Quantitative Aptitude: Inequalities Set 2

Directions: Each question below contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.

1. The price of item X is marked at Rs 10,000. After giving a discount of 4%, a gain of 20% is achieved.
   Quantity I: Cost Price of Item X.
   Quantity II:  Selling Price of item Y after two successive discount of 10% and 15% is provided on the Marked Price of Rs 10,000.
   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
   
   View Answer

      Option A
   Solution:  
    I : Mp =Rs 10,000; SP after 4% discount = Rs 9600; CP = Rs 8000
   II : SP= 10,000*90/100 * 85/100 = Rs 7650
   Hence I > II 
2. Quantity I: Simple Interest on a sum of Rs 6200 at the rate of 6% per annum for 5 years
   Quantity II:  Compound Interest on a sum of Rs 6400 at rate of 10% compounded semi-annually for one year.
   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
   
   View Answer

      Option A
   Solution:  
    I: SI = 6200*6*5/100 = Rs 1860
   II: CI = 6400 [1 – 5/100]² – 6400 = Rs 1376
   Hence I > II 
3. The price of item X is marked at Rs 10,000. After giving a discount of 4%, a gain of 20% is achieved.
   Quantity I: Cost Price of an article if it is sold making a loss of 10% given that if the cost price was 20% less, a profit of Rs 12 could be made.
   Quantity II:  Labeled Price of an article if a discount of 30% is given on it and then sold for  Rs 105 at a loss of 20%
   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
   
   View Answer

      Option E
   Solution:  
    I: Let CP = Rs 100, then at 10% loss, SP = Rs 90
   Now if CP is 20% less, means CP = Rs 80, then profit is = 90 – 80 = Rs 10
   but given profit is Rs 12
   So if profit is Rs 10, then CP = Rs100
   If profit is Rs 12, then CP is 100/10 * 12 = Rs 120
   II: Use formula MP = (100-loss%)/(100-dicount%) * CPS
   So MP = (100-20)/(100-30) * 105 = Rs 120
   Hence I = II 
4. Given that x < 0 and y > 0
   Quantity I: 14x³y²
   Quantity II:  28x²y³
   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
   
   View Answer

      Option C
   Solution:  
    I: Since x < 0, so I will always be negative
   II: It will be always positive
   Hence I < II 
5. Quantity I: Area of rectangle whose sides are in the ratio 3 : 1 and perimeter twice the circumference of a semi-circle with area 77 cm²
   Quantity II:  Curved surface area of a cylinder of radius 12 cm and height equal to the side of square of area 49cm²
   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
   
   View Answer

      Option C
   Solution:  
    I: ½ πr² = 77 [area of semicircle]
   So r = 7 cm
   So circumference of semicircle  = πr = 22 cm
   So perimeter of rectangle = 2*22 = 44 cm
   So 3x+x = 44 cm, x = 11,
   So area of rectangle = 3x<sup>2<sup> = 363cm²
   II: height of cylinder = √49 = 7 cm
   So CSA = 2πrh = 528cm²
   Hence I < II 
6. Quantity I: Time taken by A to complete 1/5^th of work if B takes 6 days to complete 3/5^th of work and together they take 5 days to complete 3/4^th of work
   Quantity II:  Time taken by a train to cross a platform of length 60 km given that it crosses a pole in two and a half hours running at 60 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
   
   View Answer

      Option A
   Solution:  
    I: A and B completes 3/4^th work in 5 days, so complete 1 work in 4/3 * 5 = 20/3 days
   B complete 3/5 work in 6 days, so complete work in 5/3 * 6  = 10 days
   So in 1 day A completes = 3/20 – 1/10 = 1/20
   So to complete 1/5^th work = 1/5 * 20 = 4 days
   II: length of train = 2.5 * 60 = 150 km
   So time taken to cross platform of length 60 km with speed 60 km/hr = (150+60)/60 = 3.5 hrs
   Hence I > II 
7. Quantity I: Age of A five years ago if 6 years hence ratio of age of A to B will be 14 : 11, and 1 year ago the ratio was 7 : 5.
   Quantity II:  Average age of 2 students included in a group of 5 students with average age 18 years given that addition of these 2 students in group makes average increase by 1.
   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
   
   View Answer

      Option C
   Solution:  
    I: (A+6)/(B+6) = 14/11
   And (A-1)/(B-1) = 7/5
   Solve the equations, A = 22, so 5 years ago = 22 – 5 = 17 years
   II: Total age of 2 students added to group = 18*2 + 7*1 = 43
   So their average age = 43/2 = 21.5 years
   Hence I < II 
8. Quantity I: x, such that 6x² + 22x + 15 = 0
   Quantity II:  y, such that 6y² – 7y – 10 = 0
   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
   
   View Answer

      Option D
   Solution:  
    I: x = -3, -5/6
   II: y= -5/6, 2
   Hence I ≤ II 
9. A shopkeeper gets a discount of 20% on an article and sells it at 10% profit.
   Quantity I: Labeled price as a percent of cost price.
   Quantity II:  Profit percent if discount given is 60%
   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
   
   View Answer

      Option C
   Solution:  
    I: MP = (100+10)/(100-20) * CP = 11/8 of CP
   So required % = MP/CP * 100 = (11CP/8)/CP * 100 = 137.5%
   II: % profit = (100+10) [(100-10)/(100-60)] – 100 = 590/4 = 147.5%
   Hence I < II 
10. Quantity I: Speed of boat in still water f it took 24 hours to cover a distance of 64 km going downstream and back. Given the speed of current to be 2 km/hr
    Quantity II:  Average speed of a boy to go from home to school if his average speed is 2 km/hr from school to home and he covers a total of 12 km in 5 hours.
    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
    
    View Answer

       Option A
    Solution:  
     I: Let speed of boat = x km/hr
    So 64/(x+2) + 64/(x-2) = 24
    Solve, x = 6 km/hr
     II : Distance from home to school is 12/2 = 6 km
    Let speed from home to school is x km/hr
    So using distance = total time * average speed
    6 = 5 *[2*x/(2+x)]
    Solve, x = 3 km/hr
    Hence I > II