Quantitative Aptitude - Inequalities
Exam Duration: 45 Mins Total Questions : 30
If m < n and a < b, then
- (a)
m - a < n - b
- (b)
ma < nb
- (c)
m/a < n/b
- (d)
m + a < n + b
If a < b and c < 0 then
- (a)
a/c < b/c
- (b)
a/c > b/c
- (c)
a/c = b/c
- (d)
a/c =0
Quartiles can be found through which graph?
- (a)
Ogive
- (b)
Histogram
- (c)
Frequency polygon
- (d)
Frequency curve
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 related by the inequality
- (a)
x + y≠ 9
- (b)
x+y\(\le \)9
- (c)
x+y\(\ge \)9
- (d)
None of these
The union however forbids him to employ less than 2 experienced persons to each fresh person. This situation can be expressed as
- (a)
x\(\le \)y/2
- (b)
y\(\le \)y/2
- (c)
x\(\ge \)2y
- (d)
Both (b) and (c)
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 hours by each product and total time available in hours per week on each machine are as follows
Machine | Product A | Product B | Product C |
M1 | 3 | 3 | 36 |
M2 | 5 | 2 | 50 |
M3 | 2 | 6 | 60 |
The constraints can be formulated by taking X1 = number of units of A and x2 = number of units of B as
- (a)
3x1+3x2\(\le \)36,5x1+2x2\(\le \) 50,2x1+6x2 \(\le \)60,x1\(\ge \)0 x2 \(\ge \)0
- (b)
3x1+3x2\(\ge \)36,5x1+2x2\(\ge \) 50,2x1+6x2 \(\ge \)60,x1\(\ge \)0 x2 \(\ge \)0
- (c)
3x1+3x2\(\le \)36,5x1+2x2\(\ge \) 50,2x1+6x2 \(\le \)60,x1\(\ge \)0 x2\(\le \)0
- (d)
3x1+3x2\(\ge \)36,5x1+2x2\(\le \)50,2x1+6x2 \(\ge \)60,x1\(\le \)0 x2 \(\le \)0
A company produces two types of leather belts, say A and B. Belt A is of superior quality and belt B is of lower quality. Each belt of type A requires twice as much as time required by a belt of type B. If all belts were of type B, the company could produce 1000 belts per day. But the supply of leather is sufficient only for 800 belts per day. Belt A requires fancy buckles and only 400 fancy buckles are available per day. For belt of type B only 700 buckles are available per day. Constraints can be formulated by assuming that the company produces x units of belt A and y units of belt B as
- (a)
x + 2y\(\ge \)1000, x + y\(\ge \)800 , x\(\ge \)400; y\(\le \)700, x\(\ge \)0; y\(\ge \)0
- (b)
2x + y\(\le \)1000, x + y\(\le \)800 , x\(\le \)400; y\(\le \)700, x\(\ge \)0; y\(\ge \)0
- (c)
x + 2y\(\ge \)1000, x + y\(\le \)800 , x\(\ge \)400; y\(\ge \)700, x\(\ge \)0; y\(\ge \)0
- (d)
x + 2y\(\le \)1000, x + y\(\ge \)800 , x\(\le \)400; y\(\ge \)700, x\(\ge \)0; y\(\ge \)0
A man makes two types of furniture: chairs and tables. Profits are Rs.20 per chair and Rs.30 per table. Both the products are processed on two machines M1 and M2. The time required for each product in hours and total time available in hours per week on each machine are as follows
Machine | Chair | Table | Available Time |
M1 | 3 | 3 | 36 |
M2 | 5 | 2 | 50 |
Constraints can be formulated by taking x = no.of chairs, y = no.of tables produced as
- (a)
x + y\(\le \)12, 5x + 2y\(\ge \) 50, x\(\le \)0; y\(\ge \)0
- (b)
x + y\(\ge \)12, 5x + 2y\(\le \) 50, x\(\ge \)0; y\(\le \)0
- (c)
x + y\(\le \)12, 5x + 2y\(\le \) 50, x\(\ge \)0; y\(\ge \)0
- (d)
x + y\(\ge \)12, 5x + 2y\(\ge \) 50, x\(\le \)0; y\(\le \)0
A small manufacturing firm produces two types of gadgets A and B, which are first processed in the foundary, and then sent to another machine for finishing. The number of man-hours for the firm available per week are as follows
Foundry | Machine-shop | |
A | 10 | 5 |
B | 6 | 4 |
Capacity per Week (man hours) |
1000 | 600 |
Let the firm manufacture x units of A and y units of B. The constraints are
- (a)
10x+ 6y\(\le \)1000,5x + 4y\(\ge \)600, x\(\ge \)0; y\(\le \)0
- (b)
10x+ 6y\(\le \)1000,5x + 4y\(\le \)600, x\(\ge \)0; y\(\ge \)0
- (c)
10x+ 6y\(\ge \)1000,5x + 4y\(\le \)600, x\(\le \)0; y\(\ge \)0
- (d)
10x+ 6y\(\ge \)1000,5x + 4y\(\ge \)600, x\(\le \)0; y\(\le \)0
Suppose a man needs a minimum of 50 units of carbohydrate, 40 units of proteins per month for good health. He is taking food at two places, viz., A and B, food at A contains 4 and 5 units of carbohydrates and proteins respectively.
Express this in the form of linear inequalities assuming the man is in good health. Let x1 and x2 represent carbohydrates and proteins respectively. Then mathematical inequalities are
- (a)
4x1+ x2 \(\ge \)50, 5x1+3x2\(\le \)40, x1\(\ge \)0;x2\(\ge \)0
- (b)
4x1+ x2 \(\le \)50, 5x1+3x2\(\ge \)40, x1\(\le \)0;x2\(\ge \)0
- (c)
4x1+ x2 \(\ge \)50, 5x1+3x2\(\ge \)40, x1\(\ge \)0;x2\(\ge \)0
- (d)
4x1+ x2 \(\le \)50, 5x1+3x2\(\le \)40, x1\(\le \)0;x2\(\le \)0
The common region (Shaded part) shown in the diagram refers to
L1:5x + 3y = 30 : L2: x + y = 9: L3: y = x/30: L4: y = x/2
- (a)
5x+3y\(\le \)30; x+y\(\le \)9; y\(\le \)1/5x; y\(\le \)x/2
- (b)
5x+3y\(\ge \)30; x+y\(\le \)9; y\(\ge \)3/5x; y\(\le \)x/2;x\(\ge \)0,y\(\ge \)0
- (c)
5x+3y\(\ge \)30; x+y\(\ge \)9; y\(\le \)3/5x; y\(\ge \)x/2;x\(\ge \)0,y\(\ge \)0
- (d)
5x+3y>30; x+y<9; y\(\ge \)9; y\(\le \)x/2;x\(\ge \)0,y\(\ge \)0
The common region (Shaded part) shown in the diagram expressed by the set of inequalities
L1:2x +y = 9 : L2: x + y = 7: L3: y = x+2y =10: L4: x+3y=12
- (a)
2x+y\(\le \)9;x+y\(\ge \)7;x+2y\(\ge \)10;2x+3y\(\ge \)12
- (b)
2x+y\(\ge \)9;x+y\(\le \)7;x+2y\(\ge \)10;2x+3y\(\ge \)12
- (c)
2x+y\(\ge \)9;x+y\(\ge \)7;y+2y\(\ge \)10;x+3y\(\ge \)12;x\(\ge \)0,y\(\ge \)0
- (d)
None of these
The inequalities x1+ 2x2\(\le \)5, x1+x2\(\ge \)1, x1\(\ge \)0, x2\(\ge \)0 represents the region
- (a)
- (b)
- (c)
- (d)
The inequalities x1\(\ge \)0, x2\(\ge \)0,are represented by one of the graphs shown below
- (a)
- (b)
- (c)
- (d)
The region is expressed as
- (a)
x1-x2 \(\ge \)1
- (b)
x1-x2\(\le \)1
- (c)
x1+x2\(\ge \)1
- (d)
None of these
The inequality -x1+2x2\(\le \)0 is indicated on the graph as
- (a)
- (b)
- (c)
- (d)
None of these
The common region indicated on the graph is expressed by the set of five inequalities
- (a)
L1:x1\(\ge \)0;L2:x2\(\ge \)0;L3:x1+x2\(\le \)1;L4:x1-x2\(\ge \)1;L5:-x1+2x2\(\le \)0
- (b)
L1::x1\(\ge \)0;L2::x2\(\ge \)0;L3:x1+x2\(\ge \)1;L4:x1-x2\(\ge \)1;L5:-x1+2x2\(\le \)0
- (c)
L1::x1\(\le \)0;L2::x2\(\le \)0;L3::x1+x2\(\ge \)1;L4:x1-x2\(\ge \)1;L5:-x1+2x2\(\le \)0
- (d)
None of these
By lines x + y = 6, 2x - y = 2, the common region shown is the diagram refers to
- (a)
x + y \(\ge \)6, 2x - y\(\le \)2, x\(\ge \)0, y\(\ge \)0
- (b)
x + y \(\le \)6, 2x - y\(\le \)2, x\(\ge \)0, y\(\ge \)0
- (c)
x + y \(\le \)6, 2x - y\(\ge \)2, x\(\ge \)0, y\(\ge \)0
- (d)
None of these
Given conditions x+y\(\ge \)5,x+y\(\le \)5, 0\(\le \)x\(\le \)4 and 0\(\le \)x\(\le \)2
then the common region under these conditions is
- (a)
ECDE
- (b)
EOABCE
- (c)
Line segment CD
- (d)
Line segment BC
When an inequation is multiplied or divide by same negative number, inequation ---direction.
- (a)
Changes
- (b)
Does Not Change
- (c)
Either (a) or (b)
- (d)
Neither (a) or (b)
-6X < -18 implies
- (a)
X < 3
- (b)
X > 3
- (c)
X = 0
- (d)
X = 3
4X > -16 implies
- (a)
X greater than equal to-4
- (b)
X less than -4
- (c)
X greater than -4
- (d)
X less than equal to -4
X > -3 implies
- (a)
-2X less than 6
- (b)
2X greater than -6
- (c)
Both (a) or (b)
- (d)
Neither (a) nor (b)
What is the largest integer value of p that satisfies the inequality 4 + 3p < p + 1?
- (a)
-2
- (b)
-1
- (c)
0
- (d)
1
In the inequality 4x + 3 < 2x + 5, all of the following may be a value of 'x' except
- (a)
0
- (b)
1
- (c)
-1
- (d)
-2
If x\(\le \)0, then 2/a + 8/x is
- (a)
2\(\le \)x\(\le \)3
- (b)
\(\ge \)0
- (c)
\(\ge \)4
- (d)
\(\le \)-1
If xy > 1 and z < 0, which of the following statements must be true?
I. x > z
II. xyz < -1
III. xy/z < 1/z
- (a)
I only
- (b)
II only
- (c)
III only
- (d)
II and III
If A = x - 2-1, B = x + 2-1, and A2 - B2 > 0, then
- (a)
x > 0
- (b)
x < 0
- (c)
x = 0
- (d)
x = A + B
Given |byx| +|bxy|\(\ge 2|\rho|\). Therefore if \(\rho\) > 0 and m = (byx + bxy)/2, then-
- (a)
m\(\ge \)p
- (b)
m=p
- (c)
m\(\le \)p
- (d)
None of these
Solve for real 'x' if \((x-4)/(2x-3)\le0\)
- (a)
x = 1/8 or 2/3
- (b)
\(1.5<x\le4\)
- (c)
x = 4 or 3/2
- (d)
\(x\ge4\)