Inequations Questions
1. Solve x>15
Solution: We cannot cross multiply since we don’t know whether x is positive or negative.
Case 1: If x > 0
⇒⇒ 5 > x
⇒⇒ x < 5
So 0 < x < 5
Case 2: If x < 0
⇒⇒ 5 < x
⇒⇒ x > 5
No solution.
So Solution set for given inequation is 0 < x < 5.
2. Solve: x+3/x+4 ≥ 1
Solution:
x+3/x+4−1 ≥ 0
(x+3−x−4)/x+4 ≥ 0
−1/x+4 > 0
[a/b > 0 and a < 0⇒ b < 0]
x (– ∞ , 4]
3. (x−3)(6−x)(x−8)2 > 0
A. 3 <x> 10 and x ≠≠8
B. 3 < x < 9
C. 6 < x < 8 and x≠≠3
D. 3 < x < 6
E. 6 < x < 8
Answer:
(x−3)(6−x)(x−8)2 > 0
(x−3)(x−6)(x−8)2 < 0
As (x−8)2 is always positive, the sign of the above inequality depends on (x−3)(x−6)<0
So x should lie in between 3 and 6
Hence option 4
4. (x−3)√(x2)−2x−15 ≥ 0
A. x ≥ 5
B. x ≤ 5
C.x ≥ −8
D.x ≤ 3
E.x ≥ −5
Answer:
(x−3)√(x2)−2x−15 ≥ 0
(x−3)√(x2)−5x+3x−15 ≥ 0
(x−3)√(x−5)(x+3) ≥ 0
The above inequality is valid for
x−5 ≥ 0
That is, x ≥ 5
Hence option 1
5. √6x−8/5−2x > 2
A. 0.75 < x < 25
B. 0 < x < 5
C. 2 < x < 5
D. 5 < X < 11
Answer:
6x−8/5−2x > 22
6x−8/5−2x > 4
6x−8/5−2x−4 > 0
14x−28/5−2x > 0
x−2/2x−5 < 0
So, we have,
2 < x < 2.5
Hence option 5
6. |x2−12x| < 45
A. −15 ≤ x ≤ 3
B. 15 ≤ x ≤ −3
C. 15 < x < −3
D. −15 < x < 3
E. −15 ≤ x < −3
Answer:
|x2−12x| < 45
We have x2−12x < 45 and x2−12x > −45
x2−12x−45 < 0--------- (1) and x2−12x+45 > 0------(2)
Solving the first equation,
(x−15)(x+3) < 0
−3 < x < 15---------(3)
For second equation, b2−4ac = (−12)2−4.1.45 = 144−180 = −36 < 0
So no real solution exists.
So only −3 < x < 15 holds good.
Option 4.
7. |x2−5x| < 4x
A. x < 9
B. 15 < x -3
C. 1 < x < 9
D. x > 1
E. -9 < x < -1
Answer: |x2−5x|< 4x
x2−5x < 4x and x2−5x > −4x
x2−5x−4x < 0 and x2−5x+4x > 0
x(x−9) < 0 and x(x−1) > 0
For x(x−9) < 0 we have x∈(0,9)
For x(x−1) > 0 we have x < 0 and x > 1
Taking the common region from above we get 1 < x < 9
Hence Option 3
8. log0.4(x2−13.5) ≥ −1
A. −4 ≤ x ≤ 4
B. x ≤ 5x ≤ 5
C. x ≥ 0 and x ≥ 0
D. x ≤ −5 and x ≥ 2
E. x ≤ −4 and x ≥ 4
Answer :
log0.4(x2−13.5) ≥ −1
x2−13.5 ≥ 0.4(−1)
x2−13.5 ≥ 2.5
x2−16 ≥ 0
(x+4)(x−4) ≥ 0
x ≤ −4 and x ≥ 4
Hence option 5.
9. 9x2−10.3x+9 ≤ 0
A. 1 < x < 2
B. 1 < x < 9
C. 0 ≤ x ≤ 2
D. 2 ≤ x ≤ 9
E. None of these
Answer:
9x2−10.3x+9 ≤ 0
Let 3x=a
a2−10a+9 ≤ 0
a∈[1,9]
x∈[0,2]
Hence option 3.
10. A and B started a business with Rs.10000 and Rs.15000 respectively. After 6 months C joined them with Rs.20000.
Quantity I: B′s share in total profit of Rs.4,00,000 at the end of 2 years.
Quantity II: Annual Salary of Druv after-tax deduction if he earns Rs.20,000 per month and pays a tax of 20% each month.
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
solution: Option A
Quantity I: A: B: C = 10000*24 : 15000*24 : 20000*18=2:3:3
B=3/8 *400,000= Rs.150000
Quantity II: Salary after deduction = 20,000*12*80/100 = Rs.192000
Hence, Quantity I < Quantity II
11. A bag contains 3 yellow, 4 black, and 2 white balls. Two balls are drawn at random.
Quantity I: Probability that none of the flowers drawn is white.
Quantity II: Fraction of work completed by Thanu in 7 days if she is 20% more efficient than Meenu who can complete the work in 12 days.
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
solution: Option A
Quantity I: 7C2/9C2=7/12
Quantity II: Meenu-> 10 days => fraction of work in 7 days = 7/10
Hence, Quantity I < Quantity II
12. Quantity I: The Simple interest on that sum at the rate of 20 % per annum for 2 years is Rs. 1000.
Quantity II: Rs. 2500
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
solution: Option E
Quantity I:
SI = pnr/100
1000= p * 20/100 * 2
p = Rs. 2500
Quantity II: Rs. 2500
Quantity I = Quantity II
13. Quantity I: Find the volume of the sphere, whose radius is 10.5 cm.
Quantity II: Find the volume of the cone whose height and diameter is 3.5 cm and 16 cm respectively?
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
solution: Option A
Quantity I:Volume of the sphere = (4/3)*πr3 = 4/3 * 22/7 * 10.5 * 10.5 * 10.5 = 4851 cm3
Quantity II:Volume of the cone = (1/3)*πr2h = 1/3 * 22/7 * 16/2 * 16/2 * 3.5 = 234.66 cm3
Quantity I > Quantity II
14. 3x+y = 81 and 81x-y = 3
Quantity I: value of x
Quantity II: value of y
A. Quantity I > Quantity II
B. Quantity I ≥ Quantity II
C. Quantity II > Quantity I
D. Quantity II ≥ Quantity I
E. Quantity I = Quantity II or Relation cannot be established
Answer: A
3x+y= 81
3x+y= 34
x +y= 4……. (i)
81x-y= 3
(34) x-y= 31
x –y= ¼ …….. (ii)
On solving, we get
x= 17/8
y= 15/8
Hence, Quantity I > Quantity II
15. Quantity I: There are three numbers in the ratio 5: 6: 10. The sum of the largest and the smallest numbers is 126 more than the other number. Find the largest number?
Quantity II: 12 % of the first number is equal to 25 % of the second number. The difference between these two numbers is 78. Then find the largest number?
A. Quantity I > Quantity II
B. Quantity I ≥ Quantity II
C. Quantity II > Quantity I
D. Quantity II ≥ Quantity I
E. Quantity I = Quantity II or Relation cannot be established
Answer: C
Quantity I:
According to the question,
10x + 5x = 126 + 6x
15x – 6x = 126
9x = 126
X = 14
Largest number = 10x = 140
Quantity II:
(12/100)*x = (25/100)*y
(x/y) = (25/12)
x : y = 25 : 12
13’s = 78
1’s = 6
Largest number = 25x = 25*6 = 150
Quantity I < Quantity II
16. A hemisphere of radius 4 cm is to be shipped in a shipping box of rectangular shape. The dimension of the box are consecutive odd numbers.
Quantity I: Minimum volume of box required to have the shipment possible.
Quantity II: 960 cm3
A. Quantity I > Quantity II
B. Quantity I ≥ Quantity II
C. Quantity II > Quantity I
D. Quantity II ≥ Quantity I
E. Quantity I = Quantity II or Relation cannot be established
Option A
Solution:
I: The minimum length of any side of the rectangular box should be greater than the diameter of the hemisphere. Diameter of hemisphere=8; So for odd number, minimum length= 9, next side =11, next side=13
Volume=9*11*13= 1287 cm3
17. Quantity I: Distance, if a man covers a distance of 22 hours. he covers first half at 15 km/hr and 2nd half at 18 km/hr
Quantity II: Distance, if a man covers a distance in three equal parts in 20 hours. He covers first part at 10 km/hr, 2nd at 15 km/hr and 3rd at 30 km/hr
A. Quantity I > Quantity II
B. Quantity I ≥ Quantity II
C. Quantity II > Quantity I
D. Quantity II ≥ Quantity I
E. Quantity I = Quantity II or Relation cannot be established
Option A
Solution:
Quantity I:
(2*15*18)/33 * 22 = 360 km
Quantity II:
Average speed will become 15 km/hr
So distance 15*20 = 300 km
I>II
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