- A can do a piece of work in 8 days which B can destroy in 3 days. A has worked for 6 days, during the last 2 days of which B has been destroying. How many days must A now work alone to complete the work?
9(1/2) days4(1/6) days7(1/3) days8(1/2) days3(1/5) daysOption C

In 6 days part of the work done by A = 6/8 = Â¾

During 2 days, part of the work destroyed by B = 2/3

Work done = Â¾ â€“ 2/3 = 1/12

Remaining work = 1 â€“ 1/12

Required no of days = 11/12*8 = 7(1/3) days - Out of his total income, Mr. Kapoor spends 20% on house rent and 70% of the rest on house hold expenses. If he saves Rs 1,800 what is his total income (in rupees)?
Rs.8100Rs.7500Rs.9000Rs.6200Rs.7000Option B

Saving % = 100 â€“ (20 + 56) = 24%

24% = 1800

100% = 1800/24*100 = Rs.7500 - A and B are partners in a business. They invest in the ratio 5 : 6, at the end of 8 months A withdraws. If they receive profits in the ratio of 5 : 9, find how long Bâ€™s investment was used?
148101612Option E

5x : 6x, Let B investment was used for y months

8*5 x : 6x *y = 5 : 9

=>y= 12 - A train is moving at a speed at a speed of 132 km/hour. If the length of the train is 110 metres, how long will it take to cross a railway platform 165 metres long?
7.5 seconds9.1 seconds8.6 seconds3.3 seconds4.2 secondsOption A

132*5/18 = (100+165)/x

=>x = 7.5 seconds - A committee of 3 members is to be selected out of 3 men and 2 women. What is the probability that the committee has atleast one woman?
5/149/106/137/128/11Option B

Probability = (2C1*3c2 + 2c2*3c1)/5c3 = 9/10 -
**I.**4x + 7y = 42

**II.**3x â€“ 11y = â€“ 1y >xy>=xx>=yx > yNo relationOption D

On solving both the equations, we get x = 7

y = 2

x > y -
**I.**3x^2 â€“ 19x â€“ 14 = 0

**II.**2y^2 + 5y + 3 = 0x>=yx > yy >xNo relationy>=xOption B

I. 3x^2 â€“ 19x â€“ 14 = 0

=>3x^2 -21 x + 2x â€“ 14 = 0

=>3x (x -7) +2(x -7) = 0

=>(x-7)(3x+2) = 0

=>x = 7, -3/2

II. 2y^2 + 5y + 3 = 0

=>2y^2 +2y +3y + 3 = 0

=>2y (y+1) + 3(y+1) = 0

=>(y+1)(2y+3) = 0

=>y = -1, -3/2

x > y -
**I.**x^2 + 14x + 49 = 0

**II.**y^2 + 9y = 0x>=yy>=xNo relationx > yy >xOption C

I. x^2 + 14x + 49 = 0

=>X^2 + 7X+7X + 49 = 0 =>x(x+7)+7(x+7) = 0

=>(x+7)(x+7) = 0

=>x = -7,-7

II. y^2 + 9y = 0

=>y = -9,0

No relation -
**I.**3x^2 â€“ 4x â€“ 32 = 0

**II.**2y^2 â€“ 17y + 36 = 0x>=yNo relationy>=xy >xx > yOption C

I. 3x^2 â€“ 4x â€“ 32 = 0

=>3x^2 – 12x + 8x – 32 = 0

=> 3x(x – 4) + 8(x -4) = 0

=>(x-4)(3x+8) = 0

=>x = 4,-8/3

II. 2y^2 â€“ 17y + 36 = 0

=>2y^2 – 8y – 9y + 36 = 0

=>2y(y- 4) -9(y-4) = 0

=>(y-4)(2y-9) = 0

=>y = 4,4.5

y>=x -
**I.**9x^2 â€“ 29x + 22 = 0

**II.**y^2 â€“ 7y + 12 = 0No relationx > yy>=xy >xx>=yOption D

I. 9x^2 â€“ 29x + 22 = 0

=>9x^2 â€“ 18x â€“ 11 x + 22 = 0

=>9x (x-2) -11(x-2) = 0

=>(x-2)(9x-11) = 0

=>x = 2,11/9

II. y^2 â€“ 7y + 12 = 0

=>y^2 -4y -3y + 12 = 0

=>y(y-4) -3 (y -4) = 0

=>(y-4)(y-3) = 0

=>y = 4,3

y >x

**Directions(6-10):** Find the values of x and y and then compare them and choose a correct option.