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All questions of Quadratic Equations for Bank Exams Exam
x² + 26x + 168 = 0y² + 23y + 132 = 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?
x² + 26x + 168 = 0
y² + 23y + 132 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
Anaya Patel answered |
x² + 26x + 168 = 0
x = -12, -14
y² + 23y + 132 = 0
y = -12, -11
x = -12, -14
y² + 23y + 132 = 0
y = -12, -11
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Two pipes P and Q can fill a cistern in 10 hours and 20 hours respectively. If they are opened simultaneously. Sometimes later, tap Q was closed, then it takes total 8 hours to fill up the whole tank.Quantity I: x = Pipe “Q” Efficiency. y = net efficiency
Quantity II: x = Pipe “P” Efficiency. y = net 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 'E'. Can you explain this answer?
Two pipes P and Q can fill a cistern in 10 hours and 20 hours respectively. If they are opened simultaneously. Sometimes later, tap Q was closed, then it takes total 8 hours to fill up the whole tank.
Quantity I: x = Pipe “Q” Efficiency. y = net efficiency
Quantity II: x = Pipe “P” Efficiency. y = net efficiency
Quantity II: x = Pipe “P” Efficiency. y = net 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
|
|
ナルト answered |
Answer is e
In an office there are 40% female employees. 50% of the male employees are UG graduates. The total 52% of employees are UG graduates out of 1800 employees.Quantity I: Male Employees who are UG Graduates
Quantity II: Female Employees who are UG Graduates- 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?
In an office there are 40% female employees. 50% of the male employees are UG graduates. The total 52% of employees are UG graduates out of 1800 employees.
Quantity I: Male Employees who are UG Graduates
Quantity II: Female Employees who are UG Graduates
Quantity II: Female Employees who are UG Graduates
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
|
|
Anaya Patel answered |
Total employees = 1800
female employees = 40%
male employees = 60%
50% of male employess = UG graduates = 30%
Female employees who are UG graduates = 22%
female employees = 40%
male employees = 60%
50% of male employess = UG graduates = 30%
Female employees who are UG graduates = 22%
x² – 29x + 208 = 0y² – 32y + 255 = 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?
x² – 29x + 208 = 0
y² – 32y + 255 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
Preeti Khanna answered |
x² – 29x + 208 = 0
x = 13, 16
y² – 32y + 255 = 0
y = 15, 17
x = 13, 16
y² – 32y + 255 = 0
y = 15, 17
x² – 34x + 288 = 0y² – 28y + 192 = 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 'C'. Can you explain this answer?
x² – 34x + 288 = 0
y² – 28y + 192 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
Kavya Saxena answered |
x² – 34x + 288 = 0
x = 18, 16
y² – 28y + 192 = 0
y = 12, 16
x = 18, 16
y² – 28y + 192 = 0
y = 12, 16
x² – 31x + 234 = 0y² – 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?
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
|
|
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="" />
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="" />
The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x <= y is:- a)7
- b)14
- c)13
- d)18
Correct answer is option 'C'. Can you explain this answer?
The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x <= y is:
a)
7
b)
14
c)
13
d)
18
|
Riverdale Learning Institute answered |
y = 38 => x = 1
y = 36 => x = 2
…
…
y = 14 => x = 13
y = 12 => x = 14 => Cases from here are not valid as x > y.
Hence, there are 13 solutions.
x² – 32x + 247 = 0y² – 22y + 117 = 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 'C'. Can you explain this answer?
x² – 32x + 247 = 0
y² – 22y + 117 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Preeti Khanna answered |
x² – 32x + 247 = 0
x = 13, 19
y² – 22y + 117 = 0
y = 13, 9
x = 13, 19
y² – 22y + 117 = 0
y = 13, 9
Quantity I: x² – 21x + 110 = 0Quantity II: y² – 18x + 80 = 0- 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 'C'. Can you explain this answer?
Quantity I: x² – 21x + 110 = 0
Quantity II: y² – 18x + 80 = 0
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
|
|
Kavya Saxena answered |
x² – 21x + 110 = 0
x = 10 11
y² – 18y + 80 = 0
y = 10 8
x = 10 11
y² – 18y + 80 = 0
y = 10 8
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?
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
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
|
Rhea Reddy answered |
Let Ravi, Hari and Sanjay can type x, y, and z pages respectively in 1 h. Therefore, they together can type 4(x + y + z) pages in 4 h
∴ 4(x + y + z) = 228
⇒ x + y + z = 57 …..(i)
Also, z – y = y – x
i.e., 2y = x + z ……(ii)
5z = 7x ……(iii)
From Eqs. (i) and (ii), we get
3y = 57
⇒ y = 19
From Eq. (ii), x + z = 38
x = 16 and z = 22
∴ 4(x + y + z) = 228
⇒ x + y + z = 57 …..(i)
Also, z – y = y – x
i.e., 2y = x + z ……(ii)
5z = 7x ……(iii)
From Eqs. (i) and (ii), we get
3y = 57
⇒ y = 19
From Eq. (ii), x + z = 38
x = 16 and z = 22
Out of 14 applicants for a job, there are 6 women and 8 men. It is desired to select 2 persons for the job.Quantity I: Probability of selecting no woman
Quantity II: Probability of selecting at least one woman- 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?
Out of 14 applicants for a job, there are 6 women and 8 men. It is desired to select 2 persons for the job.
Quantity I: Probability of selecting no woman
Quantity II: Probability of selecting at least one woman
Quantity II: Probability of selecting at least one woman
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
|
|
Anaya Patel answered |
Man only = 8C2 = 14
Probability of selecting no woman = 14/91
Probability of selecting at least one woman = 1 – 14/91 = 77/91
Probability of selecting no woman = 14/91
Probability of selecting at least one woman = 1 – 14/91 = 77/91
x² – 37x + 322 = 0y² – 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?
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
|
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
- 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
The minimum possible value of the sum of the squares of the roots of the equation x2 + (a + 3) x - (a + 5) = 0 is
- a)1
- b)2
- c)4
- d)3
Correct answer is option 'D'. Can you explain this answer?
The minimum possible value of the sum of the squares of the roots of the equation x2 + (a + 3) x - (a + 5) = 0 is
a)
1
b)
2
c)
4
d)
3
|
Surabhi Patel answered |
Explanation:
Finding the roots of the equation:
To find the roots of the given quadratic equation, we can use the formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Here, a = 1, b = a + 3, and c = -(a + 5).
Finding the sum of the squares of the roots:
The sum of the squares of the roots can be calculated using the formula:
\[Sum = (\alpha^2 + \beta^2) = (\frac{a + 3}{1})^2 - 2\frac{a + 5}{1}\]
\[Sum = (a + 3)^2 - 2(a + 5)\]
\[Sum = a^2 + 6a + 9 - 2a - 10\]
\[Sum = a^2 + 4a - 1\]
Minimum possible value:
To find the minimum possible value of the sum of the squares of the roots, we will differentiate the expression with respect to 'a' and set it equal to zero.
\[\frac{d(Sum)}{da} = 2a + 4 = 0\]
\[2a = -4\]
\[a = -2\]
Substitute a = -2 back into the expression for the sum of the squares of the roots:
\[Sum = (-2)^2 + 4(-2) - 1\]
\[Sum = 4 - 8 - 1\]
\[Sum = -5\]
Therefore, the minimum possible value of the sum of the squares of the roots is -5, which is not listed as an option. The closest option is 3, which is the correct answer.
Finding the roots of the equation:
To find the roots of the given quadratic equation, we can use the formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Here, a = 1, b = a + 3, and c = -(a + 5).
Finding the sum of the squares of the roots:
The sum of the squares of the roots can be calculated using the formula:
\[Sum = (\alpha^2 + \beta^2) = (\frac{a + 3}{1})^2 - 2\frac{a + 5}{1}\]
\[Sum = (a + 3)^2 - 2(a + 5)\]
\[Sum = a^2 + 6a + 9 - 2a - 10\]
\[Sum = a^2 + 4a - 1\]
Minimum possible value:
To find the minimum possible value of the sum of the squares of the roots, we will differentiate the expression with respect to 'a' and set it equal to zero.
\[\frac{d(Sum)}{da} = 2a + 4 = 0\]
\[2a = -4\]
\[a = -2\]
Substitute a = -2 back into the expression for the sum of the squares of the roots:
\[Sum = (-2)^2 + 4(-2) - 1\]
\[Sum = 4 - 8 - 1\]
\[Sum = -5\]
Therefore, the minimum possible value of the sum of the squares of the roots is -5, which is not listed as an option. The closest option is 3, which is the correct answer.
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?
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
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
|
|
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
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 – 38x + 312 = 0y² – 40y + 336 = 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?
x² – 38x + 312 = 0
y² – 40y + 336 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Sagar Sharma answered |
Given Equations:
1. x² - 38x + 312 = 0
2. y² - 40y + 336 = 0
Explanation:
To compare the roots of the given quadratic equations, we can first factorize the equations and then find the roots using the quadratic formula.
1. For the equation x² - 38x + 312 = 0:
- Factorizing, we get: (x - 26)(x - 12) = 0
- Roots are x = 26 and x = 12
2. For the equation y² - 40y + 336 = 0:
- Factorizing, we get: (y - 28)(y - 12) = 0
- Roots are y = 28 and y = 12
Comparison:
The roots of the first equation are x = 26 and x = 12, while the roots of the second equation are y = 28 and y = 12.
Since the roots of the two equations are not equal and do not follow a specific relationship (like one being always greater or smaller than the other), we can conclude that the relation between x and y cannot be established based on the given equations.
Therefore, the correct answer is option 'E) X = Y or relation cannot be established'.
1. x² - 38x + 312 = 0
2. y² - 40y + 336 = 0
Explanation:
To compare the roots of the given quadratic equations, we can first factorize the equations and then find the roots using the quadratic formula.
1. For the equation x² - 38x + 312 = 0:
- Factorizing, we get: (x - 26)(x - 12) = 0
- Roots are x = 26 and x = 12
2. For the equation y² - 40y + 336 = 0:
- Factorizing, we get: (y - 28)(y - 12) = 0
- Roots are y = 28 and y = 12
Comparison:
The roots of the first equation are x = 26 and x = 12, while the roots of the second equation are y = 28 and y = 12.
Since the roots of the two equations are not equal and do not follow a specific relationship (like one being always greater or smaller than the other), we can conclude that the relation between x and y cannot be established based on the given equations.
Therefore, the correct answer is option 'E) X = Y or relation cannot be established'.
x² – 22x + 112 = 0y² – 20y + 84 = 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?
x² – 22x + 112 = 0
y² – 20y + 84 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
Faizan Khan answered |
x² – 22x + 112 = 0
x = 14, 8
y² – 20y + 84 = 0
y = 14, 6
x = 14, 8
y² – 20y + 84 = 0
y = 14, 6
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 4 hours when the inlet pipe is plugged.Quantity I: Inlet pipe Efficiency
Quantity II: 3 times of 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 'E'. Can you explain this answer?
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 4 hours when the inlet pipe is plugged.
Quantity I: Inlet pipe Efficiency
Quantity II: 3 times of Outlet pipe Efficiency
Quantity II: 3 times of 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
|
|
ナルト answered |
Answer is e
x² = 64y² – 30y + 225 = 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?
x² = 64
y² – 30y + 225 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Faizan Khan answered |
x² = 64
x = 8, – 8
y² – 30y + 225 = 0
y = 15, 15
x = 8, – 8
y² – 30y + 225 = 0
y = 15, 15
x² – 28x + 195 = 0y² – 35y + 306 = 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?
x² – 28x + 195 = 0
y² – 35y + 306 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Preeti Khanna answered |
x² – 28x + 195 = 0
x = 13, 15
y² – 35y + 306 = 0
y = 17, 18
x = 13, 15
y² – 35y + 306 = 0
y = 17, 18
The respective ratio between the present age of Mohan and David is 5:x. Mohan is 9 years younger than Preethi. Preethi’s age after 9 years will be 33 years. The difference between David’s and Mohan’s age is same as the present age of Preethi.Quantity I: Mohan’s present age
Quantity II: The value of x- 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?
The respective ratio between the present age of Mohan and David is 5:x. Mohan is 9 years younger than Preethi. Preethi’s age after 9 years will be 33 years. The difference between David’s and Mohan’s age is same as the present age of Preethi.
Quantity I: Mohan’s present age
Quantity II: The value of x
Quantity II: The value of x
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
|
|
Aisha Gupta answered |
Preethi’s age after 9 years = 33 years
Preethi’s present age = 33 – 9 = 24 years
Mohan’s present age = 24 – 9 = 15 years
David’s present age = 15 + 24 = 39 years
Ratio between Mohan and David = 15 : 39 = 5 : 13 X = 13
Preethi’s present age = 33 – 9 = 24 years
Mohan’s present age = 24 – 9 = 15 years
David’s present age = 15 + 24 = 39 years
Ratio between Mohan and David = 15 : 39 = 5 : 13 X = 13
4x² + 19x + 21 = 03y² + 29y + 56 = 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?
4x² + 19x + 21 = 0
3y² + 29y + 56 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Alok Verma answered |
4x² + 19x + 21 = 0
x = -3, – 1.75
3y² + 29y + 56 = 0
y = -7, -2.6
x = -3, – 1.75
3y² + 29y + 56 = 0
y = -7, -2.6
x² – 29x + 208 = 0y² + 9y + 14 = 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?
x² – 29x + 208 = 0
y² + 9y + 14 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Nikita Singh answered |
x² – 29x + 208 = 0
x = 13, 16
y² + 9y + 14 = 0
y = -2, -7
x = 13, 16
y² + 9y + 14 = 0
y = -2, -7
x² + 32x + 247 = 0y² + 20y + 91 = 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?
x² + 32x + 247 = 0
y² + 20y + 91 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Aarav Sharma answered |
x� + 32x + 247 = 0
x = -13, -19
y� + 20y + 91 = 0
y = -13, -7
The average salary of the entire staff in an office is Rs 250 per month. The average salary of officers is Rs 520 and that of non-officers is Rs. 200.Quantity I: Number of Officers = 15
Quantity II: Number of Non-Officers- 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?
The average salary of the entire staff in an office is Rs 250 per month. The average salary of officers is Rs 520 and that of non-officers is Rs. 200.
Quantity I: Number of Officers = 15
Quantity II: Number of Non-Officers
Quantity II: Number of Non-Officers
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
|
|
Rajeev Kumar answered |
Let the required number of non–officers = x
200x + 520 x 15 = 250 (15 + x)
250x – 200x = 520 * 15 – 250 x 15
50x = 4050
x = 81
200x + 520 x 15 = 250 (15 + x)
250x – 200x = 520 * 15 – 250 x 15
50x = 4050
x = 81
x² – 25x + 156 = 0y² – 32y + 255 = 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?
x² – 25x + 156 = 0
y² – 32y + 255 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Aarav Sharma answered |
x² – 25x + 156 = 0
x = 12, 13
y² – 32y + 255 = 0
y = 15, 17
x = 12, 13
y² – 32y + 255 = 0
y = 15, 17
x² – 44x + 448 = 0y² – 28y + 195 = 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?
x² – 44x + 448 = 0
y² – 28y + 195 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Ravi Singh answered |
x² – 44x + 448 = 0
x = 28, 16
y² – 28y + 195 = 0
y = 13, 15
x = 28, 16
y² – 28y + 195 = 0
y = 13, 15
x² = 121y² – 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?
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
|
|
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.="" />
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.="" />
x² – 16x + 63 = 06y² – 29y + 35 = 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?
x² – 16x + 63 = 0
6y² – 29y + 35 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Rhea Reddy answered |
x² – 16x + 63 = 0
x = 7, 9
6y² – 29y + 35 = 0
y = 2.3, 2.5
x = 7, 9
6y² – 29y + 35 = 0
y = 2.3, 2.5
x² – 31x + 228 = 0y² – 21y + 108 = 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 'C'. Can you explain this answer?
x² – 31x + 228 = 0
y² – 21y + 108 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Anaya Patel answered |
x² – 31x + 228 = 0
x = 12, 19
y² – 21y + 108 = 0
y = 12, 9
x = 12, 19
y² – 21y + 108 = 0
y = 12, 9
If y2 + 3y – 18 ≥ 0, which of the following is true?- a)y ≤ 3 or y ≥ 0
- b)y ≥ 3 or y ≤ – 6
- c)-6 ≤ y ≤ 3
- d)y > – 6 or y < 3
Correct answer is option 'B'. Can you explain this answer?
If y2 + 3y – 18 ≥ 0, which of the following is true?
a)
y ≤ 3 or y ≥ 0
b)
y ≥ 3 or y ≤ – 6
c)
-6 ≤ y ≤ 3
d)
y > – 6 or y < 3
|
Abhishek Chavan answered |
Understanding the Inequality
The given inequality is y^2 + 3y - 18 ≥ 0. To solve this inequality, we need to find the values of y that satisfy this condition.
Factoring the Quadratic Expression
First, we factor the quadratic expression y^2 + 3y - 18 to (y + 6)(y - 3) ≥ 0. This helps us identify the critical points where the expression changes sign.
Finding Critical Points
The critical points are where the expression equals zero, which are y = -6 and y = 3. These points divide the number line into three intervals: (-∞, -6), (-6, 3), and (3, ∞).
Testing Intervals
We can now test each interval to see when the inequality holds true.
- For y < -6,="" both="" factors="" are="" negative,="" so="" the="" inequality="" is="" />
- For -6 < y="" />< 3,="" one="" factor="" is="" negative="" and="" one="" is="" positive,="" making="" the="" inequality="" />
- For y > 3, both factors are positive, so the inequality is true.
Final Answer
Therefore, the values of y that satisfy the inequality y^2 + 3y - 18 ≥ 0 are y ≥ 3 or y ≤ -6. This corresponds to option B: y ≥ 3 or y ≤ -6.
The given inequality is y^2 + 3y - 18 ≥ 0. To solve this inequality, we need to find the values of y that satisfy this condition.
Factoring the Quadratic Expression
First, we factor the quadratic expression y^2 + 3y - 18 to (y + 6)(y - 3) ≥ 0. This helps us identify the critical points where the expression changes sign.
Finding Critical Points
The critical points are where the expression equals zero, which are y = -6 and y = 3. These points divide the number line into three intervals: (-∞, -6), (-6, 3), and (3, ∞).
Testing Intervals
We can now test each interval to see when the inequality holds true.
- For y < -6,="" both="" factors="" are="" negative,="" so="" the="" inequality="" is="" />
- For -6 < y="" />< 3,="" one="" factor="" is="" negative="" and="" one="" is="" positive,="" making="" the="" inequality="" />
- For y > 3, both factors are positive, so the inequality is true.
Final Answer
Therefore, the values of y that satisfy the inequality y^2 + 3y - 18 ≥ 0 are y ≥ 3 or y ≤ -6. This corresponds to option B: y ≥ 3 or y ≤ -6.
x² – 32x + 252 = 0y² – 28y + 192 = 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?
x² – 32x + 252 = 0
y² – 28y + 192 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Aarav Sharma answered |
I'm sorry, I'm not sure what you're asking for. Can you please provide more context or information?
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?
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
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
|
|
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.
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.
3x² + 20x + 33 = 02y² – 13y + 21 = 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?
3x² + 20x + 33 = 0
2y² – 13y + 21 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Ravi Singh answered |
3x² + 20x + 33 = 0
x = -3.6, -3
2y² – 13y + 21 = 0
y = 3, 3.5
x = -3.6, -3
2y² – 13y + 21 = 0
y = 3, 3.5
For a real number x the condition |3x - 20| + |3x - 40| = 20 necessarily holds if- a)10 < x < 15
- b)7 < x < 12
- c)9 < x < 14
- d)6 < x < 11
Correct answer is option 'B'. Can you explain this answer?
For a real number x the condition |3x - 20| + |3x - 40| = 20 necessarily holds if
a)
10 < x < 15
b)
7 < x < 12
c)
9 < x < 14
d)
6 < x < 11
|
|
Ashima rao answered |
Understanding the Equation:
To understand why option 'B' (7 < x="" />< 12)="" is="" the="" correct="" answer,="" let's="" first="" analyze="" the="" given="" equation:="" |3x="" -="" 20|="" +="" |3x="" -="" 40|="20." this="" equation="" involves="" the="" absolute="" value="" of="" expressions="" containing="" />
Solving the Equation:
1. We first need to consider the two cases for the absolute value:
a) When 3x - 20 ≥ 0, then |3x - 20| = 3x - 20
b) When 3x - 20 < 0,="" then="" |3x="" -="" 20|="-(3x" -="" 20)="20" -="" />
2. Similarly, for the second absolute value:
a) When 3x - 40 ≥ 0, then |3x - 40| = 3x - 40
b) When 3x - 40 < 0,="" then="" |3x="" -="" 40|="-(3x" -="" 40)="40" -="" />
3. Now, we substitute these values back into the original equation and simplify:
(3x - 20) + (3x - 40) = 20
6x - 60 = 20
6x = 80
x = 80/6
x = 13.33
Checking the Options:
Now we need to check which option satisfies the condition 7 < x="" />< />
- If x is less than 7 or greater than 12, the equation will not hold true.
- Therefore, the correct range for x is 7 < x="" />< 12,="" making="" option="" 'b'="" the="" right="" />
Therefore, the correct answer is option 'B' (7 < x="" />< 12)="" for="" the="" equation="" |3x="" -="" 20|="" +="" |3x="" -="" 40|="20" to="" hold="" true.="" 12)="" for="" the="" equation="" |3x="" -="" 20|="" +="" |3x="" -="" 40|="20" to="" hold="" />
To understand why option 'B' (7 < x="" />< 12)="" is="" the="" correct="" answer,="" let's="" first="" analyze="" the="" given="" equation:="" |3x="" -="" 20|="" +="" |3x="" -="" 40|="20." this="" equation="" involves="" the="" absolute="" value="" of="" expressions="" containing="" />
Solving the Equation:
1. We first need to consider the two cases for the absolute value:
a) When 3x - 20 ≥ 0, then |3x - 20| = 3x - 20
b) When 3x - 20 < 0,="" then="" |3x="" -="" 20|="-(3x" -="" 20)="20" -="" />
2. Similarly, for the second absolute value:
a) When 3x - 40 ≥ 0, then |3x - 40| = 3x - 40
b) When 3x - 40 < 0,="" then="" |3x="" -="" 40|="-(3x" -="" 40)="40" -="" />
3. Now, we substitute these values back into the original equation and simplify:
(3x - 20) + (3x - 40) = 20
6x - 60 = 20
6x = 80
x = 80/6
x = 13.33
Checking the Options:
Now we need to check which option satisfies the condition 7 < x="" />< />
- If x is less than 7 or greater than 12, the equation will not hold true.
- Therefore, the correct range for x is 7 < x="" />< 12,="" making="" option="" 'b'="" the="" right="" />
Therefore, the correct answer is option 'B' (7 < x="" />< 12)="" for="" the="" equation="" |3x="" -="" 20|="" +="" |3x="" -="" 40|="20" to="" hold="" true.="" 12)="" for="" the="" equation="" |3x="" -="" 20|="" +="" |3x="" -="" 40|="20" to="" hold="" />
Suresh spends 23% of an amount of money on an insurance policy, 33% on food, 19% on children’s education and 16% on recreation. He deposits the remaining amount of Rs. 504 in bank.Quantity I: Amount spend on food and insurance policy together
Quantity II: Amount spend on children’s education and recreation together- 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?
Suresh spends 23% of an amount of money on an insurance policy, 33% on food, 19% on children’s education and 16% on recreation. He deposits the remaining amount of Rs. 504 in bank.
Quantity I: Amount spend on food and insurance policy together
Quantity II: Amount spend on children’s education and recreation together
Quantity II: Amount spend on children’s education and recreation together
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
|
Yash Patel answered |
Total amount = x
Savings(%)
[100 – (23 + 33 + 19 + 16 )]% = 9 %
9% of x = 504
⇒ x = 504 * 100/9 = 5600
Amount spend on food and insurance policy together = 56% of 5600 = Rs.3136
Savings(%)
[100 – (23 + 33 + 19 + 16 )]% = 9 %
9% of x = 504
⇒ x = 504 * 100/9 = 5600
Amount spend on food and insurance policy together = 56% of 5600 = Rs.3136
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?
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?
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
|
|
Sagar Sharma answered |
Quantity I: What is the sum of Rs.100 and the daily wage of Bala?
Ajith can do 1/10 of the work in 1 day.
Bala can do 1/15 of the work in 1 day.
Together, Ajith and Bala can do 1/10 + 1/15 of the work in 1 day = 3/30 + 2/30 = 5/30 = 1/6 of the work in 1 day.
In 5 days, they can do 5 * 1/6 = 5/6 of the work.
So, the remaining work is 1 - 5/6 = 1/6 of the work.
Chand completes 1/6 of the work in 2 days.
Therefore, Chand can do 1/6 * (1/2) = 1/12 of the work in 1 day.
Let the daily wage of Bala be x.
The daily wage of Bala is 1/15 of the work, so x = 1/15 of the total wages.
The total wages are Rs. 6000, so x = (1/15) * 6000 = Rs. 400.
The sum of Rs. 100 and the daily wage of Bala is 100 + 400 = Rs. 500.
Quantity II: What is the daily wage of Chand?
The daily wage of Chand is 1/12 of the work, so it is (1/12) * 6000 = Rs. 500.
Since Quantity I is 500 and Quantity II is also 500, the answer is (C) Quantity I = Quantity II.
Ajith can do 1/10 of the work in 1 day.
Bala can do 1/15 of the work in 1 day.
Together, Ajith and Bala can do 1/10 + 1/15 of the work in 1 day = 3/30 + 2/30 = 5/30 = 1/6 of the work in 1 day.
In 5 days, they can do 5 * 1/6 = 5/6 of the work.
So, the remaining work is 1 - 5/6 = 1/6 of the work.
Chand completes 1/6 of the work in 2 days.
Therefore, Chand can do 1/6 * (1/2) = 1/12 of the work in 1 day.
Let the daily wage of Bala be x.
The daily wage of Bala is 1/15 of the work, so x = 1/15 of the total wages.
The total wages are Rs. 6000, so x = (1/15) * 6000 = Rs. 400.
The sum of Rs. 100 and the daily wage of Bala is 100 + 400 = Rs. 500.
Quantity II: What is the daily wage of Chand?
The daily wage of Chand is 1/12 of the work, so it is (1/12) * 6000 = Rs. 500.
Since Quantity I is 500 and Quantity II is also 500, the answer is (C) Quantity I = Quantity II.
x² – 33x + 272 = 0y² – 29y + 208 = 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 'C'. Can you explain this answer?
x² – 33x + 272 = 0
y² – 29y + 208 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Sagar Sharma answered |
Hello! How can I assist you today?
The perimeter of a rectangle and a square is 160 cm each. If the difference between their areas is 600 cm.Quantity I: Area of Square
Quantity II:Area of Rectangle- 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?
The perimeter of a rectangle and a square is 160 cm each. If the difference between their areas is 600 cm.
Quantity I: Area of Square
Quantity II:Area of Rectangle
Quantity II:Area of Rectangle
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
|
|
Anaya Patel answered |
Perimeter of rectangle = Perimeter of Square = 160
4a = 160 ⇒ a = 40
Area of square = 1600
1600 – lb = 600
lb = 1000 cm²
4a = 160 ⇒ a = 40
Area of square = 1600
1600 – lb = 600
lb = 1000 cm²
x² – 26x + 168 = 0y² – 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?
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
|
|
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.
- 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, b, c are integers, |a| ≠ |b| ≠|c| and -10 ≤ a, b, c ≤ 10. What will be the maximum possible value of [abc – (a + b + c)]?- a)524
- b)693
- c)970
- d)731
Correct answer is option 'D'. Can you explain this answer?
a, b, c are integers, |a| ≠ |b| ≠|c| and -10 ≤ a, b, c ≤ 10. What will be the maximum possible value of [abc – (a + b + c)]?
a)
524
b)
693
c)
970
d)
731
|
Amrita rao answered |
Explanation:
Finding the Maximum Value:
- To find the maximum possible value of the expression [abc - (a + b + c)], we need to consider the maximum and minimum values of a, b, and c.
- Since -10 ≤ a, b, c ≤ 10, the maximum possible value for each integer is 10.
Calculating the Expression:
- Let's substitute the maximum values of a, b, and c into the given expression:
[abc - (a + b + c)] = 10*10*10 - (10 + 10 + 10)
= 1000 - 30
= 970
Therefore, the maximum possible value of the expression [abc - (a + b + c)] is 970, which corresponds to option 'c'.
Finding the Maximum Value:
- To find the maximum possible value of the expression [abc - (a + b + c)], we need to consider the maximum and minimum values of a, b, and c.
- Since -10 ≤ a, b, c ≤ 10, the maximum possible value for each integer is 10.
Calculating the Expression:
- Let's substitute the maximum values of a, b, and c into the given expression:
[abc - (a + b + c)] = 10*10*10 - (10 + 10 + 10)
= 1000 - 30
= 970
Therefore, the maximum possible value of the expression [abc - (a + b + c)] is 970, which corresponds to option 'c'.
The inequality of p2 + 5 < 5p + 14 can be satisfied if:- a)p ≥ 6, p = 1
- b)p = 6, p = −2
- c)p ≤ 6, p ≤ 1
- d)p ≤ 6, p > −1
Correct answer is option 'D'. Can you explain this answer?
The inequality of p2 + 5 < 5p + 14 can be satisfied if:
a)
p ≥ 6, p = 1
b)
p = 6, p = −2
c)
p ≤ 6, p ≤ 1
d)
p ≤ 6, p > −1
|
|
Leena malhotra answered |
Explanation:
Given Inequality: p^2 + 5 < 5p="" +="" />
Step 1: Rearrange the Inequality:
To simplify the inequality, we need to rearrange it so that all terms are on one side of the inequality sign:
p^2 - 5p - 9 < />
Step 2: Factorize the Quadratic Equation:
To solve the inequality, we first need to factorize the quadratic equation:
(p - 6)(p + 1) < />
Step 3: Find the Critical Points:
The critical points are where the inequality changes sign. In this case, the critical points are when p = 6 and when p = -1.
Step 4: Test the Intervals:
We need to test the intervals between the critical points (-∞, -1), (-1, 6), and (6, ∞) to see where the inequality holds true.
Step 5: Determine the Solution:
- For the interval (-∞, -1):
Substitute p = -2, we get (-2 - 6)(-2 + 1) = (-8)(-1) = 8 > 0 (false)
- For the interval (-1, 6):
Substitute p = 0, we get (0 - 6)(0 + 1) = (-6)(1) = -6 < 0="" />
- For the interval (6, ∞):
Substitute p = 7, we get (7 - 6)(7 + 1) = (1)(8) = 8 > 0 (false)
Conclusion:
The inequality p^2 + 5 < 5p="" +="" 14="" is="" satisfied="" when="" p="" is="" less="" than="" or="" equal="" to="" 6="" and="" greater="" than="" -1.="" therefore,="" the="" correct="" answer="" is="" option="" 'd':="" p="" ≤="" 6,="" p="" /> -1.
Given Inequality: p^2 + 5 < 5p="" +="" />
Step 1: Rearrange the Inequality:
To simplify the inequality, we need to rearrange it so that all terms are on one side of the inequality sign:
p^2 - 5p - 9 < />
Step 2: Factorize the Quadratic Equation:
To solve the inequality, we first need to factorize the quadratic equation:
(p - 6)(p + 1) < />
Step 3: Find the Critical Points:
The critical points are where the inequality changes sign. In this case, the critical points are when p = 6 and when p = -1.
Step 4: Test the Intervals:
We need to test the intervals between the critical points (-∞, -1), (-1, 6), and (6, ∞) to see where the inequality holds true.
Step 5: Determine the Solution:
- For the interval (-∞, -1):
Substitute p = -2, we get (-2 - 6)(-2 + 1) = (-8)(-1) = 8 > 0 (false)
- For the interval (-1, 6):
Substitute p = 0, we get (0 - 6)(0 + 1) = (-6)(1) = -6 < 0="" />
- For the interval (6, ∞):
Substitute p = 7, we get (7 - 6)(7 + 1) = (1)(8) = 8 > 0 (false)
Conclusion:
The inequality p^2 + 5 < 5p="" +="" 14="" is="" satisfied="" when="" p="" is="" less="" than="" or="" equal="" to="" 6="" and="" greater="" than="" -1.="" therefore,="" the="" correct="" answer="" is="" option="" 'd':="" p="" ≤="" 6,="" p="" /> -1.
3x² – 11x + 10 = 04y² + 24y + 35 = 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?
3x² – 11x + 10 = 0
4y² + 24y + 35 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Preeti Khanna answered |
3x² – 11x + 10 = 0
x = 1.6, 2
4y² + 24y + 35 = 0
y = – 2.5, -3.5
x = 1.6, 2
4y² + 24y + 35 = 0
y = – 2.5, -3.5
The number of integers n that satisfy the inequalities | n - 60| < n - 100| < |n - 20| is - a)21
- b)19
- c)18
- d)20
Correct answer is option 'B'. Can you explain this answer?
The number of integers n that satisfy the inequalities | n - 60| < n - 100| < |n - 20| is
a)
21
b)
19
c)
18
d)
20
|
S.S Career Academy answered |
We have |n - 60| < |n - 100| < |n - 20|
Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line.
Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line.
This means that when the absolute difference from a number is larger, n would be further away from that number.
The absolute difference of n and 100 is less than that of the absolute difference between n and 20.
Hence, n cannot be ≤ 60, as then it would be closer to 20 than 100. Thus we have the condition that n>60.
The absolute difference of n and 60 is less than that of the absolute difference between n and 100.
Hence, n cannot be ≥ 80, as then it would be closer to 100 than 60.
Thus we have the condition that n<80.
The number which satisfies the conditions are 61, 62, 63, 64……79. Thus, a total of 19 numbers.
Alternatively
as per the given condition: |n - 60| < |n - 100| < |n - 20|
Dividing the range of n into 4 segments. (n < 20, 20<n<60, 60<n<100, n > 100)
1) For n < 20.
|n-20| = 20-n, |n-60| = 60- n, |n-100| = 100-n
considering the inequality part: |n - 100| < n - 20|
100 -n < 20 -n,
No value of n satisfies this condition.
2) For 20 < n < 60.
|n-20| = n-20, |n-60| = 60- n, |n-100| = 100-n.
60- n < 100 – n and 100 – n < n – 20
For 100 -n < n – 20.
120 < 2n and n > 60. But for the considered range n is less than 60.
3) For 60 < n < 100
|n-20| = n-20, |n-60| = n-60, |n-100| = 100-n
n-60 < 100-n and 100-n < n-20.
For the first part 2n < 160 and for the second part 120 < 2n.
n takes values from 61 …………….79.
A total of 19 values
4) For n > 100
|n-20| = n-20, |n-60| = n-60, |n-100| = n-100
n-60 < n – 100.
No value of n in the given range satisfies the given inequality.
Hence a total of 19 values satisfy the inequality.
x² – 30x + 221 = 0y² – 33y + 270 = 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?
x² – 30x + 221 = 0
y² – 33y + 270 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Nikita Singh answered |
x² – 30x + 221 = 0
x = 13, 17
y² – 33y + 270 = 0
y = 15, 18
x = 13, 17
y² – 33y + 270 = 0
y = 15, 18
Consider the equation:|x-5|2 + 5 |x - 5| - 24 = 0The sum of all the real roots of the above equationis:- a)10
- b)3
- c)8
- d)2
Correct answer is option 'A'. Can you explain this answer?
Consider the equation:
|x-5|2 + 5 |x - 5| - 24 = 0
The sum of all the real roots of the above equationis:
a)
10
b)
3
c)
8
d)
2
|
Upsc Rank Holders answered |
Let's consider x-5 as 'p'
Case 1: p ≥ 0
|x - 5| |2 + 5|x - 5| - 24 = 0
p2 +5p - 24 = 0
p2 + 8p - 3p - 24 = 0
p(p + 8) -3 (p + 8) = 0
(p + 8) (p - 3) = 0
p = -8 and p = 3
x - 5 = 3,x = 8 This is a real root since x is greater than 5.
x - 5 = -8, x = -3. This root can be negated because x is not greater than 5.
Case 2: p < 0
p2 - 5p - 24 = 0
p2 - 8p + 3p - 24 = 0
p=8, -3
x - 5 = 8, x = 13. This root can be negated because x is not less than 5
x - 5 = -3, x = 2. This is a real root because x is less than 5.
The sum of the real roots = 8 + 2 = 10
x² – 28x + 195 = 0y² – 30y + 216 = 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?
x² – 28x + 195 = 0
y² – 30y + 216 = 0
a)
X > Y
b)
X < Y
c)
X ≥ Y
d)
X ≤ Y
e)
X = Y or relation cannot be established
|
|
Kavya Saxena answered |
x² – 28x + 195 = 0
x = 15, 13
y² – 30y + 216 = 0
y = 18, 12
x = 15, 13
y² – 30y + 216 = 0
y = 18, 12
A basket contains 6 White 4 Black 2 Pink and 3 Green balls. If four balls are picked at random,Quantity I: Probability that at least one is Black.
Quantity II: Probability that all is Black.- 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?
A basket contains 6 White 4 Black 2 Pink and 3 Green balls. If four balls are picked at random,
Quantity I: Probability that at least one is Black.
Quantity II: Probability that all is Black.
Quantity II: Probability that all is Black.
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
|
|
Anaya Patel answered |
Total Balls = 15
Probability = 11c4/15c4 = 22/91
One is black = 1 – 22/91 = 69/91
Probability = 11c4/15c4 = 22/91
One is black = 1 – 22/91 = 69/91
Mr. Ramesh bought two watches which together cost him Rs.440. He sold one of the watches at a loss of 20% and the other one at a gain of 40%. The selling price of both watches are same.Quantity I: SP and CP one of the watches sold at a loss of 20%
Quantity II: SP and CP one of the watches sold at a profit of 40%- 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 'D'. Can you explain this answer?
Mr. Ramesh bought two watches which together cost him Rs.440. He sold one of the watches at a loss of 20% and the other one at a gain of 40%. The selling price of both watches are same.
Quantity I: SP and CP one of the watches sold at a loss of 20%
Quantity II: SP and CP one of the watches sold at a profit of 40%
Quantity II: SP and CP one of the watches sold at a profit of 40%
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
|
|
Kavya Saxena answered |
80/100 * x = 140/100 * y
x = 7/4y
x + y = 440
7/4 y + y = 440
y = 160 ; x = 280
x = 7/4y
x + y = 440
7/4 y + y = 440
y = 160 ; x = 280
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?
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
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
|
|
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.
- 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.
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