Quantitative Aptitude - Arithmetic Ability Questions

P(X) = 15, P(Y) =14 , P(Z) = 13

Required probability:

= [ P(A)P(B){1−P(C)} ] + [ {1−P(A)}P(B)P(C) ] + [ P(A)P(C){1−P(B)} ] + P(A)P(B)P(C)

=14*13*45+34*13*15+23*14*15+14*13*15

= 460+360+260+160= 1060= 16

Given that, 4198+612345+34866+2411+1

Now Power Cycle of

4 is 4, 6 

6 is 6 

8 is 8, 4, 2, 6

1 is 1

4198- 198 is even number , So the unit digit value is 6 

612345 - It is always 6

34866-  66/4 the remainder is 2 , So the second value is 4

2411- 11 is odd number, So the unit digit value is 4

1 - It is always 1 

Total = 21

Therefore, The unit digit is 1

Let the required no of binders be X.

Less books , Less binders       (Direct Proportion)

More days, Less binders         (Indirect proportion)

 Books900:660Days12 : 10⋮⋮ 18:x

Therefore, (900 * 12 * x) =  (660 * 10 * 18)  

=> x = 11

Let the total population be 'p'

Given ratio of male and female in a city is 7 : 8

In that percentage of children among male and female is 25% and 20%

=> Adults male and female % = 75% & 80%

But given adult females is = 156800

=> 80%(8p/15) = 156800

=> 80 x 8p/15 x 100 = 156800

=> p = 156800 x 15 x 100/80 x 8

=> p = 367500

Therefore, the total population of the city = p = 367500

A + B= 70%   

B + C =50%   

 ∵ (A+B)+(B+C)-(A+B+C)= B

=> B= 20%     A= 50%        and   C=30%

Hence A is most efficient

Area of the room=(544 * 374)

size of largest square tile= H.C.F of 544 & 374 = 34 cm

Area of 1 tile = (34  x  34) cm2

Number of tiles required== [(544 x 374) / (34 x 34)] = 176

L.C.M. of 5, 6, 7, 8 = 840.

 Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

 Required number = (840 x 2 + 3) = 1683.