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

- 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 < III >= III = II or No relationI =< III > IIOption A

From I: S.I. = (2400*8*3)/100 = Rs.576

From II: C.I. = 2800*[(1.1)^2 – 1] = Rs. 588

I < II - 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: 24I >= III > III =< III = II or No relationI < IIOption 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 - 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 =< III > III >= III < III = II or No relationOption 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 - 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 =< III = II or No relationI < III >= III > IIOption 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 - 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 hoursI = II or No relationI < III =< III > III >= IIOption 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 - 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 < III = II or No relationI >= III > III =< IIOption 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 - 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 > III = II or No relationI >= III < III =< IIOption 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 - 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 =< III >= III > III = II or No relationI < IIOption 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 - 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 kmI > III =< III < III >= III = II or No relationOption 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 - 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 >= III = II or No relationI < III > III =< IIOption 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