Quantitative Aptitude - Arithmetic Ability Questions

M+A+S3=11111 

also S+M2=11111   (∵ M - 11111 = 11111 - S) 

=>A = 11111

Let the number be 'x'

Then, according to the given data,

 52x-25x52xx100

= 2125x100

= 84%

Let the third number be 100.

Then, the first number = 100 - 20 = 80 and second number = 100 - 30 = 70.

Difference between the first and second number = 80-70 = 10

 Required percentage = 1080 x 100% = 100/8 = 12.5%

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

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

Given taps X and Y can fill the tank in 30 and 40 minutes respectively. Therefore,

part filled by tap X in 1 minute = 1/30

part filled by tap Y in 1 minute = 1/40

Tap Z can empty the tank in 60 minutes. Therefore,

part emptied by tap Z in 1 minute = 1/60

Net part filled by Pipes X,Y,Z together in 1 minute = [1/30 +1/40 - 1/60] = 5/120 = 1/24

i.e., the tank can be filled in 24 minutes.

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3,4),(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

n(E) = 27.

P(E) = n(E)/n(S) = 27/36 = 3/4.