Quantitative Aptitude: Inequalities Set 10 (Important for Upcoming Exams)

Directions(1-10): Establish the correct relation between these two quantities and choose a correct option.

  1. The length and breadth of rectangle is 2 m more and 2 m less than the side of a square.
    Quantity I: Find the area of square.
    Quantity II: Find the area of rectangle.

    I > II
    I < II
    I >= II
    I = II
    I =< II
    Option A
    Let the side of square be ‘x’ m.
    Length of rectangle = (x+2) m
    Breadth of rectangle = (x-2) m
    From I: Area of square = x*x = x^2 m^2
    From II: Area of rectangle = (x+2)(x-2) = (x^2 – 4) m^2
    I > II

     

  2. Quantity I: A and B started a flower shop with an investment of Rs.4800 and Rs.4000 resp. After 3 months, C joined them with an investment of Rs.3600. Find the profit of C out of the total profit of Rs.4600 in a year.
    Quantity II: Find the amount of money invested for which compound interest is Rs.15 more than simple interest at rate of 10% per annum for 2 years.

    I =< II
    I = II
    I > II
    I >=II
    I < II
    Option E
    From I: Ratio of investment = A:B:C = 4800:4000:3600
    = 12:10:9
    Ratio of profit share = A:B:C = (12*12): (10*12) : (9*9) = 48:40:27
    Profit of C = 27/(48+40+27)*4600 = Rs.1080
    From II: Difference between the compound interest and simple interest for 2 years.
    CI – SI = P * (R/100)^2
    => P = Rs.1500
    I < II

     

  3. The sum of the surface area of a cube and area of a rectangle is 3240 m^2. The length and breadth of the rectangle is 50% more and 40% more than the side of the
    cube.
    Quantity I: Find the volume of the sphere that can be fitted inside the given cube such that it touches all the six faces of the cube. (pi = 3)
    Quantity II: Find the volume of cuboidal hole of dimension equal to the rectangle and depth 4.5 m.

    I =< II
    I = II
    I < II
    I > II
    I >=II
    Option D
    Let l and b be the length and breadth of rectangle and ‘a’ be the side of the cube.
    Length of rectangle = 1.5a metre
    Breadth of rectangle = 1.4a metre l*b + 6a^2 = 3240
    1.5a * 1.4a + 6a^2 = 3240
    => a = 20 m
    side of cube = 20 m
    length of rectangle = 1.5*20 = 30 m
    breadth of rectangle = 1.4*20 = 28 m
    From I: Diameter of sphere = side of cube
    radius of sphere = 20/2 = 10 m
    Volume of sphere = (4/3)*pi*r^3 = 4000 m^3
    From II: Volume of cuboidal hole = 30*28*4.5 = 3780 m^3
    I > II

     

  4. Quantity I: X travels from point A to B at the speed of 80 km/hr. and from point B to C at the speed of 65 km/hr. If total distance travelled by X is 710 km and time is taken by him to reach point B from point A is 4 hours, find the time taken by X to travel point B to C.
    Quantity II: 380 minutes

    I =< II
    I >=II
    I < II
    I = II
    I > II
    Option C
    From I: Let the time taken by X to travel from B to C be x hours.
    80*4 + 65*x = 710
    => x = 6 hours = 360 minutes
    I < II

     

  5. Yash invested in PPF, NSC and KVP. PPF and NSC are offering simple interest at the rate of 8% pa and 12% pa resp. while KVP is offering interest at rate of 10% compounded annually. The amount invested in PPF and NSC is in the ratio of 4:5 resp. and amount invested in KVP is Rs.9000. After two years she earned a total amount of Rs.4834 as interest.
    Quantity I: Find the amount invested by Yash in NSC.
    Quantity II: Find the average amount invested by Yash in PPF and KVP taken together.

    I = II
    I =< II
    I < II
    I > II
    I >=II
    Option D
    Let the amount invested in PPF and NSC be Rs.4x and Rs.5x resp.
    4x*8% *2 + 5x*12%*2 + 9000*[(1.1)^2 – 1] = 4834
    =>x = Rs.1600
    Amount invested in PPF = 4*1600 = Rs.6400
    Amount invested in NSC = 5*1600 = Rs.8000
    From I: Rs.8000
    From II: (6400+9000)/2 = Rs.7700
    I > II

     

  6. A and B started a work in partnership with a total initial investment of Rs.7000. After 3 months, C joined them with an investment of Rs.3500. If after a year, B received Rs.5950 as profit out of the total profit of Rs.13750 and it was decided that B will be given 20% of the total profit for taking care of the work and remaining profit will be distributed according to their
    investments.
    Quantity I: Find the initial investment of B.
    Quantity II: Find the inital investment of A.

    I = II
    I < II
    I >=II
    I > II
    I =< II
    Option B
    Let the initial investment of B be Rs.x.
    So, initial investment of A = Rs.(7000-x)
    Ratio of investment A:B:C = (7000-x):x:3500
    Ratio of profit share A:B:C = (7000-x)*12 : 12x : 3500*9
    = 4(7000-x):4x:10500
    Profit received by B for taking care of the work = 0.2*13750 = Rs.2750
    Remaining Profit = 13750 – 2750 = Rs.11000
    4x/{4*(7000-x) + 4x + 10500} = (5950 – 2750)/11000
    => x = 2800
    From I: Initial investment of B = Rs.2800
    From II: Initial investment of A = Rs.4200
    I < II

     

  7. Arun is 50% more efficient than Aman and Anil is 40% less efficient than Arun. The time taken by Anil alone to complete the work is 2 days more than the time taken by Aman alone to complete the work.
    Quantity I: Find the time taken by Arun to complete 60% of the work with 60% of his original efficiency.
    Quantity II: Find the time taken by Anil to complete 40% of the work with 80% of his original efficiency.

    I > II
    I = II
    I >=II
    I < II
    I =< II
    Option A
    Let the time taken by Aman to complete the work be x days.
    Time taken by Arun to complete the work = x/1.5 days
    Time taken by Anil to complete the work = (x/1.5)/0.6 = x/0.9 days
    Now, x/0.9 = x+2
    => x = 18 days
    Time taken by Arun to complete the work = 18/1.5 = 12 days
    Time taken by Anil to complete the work = (x/1.5)/0.6 = 18/0.9 = 20 days
    From I: Time taken Arun to complete 60% of the work with 60% of his efficiency = 12/0.6*0.6 = 12 days
    From II: Time taken by Anil to complete 40% of the work with 80% of his efficiency = 20/0.8*0.4 = 10 days
    I > II

     

  8. Quantity I: The sum of the ages of three friends A,B and C is 42 years, and the respective ratio of age of A and B is 3:2. If B is 7 years older than C. Find the age of C 3 years ago.
    Quantity II: The average age of family of five members five years before was 32 years. In the meantime, a new baby was born in the family and present average age of family is half year less than the average age of family of five members, five years before. Find the present age of newborn baby.

    I > II
    I =< II
    I >=II
    I = II
    I < II
    Option D
    The average age of family of five members, five years before. Find the present age of newborn baby.
    From I: Let the ages of A and B be 3x and 2x years resp.
    Age of C = (2x-7) years
    A+B+C = 42 years 3x + 2x + (2x-7) = 42
    => x = 7
    Age of C = (2*7-7) = 7 years Age of C, 3 years before = 7 – 3 = 4 years
    From II: Sum of age of family five years before = 32*5 = 160 years
    Sum of age of present age of family (without baby) = 160 + 5*5 = 185 years
    Sum of age of present age of family (with baby) = (32 – 0.5)*(5+1) = 189 years
    Age of newborn baby = 189 – 185 = 4 years
    I = II

     

  9. A thief ran away from a police station situated in Shivgarh towards Goa city at 3 am with a speed of 40 km/hr. Inspector realised the breakdown at 6 am and started chasing the thief with speed of 60 km/hr. The distance between Shivpuri and Goa is 480 km.
    Quantity I: At what distance from Goa, Inspector will catch the thief?
    Quantity II: If Inspector would have realised the breakdown at 4:30 am, then at what distance from Shivgarh, would Inspector have caught the thief?

    I < II
    I =< II
    I = II
    I > II
    I >=II
    Option A
    From I: Dist. Travelled by thief when police realised the breakdown = 3*40 = 120 km
    Time taken by police to chase the thief = 120/(60-40) = 60 hour
    Dist. Travelled by Police = 480 – 360 = 120 km
    From II: Dist. Travelled by thief when police realised the breakdown = 1.5*40 = 60 km
    Time taken by police to chase the thief = 60/(60-40) = 3 hours
    Required distance = 3*60 = 180 km
    I < II

     

  10. There are two sets of letters and we have to choose exactly one letter from each set. Set A = {A,B,C,D,E} and Set B = {M,N,O,P,Q}
    Quantity I: The probability of choosing ‘C’ or ‘P’ but not both.
    Quantity II: The probability of choosing two vowels.

    I >=II
    I < II
    I =< II
    I > II
    I = II
    Option D
    From I: Number of favorable cases = (C,M),(C,N),(C,O),(C,Q),(A,P),(B,P),(D,P),(E,P) = 8
    Total number of cases = 25 Required Probability = 8/25
    From II: Probability of choosing two vowels = (2/5)*(1/5) = 2/25
    I > II

     


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