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All questions of Chapter 3: Linear Inequalities for CA Foundation Exam
The solution of the inequality 8x + 6 < 12x + 14 is- a)(-2,2)
- b)(0,-2)
- c)(2, ∞)
- d)(-2,∞)
Correct answer is option 'D'. Can you explain this answer?
The solution of the inequality 8x + 6 < 12x + 14 is
a)
(-2,2)
b)
(0,-2)
c)
(2, ∞)
d)
(-2,∞)
|
Mrinalini Iyer answered |
There are different ways to express the solution of an inequality, but one common way is to write it in interval notation. To do this, we first isolate the variable on one side of the inequality:
8x - 6 ≤ 0
8x ≤ 6
x ≤ 6/8
x ≤ 3/4
This means that any value of x that is less than or equal to 3/4 makes the inequality true. We can represent this as the interval:
(-∞, 3/4]
Alternatively, we can write the solution using set-builder notation, which specifies the set of all x values that satisfy the inequality:
{x | x ≤ 3/4}
Either way, the solution indicates that x can be any number less than or equal to 3/4, including fractions, decimals, and negative numbers.
8x - 6 ≤ 0
8x ≤ 6
x ≤ 6/8
x ≤ 3/4
This means that any value of x that is less than or equal to 3/4 makes the inequality true. We can represent this as the interval:
(-∞, 3/4]
Alternatively, we can write the solution using set-builder notation, which specifies the set of all x values that satisfy the inequality:
{x | x ≤ 3/4}
Either way, the solution indicates that x can be any number less than or equal to 3/4, including fractions, decimals, and negative numbers.
By solving the inequality 3(a - 6) < 4 + a, the answer will be
- a)a < 9
- b)a < 12
- c)a < 13
- d)a < 11
Correct answer is option 'D'. Can you explain this answer?
By solving the inequality 3(a - 6) < 4 + a, the answer will be
a)
a < 9
b)
a < 12
c)
a < 13
d)
a < 11
|
Akshat Choudhary answered |
We cannot solve this inequality as it is incomplete. There must be a comparison operator (such as <,>, ≤, or ≥) to determine the relationship between the expression 3(a - 6) and some other value or expression.
3x2 - 7x + 6 < 0- a)0.66 <x< 3
- b)-0.66 <x< 3
- c)-1 < x < 3
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
3x2 - 7x + 6 < 0
a)
0.66 <x< 3
b)
-0.66 <x< 3
c)
-1 < x < 3
d)
None of these
|
|
Avik Choudhury answered |
At x = 0, inequality is not satisfied.
Hence, options (b), (c) and (d) are rejected. At x = 2, inequality is not satisfied. Hence, option (a) is rejected.
Thus, option (d) is correct.
Hence, options (b), (c) and (d) are rejected. At x = 2, inequality is not satisfied. Hence, option (a) is rejected.
Thus, option (d) is correct.
For all integral values of x,|x - 4| x< 5- a)-1 ≤x≤5
- b)1 ≤x≤5
- c)-1 ≤ x ≤ 1
- d)x<5
Correct answer is option 'D'. Can you explain this answer?
For all integral values of x,
|x - 4| x< 5
a)
-1 ≤x≤5
b)
1 ≤x≤5
c)
-1 ≤ x ≤ 1
d)
x<5
|
Preeti Khanna answered |
At x = 0 inequality is satisfied, option (b) is rejected.
At x = 2, inequality is satisfied, option (c) is rejected.
At x = 5, LHS = RHS.
Thus, option (d) is correct.
|x2 – 2x – 3| < 3x – 3- a)2 < x < 5
- b)–2 < x < 5
- c)x > 5
- d)1 < x < 3
Correct answer is option 'A'. Can you explain this answer?
|x2 – 2x – 3| < 3x – 3
a)
2 < x < 5
b)
–2 < x < 5
c)
x > 5
d)
1 < x < 3
|
Yash Patel answered |
x2 - 2x - 3 ≥ 0
(x-3) (x+1) ≥ 0
x belongs to (-∞,-3]∪[3,∞)
Therefore, x belongs to (-1,3)
=> x2 - 2x - 3 > 0
x2 - 2x - 3< 3x - 3
x2 - 5x < 0
x(x-5) < 0
x belongs to (0,5)........(1)
x2 - 2x - 3 < 0
x2 - 2x - 3 < 3x - 3
x2 + x - 6 > 0
(x+3)(x-2) > 0
x belongs to (-∞,-3]∪[2,∞)
x belongs to (2,3)........(2)
Taking intersection of (1) and (2)
we get,
x belongs to (2,5)
(x-3) (x+1) ≥ 0
x belongs to (-∞,-3]∪[3,∞)
Therefore, x belongs to (-1,3)
=> x2 - 2x - 3 > 0
x2 - 2x - 3< 3x - 3
x2 - 5x < 0
x(x-5) < 0
x belongs to (0,5)........(1)
x2 - 2x - 3 < 0
x2 - 2x - 3 < 3x - 3
x2 + x - 6 > 0
(x+3)(x-2) > 0
x belongs to (-∞,-3]∪[2,∞)
x belongs to (2,3)........(2)
Taking intersection of (1) and (2)
we get,
x belongs to (2,5)
On solving the inequalities 6x+ y ≥ 18, x+4y ≥ 12, 2x+y ≥ 10, we get following situation- a)(0,18), (12,0), (4,2) and (7,6)
- b)(3,0),(0,3),,(4,2) and (7,6)
- c)(5,0),(0,10),,(4,2) and (7,6)
- d)(0,18), (12,0),(4,2) ,(0,0) and (7,6)
Correct answer is option 'A'. Can you explain this answer?
On solving the inequalities 6x+ y ≥ 18, x+4y ≥ 12, 2x+y ≥ 10, we get following situation
a)
(0,18), (12,0), (4,2) and (7,6)
b)
(3,0),(0,3),,(4,2) and (7,6)
c)
(5,0),(0,10),,(4,2) and (7,6)
d)
(0,18), (12,0),(4,2) ,(0,0) and (7,6)
|
|
Ishan Goyal answered |
I'm sorry, but the inequality is incomplete. There is no inequality sign (such as <,>, ≤, or ≥) to indicate the relationship between 6x + y and some other value. Please provide more information or context so I can assist you better.
x2 - 7x + 12 < | x - 4 |- a)x < 2
- b)x > 4
- c)2 < x < 4
- d)2 ≤ x ≤ 4
Correct answer is option 'C'. Can you explain this answer?
x2 - 7x + 12 < | x - 4 |
a)
x < 2
b)
x > 4
c)
2 < x < 4
d)
2 ≤ x ≤ 4
|
Harsh Jain answered |
At x = 0, inequality is not satisfied, option (a) is rejected.
At x = 5, inequality is not satisfied, option (b) is rejected.
At x = 2 inequality is not satisfied.
Options (d) are rejected.
Option (c) is correct.
At x = 2 inequality is not satisfied.
Options (d) are rejected.
Option (c) is correct.
The rules and regulations demand that the employer should employ not more than 5 experienced hands to 1 fresh one and this fact be expressed as- a)y ≥ x/5
- b)5y ≤ x
- c)5 y ≥ x
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The rules and regulations demand that the employer should employ not more than 5 experienced hands to 1 fresh one and this fact be expressed as
a)
y ≥ x/5
b)
5y ≤ x
c)
5 y ≥ x
d)
none of these
|
|
Simran Pillai answered |
The employer must maintain a ratio of 5 experienced employees to 1 fresh employee as per the rules and regulations.
The shaded region represents:
- a)x + y > 6, 2x - y > 0
- b) x + y < 6, 2x -y > 0
- c) x + y > 6, 2x -y < 0
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The shaded region represents:
a)
x + y > 6, 2x - y > 0
b)
x + y < 6, 2x -y > 0
c)
x + y > 6, 2x -y < 0
d)
None of these
|
Rashmi Chaurasia answered |
I think option c is correct because sheded is above the eqn x+y=6 than x+y>6and second sheded blow the eqn 2x - y=0than 2x-y< o="" so="" c="" is="" correct="" o="" so="" c="" is="" />
A car manufacturing company manufactures cars of two types A and B. Model A requires 150 man-hours for assembling, 50 man-hours for painting and 10 man-hours for checking and testing. Model B requires 60 man-hours for assembling, 40 man-hours for painting and 20 man-hours for checking and testing. There are available 30 thousand man-hours for assembling, 13 thousand man-hours for checking and testing. Express the above situation using linear inequalities. Let the company manufacture x units of type A model of car and y units type B model of car. Then, the inequalities are:- a)5x+2y ≥ 1000; 5x + 4y ≥ 1300,x+2y ≤ 500; x ≥ 0, y ≥ 0,
- b)5x + 2y ≤ 1000, 5x+4y ≤ 13000,x+2y ≥ 500; x ≥ 0, y ≥ 0.
- c)5x+2y ≤ 1,000, 5x+4y ≤ 1300,x+2y ≤ 500; x≥ 0, y≥ 0.
- d)5x + 2y = 1000, 5x+4y ≥ 1300,x+2y = 500; x≥ 0, y ≥ 0.
Correct answer is option 'C'. Can you explain this answer?
A car manufacturing company manufactures cars of two types A and B. Model A requires 150 man-hours for assembling, 50 man-hours for painting and 10 man-hours for checking and testing. Model B requires 60 man-hours for assembling, 40 man-hours for painting and 20 man-hours for checking and testing. There are available 30 thousand man-hours for assembling, 13 thousand man-hours for checking and testing. Express the above situation using linear inequalities. Let the company manufacture x units of type A model of car and y units type B model of car. Then, the inequalities are:
a)
5x+2y ≥ 1000; 5x + 4y ≥ 1300,x+2y ≤ 500; x ≥ 0, y ≥ 0,
b)
5x + 2y ≤ 1000, 5x+4y ≤ 13000,x+2y ≥ 500; x ≥ 0, y ≥ 0.
c)
5x+2y ≤ 1,000, 5x+4y ≤ 1300,x+2y ≤ 500; x≥ 0, y≥ 0.
d)
5x + 2y = 1000, 5x+4y ≥ 1300,x+2y = 500; x≥ 0, y ≥ 0.
|
|
Vaishnavi Gupta answered |
1. Assembling constraint:
150x + 60y ≤ 30,000
This inequality represents the constraint that the total man-hours spent on assembling cars cannot exceed the available 30,000 man-hours.
2. Painting constraint:
50x + 40y ≤ M
The value of M is not given in the problem, but it represents the maximum number of man-hours available for painting cars. We cannot determine the value of M from the given information, so we leave it as an unknown constant.
3. Checking and testing constraint:
10x + 20y ≤ 13,000
This inequality represents the constraint that the total man-hours spent on checking and testing cars cannot exceed the available 13,000 man-hours.
4. Non-negative constraint:
x ≥ 0, y ≥ 0
This inequality represents the fact that the company cannot manufacture negative units of cars.
5. Integer constraint:
x, y ∈ ℤ
This inequality represents the fact that the company can only manufacture integer units of cars, and not fractional units.
Putting all the above inequalities together, we get the following set of linear inequalities:
150x + 60y ≤ 30,000
50x + 40y ≤ M
10x + 20y ≤ 13,000
x ≥ 0, y ≥ 0
x, y ∈ ℤ
150x + 60y ≤ 30,000
This inequality represents the constraint that the total man-hours spent on assembling cars cannot exceed the available 30,000 man-hours.
2. Painting constraint:
50x + 40y ≤ M
The value of M is not given in the problem, but it represents the maximum number of man-hours available for painting cars. We cannot determine the value of M from the given information, so we leave it as an unknown constant.
3. Checking and testing constraint:
10x + 20y ≤ 13,000
This inequality represents the constraint that the total man-hours spent on checking and testing cars cannot exceed the available 13,000 man-hours.
4. Non-negative constraint:
x ≥ 0, y ≥ 0
This inequality represents the fact that the company cannot manufacture negative units of cars.
5. Integer constraint:
x, y ∈ ℤ
This inequality represents the fact that the company can only manufacture integer units of cars, and not fractional units.
Putting all the above inequalities together, we get the following set of linear inequalities:
150x + 60y ≤ 30,000
50x + 40y ≤ M
10x + 20y ≤ 13,000
x ≥ 0, y ≥ 0
x, y ∈ ℤ
For x = 15, y = 10 and z = 9, find the value of le(x, min(y, x-z), le(9, 8, ma(x, y, z)).- a)5
- b)9
- c)12
- d)4
Correct answer is option 'B'. Can you explain this answer?
For x = 15, y = 10 and z = 9, find the value of le(x, min(y, x-z), le(9, 8, ma(x, y, z)).
a)
5
b)
9
c)
12
d)
4
|
Sheetal patil answered |
Explanation:
Given values:
x = 15
y = 10
z = 9
Calculations:
1. Find the minimum of y and x-z:
min(y, x-z) = min(10, 15-9) = min(10, 6) = 6
2. Find the maximum of x, y, and z:
ma(x, y, z) = max(15, 10, 9) = 15
3. Evaluate the expression:
le(x, min(y, x-z), le(9, 8, ma(x, y, z)) = le(15, 6, le(9, 8, 15))
4. Compare 9 and 8:
le(9, 8) = 0 (as 9 is not less than 8)
5. Evaluate the expression further:
le(15, 6, 0) = 6
Therefore, the value of le(x, min(y, x-z), le(9, 8, ma(x, y, z)) for the given values is 6, which corresponds to option 'b'.
Mary scored 200 marks in three tests. If average score of 60 is required then the lowest marks she must score for fourth test are- a)x gt; 40
- b)x gt; 50
- c)x gt; 60
- d)x gt; 70
Correct answer is option 'A'. Can you explain this answer?
Mary scored 200 marks in three tests. If average score of 60 is required then the lowest marks she must score for fourth test are
a)
x gt; 40
b)
x gt; 50
c)
x gt; 60
d)
x gt; 70
|
|
Aditi Joshi answered |
Given:
Mary scored 200 marks in three tests.
Average score of 60 is required.
To find:
The lowest marks she must score for the fourth test.
Solution:
Let the marks scored by Mary in the fourth test be x.
Total marks scored by Mary in four tests = 200 + x
Average score = (Total marks scored) / (Number of tests)
60 = (200 + x) / 4
240 = 200 + x
x = 40
Therefore, Mary must score at least 40 marks in the fourth test to maintain an average score of 60.
Answer: Option A, x > 40.
Mary scored 200 marks in three tests.
Average score of 60 is required.
To find:
The lowest marks she must score for the fourth test.
Solution:
Let the marks scored by Mary in the fourth test be x.
Total marks scored by Mary in four tests = 200 + x
Average score = (Total marks scored) / (Number of tests)
60 = (200 + x) / 4
240 = 200 + x
x = 40
Therefore, Mary must score at least 40 marks in the fourth test to maintain an average score of 60.
Answer: Option A, x > 40.
p, q and r are three non-negative integers such that p + q + r = 10. The maximum value of pq + qr + pr + pqr is- a)≥ 60 and < 70
- b)≥ 50 and < 60
- c)≥ 40 and < 50
- d)≥ 70 and < 80
Correct answer is option 'A'. Can you explain this answer?
p, q and r are three non-negative integers such that p + q + r = 10. The maximum value of pq + qr + pr + pqr is
a)
≥ 60 and < 70
b)
≥ 50 and < 60
c)
≥ 40 and < 50
d)
≥ 70 and < 80
|
S.S Career Academy answered |
The product of 2 numbers A and B is maximum when A = B.
If we cannot equate the numbers, then we have to try to minimize the difference between the numbers as much as possible.
pq will be maximum when p=q.
qr will be maximum when q=r.
qr will be maximum when r=p.
Therefore, p, q, and r should be as close to each other as possible.
We know that p,q,and r are integers and p + q + r = 10.
=> p,q, and r should be 3, 3, and 4 in any order.
Substituting the values in the expression, we get,
pq + qr + pr + pqr = 3*3 + 3*4 + 3*4 + 3*3*4
= 9 + 12 + 12 + 36
= 69
3x2 - 7x + 4 ≤ 0- a)x > 0
- b)x < 0
- c)All x
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
3x2 - 7x + 4 ≤ 0
a)
x > 0
b)
x < 0
c)
All x
d)
None of these
|
|
Rajdeep Ghoshal answered |
At x = 0, inequality is not satisfied. Thus, option (c) is rejected. Also x = 0 is not a solution of the equation. Since, this is a continuous function, the solution cannot start from 0. Thus options (a) and (b) are not right. Further, we see that the given function is quadratic with real roots. Hence, option (d) is also rejected.
What is the value of ma(10, 4, le((la10, 5, 3), 5, 3))?- a) 7
- b) 6.5
- c) 8
- d) 7.5
Correct answer is option 'B'. Can you explain this answer?
What is the value of ma(10, 4, le((la10, 5, 3), 5, 3))?
a)
7
b)
6.5
c)
8
d)
7.5
|
|
Shubhangi kumar answered |
To solve the given expression ma(10, 4, le((la10, 5, 3), 5, 3)), we need to follow the order of operations (also known as PEMDAS/BODMAS).
Step 1: Evaluate the innermost expression
The innermost expression is le((la10, 5, 3), 5, 3). Let's break it down further:
- la10 means the largest among 10, 5, and 3, which is 10.
- le(10, 5, 3) means the smallest among 10, 5, and 3, which is 3.
So, the innermost expression evaluates to 3.
Step 2: Substitute the innermost expression in the main expression
Now, we substitute the value of the innermost expression (3) in the main expression: ma(10, 4, 3).
Step 3: Evaluate the main expression
The expression ma(10, 4, 3) means the arithmetic mean (average) of the three numbers: 10, 4, and 3.
Arithmetic Mean Formula: sum of numbers/total number of numbers
Sum of numbers = 10 + 4 + 3 = 17
Total number of numbers = 3
Arithmetic Mean = 17/3 = 5.67
Rounded to one decimal place, the value of ma(10, 4, 3) is 5.7.
Step 4: Corresponding option
The correct option given is 'B' which states the value as 6.5. However, this does not match the calculated value of 5.7. Hence, the correct option should be re-evaluated or reviewed, as it does not match the calculated value.
Step 1: Evaluate the innermost expression
The innermost expression is le((la10, 5, 3), 5, 3). Let's break it down further:
- la10 means the largest among 10, 5, and 3, which is 10.
- le(10, 5, 3) means the smallest among 10, 5, and 3, which is 3.
So, the innermost expression evaluates to 3.
Step 2: Substitute the innermost expression in the main expression
Now, we substitute the value of the innermost expression (3) in the main expression: ma(10, 4, 3).
Step 3: Evaluate the main expression
The expression ma(10, 4, 3) means the arithmetic mean (average) of the three numbers: 10, 4, and 3.
Arithmetic Mean Formula: sum of numbers/total number of numbers
Sum of numbers = 10 + 4 + 3 = 17
Total number of numbers = 3
Arithmetic Mean = 17/3 = 5.67
Rounded to one decimal place, the value of ma(10, 4, 3) is 5.7.
Step 4: Corresponding option
The correct option given is 'B' which states the value as 6.5. However, this does not match the calculated value of 5.7. Hence, the correct option should be re-evaluated or reviewed, as it does not match the calculated value.
x, y and z are three positive integers such that x > y > z. Which of the following is closest to the product xyz?- a)(x - 1)yz
- b)x(y - 1)z
- c)xy(z - 1)
- d)x(y + 1)z
Correct answer is option 'A'. Can you explain this answer?
x, y and z are three positive integers such that x > y > z. Which of the following is closest to the product xyz?
a)
(x - 1)yz
b)
x(y - 1)z
c)
xy(z - 1)
d)
x(y + 1)z
|
Innovative Classes answered |
The expressions in the four options can be expanded as
xyz-yz; xyz-xz; xyz-xy and xyz+xz
The closest value to xyz would be xyz-yz, as yz is the least value among yz, xz and xy.
Option a) is the correct answer.
An employer recruits experienced (x) and fresh workmen (y) for his firm under the condition that he cannot employ more than 9 people. X and Y can be related by the inequality- a)x+y ≠ 9
- b)x+ y ≤ 9 x ≥ 0, y ≥ 0
- c)x+y ≥ 9 x ≥ 0, y ≥ 0
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
An employer recruits experienced (x) and fresh workmen (y) for his firm under the condition that he cannot employ more than 9 people. X and Y can be related by the inequality
a)
x+y ≠ 9
b)
x+ y ≤ 9 x ≥ 0, y ≥ 0
c)
x+y ≥ 9 x ≥ 0, y ≥ 0
d)
none of these
|
|
Mehul Ghoshal answered |
The inequality given is not complete. There is no inequality symbol specified. Assuming that the inequality is meant to be "x + y ≤ 9" (since the employer cannot employ more than 9 people), then:
- x represents the number of experienced workmen the employer recruits.
- y represents the number of fresh workmen the employer recruits.
The inequality "x + y ≤ 9" means that the sum of x and y cannot exceed 9, which is the maximum number of people the employer can hire. This also implies that x and y must be non-negative integers (since the employer cannot hire a fraction of a person).
- x represents the number of experienced workmen the employer recruits.
- y represents the number of fresh workmen the employer recruits.
The inequality "x + y ≤ 9" means that the sum of x and y cannot exceed 9, which is the maximum number of people the employer can hire. This also implies that x and y must be non-negative integers (since the employer cannot hire a fraction of a person).
The number of positive integer valued pairs (x, y), satisfying 4x – 17 y = 1 and x < 1000 is:- a)55
- b)57
- c)59
- d)58
Correct answer is option 'C'. Can you explain this answer?
The number of positive integer valued pairs (x, y), satisfying 4x – 17 y = 1 and x < 1000 is:
a)
55
b)
57
c)
59
d)
58
|
EduRev CLAT answered |
The integral values of x for which y is an integer are 13, 30, 47,……
The values are in the form 17n + 13, where n ≥ 0
17n + 13 < 1000
⇒ 17n < 987
⇒ n < 58.05
⇒ n can take values from 0 to 58
⇒ Number of values = 59
|x2 + x| – 5 < 0
- a)x < 0
- b)x > 0
- c)None of these
- d)All values of x
Correct answer is option 'C'. Can you explain this answer?
|x2 + x| – 5 < 0
a)
x < 0
b)
x > 0
c)
None of these
d)
All values of x
|
|
Sinjini Gupta answered |
At x = 0 inequality is satisfied.
Thus, options (a), (b), and (d) are rejected.
Option (c) is correct.
Thus, options (a), (b), and (d) are rejected.
Option (c) is correct.
The Linear relationship between two variables in an inequality :- a)ax + by ≤ c
- b)ax + bxy ≤ c
- c)axy + by ≤ c
- d)none
Correct answer is option 'A'. Can you explain this answer?
The Linear relationship between two variables in an inequality :
a)
ax + by ≤ c
b)
ax + bxy ≤ c
c)
axy + by ≤ c
d)
none
|
|
Sinjini Gupta answered |
The linear relationship between two variables in an inequality can be expressed in the form of ax + by < c="" or="" ax="" +="" by="" /> c, where a, b and c are constants, and x and y are the two variables.
If ax + by < c,="" it="" means="" that="" the="" values="" of="" x="" and="" y="" that="" satisfy="" this="" inequality="" lie="" below="" the="" line="" represented="" by="" the="" equation="" ax="" +="" by="c." in="" other="" words,="" the="" region="" below="" the="" line="" is="" the="" solution="" set="" for="" the="" />
Similarly, if ax + by > c, it means that the values of x and y that satisfy this inequality lie above the line represented by the equation ax + by = c. In other words, the region above the line is the solution set for the inequality.
If ax + by < c,="" it="" means="" that="" the="" values="" of="" x="" and="" y="" that="" satisfy="" this="" inequality="" lie="" below="" the="" line="" represented="" by="" the="" equation="" ax="" +="" by="c." in="" other="" words,="" the="" region="" below="" the="" line="" is="" the="" solution="" set="" for="" the="" />
Similarly, if ax + by > c, it means that the values of x and y that satisfy this inequality lie above the line represented by the equation ax + by = c. In other words, the region above the line is the solution set for the inequality.
|x2 - 2x| < x- a)l < x < 3
- b)—1 < x < 3
- c)0 < x < 4
- d)x > 3
Correct answer is option 'A'. Can you explain this answer?
|x2 - 2x| < x
a)
l < x < 3
b)
—1 < x < 3
c)
0 < x < 4
d)
x > 3
|
Aarav Sharma answered |
The expression |x2 - 2x| represents the absolute value of the quadratic expression x^2 - 2x.
To find the absolute value of x^2 - 2x, we need to consider the cases when the expression inside the absolute value is positive and negative.
When x^2 - 2x is positive (greater than or equal to zero):
x^2 - 2x ≥ 0
x(x - 2) ≥ 0
In this case, the absolute value of x^2 - 2x is simply x^2 - 2x.
When x^2 - 2x is negative (less than zero):
x^2 - 2x < />
x(x - 2) < />
In this case, the absolute value of x^2 - 2x is -(x^2 - 2x), which is equal to -x^2 + 2x.
Therefore, the expression |x^2 - 2x| can be written as:
x^2 - 2x, for x ≤ 0 or x ≥ 2
-x^2 + 2x, for 0 < x="" />< 2="" />
To find the absolute value of x^2 - 2x, we need to consider the cases when the expression inside the absolute value is positive and negative.
When x^2 - 2x is positive (greater than or equal to zero):
x^2 - 2x ≥ 0
x(x - 2) ≥ 0
In this case, the absolute value of x^2 - 2x is simply x^2 - 2x.
When x^2 - 2x is negative (less than zero):
x^2 - 2x < />
x(x - 2) < />
In this case, the absolute value of x^2 - 2x is -(x^2 - 2x), which is equal to -x^2 + 2x.
Therefore, the expression |x^2 - 2x| can be written as:
x^2 - 2x, for x ≤ 0 or x ≥ 2
-x^2 + 2x, for 0 < x="" />< 2="" />
x2 – 14x – 15 > 0- a)x < –1
- b)15 < x
- c)Both (a) and (b)
- d)–1 < x < 15
Correct answer is option 'C'. Can you explain this answer?
x2 – 14x – 15 > 0
a)
x < –1
b)
15 < x
c)
Both (a) and (b)
d)
–1 < x < 15
|
|
Aditi Menon answered |
At x = 0 inequality is not satisfied. Thus option (d) is rejected.
x = –1 and x = 15 are the roots of the quadratic equation. Thus, option (c) is correct.
x = –1 and x = 15 are the roots of the quadratic equation. Thus, option (c) is correct.
3x2 – 7x – 6 < 0- a)–0.66 < x < 3
- b)x < – 0.66 or x > 3
- c)3 < x < 7
- d)–2 < x < 2
Correct answer is option 'A'. Can you explain this answer?
3x2 – 7x – 6 < 0
a)
–0.66 < x < 3
b)
x < – 0.66 or x > 3
c)
3 < x < 7
d)
–2 < x < 2
|
|
Gargi Kulkarni answered |
At x = 0, inequality is satisfied. Hence, options (b) and (c) are rejected. x = 3 gives LHS = RHS.
and x = – 0.66 also does the same. Hence. roots of the equation are 3 and – 0.66.
Thus, option (a) is correct.
and x = – 0.66 also does the same. Hence. roots of the equation are 3 and – 0.66.
Thus, option (a) is correct.
On solving the inequalities 6x + y ≥ 18, x + 4y ≥ 12, 2x + y≤ 10, we get the following situation:- a)(0,18), (12,0), (4,2), &(7,6)
- b)(3,0),(0,3),(4,2),&(7,6)
- c)(5,0),(0,10),(4,2)&(7,6)
- d)(0,18),(12,0),(4,2),(0,0) and (7,6)
Correct answer is option 'A'. Can you explain this answer?
On solving the inequalities 6x + y ≥ 18, x + 4y ≥ 12, 2x + y≤ 10, we get the following situation:
a)
(0,18), (12,0), (4,2), &(7,6)
b)
(3,0),(0,3),(4,2),&(7,6)
c)
(5,0),(0,10),(4,2)&(7,6)
d)
(0,18),(12,0),(4,2),(0,0) and (7,6)
|
Ishleen Kohli answered |
Please give explanation ?
x2 – 7x + 12 < |x – 4|- a)x < 2
- b)x > 4
- c)2 < x < 4
- d)2 £ x £ 4
Correct answer is option 'C'. Can you explain this answer?
x2 – 7x + 12 < |x – 4|
a)
x < 2
b)
x > 4
c)
2 < x < 4
d)
2 £ x £ 4
|
|
Prisha Shah answered |
At x = 0, inequality is not satisfied, option (a) is rejected.
At x = 5, inequality is not satisfied, option (b) is rejected.
At x = 2 inequality is not satisfied.
Options (d) are rejected.
Option (c) is correct
At x = 5, inequality is not satisfied, option (b) is rejected.
At x = 2 inequality is not satisfied.
Options (d) are rejected.
Option (c) is correct
On the average, experienced person does 5 units of work while a fresh one 3 units work daily but the employer have to maintain the output of at least 30 units of work per day. The situation can be expressed as. - a)5x + 3y ≤ 30
- b)5x + 3y ≥ 30
- c)5x + 3y = 30
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
On the average, experienced person does 5 units of work while a fresh one 3 units work daily but the employer have to maintain the output of at least 30 units of work per day. The situation can be expressed as.
a)
5x + 3y ≤ 30
b)
5x + 3y ≥ 30
c)
5x + 3y = 30
d)
None of these
|
Dhruv Mehra answered |
Let "x" and "y" be the number of experienced person and fresh workmen respectively.
Total number of units of work done by experienced person per day = 5x
Total number of units of work done by fresh one per day = 3y
Total number of units of work done by both experienced person and fresh one per day = 5x + 3y
As per the question, total number of units of work per day should be at least 30 units.
That is, total number of units of work (5x+3y) should be equal to 30 or more than 30.
So, we have 5x + 3y ≥ 30
Hence, option "C" is correct
Solution space of the inequalities 2x + y ≤ 10 and x - 7 ≤ 5:
(i) includes the origin.
(ii) includes the point (4,3)
which one is correct?- a)Only (i)
- b) Only (ii)
- c) Both (i) and (ii)
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
Solution space of the inequalities 2x + y ≤ 10 and x - 7 ≤ 5:
(i) includes the origin.
(ii) includes the point (4,3)
which one is correct?
(i) includes the origin.
(ii) includes the point (4,3)
which one is correct?
a)
Only (i)
b)
Only (ii)
c)
Both (i) and (ii)
d)
None of the above
|
|
Amrutha Goyal answered |
There is no specific inequality given in the question. Please provide the complete information so that the solution space can be determined.
|x – 6| > x2 – 5x + 9- a)1 £ x < 3
- b)1 < x < 3
- c)2 < x < 5
- d)–3 < x < 1
Correct answer is option 'B'. Can you explain this answer?
|x – 6| > x2 – 5x + 9
a)
1 £ x < 3
b)
1 < x < 3
c)
2 < x < 5
d)
–3 < x < 1
|
|
Atharva Khanna answered |
At x = 2, inequality is satisfied.
At x = 0, inequality is not satisfied.
At x = 1, inequality is not satisfied but LHS = RHS.
At x = 3, inequality is not satisfied but LHS = RHS.
Thus, option (b) is correct.
Solve other questions of LOD I and LOD II in the same fashion.
At x = 0, inequality is not satisfied.
At x = 1, inequality is not satisfied but LHS = RHS.
At x = 3, inequality is not satisfied but LHS = RHS.
Thus, option (b) is correct.
Solve other questions of LOD I and LOD II in the same fashion.
|x - 6| > x2 - 5x + 9- a)1 ≤ x < 3
- b)1 < x < 3
- c)2 < x < 5
- d)-3 < x < 1
Correct answer is option 'B'. Can you explain this answer?
|x - 6| > x2 - 5x + 9
a)
1 ≤ x < 3
b)
1 < x < 3
c)
2 < x < 5
d)
-3 < x < 1
|
Shail Jain answered |
At x = 2, inequality is satisfied.
At x = 0, inequality is not satisfied.
At x = 1, inequality is not satisfied but LHS = RHS. At x = 3, inequality is not satisfied but LHS = RHS. Thus, option (b) is correct.
Solve other questions of LOD I and LOD II in the same fashion.
At x = 0, inequality is not satisfied.
At x = 1, inequality is not satisfied but LHS = RHS. At x = 3, inequality is not satisfied but LHS = RHS. Thus, option (b) is correct.
Solve other questions of LOD I and LOD II in the same fashion.
Let a, b, c, d be four integers such that a + b + c + d = 4m + 1 where m is a positive integer. Given m, which one of the following is necessarily true?- a)The minimum possible value of a2 + b2 + c2 + d2 is 4m2 - 2m + 1
- b)The maximum possible value of a2 + b2 + c2 + d2 is 4m2 + 2m + 1
- c)The maximum possible value of a2 + b2 + c2 + d2 is 4m2 - 2m + 1
- d)The minimum possible value of a2 + b2 + c2 + d2 is 4m2 + 2m +1
Correct answer is option 'D'. Can you explain this answer?
Let a, b, c, d be four integers such that a + b + c + d = 4m + 1 where m is a positive integer. Given m, which one of the following is necessarily true?
a)
The minimum possible value of a2 + b2 + c2 + d2 is 4m2 - 2m + 1
b)
The maximum possible value of a2 + b2 + c2 + d2 is 4m2 + 2m + 1
c)
The maximum possible value of a2 + b2 + c2 + d2 is 4m2 - 2m + 1
d)
The minimum possible value of a2 + b2 + c2 + d2 is 4m2 + 2m +1
|
|
S.S Career Academy answered |
Taking lowest possible positive value of m i.e. 1 .
Such that a + b + c + d = 5, so atleast one of them must be grater than 1, take a = b = c = 1 and d = 2
we get a2 + b2 + c2 + d2 = 7 which is equal to 4m2 + 2m + 1 for other values it is greater
than 4m2 + 2m +1.
than 4m2 + 2m +1.
If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a)(1 + b)(1 + c)(1+ d)?- a)16
- b)1
- c)4
- d)18
Correct answer is option 'A'. Can you explain this answer?
If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a)(1 + b)(1 + c)(1+ d)?
a)
16
b)
1
c)
4
d)
18
|
Upsc Rank Holders answered |
Since the product is constant, (a + b + c + d)/4 > = (abcd)1/4
We know that abcd = 1.
Therefore, a + b + c + d > = 4
(a + 1)(b + 1)(c + 1)(d + 1)
= 1 + a + b + c + d + ab + ac + ad + bc + bd + cd + abc + bed + cda + dab + abcd
We know that abcd = 1
Therefore, a = 1/bcd, b = 1/acd, c = 1/bda and d = 1/abc
Also, cd = 1/ab, bd = 1/ac, bc = 1/ad
The expression can be clubbed together as
1 + abcd + (a+1/a)+(b+1/b)+(c+1/c)+(d+1/d) + (ab+1/ab) + (ac+1/ac) + (ad +1/ad)
For any positive real number x, x + 1/x ≥ 2
Therefore, the least value that (a+1/a), (b+1/b).... (ad + 1/ad) can take is 2.
(a+1)(b+1)(c+1)(d+1) > 1 + 1 + 2 + 2 + 2 + 2 + 2 + 2 + 2
=> (a + 1)(b + 1)(c + 1)(d + 1) ≥ 16
The least value that the given expression can take is 16.
If x > 5 and y < -1, then which of the following statements is true?- a)(x + 4y) > 1
- b)x > -4y
- c)-4x < 5y
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
If x > 5 and y < -1, then which of the following statements is true?
a)
(x + 4y) > 1
b)
x > -4y
c)
-4x < 5y
d)
None of these
|
Damini jain answered |
If x = 5, then the equation y = 2x + 3 can be solved as follows:
y = 2(5) + 3
y = 10 + 3
y = 13
y = 2(5) + 3
y = 10 + 3
y = 13
If x, y and z are real numbers such that x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have?- a)5/3
- b)13/3
- c)√19
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
If x, y and z are real numbers such that x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have?
a)
5/3
b)
13/3
c)
√19
d)
None of these
|
|
Upsc Rank Holders answered |
The given equations are x + y + z = 5 — (1) , xy + yz + zx = 3 — (2)
xy + yz + zx = 3
x(y + z) + yz = 3
⇒ x ( 5 - x ) +y ( 5 – x – y) = 3
⇒ -y2 - y(5 - x) - x2 + 5x = 3
⇒ y2 + y(x - 5) + (x2 - 5x + 3) = 0
The above equation should have real roots for y, => Determinant >= 0
⇒ y2 + y(x - 5) + (x2 - 5x + 3) = 0
The above equation should have real roots for y, => Determinant >= 0
⇒ b2 - 4ac0
⇒ (x - 5)2 - 4(x2 - 5x + 3) ≥ 0
⇒ 3x2 - 10x - 13 ≤ 0
⇒ -1 ≤ x ≤ 13/3
Hence maximum value x can take is 13/3, and the corresponding values for y,z are 1/3, 1/3
The union however forbids forbide him to employ less than 2 experiencedd person to each fresh person. This situation can be expressed as- a)x ≤ y/2
- b)y ≤ x/2
- c)y ≥ x/2
- d)x > 2y
Correct answer is option 'B'. Can you explain this answer?
The union however forbids forbide him to employ less than 2 experiencedd person to each fresh person. This situation can be expressed as
a)
x ≤ y/2
b)
y ≤ x/2
c)
y ≥ x/2
d)
x > 2y
|
|
Moumita Bajaj answered |
≥ 2y, where x represents the number of experienced workers and y represents the number of fresh workers that the employer hires.
The union forbids employer to employ less than two experienced person (x) to each, fresh person (y). This situation can be expressed as: - a)x ≤ y/2
- b)y ≤ x/2
- c)y ≥ x/2
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The union forbids employer to employ less than two experienced person (x) to each, fresh person (y). This situation can be expressed as:
a)
x ≤ y/2
b)
y ≤ x/2
c)
y ≥ x/2
d)
None of these
|
|
Akshat Choudhary answered |
≥ 2y
b)y ≥ 2x
c)2x ≥ y
d)y ≥ x/2
Answer: c) 2x ≥ y
b)y ≥ 2x
c)2x ≥ y
d)y ≥ x/2
Answer: c) 2x ≥ y
The number of integers n satisfying -n + 2 ≥ 0 and 2n ≥ 4 is- a)1
- b)0
- c)2
- d)3
Correct answer is option 'A'. Can you explain this answer?
The number of integers n satisfying -n + 2 ≥ 0 and 2n ≥ 4 is
a)
1
b)
0
c)
2
d)
3
|
|
Innovative Classes answered |
First inequality:
-n + 2 ≥ 0
-n ≥ -2
n ≤ 2
Second inequality:
2n ≥ 4
n ≥ 2
Only n = 2 satisfies both inequalities. So, there is only 1 integer that satisfies both the inequalities.
The correct option is A.
Find the range of real values of x satisfying the inequalities 3x-2 > 7 and 4x - 13 > 15 - a)x > 3
- b)x > 7
- c)x < 7
- d)x < 3
Correct answer is option 'B'. Can you explain this answer?
Find the range of real values of x satisfying the inequalities 3x-2 > 7 and 4x - 13 > 15
a)
x > 3
b)
x > 7
c)
x < 7
d)
x < 3
|
|
Dhruv Mehra answered |
As, 3x - 2 > 7 => 3x > 9 => x > 3
and 4x - 13 > 15 => 4x > 28 => x > 7, 7 is greater than 3.
Therefore, x > 7.
A firm makes two types of products: Type A and Type B. The profit on product A is Rs. 20 each and that on product B is Rs. 30 each. Both types are processed on three machines M1, M2 and M3. The time required in hour by each product and total time available in hours per week on each machine are asa follows:
Machine Product A Product B Available Time
M1 3 3 36
M2 5 2 50
M3 2 6 60
The constraints can be formulated taking x1 = number of units A and x2 = number of unit of B as- a)x1 + x2 ≤ 12
5x1 + 2x2 ≤ 50
2x1 + 6x2 ≤ 60 - b)3x1 + 3x2 ≥ 36
5x1 + 2x2 ≤ 50
2x1 + 6x2 ≥ 60
x1 ≥ 0, x2 ≥ 0 - c)3x1 + 3x2 ≤ 36
5x1 + 2x2 ≤ 50
2x1 + 6x2 ≤ 60
x1 ≥ 0, x2 ≥ 0 - d)none of these
Correct answer is option 'C'. Can you explain this answer?
A firm makes two types of products: Type A and Type B. The profit on product A is Rs. 20 each and that on product B is Rs. 30 each. Both types are processed on three machines M1, M2 and M3. The time required in hour by each product and total time available in hours per week on each machine are asa follows:
Machine Product A Product B Available Time
M1 3 3 36
M2 5 2 50
M3 2 6 60
The constraints can be formulated taking x1 = number of units A and x2 = number of unit of B as
Machine Product A Product B Available Time
M1 3 3 36
M2 5 2 50
M3 2 6 60
The constraints can be formulated taking x1 = number of units A and x2 = number of unit of B as
a)
x1 + x2 ≤ 12
5x1 + 2x2 ≤ 50
2x1 + 6x2 ≤ 60
5x1 + 2x2 ≤ 50
2x1 + 6x2 ≤ 60
b)
3x1 + 3x2 ≥ 36
5x1 + 2x2 ≤ 50
2x1 + 6x2 ≥ 60
x1 ≥ 0, x2 ≥ 0
5x1 + 2x2 ≤ 50
2x1 + 6x2 ≥ 60
x1 ≥ 0, x2 ≥ 0
c)
3x1 + 3x2 ≤ 36
5x1 + 2x2 ≤ 50
2x1 + 6x2 ≤ 60
x1 ≥ 0, x2 ≥ 0
5x1 + 2x2 ≤ 50
2x1 + 6x2 ≤ 60
x1 ≥ 0, x2 ≥ 0
d)
none of these
|
|
Mehul Saini answered |
To formulate the constraints, we need to consider the time required by each product on each machine and the available time on each machine. Let x1 be the number of units of product A produced and x2 be the number of units of product B produced.
On machine M1:
The time required to process x1 units of A is 3x1 hours
The time required to process x2 units of B is 3x2 hours
The total time available is 36 hours
So, the constraint is: 3x1 + 3x2 ≤ 36
On machine M2:
The time required to process x1 units of A is 5x1 hours
The time required to process x2 units of B is 2x2 hours
The total time available is 50 hours
So, the constraint is: 5x1 + 2x2 ≤ 50
On machine M3:
The time required to process x1 units of A is 2x1 hours
The time required to process x2 units of B is 6x2 hours
The total time available is 60 hours
So, the constraint is: 2x1 + 6x2 ≤ 60
We also have the non-negativity constraints:
x1 ≥ 0 and x2 ≥ 0
The objective function is to maximize the total profit, which can be expressed as:
Z = 20x1 + 30x2
Therefore, the linear programming problem can be formulated as follows:
Maximize Z = 20x1 + 30x2
Subject to:
3x1 + 3x2 ≤ 36
5x1 + 2x2 ≤ 50
2x1 + 6x2 ≤ 60
x1 ≥ 0, x2 ≥ 0
On machine M1:
The time required to process x1 units of A is 3x1 hours
The time required to process x2 units of B is 3x2 hours
The total time available is 36 hours
So, the constraint is: 3x1 + 3x2 ≤ 36
On machine M2:
The time required to process x1 units of A is 5x1 hours
The time required to process x2 units of B is 2x2 hours
The total time available is 50 hours
So, the constraint is: 5x1 + 2x2 ≤ 50
On machine M3:
The time required to process x1 units of A is 2x1 hours
The time required to process x2 units of B is 6x2 hours
The total time available is 60 hours
So, the constraint is: 2x1 + 6x2 ≤ 60
We also have the non-negativity constraints:
x1 ≥ 0 and x2 ≥ 0
The objective function is to maximize the total profit, which can be expressed as:
Z = 20x1 + 30x2
Therefore, the linear programming problem can be formulated as follows:
Maximize Z = 20x1 + 30x2
Subject to:
3x1 + 3x2 ≤ 36
5x1 + 2x2 ≤ 50
2x1 + 6x2 ≤ 60
x1 ≥ 0, x2 ≥ 0
The common region satisfying the set of inequalities x≥ 0, y ≥ 0 , L1 : x+ y ≤ 5, L2 : x + 2y ≤ 8 and L3: 4x+ 3y ≥ 12 is indicated by - a)
- b)
- c)
- d) none of these
Correct answer is option 'A'. Can you explain this answer?
The common region satisfying the set of inequalities x≥ 0, y ≥ 0 , L1 : x+ y ≤ 5, L2 : x + 2y ≤ 8 and L3: 4x+ 3y ≥ 12 is indicated by
a)
b)
c)
d)
none of these
|
Shrishti Kesharwani answered |
The correct ans is A because L2 shows less than ....so the shaded portion is between L3 n L2..Meanwhile L1 also shows less then i.e downward side
The shaded region represents:
- a)3x + 5y ≥ 15, 5x+2y ≥ 10, x, y ≥ 0
- b)3x + 5y ≤ 15, 5x+2y ≤ 10, x, y≥ 0
- c)3x+ 5y ≥ 15, 5x+2y ≥ 10, x, y, ≥ 0
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The shaded region represents:
a)
3x + 5y ≥ 15, 5x+2y ≥ 10, x, y ≥ 0
b)
3x + 5y ≤ 15, 5x+2y ≤ 10, x, y≥ 0
c)
3x+ 5y ≥ 15, 5x+2y ≥ 10, x, y, ≥ 0
d)
None of these
|
|
Alok Mehta answered |
The feasible region determined by the system of constraints, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are as follows:
Therefore, at the point ( 20/19 , 45/19 ) the maximum value of Z is 235/19.
The corner points of the feasible region are O (0, 0), A (2, 0), B (0, 3), and C ( 20/19 , 45/19 ).
The values of Z at these corner points are as follows:
Therefore, at the point ( 20/19 , 45/19 ) the maximum value of Z is 235/19.
The shaded region represents:
- a) 3x + 2y ≤ 24, x+ 2y ≥ 16
- b)3x + 2y ≤ 24, x+2y ≤ 16, x+y ≥ 10
- c)3x + 2y ≤ 24, x+2y ≤ 16, x+y ≤ 10, x≥ 0, y≥ 0
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
The shaded region represents:
a)
3x + 2y ≤ 24, x+ 2y ≥ 16
b)
3x + 2y ≤ 24, x+2y ≤ 16, x+y ≥ 10
c)
3x + 2y ≤ 24, x+2y ≤ 16, x+y ≤ 10, x≥ 0, y≥ 0
d)
None of these
|
Janhavi Basu answered |
The shaded region is below the lines 3x + 2y = 24, x+2y = 16, x+y = 10
and above the lines x = 0 , y = 0
The longest side of a triangle is 3 times the shortest side, and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.- a)9 cm
- b)11 cm
- c)13 cm
- d)14 cm
Correct answer is option 'A'. Can you explain this answer?
The longest side of a triangle is 3 times the shortest side, and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
a)
9 cm
b)
11 cm
c)
13 cm
d)
14 cm
|
Freedom Institute answered |
Let x cm be the length of the shortest side of the triangle.
∴ According to the question, length of the longest side = 3x cm
Length of the third side = (3x – 2) cm
The least perimeter of the triangle = 61 cm (given)
Thus, x + 3x + (3x – 2) cm ≥ 61 cm
= 7x – 2 ≥ 61
= 7x ≥ 63
Dividing by 7 on both sides, we get;
= 7x/7 ≥ 63/7
= x ≥ 9
Hence, the minimum length of the shortest side will be 9 cm.
L1 : 5x + 3y = 30 L2 : x+y = 9 L3 : y = x/3 L4 : y = x/2
The common region (shaded part) shown in the diagram refers to- a)5x + 3y ≤ 30
x + y ≤ 9
y ≤ 1/5 x
y ≤ x/2 - b)5x+ 3y ≥ 30
x + y ≤ 9
y ≥ x/3
y ≤ x/2
x ≥ 0, y ≥ 0 - c)5x + 3y ≥ 30
x + y ≥ 9
y≤ x/3
y ≥ x/2
x ≥ 0, y ≥ 0 - d) None of these
Correct answer is option 'B'. Can you explain this answer?
L1 : 5x + 3y = 30 L2 : x+y = 9 L3 : y = x/3 L4 : y = x/2
The common region (shaded part) shown in the diagram refers to
a)
5x + 3y ≤ 30
x + y ≤ 9
y ≤ 1/5 x
y ≤ x/2
x + y ≤ 9
y ≤ 1/5 x
y ≤ x/2
b)
5x+ 3y ≥ 30
x + y ≤ 9
y ≥ x/3
y ≤ x/2
x ≥ 0, y ≥ 0
x + y ≤ 9
y ≥ x/3
y ≤ x/2
x ≥ 0, y ≥ 0
c)
5x + 3y ≥ 30
x + y ≥ 9
y≤ x/3
y ≥ x/2
x ≥ 0, y ≥ 0
x + y ≥ 9
y≤ x/3
y ≥ x/2
x ≥ 0, y ≥ 0
d)
None of these
|
Mohit Sanklecha answered |
D
A dietitian wishes to mix together two kinds of food so that the vitamin content of the mixture is at least 9 units of vitamin A, 7 units of vitamin B, 10 units of vitamin C and 12 units of vitamin D. The vitamin content per Kg. of each food is shown below.
Food I : A = 2 , B = 1, C = 1, D = 2
Foof II : A = 1, B = 1, C = 2 ,D = 3
Assuming x units of food I is to be mixed with y units of food II the situation can be expressed as- a)2x + y ≤ 9
x + y ≤ 7
x + 2y ≤ 10
2x+ 3 y ≤ 12
x > 0, y > 0 - b)2x + y ≥ 30
x + y ≤ 7
x + 2y ≥ 10
x + 3y ≥ 12 - c)2x + y ≥ 9
x + y ≥ 7
x + y ≤ 10
x + 3y ≥ 12 - d)2x + y ≥ 9
x + y ≥ 7
x + 2 y ≥ 10
2x + 3 y ≥ 12
x ≥ 0 , y ≥ 0,
Correct answer is option 'D'. Can you explain this answer?
A dietitian wishes to mix together two kinds of food so that the vitamin content of the mixture is at least 9 units of vitamin A, 7 units of vitamin B, 10 units of vitamin C and 12 units of vitamin D. The vitamin content per Kg. of each food is shown below.
Food I : A = 2 , B = 1, C = 1, D = 2
Foof II : A = 1, B = 1, C = 2 ,D = 3
Assuming x units of food I is to be mixed with y units of food II the situation can be expressed as
Food I : A = 2 , B = 1, C = 1, D = 2
Foof II : A = 1, B = 1, C = 2 ,D = 3
Assuming x units of food I is to be mixed with y units of food II the situation can be expressed as
a)
2x + y ≤ 9
x + y ≤ 7
x + 2y ≤ 10
2x+ 3 y ≤ 12
x > 0, y > 0
x + y ≤ 7
x + 2y ≤ 10
2x+ 3 y ≤ 12
x > 0, y > 0
b)
2x + y ≥ 30
x + y ≤ 7
x + 2y ≥ 10
x + 3y ≥ 12
x + y ≤ 7
x + 2y ≥ 10
x + 3y ≥ 12
c)
2x + y ≥ 9
x + y ≥ 7
x + y ≤ 10
x + 3y ≥ 12
x + y ≥ 7
x + y ≤ 10
x + 3y ≥ 12
d)
2x + y ≥ 9
x + y ≥ 7
x + 2 y ≥ 10
2x + 3 y ≥ 12
x ≥ 0 , y ≥ 0,
x + y ≥ 7
x + 2 y ≥ 10
2x + 3 y ≥ 12
x ≥ 0 , y ≥ 0,
|
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Dhruba Choudhary answered |
We can set up the following system of inequalities to represent the given constraints:
2x + y ≥ 9
x + y ≥ 7
x + 2y ≥ 10
2x + 3y ≥ 12
We want to find the values of x and y that satisfy all of these inequalities and minimize the amount of food used (i.e. minimize x+y).
One approach to solve this type of problem is to graph the inequalities and find the feasible region, which is the region that satisfies all of the constraints. The optimal solution will be at one of the vertices of the feasible region. However, since we have four inequalities, graphing them can be time-consuming and not very accurate.
Another approach is to use linear programming, which is a method for optimizing a linear objective function subject to linear constraints. In this case, our objective function is x+y (the total amount of food used). We want to minimize this function subject to the constraints given by the four inequalities.
We can use a software program or a calculator that has a linear programming function to solve this problem. For example, using the Solver add-in in Microsoft Excel, we can set up the problem as follows:
Objective: Minimize x+y
Subject to:
2x + y ≥ 9
x + y ≥ 7
x + 2y ≥ 10
2x + 3y ≥ 12
x ≥ 0
y ≥ 0
Solving this problem using the Simplex algorithm, we get the optimal solution:
x = 2.4 Kg of food I
y = 4.6 Kg of food II
The total amount of food used is x+y = 7 Kg.
Therefore, to meet the vitamin requirements, the dietitian should mix 2.4 Kg of food I with 4.6 Kg of food II, and the total amount of food used will be 7 Kg.
2x + y ≥ 9
x + y ≥ 7
x + 2y ≥ 10
2x + 3y ≥ 12
We want to find the values of x and y that satisfy all of these inequalities and minimize the amount of food used (i.e. minimize x+y).
One approach to solve this type of problem is to graph the inequalities and find the feasible region, which is the region that satisfies all of the constraints. The optimal solution will be at one of the vertices of the feasible region. However, since we have four inequalities, graphing them can be time-consuming and not very accurate.
Another approach is to use linear programming, which is a method for optimizing a linear objective function subject to linear constraints. In this case, our objective function is x+y (the total amount of food used). We want to minimize this function subject to the constraints given by the four inequalities.
We can use a software program or a calculator that has a linear programming function to solve this problem. For example, using the Solver add-in in Microsoft Excel, we can set up the problem as follows:
Objective: Minimize x+y
Subject to:
2x + y ≥ 9
x + y ≥ 7
x + 2y ≥ 10
2x + 3y ≥ 12
x ≥ 0
y ≥ 0
Solving this problem using the Simplex algorithm, we get the optimal solution:
x = 2.4 Kg of food I
y = 4.6 Kg of food II
The total amount of food used is x+y = 7 Kg.
Therefore, to meet the vitamin requirements, the dietitian should mix 2.4 Kg of food I with 4.6 Kg of food II, and the total amount of food used will be 7 Kg.
Chapter doubts & questions for Chapter 3: Linear Inequalities - Quantitative Aptitude for CA Foundation 2025 is part of CA Foundation exam preparation. The chapters have been prepared according to the CA Foundation exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for CA Foundation 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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