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Direction (1-10): In the given questions, two

quantities are given, one as ‘Quantity 1’ and another

as ‘Quantity 2’. You have to determine relationship

between two quantities and choose the appropriate

option:

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

1. A can do a work in 16 days. B is 60% more efficient

than A.

Quantity I: Time taken by A and B together to do the

work.

Quantity II: Time taken by A and B to do the work

together when A works at double his original efficiency

and B works at half his original efficiency.

2. Quantity I: Find amount after 4 years, if rate of

interest is 20% and the principal amount is Rs 8000

Quantity II: Find the amount after 4 years, if Rs 10,000

becomes Rs 12,000 in 2 years at compound interest.

3. Quantity I: Days in which B can complete work alone, if A and B can complete work in 40 days, B and C

in 20 days and C and A in 30 days.

Quantity II: Days in which B can complete work alone,

if A and B can complete work in 24 days and A is 50% more efficient than B.

4. 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

5. Quantity I: Find the curved surface area of the cone of radius 7 cm and height 24cm.

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

6. Quantity I: Total surface area of cylinder who radius is 7 cm and height is 10 cm

Quantity II: Total surface area of cuboid whose dimensions are 10×12×15 cm

7. 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

8. Quantity I: x, if 6x2 – 29 x – 20 = 0

Quantity II: y, if 6y2+ 13y – 15 = 0

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

10. It takes (X-2) men to do a work in 2Y days and it takes (X+6) men to do the same work in Y days.

Quantity I: X

Quantity II: Y

__SOLUTION__

__1. A__

__I: A=16 days ; B = 16 * 100/160=10 days__

__A+B together = 16*10/(26)=80/13 days__

__II: A = 16 days ; B = 10 days__

__A(double efficiency) = 8 days ; B(half efficiency) = 20__

__days__

__A+B together = 80/14__

__hence I > II__

__2. B__

__I: SI = 20*4 = 80%__

__100% = 8000__

__180% = Rs 14400__

__II.__

__10000 : 12000 :: 12000 : x__

__x = 12000*12000/10000 = Rs 14400__

__II>I__

__3. B__

__I:__

__A + B = 40……..3__

__B + C = 20………6 ……..(LCM = 120)__

__C+ A = 30……….4__

__Total = 2 (A+B+C) = 3+6+4 = 13__

__So A+B+C = 13/2__

__(A+B+C) – (B+C) = 13/2 – 4 = 5/2__

__So B can complete work in 120/(5/2) = 48 days__

__II: Efficiency A …….. B = 3 : 2__

__So days = 2 …. 3__

__LCM of 2 and 3 is 6__

__A = 2………6/2 = 3__

__B = 3………6/3 = 2__

__Total A+B = 3+2 = 5__

__So 6/5 == 24__

__So 1 == 20__

__So 3 == 60__

__So II > I__

__4. B__

__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__

__5. A__

__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__

__6. B__

__I: 2Ï€r(r+h) = 748 cm2__

__II: 2 (lb+bh+lh) = 900 cm2__

__7. 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__

__8. C__

__x = 5/6, 4__

__y = -3, 5/6__

__9. B__

__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. E__

__As the time is becoming half so means number of people__

__have doubled; So (X+6)=2*(X-2)=>X=10__

__Using M1D1=M2D2 the value of Y cannot be found. Y__

__can take any value. So we cannot determine a unique__

__value of Y__