pipe efficiency = C/1.33
- Outlet pipe efficiency = C/6

Comparing Quantity I and Quantity II:
- Inlet pipe efficiency (X) > Outlet pipe efficiency (Y) as X = C/1.33 and Y = C/6.
- Therefore, Quantity I is greater than Quantity II i.e., X > Y.
Thus, option A) Quantity I > Quantity II is the correct answer.

Chapter doubts & questions for Quadratic Equation - Quantitative Aptitude for Competitive Examinations 2024 is part of Banking Exams exam preparation. The chapters have been prepared according to the Banking Exams exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Banking Exams 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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