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Q1. Pairs of consecutive odd positive integers, both of which are larger than 10 such that their sum is less than 40, are

• (11, 13), (13, 15), (15, 17), (17, 19)
• (11, 13), (13, 15), (15, 17), (17, 19), (19, 21)
• (11, 13), (13, 15), (15, 17)
• (11, 13), (13, 15), (17, 19)

Solution

Let the pair of consecutive odd positive integers be x, x + 2 such that both are greater than 10 and their sum is less than 40     x > 10 and  x + (x + 2) < 40     x > 10 and  2x + 2 < 40     x > 10 and  2x < 38     x > 10 and  x < 19     10 < x < 19 and x is odd     x = 11, 13, 15, 17 and therefore, x + 2 = 13, 15, 17, 19     Possible pairs are (11, 13), (13, 15), (15, 17), (17, 19)   Note:  x can’t be 19 because x + (x + 2) = 19 + 21 = 40 which is not less that 40.

Q2. The region bounded by 2x + y < 6 is

• A triangle
• Square
• Bounded
• Unbouned

Solution

The region represented by the above inequality is a half plane, Hence it is unbounded.

Q3. The point that lies in the region x + 2y < 2 is

• (0,3)
• (1,2)
• (-1,0)
• (2,3)

Solution

By substituting the coordinates of the points we find that (-1,0) satisfies the equation x + 2y < 2.

Q4. Find the solution for the pair of inequations x > 1 and x < -1

• x > 1
• x < -1
• -1 < x < 1
• no solution

Solution

The pair of equations x > 1 and x < -1 have no solution since no region is common between them.

Q5. A point P lies in the solution region of 3x - 7 > x + 3. So the possible coordinates of P are

• (2,5)
• (-3,4)
• (0,7)
• (6,-1)

Solution

3x - 7 > x + 3 3x > x + 10 2x > 10 x > 5 so for a point to lie in the required region, its x coordinate must be > 5, hence (6,-1) is one such point.

Q6. A linear inequality in two variables represent a

• Point
• Line
• Half plane
• Quadrant

Solution

A linear inequation respresents a half plane as linear inequality in two variables divide the coordinate plane in two half planes each of them is called a half plane.

Q7. If 5x+6<2x-3, then

• x>-3
• x<-3
• x>3
• x<3

Solution

5x+6+3< 2x-3+3 So, continuing, we get, 3x < -9 ⇒ x < -3

Q8. What values of x satisfy -6x>24 and x is an integer?

• all integers less than -4
• all integers greater than -4
• all integers greater than -6
• all integers less than -6

Solution

-6x>24 So, -6x/-6 < 24/-6 (note the fact that the sign has inverted as we are dividing by a negative number) Thus, x < -4

Q9. The region y < -2 lies

• Below the line y = -2
• Above the line y = -2
• To the right of the line x = -2
• To the left of the line x = -2

Solution

The region lies below the line y=-2.The graph of the inequality is as shown. So,the region lies below the line y=-2.

Q10. A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to be 6 cm longer than the shortest and the third length is to be thrice as long as the shortest. If the third piece is to be at least 4 cm longer than the second, then the length of the shortest board will lie between

• 8 cm and 22 cm
• 7 cm and 21 cm
• 6 cm and 19 cm
• 5 cm and 17 cm

Solution

If x be the length of the shortest board then (x + 6) and 3x are the lengths of the second and third piece, respectively.      and                      and                           and                             and      The length of the shortest side will lie between 5 cm and 17 cm.

Q11. The origin lies on which side of the graph x + y < 5?

• Left of line
• Right of line
• Above the line
• Data inadequate

Solution

From the graph drawn below we find that the origin lies on the left side of the graph.

Q12. Find the solution for the pair of solution x > 1 and x > -1

• x > 1
• x < -1
• -1 < x < 1
• No solution

Solution

The solution to the pair of equations is x > 1 as x>1 is the region common to both.

Q13. A plumber can be paid under two schemes as given below: I:  Rs 600 and Rs 50 per hour. II: Rs 170 per hour. If the job takes n hours, then the values of n for which the scheme I will give the plumber better wages are

• 4 hours
• more than 5 hours
• less than 5 hours
• less than 4 hours

Solution

For better wages, earning in scheme I should be more than Scheme II    600 + 50 n > 170 n    600 > 170 n – 50 n    600 > 120 n    120n < 600              n < 5 hours     Values of n for which Scheme I will give the plumber better wages should be less than 5 hours.

Q14. What is the region represented by x > 0 and y < 0?

• 1^st quadrant
• 2^nd  quadrant
• 3^rd quadrant
• 4^th  quadrant

Solution

The region in which x > 0 and y < 0 is 4^th quadrant

Q15. If -5x+2<7x -4, then x is

• >1/2
• <1/2
• >3/2
• <3/2

Solution

As -5x+2<7x-4 So, -5x+2+5x < 7x-4+5x                  2  < 12x-4                  6  < 12x.     Thus, continuing further, we get,x>.

Q16. Find the pairs of consecutive even positive integers both of which are smaller than 10 and their sum of more than 11

• (6, 8)
• (6, 8) and (4, 8)
• (6, 4)(4, 2)
• (4, 8)

Solution

Even integers smaller than 10 are 2, 4, 6, 8 now the sum must be more than 11 and the integers should be consecutive so the pair of numbers is (6, 8).

Q17. Two less than 5 times a number is greater than the third multiple of the number. So the number must be

• Greater than 0
• Less than 2
• Greater than 1
• Less than 3

Solution

Let the number be x So according to question, 5x - 2 > 3x So, 2x > 2 i.e. x > 1

Q18. Pairs of consecutive even positive integers, both of which are larger than 7 such that their sum is less than 28, are

• (8, 10), (10, 12), (12, 14)
• (8, 10), (10, 12), (12, 14), (14, 16)
• (8, 10), (10, 12)
• (8, 9), (9, 10), (10, 11)

Solution

Let the pair of consecutive even positive integers be x, x + 2 such that such that both are greater than 7 and their sum is less than 28     x > 7 and  x + (x + 2) < 28     x > 7 and  2x + 2 < 28     x > 7 and  2x < 26     x > 7 and  x < 13     7 < x < 13 and x is even     x = 8, 10, 12 and therefore, x + 2 = 10, 12, 14     Possible pairs are (8, 10), (10, 12), (12, 14)

Q19. The cost and revenue functions of a product are given by C(x) = 2x + 400 and R(x) = 6x + 20 respectively, where x is the number of items produced by the manufacturer. The minimum number of items that the manufacturer must sell to realize some profit is

• 96
• 95
• 100
• 105

Solution

To realize some profit, Revenue must be more than Cost.   R(x) > C(x)   6x + 20 > 2x + 400   6x > 2x + 380   4x > 380   x > 95   Manufacturer must sell at least 96 items to realize some profit.

Q20. The longest side of a triangle is 4 times the shortest side and the third side is 4 cm shorter than the longest side. If the perimeter of the triangle is at least 59 cm, then the minimum length of the shortest side is

• 6 cm
• 7 cm
• 8 cm
• 9 cm

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