Quantitative Aptitude: Inequalities Set 9 (Based on SBI Pattern)

Directions(1-10): Compare both the equations then choose a correct option accordingly.

  1. Quantity I: Find the simple interest earned on Rs.2400 at the rate of 8% p.a. in 3 years.
    Quantity II: Find the compound interest earned on Rs. 2800 at the rate of 10% p.a. compounded annually after 2 years.

    I < II
    I >= II
    I = II or No relation
    I =< II
    I > II
    Option A
    From I: S.I. = (2400*8*3)/100 = Rs.576
    From II: C.I. = 2800*[(1.1)^2 – 1] = Rs. 588
    I < II

     

  2. Quantity I: Find the difference between the sum of roots and product of roots of the given equation.
    2x^2 – 14x + 50 = 0
    Quantity II: 24

    I >= II
    I > II
    I =< II
    I = II or No relation
    I < II
    Option E
    From I: y = 2x^2 – 14x + 50
    sum of roots = -b/a = 14/2 = 7
    Product of roots = c/a = 50/2 = 25
    Required Difference = 25 – 7 = 18
    I < II

     

  3. Quantity I: Find the curved surface area of the cone of radius 7 cm and height 24 cm.
    Quantity II: Find the curved surface area of the cylinder of radius 7 cm and height 12 cm.

    I =< II
    I > II
    I >= II
    I < II
    I = II or No relation
    Option B
    From I: CSA = (22/7)*7*[7^2 + 24^2]^1/2 = 550 cm^2
    From II: CSA = 2*(22/7)**12 = 528 cm^2
    I > II

     

  4. Quantity I: Find the compound interest earned on investing a sum of Rs.25000 at the rate of 8% per annum for 2 years.
    Quantity II: Find the simple interest earned by Amit on lending a sum of Rs.18000 at the rate of 6% per annum for 4 years.

    I =< II
    I = II or No relation
    I < II
    I >= II
    I > II
    Option C
    From I: Required C.I.= 25000[(1+8/100)^2 – 1] = Rs.4160
    From II: Required Interest = 18000*6%*4 = Rs.4320
    I < II

     

  5. Quantity I: A and B can do a piece of work in 12 hours and 18 hours resp. resp. If both started to work on alternate days with B starting the work, find the total time taken to complete the whole work.
    Quantity II: 15 hours

    I = II or No relation
    I < II
    I =< II
    I > II
    I >= II
    Option B
    From I: Total work = 36 units
    Units of work done by A and B in one hour resp. = 3 units and 2 units
    Work done by both of them in two hours = 5 units
    Work done by both of them in 14 hours = 7*5 = 35 units
    Remaining work = 36 – 35 = 1 unit
    Time taken by B to complete the remaining work = ½ hours
    Time taken to complete the work = 14.5 hours
    I < II

     

  6. Quantity I: A mixture of milk and water contains 32 litres of water. If 8 litres of water is added to the mixture, the quantity of water in resultant mixture is 83(1/3)% of the quantity of milk in the resultant mixture. Find the quantity of milk in the resultant mixture. Quantity II: 48 litres
    I < II
    I = II or No relation
    I >= II
    I > II
    I =< II
    Option B
    From I: Let the quantity of milk in resultant mixture be x.
    Quantity of water in resultant mixture = 32+8 = 40 litres
    83(1/3)% of x = 40
    => x = 48
    I = II

     

  7. Quantity I: Find the curved surface area of the cone of radius 5 cm and height 12 cm.
    Quantity II: Find the total surface area of the sphere of radius 4 cm. [use the value of pi = 3 cm]

    I > II
    I = II or No relation
    I >= II
    I < II
    I =< II
    Option A
    From I: CSA of cone = 3*5*[5^2+12^2]^1/2 = 195 cm^2
    From II: TSA = 4*3*4*4 = 192 cm^2
    I > II

     

  8. Quantity I: A shopkeeper marks up an item by 50% above the cost price and sold it after offering a discount of 20%. The profit earned by shopkeeper is what % less than the profit earned by him, if he marks up the item by 75% above the cost price and offers a discount of 20%.
    Quantity II: The area of the square built on the side of a square is what percentage less than the area of a square build on the diagonal of a square?

    I =< II
    I >= II
    I > II
    I = II or No relation
    I < II
    Option D
    From I: Let CP be Rs. x.
    MP = Rs.1.5x
    SP = 1.5x*0.8 = Rs.1.2x
    Profit = 1.2x – x = Rs.0.2x
    Also, CP be Rs.x.
    MP = Rs.1.75x SP = 1.75x*0.8 = Rs.1.4x
    Profit = 0.4x
    Required% = (0.4x – 0.2x)/0.4x*100 = 50%
    From II: Let ‘x’ be the side of the square.
    Length of the diagonal of square = (2)^1/2a units
    Area of square = a^2 units
    Area of square built on side of square = [(2)^1/2a]^1/2 = 2a^2 units
    Required% = (2a^2 – a^2)/2a^2 * 100 = 50%
    I = II

     

  9. Quantity I: The time taken by the boatman to cover 48 km upstream and 60 km downtream is same. If the speed of the stream is 1.5 km/hr. find the distance covered by the boatman in still water in 8 hours.
    Quantity II: 110 km

    I > II
    I =< II
    I < II
    I >= II
    I = II or No relation
    Option C
    From I: Let the speed of the boat in still water be x km/hr.
    48/(x – 1.5) = 60/(x+1.5)
    => x = 13.5 km/hr.
    Required distance = 13.5*8 = 108 km
    I < II

     

  10. Quantity I: A and B together started a business with an investment of Rs.2400 and Rs.3600 resp. After ‘x’ months, C joined them with an investment of Rs. 3000. If a year, B received Rs.2700 out of total profit of Rs.6000, find x.
    Quantity II: P and S together started a business with total investment of Rs. 4500 in the ratio of investment 5:x resp. If after a year, S received Rs. 3200 as profit out of a total profit of Rs. 7200, find x.

    I >= II
    I = II or No relation
    I < II
    I > II
    I =< II
    Option B
    From I: Ratio of investments of A:B:C = 2400:3600:3000
    = 4:6:5
    Ratio of profit share of A:B:C = 4*12:6*12:5*(12-x) = 48:72:5(12-x)
    Now, 72/[48+72+5(12-x)] = 2700/6000 = 9/20
    => x = 4
    From II: x/(5+x) = 3200/7200
    => x = 4
    I = II

     


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