Aptitude and Reasoning Questions

Given total number of smileys = 5 + 6 + 3 = 14

Now, required probability = 

$$\begin{gathered} 5C214C2 + 3C214C2 \\ = 1091 + 391 \\ = 1391 \\ = 17 \end{gathered}$$

Length of train A = 48 x 9 x 5/18 = 120 mts

Length of train B = 48 x 24 x 5/18 - 120

=> 320 - 120 = 200 mts.

Let the sum be Rs. x. Then

C.I=x[1+4/100)^2-x]=[626/675x-x]

S.I=x^2/25

C.I-S.I=1

Let the rquired number be 'P'

According to the question,

$$\begin{gathered} 100 x P100 = 20 \\ => P = 20 x 100100 = 20 \end{gathered}$$

Hence, 100 percent of the number 20 is 20.

Now, you can also try for

20 percent of what number is 100

$$\begin{gathered} 20 x P100 = 100 \\ => P = 100 x 10020 = 500 \\ => P = 500 \end{gathered}$$

Hence, 20 percent of the number 500 is 100.

Let the sum invested be Rs. P

Let the rate of interest be R% per annum

=> Interest earned for 5 years = (P x 5 x R/100) = PR/20

Now, given that the interest earned increased by Rs. 600 if the Rate increased to (R+2)%

=> SI = (P x 5 x (R+2))/100 = PR/20 + 10P/100

Hence,

PR/20 + 10P/100 = PR/20 + 600

=> P = 6000

Therefore, the sum invested is Rs. 6000

Water required = 4000 x 140 = 560000 lit = 560 cu.m (1 cu.m = 1000 lit)

Volume of tank = 18 x 12 x 8 = 1728 cu.m

Water of this tank will suffice for = 1728/560 = 3 days.

From the given data,

3500x7xt/100 = 500

=> t = 100/49 years

Now, in the second case

The interest per year = 49/100 x 800 = 392

 => 4900 x 1 x r/100 = 392

=> r = 8%

S = { (1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (2, 1), (2, 2),.........(6, 5), (6, 6) }

=> n(S) = 6 x 6 = 36

E = {(6, 3), (5, 4), (4, 5), (3, 6) }

=> n(E) = 4

Therefore, P(E) = 4/36 = 1/9